Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
How big of a neutron star would be needed to form a quark star inside of it?
A quark star is a hypothetical type of compact exotic star composed of quark matter. These are ultra-dense phases of degenerate matter theorized to form inside particularly massive[citation needed] neutron stars. -Wikipedia
If you add enough mass to a neutron star it forms a black hole but how much mass is needed to form a quark star? When do neutrons and protons start to break down into a mush of quarks and gluons?
| This article might be of interest to you! :)
Extract from Quark Stars
Vuorinen joined forces with Aleksi Kurkela at ETH Zurich in Switzerland and Paul Romatschke at the University of Washington in Seattle to examine how the pressure of strange quark matter depends on its density – a relationship described by the star's "equation of state". Physicists have looked at this before, but only using highly simplified models of quark interactions. Instead, Vuorinen's group has employed perturbation theory – a technique that approximates mathematical solutions in stepwise fashion, which on the whole is far more accurate.
The result may surprise other physicists. Current thinking has it that quark stars should be smaller than neutron stars, and indeed that compact stars above a certain size – typically about twice the mass of our Sun, or two solar masses – must be pure neutron stars with no quark core. However, Vuorinen's group conclude almost the opposite: that the biggest quark stars can be larger than neutron stars, perhaps up to 2.5 solar masses. In other words, as Vuorinen points out, the detection of a compact star with a mass near that limit would be a "strong indication" of a quark star
Image Source: Quark Stars
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/275744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Estimate the reaction force on each leg of a 4-legged table Assume you have a table with four legs with a weight placed somewhere inside the boundary created by the legs. The task is to determine the reaction force at each leg. Here are the problem constraints...
*
*The sum of the four reaction forces must equal the total weight set on the table.
*The table has no weight of its own.
*You cannot directly solve for the reaction of each leg using sum of forces and sum of moments because the structure is statically indeterminate.
*We are not allowed to introduce deflection in order to develop a 4th independent relationship between the four reactions
There are methods to estimate the reaction force on each leg. One such method uses ratios to determine the four leg reactions.
\begin{eqnarray}
P_1 & = & \left(\frac{y_w-y_b}{y_f-y_b}\right)\left(\frac{x_w-x_l}{x_r-x_l}\right) \\
P_2 & = & \left(\frac{y_w-y_b}{y_f-y_b}\right)\left(\frac{x_w-x_r}{x_l-x_r}\right) \\
P_3 & = & \left(\frac{y_w-y_f}{y_b-y_f}\right)\left(\frac{x_w-x_r}{x_l-x_r}\right) \\
P_4 & = & \left(\frac{y_w-y_f}{y_b-y_f}\right)\left(\frac{x_w-x_l}{x_r-x_l}\right) \end{eqnarray}
Here is my question. The above solution is only good for a square leg configuration.
Can this method be modified to work for a trapezoidal leg configuration, like this?
Better yet, can this method be modified to work for any leg configuration?
Edit 1
@ja72, This is in response to your answer. Please forgive me. It's been 15 years since I've worked with summations and linear algebra. I'm having trouble solving the two torque equations for $\psi$ and $\varphi$. Please look at what I have below and let me know if I'm on the right track. First I found $F_i$ using your second equation.
\begin{equation}
F_i=-\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w)}{3k}
\end{equation}
Now I substitute this into your two torque equations and attempt to solve.
\begin{equation}
-\sum_{i=1}^n\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w))}{3k}(y_i-y_w)=0\tag{1}
\end{equation}
\begin{equation}
\sum_{i=1}^n\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w))}{3k}(x_i-x_w)=0\tag{2}
\end{equation}
Rearrange equations 1 and 2
\begin{equation}
\frac{nW}{3k}-\sum_{i=1}^n\frac{\psi(x_i-x_w)(y_i-y_w)+\varphi(y_i-y_w)^2}{3}=0\tag{3}
\end{equation}
\begin{equation}
\frac{nW}{3k}+\sum_{i=1}^n\frac{\psi(x_i-x_w)^2+\varphi(y_i-y_w)(x_i-x_w)}{3}=0\tag{4}
\end{equation}
\begin{equation}
\begin{bmatrix}
\sum_{i=1}^n(x_i-x_w)(y_i-y_w)&\sum_{i=1}^n(y_i-y_w)^2\\
\sum_{i=1}^n(x_i-x_w)^2& \sum_{i=1}^n(y_i-y_w)(x_i-x_w)
\end{bmatrix}
\begin{bmatrix}
\psi\\\varphi
\end{bmatrix}=
\begin{bmatrix}
-\frac{nW}{k}\\
\frac{nW}{k}
\end{bmatrix}
\tag{5}
\end{equation}
I doubt this is correct and I do not know how to solve this. Can you please help me solve for $\psi$ and $\varphi$.
Edit 2
Starting with the 2x2 given by ja72. Notice I have eliminated the $k$ terms on the right hand side since they cancel out in the end.
\begin{equation}
A\enspace
\begin{array}{|c|} \psi \\ \varphi
\end{array}=
\begin{array}{|c|}
W(y_w-\overline{y})\\W(x_w-\overline{x})
\end{array}
\tag{1}
\end{equation}
where
\begin{equation}
A=\begin{bmatrix}
n\overline{x}\overline{y}-\displaystyle{\sum_{i=1}^n(x_iy_i)} &
\displaystyle{\sum_{i=1}^n(y_i^2)}-n\overline{y}^2 \\
\displaystyle{\sum_{i=1}^n(x_i^2)}-n\overline{x}^2 &
n\overline{x}\overline{y}-\displaystyle{\sum_{i=1}^n(x_iy_i)}
\end{bmatrix}
\end{equation}
Implement variable transformation
\begin{equation}
a = n\overline{x}\overline{y}-\sum_i^n(x_iy_i)\quad
b=\sum_i^n(y_i^2)-n\overline{y}^2
\end{equation}
\begin{equation}
c=\sum_i^n(x_i^2)-n\overline{x}^2\quad
d=n\overline{x}\overline{y}-\sum_i^n(x_iy_i)
\end{equation}
now A becomes
\begin{equation}
A=\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
\end{equation}
Solve for $\psi$ and $\varphi$
\begin{equation}
\begin{array}{|cc|}
\psi\\\varphi
\end{array}=A^{-1}\enspace
\begin{array}{|cc|}
W(y_w-\overline{y}\\
W(x_w-\overline{x}
\end{array}
\end{equation}
\begin{equation}
\psi=\frac{w}{ad-bc}(b(x_w-\overline{x})-d(y_w-\overline{y}))
\tag{2}
\end{equation}
\begin{equation}
\varphi=\frac{w}{ad-bc}(a(y_w-\overline{y})-c(x_w-\overline{x}))
\tag{3}
\end{equation}
Here is the equation for $z_w$. Again, I have removed the $k$ terms since they all cancel out.
\begin{multline}
z_w=\frac{1}{n}(W+
\displaystyle{\sum_{i=1}^n}\psi(x_i-x_w)- \varphi(y_i-y_w))
\end{multline}
When we substitute $\psi$ and $\varphi$ we get this ugly thing.
\begin{multline}
z_w=\frac{W}{n}(1+\frac{1}{ad-bc}
\displaystyle{\sum_{i=1}^n}(d(y_w-\overline{y})-b(x_w-\overline{x}))(x_i-x_w)- \\
(a(x_w-\overline{x})-c(y_w-\overline{y}))(y_i-y_w))
\tag{4}
\end{multline}
Finally we have...
\begin{equation}
F_i=Term1-Term2+Term3
\end{equation}
Where
\begin{eqnarray}
Term1 & = & z_m \\
Term2 & = & \psi(x_i-x_w) \\
Term1 & = & \varphi(y_i-y_w) \\
\end{eqnarray}
| To solve this problem make the legs elastic, but nearly rigid. The reaction force on each leg is going to be a spring force. There are three degrees of freedom: a) two tilt angles and b) Overall height.
Let's assume each leg attachment is at $(x_i,y_i,z_i)$ where $z_i$ is the unknown "lift" of the table leg from the horizontal plane. This lift is a function of the center of mass height $z_w$ and two angles
$$ z_i = z_w + \psi (x_i - x_w) - \varphi (y_i - y_w) $$
The force in each leg is $F_i =-k z_i$ so the total weight must be
$$ \left. W =- k \sum_i^n z_i \;\right\} z_w =- \frac{W+k (\sum \limits_i^n \psi (x_i - x_w) - \varphi (y_i - y_w) )}{3 k}$$
Now the sum of the two torque components along x and y for each leg force is zero
$$ \sum_i^n {\tau_x}_i = \sum_i^n F_i (y_i-y_w) = 0$$
$$ \sum_i^n {\tau_y}_i = -\sum F_i (x_i-x_w) = 0$$
These two equations are solved for $\varphi$ and $\psi$ and when used back in the force equation $F_i = -k (z_w + \psi (x_i - x_w) - \varphi (y_i - y_w))$ miraculously the stiffness $k$ cancels out.
Edit 1
To solve for the tilt angles you make the following 2×2 system of equations
$$\begin{bmatrix} n \bar{x} \bar{y} - \sum \limits_i^n \left(x_i y_i\right) & \sum \limits_i^n \left( y_i^2 \right) - n \bar{y}^2 \\
\sum \limits_i^n \left( x_i^2 \right) - n \bar{x}^2 & n \bar{x} \bar{y} - \sum \limits_i^n \left(x_i y_i\right)
\end{bmatrix} \begin{vmatrix} \psi \\ \varphi \end{vmatrix} =
\begin{vmatrix} \frac{W}{k} \left(y_w - \bar{y}\right) \\ -\frac{W}{k} \left(x_w - \bar{x}\right) \end{vmatrix}$$
where $\bar{x} = \frac{1}{n} \sum \limits_i^n x_i$ and $\bar{y} = \frac{1}{n} \sum \limits_i^n y_i$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/276007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Increasing water pressure to shower head I have a 200 liter water tank 3 meters above a shower head.
The tank and head are connected directly by a relatively small pipe.
The shower head has an intake with a diameter the same size as the pipe leading from the tank.
The water pressure coming from the head is too low.
I am tempted to believe increasing the size of the pipe from the tank to maybe two or three times of the intake of the shower will increase the water pressure into the shower.
I, like many people, use the thumb over the head of a hosepipe as method of how to increase pressure but after some reading up I have found you are trading reduced water flow for increased water pressure and in reality you have hindered the rate of water flow.
However this maybe the effect i require for the instant shower to work.
Do you think this will help?
| What you want is increased water flow I suppose. Pressure difference between your shower head and water tank located above is a constant (approximately, if height of water in tank is much less than the height at which tank is located). So if you want greater flow rate you need to reduce the resistance to flow, and maximum resistance is offered by valves somewhere along the pipeline if they are not fully open, or may be somewhere inside your shower head.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/276104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What would happen if a charge could travel at $c$? We know that an electron can move at sublight speed, but can we figure out what would happen if a charge moved at $c$, surfing the wave of the electric field it has produced? There does not seem to be anything that prohibits charged particles from moving at light speed. (If it sounds better, consider a Weyl fermion)
It would be pushed forward, repelled by its own field, but there are 2 possible scenarios depending on whether the speed of propagation of the wave can be exceeded or not.
| The electron has mass. Particles that have mass cannot move at velocity c. There are no charged massless particles, thus no charged particles can move at c.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/276291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Is an EMF more/same/less in an insulator than in a conductor? Is an EMF (electromotive force) more/same/less in an insulator than in a conductor?
For example: A loop of copper and a loop of plastic in a changing magnetic field.
In which will the emf be the greatest?
| The part of net emf due to external magnetic field is the same. However, net emf in the case of conductor includes also contribution due to self-induction when current is changing in time. In plastic this should be negligible. So net emf will be different if current is changing in time in the copper ring.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/276383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the physical meaning of the Schwarzschild radius for objects that aren't black holes? Earth has a Schwarzschild radius of a little less than a centimeter. What does this mean for the matter of Earth's core that is within this radius?
A related question comes up for what happens when an almost black hole accretes matter and slowly becomes a black hole. Prior to the moment of the Schwarzschild radius crossing the boundary of the object, what does the matter within the radius experience?
| (1) The Schwarzschild radius for an object that is not a black hole is the size down to which you would need to compress the object before it becomes a black hole.
(2) By the equivalence principle, absolutely nothing special happens locally at the event horizon—there's no (local) way to even tell that you've crossed it. So nothing unusual happens as the boundary of the object crosses the Schwarzschild radius.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/276610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How is heat represented on a quantum level? Heat is just a form of kinetic energy for molecules, because as temperature rises, the heated molecules are "shake" and "vibrate" more and more. But how does that show up on a quantum scale? What element actually carries the kinetic energy: the heated molecule as a whole, its atoms, the nuclei, or the electrons' orbits? (Maybe even the quarks found in the nuclei?). Or is it that the shaking described is only an analogy for a notion of energy that is more difficult to grasp as their is no real physical movement in the heated object?
| Heat energy, at a microscopic level, is stored in the degrees of freedom of atoms and molecules. These degrees of freedom are translational, rotational and vibrational. They all store different amounts of energy, depending on the geometry of the atom.
Translational degrees of freedom are the atom or molecule moving around in space, and there are always 3 for the 3 dimensions of space. The rotational and vibrational modes come from the geometry of the atom/molecule.
From quantum mechanics, we get the idea that energy stored in rotational and vibrational (and translational, if confined) modes must come in quantised packets, with a minimum size. This size depends on the form of a certain mode. For single atoms, the moment of inertia and the energy of rotation is very small. The quantum of energy that must be added to excite the rotational modes is large, and so these do not contribute to heat storage until very high temperatures.
Molecules have much higher moments of inertia around certain axes. For example, O2 has high moment of inertia around the two axes perpendicular to its bond axis and a low moment of inertia around its bond axis. It therefore stores heat energy in those two and they contribute to the heat capacity of O2.
Vibrational modes store much more energy than translational or rotational modes, and are active only at higher temperatures.
This is basically what heat is at a microscopic level. Quantum mechanics gives us that the energy stored in the modes must be quantised.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/276724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Laser spectrum range What are the spectrum limits in which a laser can be produced? Also can you focus other electromagnetic waves (microwaves, radio waves, ect…) into a laser like beam?
| Actually, the first "laser" (1954) was emitting microwaves. Charles Townes wanted to produce a microwave source with a very narrow linewidth and came up with the idea to separate ammonia molecules residing in the upper component of the inversion splitting from those residing in the lower component using a hexapole. In this way, he was able to obtain the required population inversion for laser action. He named the apparatus "Microwave Amplification by Stimulated Emmission of Radiation" or MASER. Later on (1960), stimulated emission was used in Ruby crystals by Maiman to produce an optical maser or laser. Townes wrote an interesting account on his personal experiences in "How the laser happened".
Since the early days of the laser, the frequency range has been extended quite a lot. The invention of dye lasers introduced lasers that can typically scan tens of nanometers in the near infrared, visible and near UV part of the spectrum by using population inversion in organic dyes. Also, excimer lasers using excited states of certain gasses to produce high energy radiation in the UV are commonly available. Solid state lasers like the ones using titanium sapphire crystals reach wavelengths from roughly 700 to 1000 nm. In addition, nonlinear optics allows one to double the frequency of a laser extending its domain.
Nowadays, quantum-cascade lasers (mostly in the infrared) and diode lasers are the state of the art. They are turn key and can be produced at many different wavelengths. However, their scanning capability is very narrow as compared to a dye laser.
You might find it interesting to know that masers do also exists in nature. In space, methanol and OH masers, for instance, are among the brightest radio sources in the sky.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/277070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does curved spacetime affect gravitational waves? How differently will a LIGO detector detect a gravitational wave which came directly to it with a detector which happened to have a black hole between it and the source?
| Gravitational waves will scatter off strong gravitational fields. You'd expect that just from the equivalence principle, as Lawrence Crowell states. It's basically like light lensing. You can also have more sophisticated estimates of how much.
The arxiv article from 2008 calculates scattering, absorption and reflection of gravitational waves from black holes, specifically Kerr black holes, and finds interesting relationships between the polarizations and helicities of the gravitational waves, and the black hole rotation and orientations. See https://arxiv.org/abs/0801.3805. It calculates rather exact results. The scattering cross sections and absorption and reflex ion coefficients are calculated. The main dependence, besides the polarization, helicities and relations to the black hole rotation axis and sense, is with respect to the horizon radius over the wave wavelength, and the ratio of angular momentum to mass (I.e., the normalized angular momentum with max of 1). In 2008 gravitational waves had not been directly detected yet, but they did speculate at the end, just a bit, on possible observational consequences, but with no specific gravitational strain numbers plugged into their equations.
