Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Which units should I use in molecular dynamics simulation? I have written a simple molecular dynamics simulation program. The simulation runs fine but the physical properties (in particular, I have calculated temperature) are off by many scales. I understand that this might be due to not using the proper dimensionless units for the parameters and variables. How should I choose the units? Are there different methods for doing this? I would appreciate if someone could provide a good reference regarding this.
| The possible choices of sets of units for your simulations is probably infinite, so the answer is ultimately going to be choose a set of units that fit what you need & run with it.
For instance, suppose you want to study the Argon interacting via the Lennard-Jones potential, an appropriate choice of units could be mass, $\mu$, length, $\sigma$, and energy, $\varepsilon$ (which would then allow you to define all other relevant scales via products & divisors of the three) with the scales defined as
$$
\sigma=3.4\times10^{-10}\,\rm m\\
\mu=6.69\times10^{-26}\,\rm kg\\
\varepsilon=1.65\times10^{-21}\,\rm J
$$
all of which are related to the properties of Argon & the L-J potential. If you are studying some other atom/molecule or a different potential, you'll want to scale the different quantities to the properties of the atom/molecule/potential you are interested in.
It might also be worthwhile to look into the literature for the scalings used by other people studying the same thing as you.
I had suggested in another answer that you should be defining your initial conditions in physical units, then scaling them appropriately after the IC has been set. For output/visualization purposes, you would want to re-scale the quantities to have physical values for display.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Units of vector differential operator del ($\nabla$) My book says that $\left[\nabla \cdot (\vec E \times \vec H)\right] = \mathrm{W/m^3}$. I see that $\vec E$ is in $\mathrm{V/m}$ and $\vec H$ is $\mathrm{A/m}$, so these multiplied is $\mathrm{W/m^2}$, but how does dotting with $\nabla$ give another $\mathrm{m^{-1}}$?
| The $\nabla$ operator is a spatial derivative of the $\frac{\delta}{\delta x}$ etc kind. This has units of $1/m$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Total angular momentum of electron in a magnetic field In this question: Electron in the proximity of a magnetic monopole
It is stated that for an electron in the magnetic field of a monopole,
$
\vec{B}(\vec{r}) = \frac{g}{r^3}\vec{r}
$
, that the quantity
$
\vec{J} = \vec{r} \times \vec{p} + eg\frac{\vec{r}}{r}
$
is constant. It appears that $\vec{J}$ is a form of the total angular moment (is that correct?), which should indeed be conserved here since magnetic fields do no work, but I do not understand what the $eg\frac{\vec{r}}{r}$ term in $\vec{J}$ represents or where it comes from.
Can anyone elucidate where this contribution to the total angular momentum is coming from?
| After further examining the original question, and the source for the question, which was in the book "Electromagnetic Theory" by Ferraro on p.543, I was able to understand the conserved quantity $\vec{J}$ as thus.
Considering that with $\vec{B}(\vec{r}) = \frac{g\vec{r}}{r^3}$ the Lorentz force yields:
$\begin{align}
m\ddot{\vec{r}} &= q\dot{\vec{r}} \times \vec{B}\\
\vec{r}\times m\ddot{\vec{r}} &= q[\vec{r}\times(\dot{\vec{r}} \times \vec{B})]\\
\vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\vec{r}\cdot\vec{B}) - \vec{B}(\vec{r}\cdot \dot{\vec{r}})]\\
\vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\frac{g}{r}) - \vec{B}(r\dot{r})]\\
\vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\frac{g}{r}) - \vec{r}(\frac{\dot{r}g}{r^2})]\\
\vec{r}\times m\ddot{\vec{r}} &= qg\Big[\frac{\dot{\vec{r}}}{r} - \vec{r}\frac{\dot{r}}{r^2}\Big]\\
\frac{d}{dt}\Big(\vec{r}\times m\dot{\vec{r}}\Big) &= qg\frac{d}{dt}\Big(\frac{\vec{r}}{r}\Big)\\\\
\therefore \vec{r} \times \vec{p} = qg\frac{\vec{r}}{r} + \vec{J}\\
\end{align}$
Where $\vec{J}$ is an arbitrary constant vector. Thus $\vec{J}$ is conserved in both magnitude and direction, so $J$ is constant.
$\begin{align}
\therefore \vec{J} =\vec{r} \times \vec{p} - eg\frac{\vec{r}}{r} \\
\end{align}$
Thus it follows that
$\begin{align}
\vec{J} \cdot \vec{r} =(\vec{r} \times \vec{p} - eg\frac{\vec{r}}{r} ) \cdot \vec{r} &=
(0) - egr\\
\therefore \vec{J} \cdot \vec{r} &= Jr\cos\theta = -egr\\
\therefore Jr\cos\theta &= -egr\\
\therefore \cos\theta &= -\frac{eg}{J}\\
\end{align}
$
Thus $\theta$ is constant, and $\dot{\theta}$ = 0
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to determine whether a nuclear transition would be electric octupole, or hexadecapole? The transition from one nuclear state to another is classified as quadrupole/octupole, etc, depending on the units on angular momentum transferred. But depending on the angular momentum of the two states involved, the net J can take different values. So what decides whether a nuclear transition would be electric octupole, or hexadecapole?
| The short answer is that all transitions which are not forbidden by parity or angular momentum conservation happen, though not necessarily at the same rate. When several multipolarities are allowed, the one with the lowest multipolarity dominates.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\hat{L}_{x}$ and $\hat{L}_{y}$ do not commute... or do they? So $\hat{L}_{x}$ and $\hat{L}_{y}$ do not commute:
$$ [ \hat{L}_{x}, \hat{L}_{y}] = i\hbar \hat{L}_{z}$$
But, what if we perform this operation on a state such that:
$$\hat{L}_{z} \phi_{l, m_{l}} = \hbar m_{l}\phi_{l, m_{l}},$$
where we require that $m_{l} = 0$, so
$$\hat{L}_{z} \phi_{l, m_{l}} =0.$$
Hence, for the case $m_{l}$ = 0,
$$ [ \hat L_{x}, \hat L_{y}] \phi_{l, m_{l}} = 0,$$
and thence $\hat L_{x}$ and $\hat L_{y}$ share eigenstates! Does this work?
| Elaborating on ACruiosMind's comment, assume that the matrices $A$ and $B$ are defined the following way:
$$A=\begin{pmatrix} 1 & 2 \\ 5 & 4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$
Notice that the eigenvectors of $A$ are
$$\begin{pmatrix} 1 \\ 5/2 \end{pmatrix} \quad \text {and} \quad \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$
and the eigenvectors of $B$ are degenerate and its only eigenvector is
$$\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$
However as you can easily verify the commutator does not vanish ie
$$[A,B]= \begin{pmatrix} -7 & -7 \\ 7 & 7 \end{pmatrix} $$
This shows that though one of the eigenvectors of matrices (if you want operators) are the same, they don't commute.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why specific heat at constant pressure is greater than specific heat at constant volume? I know the relation between specific heat at constant volume and pressure and I also know how to calculate it.
Thing is, I don't understand its concept
I want to know why at constant pressure, specific heat is always greater than at constant volume
| At constant volume, all the heat that goes into the system goes into raising the temperature of the system, and no external work is done.
At constant pressure, some of the heat goes into expanding the system, which does external work, and therefore leaves less energy available for raising the temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
direction of friction on an object rolling with constant speed Although there are many questions similar to this, but none of these specifically talk about the situation with constant speed.
A wheel is rolling on a horizontal plane (having some friction) with constant speed.
What will be the direction of force of friction on the contact point of the wheel ?
One of the books by Resnik Haliday says there would not be any friction since the "contact point will not have any tenancy to slide". I am unable to understand this point.
| What Haliday describes is in an "Ideal" situation, where the wheel and surface are perfectly solid and there are no other forces like air resistance. Then the only force between the wheel and surface is an upward force equal to the weight of wheel (assuming we have gravity)
Actually if there is any friction, then energy will be lost and the wheel will not move in constant speed and loose speed over time.
This looks strange to us because in real life we never get to that ideal situation, but we get close. For example a marble ball on a marble smooth surface in vacuum. In practical engineering, we always have a rolling resistance which occurs from lack of that solidness. That is low on a metal-on-metal wheel like rails , but higher on cars. It gets higher when tyre pressure is lower which causes more fuel consumption.
The rolling resistance is a complex thing when looked at from theoretical physics point of view. It is cause by molecule level friction that results in loss of energy when part of the wheel compressed under the weight rolls out of the contact point.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Smooth collar sliding down smooth shaft
For part (a), we find that the acceleration of the collar is $6.94~\mathrm{m/s^2}$. In part (b), we say that the acceleration of B with respect to A is $6.94~\mathrm{m/s^2}$, which makes sense.
However, in part C, the acceleration of C with respect to A is not $6.94$ anymore. From what I understand, someone standing on A would see the collar C coming down at $6.94~\mathrm{m/s^2}$ but they get an answer of $5.522$. What is the logic behind this value and why is it not $6.94~\mathrm{m/s^2}$ (which is the value one would see IF THEY WERE SITTING ON A)?
Edit: No matter how fast or slow A is going, someone sitting on A would see the collar C coming down at the same acceleration of $6.94~\mathrm{m/s^2}$ no?
| No.
A is accelerating away from C. Don't forget that C is not fixed to the smooth shaft, AB. The shaft is accelerating away from C, which has to catch up with A. So from the standpoint of A, C is accelerating slower.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Explanation of Michelson Interferometer Fringe Shift I have been working on an experiment where 2 glass microscope slides are pinched together at one end (so that there is a "wedge" of air between them) and placed in the path of a laser in one leg of a Michelson interferometer. When I move the glass slides (fractions of a mm at a time) so that the path of the laser is closer or further from the place where the slides are pinched, a fringe shift occurs. I cannot seem to explain why this is happening! Any help with explaining this phenomenon would be greatly appreciated! If any more specifics about the setup or dimensions of the slides are needed, please let me know.Also, a full "light to dark" fringe shift occured roughly every 4mm of moving the slides.
| In the 2 glasses there are 4 surfaces, i.e interface air/glass, and 8 surface orientations (a..h) and plenty room for interference between reflections and the main beam.
LASER (air) a1b (glass) c2d (air) e3f (glass) g4h
At each interface the is reflection that will be reflected forward again (self-interference)
look for iridiscence in thin materials.
previous answer, not important now:
Evaluate the dimensions of the screw thread that moves the clamp (distance, pitch) and how many turns are required for a given deviation.
I do not see the screw, but I imagine that it exists and influences the outcome.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Physical Meaning of Cone used in Conic Section for Orbital Mechanics Does the polar angle (complement of $\theta$ below) of a cone which intersects a plane to yield a conic section have a physical meaning in orbital mechanics?
Note that the angle of incident planes, which form parabolas, ellipses, etc., is not the polar angle of the cone.
Here's an example. Consider the initial case of an elliptical orbit illustrated below:
Now, consider the mass of the central body greatly increasing. To conserve momentum, the elliptical orbit would shrink in area. In fact, the polar angle of the cone would decrease to reflect this proportional reduction in area:
[Note the smaller orbit and more narrow cone, despite keeping the same angle of the slicing plane as before.]
Is, therefore, the polar angle all and only a function of the mass of the orbiting bodies?
| Ellipses, parabolas and hyperbolas are both:
*
*Defined as the conic sections you speak of: work out the intersection between the cone $\vec{R}.(\cos\phi\,\hat{X} + \sin\phi\,\hat{Z}) = \sin\theta\,|\vec{R}|$ and the plane $z=const$ where $\theta$ is your polar angle and $\phi$ the angle between the cone's axis of symmetry and the slicing plane and you'll find that the equation for $x$ and $y$ is a general quadric form;
*The path solving $\ddot{\vec{R}} = -\frac{G\,M}{|R|^3} \vec{R}$, i.e. the path of a point mass with a central attractive force proportional to the inverse square of the distance a.k.a. the Newtonian two-body celestial body orbit model: solve this when the initial velocity is contained in the $X\wedge Y$ plane and you'll see that its the same general quadric form.
So, most certainly, the two are precisely the same notion, just different ways of stating it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
General solution of a mass spring system This is the differential equation that describes small amplitude vertical oscillations of a mass $m$ that is hanging from a spring
$$\frac{d^2x}{d t^{2}} + \frac{b}{m}\frac{dx}{dt} + \frac{k}{m} x = 0$$
Where $x$ is defined to be the displacement of the mass from equilibrium position, $b$ is the damping constant, $t$ is time and $k$ is the spring constant.
I want to find a general solution for the case in which small damping occurs.
I understand this to be when the damping constant $b^2 < 4\,\mathrm{km}$ but I am unsure how to go about finding a general solution. What type of mathematical tools would I need to find the solution and would these tools help me to find a general solution for if there was large damping or even critical damping?
| Substitute $x$ by $A(e^{wt})$.
Thus your equation becomes:
$$
A(w^2)(e^{wt})+A(\frac{b}{m})(w)(e^{wt})+(\frac{k}{m})(A)(e^{wt})=0
$$
Simplifying:
$$
(w^2) + (\frac{b}{m})(w) + (\frac{k}{m})=0
$$
Find out the roots . Here you understand that $(b^2)<=4km$ for real values of $w$.
Let the roots be $w_1$ and $w_2$
Finally your solution to the differential equation is:
Case 1: $w_1$ and $w_2$ are distinct:
$$x(t) = A (e^{w_1t})+B (e^{w_2t}) $$ where $A$ and $B$ are arbitrary constants.
Case 2: $w_1$ and $w_2$ are identical:
$$x(t) = A (e^{w_1t}) + B t (e^{w_1t})$$ where $A$ and $B$ are arbitrary constants.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is this definition of orthohelium and parahelium incorrect?
"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)."
From HyperPhysics
When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:
$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$
Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?
| Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What would put a harddisk drive (HDD) under 350G's of force? I always see the label and it says 350G's withstandable. What would put this over 350G's? Is it even possible to hit 350Gs of force to a hard drive?
| Here's an application where an ability to withstand high shock is important.
Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research).
We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a known location with the data loggers running. We would then recover the now utterly destroyed computers: their circuit boards shredded into resin and glass fibres. But their log of seismic data was perfectly recoverable from the HDD and we could seismically map the area for minerals prospecting. We would have destroyed thousands of computers in this way. But I don't believe there was a single unreadable HDD, even those within meters of the blast.
