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Will a plastic feel less heavy when I put it in a bucket of water and carry it? If I'm carrying a bucket of water in one hand and a piece of plastic in the other, and then I decide to keep the plastic in the bucket of water (it floats). Will it feel less heavy in the second case?
I think it will feel the same because it's mass adds up to the bucket's mass and will be pulled by gravity with the same extent. But somehow I can't get my mind off from the fact that it's weight is already balanced by the up-thrust.
Is there a simple way to explain how this works? It would be clearer if you helped me with some free body diagrams or an analogy or something simple.
| I don't think you will feel less however the plastic (if could sense ) will of course feel less weight .
Let me explain.
Since the liquid applies buoyant force on the plastic (say $F_b$ ) so by Newton's third law the plastic also applies a force $ F_b$ on it in the downward direction . So the forces acting on the liquid are $Mg$ , $F_b$ and the normal force $N$ due to the bottom part of the bucket . Here's the fbd for the situation
Now what you feel is the normal force due to the bucket and this is equal to the normal force ($N$) applied by the liquid on the bucket in downward direction ( since bucket Also applied the same force on the liquid which is shown in the fbd).
So you feel more force .
Note : I have considered forces only in the vertical direction however there are forces on the liquid by the bucket in contact in other direction also.
Hope it helps .
| {
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Deriving ideal gas law from Boyle and Charles My textbook states
Notice that since $PV = \text{constant}$ and $\frac{V}{T} = \text{constant}$ for a given
quantity of gas, then $\frac{PV}{T}$ should also be a constant.
I tried to prove this, but no success:
$$PV = a$$
$$\frac{V}{T} = b$$
$$\frac{PV^2}{T} = ab$$
$$PT = \frac{a}{b}$$
But I am not able to cook up $\frac{PV}{T}$... Any help?
| $PV$ is constant for fixed $T$, and $V/T$ is constant for fixed $P$. Hence
$PV=f(T)$ and $V/T=g(P)$.
From these we can write
$V=f(T)/P=T\times g(P)$.
This implies that
$f(T)=kT$ and $g(P)=k/P$ for some constant $k$.
Hence $PV/T = k$ (constant, actually $nR$) is the required answer.
| {
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How do forces 'know' they need to move when a system is in motion? I am curious as to how forces move when a system is in motion. This was never fully explained in my physics classes at university. Let me explain:
I understand the Newtonian (classical) physics that there are equal and opposite forces in play. So when I am standing the force I am exerting on the ground due to gravity is balanced by an upward force from the ground. However, when I lift one foot (say the left foot) the force from my body is now transferred through the right foot. However, where did the upwards force that was under my left foot go?
I assume the upward force 'moved' to balance the increased force exerted by my right foot. I can understand that it general, except for one point. How did the upward force 'know' that it needed to move - and, secondly, where it needed to move to?
This same question can be applied to many dynamic situations of motion, such in a moving vehicle. (I can think of many other examples as well).
I had one physicist trying to explain it to me but, I admit, I lost his explanation when he went down the quantum mechanics rabbit hole. Is there a classical explanation as to how forces know when and where to move when a system is in motion?
| Long story short, the upward force is electromagnetic repulsion from the molecules of the floor. As soon as you raised your leg, you removed it from the floor molecule lattice, so it was not close enough any longer to feel a strong repulsion, like before.
| {
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When do two massed exert the same centripetal on a point? I was reading the mark scheme for 2020 Cambridge A-level physics when I came across a question that reads
The stars S1 and S2 rotate with the same angular velocity ωabout a point P, as illustrated in Fig. 1.2.
Point P is at a distance xfrom the centre of star S 1.
The period of rotation of the stars is 44.2 years.
And then it asked:
By considering the forces acting on the two stars, show that the ratio of the masses of the stars is given by
mass of S1 / mass of S2 = (d– x) / x.
Could anyone explain WHY the gravitational forces on the two masses S1 and S2 are equal, OR in other words, the centripetal forces about point P are the same?
| By Newton's third law, the magnitude of the gravitational forces on each mass must be the same.
"Centripetal" is just a direction indicator, just like how "horizontal" and "vertical" also indicate directions. Since gravity acts in a line between the bodies, and because the center of rotation lies on this line, the gravitational force is always centripetal. Therefore, since the forces are equal in magnitude, it must be that the centripetal forces acting on each mass are also equal.
| {
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Amplitude of superposition states I have a question about the following chart (for 3 and 4 particles case):
What does $P_{cycl}$ mean? How's that relevant to the amplitude of superposition states and their symmetricity?
| *
*The states given in your table seem to be states of a system of N spins in the $\hat{S}^z_{\text{tot}}$ and $\hat{S}^2_{\text{tot}}$ basis which are also eigenstates of the cyclic permutation operator $\hat{P}_{\text{cycl}}$ with $P_{\text{cycl}}$ being the eigenvalue of the cyclic permutation operator.
*An Example definition for the cyclic permutation Operator would be:
$$\hat{P}_{\text{cycl}}|s_1,...,s_n>=|s_n,s_1,...,s_{n-1}>$$
*You could calculate this explicitly for these states to convince yourself for example:
$$\hat{P}_{\text{cycl}}(|\uparrow\uparrow\downarrow>+e^{i\frac{2}{3}\pi}|\uparrow\downarrow\uparrow>+e^{-i\frac{2}{3}\pi}|\downarrow\uparrow\uparrow>)=
|\downarrow\uparrow\uparrow>+e^{i\frac{2}{3}\pi}|\uparrow\uparrow\downarrow>+e^{-i\frac{2}{3}\pi}|\uparrow\downarrow\uparrow>=e^{i\frac{2}{3}\pi}(|\uparrow\uparrow\downarrow>+e^{i\frac{2}{3}\pi}|\uparrow\downarrow\uparrow>+e^{-i\frac{2}{3}\pi}|\downarrow\uparrow\uparrow>)$$
$$\longrightarrow P_{\text{cycl}}=e^{i\frac{2}{3}\pi}$$
*The $P_{\text{cycl}}$ is just a third quantumnumber you could use to characterise these states. That means if you would have a N-particle Hamitlonian which commutes with the operators $\hat{S}^z_{\text{tot}}$, $\hat{S}^2_{\text{tot}}$ and $\hat{P}_{\text{cycl}}$ the usage of these states as a basis would probably reduce the complexity of your problem.
*To use these states as an orthonormal basis, you need to normalise them. Therefore, $P_{\text{cycl}}$ is in my opinion only relevant for the normalisation of the states concerning the "amplitudes".
*As already discussed these states are "symmetric" under the cyclic permutation operation with different eigenvalues.
I hope this helps you with your understanding.
Jan
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Is it possible to bend light without changing its color? It seems to me that whenever you change the direction of a wave it also affects frequency. Would this not also be true of light waves bending from, for example, gravity?
| The frequency of light "in transit" isn't well defined in special or general relativity. You can only define frequency shift for light that travels from an emitter to a receiver, and you get a single frequency ratio for the whole trip; it can't be attributed to any particular part of the trip.
In the case of the bending of light by a massive body like the sun, you can express the frequency shift approximately (but very accurately) as a product of shifts due to the emitter's and receiver's motion relative to the sun, and the quasi-Newtonian gravitational potential at the emitter's and receiver's locations. The bending angle doesn't enter into it, so it seems fair to say that the gravitational bending of the light doesn't affect its frequency.
| {
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Do "almost black holes" exist? The only things I read about so far in astrophysics are either black holes, developing black holes or not black holes at all.
So I am wondering, is it physically possible to have an object that is almost a black hole, but not a black hole. What I mean by that, is an object that would have a gravitational pull almost as strong as a black hole, but not equal, so light would be bent and considerably slowed down, among other effects yet able to escape.
I am not a physicist so I use my own words. The point of my question, if this helps, is to know if we can/could observe and study such objects as intermediary between non-blackholes and blackholes with its own properties. Again this is NOT about the formation of black holes. So maybe such an intermediary object is impossible because things are binary (like starting the process of black hole formation would not stop).
Also I know there are massive objects that are not black holes for example neutron stars but they do not seem to have "almost black holes" radiations.
| Neutron stars
(The question is written informally, so my answer is. "For simplicity" <smirk>)
Neutron stars are massive, incredibly dense, hence have very strong gravity; but not so strong that light cannot escape. It has no "event horizon"; but you wouldn't want to be near it.
Related: Quark star (a hypothetical type of star even more dense than a neutron star). – Comment added by @DevSolar - thank you!
| {
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How much energy is transferred to a human hit by lightning? Wikipedia tells me that a bolt of lightning releases roughly 1 GJ of energy, but I'm guessing that's along the entire length of the bolt and that most of it is dissipated as heat and light to the surrounding atmosphere.
Don't know much about the physics behind this, but assuming the bolt is 20km long that's about 50 KJ per meter, or 90 KJ for an average human.
Or am I WAY off on my assumptions here?
| When lightning is fatal, it is usually due to an electrical discharge-induced heart attack. Since lightning is essentially a electrostatic discharge event, we can roughly view it as analogous to a capacitive discharge.
IEC 60470-1 provides threshold values of various physiological effects of capacitive discharge current, including threshold for ventricular fibrillation due to current through the heart, as a function of capacitance and voltage. Using the relationship $E=\frac{CV^2}{2}$ you can compute the energy stored in the capacitor prior to discharge associated with the IEC various physiological effects.
However, you can't assume all of the lightning strike current and energy will be delivered to the heart. Although the available energy of the strike is high, the source cited below states that most people do survive a lightning strike. It says one reason is that lightning rarely passes through the body. Instead, a “flashover” occurs, meaning that the lightning travels over the surface of the body through the conductive sweat (and perhaps rain) on the surface of the body which provides an alternative external pathway around the body for current to flow.
For additional information, see.
https://allthatsinteresting.com/effects-of-a-lightning-strike
Hope this helps.
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Does work done on a spring = elastic potential energy? So the UK exam board specifications (AQA GCSE) clearly state
"...the work done on the spring and the elastic potential energy stored are equal"
Here's my problem,
So work done = Force x displacement
Force = Spring constant x extension
Elastic potential energy = 0.5 x spring constant x extension squared
Extension = displacement for a strecthed spring
However if I take some sample values and calculate the work done on the spring, and then the elastic potential energy stored. The Elastic potential is always exactly half the work done. This contradicts the statement in the specifications.
I have looked all over the place but can't find a satisfactory answer to this question. What am I missing? I get that elastic potential energy is equal to area under the Fx graph but why does that not equal the work done?
| Work is not "force times displacement". Work is an integral $$W=\int\mathbf F\cdot \text d\mathbf x$$
which becomes $W=Fx$ under certain conditions.
The work done by a conservative force is always equal to the negative change in potential energy associated with that force:
$$W_\text{cons}=\int\mathbf F\cdot\text d\mathbf x=\int-\nabla U\cdot\text d\mathbf x=-\Delta U$$
The area under the Fx graph should be the work done by the spring, which is the negative change in potential energy.
Your mistake is most likely thinking that $W=Fx$ holds here, but it does not because the spring force varies with displacement. You are probably getting a discrepancy of a factor of $2$.
| {
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Is it okay to for the integrand and bounds of integration to be functions of the same variable? For the sake of simplicity, say we given that $a = 6t$ and that we must find the velocity as a function of time. We would set up the following integral:
$$\displaystyle \int_{v(t_0)}^{v(t)}dv = \int_{t_0}^t 6t\,dt$$
Having $t$ in the bounds of integration and also in the integrand seems weird to me. Is this the best way to notate it?
| Very bad mathematical grammar. It's perhaps comprehensible, but can lead to horrible errors. Don't do it!
Write
$$
\int_0^t 6 \tau\, d \tau.
$$
| {
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Are humans special in that they collapse wave functions? First of all, I don't really believe that humans are special. So I know the answer must be that they are not.
But the way quantum mechanics is described is that all particles exist as clouds when humans are not looking but exist as particles when humans look at them. How is this consistent with the fact that humans are not special (I mean..humans are made of the same particles, so it's not like they can do anything special)?
|
that all particles exist as clouds
This is a miss-representation of the mathematics that describes quantum mechanically elementary particles and their composites.
Look at the orbitals of the hydrogen atom:
If one is not careful to understand the postulates and mathematics of quantum mechanics one ends up with particles as clouds, because of the above pictures.
What the image conveys is that each point is a probable point for findng an electron if one studies an excited hydrogen atom. Quantum mechanics predicts probabilities to find a particle at (x,y,z,t), not trajectories or orbits.
when humans are not looking but exist as particles when humans look at them
Any interaction with a particle "finds it", according to the probabilities that its wavefunction $Ψ$ dictates. The probability is calculated by $Ψ^*Ψ$, and particles interact with other particles according to the theory. So humans are not necessary as observers to get interactions and "collapse" the wavefunction into another wavefunction (that is what collapse means, different conditions impose different wavefunctions).
| {
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When is a quantum state pure and when mixed? Every definition of the two is always very abstract to me. Like, A pure state is located on the surface of the bloch sphere while the mixed state is somewhere within.
First of all, what is an intuitively definition?
And second of all, how do you practically recognize whether a given state is mixed or pure?
| A quantum state is pure if you know as much as one can know about a quantum state.
A quantum state is mixed if you could know more about, that is, you don't know as much as is allowed to know about a quantum state by the laws of quantum mechanics.
| {
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Contradiction in Faraday's law and Motional EMF Consider two parallel conducting frictionless rails in a gravity free rails parallel to x axis. A movable conductor PQ( y direction) of length $l$ slides on those rails. The rails are also connected by a fixed wire AB with a resistor of resistance $R$. Suppose a magnetic field exists in region which varies as $$B = cx$$The magnetic field is perpendicular to the plane of the system. Initially PQ is given some velocity $v_0$ in the x direction. Let the velocity at any instant be $v$ and the distance from AB be $x$
*
*According to the flux approach, $$\Phi=cx^2l$$
$$\frac{d\Phi}{dt}=2cxlv$$
Force on conductor $= 2c^2x^2l^2v$
*According to motional EMF approach
$$\epsilon = cxvl$$
Force on conductor $= c^2x^2l^2v$
What have I done wrong?
|
According to the flux approach,
Φ=2
This step is incorrect. If I take any dx element at a distance x from the AB, then area of element is $ldx$ and magnetic field $$B=cx\tag1$$.
Then Flux $\phi$ is given by:
$$d\phi = B dA = cx l dx$$
Integrating the expression:
$$=>\phi = \int cl xdx$$from x=0 to x=x, we get:
$$\phi = \frac12 clx^2$$
EMF $\epsilon$ is given by:
$$\epsilon=\frac{d\phi}{dt}=clx\frac{dx}{dt}=clxv\tag2$$
Further force on conductor is:
$$F=ilB$$
where $$i=\frac{\epsilon}{R}\tag3$$
Substituting the known expressions from eq(1),eq(2) and eq(3) at position x:
$$F=\frac{c^2L^2x^2v}{R}$$
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Squares of operators in QM Let $\vec{p} = p_x \hat{x} + p_y \hat{y} + p_z \hat{z}$, and also use the notation $|\vec{p}| = p$, where $p^2 = p_x^2 + p_y^2 + p_z^2$.
What is the difference between the operator $\hat{p}^2 = \hat{\vec{p}} \cdot \hat{\vec{p}}$, and the operator
$\widehat{p^2}$? And which one is the operator that correctly represents the magnitude of the momentum, squared?
|
What is the difference between the operator $\hat{p}^2$, and the operator $\widehat{p^2}$?