It is true that gravitational waves do not interact much with matter directly, but they will be affected, and with larger interferometers we might see some effects. Still, gravitational wave astronomy is probably counting more on what can be observed from detections of objects that can produce a large amount of gravitational radiation, black holes and neutron stars, supermassive black holes accreting matter or compact massive objects, cosmological grav waves, from exploding and collapsing supernova, and others. It is a burgeoning field. The wiki article summarizes many of what they will be looking for. See https://en.m.wikipedia.org/wiki/Gravitational-wave_astronomy
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/277282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving first order perturbation exactly in this situation I have this homework problem in QM Perturbation Theory
The Hamiltonian of a system is given by $$H_0 = A L^2 + B L_z$$ where
$A$ and $B$ are constants. If a perturbation $V = C L_y$ is added to
the system (where $C \ll A, B$ , find the lowest order correction to
the energy. Also solve the problem exactly
The lowest order correction to energy is
$$\langle l,m \rvert C L_y \lvert l, m \rangle = 0$$
as $L_y$ will either give me $\lvert l, m-1 \rangle$ or $\lvert l, m+1 \rangle$
But how do I solve it exactly? If I fix $l$, then $L_y$ is a $(2l+1) \times (2l+1)$, which is still more tedious than this assignment is supposed to be. Please help.
| (I assume all coefficients are real obviously.) Define $N=\sqrt{B^2+C^2}$. Using a unitary transformation $U$ representing in the Hilbert space a certain rotation, the one rotating $$\left(0,\frac{C}{N} ,\frac{B}{N}\right)$$ to $$(0,0,1)$$ which therefore leaves $L^2$ invariant, you have $$U(AL^2+BL_z+CL_y)U^∗=AL^2+NL_z\:.$$ Unitary transformations do not change the eigenvalues so that eigenvalues of $AL^2+NL_z = U(AL^2+BL_z+CL_y)U^∗$ are the same as of $AL^2+BL_z+CL_y$.
From this we conclude that the exact eigevalues of $AL^2+BL_z+CL_y$ are the ones of $AL^2+NL_z$:
$$Al(l+1)+Nm\:, \quad l=0,1,2,\ldots \quad m = \pm l$$
the same eigenvalues as for $H_0$ with $B$ replaced for $N$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/277439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The symbol $\gtrsim$ versus $\ge$ in physics Currently I am going through Thermodynamic cost of computation, algorithmic complexity and the information metric by W. H. Zurek. In the introduction, he mentioned a relationship between the change in entropy and the difference between the length of a program and its output. It is as follows:
$$\delta S (i \to o) \gtrsim |i^*| - |o^*| $$
From the $\LaTeX$ code \gtrsim, I realize that $\gtrsim$ means 'greater or similar'. My question is how is it different from $\ge$?
| There's not going to be a rigorous definition; this is how I would read it as a working physicist. Usually $\sim$ means "of about the same size," often in an order of magnitude sense. So if we say $v \sim 1$ in means that $v$ might be like, 5. But probably $v < 10$. This kind of symbol would often appear in a calculation in which I chose not to keep track of certain things--for instance, if I'm interested in whether a relation is linear or quadratic, I don't care much about prefactors.
Which is all to say: I would read the equation you wrote as saying that the LHS is greater than something of similar size to the right hand side. So it might not be greater than, say, $10^9 \left( |i| - |o| \right)$ but the idea is that there's probably some constant attached to the front that isn't important for the relation being stated, and the method we used to calculate it might not give a reliable answer for that constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/277798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Similarities between light and other frequencies of EM waves This may be a ridiculous question, but I'll learn something from it! Let's say there's a TV transmitter transmitting at 100kW. I can receive the station just fine 20 miles away. The antenna is 300m in the air. If I replicate this but with light (put a 100kW "bulb" 300m in the air), would I be able to see the light 20 miles away?
I understand there are things to take into account, particularly propagation differences at the different frequencies. But is this a meaningful analogy, or totally useless?
| Technically speaking, if there is no decay of EM waves of any particular frequency, you can measure the signal in the configuration you are describing. It all depends on how sensitive the measuring instrument (antenna, eye) is to the power under consideration. This of course being true for some large values of EM wave intensity where quantum effects would be irrelevant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/277892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
For lattice, what are the Goldstone bosons for the broken rotation symmetries? In $1$ dimension, we know that lattice breaks continuous translational symmetry into discrete translational symmetry, which generates $1$ Goldstone boson, i.e. $1$ longitudinal phonons.
In $d$ dimensions, if there are only $1$ type of atoms, then there are $1$ longitudinal phonon and $d-1$ transverse phonons. However, in $d$ dimensions, the continous symmetries of $d$ dimensional Euclidean group are broken, and in principle we should have $d+\frac{d(d-1)}{2}=d(d+1)/2$ Goldstone bosons. What are the other Goldstone Bosons?
| The Wikipedia article on "Goldstone boson" says
In general, the phonon is effectively the Nambu–Goldstone boson for spontaneously broken Galilean/Lorentz symmetry. However, in contrast to the case of internal symmetry breaking, when spacetime symmetries are broken, the order parameter need not be a scalar field, but may be a tensor field, and the corresponding independent massless modes may now be fewer than the number of spontaneously broken generators, because the Goldstone modes may now be linearly dependent among themselves: e.g., the Goldstone modes for some generators might be expressed as gradients of Goldstone modes for other broken generators.
I won't pretend to fully understand that, but apparently the usual rule - that in a relativistic system, each spontaneously broken symmetry gets an independent Goldstone mode - only applies for internal symmetries, not for spatial ones.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 2
} |
How does the size of the magnetic field vary with the wavelength of a photon? I tried a search and could not find it in a simple format. Like if the wavelength halves than the magnetic field falls off like $~\frac{1}{\lambda^2}$.
I mean the maximum value of the sinusoid for a single photon. I have read about this some where, but no idea where.
A photon has a well defined energy and wavelength and they can come as single particle. So i just wondered how one photon would induce a magnetic field. They use photons in laser traps and then make a beautfl picture of the electric field as all photons are added. So i was thinking how would the field of a single one be. I mean we can divide by the big number of photons, that are present.
I know that in QED, there is just a chance of the photon coupling to some other particle at some time. And there is little interest in what is happening between.
But in the laser trap we have both pictures.
| The size of the magnetic field for a photon wouldn't depend its wavelength. The amplitude of the fields do not depend on the frequency or wavelength, it depends on the number density of photons (and then there is the distinction between coherent and incoherent sources to consider).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Number of degrees of freedom of the coupled pendulum problem In Chapter 4 from the book Theoretical Mechanics of Particles and Continua by A. L. Fetter and J. D. Walecka, it is solved the problem of a coupled pendulum system while considering small oscillations.
There, they say the number of degrees of freedom needed to describe the Lagrangian, are the infinitesimal displacements from equilibrium $\eta_1$ and $\eta_2$, corresponding to each pendulum mass.
My question is: why there are needed two degrees of freedom? Isn't the spring that is attached to both masses a constraint of motion that reduces the degrees of freedom to only one?
Actually, they explicitly write the following equation:
$d-d_{0}=\eta_{2}-\eta_{1}$
which is the equation of the change in length of the spring.
Thank you for any answers or suggestions!
| If the spring were perfectly rigid it would reduce the number of degrees of freedom as a constraint. Because it is not, you need the positions of both pendulums to calculate the stretch of the spring.
$$\begin{align}d &= \sqrt{l^2(\cos\theta_1 - \cos\theta_2)^2 + (l\sin\theta_1 - l\sin\theta_2 + d_0)^2} \\
&\approx l(\theta_1 - \theta_2) + d_0.\end{align}$$
So, because you need two quantities to calculate the stretch, there are two degrees of freedom. You can, in principle, change variables to write one of the angles in terms of the stretch and the other variable, but that isn't likely to gain you anything.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why does increasing resistance decrease the heat produced in an electric circuit? If $H=\frac{V^2}{R}{t}$ ,then increasing resistance means decreasing the heat produced.
But, isnt it that the heat in a circuit is produced due to the presence of resistors? Moreover metals with high resistances are used as heating elements ,like Nichrome?
Why does the equation state that the heat produced is inversely proportional to Resistance
| Consider analogy with fluid flow. When there flow in a pipe say, due to friction, energy of motion is dissipated away into heat. Therefore for dissipation into heat to occur two things are necessary: flow, and resistance to flow. In the absence of either of them there is no dissipation into heat. Same is true of current in a circuit.
In the equation $H=\frac{V^2}{R}t$ you can only see what the resistance is, but not what the current is. Turns out that for constant applied voltage, if you increase resistance, current reduces more than proportionately (see @Farcher's answer). In the extreme case if there is break in the circuit and therefore no current, there would be no heat generation at all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
At what pressure does the helium undergo phase transition at room temperature? At what pressure does the helium undergo phase transition at room temperature?
I can not find phase diagram at room temperature.
| From Chemicool
At normal atmospheric pressure helium does not solidify. At 25 atmospheres of pressure helium is a solid at 0.95 K. As the pressure rises, the temperature at which solid helium exists also rises. Helium can be made solid at room temperature if the pressure rises to about 114 thousand atmospheres: that is a pressure of 1.67 million psi, or 834 tons per square inch. This is over 100 times greater than the pressure at the oceans’ deepest point, the Challenger Deep, which is almost seven miles deep (10 916 meters).
I cannot find a phase diagram at room temperature either.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How does the law of conservation of energy explain magnetism? The law of the conservation of energy describes that energy cannot be created nor destroyed however in fact only changes form.
How does this law explain the energy transferred by magnetic fields?
| Magnetic fields work like springs.
So for permanent magnets one can align the magnetic dipole moments of the involved subatomic particles and "freeze" this state. But themagnet is than under iner pressure and it is not advised to drop the magnet, it will explode in pieces and the contributed energy will be released.
In the case of the Lorentz force the magnetic dipole moment of moving electrons get aligned under the influence of an external magnetic field. During the deflection of the electron it emits photons, gets dissaligned again and by this the external magnetic field is unchanged again. After the moving electron cames to rest the external magnetic field is the same as at the beginning.
It has to be keeped in mind, that magnetic fields do not interact with electric fields and that they stay unchanged after Lorentz force applications.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Entropy and reversible paths I'm a little bit confused about calculating entropy changes along irreversible paths by integrating over a reversible path. When using the central equation I can understand the argument, entropy and all the quantities we use to calculate the entropy change are state functions that have well defined end points. So the process between these two states doesn't matter, we will get the same entropy change. I find it difficult to grasp how we can use dS=dQ/T to calculate entropy changes in irreversible processes. Since Q is path dependent I don't see how the original argument applies. We can take the free expansion as an example, if we try to use dS=dQ/T we will get zero entropy change as no heat flows into the system. But using the central equation where all quantities are state functions we get a entropy change which is larger that zero as expected.
| For an irreversible path, the integral of dQ/T is not equal to $\Delta S$. To get the change in entropy between the two end points of an irreversible path, you need to devise a reversible path between the same two end points, and calculate the integral for that path. It doesn't matter what reversible path you choose; they will all give the same value for $\Delta S$. For more details on this, see the following link:
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
This link contains contains a simple recipe for determining the entropy change for any arbitrary irreversible process, and presents several typical examples of irreversible processes encountered in thermodynamics courses, and how to apply the recipe to determine the change in entropy for each.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Formula for lens periscope I want to build a periscope.
To my great surprise even after intense googling I could find little relevant information about periscopes anywhere on the internet.
Let's say I want to build a periscope with height h, apparent height h1, diameter d and viewing angle v. What other variables should I define for a periscope?
Source: Wikipedia.
I am looking for a formula/manual/recipe for calculating the lenses and distances as well as any practical advice on building a DIY periscope.
Thanks.
| Assuming the two prisms are planar reflectors, there's no significant effect from them. If you do your refractor-telescope calculations by measuring the distances along the center of the light path, including the right angle bends, it should all work out. Then you put in the right-angle, planar reflectors (prisms or mirrors) and get the direction you want as an effect independent from magnification and focus. The only other issue of the reflectors is their width relative to the angle of view you want to get. But again, this is independent of the right angles - it's still just distance and width of the tube.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/278940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Does Gravity Depend on Spatial Dimension? Consider a line containing two point masses, $m$ and $M$. The line is a $1D$ space.
What's the gravitational force between the two masses?
Newton's formula for the gravitational force $F$ between two masses $m$ and $M$ in 3D space is
$$F=\frac{G M m}{r^2}$$
where $G$ is a constant and $r$ is the distance between the two masses.
The $r$ term is good in a $3D$ space, but in general it's $r^{n-1}$ where $n$ is the dimension of the space. So putting $n=1$ for $1D$ space we get
$$r^{1-1}=r^0=1 \Rightarrow F=GMm \, ,$$
Which means $F$ is independent of distance. Gravity has the same strength no matter how far apart the two objects are!
Of course, this calculation uses Newton's theory of gravity. Perhaps General Relativity would give a different result.
| The reason why the gravitational force is in $r^{-2}$ is because it has zero divergence in void — its flux is conserved through a close surface encompassing matter is conserved.
Let's take the example of a massive point, the flux of the gravitational force through a sphere of radius $r$ centered on the point is:
$$ \Phi = \iint\mathbf{F} \mathrm{d}\mathbf{S}.$$
If we assume that the force is isotropic (independent of the direction), then the equation is only:
$$ \Phi = F \iint \mathrm{d} S,$$
which is the norm of the force times the surface of a sphere in $n$ dimensions.
So in 3D we have $\Phi \propto F r^2$ hence $F \propto r^{-2}$ and in $n\geq 1$ dimensions, $F \propto r^{n-1}$.
Now the case of one dimension is similar: a sphere in one dimension is defined by two points located at $\pm r$. Because the force is isentropic,
$$\int_\mathrm{1D}\mathbf{F}\mathrm{d}\mathbf{S} = F(+r) + F(-r) = 2 F(r).$$
The conservation of the flux is then:
$$ \forall r\in\mathbb{R}, \quad 2F(r) = \Phi$$
So you are correct to say that the gravitational force is independent of distance in a 1D universe!
If you want to overcome this issue of infinitely propagating foce, you have to take into account the time it takes for the force to propagate and you should use general relativity and eventually cosmology (with expanding universe).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Showing that pseudorapidity equals $\ln \left(\frac{\sqrt{p_z^2 + p_t^2}+p_z}{p_t}\right)$ I found one method to calculate pseudorapidity, $\eta$, using the formula
\begin{equation}
\eta = - \ln \tan \frac{\theta}{2}
\end{equation}
And I know how we can get to this point from Lecture 7 - Rapidity and Pseudorapidity in these notes.
But how can we get from there to this formula?
\begin{equation}
\eta = \ln \left(\frac{\sqrt{p_z^2 + p_t^2}+p_z}{p_t}\right)
\end{equation}
| Oh Maybe I have found,
Another formula for the pseudorapidity $\eta$ is :
\begin{equation}
\eta = \ln \frac{p+p_z}{p-p_z} \\
\eta = \ln \dfrac{\sqrt{p_x^2+p_y^2+p_z^2}+pz}{\sqrt{p_x^2+p_y^2+p_z^2}-p_z} \\
\eta = \ln \dfrac{\sqrt{p_x^2 +p_y^2 + 2 \times p_z^2}}{\sqrt{p_x^2+p_y^2}} \\
\end{equation}
So We can find :
\begin{equation}
\eta = \ln \left( \frac{\sqrt{p_\perp^2 +p_z^2}+p_z}{p_\perp} \right)
\end{equation}
This is wrong, the transition between line 3 and 4 are illegal mathematically
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do we prove or disprove that a particle has no internal structure? In many pop physics books I have read that an electron has no internal structure.
How do we know that and how can we rigorously prove that it has no such structure at all?
| The electron has a magnetic dipole moment and an intrinsic spin. Both phenomenon need an extend and a structure and hence the electron has to have an internal structure.
That until know we haven't found this structure has to do with our capabilities to build the right instrument.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
Why doesn't electric charge immediately leak off charged objects? I will focus my question with a particular example: a metal sphere, surrounded by vacuum, is given a negative charge. I know that when this charge is great enough, electrons will be emitted from the sphere, but why is the threshold for this so high? As I understand it, the reason an electron stays on the negative sphere despite the electric repulsion is because of the metal's work function. But the work function of metals are typically ~4 eV. Wouldn't this suggest that -4 Volts would be the threshold for electron emission from the sphere in a vacuum? (Or a voltage even closer to zero, because of the thermal distribution of electron energies in the metal.) This seems way too small and I would think the threshold would concern a minimum field strength rather than minimum voltage.
| As an electron makes its way outside a metal surface, there is immediately
an image charge (positive virtual charge) on the other side of that surface
that attracts it back. The 'work function' is the kinetic energy that the
free electron needs if it is to escape to a large distance, and has
to include both surface-binding-energy and the long-range attractive force
that a conductor exerts on a nearby charge.
The sphere, if it is very large, might have a local field completely dominated
by the (attractive) image charge, and not by the net (repulsive) charge on the sphere. Net charge is one sphere radius away, and image charge is two
surface-to-electron steps away.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can the work done in bringing a charged particle from infinity to a grounded conductor be zero? Suppose we have a grounded conducting sphere (potential of sphere = 0) and we bring a charged particle to its surface.
Since work done by us would be equal to the change in potential energy of the charge, it will be equal to zero,
As potential at infinity = potential on sphere = 0 therefore change in potential energy = 0.
But this is not possible as when we move the charge, it will induce a negative charge on the sphere, thus for charge to move without acceleration, a negative force (with respect to our displacement) would be required.
Thus we would do negative work all over the path and work done can never be zero.
Am I wrong in assuming that the work done = change in potential in this case? Because only that would make sense.