At the time, this was actually cheaper than running communications links of the required datarate to all the points in the grid.
As Paul's Answer Points Out, the specification is likely for a parked HDD. Our HDDs were running, but of considerably less data density than HDDs today (they were 160MB drives), which probably explained their ability to withstand such shock whilst working.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
The notion of fixing a gauge I don't understand the notion of gauge fixing; can we choose any gauge or are there some restrictions?
For example why can we choose $\nabla\phi = 0$ here:
https://physics.stackexchange.com/q/188778/
| It's first important to note that in classical electromagnetism, the $\mathbf{A}$ and $\phi$ fields are not physical in the same way that the $\mathbf{E}$ and $\mathbf{B}$ are. We can't measure them and they aren't uniquely defined.
Gravitational potential energy is a good analogy. Suppose an object is on a table a height $h$ above the ground. We could say that the object has potential energy $U=mgh$, picking the floor to be $U=0$, or that the object has potential energy $U=0$, picking the table to be where $U=0$. In fact, we could choose the gravitational potential energy to be any $U=mgh+C$ for any $C$, as long as we use the same definition of potential energy in the same problem.
Just like we can add any constant to the gravitational potential energy, we can add some additional vector field $\mathbf{V}$ to $\mathbf{A}$. Let's define the new vector potential $\mathbf{A'}=\mathbf{A}+\mathbf{V}$. By the definition of the vector potential, we know $\nabla \times \mathbf{A'} = \mathbf{B}$. Using the properties of the curl, we find $\nabla \times \mathbf{A'} = \nabla \times \mathbf{A} + \nabla \times \mathbf{V} = \mathbf{B}$. However, we know $\mathbf{B}=\nabla \times \mathbf{A}$, so we can write $\mathbf{B}+\nabla \times \mathbf{V} = \mathbf{B}$. Therefore, we see that we can add any $\mathbf{V}$ such that $\nabla \times \mathbf{V}=\mathbf{0}$. Because of an identity in vector calculus, we know that any curl-less vector field can be written as the gradient of a scalar field, say $\psi$.
When we go through the math we find that when we change $\mathbf{A}$ to $\mathbf{A'}=\mathbf{A}+\nabla\psi$ we have to change $\phi$ to $\phi'=\phi-\frac{\partial \psi}{\partial t}$.
So we don't have infinite freedom in gauges. We have the freedom to choose any scalar field $\psi$ and then transform $\mathbf{A}$ and $\phi$ as above. The two most popular gauge choices are the Coulomb and Lorenz, which are detailed in the Wikipedia article you linked to in your question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Power of several focused laser beams on a small surface After viewing this: My Homemade 40W Laser Shotgun
Will the initial 40W power of the diodes beams be roughly transferred to the targeted surface where the 8 beams are focused, or will there be a power loss due to some kind of interferences where the beams overlap?
| If energy is lost, it must go somewhere. So, there won't be any "interference" where the beams overlap which saps the power. Unless, and this does happen, the lasers are powerful enough to ionize the air, in which case the light is blocked at that point and converted to heat, which will never make it to the surface. See this article on laser-induced breakdown.
I doubt the contraption in question has the power-density to do this, though.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What happens if gravity of all objects in the universe disappears? I've been trying to find the answer to this question for a few days and I ended up in having different answers from different sources. A website says:
Everything on the universe would move in this exact direction it was moving at the instant gravity stopped.
And another one says:
Since it is gravity that binds planets in shape, when gravity disappears, planets will lose their shape are turn into dust clouds.
Now which of the above is correct?
| The Earth is rotating at around 1000mph. There is also immense pressure inside the earth pushing the crust outwards. If gravity "stopped working" then the rotational velocity plus the internal pressure would rip the planet apart pretty quickly. However, if gravity "stopped" the sun would explode, as would all other stars, so it's debatable which would get us first - the earth falling apart or the sun burning us to a crisp.
BUT, gravity is a result of mass. If gravity "stopped working" then presumably that would be a result of all mass suddenly disappearing. The only way that could happen is if all the mass in the universe suddenly converted to energy, so the result is again a huge explosion. :-)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/188953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Flaws of Broglie–Bohm pilot wave theory? I recently learned about an oil drop experiment that showed how a classical object can produce quantum like behavior because its assisted by a pilot wave. How has this not gained more attention? What flaws does Broglie–Bohm pilot wave theory have in explaining particle behavior?
| The Copenhagen interpretation refutes pilot wave theory because it's stand is that nothing has position until measured..... It has created a long history of discourse among physisist.
You have to buy in to the concept and I don't.
Einstein once asked a proponent of the Copenhagen interpretation if the moon was in place when you are not looking at it.....
The politics of academia has resulted in no serious consideration, since 1952, of the fact that all avenues should be investigated not just the ruling elites view.
Copenhagen interpretation is WRONG, PERIOD
But you will be admonished just for saying so...
Politics is man's Achilles heel. The Copenhagen interpretation was born at the conference pictured below...
Copenhagen interpretation claims there are no tragectories, and slams pilot wave theory because the tragectories are surrealistic. Self serving rebuttal. Everything is nowhere untill the observation. Poppycock.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 3
} |
Are wavelength and the distance same thing? Can you clarify for me the following question: are wavelength and distance same?
I know wavelength is measured in terms of distance but when we have a look at the two equations:
$$
c=f\,\lambda\\
v=d/t
$$
it actually explains the same thing where $v=c$=velocity and $1/t$ is frequency. So $\lambda$ should be equal to $d$. So if $\lambda = d$, then why do we have two equations existing instead of one. Can we use any equation to calculate velocity?
| The lambda is the distance between 2 points having the same phase like two successive crests the velocity is the wave can be conceived as how many crests for example passes through a reference in a given time you can use both equations but c=f*lambda is used if you have lambda , its proof is V = distance / time , if a crest traveled a distance = wavelength then that is done in a time = T periodic time , OK that means the crest moves one cycle we have by definition the frequency = number of cycles per second so the time required for one cycle = periodic time = 1 / f substitute with it in v= d / t , you find v = lambda * f
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Could some astronomical objects have superconducting properties? The colder it is, the more efficient the superconductivity process works. And as we know, if there is no star nearby, space gets pretty cold.
I do appreciate that many condensed, burnt out, stars may take a long time to cool off, but are there any other types of known astronomical objects that may feature superconductivity to create and/or maintain a very strong magnetic field?
| This is not a complete answer to the question, rather a explanation of Kyle Oman's answer.
When we (or at least me) think of superconductivity, we have in mind the pairing of electrons to form Cooper pairs. But this pairing is quite weak, and a moderate magentic field can destroy superconductivity.
But electrons are not the only particles around! At the extreme densities found in neutron stars, both neutrons and protons can form pairs. For them, critical temperatures can be as high as $5\cdot10^8$K for neutrons (larger for protons), and critical magnetic fields of $10^{15}$G. In neutrons, this pairing leads to superfluidity, and in protons to both superfluidity and superconductivity.
The protons of the outer core are thought to be in a Type II superconductor, that is, the magnetic field is confined to vortices where the field strength can be that of a magnetar.
And how has all of this been discovered? Neutron stars where discovered which cooled down unusually fast. This cooling wasn't compatible with their X-ray emission, so they are thought to emit neutrinos as well. The pairing of two neutrons lowers their energy, and this energy difference is liberated as neutrinos. Proton superconductivity is required to suppress other cooling mechanisms.
Sources
*
*D. Page et al.: Rapid Cooling of the Neutron Star in Cassiopeia A Triggered by Neutron Superfluidity in Dense Matter arXiv:1011.6142
*P. S. Shternin et al.: Cooling neutron star in the Cassiopeia A supernova remnant: Evidence for superfluidity in the core arXiv:1012.0045
*C. O. Heinke: Superfluids and superconductors in the core of a neutron star: the highest-temperature superconductor University of Alberta
*B. Haskell et al.: Investigating superconductivity in neutron star interiors with glitch models arXiv:1209.6260
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Really how can an observable quantity be equal to an operator?
A wave-function can be written as $$\Psi = Ae^{-i(Et - px)/\hbar}$$ where $E$ & $p$ are the energy & momentum of the particle.
Now, differentiating $\Psi$ w.r.t. $x$ and $t$ respectively, we get
\begin{align}
\frac{\partial \Psi}{\partial x} &= \frac{i}{\hbar} p\Psi \\
\frac{\partial \Psi}{\partial t} &= -\frac{i}{\hbar}E\Psi \, .
\end{align}
The above equations can be written in suggestive forms
\begin{align}
p \Psi &= \left( \frac{\hbar}{i} \frac{\partial}{\partial x} \right) \Psi \\
E \Psi &= \left( i \hbar \frac{\partial}{\partial t} \right) \Psi \, .
\end{align}
Evidently the dynamical quantities momentum and energy are equivalent to the operators
\begin{align}
p &=\frac{\hbar}{i} \frac{\partial}{\partial x} \\
E &= i \hbar \frac{\partial}{\partial t} \, .
\end{align}
quoted from Arthur Beiser's Concept of Modern Physics
How can observable quantities be equal to operators?
You can't measure an "operator".
Can anyone intuitively explain how momentum and energy are equal to operators?
| Not equal, but equivalent, in the sense that they have the same effect on the wavefunction in question.
More precisely, the book is using a slight abuse of terminology. Taking momentum as an example, it's not really the case that the dynamical quantity of momentum is equivalent to the operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$, because a number can't actually be equivalent to an operator. A number is a thing all on its own, whereas an operator is something that needs to be applied to something else to have any meaning. But the operation of multiplying the wavefunction by the wave's momentum (a number) is equivalent to the operation of taking the derivative and multiplying by $\frac{\hbar}{i}$. In mathematical language:
$$\frac{\hbar}{i}\frac{\partial}{\partial x}\psi = p\psi\tag{1}$$
whereas, strictly speaking, we can't say this:
$$\frac{\hbar}{i}\frac{\partial}{\partial x} = p$$
At least, not in the way you're thinking about the notation.
What we normally do in quantum physics is always think about quantities as operators. That is, if you write just $p$ in an equation, it's implicitly understood that this should be applied to some wavefunction. For example, if you write
$$H = \frac{p^2}{2m}$$
what you really mean is
$$H\psi = \frac{1}{2m}p (p \psi)$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
} |
Antenna direction I have a router with a wifi antenna that can be turned in any angle. I wonder what difference does the direction of the antenna make to the electromagnetic signals propagation? Where is the signal strength the biggest?
| Well a router antenna is simply a dipole. It will have maximum radiation in its broad side direction and it's radiation pattern looks like a donut.
Check the link below for the illustration of the dipoles far-field
https://www.cst.com/Academia/Examples/Wire-Dipole-Antenna
Of course the router itself and any metallic objects nearby will influence the radiation pattern and also in small rooms you will have multiple reflections from walls, furniture, etc. so it is difficult to say what is the maximum.
But as long as you keep the routers antenna pointing upward you should get an optimal signal.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The speed of light/EM waves in vacuum; as if there was another one in non-vacuum? Q1: is there a speed of a photon other than in "vacuum"?
Q2: isn't "speed of light in vacuum" misleading?
If I understand, that light moves with speed of light until there is "something in between" (no matter what) (1)
What I ask for, is not a deeply explanation; it's just:
Children ask me:
*
*"But how can be light slower" if it is a constant?
My explaination is:
*
*"Till it collides" (not the deepest answer, I know)
The question (since I prob. are not at the pulls of new physics): Is there another speed of light than in "vacuum" / nothing crossing ?
(1) no discusss, what "what" is
| first the speed of light is related to the permittivity and permeability of the medium. changing either one of those values changes the speed of light.
copper has different values then free space. the speed of light through copper is 2/3rds that of free space. about 1 foot per nanosecond. slightly faster in aluminum, slower in iron. this can be measured with an oscilloscope.
freespace is not perfectly even. there are variants.
these are simple facts that can be imperially verified and are the basis of electromagnetic communication. it is a provable fact that free space has an inductive reactance of 376 ohms. and that should start you thinking ... how can something that is devoid of all things have a measurable property.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why do sea waves seem to be standing still when you look from the window of an airplane? Looking from the window of a passenger plane even at moderate altitude such that one can still recognize individual waves and even something like white foam, and small boats close to the cost line, it already looks like the water is not moving.
To make more clear what i mean, here is an examplein this video at around 10:40. In HD eye resolution it is much more intriguing, but the video shows the idea, that even when the plane is quite low, waves close at the beach appear to be "frozen".
Why is that? and does that effect have a name?
| Think of it this way: You are way up in the sky and the distance you see as a centimeter could be meters long since you see objects getting smaller as you go farther away. Assuming that an ordinary water wave travels with a velocity of 3 or 4 m/s at a maximum, it is not hard to imagine that you are seeing them as if they are standing on the ocean.
Furthermore you get to observe the phenomena for a really short amount of time, which means that the displacement of waves is really small compared to the distance from the plane to the ocean, hence the standing looking waves. I would also add that if you were on a helicopter above the ocean observing waves for a long time, you would definitely see them spreading out since that is actually what happens.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/189951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Constructing differential equation from arbitrary Hamiltonian Suppose I begin with the time-independent Schrodinger equation
$$ \left(-\frac{1}{2m}\partial_x^2 + V(x)\right)\psi_n(x) = E_n\psi_n(x), $$
ordinarily we specify the function $V$ and then solve for a set of eigenfunctions and eigenvalues. And just to be slightly more general, we do the same thing with Sturm-Liouville equations, which I'll write in terms of the momentum operator and an extra function $U$,
$$ \left(\hat{p} U(\hat{x}) \hat{p} + V(\hat{x})\right)\psi_n = E_n\psi_n.$$
Now nothing is stopping us from defining a new Hamiltonian operator with the same eigenvectors but different arbitrary eigenvalues $\lambda_n$,
$$\hat{H}\psi_n = \lambda_n \psi_n$$
Under what conditions can this eigenvalue equation for the new Hamiltonian be represented as a (not-necessarily second order) differential equation in $x$ with the same eigenfunctions? In other words when does $\hat{H}$ belong to the operator algebra generated by $\hat{x}$ and $\hat{p}$?