For the sake of clarity let me call your operator $\widehat{P^2}$ as $\hat{M}$ (and let us first work with one dimension). Then by your definition (as stated in the comments), $$\hat{M}\vert\psi\rangle = p^2 \vert\psi\rangle,$$
where $\vert\psi\rangle$ is a momentum squared eigenstate. Now, $$p^2\vert\psi\rangle = p\cdot p \vert\psi\rangle = p\cdot \hat{P}\vert\psi\rangle = \hat{P}(p\vert\psi\rangle) = \hat{P}^2 \vert\psi\rangle.$$ Which means that $\hat{M} = \widehat{P^2} = \hat{P}^2$. Note that the operators $\widehat{P^2}$ and $\hat{P}^2$ share eigenstates (Why? It is because they commute. This can be arrived at by considering the Poisson bracket of classical $P^2$ and $P$).
But typically $P$ (from classical mechanics) is always quantized to give $\hat{P}$ (of quantum mechanics).
And which one is the operator that correctly represents the magnitude of the momentum, squared?
The above discussion should have made this clear. It really doesn't matter which one you use, as long as you define your $\widehat{P_i^2}$ to obey the eigenvalue equation that returns the ($i^{th}$ compononent of the) momentum squared value for any momentum eigenstate.
| {
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How is the spin of hurricanes explained from an inertial frame? I have read that hurricanes spin because of the Coriolis effect. Since the Coriolis force is a ficticious force, which is only present in a frame that is rotating w.r.t. to an inertial one, I am wondering if an inertial observer would have an explanation which does not rely on ficticious forces.
| In an inertial frame the still (relative to the ground) air at the equator is moving very rapidly to the east and the still air at the poles is stationary with still air in between moving at some intermediate speed to the east.
In the northern hemisphere air moving to the north is going from a region of fast eastward moving air to a region of slow east moving air. So because it is going faster eastward than still air it is deflected eastward relative to the ground. For similar reasons air moving to the south is deflected westward relative to the ground. This combination gives rise to a counter clockwise rotation.
| {
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Why is the vibrational kinetic energy of a body ignored when calculating the total kinetic energy? In calculating the total kinetic energy of a macroscopic rigid body, we add the total translational kinetic energy and the total rotational kinetic energy of the constituent particles. Why is the total vibrational kinetic energy of the constituent particles left out?
| A net force or torque on a rigid body will not affect its internal energy. As it remains constant before and after the application of the force/torque, it is not relevant to the equations of motion.
| {
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Can alpha, beta or gamma particles induce fission? Nuclear weapons and reactors get their chain reactions underway via neutron radiation, but do other radiation particles ever contribute?
| Gamma initiated fission is well known, being studied from the very beginning of nuclear physics (lots of work using photofission to understand energy levels in light nuclei) . A little-known feature of the Evaluated Nuclear Data Files is that it has evaluated cross sections for gammas as well as for neutrons. I use the mirror at Brookhaven.
As an example, I enter '238U' for the Target, 'g,*' for the Reaction (not noted in the suggested list to the left - sigh), and 'sig' for the Quantity, hit return, and get a listing of 5 different types of gamma cross sections. If I select the "U238(G,F),SIG" (yes, "F" for Fission, not Fluorine) entry and plot it I get:
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What's exactly the new definition of kilogram, second and meter? Could one explain this?
Technically a kilogram (kg) is now defined:
[…] by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 × 10–34 when expressed in the unit J s, which is equal to kg m2 s–1, where the metre and the second are defined in terms of c and ΔνCs.
Does that mean that 1 kg = 1 Planck constant?
And what exactly is the new definition of the second and the meter?
| The definitions of "second" and "metre" have not changed.
One second is defined in terms of frequency. Frequency is measured in hertz $(1\ \rm Hz=1\ s^{-1}$
We take an atom of $\rm Cs$. And then, we count the frequency of its spectrum. We extract the unit "1 second" from there.
As for the meter, we set that "one metre is the distance light travels in $\frac{1}{2,997955}~\rm s".$
So the meter and the second are perfectly defined.
The new thing is that the kilogram is no longer "the mass of a weight located in Paris, France". Now we have redefined it in terms of absolute things.
If you take the actual definition of metre and second, Plank's constant is
$$h=6,626\ldots \times 10^{34}~\rm Js$$
with many decimal numbers.
So we say "okay, let's cut the decimals somewehre". Let's say that Plank's constant is now EXACTLY
$$h:= 6.626 070 15 \times 10^{-34}~\rm Js$$
And then we say "adapt the value of $1~\rm kg$ so that Plank's constant is exactly that one.
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Proof of a uniqueness theorem in electrostatics I am trying to understand problem 3.4 in Griffiths' Introduction to Electrodynamics:
Prove that the field is uniquely determined when the charge density is given and the potential $V$ is specified on each boundary surface. Do not assume the boundaries are conductors, or that $V$ is constant over any given surface.
I've tried and this is my approach:
Let there exist two potentials for the given constraints as $V_1$ and $V_2$.And let $V_{3} \equiv V_{1}-V_{2}$.
Both of the potentials$V_1$ and $V_2$ obey :$\nabla^{2} V=-\frac{\rho}{\epsilon_{0}}$. So $\nabla^{2} V_{3}=\nabla^{2} V_{1}-\nabla^{2} V_{2}=0$.
Also as the potential at every boundary is specified, then at the boundaries $V _3=0$ .
So we have$\nabla^{2} {V_3}=0$ and
$V_{3}=0$ at boundaries.
Since $V_3$ is a harmonic function so we can deduce from above that $V_3=0$ everywhere and hence $V$ is specified uniquely which implies that the field is unique as well because$-\nabla V=\vec{E}$. Which Completes the proof.
The textbook has a different and more difficult proof, and I feel that I'm incorrect. Please hint me how to do this problem or where I'm wrong. I don't have any teacher to go to. Thank you
| Below is an attempt to convince myself (and sorry I literally don't know how to use latex without the physics package).
Here is a link to the Overleaf document
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4-dimensional Fourier transform of $(k\cdot v)^{-1}$ I have been trying to compute, without much success, the following Fourier transform in 4-dimensional Minkowski space
$$
I=\frac{1}{(2\pi)^4}\int d^4 k \,\frac{e^{ik\cdot x}}{k\cdot v},
$$
where $v^\mu$ is any constant vector. $v^\mu$ would be spacelike in my case, if that's helpful. Do you have any ideas on how to do this? I suspect $I$ does not have a closed form.. but maybe can be expressed in terms of Bessel functions or something similar.
Disclaimer: although it's practically a mathematical computation I felt asking the physics community would be more fitting, since this type of integrals are typical in QFT.
| If $v$ is spacelike, choose the $k$ axes so that $v$ has only a $z$ component, so that the integral written out explicitly is
$$\begin{aligned}
I &= \frac{1}{(2\pi)^4} \int dk^t\, dk^x\, dk^y\, dk^z\, \frac{e^{i(k^t t - k^x x - k^y y - k^z z)}}{-k^z v} \\
&= -\frac{1}{2\pi v} \delta(t) \delta(x) \delta(y) \int dk^z\, \frac{e^{-ik^z z}}{k^z},
\end{aligned}$$
e
where $(t, x, y, z)$ are the components of $x^\mu$ (with an abuse of notation in repeating $x$) in an orthonormal basis $\{e_0, e_1, e_2, v\}$ , so that
$$\begin{gather}
t = x \cdot e_0 \\
x = -x \cdot e_1 \\
y = -x \cdot e_2 \\
z = -x \cdot v
\end{gather}$$
The last integral can be computed by a variety of methods; the simplest is to use the principal value to discard the cosine part, so that
$$\int dk\, \frac{e^{-ikz}}{k} = -i \int dk\, \frac{\sin(kz)}{k} = -i\pi \operatorname{sgn}(z)$$
(The second link has a mysterious extra factor of $1/2$.)
Putting it all together, we have
$$I = \frac{i}{2} \delta(t)\delta(x)\delta(y) \operatorname{sgn}(z),$$
or, written covariantly,
$$I = -\frac{i}{2} \left( \prod_{i=0}^2 \delta(x \cdot e_i) \right) \operatorname{sgn}(x \cdot v).$$
This last form emphasizes that the triple delta function is invariant under Lorentz transformations that leave $v$ fixed.
If $v$ is timelike, there are three sign changes: one from $k\cdot v$, one in the exponent which ends up in front of the sign function, and a third from the expression of the $t$ inside the sign function (which used to be $z$ in the spacelike case) as a scalar product $t = x \cdot v$, so the integral changes sign.
| {
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Electromagnetic field in the Casimir effect So, I read, that the Casimir effect arises from the ground state of the electromagnetic field. But I don't understand where the electromagnetic field in the Casimir effect comes from, since we are considering neutral metallic plates.
| The Casimir force has a mystique that it doesn't deserve. It's just an ordinary electromagnetic force between charged particles.
If the plates were made of uncharged particles, there would be no Casimir force. But metals are made of charged electrons and nuclei. At large distances the electromagnetic field of the positive and negative charges cancels almost perfectly, but at small distances the separation between the charges is relatively large and there are detectable electromagnetic effects.
| {
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How does information of the parent neutron star get encoded on the event horizon of a black hole? I was watching a video on a typical black hole formation from a neutron star and it suggested that the event horizon appears instantaneously at the surface of the star even as the stellar matter inside starts to converge into a singularity.
I have also read that the information that gets encoded on an event horizon is because of the fact that all the infalling matter gets sort of smeared on the black hole's surface from the point of view of an outside observer. So what happens to the information of all the original stuff in the neutron star if the stuff is inside the event horizon to begin with and so never gets smeared over it?
( And if we say that the event horizon starts out small from the center of the neutron star then shouldn't Hawking radiation destroy or at least impede the formation of the nascent and tiny black hole? )
| Let me begin to state that the neutron star must have a mass of at least 1,43 times the Solar mass to form a black hole. When the neutron star starts to contract it will become a quark star, which contracts further. So the event horizon can never be formed at the surface of the neutron star.
The radius of the event horizon (Schwarzschild radius) is $\frac{2MG}{c^2}$, but that's not of importance here. The question about the information of the stuff inside the star was a long time called "the information paradox". In my humble opinion, this paradox hasn't been resolved yet. Süsskind says he has, but his evidence is based on string theory (ADS/CFT correspondence) which I don't believe to correspond to reality.
For an observer falling in the material doesn't get smeared out. All material is falling in according to him.
| {
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What if Jupiter or the Sun was made of rock, like Earth and Mars, rather than gas? Jupiter is a "gas giant". If it was (significantly) bigger the pressure from gravitation would ignite a fusion process and it would become a star, which is basically what happened to the sun.
However, what if a body the size of Jupiter or the Sun was made of rocks like Earth and Mars are - what would happen then? Somewhere around iron (lead?) fusion can no longer take place and there is plenty of heavier-than-iron material on Earth.
Or is there something that would prevent such a large body of rocks to form?
| As far as I understand, rocky planets can only grow up to a certain size. This has to do with planetary formation period. A planet cannot grow indefinitely. It can grow only as long as there are particles around the star that can contribute to its increase of mass. During the formation period, dust particles collide and coalesce to form chunks, which further grow in size by gathering more dust particles or by combining with other chunks. This can go on only as long as there are supply from the dust disk surrounding the star. Eventually this gets depleted and planet can no longer grow much. Also note that the rocky planets are usually found to be closer to the star. Due to this, the amount of material available to form the planet is relatively smaller than the gas giants. The gas giants being formed at the outer regions, have larger circumference and thus more material gets fed into it. This cannot be the case with inner rocky planets.
| {
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In what case can we describe forces by potential? Let's consider a particle in an $N$-dimensional space and let's assume that acceleration of this particle depends on its position. So, one can say that we have an $N$-dimensional vector field in an $N$-dimensional space (a mapping from position to acceleration)
What allows us describe this "acceleration field" as gradient of some potential?
Should dependency of acceleration on position have a special form to make it possible to introduce a potential?
| If a force ${\bf F}=\sum_{i=1}^N F_{i}({\bf x})\mathrm{d}x^i$ is given as a co-vector field/1-form on an $N$-dimensional manifold, then Poincare Lemma states that there exists locally a potential $V({\bf x})$ such ${\bf F}=-\mathrm{d}V$ if and only if the 1-form ${\bf F}$ is closed $\mathrm{d}{\bf F}=0$.
| {
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Sand leaking out of bag and skater "A person skating on a frictionless icy surface is holding a sandbag. The sandbag has a small hole at the bottom, from which the sand starts to leak. As the sand leaks from the sandbag, the speed of the skater..."
The answer was that the speed of the skater stays the same. I thought that the speed of the skater must increase because now the mass of the sandbag is less than the mass before it leaked. What is wrong with my intuition here? The system I'm considering is the skater and sandbag together.
$p_{system}(t_i) = (m_{skater}+m_{sandbag})v_i \\
p_{system}(t_f) = (m_{skater}+m_{sandbag})v_f$
Since the mass of the sandbag is smaller, the $v_f$ must increase, no?
| So what you have done is proceeded with "momentum is always conserved. So $m_i v_i = m_f v_f$ so that $v_f = \frac{m_i}{m_f} v_i$. If you know what the ratio $\frac{m_i}{m_f}$ is, this should answer your question.
But
The net force on skater-bag system is zero. You originally proceeded to apply momentum conservation and therefore there'd be an increase/decrease in velocity to the skater, but this is not the case. The sand still has its initial momentum when it is released. The mass that is being lost (leaking sand) has the same horizontal velocity as the skater-bag. So the sand carries away horizontal momentum from the system. At the end there is less moving mass but also less horizontal momentum remains in the system. That is, as the bag leaks sand, the sandbag decreases in mass and the total momentum carried by the skater-bag system therefore decreases.
Overall the velocity for the skater must stay the same (since no external forces are acting in the direction of the skater-bag system).
| {
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Does the oscillating electric and magnetic field of a photon generate gravitational waves? From my understanding, little as it may be, because photons have energy they warp spacetime. The energy is expressed as an oscillating electric and magnetic field. Would this mean that the energy is also oscillating and would generate a gravitational wave?
| Photons are massless and you need massive particles to generate gravitational waves. So clearly it cannot generate gravitational wave. We need massive particles for gravitational waves as, gravitational waves are periodic oscillation of the spacetime framework. Only massive particles can create curvature in spacetime and not massless ones. Massless particles only follow the trajectory set by the curvature of spacetime. As far as I know, the massless particles cannot effect the curvature of the spacetime.
In case you are unfamiliar with the term massive, it is used for any particle with mass, no matter how small or big it may be. By massive, we mean it has mass and it has nothing to do with its size.
| {
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How does radiation transfer of heat fit in with the Clausius statement of second law of thermodynamics? The Clausius statement of the second law of thermodynamics says that heat flows from a hotter body to a colder body. Heat can flow in many different mechanisms. In the mechanism of radiation for transferring heat, the body emits radiation though there may not be a temperature difference between it and outside. A simple example: consider a body in a complete vacuum, the vacuum has no defined temperature (acc. to this stack) but it still should emit radiations.
And this radiation that is emitted may travel through space and strike another body which may have hotter temperature than the body emitted it and then cause it to heat up. So, this seems like a violation of the second law.