Any help would be appreciated.
| Lets proceed like this. The surface of the sphere S1 being grounded is at a potential we are labelling as V1 and an imaginary system boundary S2 at infinity is as V2(imagine a colossal spherical surface S as the boundary, this S2 is basically the sphere of influence of the grounded sphere.) WHen we start bringing a positive charge, it is initially outside S2(no interaction initially). So initially, the boundaries of my system S1 and S2 are at potential V1 and V2(fixed). After i bring the charge onto the conductor. The potential V1 should've changed, but 'grounding' essentially means a technique to keep the potential constant no matter how large a finite charge is added. Technically, the test charge should also be small enough for this assumption. So now V1 is unchanged. Obviously V2 is unchanged too, and therr is no charge in between S1 and S2(my system). So the Neumann boundary conditions for thr laplace equation are valid. The field E between S1 and S2 must therefore be unchanged. Since there is no change in $E^2$dv, the net field energy is same. Hence no work has been done on the system in consideration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If you are vacuuming your carpet and you wrap the cord around your body do you become a magnet? If you wrap an active electric cord around your body, do you become an electromagnet?
| The answer is no, because the electric field of each wire cancels the field of the other wire. But even if you wrap yourself many times with one wire and connect the ends to electric power, you will not become a "measurable" electro-magnet. The reason for this is that the amount of ferromagnetic material in your body (blood iron, etc.), is very small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "85",
"answer_count": 4,
"answer_id": 3
} |
Why does this planet (J1407 b) and Saturn's ring center on its equator? Why is the material in J1407 b's or Saturn's rings stay in a disk not scattered? Is it the gravity and/or magnetic field that causes this? Does it differ with other large bodies that might not have a electromagnetic field?
| Saturn's rings are rather interesting. From what I've read, the different layers of dust and rocks have been formed over millions of years, by a phenomenon known as orbital resonance. Orbital resonance is something like an increased(or decreased) gravitational effect due to 2 bodies moving with certain time periods. This effect causes stability in some cases, and instability in others. When there is instability, there is an effect known as 'clearing the neighbourhood' , in which the unstable objects undergoing orbital resonance, are flung out of their current orbit, unless they find a more stable one. That's how the rings arranged themselves in bands, and are not continuous.
For more info, read this : orbital resonance
Correction: clearing the neighbourhood is not the name used in this case. It is used for the removal of unstable orbits in larger systems like the solar system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Combination of more than two lenses? I know the formula that when two lenses are separated by a distance $d,$ then the resultant focal length is given by :-
$$\frac 1f = \frac1f_1 + \frac1f_2 - \frac d{f_1\cdot f_2}$$
However , there are two doubts that arise in my mind.
1) My book says that a combination of two thin lenses can only be replaced with a thick lens. What exactly is a thick lens? Also, how do we find out the position of the lens i.e. the position of the pole of this single replaced lens?
I assume it to be the half of the distance between the two lenses($d,$ in the above formula). Am I correct?
2)Also , what if we have more than 2 lenses? In that case do we first replace two lenses by a single lens, and then use that new lens with the other lens/lenses to find its focal length and keep repeating it until we run out of lenses? In this approach, I still need some distance of the pole of the new lens with respect to something OR any distance in order to find out the new lens's position (i.e. the ans to my 1st question).
Or is there a formula to find that in one go Like $1/f_1 + 1/f_2 + \ldots + 1/f_n - d/{f_1f_2\ldots f_n}\;?$
Or is it something more complicated? I was able to prove the formula for 2 lenses but not for any $n$ number of them. I guess that is the case because my book says that the theory of a thick lens is very complicated and far beyond the scope of the book.
| *
*A thick lens is a more complicated device than a thin lens. Theoretically, it is defined by the curvature radius of each of the two sides of the lens, the thickness and the refractive index of the material (4 variables). Note that your equivalent thick-lens is under-constrained: 4 variables > 3 parameters. By playing with these variables, you can get the thick lens to behave like your two thin lenses or any optical system in air, because the determinant of the ray-transfer matrix is 1 for a thick lens. Therefore, it might be possible to replace the two lenses by just one thick lens half way in between the two lenses.
*There is a way to obtain the focal length of an array of lenses as described here. The trick is to keep track of the front and back distances of the equivalent lens. Note that the formula you provided is incorrect because the last term $d/f_1 ... f_n$ does not have units of 1/distance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/280295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating Centripetal Acceleration of Sun about Galaxy This is a question from my Mechanics book:
Calculate the centripetal acceleration relative to the acceleration due to gravity of:
c) the Sun in its rotation about the center of the galaxy (the radius of the Sun's orbit about the center of the galaxy is $2.8 * 10^4$ LY and its orbital speed is 220 km/s)
My attempt at an answer: I convert the radius of orbit and orbital speed to meters. To get angular velocity, I then divide the orbital speed by the radius. To get centripetal acceleration, I square the angular velocity and multiply it by the radius. I then divide by $g = 9.8$ to get the centripetal acceleration relative to gravity.
The problem is, the figure I get is about $1.8 * 10^{-11}$, whereas the book's answer is $1.5*10^{-12}$. I tried to use the Sun's gravity instead (274, right?) and got about $6.437 *10^{-13}$. Neither answer matches the book's, and I have no clue how they got it.
I was wondering if there were any flaws in reasoning I was making here, or if the book's answer is incorrect.
| Can't seem to work out how the book got that answer either and my conclusion would be that the answer is wrong or that the question is poorly worded/explained. I also computed about $1.8(6)*10^{-11}$ but did notice some lack of optimisation in your approach...
Note that the centripetal acceleration can be calculated using the angular velocity and radius of orbit as you did:
$a = rw^2 $
note that $w = v/r$ where $v$ is the linear velocity. Thus the centripetal acceleration can be written:
$a = r(v^2/r^2) = v^2/r$
hence there was no need to convert to angular velocity.
All the best
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/280389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are electric fields produced by static electric charges different from those produced by time-varying magnetic fields? I came across an interesting yet perplexing statement in my physics textbook:
However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields.
My questions are, simply put:
What exactly are these properties? And why do they differ?
| $$\begin{array}{|c|c|} \hline\textrm{True in Statics} &\textrm{True in General}\\ \hline \mathbf F ~= \dfrac1{4\pi\varepsilon_o}~ \dfrac{q_1q_2}{r^2}~\mathbf{\hat r} & \mathbf F= q(\mathbf E+ \mathbf v\times \mathbf B)\\ \hline \nabla \cdot \mathbf E = \dfrac{\rho}{\varepsilon_0} & \nabla \cdot \mathbf E = \dfrac{\rho}{\varepsilon_0} \\ \hline \nabla\times \mathbf E= \mathbf 0& \nabla \times \mathbf E = ~-\partial_t\mathbf B\\ \hline \mathbf E= -\nabla\varphi & \mathbf E= -\nabla \varphi - \partial_t \mathbf A\\ \hline \nabla^2\varphi = -\dfrac{\rho}{\varepsilon_0} & \nabla^2\varphi -\dfrac1 {c^2}\partial^2_t\varphi = -\dfrac\rho{\varepsilon_0}\\ \hline \varphi(\mathbf 1)= \dfrac1{4\pi\varepsilon_0}\displaystyle\int\dfrac{\rho(\mathbf 2)}{r_{12}}~\mathrm dV_2 & \varphi(\mathbf 1, t)= \dfrac{1}{4\pi\varepsilon_0}\displaystyle \int \dfrac{\rho(\mathbf 2, t^\prime)}{r_{12}}~\mathrm dV_2\\\hline U= \dfrac12 \left(\displaystyle\int \rho\varphi~\mathrm dV \right) & U= \dfrac{\varepsilon_0}2 \left(\displaystyle\int \mathbf E\cdot \mathbf E~\mathrm dV + {c^2}\displaystyle\int \mathbf B\cdot \mathbf B~\mathrm dV\right)\\ \hline\end{array}$$
where
\begin{align}\mathbf F& \equiv \textrm{Total force},\\ \mathbf E& \equiv \textrm{Electric field},\\ \mathbf B &\equiv \textrm{Magnetic field},\\ \nabla &\equiv \textrm{del operator},\\ \rho &\equiv\textrm{Charge-density},\\ \varphi &\equiv \textrm{Scalar-potential},\\ \mathbf A &\equiv \textrm{Vector-potential},\\ \partial_t &\equiv \dfrac{\partial}{\partial t},\\ \nabla^2 &\equiv \textrm{Lapalacian operator},\\ \partial^2_t &\equiv \dfrac{\partial^2}{\partial t^2},\\ U &\equiv \textrm{Potential-energy}, \\ t^\prime &\equiv \textrm{retarded-time} = t-\dfrac{r_{12}}c\;. \end{align}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/280787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Lagrangian and finding equations of motion I am given the following lagrangian:
$L=-\frac{1}{2}\phi\Box\phi\color{red}{ +} \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$
and the questions asks:
*
*How many constants c can you find for which $\phi(x)=c$ is a solution to the equations of motion? Which solution has the lowest energy (ground state)?
*My attempt:
since lagrangian is second order we have the following for the equations of motion:
$$\frac{\partial L}{\partial \phi}-\frac{\partial}{\partial x_\mu}\frac{\partial L}{\partial(\partial^\mu \phi)}+\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=0 $$
then the second term is zero since lagrangian is independent of the fist order derivative. so we will end up with:
$$\frac{\partial L}{\partial \phi}=-\frac{1}{2} \Box \phi+m^2\phi-\frac{\lambda}{3!}\phi^3$$
and:$$\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=-\frac{1}{2}\Box\phi$$
so altogether we have for the equations of motion:
$$-\frac{1}{2}\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3-\frac{1}{2}\Box\phi=0$$
and if $\phi=c$ where "c" is a constant then $\Box\phi=0$ and then the equation reduces to $$m^2\phi-\frac{\lambda}{6}\phi^3=0$$ which for $\phi=c$ gives us 3 solutions:$$c=-m\sqrt{\frac{6}{\lambda}}\\c=0\\c=m\sqrt{\frac{6}{\lambda}}$$
My question is is my method and calculations right and how do I see which one has the lowest energy (ground state)? so I find the Hamiltonin for that?
| You have a minor error from a missing minus sign here:$$\frac{1}{2}\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3-\frac{1}{2}\Box\phi.$$ It should be (after combining terms):
$$-\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3.$$
Now, for finding the Hamiltonian you might find it easier to integrate the term $\phi \Box \phi / 2$ by parts to get $-\partial_\mu \phi \partial^\mu \phi / 2 + \mathrm{surface\ term}$. That way you can use standard formulae for constructing the Hamiltonian using canonical momenta. That is, assuming you want to construct the Hamiltonian. This Lagrangian has a fairly simple structure with a kinetic energy term (time derivatives of $\phi$), and every other term is potential energy. So, since these states are constant in time and space, their energy will be just potential energy:$$E = \int \left[-\frac{m^2}{2} \phi^2 + \frac{\lambda}{4!} \phi^4\right] \operatorname{d}^3 x.$$
Edit: fix my own sign error.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/280904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solder Phenomenon I recently saw an interesting phenomenon. I was soldering a piece of electronics and once I was at the end of my wire I could simply let go and the (small) solid piece of wire would be sucked in to the molten solder blob.
I am curios what causes this force. Perhaps it is pressure difference?
| This is due to surface tension. This effect is exploited in reflow soldering of surface-mount (SMT) devices on printed circuit boards (PCBs)
*
*Example video, look at R2 at 0:13
*Example video
*Discussion of undesirable effects
*University of Bristol
Don’t worry if the components aren’t prefectly aligned at this stage: When you reflow the surface tension of the molten solder will pull them into place.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Space bends relative to what? We all are aware of gravitational waves, as it bends space and time, black hole squeeze space, but the squeezing, bending, expanding happens reference to what? Since the observable universe is the universe existing within itself, so it bends in reference to whose perspective?
| This is a fundamental question explored in non-Euclidean geometry. Here are two easy to imagine consequences of the bending of space-time, imagined as just effects on space.
First, as is used in the expanding universe model, the size of the universe, it's scale factor, can change with time. In that scenario the wavelength of waves get stretched when measured compared to quantities that arise from oscillations that do not involve spatial movement, like the radius of an atom. The Bohr radius, for example, is fixed by the mass of the electron ($m_e$), the speed of light ($c$), the conversion between energy and frequency (Planck's constant, $h$), and the strength of the interaction between photons and electrons (the fine structure constant, $\alpha$).
Second, is to answer the question: what is the relationship between the radius of a circle and it's circumference? When space is flat, unbent, the answer is $C = 2\pi r$. If space becomes bent, though, the measured value can be changed by a small amount in a way that depends on the size of the circle. This scenario is easiest to visualize by considering two dimensional surface from the outside. In a geometry like the surface of a sphere, the circumference of circles will be smaller than $2\pi r$, by more and more the bigger the circle. In the extreme case, if $R$ is the radius of the sphere then when $r = \pi R$ the circumference of the circle is $0$. The general formula for the surface of a sphere is $C = 2\pi R \sin\left(\frac{r}{R}\right) \approx 2\pi r \left(1 - \frac{r^2}{6 R^2}\right).$ The quantity measured by WMAP and Planck that is proportional to $R$ is known as the spatial curvature density, $\Omega_k$.
So, it is possible to measure the curvature of space-time without reference to any external standards.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Understanding tension based on assumptions of pulley system If we consider a simple pulley system with two masses hanging on each end of a MASSLESS and INEXTENSIBLE string around a MASSLESS and FRICTIONLESS pulley, how then can one reason that the tension at each end of the string must be the same?
My own reasoning:
MASSLESS ROPE means that for any segment of the rope with tension $T_1$ and $T_2$ we have that $\sum F = T_ 2 - T_1 = 0$ (since $m = 0$) and thus the tensions must be the same, on a non curved rope at least!
INEXTENSIBLE means that no energy can be stored in the string, however I fail to see how this is a neccesary condition (for equal tension)
MASSLESS PULLEY means that no rotational inertia exists, and thus no force can alter the tension of the string (?)
FRICTIONLESS PULLEY is hard for me to figure.
Needless to say, I feel quite at a loss conceptually!
| What we mean by a frictionless pulley is that the friction in the bearings of the pulley is negligible, and the pulley is free to rotate without any resistance. We don't mean that the friction between the string and the pulley surface is negligible. In fact, we assume that there is enough static friction between the string and pulley surface to prevent the string from slipping. But, in this case, if you do a moment balance on the pulley, you must then conclude that the tensions in the string on either side are equal (even if the pulley has angular acceleration), since the moment of inertia of the pulley is assumed to be zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What does the constant $(\mu)$ mean in the formula $B=\mu(N/L)I\,?$ In the formula $B=\mu (N/L)I,$ what is the number of the constant $\mu$ for copper?
I need to know because I did an experiment where we created an electromagnet and changed the diameter of the wire. We measured the magnetic field using a Pasco sensor. I want to prove this data is correct by manually calculating the strength of the magnetic field. Bear in mind this is year 11 physics so it isn't that complex yet.
| In this case, $\mu$ is a stand-in for $\mu_0\,k$. Here $\mu_0$ is a constant $4\pi\times10^{-7}\:\mathrm{T/(A\:m)}$ and $k$ is the relative permeability of the material used for the core of the solenoid.
For example, with Fe, it is around $200\:\mathrm{N/A}^2$. Whatever material your core is, just look up the relative permeability for that substance and multiply by $\mu_0$. That will give you the $\mu$ value in your equation B = $\mu(N/L)I$. (From what I've found $\mu_0$ for Cu is $0.9999\:\mathrm{N/A}^2$.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help needed with applying pseudo force I have read in my book that to apply a pseudo force,
We make sure that the object ( our reference frame ) is accelerating and then we add the negative of it's acceleration vector to the object that we are trying to apply the pseudo force at.
Now take a look at this picture where B and A are fixed points on a circular plate.
I want to apply a pseudo force on A w.r.t. B . I see, that B has a centripetal force $F_{c2}$ and it's direction is clearly along $BO$.
Obviously, according to B, A will be at rest. And B knows that on A there is a centripetal force $F_{c1}$ and thus there must be some force on A ( the pseudo one ) in the $O$ to $A$ direction which should be equal to $F_{c1}$ to make A look stationary.
Question- Why doesn't B apply a pseudo force on A which is equal to [ $-$ $F_{c2}$ ] ( according to the definition of pseudo force in my book ) but it applies a force of [$-$ $F_{c1}$] on A ?
| I think your basic confusion starts when you say
... the object ( our reference frame ) ...
The "object" and "the reference frame" are two different things. Pseudo forces are "caused" by acceleration of the reference frame, not the acceleration of the objects themselves.
In general, the objects can also be moving and accelerating relative to the moving and accelerating reference frame.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Energy dissipated when two charged capacitors are connected in parallel The question at hand is:
"Two capacitors of capacitances $C_1$ and $C_1$ have charge $Q_1$ and $Q_2$. How much energy, $\Delta w$, is dissipated when they are connected in parallel. Show explicitly that $\Delta w$ is non-negative."