I see if I define the new eigenvalues by some $n$-independent function $f$ of the original eigenvalues $\lambda_n = f(E_n)$, I can come up with a new differential equation, but does this exhaust the possibilities?
| After thinking about it, as long as the original eigenvalues are non-degenerate it should be possible to have the new Hamiltonian be represented by a differential equation of arbitrarily high order. The key is that the projection operators $P_n$ onto the eigenfunctions exist in the algebra generated by the original Hamiltonian $\hat{H_0}$.
For instance say the nth eigenvalue is $E_n=2$, and there are no other eigenvalues between 3 and 1. Then we can choose an indicator function $f_n(x)$ such that $f_n(2)=1$ but $f_n(x)=0$ if $x$ is less than 1 or greater than 3. Given sufficient continuity the Stone-Weierstrass theorem applies and we can represent $f$ by a polynomial basis
$$ f_n(x) =\sum_k c_{n,k} x^k.$$
Then the operator
$$ P_n \equiv f_n(\hat{H}) = \sum_k c_{n,k} \hat{H_0}^k $$
will project onto the eigenfunction with eigenvalue 2. The details that this works even though we are dealing with infinite sums comes in the proofs of Gelfand duality.
Since the projectors are in the algebra generated by $\hat{H}_0$, the arbitrary Hamiltonian $\hat{H}$ is also in the algebra
$$H=\sum_{n} \lambda_n P_n=\sum_{n,k} \lambda_n c_{n,k}\hat{H_0}^k,$$
and since the original Hamiltonian can be expanded in terms of functions of $\partial_x$ and $x$, the Hamiltonian $\hat{H}$ also can, although now in general the differential equation will be of arbitrarily high order.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Quantum entanglement and the big bang Prior to the Big Bang all matter was compressed into a point of high density. Why isn't all matter already entangled?
| I actually happen to believe that they can be, and not only that but I think it would be very likely that all particles in the universe were entangled prior to, during and after the big bang. The problem would be verifying this entanglement in experiments as I can see no way of deducing which particle any given particle would be entangled with.
Entanglement has been proven in the lab but can only be done so with a high level of quantum coherence which is unfortunately lacking in nature. If you were to observe a particle in nature change the direction of its spin spontaneously you could put it down to entanglement, or to its interaction with other particles in its environment. Otherwise I can't see why this notion of entanglement at the beginning wouldn't be possible if not somewhat likely.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
I don't understand black body radiation graphs
Let's look at the above graph.
*
*This black body graph is for the temperature of 5000K. Each temperature has a different black body graph?
*How am I supposed to read this graph? Do I start from the left, the right, or the peak?
*As wavelength is approaching zero, intensity is approaching zero. This is what my textbook says (experimentally). This is explained by Planck's hypothesis regaridng quantised energy. However, if energy is quantised, and thus the all or none principle applies, should it then simply just, with the shape of a straight line, collapse? i.e. the graph should approach its peak, then instantly drop down in a straight line because it can either absorb ALL or none. The graph does not depict this. Why?
*A black body of ANY temperature will emit all types of radiations (eg UV, X-rays, visible light etc). BUT, at higher temperatures eg 60000K, certain transitions in quantum states are favoured more, thus we would expect more intense radiation to prevail eg gamma.
| The x-axis on the graph shows the wavelength, while the y-axis shows the corresponding intensity for that wavelength (all for a given temperature of the black body, in this case 5000K). Also note that this graph shows a distribution of the various wavelengths and their intensities at that temperature. It is a distribution graph like the "bell curve" as opposed to a graph like distance vs time.
In your diagram, we can see that when the black body is heated to 5000K, most of the radition emitted from the black body has a wavelength of about 500nm (Since this is where the peak of the graph is, therefore most of the intensity of the light comes from 500nm wavelengths).
For higher temperatures, you will see that the peak of the graph will be more to the left than the one shown in your diagram, because since E = hf (and c=wavelength*f), radiation of lower wavelength will predominate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Can we find actual rest mass of things on Earth Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered.
Can we know the real rest mass? If so, can we deduce our speed in the universe?
| When the object is an elementary particle or a charged ion we can use electromagnetic interactions to measure its rest mass, given the charge in an e/m experiment. One can get the charge with Milikan's oil drop experiment. Here is a setup for the lab.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
} |
Is the scalar magnetic potential continuous? If we have two current-free spaces and separated by a surface current, we can solve the magnetic problem by solving two magnetic scalar potentials and then using matching conditions. My question is, is the general scalar magnetic potential continuous? Why?
| A potential is essentially an integral of work to carry a particle (magnetic or electric one) from an infinite distance up to the point you need to know the potential value. If the force making the work is not a Dirac's Delta, then the integral (i.e. the potential) must be a continuous function.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bottle stability optimization A few days ago some friends and I played a game called "flunkeyball" where you need to upset a bottle with a ball. Then a question occurred: "How much water do we need to put into the bottle that its stability is optimal?"
My first thoughts were that the centre of mass must be as low as possible and the mass as high as possible compared with that of the ball. My question is: Can we calculate the optimum amount of water we need to fill into the bottle to optimize its vertical stability and if so, how?
For simplicity we can assume that:
*
*the shape of the bottle is a cylinder with a cone on top
*we can fill the bottle with a solid instead of a liquid
*the ball flies parallel to the ground and hits the bottle at the top
*the material of the bottle has no weight
Sorry for my English and if you want to improve the question please do it. Also if there are any helpful links out there please point me to them.
Thanks in advance!
|
simple force body diagram shows rotational motion of the force of weight on a pivot straight edge bottom
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/190845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
What does it mean to say that "remembering the future and not the past?" I encountered a rather stupid question which I don't quite understand. - "Why can we remember the past but not the future?" It sounds cool when I first read about it but I think about it more, about how to explain it to a person who doesn't know physics, this questions starts to become rather dumb, because "remembering" implies something that happen in the past or has happened, so in order to remember the future, one has to have a predetermined future, so the only way that one can remember the future is that one has his future set. So what am I misunderstanding here. Does psychological arrow of time implies that? Instead of understanding why one can remember the past but not the future, I would like understand how will life be like, if we can only remember the future (which I find it to be absurd logically and semantically). Can someone give me a scenario of this?
How do physicists define time? I mean the normal sense of time only flows forward. The moment one talks about time flowing backwards, the person must be using a different definition.
| In this case one could argue that 'remembering the past' would be a methaphorical expression to describe being able to predict the future. Experimental evidence says that people that have declared being able to predict the future are either scammers, delusional, self-fulfilled prophecies or just extremely intuitive. As far as I can tell, there hasn't been conclusive scientific evidence of precognition.
Assuming that as a fact, then the absence of precognition (or deterministic precognition) is a observational fact. The mainstream scientific theory of time and cognition says that the brain is a system that creates memories upon facts that happen as consequence of thermodynamical time flow, and because of that future events are fundamentally indeterminate. There is no evidence against this theory, which is the reason is mainstream
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Why can microphone be much smaller than wavelength of sound? For sound from 20Hz to 20kHz, wavelength is 17m to 17mm, for sound at 2kHz, wavelength is 17cm.
And I saw tiny microphone which is much smaller than that. In electromagnetic, there is a smallest size for antenna of each wavelength (half wavelength???). And there is a law (IIRC) that if sampling frequency is smaller than half of the signal, than it is not possible to reconstruct the signal.
How can microphone size is much smaller than the wavelength? E.g. the head of the mic of singer, I think it is smaller than 1 cm
| Microphones transform the pressure wave of sound to an electric signal.
The wavelength of the sound wave tells us the distance over which the wave's shape repeats itself in space. The frequency measures the changes in the medium in time.
As the sound wave passes, the molecules of the microphone vibrate in place ,according to the frequency, like a harmonic oscillator, and this vibration can be utilized to generate an electromagnetic signal. The spatial dimensions are not limited by the sound's wavelength but by the elasticity and the molecular bonding that responds to the frequency of the incoming wave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Does isotropy imply homogeneity? This question comes from exercise 27.1 in Gravitation by Misner, Thorne and Wheeler. They required the following:
Use elementary thought experiments to show that isotropy of the universe implies homogeneity.
I know homogeneity as the universe is the same everywhere at a given time, and isotropy is related to direction.
I wonder how the isotropy of the universe implies homogeneity.
| A few years late here, but I think a clear way of thinking about this is any two points in the universe, A and B, will be connected by a great circle drawn around C. If the universe is isotropic at point C then the points A and B must look the same. This logic can then be extended to any two points in the universe.
This logic clearly relies on that which was pointed out by John Rennie, that the universe must be isotropic everywhere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Is it possible for two events happen at the exact same time? Is it possible for any two events to occur at the exact same time?
As I see it, because time intervals can always be split up into smaller units (it is infinitely divisible), we can always be more and more exact with measuring the time at which something happened, until we find out that the two things did not in fact occur at the same time. Is this correct, or is it actually possible for two things to happen at the exact same time?
| Events are points $(x,t)_S$ onto a chart $S$ on some space-time manifold and in this respect whenever two such points $P_1 = (x_1,t), P_2=(x_2,t)$ have the same $t$-coordinate in that reference frame then yes, they do occur at the same time for the observer described by the chart $S$. For another observer, represented by a different chart $S'$, the $t$-coordinate may transform and give two different times $P_1'=(x'_1,t'_1), P_2'=(x'_2,t'_2)$ with $t'_1\neq t'_2$; therefore, although the two events were simultaneous in the first reference frame, they are not in the second one.
As I see it, because time can always be split up into smaller units
does not really make much sense, because you do not split coordinates into smaller units. What you perhaps have in mind is splitting time intervals into smaller pieces, but that is a totally different thing. But even in that case, however you split your interval up, the total interval length (i. e. the integral) is always the same, no matter the procedure you use; hence you can exactly compare different intervals: if they are the same then they are the same, if not then they are not.
How to experimentally measure time intervals in order to reduce the uncertainties is anyway a totally different question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Does paint affect the thermal conductivity of a metal? We noticed that in our experiment that painting a metal increases its thermal conductivity; is this true? If so, can you guys send me a link to a research paper to support this claim.
|
painting a metal increases its thermal conductivity; is this true?
No it is not true.
Thermal conductivity is a bulk property of the material. It expresses how well the metal conducts thermal energy through the bulk of an object made of that metal.
As CuriousOne commented, surface treatments do not affect how heat is conducted in the interior of the object - they can affect emissivity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Probability flux I was reading a text on Quantum Mechanics in which it said that
$$\int{d^3 x \, j(x,t)} = \frac{\langle p\rangle}{m},$$
where $\langle p\rangle$ is the expectation value of the momentum operator at time $t$.
I tried using $$\langle p\rangle = \int d^3x \, \psi^* (-i\hbar \nabla) \psi.$$
Either I am using the wrong $\langle p\rangle$ or carrying out the following steps wrong but I keep missing out the second term of $j$ given below:
$$j(x,t) = -(i\hbar /2m)[\psi^*\nabla\psi-(\nabla \psi^*)\psi]$$
Have I misunderstood meaning of $j$ or $\langle p \rangle$?
| I was stuck on this same problem. ACuriousMind does give a good hint though. Using integration by parts on the second term for $j(x,t):$
\begin{align*}
\int d^3x ~ j(x,t)&=-\frac{i\hbar}{2m}\int d^3x~ \left(\psi^* \nabla\psi-(\nabla\psi^*)\psi\right)\\
&=-\frac{i\hbar}{2m}\left(-\left[\psi^*\psi\right]^{+\infty}_{-\infty}+\int d^3x~(\psi^*\nabla\psi+\psi^*\nabla\psi)\right)\\
\end{align*}
The boundary term goes to zero and we get the result:
\begin{align*}
\int d^3x ~ j(x,t)&=-\frac{i\hbar}{2m}\int d^3 x ~ 2\psi^*\nabla\psi\\
&=\frac{1}{m}\int d^3 x~\psi^*(-i\hbar\nabla)\psi\\[3pt]
&=\frac{\left<p\right>}{m}
\end{align*}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Velocity of Rocket Exhaust i recently learned a bit of rocket propulsion.It wasn't much complex but was explained in simple terms.The only problem i had understanding it was that in calculating the thrust of rocket the velocity of the exhaust was taken relative to the rocket.My problem is : Shouldn't the velocity of the exhaust be taken as the relative to earth.In all the previous examples we had done so , why don't we use the velocity relative to the rocket.Thanks
PS:A simple explanation would be much appreciated
| To see why the exhaust speed is important, let's do a calculation.
*
*Let's start with a rocket of mass $m$ going at speed $u$. (We measure all speeds with respect to some inertial reference frame.)
*Now, suppose it exhausts a tiny amount of propellant of mass $\delta m$ and the propellant is traveling at speed $u_P$. After it exhausts that fuel, the rocket now has mass $m-\delta m$ and is now going at speed $u+\delta u$.
Conservation of momentum says:
$$ (m - \delta m)(u +\delta u) + \delta m\ u_P = m u$$
Simplifying the above and keeping only first order terms, we obtain:
$$ m \delta u = \delta m (u-u_P) $$
In other words, for a given amount of mass, the increase in speed of the rocket, $\delta u$, depends only on the difference between the speed of the rocket, $u$, and the speed of the propellant, $u_P$. The absolute value of either speed is irrelevant.
The difference between the rocket speed and the propellant speed is called the exhaust speed.
For the best chemical rockets, the exhaust speed is around 3,000 meters per second. When electric propulsion is used, exhaust speeds can be up to 20,000 meters per second or more.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/191949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How are determined experimentally the energy levels of the atoms ? How is the calibration done to several decimal points? I see discrepancy for the absorption edges for the atoms in the X-ray ?
For example K-absorption edge of carbon can be anywhere between 282 to 284eV according to different sources.
My question is how is energy measured and calibrated ?
Are the energy levels determined by electron collisions with specific eV ?
| It is possible to measure wavelengths of light to many decimal places. When you see accurate determinations of atomic energy levels, they were done spectroscopically, looking at absorption or, more commonly, atomic fluorescence. Since $E=hc/\lambda$, one can accurately convert between wavelength and energy.