A possible resolution: The lightwave radiated out by the body will redirect itself (somehow?) to strike only bodies colder than it... but this almost seems ridiculous to think about.
| Clausius's differential inequality $dS\ge \frac{\delta Q}{T}$ can also be written as an inequality between rates as follows
$$\frac{dS}{dt} = \dot S \ge \oint_{\partial \mathcal B} \frac{\dot q}{T} dA \tag{1}\label{1}.$$
In $\eqref{1}$ $\mathcal B$ is the body of the system receiving heat through its boundary $\partial \mathcal B$ at a rate $\dot q$ and the temperature of the surface element $dA$ is $T=T(dA)$. As written this inequality has only "surface heat sources" but it can be generalized to include "volume heat sources"; Truesdell calls it the Clausius-Duhem inequality[1] :
$$\frac{dS}{dt} = \dot S \ge \oint_{\partial \mathcal B} \frac{\dot q}{T} dA + \int_{\mathcal B} \frac{\dot s}{T} dm\tag{2}\label{2}.$$
In $\eqref{2}$ the quantity $\dot s$ represents the heat supply per unit mass $dm$ and per unit time (it is a rate) at temperature $T=T(dm)$. When the process including the heat transfer is reversible one has equality in $\eqref{2}$. This is a very natural generalization of Clausius's inequality and it also includes radiation that is absorbed "bodily". Just as with $\dot q$ the sign of $\dot s$ tells you in what direction does "heat", i.e., energy and entropy may flow; more specifically when $\dot s$ is the radiated heat supply between two bodies then depending on their relative temperatures one body may be the source while the other the sink, or vice versa. Of course, if they have the same temperature then there is no net flow between them, for whatever one absorbs it will also radiate it out.
[1] Truesdell: Rational Thermodynamics, page 117
| {
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Is a Lagrangian term $L_{kin}=(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)$ equivalent to $L_{kin}=\phi^{*}\partial^{\mu}\partial_{\mu}\phi$? In looking at the Lagrangian of a (free for simplicity) complex scalar field $\phi$, we have a kinetic term that goes like:
$$L_{kin}=(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)$$
Given instead, a kinetic term of the form:
$$L_{kin}=\phi^{*}\partial^{\mu}\partial_{\mu}\phi$$
Is this not equivalent to the first term? Just by simply varying each term with respect to $\phi^{*}$ and it's derivative we arrive at the same equation of motion (up to some arbitary multiplacative factor) namely:
$$\partial^{\mu}\partial_{\mu}\phi$$
I'm sure this can be expressed in terms of formal adjoints more precisely, but if they are equivalent, why do we always write scalar fields in terms of first order lagrangians (boundary conditions maybe)?
| Almost, you're off by a minus sign, but otherwise yes. You're supposed to use (higher-dimensional) integration by parts on the action
$$
S = - \int_{\Omega} d^4 x\ \big( \partial^\mu \phi^\ast(x) \big) \big( \partial_{\mu} \phi(x) \big) = - \int_{\partial \Omega} d^3 \Sigma\ \phi^\ast(x) n^{\mu} \partial_{\mu} \phi(x) \big) + \int_{\Omega} d^4 x\ \phi^\ast(x) \partial^\mu \partial_{\mu} \phi(x)
$$
where $\Omega \to \mathbb{R}^{4}$ and $n^{\mu}$ is the outward unit normal to the boundary $\partial \Omega$ (and $d \Sigma$ is the volume form on the boundary). The standard thing to do is assume that the boundary term vanishes and you find that $\partial^\mu \phi^\ast \partial_\mu \phi$ and $-\phi^\ast\partial^\mu \partial_{\mu} \phi$ can be used interchangebly.
| {
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Why is each wavelength of light diffracted differently?
Why is do we see different colors, instead of white spots like in the middle?
Why is the single slit less intense the further away from the middle you look?
| This occurs because the wavelengths are different for different colours, using the relation between diffraction strength as $S=S(h/\lambda)$ ($h$ is slit size and $\lambda$ is wavelenght), it is apparent that when you keep the slit size constant and compare the different wavelengths then diffraction strength changes so does the separation between the colours emerges. The equation can be driven by the propagation of waves.
| {
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Is there any physical interpretation of the constant which is seen in the constraint curve of an adiabatic process? What is the $C$ in $ PV^{\gamma} = C$? I always saw it as a result out of the mathematical calculations that we do but I recently saw this video which made me think that the constant may have more meaning that meets the eye.
See this video at 4:11
He writes $ S = PV^{\gamma}$ .. but where exactly is this equation from? I don't think I've seen it anywhere else.
| $C$ is a constant. It simply means that the relationship on the left between pressure, volume and the ratio of specific heats of constant pressure and volume yields the same number at any equilibrium state during the process. The physical interpretation in the case of this process is that it is a reversible adiabatic (isentropic, or constant entropy) process for an ideal gas.
The derivation of this equation is based on no change in entropy, coupled with the ideal gas equation, definition of enthalpy and internal energy, and assumption of constant specific heats. For a derivation see:
http://www.mhtl.uwaterloo.ca/courses/me354/lectures/pdffiles/ch2.pdf
As @Knzhou pointed out although the process is constant entropy, $C$ is not the value of the entropy.
A similar situation exists for a reversible isothermal (constant temperature) process. The equation is
$$PV=C$$
Where again $C$ is a constant (not the same constant as the constant entropy process) but it is not the value of the constant temperature. Here
$$C=nRT$$
Both the isentropic and reversible isothermal process are specific cases of the more general reversible polytropic process for an ideal gas, where
$$PV^{n}=C$$
For the isentropic process, $n=C_{p}/C_{v}$. For the isothermal process, $n=1$. For a constant pressure (isobaric) process $n=0$.
Hope this helps.
| {
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If objects in motion experience time differently, how does my body stay synced when I move my legs or arms? If I move my swing my arm really fast, the matter in my arm should experience time slower than the matter in my body.
So how does my body still sync with each other?
And a more general question that derives from this: A lot of matter move at different speeds inside our body, how does anything ever stay synced?
|
how does anything ever stay synced ?
Not sure what you mean by "stay synced". Different parts of your body maintain their structural integrity at the atomic level because of the electromagnetic forces between atoms and molecules. This simply involves the exchange of photons (the force carrier for the electromagnetic force) over very short distances - no "syncing" is required. Similarly, nerve impulses to and from different parts of your body are chemical signals sent down nerves, which also ultimately depends on the exchange of photons at an atomic level. Again, no "syncing" required.
In computer science terms, the body is an asynchronous system. There is no master clock in the body that says "hey, arm, you're a femto second behind everyone else".
| {
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Why don't we calculate spin number in classical mechanics? We all know classical mechanics deal with bigger objects and
quantum mechanics deal with very tiny particles.
I hear spin number in quantum mechanics, but I don't see anything
like that in classic mechanics.
Why don't we calculate spin number in classical mechanics?
| Loosely speaking:
Classical mechanics is the approximation of quantum mechanics
in the limit of $\hbar \to 0$ (where $\hbar$ is Planck's constant).
In quantum mechanics the electron's spin is given by
$$\vec{S}=\frac{1}{2}\hbar\vec{\sigma}.$$
where $\sigma_x$, $\sigma_y$, $\sigma_z$ are the Pauli matrices.
The classical limit (for $\hbar \to 0$) of this is obviously:
$\vec{S}=\vec{0}$.
That means: There is no spin in classical mechanics.
Spin is an entirely quantum mechanical phenomenon.
| {
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Symmetry for dipole conservation in field theory In article The Fracton Gauge Principle complex scalar field is considered.
There's statement, that for conservation of charge one needs usual U(1) global symmetry:
$$
\phi \to e^{i\alpha}\phi \Rightarrow Q =\int d^Dx \rho
$$
For conservation of dipole moment:
$$
\phi \to e^{i\vec x \cdot \vec \lambda}\phi \Rightarrow Q^i =\int d^Dx x^i\rho
$$
Why for dipole moment we need such symmetry?
| I think the right way to think about it is to first ask what the symmetry
$$
\phi \longrightarrow e^{i\theta} \phi
$$
means. The actual Unitary operator acting on the many-body Hilbert space that corresponds to this transformation is
$$
U(\theta) = \exp\left( i\theta \sum_r \phi^\dagger_r \phi_r \right)
$$
It is easy to see that $\phi$ transforms as shown in the first equation upon the adjoint action of $U$. So you see that the the $U(1)$ symmetry is generated by the total particle number operator. So, a system that has this symmetry conserves particle number and vice versa.
Now we consider a system with conservation of dipole moment. Dipole moment is given by the vector operator $\sum_r \boldsymbol{r}\phi^\dagger_r \phi_r $. Therefore, conservation of dipole moment actually refers to three independent conservation laws, whose corresponding unitary operator will be parametrised by a 3-vector $\boldsymbol{\lambda}$ as
$$
U(\boldsymbol{\lambda}) = \exp\left( i\boldsymbol{\lambda} \cdot \sum_r \boldsymbol{r}\phi^\dagger_r \phi_r \right)
$$
Now one can ask how does this act on the particle annihilation operator at a particular site $\phi_{r_0}$. We can simply look at the adjoint action
\begin{equation}
\begin{split}
\phi_{r_0} \longrightarrow U^\dagger\phi_{r_0} U &= \exp\left( -i\boldsymbol{\lambda} \cdot \sum_r \boldsymbol{r}\phi^\dagger_r \phi_r \right) \phi_{r_0} \exp\left( i\boldsymbol{\lambda} \cdot \sum_r \boldsymbol{r}\phi^\dagger_r \phi_r \right) \\
&=\exp(i\boldsymbol{\lambda} \cdot \boldsymbol{r_0})\phi_{r_0}
\end{split}
\end{equation}
Thus we see that what you stated is indeed the transformation that is generated by the total dipole moment operator, so it is the corresponding symmetry to it. Note that we can easily generalise this argument to the continuum.
| {
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Why is kinetic friction present when two objects are slipping across each other my mechanics sir was telling us the other day that when the relative velocity is zero between the surface and the object then the object starts sliding, eg in rolling motion without slipping, the v rel is zero at the bottom most point .how does kinetic friction come into play here. I understand that since the object has started moving then kinetic friction will come into play. Is this the one and only reason?
he, secondly, mentioned that slipping and sliding are essentially the same phenomena. I don't quite agree with this statement. can someone pls explain this as well?
Also, this is not really a question but a few thoughts that came to my mind that I wanted to discuss.
| When you send a bowling ball down the alley, it starts by sliding. The kinetic friction force (predicted by the coefficient of kinetic friction) on the bottom of the ball produces a torque which causes an angular acceleration. When the backward tangential velocity of the bottom of the ball (measured relative to the center of mass), matches the forward velocity of the center of mass (measured relative to the alley), the ball is rolling, and the bottom is in static contact with the alley. There may be a small “rolling friction” due to the deformation of the two surfaces. If the ball then rolls up an incline, there can be a static friction force acting up the incline (which matches the loss of angular velocity to the loss of linear velocity). The static force has an upper limit predicted by the coefficient of static friction. Static and kinetic friction originate at the microscopic level where the peaks and valleys of the two surfaces inter-mesh. Generally, kinetic friction is a little smaller than the maximum allowed static friction (the moving surfaces are bouncing peaks off of peaks).
| {
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Antiproton synthesis For a pion minus hitting a stationary proton, what are the other particles if an antiproton is to be created among them? A positive pion is possible but the total rest mass energy of the final state is comparable to the initial particles.
| I'm not sure you have spelled out the conditions of your problem.
Does
$$
\pi^- + p \to n + \bar p + p ,
$$
that is,
$$
d\bar u + uud \to ddu +\bar u \bar u \bar d + uud
$$
meet them? To conserve baryon number, this is your most economical option.
| {
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Can the universe ever contract? I am currently going through this answer related to the Big Bang theory and from there a question arose in my mind:
*
*Can the universe ever contract?
*Can it ever contract to singularity?
I wonder, if it is possible, how it would happen? Is there any chance at all?
I am unaware about the physical reality of this question.
| Whether it will or it won't, we don't know for sure. Unless we understand the dark energy and dark matter, only time will tell.
It was one of the earlier theories but current evidence seems to suggest that won't be the case. According to the most prominent contemporary physicists, everything in the universe will spread apart so far that it will result in a cold death of the universe where you won't be able to see any star in the night sky.
| {
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Why is there a need for the concept of energy if we have the concept of momentum? We have two concepts that are energy and momentum. To me, momentum is more fundamental than energy and I think that momentum was the thing which we wanted to discover as energy.
Now momentum can describe several things that that energy does and if it is not able to describe it then it can be somehow extended to describe that thing.
For example, momentum can not describe the potential energy let's say due to the gravitational field of Earth on an object but it can be easily twisted to be able to describe it.
Momentum can describe quantum stuff as well. Also, we know that momentum is conserved just as energy.
In short, I want to know the physical difference b/w momentum and energy.
| Here is a thought experiment that should convince you that it's necessary to consider energy and not just momentum. Suppose you stand on a train track in between two identical trains traveling at the same speed. One of them approaches you from your left, and the other from your right. The trains are moving in such a way that they will both collide with you at the same instant.
Since the trains have the same mass and speed but are traveling in opposite directions, they carry zero total momentum. When they hit you, they will not tend to accelerate you to the left or to the right — you will stay right in place. In spite of their lack of net momentum, the trains do carry kinetic energy. When they collide with you, they will transfer some of this kinetic energy to your body. Anyone watching will clearly see the effect of this energy, although you yourself will probably not be able to observe it.
| {
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What does it mean for particles to "be" the irreducible unitary representations of the Poincare group? I am studying QFT. My question is as the title says. I have read Weinberg and Schwartz about this topic and I am still confused. I do understand the meanings of the words "Poincaré group", "representation", "unitary", and "irreducible", individually. But I am confused about what it means for it to "be" a particle. I'm sorry I'm not sure how to make this question less open-ended, because I don't even know where my lack of understanding lies.
| Irreducible representations of the Poincare group are the smallest subspaces that are closed under the action of the Poincare group, which includes boosts, rotations, and translations. The point is that we should interpret these subspaces as the set of possible states of a particle. For example, if you start with a state representing a particle at rest, then you can boost it (so it starts moving), rotate it, translate it, and so on. But all the states you can reach represent, by definition, the same kind of particle, just in different states of motion.
The requirement that the representation be unitary just means these operations keep states normalized.
| {
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Computing a quality factor of multiple measurements Suppose I measure the same quantity twice with two methods, first I get 0 with 0.001 uncertainty, then I get 1 with 0.000001 uncertainty. We can see from this most likely there is something wrong with the uncertainties or measurements. I am faced with such a problem, I have multiple measurements of the same quantity with prescribed uncertainties, and I would like to calculate how compatible these measurements are, or if they are correct, how unlikely it is to get those values. A quality factor or something is what I'm looking for.
What is the most sensible way to do this?
The goal is to know if there is something wrong with the uncertainties or not.
| Measurements that use the same method can be assumed to have the same systematic error, affecting all measurements equally. So you can, for example, take an average value of these measurements or discuss the consistency of different measurements without knowing the systematic error.
However, it is not possible to compare measurements made with different methods and hence different systematic errors without knowing what these systematic errors are. For example, the two measurements in your example could be consistent with each other if one has a systematic error of $-0.5$ and the other has a systematic error of $+0.5$.
| {
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Why two orbitals having same phase is not a random phenomenon? I have been reading about Molecular Orbital Theory for Chemistry.