I'm confused about what the physical situation is. I took the assumption that these capacitors were somehow pre-charged, and then connected to each other in parallel without a voltage source. However, I don't understand how this would work. If a circuit is composed of only 2 elements, I don't see how they could be in any arrangement but series. Nevertheless, I tried solving it that way:
Two capacitors in parallel have the same voltage drop. Charge will be redistributed to make it the same voltage for both. Let $Q_1'$ and $Q_2'$ be the charges on the capacitors after they are connected. Now, picture the equivalent capacitor
$C_{eq} = C_1 + C_2 =$ $\frac{Q_1' + Q_2'}{V_f}$
conservation of charge:
$Q_1' + Q_2'= Q_1 + Q_2$,
$C_{eq} =$ $\frac{Q_1 + Q_2}{V_f}$
$V_f = $$\frac{Q_1 + Q_2}{C_1 + C_2}$
The initial energy of the capactiors is:
$U_0 = $$\frac{Q_1^2}{2C_1}$$+\frac{Q_2^2}{2C_2}$
$U_f = $$\frac{1}{2}$$(C_1 + C_2)V_f^2 = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$
$\Delta U$$ = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$-$\frac{Q_1^2}{2C_1}$$-\frac{Q_2^2}{2C_2}$
However, I don't see how this is necessarily non-negative.
| loss of energy when 2 capacitors are connected in parallel( -ive terminal with
-ive terminal of capacitors and +ive terminal with
+ive terminal of capacitor)
let, C1 capacitor is charged up to V1 potential.
C2 capacitor is charged up to V2 potential.
Q=CV
initial total charge on the capacitors= (C1*V1)+(C2*V2)
common potential V= total charge/total capacity,
V=((C1V1)+(C2V2))/C1+C2
energy stored, E=(CV^2)/2
final energy= ((C1+C2)*V^2)/2
initial energy= (C1V1^2-C2V2^2)/2
FINAL ENERGY-INITIAL ENERGY=-(C1*C2(V1+V2)^2)/2*(C1+C2)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Time dilation for a clock in orbit Suppose that we want to compute the total time dilation for a clock located in an orbiting satellite relative to the clock in our cell phone on the ground.
Consider two different approaches below.
*
*Use special relativity and compute time contraction due to the relative velocity. Use approximation of General Relativity in the Newtonian limit and compute time dilation due to lower gravity and then find the total time dilation.
*Don't use Special Relativity. Stick to the approximation of General Relativity based on the symmetry and find Schwarzschild metric and the geodesic for the Earth limits. Find the time dilation assuming a relative velocity in the metric.
The question is:
Which of them are more justified and provide a better approximation? Are they equivalent? What happens when the relative velocity of the satellite is zero?
How good is the approximation in either of the two approaches above.
When we pick the second approach and use the Schwarzschild metric we get this equation:
$$ dt' = \sqrt{1-\frac{3GM}{c^2r}}dt = \sqrt{1-\frac{3r_s}{2r}}dt $$
where $r_s$ is the Schwarzschild radius: $r_s = 2GM/c^2$.
Here we not only assume the asymptotic flat metric to measure $r$ but also switch to Newtonian gravity when we want to cancel $v$:
$$ v = \sqrt{\frac{GM}{r}} $$
So it appears that in the second approach there are many more approximation assumptions.
| Following on from @ProfRob's method, instead of considering the SR and GR "time dilations" separately, we can work directly from the Schwarzchild metric.
(Using signature $+,-,-,-,-$)
$$\boxed{ds^2 = c^2\left(1-\frac{r_s}{r}\right)dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}dr^2 - r^2\sin^2\theta\,d\phi^2 - r^2d\theta^2}$$
Circular orbit, so let $\theta=90^\circ \implies d\theta = 0, \sin\theta=1$. We also have $d\theta = \omega dt$, so $\displaystyle r^2d\theta^2 = v^2dt^2 = \frac{GM}{r}dt^2 = c^2\frac{r_s}{2} dt^2$.
Therefore:
$$ds^2 = c^2\left(1-\frac{3r_s}{2r}\right)\,dt^2$$
and comparing with an instantaneous rest frame (IRF) $ds^2 = c^2\,d\tau^2$, we have that:
$$d\tau = \sqrt{1-\frac{3r_s}{2r}}\,dt$$
Note we have not have to consider the Geodesic equation, or make any further approximations than those in deriving the Schwarzchild metric itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/281963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Average Velocity: $( v_1+v_2)/2$ While searching for the answer regarding, why acceleration is needed to be constant for using the formula $(v_1+v_2)/2$ , I found many simple and easy proofs regarding this, here in this Physics.SE website, one of which is ,
But can anyone come up with a daily life simple explanation for understanding why acceleration is needed to be constant for using the formula $(v_1+v_2)/2$ , for a freshman student in physics like me.
| Here's one way to think of it that might help.
If the acceleration is not constant you could have a case where something moves at a slow velocity for a long time and then accelerates briefly at the end of its motion to a higher velocity. Intuitively the average velocity should be closer to the initial slower velocity because it was travelling at that velocity for longer but the formula always puts halfway between initial and the final.
If there is a constant acceleration then half the time the velocity is slower than the average and half the time it's faster and the formula works.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why does the cross section for Compton scattered photons decrease with increasing energy of incident photon? The cross section is independent of energy for energies less than electron's mass, but for greater energies the cross section of compton scattered electrons decreases, why?
| Think about what happens as you increase the photon energy, you begin to move away from merely dislodging an electron to an energy regime in which pair production becomes more predominant.
As you probably know, in the pair-production reaction, we need a third body for momentum conservation. Not surprisingly, that heavy body is
one of the nuclei in the surface of the metal. It recoils just a little, due to it's relatively high mass, so that leaves the majority of the incident photon's energy available for pair production.
We can also discover pair production in the region of an atomic, as opposed to free, electron within the metal. Obviously in this situation, much more energy is absorbed by recoil effects. In fact, the normal cutoff turns out to be $4mc^2$.
So the combined cross section required for pair production is the sum of the cross sections associated with the above two reaction pathways. The total pair-production cross section is the sum of the two components, nuclear and electronic. These cross sections depend on the energy of the incoming photon. At high energies, approximately equal to or greater than 100 MeV, pair production is the dominant mechanism of radiation interaction with matter.
So as you increase the incident energy, you begin to see less of the Compton effect and more pair production, Compton scattering is still a factor, at low energy it can be related to the photoelectric effect and when you increase the energy, it competes with pair production.
Energy Dependent Photoelectric Effect, Compton Scattering & Pair Production
A good example of this is incident photon energy on lead. Below 0.1 MeV, we find almost totally photoelectric processes occurring , then between between 0.1 MeV and 2.5 MeV we can detect both photoelectric and Compton processes and finally, as the energy level rises above to between 2.5 MeV and 100 MeV, we can observe both Compton scattering and pair production.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Could the universe be shrinking? It is thought the universe is expanding because of the red shift of most galaxies but if all the matter in the universe was actually falling into a massive black hole wouldn't most galaxies still be red shifted because the black hole would accelerate every galaxy and every galaxy that was nearer the black hole would be accelerating away from us and we would be accelerating away from any galaxy that was further from the black hole than us. There must be something wrong with this thought can you tell me what it is?
| The red shift is in all galaxies regardless of direction. If the galaxies were falling into a black hole the red shift would happen, but considering the fact that all of them are red shifted, and all are moving away from us, the black hole would need to have the structure of a spherical shell of enormous radius or there would need to be a large number of them farther away than we can see.
Let's consider the second possibility first. If there was more than one, we could detect this because we would see lateral movement of galaxies as they move toward them. This is not seen in the data. For the first possibility, if there was possible to create a black hole as a sphere, we would not feel its effects. This is because inside a spherical shell of mass, the sum of the gravity is zero. So that would not work either.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How would you include gravity in a momentum problem?
Say you have a big ball of mass $m_1$ and a little ball on top of that of mass $m_2$ (assume they are a small distance apart, like $1~\mathrm{mm}$). Now lets drop these from a height of $h$ so that the big ball will bounce off the ground and collide into the little ball in an elastic collision.
Now I know gravity would play a key role in this example but how would one perform calculations with it? I know $F=p/t$ and momentum will not be conserved since there is an external force (gravity). So, knowing this how can one determine the height each ball will rise after the collision?
| Since both balls suffer same acceleration due to gravity, their relative acceleration is zero. This means that to analyse relative motion of balls you may pretend that there is no gravity. This is equivalent to switching to a free-fall frame, which incidentally is not an inertial frame, but that does not matter so far as you want to solve kinematic problems.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Equations of motion for a free particle on a sphere I derived the equations of motion for a particle constrained on the surface of a sphere Parametrizing the trajectory as a function of time through the usual $\theta$ and $\phi$ angles, these equations read:
$$ \ddot{\theta} = \dot{\phi}^2 \sin \theta \cos \theta $$
$$ \ddot{\phi} = - 2 \dot{\phi} \dot{\theta} \frac{1}{\tan \theta} $$
I've obtained them starting from the Lagrangian of the system and using the Euler-Lagrange equations.
My question is simple: is there a way (a clever substitution, maybe), to go on and solve the differential equations? I would be interested even in a simpler, partially integrated solution. Or is a numerical solution the only way?
| In case the particle is not subject to any external forces except those maintaining the constraint, there is no need to write and solve the equations of motion in particular system of coordinates. The particle will move with constant speed around some great circle on the sphere. Which circle it will be and the speed of motion are determined by the initial position and velocity of the particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Concentric Circular Loops We have two concentric circular wire loops. The inner loop has a stable, clockwise current. The outer loop does not have current. If the current of the inner loop increases, what is the direction of the current of the outer loop?
My first thought was "counter-clockwise" using the RHR, increasing flux, and so on. But after thinking a little more, I was wondering if the answer is "cannot be determined". I imagined the inner loop being very very small compared to the outer loop. In that case, the space between the outer and the inner loop would see an increase in B-field out of the page, and since that part has a much bigger area than the inner loop, it would have a bigger flux. The resulting current would be "clockwise" to counter that change. Is my logic in this case correct?
| Yes the area in between the outer and inner loop will be large, but the magnitude of the magnetic field inside the inner loop will more than compensate. (1) magnetic fields near a wire are larger than at points farther away from the wire, and (2) the points inside the inner loop are receiving nearly-aligned magnetic fields from a lot of points nearby. That's a qualitative explanation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What can Maxwell's Equations tell us about permanent magnets/ how are permanent magnets and electromagnets related? It makes sense that Maxwell's equations tell us that there are no monopoles, but can the equations tell us anything else about the magnetic fields of permanent magnets on their own, i.e. without interactions with a wire/current, or how such fields arise?
I only have a facile understanding of Maxwell's equations and I was wondering if someone who knows more than me can elaborate a bit. Permanent magnets and electromagnets must be intimately related somehow, but it seems a lot of the introductory literature emphasizes their differences.
|
but can the equations tell us anything else about the magnetic fields of permanent magnets on their own, i.e. without interactions with a wire/current, or how such fields arise?
Permanent magnets have a nonzero magnetization $\textbf{M}$ which gives rise to bound volume and surface currents $\textbf{J}_b=\nabla\times\textbf{M}$ and $\textbf{K}_b=\textbf{M}\times\textbf{n}$ which in turn contributes to a vector potential $\textbf{A}(\textbf{r})$. Maxwell's equation (without displacement current) $$\nabla\times \textbf{B}=\mu_0\textbf{J}$$ then, in principle, gives the magnetic field. There is nothing in classical electrodynamics that is beyond Maxwell's equations and the Lorentz force law.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Possible Error in Assumption - Griffiths Quantum Mechanics In "Introduction to Quantum Mechanics" by Griffiths, right at the beginning of section 9.1.1 (Time-Dependent Perturbation Theory, The Perturbed System), Griffiths states:
Now suppose we turn on a time-dependent perturbation, $H'(t)$. Since $\psi_a$ and $\psi_b$ constitute a complete set [of the two-level system], the wave function $\Psi (t)$ can still be expressed as a linear combination of them. The only difference is that $c_a$ and $c_b$ are now functions of t:
I don't understand. You modify the Hamiltonian, you modify the solution basis - easy as that. Why on earth does he assume that if you add a time-dependent perturbation to the Hamiltonian the basis (for the two-level system that he considered in the section right before) will remain the same? And if this is indeed a mistake, then how valid is the assumption that the true wave function $\Psi (t)$ is merely a time-dependent linear combination of the two states $\psi_a$ and $\psi_b$?
| That's why it's called perturbation. You use the Hamiltonian $H_0$ and you get a set of eigenfunctions. Then you add a perturbative Hamiltonian $H'$. Though you tweaked the Hamiltonian, the original eigenfunctions remains the same. You can always calculate the perturbative eigenfunctions using iteration method, but your original eigenfunctions is still related to $H_0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/282836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why is it that the change in internal energy always uses the formula with Cv in regards to pressure/volume/temperature changes on a gas? Normally I would associate the use of $C_v$ with finding the energy taken into or leaving a system when the volume is kept constant. However, the formula to find $\triangle E_i$ (change in internal energy) is $nC_v \triangle T$. Why $C_v$? Also, does this apply to pretty much anything? Or are there limitations?
| We call $C_v$ the heat capacity at constant volume because that is how it can be measured experimentally, by measuring the amount of heat Q added in a constant volume test and dividing by the temperature change. But, this physical property that we call $C_v$ has a more general meaning and applicability than that. In particular, in general, $C_v=(\partial U/\partial T)_V$. For an ideal gas, U(T,V) is a function U(T) only of T, and not V. So, the partial derivative becomes a total derivative, and thus, for an ideal gas, we always have $C_v=dU/dT$, irrespective of whether the volume is changing. But we can still measure Cv directly by measuring the amount of heat Q added in a constant volume test.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/283054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
One straw drinking from many containers of liquid One of my friends brought up a photo:
Which sparked a debate about whether the containers closest to the end of the straw would empty first. I was just wondering if someone could explain if the closest two containers would be empty before the furthest.
| The containers would empty at a different rate with a restrictive orifice separating the containers. These straws may very well qualify. Without knowing the plastic viscosity and yield point of the fluid and an exact measurement of the ID, length and exact suction on the straw it can not be answered accurately. BUT...Looking at the straws and knowing about how thick those things are and assuming an average sucker if you will I strongly suspect with this contraption the last containers would empty a bit more slowly. Also the side with the thicker fluid properties probably would be a bit slower than the thinner ones too.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/283183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Explain difference between internal energy and enthalpy I read that $U=q+w$ and $H=u+PV$ so aren't $PV$ and $w$ same?
If they are, can't we write $H=q+2w$?
Also,
$dU=$ change in internal energy wrt change in temperature keeping $V$ constant times $dT+$change in internal energy wrt change in volume keeping $T$ constant times $dV$
Here is the second term $=w$?
And,
$dH=$ change in enthalpy wrt change in temperature keeping $P$ constant times $dT+$ change in enthalpy wrt change in pressure keeping $T$ constant times $dP$.
Here is the second term $=PV$?
| According to the sign convention you are using, $$\Delta U=q+w$$where, for a reversible expansion or compression, $$w=-\int{PdV}$$So, $$\Delta U=q-\int{PdV}$$and$$\Delta H=\Delta U+\Delta (PV)=q-\int{PdV}+\Delta (PV)$$Integrating by parts, we get$$\Delta H=q+\int{VdP}$$
The quantity $$\int{\left(\frac{\partial U}{\partial V}\right)_TdV}$$ is not equal to the work. In fact, for an ideal gas, this quantity is always equal to zero. Similarly for $$\int{\left(\frac{\partial H}{\partial P}\right)_TdP}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/283283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If mass density curves space-time, then why isn't density (at each $x$, $y$, $z$) considered a dimension in space-time? From http://science.howstuffworks.com
"Theodor Kaluza theorized that a fourth spatial dimension might link general relativity and electromagnetic theory. But where would it go? Theoretical physicist Oskar Klein later revised the theory, proposing that the fourth dimension was merely curled up, while the other three spatial dimensions are extended."
We know high mass density creates gravitational waves that compress length and time (relative to an observer at lower ambient gravitational waves). Why not include density as a dimension in space-time?
| Because it doesn't add another term to the length element?
One way to understand the dimensionality of space is to examine the nature of the length element. In three dimensional space (and with a Cartesian basis) it is
$$ (\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \;, $$
which you will note has three terms. In four-space it would be
$$ (\Delta s)^2 = (\Delta w)^2 +(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \;. $$
Density doesn't add a term to the length element, instead it effects the coefficients of the sum. Something like
$$ (\Delta s)^2 = c_x(\Delta x)^2 + c_y(\Delta y)^2 + c_z(\Delta z)^2 \;, $$
with no guarantee that the $c$'s are unity.
That's a different kind of modification to the meaning of space.
Aside: This kind of definition has a couple of nice features. First, it extends smoothly to include intervals in Minkowski space or the more generalized space of general relativity. Second it allow a clear explanation of why the proposed compact dimensions of string theory don't appear to effect physics at human (or even nuclear) length scales.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/283377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Theoretically, could there be different types of protons and electrons? Me and my friend were arguing. I think there could theoretically be different types of protons, but he says not. He says that if you have a different type of proton, it isn't a proton, it's something else. That doesn't make sense to me! There are different types of apples, but they're still called apples!