When excited, atoms emit many wavelengths of light. By examining patterns in these wavelengths, one attempts to assign them to transitions between particular atomic energy levels. As an example, one can look for wavelengths that fit the pattern of a rydberg series, that is of transitions to some particular lower level from a series of highly-excited upper levels. Because highly-excited levels are hydrogen-like and therefore well-understood, one can immediately deduce the energy of lower level from from the observed wavelengths. For much more information on this process, see the NBS monograph entitled Atomic Energy Levels As Derived From the Analyses of Optical Spectra by Charlotte Moore (NBS Circular 467).
As an example of accurate wavelength measurements, you can read here about a commercial instrument which measures wavelengths accurate to 1 pm or better (1pm=$10^{-12}m$).
Beams of electrons cannot be created with anything like that level of precision. Consequently, electron-atom collisions are rarely, if ever, used for atomic energy level determination.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is the Energy of an absorbed photon exactly the energy of the band gap? I was wondering, if the Energy of a Photon which is absorbed by an Electron, hast to be exactly the Energy of the bound gap.
So if i have two energy levels in an atom $E_2$ and $E_1$, does my Electron have to have exactly the Energy
$$h\nu = E_2 - E_1$$
or is it sufficient if the photon has a bigger Energy than that ?
I was wondering because if one assumes the spectrum to be continuous the chance of finding a photon with just the right energy of lets say $h\nu=10.2\text{eV}$ should be rather small.
| No, it is sufficient for the photon energy to exceed the band gap. Any excess energy is transformed into kinetic energy for the electron in the new band. You get exactly the same effect when ionizing an atom - the excess energy simply powers the electron into a faster continuum state.
You should also take into account that photon energies are never exactly defined except for monochromatic beams with infinite temporal duration. This is exactly because of the energy-time uncertainty relation: the only way to have a perfectly defined photon frequency, and hence energy, is to observe it for an infinitely long time. Thus, the photon energy is always spread out over a finite bandwidth.
A similar effect holds for atomic bound-bound transitions, which will always have a finite natural linewidth. This is caused by spontaneous emission, which means that if you leave the atom in an excited state for long enough then it will eventually return to the ground state. This then limits the amount of time in which you can coherently probe the frequency of the transition, and in turn limits the precision to which you can measure this frequency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Do electrostatic fields really obey "action at a distance"? In an electromagnetic theory class, my professor introduced the concept of "action at a distance in physics".
He said that:
If two charges are at some very large distance, and if any one of the charge moves, then the force associated with the charges changes instantaneously.
But according to Einstein, no information can travel faster than the speed of light.
So photons (the information carriers in electromagnetic force) cannot instantaneously deliver information.
So that we associate a field with the two charges and if any charge moves, there is a deformation in that field and this deformation travels with the speed of light and conveys the information.
If the field deformation information cannot travel more than the speed of light, how does the force instantaneously change at very large distances?
| The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
current splitting in the presence of superconductors When you have two resistors in parallel, the current splits up based on the resistances. What will happen if we have two superconductors in place of the resistors? What will happen to the current?
| Current going through a superconductor (or otherwise) will form a magnetic field. The potential energy of the magnetic field depends on its size, and the permeability of its surrounding environment.
The current will be divided between the two superconductors such that the total magnetic field energy is minimal.
Actually, this effect is observable using inductors instead of superconductors. Perhaps the better answer is simply that the superconductors will behave like ideal inductors.
If you connect a big inductor in parallel with a small one, the small one will tend to short-circuit the big one. The total inductance is determined by $1/(1/L_1+1/L_2)$, similar to the parallel resistance formula. Likewise, the current through each inductor is inversely proportional to its inductance. If you leave the parallel inductors connected to a DC source, then the current will gradually be redistributed according to their resistances. This effect is usually neglected because inductors are typically used in an AC frequency domain where $IR << LdI/dt$. But it's always there.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does phonon scattering change the distribution function? For a one-dimensional structure, we know that the modified distribution function has the following energy dependency in equilibrium:
\begin{equation}
Z(\varepsilon)\,f(\varepsilon) = \dfrac{N_\text{1D}}{\sqrt{\varepsilon-E_\text{C}}}
\times \exp\left( -\dfrac{\varepsilon-E_\text{F}}{k_\text{B}T} \right)
\,\Theta(\varepsilon-E_\text{C})
\end{equation}
with $E_\text{C}$ being the conduction band minimum and $E_\text{F}$ the Fermi level.
$Z(\varepsilon)$ is the 1D density of states, and $f(\varepsilon)$ is the equilibrium distribution function (excluding the Pauli principle, of course).
How does the presence of elastic, inelastic, optical and acoustic phonon scattering mechanisms change the above mentioned distribution function? Do we expect to see sharp peaks and spikes in certain energies?
I would be thankful if you could also recommend references on the subject.
| First off, let's answer the question: what part of that equation would change?
$f\left(\epsilon\right)$ is always the same if you're in equilibrium. Scattering won't change that, because scattering alone won't move you out of equilibrium. If anything, scattering has the opposite effect.
What can change is $Z\left(\epsilon\right)$. How it would change depends on your system, which you haven't really specified. That said, scattering mechanisms are usually assumed to be small perturbations on the known system. That's usually a good assumption, so unless you have a crazy system, I'd say that things aren't likely to change much.
Electrons and Phonon by J. M. Ziman is a classic text on the subject.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/192984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Free fall into circular motion If I'm on a roller coaster free falling from height $h$ and then suddenly start going into horizontal motion with a radius $r$ of turn what is the $g$-force I experience?
I worked out the equation like this but am not sure if it is correct:
*
*(1) instant velocity of free-fall $v=\sqrt{2 g h}$
*(2) uniform circular motion acceleration $a = \frac{v^2}{r}$
*(3) $g$-force $ gf = \frac{a}{g} = \frac{v^2}{g r}$
My doubts are:
*
*I don't know if I can use uniform circular motion equation since $v$ is not constant
*Where is the g-force directed towards? The center of the turn?
| You asked two simple questions - I will give two simple answers.
I don't know if I can use uniform circular motion equation since v is not constant
At the very instant that the curve starts, the velocity is given by $\sqrt{2gh}$ - and for that first instance it is constant. So yes, you can use uniform circular motion
Where is the g-force directed towards? The center of the turn?
It depends on how you define "G force". Usually, it is the "non gravitational acceleration experienced". If that is so, then it is pointing at the center of the circle at the moment you start moving around the circle.
If you accept that a person experiences "1 g" when standing still, then the g force due to gravity will depend on the angle of the rail - it will increase with the $\sin$ of the angle of the radial vector, and will cause the force to point slightly above the center of the circle.
Of course real rollercoaster tracks describe a spline - that is, the rate of change of curvature is continuous. Otherwise the sudden change in g force would be most unpleasant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
What happens if the load on the electrical generator exceeds its generation power? And why? What happens if the load on the electrical generator exceeds its power generation? and why?
To be more precise, suppose we have a standard induction generator operating at frequency $\nu=50\:\mathrm{Hz}$ and voltage $V_0$, and rated to produce a maximum power $P_0$, and that we connect this to a load $R<V_0^2/P_0$, which will try to draw more power than the generator's capacity. Obviously the details will depend on the type of generator, but, generally speaking: what will be the generator's response, and what physical processes are involved?
| If the generator's power source exceeds the generator's capacity, and if a load is placed on the generator that also exceeds the generator's capacity, and if all safety devices are disabled; the generator would heated up to a point where the weakest link would burn out like a fuse and thus remove the electrical load.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How do gauge boson interact with elementary particles? We know that gauge bosons are the force carriers of fundamental interactions, but how do the gauge bosons themselves interact with particles?
| Gauge bosons, such as the photon, are indeed the force carriers of fundamental interactions. The interactions are built upon local gauge symmetries of a Lagrangian. A Lagrangian is a list of interactions in our theory. A symmetry is an operation (such as a rotation) that leaves the Lagrangian invariant.
Local gauge symmetries are very restrictive. If we wrote down any old random interactions, they would violate a symmetry. In fact, local gauge symmetries exactly specify the allowed interactions between gauge bosons and fermions and the allowed self-interactions between gauge bosons. That's how we know how the gauge bosons interact.
We cannot deduce everything from gauge symmetry though: we still have to measure/postulate the strength of the interactions and the charges of the fermion fields. Once we've done that, the gauge symmetry tells us the interactions. For fermions in the Standard Model, like electrons, muons and quarks, the interaction is indeed in the form
$$
\text{fermion}\quad\leftrightarrow\quad\text{fermion and photon}
$$
or simiarly,
$$
\text{fermion and anti-fermion}\leftrightarrow\quad\text{photon}.
$$
However, these interactions only occur as part of a bigger series of interactions, otherwise they would violate energy-momentum conservation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Free energy of coupled classical harmonic oscillators I'm looking to find the thermodynamic (NVT) free energy of a classical coupled harmonic oscillator system such as the one below:
(image taken from http://openmetric.org/StatisticalPhysics/equilibrium/week3.html)
I would like a solution that allows an arbitrary $N$ number of masses, and ideally I would like to have a fully general expression with arbitrary (different) spring constants and arbitrary (different) masses.
I tried to compute this free energy by hand by computing the partition function:
$$ Z = \int_{-\infty}^{\infty}d\vec{p} \int_{-\infty}^{\infty}d\vec{x}\; e^{-\beta H}$$
where
$$ H = \sum_{i=1}^{N} \frac{p_i^2}{2m_i} + \sum_{i=0}^{N} m_i\omega_i^2(x_{i+1}-x_i)^2 $$
and $x_i$ denotes the displacement from equilibrium of the $i$th block, and $x_0=x_{N+1}=0$ represent the walls at the ends.
I was able to derive the expressions for the free energy $F = -\frac{1}{\beta}\ln Z$ for one, two, and three blocks with identical masses and identical springs (with the hope of seeing an extendible pattern) but sadly no obvious patterns emerged. The calculations are also rather tedious.
I don't doubt that this has already been done many times before -- does anyone have a reference to a solution?
| Make the change of variable to $\delta_i = x_{i+1} - x_i,\; i=0\dots N-1.$ Then the system is uncoupled and $Z = \prod_i z_i$ with
$$ z_i = \int e^{-\beta p^2/2m_i}dp \times \int e^{-\beta m_i\omega_i^2\delta^2}d\delta = \frac{\pi\sqrt{2}}{\beta\omega_i}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Why aren't the weights of the beads considered in this equation? I was solving this problem:
A ring of mass $M$ hangs from a thread and two beads of mass $m$ slide on it without friction.The beads are released simultaneously from the top of the ring and slides down in the opposite sides.
We are asked to find the condition on $m$ such that the ring will move up during the motion of the beads.
Now I wrote down the equation
$$ N + mg\cos\theta =\frac{mv^2}{r} $$
where $N$ is the normal reaction force provided by the ring (I am working in the frame of reference of the bead)
and by using the work energy theorem I get
$$ \frac{mv^2}{r} = 2mg(1-\cos\theta)$$
After that, by solving for $N$, I take the downward component of $N$ and multiply it by $2$ for the two beads so it becomes $2N\cos\theta$ which provides force to lift the ring up. Now differentiating and finding maximum force for corresponding $\theta$, we get
$$F_\text{max} = \frac{2mg}{3}$$
Now my question is, I will get the correct answer which is $m>3M/2$ if I use $F>Mg$ where $F = 2N\cos\theta$, but shouldn't I write it as $F>(M+2m)g$ considering the weight of the other two small beads sliding upon the ring?
| Your confusion will be removed if you consider the FBD of the ring itself.
(Mg acts from center of mass of the ring.)
I think you can now see why the weight of the beads is inconsequential while determining equilibrium condition of the ring: regardless of the various forces felt by the beads, the ring itself feels only the following forces:
1. Weight
2. Tension of the ring
3. Normal reaction from both the beads
Hence the weight of the beads is not to be used in the comparison: F>Mg
NOTE:
Do, as an additional exercise, check whether the beads will tend to "fly off" from the ring before or after the ring will tend to move upwards(though logic says it WILL happen so).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A simple derivation of the Centripetal Acceleration Formula? Could someone show me a simple and intuitive derivation of the Centripetal Acceleration Formula $a=v^2/r$, preferably one that does not involve calculus or advanced trigonometry?
| In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant.
So, when the object has to transverse a curve trajectory, the main requisite is the introduction of a force that manipulates the direction of velocity such that the resultant locus is the required curvilinear path, otherwise the object would travel straight.
The direction of the force is evidently the direction of the acceleration or the limit of change of velocity with respect to time. So, to find the direction, let's think of an infinitesimal situation.
Let for a short amount of time $\Delta t$, the distance traveled is $v(t)\Delta t$ along a circular arc of radius $r$. The angle transversed is then $$\Delta\theta = \dfrac{v(t) \Delta t}{r}$$.
Imagine the bisector of the angle. Now, consider the changes in velocity parallel & perpendicular to this bisector. Initially the velocity has a component $v\sin(\frac{\Delta\theta}{2})$ away from the center & $v\cos(\frac{\Delta\theta}{2})$ transversely.Afterwards, it has a component $v\sin(\frac{\Delta\theta}{2})$ toward the center & $v\cos(\frac{\Delta\theta}{2})$ transversely as before. Thus the change of velocity is of magnitude $2v\sin(\frac{\Delta\theta}{2})$ toward the center of the arc.
As $\Delta\theta$ is vanishingly small, $\sin(\frac{\Delta\theta}{2})$ becomes indistinguishable as $\dfrac{\Delta\theta}{2}$. Thus, we can put $$|\Delta v(t)| = v^2 \dfrac{\Delta\theta}{r}$$. And the direction is towards the center. Thus the force is rotating the position vector along the curved trajectory and the change is radially inward regardless of whether it is traced clockwise or counter-clockwise.
The picture becomes more vivid if we calculate using polar co-ordinate.
First, we write the position vector as $\mathbf{r} = r\cdot \mathbf e_r$ . Now consider the change of $\mathbf{r}$ with time. Its change during $\Delta t$ is $r\Delta\theta \cdot \mathbf e_{\theta}$. $e_r \, \textrm{and} \, e_{\theta}$ are mutually perpendicular, the first being outward radially from the center . Therefore velocity is $$\mathbf v = \dfrac{\mathrm d\mathbf{r}}{\mathrm dt} = r\frac{\mathrm d\theta}{\mathrm dt}\cdot \mathbf e_{\theta} = \omega r \cdot \mathbf e_{\theta}$$.