*
*I tend to believe that when two Hydrogen atoms approach each other, whether the $1s$ orbitals are in-phase or out-phase is a random phenomenon. However I know that this is not so. Please provide some arguments to counter it.
Please try not to indulge in complex Mathematics as I am relatively new to the subject. My mathematical understanding is "negligible".
| I have been going through several articles on the matter. Based on that, I have found that the Linear Combination of Atomic Orbitals is a mathematical model with little correlation to the actual process of formation of Molecular Orbitals. Hence the concept of phase/sign of atomic orbitals are simply for the process of denoting that whether our mathematical model will "add" these orbitals or "subtract" these orbitals.
In short, what I am trying to look for in my question (and I suspect a lot others who are new to the subject) is simply not there "physically" but only mathematically.
| {
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Are wormholes evidence for traversal of a higher dimension? Warning, pop science coming.. please correct what I’m getting wrong. Einstein’s equations of relativity showed the potential for existence of wormholes that can connect different points in space time. I understand the mechanisms for their practical implementation are nothing near feasible. However, based on the equations of gravitational “tunneling”, I can traverse back and forth between times and locations. Wouldn’t this require a higher dimension than 4d space time?
That is, we’re moving from a point that we would think of as the present to another point we would think of as the present. If this were feasible, Would These “presents” need to be on a traversable continuum?
To my lay brain, This seems as though there are points along a higher dimension where what we would consider the future is currently present, and what we consider the past is also present. That the world we see is determined and laid out as slices in a higher dimension that would be traversed with a wormhole, and that we normally traverse in a single direction.
| Wormholes in GR do not require higher dimensions. It easier to imagine curved spacetime as being embedded in higher dimensions, but the usual mathematical description of curved spaces does not require that.
| {
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Direct and in-direct measurments I have seen and heard this many times that a certain (physical) measurement is "direct" or "indirect". What is the characteristic or definition that sharply separates these two notions?
| I'm guessing a direct measurement is one you would read directly from your measuring instrument, like a weight on a scale or the size of an object read from a tape measure. An indirect measurement might require a calculation, like the mass of a planet from the orbital period of a moon.
| {
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About parallel and intersecting timelike worldlines Suppose that two straight timelike worldlines are (not) parallel with respect to some frame $S$. Will these worldlines remain (not) parallel with respect to any other system $S^{\prime}$ related to $S$ by some Lorentz transformation? Whatever the answer, how can this be formally demonstrated?
| Two parallel lines must always remain parallel. Let us say two objects are stationary in some reference frame at $x_{0}$ and $x_{1}$, then performing a Lorentz transformation to a primed frame with velocity $\beta$ yields $x'_{0} = \gamma x_{0} - \gamma \beta t$ and $x'_{1} = \gamma x_{1} - \gamma \beta t$.
Further, since $t’ = \gamma t- \gamma \beta x_{0}$, we have equivalently $t = t’/\gamma + \beta x_{0}$, and hence $x'_{0} = \gamma x_{0} - \gamma \beta^2 x_{0} - \beta t’$ and $x'_{1} = \gamma x_{1} - \gamma \beta^2 x_{1} - \beta t’$ which are clearly two parallel lines in the $t‘-x'$ plane, with angle $\arctan(-\beta)$ relative to the $x'$ axis, with the domain restricted to $[0, \pi]$.
From here it is easy to generalize to the case where they are not stationary to begin, but are parallel with some angle $\theta$ with respect to the $x$ axis. Let us say your transformation is $\Lambda_{f}$. Use the approach in the previous paragraph (working backward) to go back to the stationary frame (with a transformation $\Lambda_{inv}$, say), and then applying $\Lambda_{f} (\Lambda_{inv})^{-1}$, we get our result. Note that the fact that $(\Lambda_{inv})^{-1}$ is also a Lorentz transform and that the product of two Lorentz transforms is still another transform is guaranteed by the group structure.
A remark: you can think that if two lines were parallel in one frame but not another, then the act of changing the reference frame has somehow introduced one to accelerate with respect to the other, which is grossly against the spirit of relativity.
| {
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Am I understanding the concept of rotations correctly? Consider a ball placed at distance $r$ from an origin point $O$ on a horizontal plane and remains stationary.
*
*When the ball is viewed from a coordinate system which is rotating
anticlockwise about $O$ with an angular velocity $\Omega$ , what is
the apparent motion of the ball?
*What are the forces viewed from the non-rotating coordinate system?
*What are the apparent forces viewed in the rotating coordinate system?
From what I understand.
Let r be the position vector between origin $O$ and the ball, and r' be the position vector between the ball and the origin of the rotating frame, $O'$
*
*Since the ball itself is in the non-rotating frame and remains stationary, I am tempted to say no motion. However, considering the change in position vector r' between the ball to the rotating frame, so is the apparent motion moving in a curved path? I am having trouble to figure out who exactly observes this change in position vector. Perhaps, since the rotating frame itself is moving, it should not be observing any form of motion?
*There are no motions in the non-rotating frame. Thus, there should be no forces here.
*In general, one observes Coriolis force and centrifugal forces in a rotating frame of refence. So both forces.
| *
*The vector as an abstract object which points from the origin to the ball is independent of the coordinate system and remains constant in time. However, the coordinates of the vector do depend on the coordinate system and thus are not constant if the basis vectors are not constant.
*Exactly!
*Nope, you have the centrifugal force only, no Coriolis force. You would get a Coriolis force, if the ball had a velocity in the rest frame though.
| {
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Generalized Pauli matrices I wanna know the generalized form of Pauli matrices, for example for $3\times 3$. And do they satisfy all of the properties of Pauli $2\times 2$ matrices?
I wrote $3\times 3$ but I couldn’t write all Hermitian $3\times 3$ matrices with those.
| Pauli matrices (plus the identity matrix) are just a choice of matrices that allow decomposition of an arbitrary 2-by-2 matrix - i.e. a matrix with 4 independent parameters. One could choose them differently, so this particular choice is more due to the tradition and the fact that all the three matrices are already Hermitian. It is not uncommon to use $\hat{\sigma}_\pm$ instead of $\sigma_{x,y}$.
For 3-by-3 case one needs in principle 9 matrices, one of which can be the identity matrix. They also can be all chosen Hermitian. However, one will have more freedom than in the 2-by-2 case. This freedom might be further restricted by a particular application - e.g., describing certain type of rotations.
| {
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The derivative of the unit velocity vector
The set up:
An intertial frame Y-X used to describes trajectory of an insect on some rigid body using some relative vectors. Symbols: $ \vec{r_a}$ is is the vector connecting the origin to some point on the rigid body, $ \vec{r_b} $ is the vector connecting origin to the insect and $ \vec{r } $ is the vector connecting the reference to the insect. The relation between vectors:
$$ \vec{r_b} = \vec{r_a} + \vec{r } $$
In a video lecture about corollis force by professor M.S.Sivakumar, I don't get a formula at 19:12 which is used:
$$ |v_{rel} | \frac{ d \hat v_{rel} }{dt} = \vec{\omega} \times \vec{v_{rel}} $$
With,
$$ v_{rel} = \frac{ d|r| }{dt} \hat{r}$$
Where $ \hat{r} $ is a unit vector connecting the reference to the insect $|r|$ is the length of the whole vector connecting the reference to insect.
In a previous post, I had it explained to me the relation about the time rate change of basis is related to the angular velocity by the equation $ \frac{d}{dt} \hat{u} = \omega \times \hat{u}$. However, I do not understand how that idea extends to this case as we are talking about the basis of velocity since $ \omega$ which was used initially was regarding the angular change of the position vectors.
References:
Previous stack post
Lecture Series on Mechanics of Solids by Prof.M.S.Sivakumar, Department of Applied Mechanics, I.I.T.Madras.
| Since $\hat{v}_{rel}$ is a unit vector $\dot{\hat{v}}_{rel} = \vec{\omega} \times \hat{v}_{rel}$. Multiplying by $|\vec{v}_{rel}|$ on both sides gives you that equation.
| {
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Superfluid Stiffness Definition I am currently reading S. Sachdevs Book on Quantum Phase Transitions focusing on the Bose-Hubbard Model (Chapter 9) and especially the Dilute-Boson Field Theory (Chapter 16).
When describing the fluid phase of the one dimensional model Sachdev says that this phase has quasi-long range order in the superfluid order parameter, intermediate boson occupation number and a non-zero superfluid stiffness.
I could not find any definition of a superfluid stiffness in the entire book and also doing some research on the internet I was not able to find a clean definition of superfluid stiffness in this context (Most likely because of my incapability :D to find something).
Therefore my question:
*
*Could somebody provide a definition of a superfluid stiffness in the context of the Bose-Hubbard Model?
*Any further explaination of this quantity in the "quasi long-range ordered" phase of the XY-chain would also be very kind?
Thank you all in advance.
| There are probably different conventions that lead to definitions that differ by some numerical factor or factors of the mass density, but essentially the superfluid stiffness is the coefficient $\alpha$ in the expressions for the energy density
$$
E[\theta]= \int d^3 x\frac 12 \alpha |\nabla \theta|^2,
$$
where $\theta$ is the phase of the superfluid order parameter. A non-zero $\alpha$ means that it costs energy to have a space-varying phase, hence "stiffness". The superfluid particle-number current is then
$$
\rho_s{\bf v}_s = \alpha \nabla \theta,
$$
where $\rho_s$ is the superfluid (number) density.
As
$${\bf v_s}=\frac 1 m \nabla\theta
$$ where mass is $m$ of the superfluid particle one often writes
$$
E= \int d^3 x\frac {\rho_s}{2m} |\nabla \theta|^2,
$$
so $\alpha= \rho_s/m$. At finite temperature, the "energy" should be understood to be a local free energy $F=E-TS$.
| {
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Interpretation of normal modes from the mathematical formula In the topic of small oscillations, the system below has a normal mode described by:
$$n_{1} = \frac{x1+x2}{2}.$$
This normal mode is represented as the symmetric mode:
In that case, the center of mass moves as a simple harmonic oscillator. However, the picture also shows that both of them start in the same initial conditions and move in phase. My question is where that information is on the normal coordinate $n_{1}$ since I can not relate the normal mode with the picture representing it. Where does it say the blocks must be strechted the same distance in the same direction in the formula of the normal coordinate?
| I) Equations of motion
Kinetic energy :
$$T=\frac{m}{2}\left(\dot{x}^2_1+\dot{x}_2^2\right)$$
Potential energy
$$U=\frac{k}{2}\left(x_1^2+(x_2-x_1)^2+x_2^2\right)$$
with Euler Langrage you get:
$${\ddot x}_{{1}}+{\frac {2\,kx_{{1}}-kx_{{2}}}{m}}=0\tag 1$$
$${\ddot x}_{{2}}+{\frac {2\,kx_{{2}}-kx_{{1}}}{m}}=0\tag 2$$
II) Equations of motion: Normal mode
In normal space the equations of motion will be:
$$\ddot n_1+\omega_1^2\,n_1=0\tag 3$$
$$\ddot n_2+\omega_2^2\,n_2=0\tag 4$$
to obtain equations (3) and (4) we have to transformed the coordinates $~x_1~,x_2$ to $~n_1~,n_2$
this can be done with those equations
$$n_1=\frac 12(x_1+x_2)$$
$$n_2=\frac 12(x_1-x_2)$$
$\Rightarrow~$
$$x_1=n_1+n_2$$
$$x_2=n_1-n_2$$
with this transformation you get:
$$\ddot n_1+\frac km\,n_1=0\tag 5$$
$$\ddot n_2+\frac{3\,k}{m}\,n_2=0\tag 6$$
Remark:
you get the same results equation $(~5~,6~)$if you obtain this transformation :
the center of mass coordinate for n_1:
$$n_1=\frac{m\,(x_1+x_2)}{2\,m}=\frac 12(x_1+x_2)$$
and
$$n_2=\frac 12(x_1-x_2)$$
| {
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How we feel (perceive) exact size of object through our eyes? Light after getting reflected from objects gets focused on retina by our lens. The images formed on retina is small, which is then sensed by our brain and depending on distance we feel size of that object.
If an object is at particular distance from us, the image on retina is not going to be the exact size which we feel by our eyes after sensed by our brain. Our brain predicts size depending on the distance that brain sensed. But why the size we feel by our eyes is same as the size we feel by touch. Like if we see an object we can feel its boundary, and if we touch it and sense the boundary it does not extend or change.
Why the boundary of object we feel by eyes doesn't contradict reality?
| As R.W. Bird said, our 2 eyes give us a perception of distance for not too far objects. Our brain uses that information and the apparent size to estimate the real size.
But it doesn't work for distant objects. The moon and the sun have almost the same (apparent) size. When guided only by perception, we are completely unable to estimate their real sizes.
| {
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Heating cup in microwave? I heated my milk cup in the microwave today and noticed that the cup was hot but not the handle. Even if I heat it too much , cups handle temperature remains the same. How is that possible?
| It's probably because your milk cup is made of a material that is a relatively good thermal insulator.
First of all, the microwaves directly heat the milk, and not the cup, as long as the cup is made of material that microwaves pass through without being absorbed.
The heated milk, in turn, being in contact with the sides of the milk cup directly heats the sides but does not directly heat the handle because the milk is not in direct contact with the grasped portion of the handle. For the handle to get warm there needs to be heat transfer by conduction from the sides of the cup to the handle.
The heat transfer from the sides of the cup to the grasped portion of the handle will depend on the thermal conductivity of the cup material, the cross sectional area of the handle part of the cup, and the length of the path from the side of the coffee cup to the part of the handle being grasped. If your milk cup is made of glass, it's thermal conductivity is relatively low (roughly 1 W/m K) making it a reasonable thermal insulator.
All of the above factors can keep the grasped portion of the handle from getting too hot regardless of how hot the milk is, at least for a reasonable amount of time.
Hope this helps.
| {
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Cross section in Coulomb scattering of relativistic electrons i'm currently reading about electron scattering and i cant understand the following statement of the book im reading. I quote the book(translated by me):
Many scattering experiments are done with unpolarized beams, and the polarization of the scattered particles are not measured. It is required that we take the mean value of the absolute square of the Matrix element, summed over both the spin states of the incoming electrons and the outgoing electrons.