He says that's how protons work, but can we really know that?
| There are in fact 2 types of Protons - Protons and anti-Protons.
They interact and behave exactly the same way and are completely indistinguishable from one another. However they will mutually annihilate on contact.
Other than that refer to the other answers here. If a particle looks like a Proton, acts like a Proton, reacts like a Proton - then it's a Proton.
If you have a particle that shares some similarities with a Proton (e.g. a previously unknown stable Hadron that has a positive charge); but is different in some way - mass, charge quantity, etc... well then it won't be called a Proton, it would be called something else.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/283682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 11,
"answer_id": 9
} |
Mass of the asymptotic fields: physical or bare? If I understand it correct, then the physical mass $m$ of a particle, is the mass in presence of the interaction (i.e., the mass of the dressed particle) where as the bare mass $m_0$ is the mass in absence of interaction. However, in the derivation of LSZ reduction formula, as given in Bjorken and Drell, it is said in Eqn. 16.6 that the 'in' state $\phi_{in}(x)$ at the asymptotic past $t\rightarrow -\infty$ (and similarly, the 'out' state $\phi_{out}(x)$ at the asymptotic future $t\rightarrow +\infty$) obeys free Klein-Gordon equation with the physical mass m.
But since the 'in state' is a free-particle state, shouldn't the KG equation be written in terms of th bare mass $m_0$?
| The physical mass can be defined as the pole of the propagator
$$\int d^4x e^{-iq\cdot x} \cdot\langle \Omega|T\{\Psi_l(x)\Psi^\dagger_{l^\prime}(0)\}|\Omega\rangle$$
where $\Psi_l$ are renormalized fields.
The LSZ reduction formula says that if a one-particle state with momentum squared $m^2$ has non-vanishing matrix elements with the states $\Psi^\dagger_l|\Omega\rangle$, then $m^2$ is the pole of the propagator. So the square of the physical mass of a field equals to the momentum squared of the in state (or out state), which is just the mass squared present in the asymptotic K-G equation.
Note that the in state is not a free state (they are energy eigenstates belonging to two different systems, which have a one-to-one correspondence, but it doesn't mean they have the same mass) and that the mass can only be measured via interaction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/283862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
What is the name of a quantity expressed in Tm? Consider the quantity defined as a magnetic field times a length is expressed in tesla*meter:
$Bl\ [\mathrm{Tm}]$
that arises for instance in the expression $F=BlI$.
What is the appropriate terminology for this quantity $Bl$?
| A common quantity is the magnetic flux, $\int \mathrm dA \, B$. That is measured in $\mathrm{T \, m^2}$.
Your quantity seems to be a linear flux density, so the flux density integrated along a single dimension. That would be a little strange though.
It could also be a line integral, like $\oint \mathrm d\vec l \cdot \vec B$. This is used in Maxwell's equation in Ampère's law. The quantity could be called “magnetic line integral” or so, I do not know a canonical name for it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Muon Decay Lagrangian I am working on figuring out the following problem
The muon decays to an electron and two neutrinos through an
intermediate massive particle called the W$^-$ boson. The muon,
electron and W$^-$ all have charge -l.
(a)Write down a Lagrangian that would allow for $\mu^-\rightarrow
e^-\bar{\nu_{e}}\nu_\mu$ Assume the W and other particles are all
scalars, and the e$^-$, $\nu_e$ and $\nu_\mu$ are massless. Call the
coupling g.
So I know that I will have some free L for all particles
$$L_{free}=\frac{1}{2}(\partial \phi([\mu])^2+\frac{1}{2}(\partial \phi[W])^2+\frac{1}{2}(\partial \phi[e])^2+\frac{1}{2}(\partial \phi[\nu_e])^2+\frac{1}{2}(\partial \phi[\nu_\mu])^2+\frac 12 m_\mu^2\phi[\mu]+\frac 12 m_W^2\phi[W]$$
But for the interaction, is it enough to say that
$$L_{int}=-g\phi[\mu]\phi[W]\phi[e]\phi[\nu_e]\phi[\nu_\mu]$$
Or do I have to have different interactions for the decay shown in the image below?
| Note that the Feynman diagram includes two different types of interaction vertex, so each one will need a term in the Lagrangian:$$-\mathcal{L}_{int}=g \bar{\nu}_\mu W^+ \mu + g \bar{\nu}_e W^+ e + \mathrm{h.c.}$$
Both terms have the same coupling $g$ because of the universality of weak interactions.
Usually, the energy of the initial and final particles is very small compared to the mass of the $W$ boson (around 80 GeV). Therefore, we can integrate out the $W$ boson in order to obtain an effective Lagrangian that describes only the low energy degrees of freedom of the theory: $$-\mathcal{L}_{eff} = G [\bar{\nu}_\mu \mu][\bar{e}\nu_e]$$ where $G$ is a dimensionful coupling constant (while $g$ is dimensionless) which depends on the mass of the $W$ boson as $G\sim M_W^{-2}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the shape of the detector pattern in a Stern-Gerlach experiment with a beam source (instead of a fan)? I've been delving into Bell test experiments and woefully most sources fall into either dense physics papers, or very low level simplifications with lots of handwavium.
One thing in particular I'd like to understand a bit better is the Stern-Gerlach device. The original experiment used a rectangular or fan-shaped beam, which results in a pattern on the detector kind of like a Gaussian shape, with a mirror image, like this:
Accounting for blurring/uncertainty and such, if you trace the local maxima you end up with this cat-eye shape.
Now as I understand it, spin magnetic moment $\vec{\mu}_S$ is a vector in 3-space, and firing something with net SMM (i.e. silver atom) through the SG device "measures" $\vec{\mu}_S$ along the up-down axis of the device, which we call $z$. Atoms which deflect maximally up or down are said to have spin aligned to that axis. But what happens to particles that don't have $z$-aligned spin? In other words, if the aperture were beam-shaped, how would the pattern on the screen appear, and why?
My intuition is that it would be ring-shaped or elliptical, since particles with non-perfectly-aligned spin would have to deflect less (along the SG device's major axis). But intuition and QM rarely mix.
Edit: additional question: Does the SG device alter the spin moment of the particle (much like torque on a gyroscope causes procession)?
| From a practical point of view the reason for using of a broad flat beam may be as simple as getting easy alignment and decent rate while not blurring the signal with a significant z-direction dispersion.
No need to over-complicate things.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
What does it mean for a system to be integrable? I am reading this paper: http://aleph.physik.uni-kl.de/~korsch/papers/JPA_24_45.pdf and in section 3 they talk about the integrability of the system. What does that mean?
| The paper refers to the notion of complete integrability, aka. Liouville integrability, i.e. the property that there exist $n$ independent globally defined Poisson-commuting constants of motion. Here $2n$ is the dimension of phase space. In the paper $n=2$. See also this related Phys.SE post.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Motion of one body with reference to another I studied that Galileo was punished by the church for teaching that the sun is stationary and the earth moves around it. His opponents held the view that earth is stationary and sun moves around it.
The question i want to ask is if the absolute motion has no meaning,are the two viewpoints not equally correct or equally wrong?
Thanks in advance for any help.
| *
*Here the important thing is the frame of reference. When we look from the earth frame, the sun is rotating around the earth.
*When we look from sun frame, the earth is rotating around the sun.
*When we look at them from outside of both frames, like when we are looking from the galaxy of sun system, then we come to know that the earth is rotating around the sun.
So both points of view are different. Galileo looked at it from outside of the two frames. The church looked at it in the earth frame.
That is a nice question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Measuring very small temperature differences Can one use a thermometer with $\pm$5 mK accuracy to measure a temperature difference of 2 mK (the measurement is near 100 mK temperature on a sample on an ADR)? Using the same thermometer, I am thinking to measure temperature of the sample, heat the sample slightly, measure temperature again, and take the difference. Does the $\pm$5 mK uncertainty cancel out when I take the difference? My thermometer is sensitive enough, my AC resistance bridge is capable of resolving such small temperature differences, but I want to know if the $\pm$5 mK is really an issue here.
| Assuming the accuracy reported does not include any nonlinear factors, it refers to the inaccuracy in a single measurement, and will not vary much over the time of your experiment, then you can increase precision by integrating multiple measurements. Signal (the temperature) will add proportional to the number of measurements and noise (the inaccuracy) will add proportional to the square root of the number of measurements. Also, it is typical that inaccuracy will be fairly correlated between measurements and imprecision will be substantially less and uncorrelated. It sounds like you care a lot more about precision than accuracy--so that should help tremendously.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Partition function - q-number or c-number, classical definition, etc Why is the partition function
$$Z[J]=\int\ \mathcal{D}\phi\ e^{iS[\phi]+i\int\ d^{4}x\ \phi(x)J(x)}$$
also called the generating function?
Is the partition function a q-number or a c-number?
Does it make sense to talk of a partition function in classical field theory, or can we define partition functions only in quantum field theories?
Is the source $J$ a q-number or a c-number?
| It is called a generating function, because one can use it to generate $n$-point functions with the aid of functional derivatives with respect to the source $J$. For instance, one can compute the two point function as follows:
$$ \langle\phi(x_1)\phi(x_2)\rangle = \frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_2)} Z[J]_{J=0} . $$
The result would be a $c$-number. Hence, the generating function itself is also a $c$-number.
Formally one can treat classical field theories also with the aid of such generating functions, provided that, if you set $J=0$, you recover the original theory.
These generating functions are based on the path integral approach in which fields can be interpeted as $c$-numbers as apposed to the $q$-numbered operator-valued fields used in the second quantization approach. As a result, the source $J$ is also interpeted as a $c$-number.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/284874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Moment of Inertia of Annular Quadrant I am measuring the moments of inertia for various numbers of annular quadrants placed on a torsional oscillator. I know $\displaystyle{I=\frac{1}{2}M(R^2+r^2)}$ for a whole annulus. If I want the moment of inertia of only an annular quadrant, would I divide the formula above by four?
| *
*The MMOI about the geometric center is still $$I_{center} = \frac{m}{2} (r_1^2+r_2^2) $$
This is a result of the relationship ${\rm d}m = \rho z r \,{\rm d}r {\rm d}\theta$ and ${\rm I} = r^2 {\rm d}m$
$$ \begin{aligned} m & = \rho z \int \limits_{r_1}^{r_2} \int \limits_{0}^\Theta r\, {\rm d}\theta {\rm d}r & I_{center} & = \rho z \int \limits_{r_1}^{r_2} \int \limits_{0}^\Theta r^3\, {\rm d}\theta {\rm d}r \end{aligned} $$
*
*The MMOI about the center of mass is found from the center of mass distance $c$ and the parallel axis theorem $I_{cm} = I_{center} - m c^2$
$$ \begin{aligned} c &= \frac{4 (r_1^2+r_1 r_2 + r_2^2) \sin \left(\tfrac{\Theta}{2}\right)}{3 \Theta (r_1+r_2) } \\ \\
I_{cm} & = \frac{m}{2} (r_1^2+r_2^2) - \frac{8 m (r_1^2+r_1 r_2 + r_2^2)^2 (1-\cos\Theta)}{9 \Theta^2 (r_1 +r_2)^2 }
\end{aligned} $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is difference between operating wave function with operator of an observable and measuring for an observable? People say operator of an observable helps in measuring for an observable. We also know that measuring leads to collapse of wave function. But operator on wave function gives a number times same wave function (which of course is not a collapsed wave function!). All intuitions I made about operator, wave function, measures, collapse are all seeming to be inconsistent. If operator doesn't collapse a wave function then what it is for. Is it just for calculating expectation value of observable. What in physical sense it is?
| There are a few different points/distinctions that need to be made here.
1. The expectancy value when measuring an observable to which the operator $A$ can be assigned is $\langle A \rangle = \langle \Phi | A | \Phi \rangle$ for a system described by a wavefunction $|\Phi\rangle$.
2. If the value $a$ is measured, the collapse of the wavefunction means that we project $ |\Phi\rangle$ onto the eigenspace of the eigenvalue $a$ of the operator, i.e., the wavefunction is changed.
3. $A | \Phi \rangle = a | \Phi \rangle$ for $a \in \mathbb{C}$ holds only if $| \Phi \rangle$ is an eigenfunction of the operator $A$. In general, $A | \Phi \rangle$ does not need to be proportional to $| \Phi \rangle$ but can be a different wavefunction.
4. An operator $A$ determines the possible values of its measurement variable by its eigenvalues and determines the possible wavefunctions after it has been measured by its eigenstates/eigenspaces.
5. Applying an operator to a wavefunction does not describe the measurement process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is Del (or Nabla) an operator or a vector? Is Del (or Nabla, $\nabla$) an operator or a vector ?
\begin{equation*}
\nabla\equiv\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k}
\end{equation*}
In some references of vector analysis and electromagnetism, it is considered as an operator (and noted as $\nabla$), and in other ones, it is considered as a vector (and noted as $\vec\nabla$).
| Both. It's an operator that transforms as a covector under rotations. What this means is that if you rotate the coordinate system the gradient in the new coordinate system, $\nabla'$, can be written as:$$\nabla'_i = \sum_{j} R^{-1}_{ij} \nabla_j,$$ where $R^{-1}$ is the inverse of the rotation matrix, $\nabla$ is the gradient in the original coordinate system, and $\nabla'$ is the gradient in the rotated coordinate system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Is there a notion of causality in physical laws? I was reading "A Few Useful Things to Know about Machine Learning" by Pedro Domingos and towards the end of the paper he makes this statement:
"Many researchers believe that causality is only a convenient fiction. For example, there is no notion of causality in physical laws. Whether or not causality really exists is a deep philosophical question with no definitive answer in sight..."
I was surprised by this statement because apart from Heisenberg's uncertainty principle, everything else (that I know of) in physics seems to operate under the assumption of causal relations. If you have an equation that describes a definitive outcome as the result of some input factors, then it is describing a causal relationship, is it not?
| Saying that there is "no notion of causality in the physical laws" is likely a gross misinterpretation of the lack of time-asymmetry in the majority of physics. This is to say that many physical laws hold equally well given a flow of time in either the forward or reverse directions with respect to the flow of time we experience.
The concept of entropy, as described in the Second Law of Thermodynamics, serves somewhat to rectify this apparently non-physical balance in the laws of nature as described by science. Really, I think we need take nothing more from this than that the majority of physical laws are not but convenient mathematical approximations whose scope of applicability dare not exceed their reach. The science will go wrong long before the philosophy when you start trying to draw existential conclusions from mathematical relations forged of rather more modest constraints.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 10,
"answer_id": 3
} |
Why doesn't a permanent magnetic field not make a florescent light illuminate like an electromagnet field? Similar to this question: What is the difference between the magnetic field of a permanent magnet, and that of an electromagnet? mine is different because I want to know why a permanent magnet with the same field intensity does not light up a fluorescent light like these pictures below with an electromagnetic field? Can a permanent magnet be made to make a florescent light illuminate by adding a current to the permanent magnet?
| A permanent magnetic field does not transfer kinetic energy to charge carriers like an applied electric field in a gas discharge tube. Thus electrons are not able to impact ionize or excite atoms so that they can emit light that would bring the fluorescent material in a discharge tube to fluorescence.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the warped space around the Earth push objects towards the Earth? In the following video, the theoretical physicist Michio Kaku states that space is pushing objects towards the earth. https://youtu.be/fEZupmpTcOU?t=1m59s
Is the warped space around the earth providing a force that is accelerating objects to the earth?
Is this the correct interpretation of Einstein’s mathematical model of gravity?
EDIT:
I do not understand the mathematics of general relativity, but I do have a basic understanding of how the model of general relativity works.
In Newton's theory, gravity makes particles leave their straight paths. In Einstein's theory of general relativity, gravity is a distortion of space-time. Particles still follow the straightest possible paths in that space-time. But because space-time is now distorted, even on those straightest paths, particles accelerate as if they were under the influence of what Newton called the gravitational force.
This quote is from the website einstein-online.
Michio Kaku actually said that space is pushing objects towards the Earth. Did he misspoke?
Frank Wilczek said, “We can describe general relativity using either of two mathematically equivalent ideas: curved space-time, or metric field.”
Kip Thorne said, “You can reformulate Einstein’s laws in a sort of Newtonian way” https://youtu.be/rHsBDTy3yEE?t=5m7s
I am wondering if there is more than one valid interpretation of what Einstein’s mathematical model of gravity actually represents.
| No, warped space is not "providing a force". Gravitation warps the space. In other words, the causality is gravitation->warped space, so you/Kaku have it backwards. Notice that, as with other causal effects, the warping of space travels at the speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding an approximate expectation value $\langle E_0|\hat{O}|E_0\rangle$ when i dont know the ground state? I'm assuming that i know the hamiltonian although i don't know it's ground state $|E_0\rangle$ and that i have a way to find $|\psi(s)\rangle\equiv e^{-\hat{H}s}|{\psi}\rangle$, $\forall s\in\mathbb{R}$ with units one over energy. I want to find an approximate expectation value $\langle E_0|\hat{O}|E_0\rangle$ for some hermitian operator $\hat{O}$.