By putting $r = 1$, we get $$\frac{\mathrm d}{\mathrm dt} (\mathbf e_r) = \omega \mathbf e_{\theta}\;.$$
Similarly, a change of $\theta$ implies a change of $\mathbf e_{\theta}$.It can be seen that $$\frac{\mathrm d}{\mathrm dt} (\mathbf e_{\theta}) = -\omega \cdot \mathbf e_r\;.$$ Now we differentiate the velocity, $$\mathbf{a} = \omega r \dfrac{\mathrm d}{\mathrm dt} (\mathbf e_{\theta}) = -{\omega}^2 r\cdot \mathbf e_r \;.$$ This result falls down automatically with the correct direction which is opposite of $\mathbf e_r$ i.e. toward the center radially.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Why are angles dimensionless and quantities such as length not? So my friend asked me why angles are dimensionless, to which I replied that it's because they can be expressed as the ratio of two quantities -- lengths.
Ok so far, so good.
Then came the question: "In that sense even length is a ratio. Of length of given thing by length of 1 metre. So are lengths dimensionless?".
This confused me a bit, I didn't really have a good answer to give to that. His argument certainly seems to be valid, although I'm pretty sure I'm missing something crucial here.
| Meter refers to something quite physical. Two people should be able to measure something called a "meter" and agree they are the same. NIST says:
The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.
Angles come in units e.g. degree or radian. $1^\circ$ is $\tfrac{1}{360}$ of a full rotation of a circle. Hopefully we can all agree on what that means. Possibly not.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 12,
"answer_id": 9
} |
Why are heavier nuclei unstable? If you have more neutrons than protons, then there will be more strong force present to counteract the repulsive forces between protons. Why is it that above bismuth, no nucleus is stable, regardless of its N:Z ratio?
| The reason is because the strong force isn't cumulative but the electromagnetic force is. Now, the strong force is a bit more complicated as it does change based on the number of protons and neutrons, but it doesn't build continuously as more protons or neutrons are bound to the nucleus, but the electromagnetic force does.
Say you have a helium atom, 2 protons, 2 neutrons, each is tightly bound by the strong force and the 2 protons are only repelled only by each other. So, it's 1 strong force of attraction, and 1 electromagnetic force of repulsion and the strong force wins. The strong force is 137 times stronger
Now, take Uranium, 92 protons. Each Proton and Neutron is bound to the nucleus by the strong force, but it's only 1 strong force attraction but each proton is now repulsed by 91 other positively charge protons. Hence you have 91 little forces pushing it away. This is much less stable.
Quantum instability always happens at 83 or more protons (1-82 are mostly stable, except for Technetium and Promethium, with 43 and 61 protons respectively), which in and of itself is rather curious. The strong force binds tighter with specific combinations, and as a general rule, even numbers of protons are more often stable than odd numbers. I'm not sure why that is but it seems consistently true.
More on that here: https://en.wikipedia.org/wiki/Even_and_odd_atomic_nuclei
and, loosely related, here: http://io9.com/the-oddo-harkins-rule-shows-the-universe-hates-the-odd-1446581327
Also, please see Rob Jeffries answer, as he mentions the weak force - I think his answer is more correct than mine.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Do you encounter more photons (per unit time) when moving forwards at a constant velocity? Let's say you have rain hitting you evenly on all sides (not very realistic, I know). If you were to move forwards at a constant speed, there would be more droplets of rain hitting you per second on your front, since the relative speed of droplets in front of you has increased.
Now, if you were to have photons 'hitting' you evenly on all sides and you move forwards at a constant speed, surely the relative speed of the photons in front of you WON'T increase (since light travels at the same speed to all observers), and therefore photons will still be hitting you evenly on all sides (per second).
However the searchlight effect seems to disagree with my conclusion. What have I done wrong?
| You are right in that the speed of light doesn't change. It is a completely different effect to the rain drop analogy. If you had only light hitting you directly from the front and directly form the back, you would observe the same intensity in the moving frame (only blue/red shifted). But for light coming at you from an angle $\theta_s$ in the rest frame, the angle changes when you move with velocity $v$, the new angle in the moving frame is:
$ \theta_o=arccos(\frac{\cos \theta_s-\frac{v}{c}}{1-\frac{v}{c} \cos \theta_s})$
such that the angle shifts toward the front direction. And if you now had a uniform intensity in the rest frame, in the moving frame more light comes from the (general) forward direction and less from the (general) behind direction. Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Kinetic Theory of Liquids I am familiar with the Kinetic Theory of a gas, where atoms or molecules are in relatively high-speed, random motion, and the bulk properties of the gas are determined by aggregations of these particles - eg. averaging the particle velocities to determine bulk velocity.
I am curious of if, and how, this model applies to fluids, where the particles are much closer together, and intuitively shouldn't have nearly as much space to fly around and past each other. Can this theory still be applied, or is it no longer valid?
EDIT: Further reading shows that the kinetic theory is based on the Boltzmann equation, which assumes only binary collisions between particles (dilute). Can the Boltzmann equation be used to model liquids also? If so, does the binary collision assumption affect things?
| Max Born and Herbert S. Green developed a kinetic theory of liquids in the late 1940s. However, as they say in the introduction to their first paper on the topic, the kinetic theory of liquids cannot use the simplifying conditions of the kinetic theory of gases (low density) or solids (spatial order). As a result, their theory is much more difficult than either, and I do not know if it is used in practice.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Does center of mass affect how an object falls? Suppose you drop an object which has two ends, of which one is heavy and the other is pretty light. Will the object fall with its heavier end downward or with the lighter one? Why does it happen?
| Drop a piece of paper and it glides sideways as well as flips. So aerodynamics (and hence the shape) affect the way things fall.
Specifically aerodynamic forces have a center of pressure, which when ahead of the center of mass the body would rotate and flip, but if behind it will swing and stabilize at this orientation. This is the reason arrows, darts and rockets have fins.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Can anyone explain the unit for rate of expansion of universe? If you google for 'what is rate of expansion of universe' you get
Space itself is pulling apart at the seams, expanding at a rate of 74.3 plus or minus 2.1 kilometers (46.2 plus or minus 1.3 miles) per second per megaparsec (a megaparsec is roughly 3 million light-years).
Can anyone please explain what does the unit mean? what is 'per second per megaparsec'?
I believe x km/s would suffic the rate of expansion. Not sure what 'per megaparsec' specifies?
| The hubble relation is:
$$v = H d$$
where $v$ is the velocity of the galaxy relative to the Milky way, and $d$ is the distance of the galaxy relative to the milky way. The velocity is measured using redshift. The distance is measured through a complicated series of standard candles, along with the relationship $I = \frac{I_{0}}{4\pi r^{2}}$.
If you notice, these are related by Hubble's constant $H$. This tells us how fast something is moving apart, based on how far away it is. Namely, it's the rate of expansion of the universe. Since astronomers measure distance in a unit called a parsec, and velocity at these high speeds is naturally measured in km/s, we can see that the "natural" units astronomers use for $H$ are (km/s)/parsec., since that's the only way to get km/s out when you multiply by something measured in parsec.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Where does the $(\ell + x)^2\dot\theta^2$ term come from in the Lagrangian of a spring pendulum? I am reading some notes about Lagrangian mechanics. I don't understand equation 6.9, which gives the Lagrangian for a spring pendulum (a massive particle on one end a spring).
$$T = \frac{1}{2}m\Bigl(\dot{x}^2 + (\ell + x)^2\dot{\theta}^2\Bigr)\tag{6.9}$$
I don't understand where the component $(\ell + x)^2\dot{\theta}^2$ is coming from. If we say the $x$-component is radial and $y$ is tangential, so we have according to this $\vec{v}^2 = v_{x}^2 + v_{y}^2$, then $y = (\ell + x)\sin\theta$ by small angle approximation we have $y = (\ell + x)\theta$, but then if we choose this coordinate system then $V(x,y)$ equation doesn't make sense specifically the potential from gravity! If someone could shed some light into this that would be nice.
| In general you can write the kinetic energy of a free particle as:
\begin{equation}
T = \frac{1}{2} m \,\vec{v}\cdot\vec{v}
\end{equation}
which holds whatever coordinate system you choose (could a physical quantity such the trajectory of a particle depend on the coordinate system that you choose?).
We can rewrite this:
\begin{equation}
T = \frac{1}{2} m \,\vec{v}\cdot\vec{v} = m/2 \sum_{\mu ,\nu} v_{\mu}v_{\nu}\left( \vec{e}_\mu \cdot \vec{e}_\nu\right)
\end{equation}
where $\vec{e}_\mu$ is the $\mu^{th}$ basis vector. In particular, remembering that in polar coordinates $\vec{e}_r \cdot \vec{e}_r= 1$, $\vec{e}_r \cdot \vec{e}_\theta= 0$ and $\vec{e}_\theta \cdot \vec{e}_\theta = r^2$ you get:
\begin{equation}
T= m/2 \left( v_r^2 +r^2 v_\theta^2\right)
\end{equation}
which is your expression if you call the $r$ coordinate $x$ and the $\theta$ coordinate $y$ and you make a radial translation of the origin of your coordinate system of a quantity $l$ in order to center it properly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is the most efficient way to use a blow torch? Let's start with a torch and a piece of titanium. What is the fastest way to get the titanium up to red hot? I am not going for getting the Titanium malleable, just red hot.
Whenever I ask my science-y friends they like to point out that the bright blue tip is the hottest point in the torch flame.
But this doesn't take into account heat transfer. Aren't we better off with having more flame spread out on the metal transferring heat over a larger area even if that flame isn't at the most extreme possible temperature? What is the optimum balance between heat transfer area and temperature?
| That's not an easy one...
First of all you must know the final temperature you are trying to achieve, then you need to choose a heat source and a way to trap heat where you need it
I'd sugest you build a "soup-can forge" or something like that and use a MAP-gas torch.
A J23 ceramic hoven brick might also be a good choice, maybe even easier to make.
With that setup I manage to get steel to about 800°C to heat treat small knives I make.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Why are tidal forces pointing away from the Moon?
I am currently reading The Science of Insterstellar, which explains most things very well, but some things leave me confuzzled, which I hope to get answers to here. I am no physicist, but highly interested and eager to learn.
I get tidal forces in principle, but one thing never clicked for me, and that is why tidal forces are the reason that the opposite side to the other object (moon, black hole, whatever) point away from said object.
Take this picture from wiki;
https://en.wikipedia.org/wiki/File:Field_tidal.svg
I understand all vectors up to the middle of the circle, where they point inwards and ever so slightly to the right.
Everything left from that is a mystery to me. How can the sum of all the forces pointing towards the object on the right result to vectors pointing away from it?
Even when I remembered that the object itself also has gravity in itself, all those vectors should also point inwards to the object.
How can the vectors (which I understand to be the sum of all gravitational forces ... ?) point away from the mass?
| This link explains it:
The Earth experiences two high tides per day because of the difference in the Moon's gravitational field at the Earth's surface and at its center. You could say that there is a high tide on the side nearest the Moon because the Moon pulls the water away from the Earth, and a high tide on the opposite side because the Moon pulls the Earth away from the water on the far side. The tidal effects are greatly exaggerated in the sketches.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can all the theorems of classical mechanics be deduced from Newton's laws? As above, is the whole edifice of Newtonian mechanics built upon Newton's three laws of motion? Can I deduce all the theorems without referring to further assumptions?
| If "for every action there is an equal and opposite reaction" (F12=-F21) allows friction as a legitimate reaction, then no. Non-conservative forces like friction are not fundamental and would allow non-conserved energy and momentum if additional constraints are not included (like "heat" from thermo). They are not time-reversible. You can't derive Langrangian and Hamiltonian mechanics unless you use the modern mathematical form of Newton's 2nd law that forces conservation of forces. The 4th mathematical expression of mechanics, stationary action, is the only one that can solve certain non-equilibrium problems, according to a paper I saw. So many places on the internet want to say Langrangian and Hamiltonian mechanics can be derived from "Newton's laws" and do not even mention stationary action, but there are some important details not mentioned. See Feynman's "Principle of Least Action" chapter in the red books.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Does anti-matter increase or decrease in entropy over time? Antimatter is matter going backwards through time. From the perspective of a matter-based observer does antimatter:
*
*Increase in entropy (and therefore decrease in entropy in its own time)
OR
*
*Decrease in entropy (and therefore increase in entropy in its own time)
Option 2 would seem to explain why we don't see much anti-matter (it all went into energy.)
Which is correct? Does antimatter increase or decrease in entropy over time?
| The answers given are speculative, as the answer to your question is unknown. The thermodynamics of antimatter is an open question in physics, and future experimentation will be required to reach a definitive answer. If CPT (charge, parity, time) symmetry is true, then antimatter would be expected to behave has though time is reversed, that is if we observed a system of only antimatter it would appear as though its entropy change over time was strictly less than or equal to zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Noether's theorem: meaning of transformation of coordinates I have a question regarding Noether's theorem. In our introductory QFT class (which is based on the book by Michele Maggiore) we have derived the Noether currents in the same form as displayed in this post: Question about Noether theorem
In this formula, there are contributions from two different kinds of transformations: a transformation of the field alone and a transformation of the coordinates.
My problem is: I don't understand the meaning of the transformation of coordinates. I have tried to understand the derivation from different QFT books (and I haven't found the same derivation twice, which doesn't make it easier) in the hope that I then would better understand the premises, but unfortunately I have not succeeded so far.
Also Peskin/Schröder for example only discuss transformations of fields and don't mention the transformation of coordinates at all. Poincare symmetry, which is in most books treated as a transformation of the coordinates, can be treated also as a transformation of fields, as shown in the answer to the following question for pure translations: Noether's Theorem: Foundations.
Like the guy who asked that question, I think that the coordinates entering the action are only dummy variables. So what is then the meaning of the coordinate transformations in the prevalent formulation of Noether's theorem?