$\frac{d\sigma}{d\Omega}\sim \frac{1}{2}\sum_{s_i,s_f}|\bar{u_f}\gamma^0u_i|^2$
I can't understand the factor of $\frac{1}{2}$ in from of the summation. The way im thinking it you need to replace it with $\frac{1}{4}$ because of the four different combinations of the initial and the final spin, which are:
$+ \rightarrow + $,
$+ \rightarrow - $,
$- \rightarrow + $,
$- \rightarrow - $
where $+$ is spin up and $-$ is spin down.
| The usual way to do this is to average over the possible incoming states, but sum over the outgoing states. In that case the 1/2 is correct as there are 2 possible incoming states. This would be correct for the scattering cross section, since you don't know what particle you started with (thus the average) and you end up with all possible combinations (thus the sum). That is likely to be the reason for the 1/2.
| {
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Goldstone counting from symmetries and from the expansion in the Lagrangian Goldstone theorem states that when a continuous symmetry is broken there is a massless mode for each broken generator. To exemplify the theorem, many references consider the complex scalar theory with an $U(N)$ symmetry. The potential for the theory is
$$-m^2H^\dagger H+\frac{\lambda^2}{2}(H^\dagger H)^2$$
when $m>0$ the vacuum expectation value of the theory $v^2\equiv\langle H^\dagger H\rangle=\sqrt{m/\lambda}$ will be non-zero and will spontaneously break $U(N)$. If we expand the field $H$ around its vacuum as
$$H=\begin{pmatrix} v+\chi_1+i\eta_1 \\ \chi_2+i\eta_2 \\ \vdots \\ \chi_n+i\eta_n \end{pmatrix}$$
we will find that only $\chi_1$ have a mass term in the expanded potential, and that $\chi_2,...,\chi_n,\eta_1,...,\eta_{n}$ remain massless. This agrees with the general statement of the Goldstone theorem since we are breaking $U(N)$ to $U(N-1)$ and therefore we should have $N^2-(N-1)^2=2N-1$ massless modes. However, if instead, I expand the field as
$$H=\begin{pmatrix} v/\sqrt{2}+\chi_1+i\eta_1 \\ v/\sqrt{2}+\chi_2+i\eta_2 \\ \chi_3+i\eta_3\\ \vdots \\ \chi_n+i\eta_n \end{pmatrix}$$
then both $\chi_1$ and $\chi_2$ will have mass terms in the expanded potential. More generally, as I split $v$ among the different components of $H$ the respective component will get a mass term.
So I want to understand better what is going on. I am changing the symmetry breaking pattern as I split $v$ among the different components of $H$? If so, what are the symmetry breaking patterns? Alternatively, maybe this is not the right way to see if I have massless modes in the theory. If so, what is the right way and what is the relation with this way?
| If you work out the mass term using your second $H$, you will see that it is of the form
$$m^2(\chi_1+\chi_2)^2$$
Only the combination $\chi_1+\chi_2$ has a mass, and in particular the combination $\chi_1-\chi_2$ is massless.
| {
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What conditions are necessary to guarantee uniform circular motion? Suppose an object is subjected to a force of constant magnitude, which is always directed to the origin. And suppose we know the initial position of the object relative to the origin, and the initial velocity of the object, can we determine if the object will perform uniform circular motion? If so, what conditions are necessary? Can we determine its position as a function of time from these givens?
I know that if we know that an object performs uniform circular motion, and we have the equations which describe its motion, for example $$ \mathbf r= \begin{bmatrix} \cos(t)\\ \sin(t)\\ \end{bmatrix} $$
we can find the velocity, and acceleration simply by taking derivitives. But can we go the other way around and deduce the equation of motion as I described above? Perhaps by solving the differential equation $$ m \ddot{\mathbf r} = - \lVert \mathbf F \rVert \frac{\mathbf{r}}{\lVert \mathbf r \rVert}$$
where $\lVert \mathbf F \rVert$ is constant?
| These are the equations of motion
$${\ddot{r}}\,m-m{\dot\varphi }^{2}r+F=0\tag 1$$
$$ \ddot\varphi \,r+2\,{\dot r}\,\dot\varphi=0\tag 2 $$
equation (2) is also;
$$\frac{d}{dt}\left(\,r^2\dot\varphi\right)=0$$
thus :
$$\dot{\varphi}=\frac{L}{r^2}$$
where L is a constant.
substitute $\dot\varphi$ in equation (1)
$$m\,\ddot{r}-m\,\frac{L^2}{r^3}+F=0\tag 3$$
the conditions for a uniform circular motion are: $~\ddot r=0$ and $r=\text{constant}=r_0~\Rightarrow $ the initial conditions are
$$r(0)=r_0,~D(r)(0)=0$$
and the force F is:
$$F=m\,\frac{L^2}{\,r_0^3}$$
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Where is the energy involved in osmosis? Osmosis creates pressure on the side of the membrane with higher concentration. But where does the energy for this come from?
| All things naturally tend towards equilibrium, moving from high energy to low energy. Osmotic pressure is the force that helps achieve osmotic equilibrium, so it is really just a manifestation of that natural tendency. You don't really need energy to create osmotic pressure, the osmotic pressure will be present until equilibrium is reached. It comes from the higher concentration itself, not some other force. Osmosis is a sort of 'gradient-driven' process, which comes about through entropy.
https://courses.lumenlearning.com/introchem/chapter/osmotic-pressure/
https://simple.wikipedia.org/wiki/Osmosis
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Can quantum measurements be the origin of thermodynamic arrow of time? We can practically consider that the microscopic interactions are symmetric with respect to time(as we can neglect weak force for many cases which is the only interaction that can violate $T$ symmetry). So I thought that the asymmetry might be due to the irreversibility of quantum measurements. But this is only applicable for interpretations where wave function collapses like Copenhagen etc. What is the answer to this question in Many-worlds interpretation, Consistent histories, etc? Also in this page, they gave that the initial conditions of the universe are the reason for $T$ asymmetry in the 2nd law of thermodynamics. But I am not sure what they mean. Do they mean that the universe had a very low entropy at the beginning?
| Entropy is Macroscopic
As RogerJBarlow mentions in a comment, there is no need to invoke QM to explain the "arrow of time". To use the exact analogy, consider an "ideal billiard table" with the standard collection of 15 balls. Now, the macro state in which all balls are arranged in the starting triangle is very improbable, because there are only a few "micro-states" (permutations and rotations of the balls) which correspond to this macro state. Further, with typical energy inputs to the table, there are many ways to move from this macro state to other macro states, but not very many ways to move from other macro states to this one. That makes it a "high entropy" state.
On the other hand, if we consider the macro state "every pocket has at least 2 balls closer to it than any other pocket", we see that there are very many micro states in this macro state. There are also many transitions which lead into this state, and the sheer number of micro states means that many transitions will cause you to never leave this macro state. Thus, this is a "low entropy" state, and one of the more likely outcomes in a typical game of pool.
There is no entanglement, superposition, tunnelling, wave function collapse or any other QM phenomenon required to analyze the entropic behavior of this system, or to guess between a sequence of snapshots which direction indicates the arrow of time. Replace the billiard balls with gas molecules, and you start to look much like thermodynamics.
| {
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How does the whole universe agree on the laws of physics? How is it possible that the every particle in the universe agrees on the laws of physics?
What enforces those laws? Might the laws change slightly across the universe in the same way the cosmic microwave background radiation (CMBR) does?
| Let's consider an example of this violation. Consider the electrostatic attraction strength (Coulomb's constant). If we compute it and get some value $k_1$ in our labs, we might conclude that it is a law of nature that $F=k q_1 q_2 /r^2$. If we look at another part of the universe we might find some $k_2 \neq k_1$. Is this a violation?
One might say that this isn't a violation of the laws of physics, because we just found that what we thought was the law was just an approximation. The full theory would have $k(x)$ with $x$ being positions in space (or maybe space-time to include time-variation). We might not have the full theory, but surely some law exists, which upon giving initial conditions produces the function $k(x)$.
In this point of view, the law was never violated. Only our initial idea of a law was violated and was replaced by the true, more general law.
In short, if you can find a spatial variation of some "law", I would assume that you can in principle find the dynamics controlling this variation - and call that the law. That law is not violated in different locations.
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Positron decay direction from muon I read that in the rest frame of a positive muon, decay positrons are preferentially emitted in the direction of the muon spin. Why is that the case?
The decay is $\mu^+\to e^+\nu_e\bar{\nu_\mu}$. Assuming that the positron is emitted at almost the speed of light, it will be a right-handed particle. The 2 neutrinos are right-handed and left-handed. So we have a 3-body decay, with the 3 particles having their spin aligned or anti-aligned with their momentum. But I am not sure how can I get from here that the positron is more likely to be along the muon spin.
| This is the cleanest application of chirality in the weak interactions, and your instructor should have drilled it in your SM course.
The positron must be right-handed, as an antiparticle coupling to the charged weak current. The neutrino/antineutrino pair are left/right handed, respectively. All products are fast enough that to lowest order
chiralities amount to the dominant helicities.
Go to the antimuon rest frame, so take its spin pointing up. Consider the limiting case where the two neutrinos are collinear, so they are both emitted back to back to the positron emitted at maximum possible momentum (hence energy). The net spin projection of the neutrino pair is then zero, so the spin projection sum will be the spin of the right-handed positron pointing in its direction of motion, so, then, its momentum will be preferentially in the direction of the antimuon's spin. (This will be a maximum of the corresponding spin 1/2 d-rotation matrix. All such helicity arguments pick a maximum along a spin axis, and rotations off it introduce diminution of the superposition component involved!)
Might consider my old notes of my teaching muon decay in the past. The incomparable booklet of L Okun, Leptons & Quarks, which, needless to say, you should be reading regularly, does all this in Sec. 3.3, Figures 3.2 and 3.3.
*
*(Geeky) Discussion of angular distribution.
From my notes or Okun you may glean the full angular distribution
so $(3-2\varepsilon -\cos\theta (1-2\varepsilon ))/2$ for $\varepsilon\equiv 2E_e/m_\mu$ the energy of the positron as a fraction of its maximum. You then see that, near this maximum, $\varepsilon \to 1$, the angular distribution is but $(1+\cos\theta)/2= |d^{1/2}_{1/2,1/2}(\theta)|^2$, "preferring" small angles and "abhorring" π !
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Why does a capacitor act as a frequency filter? What is it about a capacitor which allows it to filter frequencies?
I understand the construction of a high-pass RC filter, and the mathematics behind it, but I'm struggling to find an explanation of the physics behind the phenomenon.
In my mind I can picture the broad spectrum signal hitting the capacitor, but I feel like the "output" behaviour would be mush, not a controlled and predictable behaviour. I'm not a physicist, but I'd like to understand this problem better.
What is the physical behaviour which allows a capacitor to act as a high or low pass filter?
| A capacitor is an open circuit. Direct current can't flow through it because the plates of the capacitor don't contact.
However, when the current is alternating (or a signal), the electric field induced by one plate induces a current in the other plate.
That current is proportional to the capacitance and the time rate of change of the voltage.
I = C * dV/dT
Capacitors are sometimes used with DC power sources to remove high frequency noise - they don't conduct the low frequency DC because they are open.
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Why is the electrostatic force felt in straight lines? When two positive charges are kept close, they get repelled in the direction of a line joining both the charges. Why is it so?
Also, why is the repulsion in a straight path?
In both the cases, the potential energy of the charge which gets repelled decreases. What makes it repel in a straight line such that the line passes through both charges?
| One answer is that forces are determined by the gradient of -U, where U is the potential energy. Of course, that just raises the question of why that is so. I believe that the Principle of Least Action prescribes that particles move along the gradient, but that similarly leads to the question of why particles are constrained to follow that. I believe that if you look at a Hamiltonian, the sum over histories of the paths other than following the gradient results in destructive interference, but I'm far from clear on that.
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What is the equation for gravitational time dilation? I have been studying relativity for a while now, but I am still unsure what the equation is to determine a gravitational time dilation. I am more than aware of the speed time dilation equation, but I would like the gravitational one.
| In general, the time dilation $\gamma$, is given by $\frac{d\tau}{dt}=\frac{1}{\gamma}$ where $dt=dx^0$ is the time coordinate and $d\tau^2=-\frac{1}{c^2} ds^2=-\frac{1}{c^2}g_{\mu \nu} dx^{\mu} dx^{\nu}$ is the proper time on the clock that you are calculating time dilation for. This expression is very general. It works for any clock motion (at rest, moving inertially, accelerating, etc) and any spacetime (special relativity flat spacetime, black holes, cosmology, etc). It only requires that you use a coordinate system with a time coordinate $dx^0=dt$.
Now, gravitational time dilation, is specifically for a clock at rest. So that means $dx^1=dx^2=dx^3=0$. Substituting that into the above expression we get $d\tau^2 = -\frac{1}{c^2} g_{\mu \nu} dx^{\mu} dx^{\nu} = -\frac{1}{c^2} g_{00} dt^2$. So then $$\gamma_{grav}=\sqrt{-\frac{c^2}{g_{00}}}$$
Let's try a couple of examples:
In flat Minkowski spacetime $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$. So $g_{00}=-c^2$ and therefore $\gamma_{grav}=\sqrt{-\frac{c^2}{-c^2}}=1$ meaning that in flat spacetime there is no gravitational time dilation. This is what we expect.
In Schwarzschild coordinates: $$ds^2 = -\left(1-\frac{2GM}{c^2r} \right)c^2 dt^2 + \left(1-\frac{2GM}{c^2r} \right)^{-1} dr^2 + r^2 \left(d\theta^2 + \sin(\theta) d\phi^2 \right)$$ So $g_{00}=-\left(1-\frac{2GM}{c^2r} \right)c^2$ and therefore $$\gamma_{grav}=\left(1-\frac{2GM}{c^2r} \right)^{-1/2}$$ which you can confirm is the expected result for the Schwarzschild coordinates.
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Partial derivative of Dirac Lagrangian with respect to derivatives of fields Why is $\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = 0$, for the Dirac Lagrangian $\mathcal{L} = \bar{\psi}(i \gamma^\mu \partial_\mu - m)\psi$?
This comes up in deriving the Noether current for $\psi \rightarrow e^{i\alpha}\psi$ for example.
My confusion comes from the fact that we can write the following term in the Lagrangian $i\bar{\psi}\gamma^\mu\partial_\mu\psi = -i(\partial_\mu \bar{\psi})\gamma^\mu\psi$ by integrating by parts which makes it look like $\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = -i \gamma^\mu \psi$.
In fact, this is how we get the equations of motion for $\bar{\psi}$.
| *
*$\psi$ and $\bar \psi$ are thought as two independent variables in the Lagrangian.
*If you write a Lagrangian as $\mathcal{L}_1 =\bar\psi(...)\psi$, you should use it to calculate the Noether current or equation of motion. If you have the other one, $\mathcal{L}_2 =\psi(...)\bar\psi$, you have to perform the derivatives based on this one.
*The two results will be equivalent, they are Dirac dual to each other.
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The possible orbits of a body about the Sun in terms of their total mechanical energy If we assume that the Sun is at rest in an inertial reference frame, the total mechanical energy ( $E$ ) of the Sun and the orbiting body is constant and equal to the sum of the kinetic energy ( $\mathcal K$ ) and the gravitational potential energy ( $\mathcal U$ ).
Is there a mathematical-physical explanation, for students of an high school, with the specific steps because:
for a high speed object with $$\mathcal K>|\mathcal U|, \text { and } E>0$$ the orbit is unbounded and its trajectory is open or hyperbolic. When $$\mathcal K=|\mathcal U|, E = 0$$ the orbit is still unbounded but the trajectory is parabolic. For $$\mathcal K<|\mathcal U|,
E >0 $$ and the orbit is termed bounded with an elliptic trajectory. For $\mathcal K=0$ there is no orbit.
| The only way I know to show analytically that the trajectories are ellipses, parabolas, or hyperbolas involves solving a differential equation. In a comment, you explained that your students have only had some precalculus, so I don’t think you can demonstrate this to them, although you could certainly tell them that “it can be shown”.
If they understand that force determines acceleration, acceleration determines the change in velocity, and velocity determines the change in position, and if they know how to do some programming, they could write a computer program to numerically simulate trajectories. However, you might then get into problems with numerical error accumulation with the simplest algorithm. The trajectories may not be sufficiently accurate; for example, the elliptical orbits won’t close.
A very simple approach is just to talk about the energy equation,
$$E=\frac12mv^2-\frac{GMm}{r}=\text{const}$$
for a small mass $m$ moving in the field of a large mass $M$.