I have, so far, taylor expanded the exponential operator and truncated the taylor expansion with an error that goes like $\mathcal{O}(s^2)$, that is
\begin{align*}
e^{-\hat{H}s} \approx \mathbb{I} - \hat{H}s + \mathcal{O}(s^2)
\end{align*}
Also, if we expand $|\psi\rangle$ in terms of the energy eigenstates we can find try to find the expectation value of some hermitian operator $\hat{O}$, or try to find the norm of $|\psi(s)\rangle$, but to be honest, I'm a bit stuck. We haven't gone through the Variational principle yet, but is that what we're asked to work with?
| Can we assume that all eigenvalues of $\hat H$ are bounded from below and/or positive and there is a gap $\Delta$ between the lowest and second lowest eigenvalue and every $| \psi \rangle$ can be expanded into sum of eigenvectors? If yes then we could explore the limit
$$
\lim_{s \rightarrow \infty} \frac{\langle \psi(s)| \hat O | \psi(s) \rangle}{\langle \psi(s)|\psi(s) \rangle}.
$$
The underlying observation is that $|\psi(s)\rangle \approx e^{-sE_0} \left(\alpha_0|E_0\rangle + e^{-s\Delta}\alpha_1|E_1\rangle + \dots \right)$. Knowing the value of $\Delta$ also allows to quantify how big $s$ has to be to provide sensible approximation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/285778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why the clock at rest runs faster, while another clock slows when moving? I have observed from my first question that it is hard for me to study the special relativity from every frame of reference. But, there is one most important question in my head right now that time runs slower for moving body if observe from rest and time runs faster in clock at rest if observe from that moving body. But, the rate at which the ticks slower for one and faster for another is different. Why it is not the same rate? Please answer in brief and simple language.
|
Please answer in brief and simple language.
If Alice and Bob are moving relative to each other with constant speed $v$, both of the following statements are true:
(1) Alice observes Bob's clock to run slow by a factor of $\frac{1}{\gamma_v}$
(2) Bob observes Alice's clock to run slow by a factor of $\frac{1}{\gamma_v}$
This is an elementary result of the Lorentz transformations that relate Alice's and Bob's spacetime coordinates. See, for example, this answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Gravity between two Photons (I searched for an answer online already but I couldn't quite find what I was looking for...)
I thought about this for a long time now. If two Photons fly in the same direction, one behind the other one, for my understanding the one behind the other one should be pulled towards the photon in front of it due to it's gravity, and because it cant get faster it should increase it's frequency and therefore gain energy. The one in front cant be pulled backwards though because gravity travels with the speed of light itself(?) and therefore the gravity of the rear photon cant reach the one infront of it, which would therefore not lose energy.
But that would break the law of conservation of energy, wouldn't it? So I'm confused...
Am I thinking something wrong? Or how does it work/what would actually happen in this scenario?
Thanks for answers in advance!
| From my understanding:.
Photons have a rest mass of zero, but the electromagnetic waves do carry energy which causes gravitational pull. Since light isn't affected by gravity because its rest mass is zero, but the space light is moving on gets curved. This is what causes light beams to curve around massive objects. The two light beams wont be pulled into one another, but will start 'bending' if they are not moving parrelel to each other. This could get them closer to each other without changing speed.
A better explanation can be found in this Physics.SE post
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
Inclined Plane and Center of Mass Say there is a block sliding down an inclined plane that rests on a frictionless table. There is kinetic friction between the block and inclined plane. If the block slides downhill, then the kinetic friction acting on it points uphill. By Newton’s third law, the inclined plane will experience a friction force pointing down hill, in the direction of the block’s velocity/acceleration. Shouldn’t the plane want to move with the block, then? If it does, then wouldn’t the center of mass move with the block and plane too? There’s no friction on the block–plane system, however, so the center of mass should not move, but my analysis claims that it does. Where have I gone wrong?
| You are forgetting the normal force between the block and the wedge (inclined plane). This force has a horizontal component, pushing the wedge to the left as the block slides down to the right.
The 2 forces acting on the wedge due to the block are the normal reaction $N$ and the friction force $F=\mu |N|$ (see diagram). Both have horizontal components : $N\sin\theta$ and $F\cos\theta$ respectively. The net horizontal force acting to the left on the wedge is $N\sin\theta-F\cos\theta=(\tan\theta -\mu)N\cos\theta$.
The condition for the block to start sliding down the incline is $\tan\theta \gt \mu$. So if the block slides to the right then the horizontal force on the wedge is always +ve to the left.
Note that in this situation the normal force is not $N=mg\cos\theta$. The wedge accelerates to the left away from the plane of contact with the block, so the normal force is reduced from this value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is water evaporated from the ocean not salty? I am wondering about this. When salty water in the ocean evaporates we are getting the clean distilled water. Why is that?
I was trying to think on this and maybe the comparative size/mass of water molecules to the size of different salts molecules plays a role here, but it is not likely. The size/mass of e.g. alcohol molecules is much higher but they still evaporate quite well.
Could you explain intuitively?
| Alcohol will boil off first
The boiling point is 78.37 °C
There will be some water but the ratio of alcohol in the vapor phase will be higher than the liquid. Water and alcohol are azeotropes so things flop at about 70% alcohol.
The size is not that much of a factor - hexane is volatile.
wiki
Table salt (NA CL) is not much more volatile than a rock
melting point 801 °C (1,474 °F)
boiling point 1,413 °C (2,575 °F)
wiki
The lowest vapor pressure I could find is at 759.88 °K = 486.73 °C
1.88322997019E-006 kPa
Vapor Pressure of Sodium chloride
The vapor pressure of water is 101.32 kPa at 100 °C
Water is 53,801,182 times more volatile at 100 °C than salt at 486.73 °C
Salt is dissolved in water but that does not make the salt more volatile.
Basically no salt evaporates. With very rapid boiling you might get some entrainment. The vapor phase is going to have essentially not salts or minerals as they are solids are at 100 °C.
Take a quart of water and add a cup of salt. Put it on the stove and let all the water boil off. You will have a cup of salt.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 4,
"answer_id": 2
} |
Problem with Bogoliubov transformations of an operator I have a set of Bogoliubov transformation as follows:
\begin{equation}
a(\beta) = a \cos \theta (\beta) - \tilde{a}^\dagger \sin \theta (\beta)\\
\tilde{a}(\beta) = \tilde{a} \cos \theta (\beta) + a^\dagger \sin \theta (\beta)
\end{equation}
And their corresponding equations obtained by taking the adjoint of both equations. Now, we define the following:
\begin{equation}
A = \begin{pmatrix}a\\\tilde{a}^\dagger \end{pmatrix}
\end{equation}
And say that $A$ transforms according to the following unitary transformations:
\begin{equation}
U(\theta)AU^\dagger(\theta)=\bar{U}(\theta)A\\\\
\text{where}\end{equation}
\begin{align}U(\theta) &= \exp{\left(-\theta\left(\tilde{a} a - \tilde{a}^\dagger a^\dagger\right) \right)}\\
\bar{U}(\theta) &=\begin{pmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}
\end{align}
$a,\tilde{a} $ are the annihilation operators of the particle and antiparticle respectively. Problem here is, I'm not sure of the correctness of the transformation equation. If the creation and annihilation operators are $n\times n$ matrices, then the dimension of $U$ is $n\times n,$ and the left hand side of the transformation equation isn't valid since an $n\times n$ matrix can never left multiply with another $2n\times n$ matrix.
I've been trying to resolve this dilemma but very little progress. The same sort of unitary operator transformations are given everywhere, but are valid only if the operator is a matrix of the same dimensionality as the unitary matrix. These sort of equations are also given in group theory in order to establish the lie algebra of any group. But the same problems are encountered over there as well. Where am I going wrong?
Reference:
Ashok Das: "Finite Temperature Field Theory", chapter 3, equations (3.29)-(3.32)
| Based on some of the expressions that are provided, it seems the issue is one of notation. So, I'm going to be ratherpedantic. Let's define
$$ \{a, \tilde{a}\} \equiv \left\{N^{(1)}_{ij}, N^{(2)}_{ij}\right\}$$
so that
$$ A \equiv N^p_{ij} $$
where $p$ is an index that selects either $a$ or $\tilde{a}$ and $ij$ are indices that run from 1 to $n$ and represent the indices of the creation and annihilation operators. Now we can say that $U$ operates on the $ij$-indices,
$$ U\equiv U_{ij} $$
whereas $\bar{U}$ operates on the $p$ index
$$ \bar{U}\equiv \bar{U}_{pq} $$
So then the expression
$$ U(\theta)AU^\dagger(\theta)=\bar{U}(\theta)A $$
in my pedantic notation becomes (dropping the $\theta$)
$$ \sum_{ij} U_{ki}N^p_{ij} U^{\dagger}_{jl}= \sum_q \bar{U}_{pq}N^q_{kl} . $$
Hope this clarifies the issue and that I've interpreted the equations correctly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What parts of a physics undergraduate curriculum have been discovered since 1966? What parts of an undergraduate curriculum in fundamental physics have been discovered since, say, 1966? (I'm choosing this because it's 50 years ago.)
Physics textbooks have moved on since 1966 (though even quantum mechanics was already in textbooks in a form pretty close to the present, as e.g. Landau and Lifshitz was in its second edition in 1965). However, a strong case can be made that a large fraction of the material that has been included after 1966 was discovered before that - i.e. that the physics textbooks have simply been catching up to the state of research in the 60s.
(As an example, the Standard Model of electroweak interactions took on its modern form in 1967, and QCD, which is rarely studied in detail in undergraduate courses, followed shortly afterwards.)
I would like to disallow applications. I'm much more interested in finding out what (if any) fundamental shifts there have been in physics at the undergraduate level.
One reason I am asking is my suspicion is that there is very little or nothing which physics undergraduates learn which has been discovered since the 1960s, or even possibly earlier. Am I wrong?
This question is based on a similar one on Math Stack Exchange.
| Maybe high-temperature superconductivity?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Should zero be followed by units? Today at a teachers' seminar, one of the teachers asked for fun whether zero should be followed by units (e.g. 0 metres/second or 0 metre or 0 moles). This question became a hot topic, and some teachers were saying that, yes, it should be while others were saying that it shouldn't be under certain conditions. When I came home I tried to find the answer on the Internet, but I got nothing.
Should zero be followed by units?
EDIT For Reopening: My question is not just about whether there is a dimensional analysis justification for dropping the unit after a zero (as a positive answer to Is 0m dimensionless would imply), but whether and in which cases it is a good idea to do so. That you can in principle replace $0\:\mathrm{m}$ with $0$ doesn't mean that you should do so in all circumstances.
| The question cannot be answered generally because it depends on the situation - on what exactly you mean. If you mean "zero mass" then writing $0 \textrm{g}$ or $0 \textrm{kg}$ or something like that is very reasonable. If you mean an abstract, unitless zero from $\mathbb{R}$ - well, that one is unitless and should be written without units.
It strictly depends on what your numerical value wants to express. Numerical values wanting to express some physical quantity which has units should be followed by the appropriate unit, unitless quantities and abstract numbers should not be.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/286964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "72",
"answer_count": 10,
"answer_id": 7
} |
Where does the extra kinetic energy of the rocket come from? Consider a rocket in deep space with no external forces. Using the formula for linear kinetic energy
$$\text{KE} = mv^2/2$$
we find that adding $100\ \text{m/s}$ while initially travelling at $1000\ \text{m/s}$ will add a great deal more energy to the ship than adding $100 \ \text{m/s}$ while initially at rest:
$$(1100^2 - 1000^2) \frac{m}{2} \gg (100^2) \frac{m}{2}.$$
In both cases, the $\Delta v$ is the same, and is dependent on the mass of fuel used, hence the same mass and number of molecules is used in the combustion process to obtain this $\Delta v$.
So I'd wager the same quantity of chemical energy is converted to kinetic energy, yet I'm left with this seemingly unexplained $200,000\ \text{J/kg}$ more energy, and I'm clueless as to where it could have come from.
| Another much clearer way to see the effect of the Oberth effect is when you add the Potential energy to the equation.
When you perform a rocket burn inside the gravity well of a massive body, the propellant ends up in a lower orbit than if you perform the rocket burn outside of the gravitational well.
The difference in the Potential Energy of the propellant will equal the difference in the Kinetic Energy of your space probe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/287101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 0
} |
Field of a Polarized Object In Griffith's Electrodynamics, in the section 4.2, just after the equation 4.9, he writes "sleight-of-hand casts this integral into a much more illuminating form"...
I have a doubt in that. If the Gradient (or differentiation if carried out) is with respect to primed coordinates, how can variable r be differentiated as r' ? It would be of great help if someone clarifies this point.
| The point is that the 'scripty r' (i dont know how to write it here) depends only on the difference between the coordinates; note that
($\frac{\partial}{\partial x}$) $f(x- x')$ = -($\frac{\partial}{\partial x'}$)$f(x- x')$. What i am trying to say is that the 'scripty r' is a vector joining the primed co-ordinates to the unprimed co-ordinates.....hence the gradient is taken wrt either of the co-ordinates....ie, gradient wrt unprimed is negative of gradient wrt primed co-ordinates
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/287383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving a problem using Newtonian mechanics and D'Alembert principle yI have to solve that problem with two methods (applying Newtonian mechanics and the D'Alembert principle.
The problem consists in two balls inside a spherical cylinder, it consists in determine the minimum value of $M$ making the tube not to knock down (where $M$ is the mass of the cylinder and $m$ the masses of the two spheres).
I have issues with both methods. With Newtonian method, I don't know what influence has $M$ on the problem, because I can choose a reference point in the center of the cylinder and there will be no torque.
With D'Alembert principle, the problem is I have no idea what virtual displacement I have to choose.
The Newtonian process brings me to this meaningless expression if the normal force acts on the lower right corner.
| If there is no friction between any contact surfaces, then the centres of the spheres, the points of contact and the axis of the cylinder will all lie in the same plane. So this is a 2D problem.
The 2 spheres exert horizontal forces on the cylinder at their points of contact with it. These forces are not aligned, so there is a clockwise torque. The cylinder will (potentially) topple about the lower RH corner. The weight of the cylinder exerts an anti-clockwise torque. Balancing these two torques, it should be possible for you to work out what conditions are required for stability.
You can work out the force $F$ which the cylinder exerts on the upper sphere (and the sphere exerts back on the cylinder) by balancing moments of $F$ and the weight $W$ of the upper sphere about the point of contact between the 2 spheres. If all contacts are frictionless the horizontal force exerted on the lower sphere by the cylinder must also equal $F$. These are the only 2 horizontal forces on the 2 spheres, which are in equilibrium, so they must be equal and opposite (but not necessarily aligned).
For the D'Alembert Principle, any small displacement of the structure (subject to the given constraints) should require no work when the structure is in equilibrium. You could (I think) displace the walls of the cylinder by increasing or decreasing radius R by a small amount.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/287465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why are planets not crushed by gravity? Stars can be crushed by gravity and create black holes or neutron stars. Why doesn't the same happen with any planet if it is in the same space time?
Please explain it in simple way. Note: I am not a physicist but have some interest in physics.
| Planets are crushed by gravity! That's why, for example, Earth is a densely packed spherical rock rather than a loose cloud of dust.
There's just not enough crushing 'force' to do more than that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/287622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 6,
"answer_id": 1
} |
The central field approximation and the quantum number $n$ Under the central field approximation the Hamiltonian of a multi-electron atom is approximated as:
$$ \hat H= \sum_i \left( -\frac{\hbar^2}{2m} \nabla_i^2 +V(r_i) \right)$$
for some central potential $V(r_i)$. This can be solved separately for each electron, thus reducing the problem to solving:
$$-\frac{\hbar^2}{2m} \nabla_i^2\psi_i +V(r_i)\psi_i=E\psi_i$$
This can be further solved by separation, with the angular part been given (as usual) by the spherical harmonics $Y^m_l(\theta, \phi)$ and the radial part been given by $R(r)$. Now it is commonly accepted that $R(r)$ depends on two quantum numbers $n$ and $l$ analogous to hydrogen. Clearly it must depend on $l$ due to the presence of the spherical harmonics, but given that $V(r_i)$ is a rather arbitrary potential how can we be sure that $R(r)$ depends on one and only one further quantum number $n$?
| The inter-electron repulsion contains a large spherically symmetric component. So, it is possible to construct a potential energy function $V(r_i)$ which is spherically symmetric. The Hamiltonian for such a potential is written as follows using perturbation theory.
$H$ = $H^*$ + $H'$
where $H^*$ = $\hat H= \sum_i \left( -\frac{\hbar^2}{2m} \nabla_i^2 +V(r_i) \right)$
$V(r_i)$ = $-\sum_i ^N\frac{Ze^2}{r_i}$ + $\langle \sum_{i<j}\frac{e^2}{r_{ij}} \rangle$ (Not arbitrary; spherically symmetric; still a function of $\frac{1}{r}$ only)
and $H'$ = $\sum_i ^N\frac{e^2}{r_{ij}}$ - $\langle \sum_{i<j}\frac{e^2}{r_{ij}} \rangle$ (Clearly, no spherical part)
The $\langle \sum_{i<j}\frac{e^2}{r_{ij}} \rangle$ term, which is added and subtracted is the average over a sphere of the electron repulsion. In $V(r_i)$, it is added to mean that the attraction experienced by the $i^{th}$ electron is reduced by this amount , and in the perturbation term, it is subtracted to nullify its addition in $V(r_i)$. Dont be troubled too much with this term since it is just a number (not an operator)
Hence, $H'$ is independent of the angular coordinates; then $H'$ becomes the Hamiltonian which contains the non spherical part of the electron repulsions, whereas $H^*$ contains the K.E, P.E in the field of the nucleus, and the spherical average electron repulsion energy.