Maybe someone can give a concrete example to illustrate the idea.
| Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. OP is in this terminology essentially asking
Q: What is the meaning of horizontal transformations? | {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Does a gas in a container lose kinetic energy? When a gas is in a container, it frequently collides with the container wall, exerting pressure. However, with a collision, kinetic energy ought to be transferred from the gas molecule to the container wall. Does that mean a gas isolated in a container will lose kinetic energy on standing, and its temperature will gradually decrease?
| The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Current constraints on lightest neutrino mass? This paper from 2005 claims that the mass of the lightest neutrino is unconstrained. (see p9)
Oscillations are only able to constrain the differences in squares as far as I know, but perhaps constraints could come from cosmology or beta decay experiments.
Is there still no constraints today or are there lower limits on the lightest neutrino mass?
| there are many experiments that put different contraints on the neutrino masses. Here is a good collection from the particle data group.
To summarize: There are lots of experiments that put upper bounds on the neutron mass. The PDG groups estimate is that $\nu_e < 2eV$, $\nu_{\mu} < 0.19eV$, $\nu_{\tau} < 18MeV$. All with a confidence level of about $90\%$, that means that with a probability of $90\%$ will a measured value be below the given bound.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What would happen if we tried to run a motor in space when it is not attached to anything to provide support to it? I know that a when a motor runs it generates torque and that torque can be used to do useful work. On the other hand, the motor needs strong support that absorbs the reaction torque. In our case let us assume that that support is provide by a workshop floor on which the motor is firmly attached ( The workshop floor is essentially the earth) The earth receives the reaction torque and being the massive object it is, it doesn't move.
Now let us imagine that we took our motor into a space where there is no gravitational field. What would happen if we tried to run the motor. Assuming the motor is powered by a battery pack. The battery pack and its control electronics are neatly packed around the stator.
Would the motor rotate at all? Would the rotor and the stator rotate in opposite directions?
Would there be a transfer of energy from the batteries to rotational mechanical energy
| The astronauts working on the Hubble space telescope had to bring special low torque wrenches to counteract the effect of the torque of the motor spinning them around, due to conservation of angular momentum, although this meant far more use of muscular power to hold them in place. And also to avoid damaging the equipment they worked on, such as screws within the HST.
So the motor driven screwdriver will work, as on Earth, but it requires more care in design, better hand, foot or body positioning to brace yourself and far more physical effort to get the job done.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is Gauss' law valid for time-dependent electric fields? The Maxwell's equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ is derived from the Gauss law in electrostatics (which is in turn derived from Coulomb's law). Therefore, $\textbf{E}$ must be an electrostatic field i.e., time-independent. Then how is this equation valid for the electric field $\textbf{E}(\textbf{r},t)$ which is time-dependent (for example, the electric field of an electromagnetic wave)? Can we prove that $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ ?
EDIT: I have changed $\boldsymbol{\nabla}\cdot \textbf{E}=0$ to $\boldsymbol{\nabla}\cdot \textbf{E}=\frac{\rho}{\epsilon_0}$ in the question.
| Maxwell derived his equations from 1) charge conservation law; 2) Coulomb's law; 3) Bio--Savart--Laplace law; 4) Faraday's law of induction. The equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ was indeed derived from Coulomb's law and in its differential form is written using Gauss--Ostrogradskiy theorem. Maxwell made one step further and suggested (postulated) that the same law is true when $\mathbf{E}$ and $\mathbf{\rho}$ are functions of space AND time. It turned out to be a correct guess and it doesn't contradict the other equations. Indeed, from Bio--Savart--Laplace law Maxwell derived $\nabla\times \vec{H}=\vec{J}$. If you take the divergence of both sides of this equation it contradicts the charge conservation law, so an additional term must be added to this equation (the so-called displacement current). And then, the resultant equation will not contradict the equation we get after differentiating $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ with respect to time and, in fact, the comparison of these two equations enables us to find the displacement current.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Polarization of the Sky? I was hiking and noticed as I tilted my sunglasses (which are probably polarized), right at the horizon where the sky was against a mountain, it would change from lighter to darker blue.
This only occured right above the horizon, and I didn't see a noticeable change when looking directly up at the sky.
Is there more of a certain polarity of light coming from the horizon? If so, how can this be explained?
| Where was the Sun at the time?
The sky is viewed through Rayleigh scattered light. Unpolarised light (or rather, the electric field in the electromagnetic radiation) from the Sun can be imagined to cause oscillations in the bound electrons of atoms and molecules in the atmosphere. These electrons then remit light as oscillating electric dipoles, with a polarisation in the direction of the oscillation. The greatest degree of polarisation occurs when the scattering angle is through 90 degrees (since one of the polarisation states of the incoming solar light is effectively "unseen"). The exact relationship between viewing angle and Sun angle will be a little more complicated since the scattered light arrives from different heights in the atmosphere.
Anyway, when the Sun is near zenith, the biggest polarisation effects will be seen near the horizon.
The details are readily found here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why does humidity cause a feeling of hotness? Imagine there are two rooms kept at the same temperature but with different humidity levels. A person is asked to stay in each room for 5 minutes. At the end of experiment if we ask them which room was hotter, they will point to the room with the higher humidity. Correct right?
How does humidity cause this feeling of hotness?
| It because the higher humidity makes it more difficult to cool the body. Even though the room temperature is below body temperature, you generate more heat than needed to maintain your body temperature. To help cool you down, you sweat, and the water evaporates from your skin. The evaporation of the water cools you down.
When the humidity is higher, the sweat does not evaporate as fast, so your skin temperature goes up. This makes you think the temperature is higher when the humidity is higher.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 2
} |
In general, why do smaller guns have more felt recoil? Why is recoil easier to control on a more massive gun compared to a smaller gun with the same bullet. Presumably the bullet leaves both guns with the same momentum, but the larger gun seems easier to control. Since the momentum you have to control is the same in both cases, why do we perceive less recoil on a bigger gun?
| The larger firearm has more mass, and therefore more inertia for the recoil momentum of the bullet to overcome.
Also, small firearms may be more difficult to secure a good grip on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How and why the phrase "quark force increases with distance"? I have seen that phrase "force between quarks increases with distance" at many resources, some even relatively credible (albeit written for general audience).
What is the reason behind that, when the area law for the confining phase clearly gives the potential energy as increasing linearly with distance and hence the force being a constant?
| This is the strong interaction which sometimes called strong 'force'. It is one of four forces which
100 times stronger than electromagnetism, a million times stronger than the weak force interaction and $10^{38}$ times at a distance of femtometer.
This strong force actually explains how nucleus of atom stick together despite charge of protons. Quantum Chromodynamics QCD is a quantum field theory of the strong interactions between quarks. Gluons are the mediator of the strong force such as photons in the electromagnetic interaction. Gluon comes from word 'glue' since they glue quarks inside hadrons(protons, neutrons, pion, kaon etc.).
Color charged particles such as quarks can not be isolated, gluons always tend to bound them together. This is the phenomenon called color confinement as Ernie mentioned. That's why you can not observe any colored particles directly. Such a quark confinement by gluons makes them colorless. And all particles we observe are color-neutral. So to take quarks apart, you need to give energy however the farther apart, the more energy you give! In particle accelerators you can go beyond this energy then what happens? No they do not fall apart, hadrons generate 'jets' and color charged particles such as quarks, immediately interact with other color charged particles around to create bound states of hadrons which always color-neutral.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Conical train wheels I've been reading about how the conical shape of train wheels helps trains round turns without a differential. For those who are unfamiliar with the idea, the conical shape allows the wheels to shift and slide across the tracks, thus effectively varying their radii and allowing them to cover different distances while rotating at the same angular velocity.
A cross-sectional view of the tracks and wheels generally looks something like:
But what about a configuration like the following?
I read in an online article that wheels in the second configuration may more easily slip and derail from the tracks (assuming there are no flanges to prevent them from doing so). But I can't convince myself using physics why that might be.
Is one of these two configurations actually more reliable than the other?
| In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, it will start to lie diagonally across the track, rather than perpendicular to the two rails.
Suppose we're using the wheel profile in the first diagram. As the axle gets farther from perpendicular, the right wheel moves ahead and drops down to have a lower-radius part in contact with the rail. This means that the axle is self-straightening since the more it steers, the more it tends to push itself back towards being perpendicular.
However, if we use the second wheel profile, as the axle gets farther from perpendicular, the right wheel climbs up the rail, causing the contact point to move to an area of higher radius. That means the right wheel goes even farther in one revolution, so it turns even more. That's completely unstable.
In the second diagram, the only way the axle can return to running straight is for the whole axle to rotate clockwise about a horizontal axis, with the wheels sliding perpendicularly across the rails. That would cause a huge amount of wear to both the wheels and the track, assuming the thing didn't just derail.
There's a Numberphile video demonstrating all of this with a few taped-together espresso cups.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 6,
"answer_id": 2
} |
Obtaining central force from the potential of an off-orgin object I was trying to go through a simple exercise and was getting tripped up on the mathematical intuition of my elementary physics. I thought I should be able to get any old central force from the central potential by noting the curl of the force is zero thus there is some potential field which satisfies:
$F_C = -\nabla U_C$.
Where the C is a subscript for central. Now let's take the basic central potential and account for the central object being displaced from the origin by a vector $\vec{x}$. That is to say that
$U_C(\vec{r}) = \frac{k}{|\vec{r}-\vec{x}|}$.
Where k is just some factor. Since this is only a radial potential then I only need concern myself with the $\hat r$ component of the gradient. However going through the exercise as described you arrive at
$F_C = \frac{|\vec{r}|-|\vec{x}|cos(\theta)}{|\vec{r}-\vec{x}|^3} \hat r$
I believe I stumbled upon the subtly of the problem. The $\nabla$ is not the $\nabla$ of my coordinate system, but the $\nabla$ of the central bodies coordinate system, which I will now denote as $\nabla'$, thus it is only correct to say
$F_C = -\nabla' U_C$.
Could anyone comment on the issues at hand. The potential should not change from being conservative to no longer being conservative by simply shifting frames, so why do I have to use a special frame's $\nabla$?
| ( I turned my comment into an answer to signal that this is question is answered)
if in doubt (and unexperienced) work using Cartesian coordinates. That being said: You simply miscalculate the gradient. $F_c$ is not directed in the direction $r$ but $r−x$. If you want to compute the Gradient in spherical coordinates you have to expand the absolute value in the potential which includes a term $\cos\theta$ and thus an additional contribution. A simpler way would be to work in appropriate curvilinear coordinates, but you dont have to do that. Given that you do the calculations right (for reference: en.wikipedia.org/wiki/Curvilinear_coordinates )
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What determines the probability of creating a particular particle in a collision? When discussing events at the quantum level, we deal in probabilities and not absolutes. Articles I've read on particle physics state that a particle has a probability of being created in a collision. What determines this probability?
Assuming we have the energy and other criteria met, which would allow us to create range of particles (please feel free to expand what particles would make sense for an example), why do certain particles have a higher probability of being created than others?
| This is actually an open question.
So far what we are able to state from theoretical considerations are restrictions in terms of the conservation laws that we have observed. These tells you for example that the whole momentum in a reaction is conserved, an the mass-energy, or some quantum numbers. And this already constraints much of what can come out from certain reaction: for example a photon interacting with charged matter can originate a positron-electron pair, and while you cannot say how probable is this without measuring experimentally, you can already say that measuring the probability for electrons you will have the one for positrons because they only can come in pair and due to charge and momentum conservation.
But when it comes to predicting ratios of outgoing particles, theoretically this would need as well the full knowledge of the coupling constants among all particles, or in other words, the exact form of the interactions, and this cannot be provided so far without experimental values.
We are more or less able to state the form of our fundamental Lagrangian, writing terms for all kinds of interactions, but we are unable to give values to the constants without experimental evidence.
So what we do so far is: we elaborate constraints from theory based on our current understanding, this helps processing the experimental information and deducing unknown yields based on the known ones, and find relative yields between particles which gives these probabilities.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can a magnet damage a compass? I've heard the claim before that a magnet can ruin a compass, and was about to repeat it to my son when I realized it sounds like complete nonsense. Googling turned up such unsubstantiated and illogical answers as this one and unanswered questions as this one but nothing that sounded reasonable to me and gave a convincing explanation. Perhaps my Google bubble is at work.
Anyway, since SE is generally very reliable, I thought this was the right place to ask, before I pass on untested nonsense to my son. Help me break the chain of untested pseudoscience via oral tradition: does a magnet actually do permanent damage to a compass, or just temporarily prevent it from detecting magnetic north?
If it actually does do this, please explain how that is so.
| A magnet is made by aligning the magnetic poles of all the molecules in a magnetic material so they all point the same way.
If you have a look at how a magnet is made, traditionally
http://www.princeton.edu/ssp/joseph-henry-project/permanent-magnet/
or commercially
http://www.arnoldmagnetics.com/Magnet_Manufacturing_Process.aspx
you will see it is made by applying a magnetic field to a magnetic material in which these molecules are able to move (either a fine powder in modern manufacturing or a ductile/malleable material).
Equally you can reverse the process, immersing a material whose molecules are able to move (like a metal, but not most sintered magnets which are held in crystal lattice) in a magnetic field that is not aligned with its existing field will cause the molecules to move, REMOVING the alignment and hence the existing magnetization.
Or in other words - yes - you can even reverse the compass if you try hard enough.
Edit - i noticed a comment that the compass needle can move and hence align itself with the external magnetic field, but of course a compass has only one degree of freedom of motion and can only align itself on that axis. Many are also unable to rotate freely unless held in an upright position.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/196996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Wavelength vs Wavenumber etiquette When am I supposed to use the terminology of EM "wavenumber", instead of "wavelength" (or frequency)?
The concepts of wavelength and frequency are no problem for me, but wavenumber (number of wavelengths per unit length) seems redundant to me as a student engineer and proto-physicist. And then there's use of energy levels at higher frequencies.
| I don't think there's much to say beyond the obvious: You should use whatever terminology is most helpful in communicating the information that you want to communicate.
*
*That has to do with the audience you're talking to. Just like how you use °F when talking to Americans and °C when talking to non-Americans ... similarly it's often wise to use cm^-1 when talking to IR spectroscopists, and nm when talking to visible-light photographers, and eV when talking to atomic spectroscopists and so on.