You can explain that when $E=0$, the small mass can just barely get to $r=\infty$ with zero velocity. And when $E<0$ it cannot get to $r=\infty$, because the kinetic term cannot be negative, so it must be in a bound orbit.
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Why does the $\mathbf{B}$-field of a cylindrical magnet have no $\phi$-component I have a cylindrical permanent magnet with uniform magnetization $\mathbf{M}=\mathbf{a_z}M$, length $L$ and Diameter $D$.
I'm wondering why the $\mathbf{B}$-field created by this uniform magnetization has no $\phi$-component, that is, the field lines don't "circulate" in the magnet.
The field only has a $r$-component and a $z$-component, so the field lines only point inward towards the center, outward away from the center and up or down along the $z$-axis. I believe it has something to do with
$$\displaystyle\oint \mathbf{B} \cdot d\mathbf{l}=\mu_0I$$
but it would be really helpful if someone could explain why this is the case.
| I tried explaining this before but I maybe wasn't so clear so here is my second attempt:
This is entirely do to symmetry, the magnet is cylindrical and thus viewed from above appears to be round and entirely symmetric.
Thus we have that the outcome would be exactly the same if rotated by a factor $\phi$
as the magnetisation is uniform (and thus equally dense for all r) and doesn't have a $\phi$ dependency (as it is given by $\vec{M} = \vec{a_z}M$) it is the same for any given $\phi$, it doesn't have a $\phi$ component and it doesn't depend on $\phi$ because the magnetic field doesn't change with a different $\phi$
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Gravitational potential energy sign Following is a small derivation just so I can explain my question. The gravitational potential energy is:
$$(*)U_g = -\frac{GMm}{r}$$
And:
$$ \Delta U =-GMm(\frac{1}{r_{final}} - \frac{1}{r_{initial}}) $$
If some mass $m$ is taken a height $h$ above the ground, we get:
$$ \Delta U =-GMm(\frac{1}{R+h} - \frac{1}{R}) = \frac{GMmh}{R(R+h)} $$ approximating $h\ll R$ :
$$ \Delta U = \frac{GMmh}{R^2} $$ and if we denote $g=\frac{GM}{R^2}$ we get the familiar $$ \Delta U = mgh$$
That indeed goes hand-in-hand with (*), since the object went further from the center of the earth and therefore gained PE.
Now to the question: Does that mean we should always express the PE to be "more negative" the closer we are to Earth? I see some texts that present PE that gets bigger when you get closer to the Earth and that quite confuses me.
| Yes, potential energy decreases in the direction of the force. So potential energy decreases as you move closer to the Earth. Any texts that say potential energy increases closer to the Earth shouldn't be taken seriously. The texts could be talking about the magnitude of the potential energy, but that really isn't a useful concept at all.
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Why doesn't the state of matter oscillates between solid , liquid and gaseous phase? I was reading about the London dispersion forces and it is written that it arises due to an asymmetric distribution of charges in an atom at an instant which gives rise to a temporal dipole and this temporal dipole then induces a dipole in nearer atoms and thus they experience the intermolecular forces.
This site says that $Cl_2$ and $Br_2$ can exist as solid because of this force at lower temperature and can exist in liquid form too.
So , my question is that since this force is due to temporal dipole moment then why doesn't the state of the compound or molecule (held together by dispersion forces) keep on oscillating between solid , liquid and gaseous phase ?
One thing which I think for the above question is that we need a repulsive force to separate two closer molecules and this is the reason why we don't notice oscillations . Am I right ?
Okay if my intuitions are correct then again there is a question.
If the forces are temporal , then surely the strengths of the compound in solid phase must be changing i.e. if it is harder at an instant then after sometime it must be softer and can be easily converted into gas with a negligible effort or with a single blow of hand or even with a beam of photons.
But I don't think we notice such strange things in our daily life.
So why don't we notice such transformations ?
| The dipole interactions may oscillate (at very high frequencies) but they average out over time (and billions of atoms) into a steady force. The ability of this force to hold atoms together against thermal agitation generally depends on the temperature. A change of state is associated with a significant amount of energy (the heat of fusion or vaporization) and time. You wont see it occurring in a random manner.
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Spin conservation in circuits with a spin filter If we have a circuit made of a battery and a resistor the number of electrons with spin up will be equal to the electrons with spin down. If we put the resistor in series with a piece of ferromagnetic material which is magnetize either up or down, let's say it is magnetized up will it produce a spin polarized current? (spin down electrons wouldn't now flow in the circuit). And if current becomes spin polarized happens wouldn't it increase electric resistance?
| Since you want to influence the spin of the electrons by a magnetic field, why don't we talk about the magnetic dipole of the electrons instead of their spin. At least both are connected by their alignment; under the influence of a magnetic field they get aligned with this field.
The next effect is the acting Lorentz force, in your case better called the Hall effect. The difference lies in the movement of the conductor with its electrons (Lorentz force) against the displacement of the electrons within an immovable conductor. However, in regions behind the magnetic field the forced alignment is lost immediately.
I can only think of one exception. The conductor behind the magnetic conductor is influenced by the incoming spin-polarised electrons to such an extent that its electrons align themselves with the incoming electrons and the magnetic field generated increases to such an extent that self-inductance (self-maintaining of the field) occurs. We obtain a magnetised material. Then the electrical resistance actually increases.
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Gravitational Field Strength of a point in between Two Planets In my textbook, under the topic of gravitation, it states that if the centres of 2 planets, each of mass $M$ and separated by a distance $r$ and you have a point halfway between the centres of the planets, the gravitational field strength at that point is $0$.
I don't fully understand why that is. Is it because the point feels an equal force in each direction so the resultant gravitational force is $0$, resulting in $0$ gravitational field strength at that point?
Surely however gravitational field strength is a measure of how many Newtons of gravitational force a body feels per kg. In this case, shouldn't it be equal to
$$2\times \frac{GM}{(0.5r)^2}=\frac{8GM}{r^2}$$
as it feels
$\frac{GM}{(0.5r)^2}$ Newtons of force from 2 planets?
| You are correct. The two forces are in opposite directions and cancel each other.
Force is a vector quantity. When adding vectors the directions are as important as the magnitudes.
Perhaps you are confusing gravitational field strength $g=GM/r^2$ and gravitational potential $V=-GM/r$. The former is gravitational force per unit mass, so like force it is a vector. When adding field strength you use vector addition (eg the parallelogram rule). The latter is the work done in moving a unit mass from infinity to a point at distance $r$ from the mass $M$, so like work it is a scalar. When adding potentials due to several masses you do so algebraically, regardless of the direction of the mass which is creating the potential.
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What is the equivalent to $\Box A^\alpha =- \mu_0 J^\alpha$ using differential forms? The set of equations $$\Box A^\alpha = -\mu_0 J^\alpha$$
can be found in section 12.3.5 of Griffiths's book. From what I understand, the real-valued functions on both side of the equations are the coefficients of some $1$-forms with respect to a chart. Thus, I am wondering how the equivalent index-free equation involving differential forms looks like.
| Recall the inhomogeneous Maxwell equation in natural units:
$\newcommand{\dif}{\mathrm{d}}$
\begin{align}
\mathrm{d}*F={}*J\in\Omega^3
\end{align}
Since $**=-1$ on $\displaystyle{\Omega^1}$, the equation is equivalent to
\begin{align}
{}*\mathrm{d}*F=-J\in\Omega^1.
\end{align}
Now consider some one-form $A$ such that $\dif A=F$, then
\begin{align}\tag{1}
{}*\dif*F={}*\dif*\dif A=-J\in\Omega^1.
\end{align}
In components, this should turn out to be equation $(12.134)$ in Griffiths's book$^1$ and as Griffiths explains in section $12.3.5$, $(12.134)$ can be reduced to $(12.137)$ - the equation in the title of my question - by exploiting the gauge invariance.
To put this into a nutshell, $(1)$ is the equation I was searching for.
$^1$ The PDF is freely available; This answer should be helpful if someone wants to check this
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The depicted shape of subatomic particles I physics books and such, I understand why they show atoms as spheres because they have the electron cloud. But why are protons, neutrons, and electrons spheres? I guess people say that because of calculations in nuclear physics, they turn out to be a sphere, but why naturally. Also, are quarks and neutrinos also spheres it are they string-like things?
| What about the colours in the pictures?
As the poet says
"Protons are Red,
Electrons are Blue,
Neutrons are Grey,
And so are You"
Colours and shapes are just artist's illustrations. Pictures don't mean anything. If they used a dot it would be too small to see it; if they used a cube it would be misleading so what else can you use?
Elementary particles don't have shapes or colours. They just have scattering cross sections.
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Would an astronaut in this spacecraft feel weightless?
A spacecraft is placed in orbit around Saturn so that it is
Saturn-stationary (the Saturn equivalent of geostationary – the
spacecraft is always over the same point on Saturn’s surface on the
equator).
Information the question provided:
mass of saturn = $5.68\times 10^{26} kg$
mass of spacecraft = $2.0 \times 10^{3}kg$
period of rotation of Saturn = $10$ hours $15$ minutes
Information I calculated:
radius of orbit = $1.1 × 10^8m$
Now part d)
Would an astronaut in this spacecraft feel weightless? Explain your answer.
I am unsure how to answer this question. I guess I first need to define "weightlessness"? From what I know the sensation of weightlessness is the absence of normal force? Or its the sensation that you feel that you weight less than your normal weight? I also calculated $g = 3.13m/s^{2}$ if thats any useful?
| We all know a fundamental principle from physics which says, " freely falling body are weightless"
That is, their Apparent weight is zero. So what is a freely falling body.....?
Any object on or around a planet (like earth or Saturn), whose acceleration is equal to acceleration due to gravity at that point (in magnitude as well as direction BOTH), is a freely falling body.
So in the given example, first you have to calculate the acceleration due to gravity at that point where the body is located. Then you have to find the centripetal acceleration of its circular motion. The direction of both of these acceleration is a towards the centre of the planet. If their magnitudes are equal, then that spacecraft is a freely falling body and therefore it will be in a state of weightlessness.
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Can we prove the Earth rotation with a disk mounted in its center on a frictionless axis? Can somebody prove the rotation of Earth if it places an uniform disk with a hole in its the center on an axis and orient it paralelly to Sun ecliptical disk?Just assure himself that the disk on the axis is in equilibrium and the friction between the disk and axis at the center hole of the disk is negligable. The disk should preserve its absolute orientation with time while the Earth would change its orientation due rotation and this should be visible after several minutes or hours by looking a labeled part of the disk regarding the floor?
| What you describe is reminiscent of a 1913 experiment by Arthur Compton. This setup is referred to as a Compton ring
A circular tube is filled with water (with suspended particles in the water to allow tracking of motion of the water).
The initial position of the tube is perpendicular to the local level surface. The water in the tube is allowed to come to complete rest. This rest state is a state of co-rotating with the Earth rotation. Then the tube is flipped 180 degrees. After that flip the water is seen to have been set in motion, the magnitude of the velocity can be observed with a microscope.
This setup will show the strongest effect at the Equator, and a smaller effect on higher latitudes.
So, contrary to assertions in comments and answer to this question using gyroscopic effect is not the only way to demonstrate the Earth's rotation. However, if a disk is used that is initially co-rotating with the Earth then the setup does need to execute a flip in order to obtain any data
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Maxwell equations in non-transparent media There are many examples of media non-transparent for the light of visible spectrum.
What does it mean in terms of the Maxwell equations in such media? More precisely what does it mean for dielectric permittivity and magnetic permeability (which depend on the medium and on the frequency of the light)?
| Non-transparent media is a vague term. Let me first note that when we talk about transparency, we have in mind propagation of electromagnetic waves. Non-transparent means that the waves do not propagate in this medium. Another important point: the medium does not enter directly into the Maxwell equations. However, these equations are not complete - they must be completed by the material equations, which express how the currents/polarization/magnetization/charge in the media respond to the electromagnetic field. Dielectric and magnetic permittivities are the simplest type of such material equations.
*
*One way a media may be non-transparent is because it is reflecting, as an ideal metal. One says sometimes that the dielectric permittivity of such a media is infinite, which means that the electric field inside the media is zero.
*Another option is a non-ideal metal, i.e., a metal with finite resistivity. In this case electromagnetic waves do penetrate the media, but they decay via generating currents and the Joule heat. In Fourier domain one can then formally include the conductivity of a metal as an imaginary component of the dielectric permittivity.
*A media can be absorbing but non-conducting. One can again mathematically model this situation via complex parts of the permittivities, which mean that the wave vector has a complex part, i.e. the wave decays.
*Finally, a media can be non-transparent, because it is very inhomogeneous and the light is scattered in all possible directions. This is usually modeled via the permittivities randomly varying in space.
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Confution on UV cut-off in the calculation of effective action and Beta function I am reading David Tong's gauge theory notes and meet some difficulties.
In section 2.4.2, he uses background field to calculate effective action $S_{eff}$ and Beta function. Simply like follows:
Writting gauge field $A_{\mu}$ as $A_{\mu}=\bar{A}_{\mu}+\delta A_{\mu}$ for $\bar{A}_{\mu}$ being a fixed field, $\delta A_{\mu}$ being fluctuation. Also we introduce Faddeev-Popov ghost fields $c$ and $c^{\dagger}$. The path integral of gauge field $A_{\mu}$ goes like
\begin{equation}
e^{-S_{eff}}=Z=\int \mathcal{D}A \mathcal{D}c \mathcal{D}c^{\dagger}e^{S(A,c,c^{\dagger})}=(\det[\Delta_{gauge}])^{-1/2} \det[\Delta_{ghost}]e^{(-1/2g^2) S_{YM}(\bar{A})}
\end{equation}
where
\begin{equation}
S_{YM}(A)=\int d^4x tr(F^{\mu\nu}F_{\mu \nu}) .
\end{equation}
So the effective action $S_{eff}$ is then
\begin{equation}
S_{eff}=\frac{1}{2g^2}S_{YM}(\bar{A})+\frac{1}{2}Tr\log\Delta_{gauge}-Tr\log\Delta_{ghost}
\end{equation}
See eq(2.63) in the notes.
Now he gets the contribution of $-Tr\log\Delta_{ghost}$ (see page 71 in the notes) as
\begin{equation}
-Tr\log\Delta_{ghost}=constant \times \int \frac{d^4k}{(2\pi)^4}tr(\bar{A}_{\mu}(k)\bar{A}_{\nu}(-k))(k^{\mu}k^{\nu}-k^2\delta^{\mu\nu})\log\bigg(\frac{\Lambda^2}{k^2}\bigg).
\end{equation}
Here is what I don't understand: how does this $\log\bigg(\frac{\Lambda^2}{k^2}\bigg)$ appear in our integral in terms of $\log$? Is it a regularization? If it is, then the integrand cannot turn back to $tr(\bar{A}_{\mu}(k)\bar{A}_{\nu}(-k))(k^{\mu}k^{\nu}-k^2\delta^{\mu\nu})$ under limit $\Lambda=\infty$.
| Tong is using a cutoff regularization in this calculation, the cutoff being this $\Lambda$. Specifically this factor appears because he has just completed the loop momenta integral (integral over the momenta $p$). This integral is divergent without a regularization scheme, so of course taking $\Lambda\rightarrow\infty$ (removing the cutoff) will cause the what you have written to diverge.