It is assumed that $H^*$ contains most of the inter electron repulsion such that the remaining term $H'$ is small enough to be treated as a perturbation.
Hence there is no reason to think that the solution to this hamiltonian will yield a radial part of the wavefunction $R(r)$ that depends on other quantum numbers besides $n$ and $l$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/287977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Since quantum mechanics give you that photons have (relativistic) mass $m=\frac{hf}{c^2}$, why gravity does not accelerate it? Since quantum mechanics give you that photons have (relativistic) mass $m=\frac{hf}{c^2}$ why gravity does not accelerate it?? I know it changes its energy hence its frequency hence its wave length-colour.But why it does not speed up? If you consider photons massless then it is obvious but then you would not take in consideration that Energy equals mass*$c^2$ and since photons have energy they can't have zero mass. (I'm a mathematics undergraduate took a course on an introduction to quantum physics so try to give a more intuitive answer than or if you use math please be rigorous to the interpretation of the quantities.) (h=planck's constant ,c=speed of light constant , f=frequency).I partly found the answer .I can accelerate without changing speed just direction hence i have bending of light in directions thus acceleration.But what if i send light Straight to the centre of the mass.Nor its speed will change nor its direction.How will i explain acceleration then?
| For a photon travelling toward a gravitational source, MC Physics would suggest that gravity would increase the kinetic energy of the rotation of a photon (ie., increase its frequency) and not its linear KE and velocity, which is limited by relativity.
For a photon travelling away from a gravitational source, MC Physics would suggest that gravity would decrease KE rotation (ie., cause a red-shift in frequency). That is an clear and simpler explanation of Hubble's red-shift of distant stars and galaxies than of an 'ever expanding and accelerating universe'.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/288170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Why is nuclear force spin dependent? Why nucleons with parallel spins have greater nuclear force than the ones with anti-parallel spins?
I just want a clear and easy explanation. Thank you!
| Because of nature. Together with their positions, their isospins etc., the spins of the nucleons are degrees of freedom you might want to consider.
Since the aim is to reproduce experimental results, as @AMS said, using a potential which has the right spin dependence you are able to get some reasonable values for the various observables.
For instance, in the specific case of the deuteron, in a simplified version, you know that the total spin is $S=1$ and the total isospin is $T=0$.
You can describe the nucleon-nucleon interaction with a potential of the form
$$
V_{NN}=V_C(r)+V_\sigma(r)\vec{\sigma}_1\cdot\vec{\sigma}_2+V_\tau(r)\vec{\tau}_1\cdot\vec{\tau}_2+V_{\sigma\tau}(r)\vec{\sigma}_1\cdot\vec{\sigma}_2\vec{\tau}_1\cdot\vec{\tau}_2.
$$
Where the first term is the Culombian part, the second depends on the spin, the third on the isospin and the last one on both.
Since $S=1$ and $T=0$ you get that
$$
\vec{\sigma}_1\cdot\vec{\sigma}_2|\psi_d\rangle=|\psi_d\rangle,\;\;\;\;\vec{\tau}_1\cdot\vec{\tau}_2|\psi_d\rangle=-3|\psi_d\rangle\;\;\;\;\text{and}\;\;\;\;\vec{\sigma}_1\cdot\vec{\sigma}_2\vec{\tau}_1\cdot\vec{\tau}_2|\psi_d\rangle=-3|\psi_d\rangle,
$$
where $|\psi_d\rangle$ is the wavefunction of the deuteron.
Now that you have the potential for this simplified model of the deuteron (with only S wave), you can solve the Schrödinger equation and get a value for the binding energy. In this way you can test your phenomenological potential and verify that a spin dependence is needed.
Maybe the answer is the the first three words, but I hope that looking at an example might be more clarifying.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/288357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Are White Holes the inside of Black Holes? I read about a theory that says that The Big Bang could be actually considered a White Hole.
Than I started thinking. White Hole: an unreachable region from which stuff can come out. Black Hole: a reachable region from which no stuff can come out.
Well one seems to be the boundary of each other. If i am inside a black hole, the surface outside does fit in the White Hole description. So what if a white hole is actually not a hole but is the external surface of a black hole from inside?
What if its true that our Big Bang is actually a white hole, or actually the inside of a Black Hole?
Can there exist a universe in a Black Hole? and Are we in a Black hole ourselves?
| A black hole is very reachable from the outside. Once inside you can't get back. You quick fall to the center and are crushed at a singularity.
A white hole is a black hole running backwards in time. If you are inside, you can't avoid being expelled away from the singularity to the outside. Once outside, you can't get back in.
The inside of a black hole is not another universe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/288570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A question about the definition of proper length As I understand it, the proper length of an object is defined as the length of the object in its rest frame. In terms of the metric it is defined as the length of the spacetime interval between two space-like separated events, i.e. $$dl^{2}=\sqrt{ds^{2}}$$ (with the "mostly plus" signature).
Now, suppose that an observer is at rest in an inertial frame that is itself at rest with respect to a given object that the observer wishes to measure. Why is it the case that one considers the "simultaneous" length of the object, i.e. when $dt=0$, such that $$dl^{2}=\sqrt{dx^{2}+dy^{2}+dz^{2}}$$ Is it simply so that it agrees with the definition of spatial distance in Euclidean geometry or are there other intuitive reasons for why it must be the case (analogous to the definition of proper time in which the proper time of an object is equal to the coordinate time of an observer who is at rest [i.e. $dx=dy=dz=0$] with respect to the object, such that $d\tau=\frac{1}{c}\sqrt{-ds^{2}}=dt$)?!
| When two guys are in the same reference frame then events for both of you occur at the same time coordinate - ie they are simultaneous. What I mean by this is that both guys will ascribe the same time coordinate to when an event A occurred. Of course depending on how far they are from A means they have to use different values for their measurements but they should both agree.
Now imagine a meter stick. In the meter sticks frame of reference let's suppose there are two events. Event A is one end of the meter stick being at x = 1m at t = 0 and event B is the other at x = 2m at t = 0. According to your formula of the space time interval we can find the distance between these two events. Since t = 0 for both A and B in the sticks frame, we then get your formula for the length. (Edit: dt = 0 because the time for both events is t = 0)
Now if I am in the meter sticks frame, according to my first paragraph, A and B should be simultaneous events for me, because these are simultaneous events for the meter stick - and we are in the same frame.
I guess it can be kind of confusing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/288667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is Moseley's law related to the Bohr model? Moseley's rule states
$$\frac{hc}{\lambda} = (Z-\sigma)^2 R (\frac{1}{n^2}-\frac{1}{m^2})$$
In what way is this linked to/derived from the shell model (which I assume means the Bohr model)? My textbook says they are related.
| First, consider an atom consisting of one electron orbiting a nucleus with atomic number $Z$. Since this is a hydrogenic atom, the energy levels of this electron are equal to
$$
E = \frac{Z^2 R}{n^2}
$$
for $n = 1, 2, 3, ...$, where $R$ is the Rydberg energy. If the electron goes from level $n$ to level $m$, it will emit a photon of energy
$$
E_\text{photon} = \Delta E_\text{electron} = Z^2 R \left( \frac{1}{n^2} - \frac{1}{m^2} \right).
$$
Hopefully you can see that this equation is very similar to the one you have above. But why is $Z$ replaced with $(Z - \sigma)$ in your equation? This is where the shell model comes in. The idea is that the other electrons in filled lower shells (which are negatively charged) partially screen the charge of the nucleus. In other words, to the outer electrons, it "looks like" there's a nucleus with a charge $(Z - \sigma)e$ at the center, rather than the "bare" nuclear charge $+Ze$. We can therefore justify replacing $Z$ with $(Z - \sigma)$ in the above expression.
This isn't a rigorous proof by any standard, of course, but hopefully it shows the connection to the shell model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/288761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the potential in a circuit? I have learnt that the potential in a point in an electric field is defined as being numerically equal to the work done in bringing a unit positive charge from infinity to the point. However, this is in the case of an electric field. What is the potential in a circuit say, consisting of a battery and simple capacitor, at one of the plates? Is it numerically equal in the work done in moving a unit charge from the 'positive' plate to the positive pole of the battery? (by having to do work in overcoming the attractive forces of the nucleus on the electrons of that plate) But from definitions this charge is a unit positive charge. This is all confusing to me and it would help for simple explanations.
| Usually one doesn't discuss potentials at a plate of the capacitor. One discusses the "potential difference" between the two plates of the capacitor. "Voltage" is always a difference between one point and another.
The voltage on the electricity supply in your house is the difference in potential between the line on the socket and the ground outside your house. What potential any of these things are with relation to a point at infinity is too complex a problem for practical purposes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/288888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Resources for astrostatistics Being an astrophysic grad student with mathematical background, I have recently become interested in the field of astrostatistics. I know that this is a quite new field that combines astrophysics, statistics and data mining due to the huge amount of data that our telescopes are able to generate. We need sophisticated tools to be able to process it.
I want to learn more about this topic, so I want to ask for some good books or resources on this subject. I have a course on general statistics and I am not at all afraid of abstract mathematical reasoning.
Could you please recommend something in this line?
| Profs. Eric Feigelson and Jogesh Babu at Penn State are widely seen as "Godfathers" of astrostatistics. Every year, they run the Summer School in Statistics for Astronomers which I attended in 2014.
The summer school's web page is publically accessible, and lecture slides and materials for exercise sessions in R with code and instructions. These sessions are connected thematically to some of the lectures, and all exercises consist of functioning code which can be run at home in e.g. RStudio. The lectures are a great introduction to a wide variety of techniques in astrostatistics, and it is easy to get one's hands on the gears in the exercises, but of course lecture slides are a bit superficial.
If one wants to go more in depth, the course materials also contain a long suggested reading list. Feigelseon & Babu are also the authors of a graduate-level astrostatistics textbook, which is built around real-life examples often encountered in astronomy, and which also contains a number of exercises in R.
EDIT: If you are more interested in the data mining/machine learning side of things, try AstroML, a combination of software module, text book and online resources like Jupyter notebooks etc. - the book costs what books cost, but many online resources are free.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difference between "Periodic motion" and "Oscillating Motion" So far I know one of them is a special case of the other: The Oscillating motion being the special case of Periodic motion. But I don't know the precise "Kinematical definition" of each one. I mean when you have an "Equation of motion" for a particle, how will you determine it's either a "Periodic motion" or an "Oscillating motion"? If some periodic functions appear in an equation of motion, can we call it a "Periodic motion"? If so how can we recognize it from "Oscillating motion"?
| From a physical stand point they are very similar. When we consider the orbit of a planet around a star, we consider its period. Hence, orbits are periodic in nature. We don't refer to the oscillations of the planet.
In contrast, say we have a spring with a mass attached, and set it out of equilibrium. We don't typically talk about the periodic nature of the system, but rather its oscillatory behavior (the oscillation of the system). This is not to say that the system does not exhibit periodic motion, as the system will oscillate with some period.
Having an equation of motion, we are able to discern different types of oscillatory behavior such as forced, damped and coupled oscillators. These are all cases of oscillation being a 'special case' of periodic motion, as they contain elements of periodic motion (sin, cos, etc)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Breaking down sawdust pellets causing heat We just discovered that pellets (sawdust, animal feed, organic constituents) when re-hydrated give off heat. We understand this at molecular levels but are having a hard time sussing this out at a macro level. Do these pellets (pellet stove sawdust type in particular) absorb energy when compressed to release it when water is added and they break down into the particles they were made from?
| Nono. What you're assuming is wrong.
Hydration enthalpy is what is behind the scene, and its macro because it's Thermodynamics. Whenever you hydrate the sawdust particles, hydration enthalpy is given off which is what you've observed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given that ice is less dense than water, why doesn't it sit completely atop water (rather than slightly submerged)? E.g.
*
*If we had a jar of marbles or something else of different densities and shook it, the most dense ones would go to the bottom and the less dense ones to the top.
(Image Source)
*If I put a cube of lead in water it would sink all the way to the bottom.
But for ice : what I am trying to understand is why doesn't the water (being denser than the ice) seek to reach the bottom, and the ice sit flat on top of it (as in the left image)? Instead, some part of the ice is submerged in the water (as in the right image), and some sits on top it.
| People have been answering based on force balance, but there's an equivalent (and, in my opinion, more satisfying) answer based on energy conservation.
Suppose the ice cube is initially just on top of the surface of the water. If you lower it a tiny bit into the water, a tiny bit of water will be forced to move up, increasing the gravitational potential energy. But the entire ice cube will move down, significantly decreasing the energy. So the minimum energy configuration always has the object sinking into the water at least a little bit.
Note that this argument doesn't apply to two fluids, where the equilibrium configuration really does have the lighter fluid on top. Why? Consider moving a cube of the lighter fluid down. The same arguments above hold, except that when the heavier fluid moves up, all of the lighter fluid is forced to move up too. So the completely separated configuration really does have minimum energy.
In the ice cube example, the 'lighter fluid' being pushed up is air, not ice. Since air is much lighter than water, the ice cube sinks.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "58",
"answer_count": 8,
"answer_id": 4
} |
Magnetic focusing of microfocus X-Ray tube I've got this Russian X-Ray tube - BS-5. It has specified focus spot size of $\approx 6~\mathrm{\mu m}$.
It works, but I've seen brief mentions that in order to reach specified $6~\mathrm{\mu m}$ spot size one should use magnetic focusing. Does anyone have any guideline/links on how this magnetic focusing should be applied in principle to the tubes of this shape/type?
| Looking around on the internet a bit, I found several resources that may be of interest.
*
*A patent, online here, that is for the magnetic focusing of x-ray tubes. It includes a nice description, as well as references and diagrams. According to the description,
A primary object of the invention resides in the provi sion [sic] of an improved method f0. [sic] the reduction of the focal spot size within an X-ray tube which is adapted to deliver an X-ray burst of extremely short time duration.
This seems to be exactly what you are looking for. It also includes a couple of diagrams showing the x-ray tube this system is intended for, and I'm not an expert but it doesn't look too different from what you've got. So that may be of assistance.
*
*Another patent, online here, describes a magnetic focusing lens that might be somewhat related to what you are looking at.
*This paper is about electrostatic focusing, not magnetic focusing, but may also be useful in providing other methods to do what you want to do.
*Finally, this book excerpt may be useful; it goes more in depth on magnetic focusing lenses.
Hope this helps! I'll be adding more resources as I find them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Variation of electromagnetic action to obtain Maxwell's equations The electromagnetic action is given in the language of differential forms by
$$S[A]=-\frac{1}{4}\int F\wedge \star F$$
The variation of the electromagnetic action $S$ gives us Maxwell's equations
$$d\star F=0.$$
How do you take the variation $\delta S = S[A+\delta A]-S[A]$ of the above action $S$ to obtain Maxwell's equations?
| Here we only consider the abelian case. Let $A\to A+\delta A$, and $F\to d(A+\delta A)=dA+d\delta A$.The action becomes
$$S[A+\delta A]=-\frac{1}{4}\int (dA+d\delta A)\wedge\star(dA+d\delta A).$$
So up to terms linear in $\delta A$,
$$S[A+\delta A]-S[A]=\frac{1}{2}\int \delta A\wedge d(\star F)+\mathcal{O}(\delta A^2),$$
where we integrated by parts and used the symmetric property of the inner product on $p$ form $(\alpha,\beta)=\int\alpha\wedge\star\beta=\int \beta\wedge \star\alpha=(\beta,\alpha).$
Now $\frac{\delta S}{\delta A}=0$ gives you $d\star F=0$ as $\delta A$ is arbitrary.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem with physical application of Dirac Delta Consider the problem of projectile motion in 2 dimensions. Launch angle is constant. Range of projectile, $x$, then depends only on launch speed, $v$, and is given by
\begin{equation}
x=v^2, \quad v\in [0,1] \tag{1}
\end{equation}
Above equation has been non-dimensionalised (by taking maximum range as our length scale, and maximum launch speed as our velocity scale), so all quantities are dimensionless. Probability density function for launch speed is assumed uniform over the interval $[0,1]$:
\begin{equation}
f(v)=1, \quad \textrm{if}~v\in [0,1]\tag{2}
\end{equation}
and zero otherwise. I want to find p.d.f for range of projectile, $x$. An easy way of doing this
\begin{equation}
f(x)=\left| \frac{dv}{dx}\right|f(v)=\frac{1}{2\sqrt{x}}, \quad x\in [0,1]\tag{3}
\end{equation}
However I wanted to solve the same problem using Dirac delta function:
\begin{align}
f(x) & =\int_0^1 dv~f(x|v)~f(v) \\
& = \int_0^1 dv~f(x|v) \\
& = \int_0^1 dv~\delta(v^2-x)\tag{4}
\end{align}
Here $f(~|~)$ denotes conditional p.d.f.. Last line was arrived at because for given value of $v$, it is certain that we shall obtain that value of $x$ that satisfies the equation $v^2-x=0$. Now I make use of the identity for delta function
\begin{align}
\delta(g(x))=\sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}\tag{5}
\end{align}
Here $x_i$ are roots of function $g(x)$, and $g'\equiv \dfrac{dg}{dx}$. Now $g(v)=v^2-x$, whose roots are $\pm \sqrt{x}$. We reject the negative root because $v\geq 0$. $g'=2v$. Hence
\begin{align}
f(x) & =\int_0^1 dv~\delta(v^2-x) \\
& = \int_0^1 dv~\frac{1}{2\sqrt{x}}\delta(v-\sqrt{x}) \\
& = \frac{1}{2\sqrt{x}}\tag{6}
\end{align}
which is correct.