*It also has to do with the concept you're talking about. Just like how you use frequency when talking about aliasing and the Nyquist limit but time-durations when talking about leap-seconds, and meters when talking about height vs diopters when talking about lens design ... similarly it's often wise to use wavenumbers when talking about the Laue equation for diffraction, and wavelengths when talking about the size of a resonant cavity or etalon, and so on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/197085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Do we regularly measure and update our physical constants (By using the LHC)? This question is motivated by sheer curiosity. I certainly do not expect that the free parameters we use in the standard model have changed in value since we started measuring them with a "modern" degree of accuracy.
It would seem to me however, that as our experimental data accumulates in both quality and quantity, that it is important to know the physical constant values we input into our models as precisely as we can.
My question is, as I know very little about experimental physics, (and to working physicists this is probably an obvious question I am asking), but is there a law of diminishing returns in knowing the value of say, the mass of an electron to the 21st decimal point?
Is it important to continually refine the constant values to make the most, for example, of seemingly minor discrepancies in the results from the Large Hadron Collider?
I hope I am making myself clear and I apologise if I am not.
My point is: if the LHC produces an unexpected, repeated but extremely subtle result in an experiment, are we confident enough in the correctness of our present measurements of physical constants that we can draw conclusions other than measurement errors in them?
I am pretty sure the answer is yes, and that perhaps some of the results from the LHC actually serve to refine the values of the constants to reduce the error bars of the current values even more.
| Every few year the Committee on Data for Science and Technology (CODATA) publishes recommended values of the fundamental constants, see http://www.codata.org/. They use the most accurate experimental results available, so yes, the values of the fundamental constants can -- in principle -- change. But what is more likely, is that the number of significant digits in these values increases.
What is important to note is that not only high-energy experiments are used to determine these constants, but also "low-energy" spectroscopy experiments or a combination of the two.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/197492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How light splits up into different colours when passed through prism using QED? I want to know how light gets splits up into different colours when it is passed through prism? How light interacts with atoms and electrons of the prism? Can someone explain this to me using Quantum Electrodynamics?
| As you know white light is made of many different wavelengths. Each wavelength of light takes a slightly different path through the prism. In QED we have a spinning clock which turns at different rates for different wavelengths of light. We calculate the probability of a certain path the light can take using these clocks which have associated probability amplitudes. It turns out that because blue light has a clock which turns at a different rate than red that it bends more or rather is more likely to travel through the prism along a more bent path.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/197677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I know what variable to use for the chain rule? In my textbook the tangential acceleration is given like this:
$$a_t=\frac{dv}{dt}=r\frac{dw}{dt}$$
$$a_t=rα$$
I understand that the chain rule is applied here like this:
$$a_t=\frac{dv}{dt}=\frac{dv}{dw}\frac{dw}{dt}=rα$$
What I don't understand is why we have to apply the rule in this specific way. Say I write like this:
$$a_t=\frac{dv}{dθ}\frac{dθ}{dt}$$
This way, I end up with entirely different result. How do I know how the chain rule must be applied?
| In the case of a circular motion, the tangential velocity $v$ can be expressed in terms of $\omega$ and $r$, i.e. $ v= \omega r$.
The formula in your textbook $$a=r\frac{d\omega}{dt} $$is obtained by taking the derivative of the previous equation, while taking into account that $r$ is constant.
Of course, you could consider $v$ as a function of $\theta$, i.e. $v=v(\theta)$. However, there is no simple, generally applicable rule like the one above relating $v$ and $\theta$. So, the formula $$a=\omega\frac{dv}{d\theta}$$ is correct but does not help us a lot.
In case of a circular motion with uniform acceleration, we get $$\theta = \frac{1}{2} \alpha t^2$$ or $$v=r \sqrt{2\alpha \theta}$$
Applying now your alternative method (derivative to $\theta$), leads to $a=r\alpha$, just like your textbook formula does.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/197783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why the optical gap is not identical to the charge gap? The optical gap is the photon energy required to create an exciton (in a solar cell for example). The charge gap (aka electrical gap) is the energy (voltage) required to create a photon (in an LED for example). Why are these two gaps not identical?
Ref:
Undoped (neutral) conjugated polymers are semiconductors, with optical
gaps of ~ 2-3 eV and charge (or band) gaps typically ~ 0.5-1.0 eV
higher in energy, reflecting the large exciton binding energies in
polymers. (source: Barford, 2013)
(source)
See also: What is the basic difference between optical band gap and electrical band gap?
| The difference between the fundamental gap (following IUPAC definitions - and your diagram ) (this is what you refer to as your electronic gap in your question) exists because the optical gap corresponds to the energy of the lowest electronic transition accessible via absorption of a single photon. The optical gap is generally substantially lower than the fundamental gap.
The reason for this is that that in the excited state the electron and the hole remain electrostatically bound to one another. This means the magnitude difference between the fundamental gap and the optical gap is related to the binding energy.
The fundamental gap corresponds to the different energy caused due to the sharp bound between eigenvalues in the Schrodinger operator.
So in the diagram you have provided, the fundamental gap is labelled $E_{fund}$, the electron-hole pair binding energy $E_{B}$ is given by $E_{fund}-E_{opt}$.
Non Technical Summary
The optical gap only lists an electron over the band gap and you must account for the electrostatic effects between your hole and the electron.
The fundamental/electronic gap is the proper quantum energy difference between the energy levels..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/197940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Will a gas rotate as fast as the spherical container it is contained within? Let's say I have a sealed spherical glass container 30 cm in diameter which contains plain air.
The glass container is rotated about its axis at 1 revolution per minute.
My question is, would the gas also rotate at 1 revolution per minute within this sealed glass container?
| Given viscous effects, that is indeed the case if:
*
*We are looking at a steady state system
*It is not a rarefied gas (i.e. at pressures much lower than atmospheric)
*The walls of the container are rough enough such that the gas molecules don't 'slip' over the surface but can be assumed at the same angular velocity (i.e. the no-slip condition)
Point 2 is important because when rarefied, gas molecules will simply bounce around without interacting with each other which is not a very effective method of momentum transfer.
In general, it is rarefied if the so-called Knudsen number:
$$\text{Kn}=\frac{\lambda}{L}$$
where $\lambda$ is the molecular length scale (mean free path of the molecules) and $L$ is the characteristic length scale of the system. If these are comparable (i.e. $\text{Kn}\sim1$), molecules rarely interact through collisions and we consider the gas rarefied. If the mean free path is much smaller than the system (i.e. $\text{Kn}\ll1$), then molecules often interact with each other through collisions in which they exchange momentum. In that case we can consider the continuum approximation to fluid dynamics; we can consider flow fields rather than individual molecular motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/198007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to estimate if an image is in focus I am building a test measurement (optics) to look a rectangular slit opening (1mm x 15 microns). The slit opening is illuminated by a white LED and using a microscope objective to magnify it to 10X on an image sensor.
What would be the possible method to estimate if the image is in it's best focus?
Current method, using a line profile and look for peaking position.
Any other ideas ?
| You could try using contrast. The contrast in the image is highest when the image is in best focus. You can analyze the contrast information and move the object, lens, or sensor into a position that gives the maximum contrast value.
The downside to this is that you will need to typically adjust your opto-mechanics through the full range of motion to find the peak contrast. In other words, you may need a brute force search to find the maximum contrast.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/198098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is heat conduction impeded at interfaces between dissimilar materials? Sound in air essentially echoes off concrete walls, rather than penetrating them, because of the difference in the material properties of air and concrete.
By analogy, are there pairs of solid materials where their interface would be very inefficient at propagating heat? Perhaps one material has heavy atoms and soft bonds and the other has light atoms and stiff bonds, and neither has free electrons.
If this phenomenon exists could it be used to create super-insulators, by laminating together large numbers of very thin layers of the two materials?
| Conduction across the interface between two materials mostly depends on how well they are in contact. If you want to make a very good insulator you try and ensure the surfaces aren't in good contact. The ideal is no contact at all - hence the thermos flask.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/199357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Infinite dimensional manifolds in general relativity In GR the concept of a manifold is very useful. However, all of these manifolds are of finite dimension. Is it possible to define a manifold with infinite dimension (ie much like Hilbert space in QM) such that we can still define terms such as curvature, tensor fields and other such concepts that appear in GR on this manifold?
If such a manifold exists does it have physical applications?
| Technically speaking, manifolds are by definition topological spaces, which resemble locally an inner-product space. Since there are vector spaces (with a dot product) of infinite dimension, then there shall be infinately-dimensional manifolds as well. The infinity of the dimension is not a problem for the tensors as well - each multi-linear function over a vector space can be called a tensor... The same holds for the connection, the curvature, the holonomy, and others - the only downside is that we are no more able to write them via a finite set of coefficients.
(PS Furthermore, any topological space that locally resembles some space A can be called a "manifold of type A" - e.g. Hilbert manifolds, Banach manifolds, etc. So definitely, the answer to your question is affirmative.)
Regarding whether those generalised manifolds are applicable in physics - I guess so, but unfortunately I cannot give you any examples.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/199422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why is it easier for the ocean to push someone over by exerting force on their front side than by exerting force on their left or right side? If I put two clones of normal weight on a beach in the ocean, with one standing perpendicular to the waves, and the other standing parallel to them, the one standing perpendicular to the waves will be a lot harder for the waves to push over than the one standing parallel to the waves. Why is that?
Is it because there's more "stuff" from shoulder to shoulder than from bellybutton to backside?
| The main reason for this difference in stability is the orientation of the feet of the clones. If you were to do this experiment on dry land by simply putting sandbags on someone's feet (the water makes it hard to move your legs fast enough to catch yourself, and so do the sandbags) and pushing on them from different angles, you would find that they fall over much easier forward and backward than side to side. This is because gravity exerts more restoring torque about a foot spaced out to the side than about the ankle hinge, which is directly below the clone's center of gravity.
If the clones are standing within the point where the waves break, then there is also the effect that you hypothesized, that the flowing water of the wave exerts more force on the larger area of your front/back than on the smaller area of your sides. This effect is relatively small, however. Also, if the clones are standing in the swell outside where the waves break, this effect is hardly present at all, since most of the force toppling the clones in the swell comes from variations in the depth and pressure of the water rather than its lateral flow. These forces are unaffected by the orientation of the clones.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/199655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Can there be eternal stars? the question is quite straightforward:
*
*Can there be stars that shine forever without ever collapsing nor growing?
*Do we know some really, really old stars? (whatever age that might be)
I hope to get answers from physicists, as for the nuclear reaction constraints involved; but I'm also looking for the point of view of cosmologists and astrophysicists.
| The word thing you are looking for is "Black Dwarf" stars. Which are White Dwarf stars which have cooled to match the temperature of the cosmic background. Since this is likely to take more than the current age of the universe, there aren't any. These will exist forever, unless hypothetical proton decay finishes them off or a hypothetical Big Rip due to Dark Energy does so.
Shining forever isn't going to happen because mass into energy would still destroy them no matter what mechanism existed
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/199742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Adding fluctuations to a hydrodynamic simulation to trigger instabilities I'm doing a 2D combustion hydrodynamic simulation and there's a hydrodynamic instability that should be triggered because of the particular physical properties of my system. The key to the instability are temperature gradients perpendicular to the flow direction of my problem. What is a good and "consistent" way to artificially add fluctuations in this perpendicular direction to trigger this instability?
| I think the answer would have to depend on the nature of the simulation. I'm guessing it is some sort of flow simulation in which fluid "enters the system" at some point and exits at some other point. If so, there must be parameters to do with the fluid at the intake which you have control over. Is $\rho(\mathbf{x})$ (fluid density as a function of position) at the intake something you can control? In that case you could create an oscillation where $\rho$ is higher on one side of the intake at one time and higher on the other side of the intake half a period later.
The resulting density gradients (perpendicular to the flow) would result in corresponding temperature gradients as long as the fluid is modelled as being in local thermodynamic equilibrium. In real life density fluctuations like this occur when fluids enter apertures at high (but subsonic) speed. It is probably related to whichever instability makes the thumping sound that you hear when you drive on the highway with only one window open in your car.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is really a light ray? I've been studying geometric optics and I'm still a little confused with this idea of light ray. In the book I'm studying everything is being done starting from Fermat's principle which states that the ray path is the one which extremizes the optical length.
Now, I thought that because of this, geometric optics viewed light as particles and I thought that a light ray then would be a trajectory of those light particles. But my question here confirmed that geometric optics doesn't care if light is made of particles or waves.
On the other hand, if we think of light ray as "the path taken by light between two points", this is no good, because the idea of trajectory in this sense is not well defined for waves. Indeed, this idea of light ray doesn't even seen clear to me, because as we know if we have some light source, the light doesn't go along one single path, but as a wave it spreads out in all of the directions at once.
So, if geometric optics doesn't care about if light is made of particles or waves, what really is a light ray? How should we understand light rays and ray paths?
| In a wave picture we can put an arrow at each point of the wave that points in the direction of propagation which is (normally) at right angles to the wavefront. If we string a bunch of these arrows together into a line, we get a ray.
I tend to think of this as being analogous to field lines in electromagnetism. In fact you can take this analogy a bit further and see that wavefronts are behaving a lot like equipotentials and the phase of the wave is playing the part of voltage.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Definition of a Supercluster A group of astronomers in September 2014 redefined what classifies a supercluster. Before this, the supercluster where the Milky Way resides was the Virgo Supercluster. Now, the Virgo Supercluster & three other previously defined superclusters are simply 4 lobes of the newly-defined Laniakea Supercluster.
To my understanding, this redefinition was prompted by the new information that outside of the Virgo Supercluster, the other 3 nearby superclusters are themselves being gravitationally pulled by & have "peculiar" velocity in the direction of the Great Attractor (in the center of the Virgo Supercluster). Therefore these 4 superclusters together form a loose structure that has been deemed a more fitting definition of a supercluster.
While I'm comfortable with that logic, I don't understand why the Virgo Supercluster & the other 3 still are known as superclusters. Laniakea essentially is a supercluster that contains superclusters, by that reasoning. It seems contradictory to make a new definition that overrides the last, yet have the last definition still apply.