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Might the Kaluza-Klein scalar provide a solution to the dark puzzles? Kaluza-Klein theories of a five-dimensional spacetime yield not only the equations of general relativity and electromagnetism, but also a scalar field. This scalar field, sometimes quantised as the radion or dilaton, is thought not to exist.
Given today's twin puzzles of Universal expansion, dubbed dark energy, and gravitational anomalies on the galactic scale, dubbed dark matter, (how) can we be sure that the Kaluza-Klein scalar is not involved in either of them?
| Yes, Kaluza-Klein excitations might be the dark matter. See e.g. this search on arxiv.org for some papers making the connection. In particular the earliest references there (on the second page) might be most useful for you.
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Why is this paper stack not flat? So we have lots of paper for photocopy. They come in a bundle. What I have noticed is that There are sine curves or maybe cosine when I saw them from the side. That's Look like:
Is there any explanation for these curves?
| This is an artifact of the so-called conversion process, by which a continuous ribbon of paper hundreds or thousands of feet long and four feet wide on a huge roll is slit to width and then sheared to length and the cut sheets then stacked and wrapped into packages for shipment.
Since the initial spooling process by which the master roll is made occurs when the paper is still hot from the drying process and occurs under significant tension, the paper takes a set on the roll and the sheets cut from it exhibit "curl" after being cut.
That curl becomes cyclic when the master roll is left to rest on a concrete floor and deforms slightly under its own weight as it cools. The net effect is small for sheets that originate in the outermost layers of the roll (which is typically 4 feet in diameter at full size) and becomes progressively of shorter period by the time you get towards the end of the roll (which is wrapped onto a core that is only ~4 inches in diameter).
If the package contains a mix of sheets cut from a variety of rolls, then some of the sheets will have cyclic curl and others will be relatively flat in comparison.
Inkjet printers re-humidify the paper, which greatly magnifies the curl- in fact, paper sheets which start out almost perfectly flat will curl like crazy because the water-based ink is applied to only one side of the sheet, which then tends to roll itself up into a tube when the residual curl coincides with the wetter side of the sheet. In injket printing circles, this is known as the "diploma effect".
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What does intrinsic and extrinsic contribution to Hall effect mean? In the context of transportation theory (electron and thermal conductivities), what is the physical meaning of extrinsic and intrinsic contributions to the Hall effect (i.e. transverse conductivity)?
For example, in the article "Anomalous Hall effect in the Dirac electron system with a split term", they define the Hall conductivity as
$$
\sigma_{xy} = \sum_n \frac{e^2}{\hbar}\int \frac{d\vec{k}}{(2\pi)^d} \Omega_{k_x,k_y}^n (\vec{k}) f(E(\vec{k}))
$$
where $\Omega$ is Berry curvature, and $f(E)$ is Fermi distribution function. They call it intrinsic Hall conductivity. What exactly does "intrinsic" mean here? What could be extrinsic?
Just another confusion, I have seen usually the formula for Hall conductivity does not contain $f(E)$ term in it (unlike in the above formula), does $f(E)$ has to do something with "intrinsic"?
| In this context, "intrinsic" means that the Hall conductivity comes from the Berry curvature. That is, it's a contribution intrinsic to the band structure. Disorder can produce "extrinsic" contributions to the anomalous Hall effect through so-called side-jump and skew-scattering mechanisms. If you're interested, you can see this answer of mine for some references.
The Fermi distribution function is there to account for temperature effects, and does not have anything to do with intrinsic vs. extrinsic. At $T=0$ it can be set to unity below the Fermi energy (as is stated in the paper you linked), and zero above it. At finite $T$ quantization in a filled band is no longer exact, which the $f(E)$ factor accounts for.
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Why are we allowed to make algebraic calculation on units Is there math proof that we can cancel out units in Physics? For example:
$\require{cancel}distance = \frac{meters}{\cancel{second}} * \cancel{second}$.
So we see that seconds cancel out and we left with meters which is correct but how is it possible if it is not actually division (meters per second) it is only our interpretation of speed. Why are we allowed to make algebraic calculations on units? Also, we can use dimensional analysis to check that we do everything correctly but why? Why does it always work?
| The question seems trivial, but it isn't at all!
Some strange formal properties of algebraic calculations on units
(e.g. the elusive rad unit which appears or disappears)
should suggest that "something is wrong" instead of the
more usual W.Allen's "Whatever works".
The problem concerns the logic (syntax and semantics) of the ''mathematical'' language of physical quantities.
What does mean "the product of a force times a distance" or "the division of a distance by a time"?
And what is the rigorous logic of the dimensional algebra?
Units of measure are samples of quantities, but the calculations made with symbols of units are a particular form of dimensional calculus.
This (little-shared) statement is not the result of an improvisation:
here I am forced to be self-referential by quoting the document:
http://pangloss.ilbello.com/Fisica/Metrologia/grf.pdf
Unfortunately the text is not in English language.
I emphasize S.3.7 ''Algebra of units of measure'' and
the last sentences:
The usefulness of dimensional monomials consists in making the calculation of the relationships between the various units of measurement of physical quantities algebrically intuitive, conforming it to the ordinary rules of algebra (as illustrated in the examples).
The ordinary calculations with units of measure are based on dimensional relations between classes of quantities, they are not univocal relations between units of measure. We must not deceive ourselves to be able to obtain in this way the exact conventional name or even the definition of the calculated unit!
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Why doesn't a mercury thermometer follow the rules of volume dilatation? let's consider a classic mercury thermometer.
I do not understand why it does not behave like a "normal" thermometer which exploits volume dilatation. In a normal thermometer, I'd say that the mercury length would be proportional to its temperature.
Therefore, I should be able to measure, for instance, 37 of body temperature, also starting with the thermometer at 38: there would be a contraction, but the measure would be correct! Why does this not happen? And why if I measure for instance, 38, and I try to cool the thermometer by putting it inside cold water, it does not become cooler? Why should I cool it by shaking it?
It seems a very not ideal thermometer... but what are the causes of these non-idealities?
| That's because it's a maximum thermometer, which works by pushing the liquid past a restriction in the tube, preventing the liquid from returning into the reservoir upon cooling. If you look closely, you might see the separation in the liquid column between the restriction and the reservoir, which may be bigger or smaller depending on the model.
You don't cool the thermometer down by shaking it - you push the liquid back to the reservoir with an additional force coming from accelerated motion.
As a side note, thermometers which have a secondary reservoir (such as in your picture) are often calibrated to provide accurate reading at room temperature. Reading them while they are still hot gives you a falsely high value.
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Is the torque on a gyroscope a conventional virtual force perpendicular to rotation? So I am having some difficulty understanding gyroscopic precession. I understand that mathematically by convention torque is perpendicular to the force and so is angular momentum but surely that force is a true force acting outwards as this is what occurs in gyroscopic precession. My question is is this torque a conventional virtual force perpendicular to rotation?
| I cannot add to the excellent responses earlier by others. Here are a few comments. Gyroscopic motion is hardly intuitive to anyone. My problem with basics physics discussions is they just discuss the effect of torque on the spinning gyro but do not really provide a good intuitive explanation; also, some of these discussions fail to mention that their simple evaluation of precession assumes the gyro is initially spinning rapidly.
More advanced treatments of the motion use a Lagrangian approach, but that does not help me really visualize what is going on either.
The earlier answers provided by others are as good an intuitive explanation for the motion as I have seen; I will save these responses in my files:)
There are many variations of such motion that have surprising answers that you can find on the web, such as the practical joke played on a hotel porter by the physicist R. W. Wood.
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Meson as hadron and boson In wikipedia page about hadrons the following image appears:
I can understand why the intersection between hadrons and fermions are baryons, as a way to say a baryon is a kind of hadron composed of several quark fermions.
However, what is the meaning of the intersection between hadron and bosons labeled in the picture as mesons? If I understand correctly, a meson consists of one quark and one antiquark, nothing related to any boson.
| Since a meson is composed of two spin 1/2 particles, its total spin must be an integer, which makes it a boson.
| {
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How exactly do quantum numbers increase in relation to energy levels in more than one dimension? When you have increasing energy levels in 2 or 3 dimensions how does the values of quantum numbers $n$ increase for each dimension?
For example if you have ground state $E_1$ then you have for $(n_x,n_y,n_z)$ is $(1,1,1)$ in 3 dimensions.
So does this mean for energy $E_2$ you have $(2,2,2)$ ?
I've only learnt it in 1 dimension so i don't understand how it works in higher dimensions?
| For a particle in a box (also known as a particle in an infinite potential well) in $d$ dimensions, the Hamiltonian inside the box is given by $$\hat{H}=\frac{1}{2m}\sum_{i=1}^d{p}^2_i$$where $p_i$ is the momentum operator in the $i^{\text{th}}$ direction. As you already know (I infer so from your question) that using the boundary conditions, we arrive at the conclusion that $p_i$ must be of the form $$p_i=\frac{n_i\pi}{L_i}$$where $L_i$ is the length of the box in the $i^{\text{th}}$ direction and $n_i\in\mathbb{N}$. Thus, the energy of an eigenstate is given by $$E=\frac{\pi^2}{2m}\sum_{i=1}^d\frac{n_i^2}{L_i^2}$$.
Now, in the case of one--dimensional box, $E$ only depends on $n_x$ and is monotonically increasing with $n_x$. This means that the spectrum of a one--dimensional particle in a box is non--degenerate, and the value of the energy depends only on one quantum number, namely, $n_{x}$. One can simply write that $E=E_{n_x}$
However, in the case of a multidimensional box, $E$ depends on the multiple quantum numbers, namely, $\{n_i\}$ where $i>1$. And, we have to write $E=E_{\{n_i\}}$ to signify that $E$ depends on multiple quantum numbers.
Moreoever, multiple different sets of values of $n_i$ can produce the same value of energy. For example, let's consider the case where $L_i=L$ and $i=3$. All the following configurations of $n_i$ produce the same value of energy:$$n_x=2,n_y=1,n_z=1$$ $$n_x=1,n_y=2,n_z=1$$ $$n_x=1,n_y=1,n_z=2$$
Thus, we see that the spectrum of a particle in a box can be degenerate in a higher dimension.
So, to answer your question, the first excited energy state in a three-dimensional particle in a box (with equal sides) can correspond to all of the above configurations of quantum numbers.
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Why does particle measurement cause quantum wavefunctions to collapse When we attempt to measure a certain property of a particle, how and why does its wave function collapse? I've tried to find answers on my own, but they've been far too complicated for me to comprehend. Would appreciate any answer with limited complex jargon, and more simplistic explanation, if possible.
| Not only is wave function “collapse” a misleading term, it is also not an intrinsic part of quantum mechanics. There are interpretations of quantum mechanics, such as the many-worlds interpretation, in which a wave function never collapses - it only appears to, due to our limited knowledge. If you loose one sock from a pair you do not describe this as “sock collapse” - the other sock is still out there somewhere, you just don’t know where.
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Mechanical Advantage of Static Double Pulleys For a set of two double pulleys, the force acting on each should be equal to the tension in the rope times the number of ropes connected to the individual pulley if I understand correctly. In this pulley system on mcmaster, it looks as though the force acting on the upper pulley should be 5T (with T being tension) and the force acting on the lower 4T. Can anyone explain why the specs give an effort (which should be the tension) of 85 lbf to give a lifting force of 500 lbf. Why is it not 340 lbf (85*4).
Is my understanding of static pulley systems flawed?
| I am pretty sure you are correct.
I tried to find the mechanical advantage by using a FBD and got it to be 4:1. But just to be sure I poked around the internet a bit and found this great site. The MA is given in section 6-8.3 .
| {
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Can coldness be converted to heat energy? We know that the heat can be converted into heat energy with the help of thermoelectric generators, but why can't we generate energy from coldness?
Like the temperature of the universe in 1 K, can this be used in the near future to be used as an energy resource for probes or satellites?
Here is the link to the article that made me think about this. Somewhere in the middle it is written that scientists can harness the cold energy using some active input method.
I think this article is poorly written.
| If you want to transfer heat from a cold environment to a warm environment, you need a heat pump but then part of the provided heat will come from the work done by the heat pump, so it comes from the fuel that drives the heat pump. To get all the heat from only the cold environment, you need to use a heat pump that exploits temperature differences in the cold environment. This means that a pure transfer of heat from only a cold environment to the warm environment requires at least two different cold environments at two different temperatures.
Suppose then that we have 3 heath baths at temperatures of $T_1<T_2<T_3$. If due to some process the heat added to heath bath $i$ by the other heath baths is $q_i$, then by the First Law of thermodynamics (conservation of energy), we have:
$$\sum_{i=1}^3 q_i = 0$$
If this is a reversible process, the total entropy does not change, we then have:
$$\sum_{i=1}^3 \frac{q_i}{T_i} = 0$$
We can then compute what fraction of the heat extracted from heath bath 2 ends up in heath bath 3. This is then ratio $-\frac{q_3}{q_2}$, solving for this using the above two equations yields:
$$-\frac{q_3}{q_2} = \frac{T_3}{T_2}\frac{T_2 - T_1}{T_3 - T_1} $$
So, we see that heat can flow from cold to warm, without having to rely on external work or having to rely on some other system at a temperature that is higher than the system where the heat is flowing to.
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Number and Types of States of Matter I wanted to know if there were more than 5 states of matter (man-made or natural) and so I searched it up. Other than solid, liquid, gas, plasma, and Bose-Einstein state, these were varying results from 7 to 15 different states. I want to know what the real answer is. How many states are there and what are they? Also, what is a condensate?
| As long as we are lacking any fundamental physical definition to a "states of matter", we will have arbitrary amount of these states of matter. This is well discussed in this question;
What determines a state of matter?
Condensate is practically a matter, where molecules holds their relative positions to each other in same order atleast over short period of time. Ie, kinetic gas theory doesn't apply.
| {
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Is Light our limit? Suppose something existed faster than light will we be able to perceive it?
And even if we encounter it wouldn't seem to travel with speed of light?
| Superluminal objects have a big problem with the relativistic mass equation:
$$
m = {m_0 \over {\sqrt{1-{v^2\over{c^2}}}}}
$$
If you set $v=2c$ in the equation, you get
$$
m = {m_0\over{\sqrt{-3}}} \approx 0.577im_0
$$
So you get an imaginary mass.
Goodness knows what that might mean... I'll have imaginary two kilos of potatoes please.
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Why does ponytail-style hair oscillate horizontally, but not vertically when jogging? Many people with long hair tie their hair to ponytail-style:
Closely observing the movement of their hair when they are running, I have noticed that the ponytail oscillates only horizontally, that is, in "left-right direction". Never I have seen movement in vertical "up-down" direction or the third direction (away-and-back from the jogger's back). Why is the horizontal direction the only oscillation?
| Very simple: The frequency for the up and down movement is twice that of the left and right swing.
Plus the up and down excitation (acceleration) is not symmetric, a good part of it consists of overtones of even higher frequency. Actually the hair and attached head are in free fall between steps.
So to get to the same amplitude, the hair would need to move twice as fast in up and down direction, and reach even higher accelerations.
That's only possible for a very short bun (see comment about the British marathon runner above), but not with a ponytail. The damping by air and internal friction is much too large.