However instead of $f(x|v)=\delta(v^2-x)$, we could equally well have begun with the equation $f(x|v)=\delta(v-\sqrt{x})$, because at least according to me, physical content of both equations is identical. However the last choice yields a completely different p.d.f.:
\begin{align}
f(x) & =\int_0^1 dv~\delta(v-\sqrt{x})=1\tag{7}
\end{align}
I don't think I have done anything wrong mathematically (if I have, please point out). To a mathematician of course the two functions are different, and so the fact that they yielded different p.d.f.s is not surprising. But when the equations are put in their proper physical context, both have identical physical content (as far as I can see). This example makes me wonder if Dirac Delta function may be used unambiguously in solving physical problems. While this was a simple problem where a second method of solution was available and so we could compare, what does one do in more complicated situations where such a comparison is not possible?
| Answer offered by @Qmechanic, that Dirac delta function depends also on the functional form of its argument, is correct in regard to $\textit{dimensionless}$ equations. Answer offered by @ACuriousMind, that dimensions of Dirac delta correctly dictates the functional form of its argument, seems to work when dealing with $\textit{dimensional}$ equations. In dimensional form conditional p.d.f. $f(x|v)$ written as either $\delta (\frac{v^2\sin(2\theta)}{2g}-x)$, or $\delta (\frac{v^2}{2g}-\frac{x}{\sin(2\theta)})$, or $\delta (\frac{v^2}{g}-\frac{2x}{\sin(2\theta)})$ etc., all of them seem to work, so far the argument has dimensions of inverse length. This means that if one is going to use Dirac delta functions, one had better work with dimensional equations. However there is going to be a problem when dealing with intrinsically dimensionless quantities such as angle or function of angle.
To summarize, this means that $\delta(f(x))=\delta(g(x))$ if and only if: (1)$f,g$ have dimensions and their dimensions is identical, (2)$f(x)=0\Leftrightarrow g(x)=0$, i.e. one can obtain equation $g(x)=0$ by manipulation of equation $f(x)=0$ and vice versa. I could not prove it, but examples I have worked out so far seems to support this conclusion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/289904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
What's a "colour triplet fermion"? I'm not a big fan of Science Alert, but this recent piece about the so-called SMASH model, whose gory details are apparently presented in arXiv:1608.05414 seems reasonable. I'm curious about this "Standard Model Axion Seesaw Higgs portal inflation" model, which they summarize it as follows:
Now, the team led by French [sic] physicist Guillermo Ballesteros from the University of Paris-Saclay says we can add these three right-handed neutrinos to the three existing neutrinos in the standard model, plus a subatomic particle called a colour triplet fermion, to solve the first four problems listed above.
For the benefit of those of us with some rudimentary understanding of group theory and particle physics, but no in-depth understanding of quantum chromodynamics - what's a "colour triplet fermion"?
| A colour triplet fermion is simply a fermion that behaves like a quark with respect to the strong force, i.e. it transforms in the triplet (or fundamental) representation of the $\mathrm{SU}(3)$ gauge group of the strong force, often denoted $\mathbf{3}$. That this is indeed the meaning here can be seen on the bottom of page 1 of the linked paper:
[...] two Weyl fermions $Q$ and $\bar{Q}$ in the $\mathbf{3}$ and $\bar{\mathbf{3}}$ representations of $\mathrm{SU}(3)_c$ and with charges $-1/3$ and $+1/3$ under $\mathrm{U}(1)_\text{Y}$[...]
After the Higgs-like process proposed for this model happens, the two Weyl fermions will combine into a single Dirac fermion because they develop a mass, so effectively, this "colour triplet fermion" is a very quark-like particle with the same electrical ($\pm 1/3$) and color charge as a quark and no weak force coupling in either handedness (this is different from quarks) as far as I can tell.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/290014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where does gravitational energy come from? We've all heard mass tells space how to curve and curved space tells matter how to move. But where does the energy to curve space come from? Likewise where does the energy that curved space uses to push planets around come from? I mean if I tell my son to clean his room, and he does, then I did not provide him the energy to do so.
| The theory of General Relativity works with the energy momentum tensor and one has to work with the mathematics of it in order to really understand what is happening, not handwaving. It is a fact that all cosmological and astrological data follow the general relativity equations, as one can see in this link
The cosmological model accepted now is the Big Bang model, based on general realtivity and its constants adjusted with observational values. In this model all the energy of the universe, the one the energy momentum tensor describes, came at the Big Bang singularity:
The expanding universe utilizes this original energy . (Special relativity is part of General relativity, and thus the equivalence of mass with energy is accounted for). As the Big Bang model is quite successful , the answer is : the energy for everything comes from the original singularity. At the planetary and galactic level it is described well with Newtonian mechanics. At the atomic and nuclear the laws of quantum mechanics and special relativity are adequate to describe energy transformations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/290230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Is time absolute? If I take a spaceship and park it near the event horizon of a black hole and then measure the age of the universe by observing SNe Ia, then travel back out to normal space (no gravitational forces, at rest with respect to CMB), will the dates agree? That is, if the measured age of the universe is 13.8 billion years near the event horizon, and it takes me 100 million years (proper time) to travel back out to normal space, will my new measurement of SNe Ia agree with a date of 13.8 + 0.1 = 13.9 billion years? If that is true, can we say that time is absolute (i.e. all observers will agree on the age of the universe when using SNe Ia when coordinate systems are normalized)?
| Well, no. We can construct a much simpler example to see this: fix a point in Minkowski spacetime, and consider two observers following worldlines from that point with a relative velocity. They can even both be inertial. At fixed Minkowski time the two observers measure different proper time.
The FLRW universe, however, is sort of special in that there is a sort of preferred reference frame: that in which the coordinates follow the expansion. This frame is "preferred" in the sense that geodesics which have some relative velocity compared to the expansion asymptote towards it. But you're still allowed to view the solution from an arbitrary coordinate system (for example by considering non-geodesic observers), so you can come up with any "age of the Universe" you like.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/290357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Jordan-Wigner transformation v.s. Bosonization Jordan-Wigner transformation is a powerful tool, mapping between models with spin-1/2 degrees of freedom and spinless fermions. The key idea is that there is a simple mapping between the Hilbert space of a system with a spin-1/2 degree of freedom per site and that of spinless fermions hopping between sites with single orbitals. One can associate the spin-up state with an empty orbital on the site and a spin-down state with an occupied orbital.
Bosonization/fermionization is also a powerful tool, mapping between 1+1d bosonic field theory to 1+1d fermionic field theory. There is a nontrivial correspondence between operators of two sides in 1+1d.
Question:
*
*Are we aware the exact relations between the twos in 1+1d: Jordon-Wigner transformation v.s. Bosonization?
*Can one use one to prove the other?
*Do both have subtle restrictions for 1+1d open chain or in 1+1d closed ring?
*Higher dimensional analogy in $d+1$d in general?
| I take a different stance than Qmechanic: bosonization is `simply' the continuum version of the Jordan-Wigner transformation. Of course Qmechanic is right in that field theories are much more subtle than lattices theories. Nevertheless, the fact that JW is so simple does not mean it is not relevant when thinking of bosonization, in fact the opposite holds: it makes bosonization much easier to follow, as for example discussed by Fisher and Glazman.
To make my statement more concrete, I would say the following diagram commutes:
$$ \begin{array}{ccc}
\textrm{fermionic chain} & \xrightarrow{\textrm{Jordan-Wigner}} & \textrm{spin chain} \\
\downarrow \small \textrm{continuum} & \circlearrowleft & \downarrow \small \textrm{continuum} \\
\textrm{fermionic field theory} & \xrightarrow{\textrm{bosonization}} & \textrm{bosonic field theory} \end{array}$$
(where in the spin-case the continuum limit would be taken using spin coherent state path integrals)
Of course one might end up with different field theory descriptions, but they would describe the same field theory. More exactly, there is a local mapping relating one to the other.
As an example, let us instead start from a spin chain. In particular, take the gapless spin-$\frac{1}{2}$ Heisenberg Hamiltonian $H = \sum \mathbf S_n \mathbf{\cdot S}_{n+1}$. Then:
$$ \begin{array}{ccc}
\textrm{interacting fermions} & \xleftarrow{\textrm{Jordan-Wigner}} & H = \sum \mathbf S_n \mathbf{\cdot S}_{n+1} \\
\downarrow \small \textrm{continuum} & \circlearrowleft & \downarrow \small \textrm{continuum} \\
\textrm{interacting fermionic field theory} & \xrightarrow{\textrm{bosonization}} & \begin{array}{c} \textrm{Wess-Zumino-Witten }SU(2)_1 \\
|| \\ \textrm{Luttinger liquid $K=\frac{1}{2}$ } \end{array} \end{array}$$
(where the LL description comes from the bosonization and the WZW description comes from the continuum limit of the spin model) and both resulting field theories are indeed equivalent after a local re-identification of operators. In particular they have the same scaling dimensions for local operators, e.g. the smallest scaling dimension for WZW $SU(N)_1$ is $\frac{N-1}{N} = \frac{1}{2}$ and for the LL is $\frac{1}{4K} = \frac{1}{2}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/290709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Physics of variable settings on microwave ovens I've been learning how microwave ovens work and haven't been able to find any information on the practical way manufacturers create variable power settings.
If I understand magnetrons correctly they are only designed to emit one wavelength/frequency, so is there something I'm missing about how the same wave can just be emitted at a lower power? I was thinking it might be similar to the difference of volts vs amps but that doesn't really explain anything.
Or would one just oscillate power to the magnetron, like a separate duty cycle kind of thing, to lower the average molecular effect?
Really trying to understand the physics of what determines a particular power setting.
| Most microwave ovens simulate variable power by cycling full power on and off. However, some microwave ovens have truly variable power that is applied uniformly during the entire cooking time. These ovens are sold as having "inverter" technology. What they do is first change the 120 V / 60 Hz input power to something that can be better controlled. Then they use that controlled power to create the microwaves.
Which kind you have matters little when heating large items, but it is easy to see the advantage when trying to heat a small amount of food, especially when it is in small pieces or odd shapes. A normal oven can cause selective overheating, arcing (sparks) or pops, even on the lowest power setting. In contrast, a microwave oven with inverter technology can heat such things without any of these problems. For this reason, I have preferred microwave ovens with inverter technology for many years. For more info, search for "Microwave oven inverter technology".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/291102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How much information can you obtain from a pulsar-black hole system? Imagine that we have detected an interesting source in the sky that we believe is generated by a pulsar orbiting a black hole.
The challenge here is the following:
What physically relevant information could you extract from the observation of this system?
Note: I am posting an answer with some possible information that we could obtain, but I will NOT mark my answer as the correct one.
| Well, you could try measuring the gravitational waves emitted from the system. If we assume an order-of-magnitude approximation for the strain $h$ as
$$h\sim\frac{1}{R}\frac{GM}{c^2}\left(\frac{v}{c}\right)^2$$
where $R$ is the distance to the source, $M$ is the sum of their masses, and $v$ is the orbital speed. For a low-mass black hole of about $5M_{\odot}$ and a neutron star of about $1.5M_{\odot}$ with $v\approx0.01c$, then at a distance of about 100,000 light-years, we should see $h\sim10^{-21}$. Obviously, a more massive black hole could increase $M$ by an order of magnitude, but we still would have difficult observing the waves from outside the Milky Way. Inside the galaxy, though, detection would be possible by LIGO. Using numerical models, we could attempt to determine some of the above characteristics using strain measurements. See this LIGO paper for information on numerical fitting.
You could attempt to derive an estimate of the age of the system using the relation
$$T\simeq\frac{P}{2\dot{P}}$$
with $P$ being the period of rotation. This is simply an approximation that holds when the period is much greater than the initial period, $P\gg P_0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/291188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Nuclear Shell Model - Spin of Nucleus I want to define the spin of the following nuclei: $^{15}_{\phantom{1}7} \textrm{N}$, $^{27}_{12} \textrm{Mg}$ and $^{47}_{20} \textrm{Ca}$. I have a scheme for the niveaus of the energies, see below:
Where left is for protons and right column for neutrons. I am not really sure how to work with such a pattern.
For $^{15}_{\phantom{1}7} \textrm{N}$ we have 7 protons and 8 neutrons. The neutrons completed two shells - so there is no contribute to the nucleus spin. But 1 proton is "left" in the niveau $1p_{1/2}$.
So the total spin is $1/2$?
$^{47}_{20} \textrm{Ca}$: 20 Protons (shells full; no contribution). But 7 Neutrons on the $1f_{7/2}$-Niveau. What's the total spin here?
And another question: What do to the numbers on the left represent? Principal quantum number $n$?
Thanks in advance.
| The scheme is:
For a nucleus at the ground state, when the shells are full for one kind of nucleons, and
*
*The other has the last shell with only one nucleon, the nucleon
total angular momentum (in $1p_{1/2}$, this is 1/2) is the nucleus spin.
The reasoning is that all the others give angular momentum $0$, so $| \vec 0 + \vec J_p | = | \vec J_p |$.
*The other has
the last shell with only one nucleon missing from being full, the missing nucleon total angular momentum (in $1 f_{7/2}$, this is 7/2) is the nucleus spin.
The reasoning is that the full shell would give $0$. You subtract the angular momentum of the nucleon you are removing. $| \vec 0 - \vec J_n | = | \vec J_n |$.
You can make this more general considering that pairing favours $0$ angular momentum couples of nucleons. Given this, any even-even nucleus at ground state has spin $0$, any even-odd has spin equal to the lone nucleon angular momentum.
For "the number on the left", if I undestand which one you refer, yes, it is the principal quantum number (notice that the exact way you number the levels can vary, but they are just labels).
Edit: I see they explain this here on Wikipedia
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/291345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why are angular frequencies $\omega=2\pi f$ used over normal frequencies $f$? When we first studying vibrations in crystals we begin by studying the monoatomic chain, and then go onto the diatomic chain with a series of alternating masses. In studying these we look to calculate the dispersion relation, which is the angular frequency as a function of the wave vector.
For example, in the monoatomic chain we can derive the dispersion relation as $$\omega=\sqrt{\frac{4C}{M}}\sin^2\Big(\frac{ka}{2}\Big),$$ where $C$ is a 'spring' constant inherent in the crystal structure, $M$ is the mass of the atoms on the chain, $k$ is the wave vector and $a$ is the atomic spacing in the chain.
When studying the diatomic chain, we get two solutions corresponding to the optical (diatomic only) and acoustic (diatomic and monoatomic) waves.
What I don't understand is exactly why we are concerned with an angular frequency. What has the property of angular frequency? As far as I know there is no rotational motion, and the intrinsic frequency of a wave is surely more useful?
In addition to this question, how can we calculate the frequency, $f$ of, say, an optical wave of a diatomic chain given the angular frequency from the dispersion relation, $\omega$?
| You are right in noting that $f$ is the more "physically intuitive" quantity, and at the end of the day measurements typically done in $f$, not $\omega$. However, the relationship between $f$ and $\omega$ is always $2\pi f = \omega$, so it is a very simple conversion to the point where people operationally don't really think of them as different. The reason $\omega$ is typically preferred over $f$ is because it is more convenient to write in equations: $\sin(2\pi f t)$ is much more cumbersome to write than $\sin (\omega t)$. This essentially has to do with the fact that sinusoids have a period of $2\pi$, not $1$. For similar reasons, people tend to use $\hbar$ and not $h$ in many equations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/291481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Degeneracy of $\left|\pm k\right\rangle $: because of reflection symmetry or time reversal symmetry? Consider a free particle in one dimension. The Hamiltonian is
$$ H = p^2/2m . $$
It is well known that the momentum states $\left|\pm k \right\rangle $ are degenerate.
The problem is, are they degenerate because of the reflection symmetry or the time reversal symmetry?
| In general (i.e. ignoring coincidental degeneracy), eigenvalues of Hamiltonian are degenerate if there exist two operators which commute with Hamiltonian (so their corresponding observable is conserved), but which don't commute with each other.
A free particle has two non-commuting conserved quantities: parity and momentum. Eigenfunctions of the former are standing waves (sines and cosines), while eigenfunctions of the latter are running waves (complex exponentials). Both types of eigenfunctions can be chosen with given energy, and it's this freedom what gives you degeneracy.
As a contrast, a particle in the box has reflection symmetry (so parity is conserved), but its energies are all non-degenerate, because there's no such other conserved quantity as momentum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/291565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.