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Can spacetime be curved even in absence of any source? Einstein's equation in absense of any source (i.e., $T_{ab}=0$) $$R_{ab}-\frac{1}{2}g_{ab}R=0$$ has the solution $$R_{ab}=0.$$
But I think $R_{ab}=0$ does not imply that all components of the Riemann-Christoffel curvature tensor $R^c_{dab}$ be zero (or does it?). From this can I conclude that spacetime can be curved even in absence of any source?
| That's right. But it doesn't mean that the curvature is from nowhere. The Field equation describes the curvature (locally) at a point only from $T_{\mu \nu}$ at the same point (Since it's all built in a differential manifold and tangent spaces at each points aren't related to each other). If $T_{\mu \nu}$ is zero at a point, then you end up deriving a vacuum solution.
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Time dependence of operators In Griffiths's Introduction to Quantum Mechanics, while studying the time evolution of the expectation value of position, the author wrote:
$$\langle x\rangle=\int_{-\infty}^{+\infty}x|\Psi(x,t)|^2\,dx.$$
So
$$\frac{d\langle x\rangle}{dt}=\int x\frac{\partial}{\partial t}|\Psi(x,t)|^2\,dx.$$
Did he just assume that $x$ has no time dependence? And why?
|
Did he just assumed that x has no time dependence? And why?
Yes. The outcome of an integral of the form
$$\int_{\mathbb{R}} f(x,t) \, \text{d}x \tag{1}$$
is a function of time $t$; that is, a function of one real variable (or, loosely speaking, the integral will evaluate to a quantity that will not depend on $x$, only on $t$). Thus, upon differentiating $(1)$, one would get:
$$\frac{\text{d}}{\text{d}t} \int_{\mathbb{R}} f(x,t) \, \text{d}x = \int \frac{\partial f}{\partial t}(x,t) \, \text{d}x$$
as dictated by Leibniz Integral Theorem (do note that I've assumed some weak assumptions on the behaviour of $f$, but it is not of incredible interest here). A trivial application of this in
$$\langle x \rangle := \int_{\mathbb{R}} x \, |{\Psi(x,t)}|^2 \, \text{d}x$$
yields the desirable result.
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Increase in temperature in a copper wire I am creating a simulation of a magnet moving relative to a copper coil generating emf in it. So far I am working with emf generated, current, resistance, resistivity, turns, coil length, and the cross-sectional area of the coil. The next step is to add the increase in temperature in the coil as a result of flow of electrons. How should I go about including that. What additional quantities do I need, including assumptions, to model that in
| With those quantities, you can only calculate heat generation. In steady state (when the temperature is constant), it will be equal to heat loss.
There are three mechanisms for heat loss: conduction, convection, and radiation. As long as the temperature is not too much higher than room temperature, the heat loss would be approximately proportional to the temperature difference. You can model it with an effective heat conductance. Time dependence depends on the thermal mass (heat capacity).
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Feasibility of the Aharonov-Bohm effect The Aharonov-Bohm effect is often presented via a region in space where the magnetic field $\textbf B=0$, but the vector potential $\textbf A \neq 0$. Usually, this is motivated via an infinitely long solenoid, which is said to have a vanishing magnetic field outside the coil.
However, if the solenoid becomes infinitely long, it appears to be more and more like an infinitely long wire, which does have a magnetic field surrounding it. So my question is: How can we really prepare a region where $\textbf B=0$, but $\textbf A \neq 0$, considering that an infinitely long solenoid coil looks like a wire?
| Experimentally, you're never going to get field that's exactly zero. However, it would help to have 2 coaxial solenoids, one wound right on top of the other. Current flows up through one and back down through the other. The solenoid-like field is the same for both, but the wire-like field is opposite and cancels out.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does radiation (heat) take away momentum? In another post, I was taught that when we are moving (running, for example), radiation (in the form of heat, both from our muscles and friction with the ground) takes away from our momentum. That makes perfect sense and I understand.
I was also taught that heat/radiation is directional.
My question is: since radiation is directional, is that why it takes away from momentum (which is also directional)? Is it correct to say that? In other words, does radiation take away from momentum because they are both directional concepts?
Also, does radiation, since it is directional, dissipate in the same direction of momentum, in the opposite direction, or no particular direction?
I am not an expert, hoping just for simple clarification, if possible.
| Suppose that we have a lone body moving freely across space, his momentum then is conserved, it has to stay the same through time. This derives directly from the laws of dynamic. End of the story. But if the body emits radiation then some of the momentum of the body can be stolen by the photons that it emits, in fact photons carry momentum. But as long as the radiation is emitted in the same way in every direction, the momentum has to be conserved. In fact if the radiation is emitted in the same way in every direction any momentum given in one direction will be compensated by the photon emitted in the opposite direction. If the radiation is not emitted equally then you can have a sort of jet effect and the momentum of the body could indeed vary.
It can be tempting to state: since the body emits photons and the photons carry momentum, then the body must lose some momentum, because it gave it to the photons. Unfortunately this intuitive line of reasoning is wrong: we have to remember that momentum is not a scalar quantity but a vectorial one, so two particle with momentum equal in module but opposite in direction amount to a total of zero momentum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why current rating of a fuse wire does not depend on its length? If legth is increased, resistance is increased and heat produced is directly proportional to the resistance. So why current rating is still independent of its length? Please give answer assuming that there is no heat loss
| A wire will melt if it exceeds a certain temperature. If resistive heating is happening in the wire, its temperature will increase until the rate of heat loss (via conduction to the air, largely) balances the rate of heat generation, or until it melts, whichever comes first. The rate of heat loss is proportional to the difference in temperature between the air and the wire, and to the surface area of the wire.
Now, if we double the length of the wire, the surface area of the wire is also proportional to the length, and the rate of heat transfer is proportional to the surface area. This means that a wire that is twice as long will generate heat at twice the rate. But at a given temperature of the wire, the rate at which the heat is lost will still balance out the rate at which heat is generated, since both numbers are twice as large when we double the length of the wire. The net effect is that the equilibrium temperature will be the same.
(All of this assumes that the wire is not coiled up or enclosed, which would reduce the rate of cooling and lead to a higher equilibrium temperature.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/595973",
"timestamp": "2023-03-29T00:00:00",
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How do we know neutrons and electrons are distinct particles on the same scale as protons? I'm aware my question may not even make sense when considering quantum physics, but please excuse my ignorance. We are taught in school that there are basically 3 particles that make up atoms (protons, neutrons, electrons) and that broadly for each 1 proton there will be 1 neutron and 1 electron. My question is how do we know that neutrons and electrons are not much smaller particles that just happen to lump together in predictable quantities due to their interactions with protons?
Let's say for instance that for each 1 proton you have 1000 neutrons and 1000 electrons. In this case, each neutron has 1/1000th the mass of a proton and each electron has 1/1000th the charge of a proton. It would seem to me that all the math still works just fine and our observations would still be the same.
We are also taught that you should think of an electron as something like a cloud rather than 1 distinct particle which is very counter-intuitive. In contrast, if 1 electron is really 1000 or a million, it is not just intuitive, but obvious that they would form as kind of cloud around the nucleus of an atom.
| If the electron were made up of smaller particles then we would expect to see some evidence of these smaller particles in particle collisions observed at CERN and other particle collider. We have never seen any such evidence. So as far as we know the electron is an elementary particle.
In the case of the proton and the neutron we have seen smaller particles that make up protons and neutrons. These particles are called quarks and each proton and neutron contains three quarks. Individual quarks are very difficult to observe because they are bound together so strongly, but we definitely know that they exist. So we know that protons and neutrons are not elementary particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Isn't $\epsilon_{ij}$ an isotropic, rank-2 tensor? Definition of isotropic tensor: components are unchanged after rotation: $T_{ij}\rightarrow T_{ij}' \equiv R_{ia}R_{jb}T_{ij} = T_{ij}$
MathWorld says there is only one rank-2 isotropic tensor, $\delta_{ij}$.
But with
$$\epsilon_{ij}=\left(\begin{matrix}0&1\\-1&0\end{matrix}\right)$$
there is no change either:
$$\epsilon_{ij}\rightarrow\epsilon_{ij}'=R_{ia}R_{jb}\epsilon_{ab}=\epsilon_{ij}$$
So it seems to me that $\epsilon_{ij}$ is also a rank-2 isotropic tensor, in addition to $\delta_{ij}$.
What am I getting wrong?
Notes:
*
*$R_{ij}=\left(\begin{matrix}\cos a&-\sin a\\ \sin a&\cos a\end{matrix}\right)$
*I asked at math.stackexchange, but got no answer. Maybe this is more Phys Math Met as in Boas, which I was reading when this question came up.
| This is only true for 2D space not in general
3D space:
$$S=\left[ \begin {array}{ccc} \cos \left( a \right) &-\sin \left( a
\right) &0\\ \sin \left( a \right) &\cos \left( a
\right) &0\\ 0&0&1\end {array} \right]
$$
and
$$\epsilon=\left[ \begin {array}{ccc} 0&-1&1\\ 1&0&-1
\\ -1&1&0\end {array} \right]
$$
$\Rightarrow$
$$\epsilon'=S\,\epsilon\,S^T=\left[ \begin {array}{ccc} 0&-1&\cos \left( a \right) +\sin \left( a
\right) \\ 1&0&\sin \left( a \right) -\cos \left( a
\right) \\ -\cos \left( a \right) -\sin \left( a
\right) &-\sin \left( a \right) +\cos \left( a \right) &0\end {array}
\right]
$$
thus only if $~a=\pi/2$ you get $\epsilon'=\epsilon$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Reason for negligible chemical potentials of different particles in early universe For the early Universe at high temperatures, the chemical potential is assumed to be zero for all types of particles is negligible. Why is this true?
| In the current universe baryon and lepton number are conserved. This means that there are two chemical potentials, $\mu_B$ and $\mu_L$. We believe that the universe has a net-baryon density, and $\mu_B\neq 0$. We don't know the net lepton number of the universe, because the net lepton number of the neutrino background cannot be measured.
The net-baryon number of the universe is non-zero, but small. The current baryon-to-photon ratio is of order a few times $10^{-10}$. Compared to baryon (proton and neutron) rest mass the current temperature is essentially zero, so the current baryon chemical potential is a tiny amount larger than the baryon rest mass.
We can take this information and extrapolate to the past. In the past the universe was much hotter, but baryon and lepton number are conserved. At temperatures above $T\simeq 200$ MeV the baryon number is no longer in protons/neutrons, but in approximately massless quarks, which carry 1/3 baryon number. Since all species of particles (leptons, quarks, photons) are approximately massless, the total densities of quarks, leptons and photons are now comparable. This means that the conserved net-quark density is much smaller than the total density of quarks. This implies that $\mu_B<<T$, and the effect of $\mu_B$ on the pressure and energy density of the early universe can be ignored.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the work done by a force that changes with time $F(t)$? There is a force that changes with time. (F(t))
And the position vector is also given as a function time. (r(t))
Here how do we find the work done by F(t) between, lets's say t=0 and t=1?
This is my actual time-dependent force:
And this is the position vector:
sometimes I get confused because of those i and j s...
| The work done during a short interval of time, $[t, t+\Delta t]$ is given by usual formula
$$
\Delta W = \mathbf{F}(t)\cdot \Delta \mathbf{r}(t),
$$
where $\mathbf{F}(t)$ and $\Delta \mathbf{r}(t)$ are the force and the displacement at the beginning of the interval. The total work is then approximately a sum over all the intervals, and this approximation becomes exact as the length of the interval goes to zero:
$$
W = \sum_{all intervals} \mathbf{F}(t)\cdot \Delta \mathbf{r}(t)
=
\int_{\mathbf{r}(0)}^{\mathbf{r}(1)} \mathbf{F}(t)\cdot d \mathbf{r}(t)
=
\int_{0}^{1} \mathbf{F}(t)\cdot \dot{\mathbf{r}}(t)dt
=
\int_{0}^{1} \mathbf{F}(t)\cdot \mathbf{v}(t)dt
=
\int_{0}^{1} P(t)dt,
$$
where
$$
P(t) = \mathbf{F}(t)\cdot \mathbf{v}(t)
$$
is the instantaneous power of the force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Can two photons excite an electron consecutively? I know that a photons energy is quantized and that it can excite a bounded electron from one energy state to the other whic depends upon the energy the photon carries my question is that can two photons consequtively excite an electron from an initial state (say E1) to E2 and then E3 without the electron falling from E2 to E1 in between.
| Short answer : yes.
Why can't it? An electron can always continue to gain energy from compatible photons and move to higher and higher energy levels (till it finally leaves the atom itself!) as long as the time between the consecutive excitations is not sufficient enough for the electron to drop back to its ground state.
Relaxation (moving from an excited state to a lower energy state) happens through three pathways:
*
*Spontaneous emission (time : $\sim 10^{-8}$s)
*Stimulated emission
*non-radiative decay
All of these take some time and do not happen instantaneously. So if the electron is re-excited before it relaxes to a lower energy state, then for sure it will go to an even higher energy state $E_n, \ n>1$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to know what voltage to apply in hydrogen gas tub to obtain the spectrum? Generally to detect the hydrogen spectrum people uses the hydrogen gas tube as a light source. How people knows the right voltage to apply to the tube in order to obtain the hydrogen atomic spectrum.
What I mean is this, if you apply low voltage then we would not be able to dissociate the $H_2$ molecules and so so atomic spectrum. If the voltage is to high the atoms will be totally ionized. How people knows the right voltage?
| For any gas, there is a well-known breakdown voltage at which it switches from being a good insulator to a good conductor. This "starting voltage" is the minimum voltage required to start the glow discharge. You can find breakdown voltages for different gases in a handbook of high-voltage engineering.
Then, to prevent the gas from immediately turning into a fully-ionized plasma at high temperature, the current flowing into the gas through the electrodes is limited electronically by something called a ballast circuit. It takes less voltage to sustain a glow discharge that it did to start it, and the ballast circuit's job is to allow the source voltage to drop down to the sustaining voltage after the glow is started.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is angular velocity in 3-dimensional space? (Revised) If I'm not mistaken, there are analogies between the translational dynamics of a rigid body and the rotational dynamics of that body. For example, the position of a rigid body is analogous to the orientation of that body. What property of a rigid body, from a rotational aspect, is analogous to the rigid body's velocity? (I would ask the similar question with respect to acceleration, but then this question would get disqualified by the moderators.)
In 2-dimensional space, I'm guessing that angular-velocity is analogous to velocity. But in 3-dimensional space, the way an object can be rotating can be much more complex. For example, the body may be rotating about an axis which itself is also rotating about some other axis. I imagine these axes all stemming from the center of mass of the rigid body.
Note that a single axis/angle pair is enough to describe a body's orientation at any given time, but it is not always sufficient to describe how that body is rotating in 3-dimensional space.
| Before getting to the question, let me first clarify a couple of points. Since the rotation group is three dimensional, it takes three angles to specify the orientation of a body in three dimensions with respect to some reference orientation. For example, we could specify the three Euler angles of the rotation.
Furthermore, at any given instant in time, the body will be rotating about a single axis, known as the axis of rotation. This axis may simply be time varying.
Now for angular velocity. I will first point out that this is not a scalar, as many introductory sources might lead one to believe. It is, in fact, a three dimensional vector which points along the axis of rotation and has magnitude equal to the radians per second spun by the body. Negating the angular velocity is equivalent to changing the direction of rotation about the axis.
Angular acceleration, since you mentioned it, is defined to be the time derivative of the angular velocity, understood in the way I described as a 3-vector.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/596909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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