So, what was the previous definition of a supercluster, & is there a verbatim updated definition for new superclusters somewhere? Is it possible for these both to apply simultaneously?
| According to here, there was no precise definition before this group redefined what it meant for a group of galaxies to constitute a supercluster--before their redefinition, it seems it was just loosely defined as "extended regions with a high concentration of galaxies." They now define a supercluster to be a volume in which "the motions of galaxies are inward after removal of the mean cosmic expansion and long range flows"--basically, their new definition says a supercluster is a collection of galaxies that are tending to move closer together. When referencing the Virgo Supercluster (in their paper called the Local Supercluster) they refer to it as the "historical Local Supercluster," implying that it indeed should no longer be classified as a supercluster and instead just as a lobe of a supercluster. The people who wrote the paper have a nice video here!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Collision of Inelastic ball above the ground
An inelastic ball of mass $m$ is dropped from a height $h$ above the ground and at the same time a second ball of mass $m_1$ projected vertically upwards to meet the former. Show that in order that immediately after collision, the balls may be at rest, the 2nd ball must be projected with a velocity $$\sqrt{\frac{m+m_1}{m_1}gh}$$
Attempt:
Let $u\implies$ Vel with which the 2nd ball is projected.
Let $v_1, v_2\implies$ Vel of 1st and 2nd ball just before impact at a height $d$ above the ground.
Then $v_1^2=2g(h-d)$ and $v_2^2=u^2-2gd$
From the principle of conservation of linear momentum, $mv_1+m_1\times (-v_2)=0\implies mv_1=m_1v_2$
We have to find $u$.
I am unable to get $u$ as provided in the question. Always $d$ or $v_1$ are coming in the answer and unable to eliminate.
For details,
$mv_1=m_1v_2\implies m^2v_1^2=m_1^2v_2^2\implies m^2\times 2g(h-d)=m_1^2(u^2-2gd)$
$d$ is there ???.
Please help.
| Let $t$ be the time from the dropping the fist ball until the collision of the balls. Then, $v_1=gt$ and $v_2=u-gt$. Moreover, $d=ut-\frac{1}{2}gt^2$ and $h-d=\frac{1}{2}gt^2$, so that $ut=d+(h-d)$, which gives $t=\frac{h}{u}$. Because $mv_1=m_1v_2$, we must have $m\left(\frac{gh}{u}\right)=m_1\left(u-\frac{gh}{u}\right)$, or
$$mgh=m_1\left(u^2-gh\right)\,.$$
The rest is your job.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Poincaré Recurrence and Immortality If, as Luboš Motl says, Poincaré recurrence is relevant for our universe, does this mean (1) that, after I die, I'll one day live through my life again after the same physical pattern that is currently me reconfigures and (2) that I'm thus immortal because this reconfiguration of my pattern will happen endlessly? (Please see Is Poincare recurrence relevant to our universe?)
| This is more a philosophical answer, but it is also more a philosophical question since it is actually the question how you define Immortality.
If you were really reborn after $10^{10^{10^{120}}}$ years, would it still be you? Of course, it will be an identical version of you, but is it really the same you as it was $10^{10^{10^{120}}}$ years ago? This problem has been discussed in philosophy a lot of times, and became known as the Theseus' paradox, but that was about rebuilding a ship rather than a person.
In this case, your reborn is qualitatively identical, but not numerically identical.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Car Crash Scenario Two vehicles travelling at 80mph in the same direction.
Vehicles are directly behind each other.
12 meter distance between them.
Front door of car in front rips of and hits the front window of the car behind.
At what speed did the door make contact with the car?
How would one go about calculating this?
====
My current guess is as follows: The car behind would take approximately 0.34 seconds to reach the door in front, the door in front would have a forward velocity starting at 80mph but would quickly be decelerating due to no longer being attached to the car in front. Considering the car behind reach the door in approximately 0.34 seconds, i would guess that the speed of the door would be still very close to 80mph.
But this is a pure guess, is there a formula i can use here to accurately calculate this?
| Once the door is detached from the front car, it will start decelerating, because of air drag, friction with the floor, etc. You can start assuming that this decelaration is constant, it is not perfectly accurate, as the door will likely be rotating, and being an irregular form, the drag will vary chaotically. Also the drag vary with velocity.
If you assume this door (de-)acceleration constant, then the calculation is easy, particularly, if you do the computation from the frame of reference of the moving cars (lowercase relativity!). The distance travelled by the door is:
$$ D_d = D_a * t^2 $$
being, $D_d$ the distance, $D_a$ the door acceleration (to be determined experimentally) and $t$ the time.
Now the rest is easy: the crash will be when $D_d = 12$.
You ask how to calculate this accurately, but that is impossible. Even if you were to do the actual experiment (please use a wind tunnel, not real roads!), you would get quite different results from one execution to the next, simply because the movement of the door in the air will be chaotic and unpredictable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the time period of an oscillator with varying spring constant? It is well known that the time period of a harmonic oscillator when mass $m$ and spring constant $k$ are constant is $T=2\pi\sqrt{m/k}$.
However, I would be interested to know what the time period is if $k$ is not constant. I have searched hours after hours for right answers from Google and came up with nothing. I am looking for an analytical solution.
| Alright, according to my knowledge there are some cases with a time dependent spring constant where a closed form solution is known. One of my favorites is the following where the spring constant is a power function. Assume that $k/m = \omega^2/t^\beta$ where $\omega \in {\mathbb R}$ and $\beta \ge 0$. Then the ODE in question take a form:
\begin{equation}
\ddot{x}_t + \frac{\omega^2}{t^\beta} x_t = 0
\end{equation}
and it has the following linearly independent solutions:
\begin{equation}
\sqrt{t} J_{\pm \frac{1}{2 \beta-2}}\left[\frac{\omega}{\beta-1} t^{1-\beta} \right]
\end{equation}
where $J_\cdot[]$ is a Bessel function.
The result is proven in several different ways in https://math.stackexchange.com/questions/673737/solution-to-a-second-order-ordinary-differential-equation .
A quick and dirty "proof" is given by the following Mathematica code:
In[166]:= w =.; b =.; t =.;
Table[FullSimplify[(D[#, {t, 2}] + w^2/t^(2 b) #) & /@ {Sqrt[
t] BesselJ[eps/(2 b - 2), w/(b - 1) t^(1 - b)]}], {eps, -1, 1,
2}]
Out[167]= {{0}, {0}}
Now, having said all this it would be natural to generalize our ODE by by replacing the power function by a linear combination of two power functions. In other words we seek solutions to the following ODE:
\begin{equation}
\ddot{x}_t + \left(\frac{\omega_1^2}{t^\beta_1} + \frac{\omega_2^2}{t^\beta_2}\right)x_t = 0
\end{equation}
I have long been struggling to solve this problem without any considerable success (however see this https://math.stackexchange.com/questions/1018897/an-ordinary-differential-equation-with-time-varying-coefficients). The usual way "plug the result into Mathematica" or "ask a mathematician" has led to nothing. As it seems mathematics as usual lags several decades behind physics and it will never catch up...;-).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Ligth clock with spaceships side-by-side In the reference frame of an observer, two spaceships travel in a straight direction (e.g. x axis) at a very high velocity and side-by-side; the distance between them is always d (km) = c (km/s) x 0.1 (s).
At its time zero, spaceship one begins the emission of one photon each 0.1 (s) to the spaceship two (parallel to y axis).
The wavelength of one photon is λ(i) = 0.001 x λ(i-1) and the first photon has a wavelength of λ(0).
Question 1: Find the wavelength of the first photon detected by spaceship two.
Question 2: Suppose that the observer can see each photon. What does he will see?
This is not homework.
| To answer question 2:
If it fires a photon orthogonal to its heading, it will travel orthogonal to the source's heading from the POV of the source. From the POV of an outside observer, it will travel not orthogonal to the motion of the source, but slightly along its direction of motion.
This can be easily intuited because in its own frame, the source is at rest. This means when one spaceship fires a photon to the adjacent one, the photon will hit the other ship. From the point of view of an outside observer this must still hold true, but since both ships are moving, the outside observer will not see the photon as fired orthogonal to the direction of motion; it will be angled such that it has the same velocity as the spaceships in their direction of motion, and thus can eventually hit the second ship.
The speed of the photon must, obviously, still be $c$. To find the velocity of the photon orthogonal to the direction of motion of the spaceships, all you need is a little trigonometry.
$$c^2-v^2=x^2$$
If you know $c$ and $v$ (the ship's velocity), then solve for $x$ for the component of the photon's velocity orthogonal to the direction of motion as seen by the observer. Note, this only works out so well when the moving ship emits the photon perpendicular to its heading
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the work of the electrostatic forces in a parallel-plate capacitor? The expression of the energy stored in a parallel-plate capacitor is:
$$U = \frac{e_0\cdot A \cdot V^2}{2d}$$
with $e_0$ the vacuum permittivity, $A$ the surface of the capacitor, $V$ the applied voltage and $d$ the distance between the two plates.
From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d.
But now, let me try to actually calculate the work of the electrostatic forces.
The electrostatic force applied to a plate is expressed as:
$$F = \frac{e_0 \cdot A \cdot V^2}{2d^2}$$
So the work done by one plate on a distance d can be expressed as:
$$W = \int_0^d{Fdx}=\int_0^d \frac{e_0\cdot A\cdot V^2}{2x^2}dx=\frac{e_0\cdot A \cdot V^2}{2}\int_0^d\frac{dx}{x^2}$$ which obviously diverges as the integral of $\dfrac{1}{d^2}$ is $-\dfrac{1}{d}$ and we need to calculate it for $d = 0$...
So what is wrong with my reasoning? How to calculate the work of the electrostatic force so as to obtain the actual expression of the electrostatic energy $U$?
| I think your approach isn't wrong; however in your calculations you're making the assumption that the potential difference between plates, $V$, is constant: What remains constant is the charge on each plate. So the equation becomes:
$$W=\int_0^d {{q^2} \over {2\epsilon_0A}} \;\mathrm{dx}={{q^2d} \over {2\epsilon_0A}}$$
Since $C=\epsilon_0A/d$ we obtain
$$W={q^2\over{2C}}={1\over2}CV^2$$
When you implicitly assume $V$ is constant, electric field when $d=0$ becomes infinite, hence the integral does not converge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is stress a property only relevant on surfaces? I saw that,
$$dF=\sigma \cdot dS$$
Where $dF$ is the differential force, $\sigma$ is the stress tensor, and $dS$ is the differential surface. This equation confuses me a bit. I'm under the impression that stress acts over a volume rather than a surface, so rather than $dS$ we should have.
$$dF=\sigma \cdot dV$$
Does stress act over a surface or a volume element?
| Both.
You can have the standard
$$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$
in the bulk, but you can also have a reduced tensor, a surface stress. This surface stress is the projection of the regular stress
$$\sigma_{\alpha \beta}^s = {\boldsymbol P} \sigma {\boldsymbol P}$$
where ${\boldsymbol P} = 1 - {\boldsymbol n}({\boldsymbol r}) \otimes {\boldsymbol n}({\boldsymbol r})$ is the projection operator constructed from the orthonormal basis using the tangent vectors from the surface normal with $\alpha, \beta = 1,2$.
Typically in large systems, the surface stress contribution to the total energy of the system is negligible. However, as the size of the system is brought smaller, the surface plays more of a role in determining the overall elastic configuration of the system and can modulate some of the elastic dependent properties.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does a moving fan seem transparent? We all know when fan starts moving faster, we cannot see its blades. Why is this?
First I assumed persistence of vision may be the reason. But that can happen with blade also right? Image of blade can remain in our memory and moving fan can appears as a circular plane with blade color.
Why only image of rear side of fan is remaining in our memory?
Note: I tried with fan whose blade area is almost same as that of non blade area
| Because human eyes and brains are slow, they cannot resolve the motion of the blades, but only see the average of the moving blades and the image in the background (this is actually primarily really due to the slow reaction time of the cones, which is slow, as is demonstrated by the fact that a 24 frames per second video does not appear as single images but in motion). Due to their motion the blades average more or less to a uniform colour, so only the image in the background is seen.
If your eyes follow the motion of the blades for a short time (eyes can track pretty fast objects), it is nevertheless possible to resolve the shape of the blades for a short time.
But summarized: The image of the blades is not even resolved by the retina, so it will certainly not be processed by your brain and leave a memory.
As to the perception of the image in the background vs. the moving blades as uniform disk of colour, those are the attention mechanisms of the human image processing. It is for the same reason that you can decide to watch the bottom of a pond or the reflection on the water surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 4,
"answer_id": 1
} |
Double slit experiment experiment with convex lens? In one version of the double slit experiment, the experimenters placed a convex lens far away from the two slits. After the individual photons passed through the two slits they then travel to the the lens, then after that they hit the back wall. However the detector screen at the back wall is placed after the focal point of the lens. And the detector screen shows that the photons hit the detector screen in a particle fashion corresponding to the two slits. My question is, what happens the individual photons after they pass the focal point after the lens that destroys the interference pattern and makes a particle pattern?
| The interference pattern is not destroyed by the lens or the focal point, because in this case it never existed. The moment the lens is placed between the slits and the detector, all particles traveling through the slits will only behave like particles passing through a slit then a mirror and then hitting one of 2 points on the detector. Designing the experiment in any way that allows the path of the particles as they are travelling between the slits and detector to be known is enough to prevent the interference pattern from showing. The particles only behave like waves when you are not looking at their path.
The link below explains it clearly. Paragraph 3 talks about the experiment with the convex lens, but the whole paper is worth a read.
http://arxiv.org/pdf/quant-ph/0701109.pdf
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the ratio of the acceleration in the two cases (a) and (b) What is the ratio of the acceleration in the two cases (a) and (b)?
I thought that the ratio would be 1:1 but it my textbook says its 1:3, so can someone explain to me how that's possible.
| $1)$ Let's make all the forces that would be acting on the blocks, in the first case.
Now we apply newtons' laws of motion assuming that block of mass 2m accelerates downward with $a_1$ acceleration and the block of mass m accelerates upward with same magnitude of $a_1$ acceleration( because they are constrained to have same acceleration till the string is tight).
So,
\begin{align}
&2mg-T=2ma_1\qquad\quad\cdots (1)\\
&T-mg=ma_1\qquad\qquad\cdots (2)\\
\end{align}
adding (1) and (2),
$$\implies(2mg-T)+(T-mg)=2ma_1+ma_1$$
$$\implies mg=3ma_1$$
$$\implies a_1=\frac{g}{3}\qquad\qquad\cdots(3)$$
Thus, acceleration of block of mass m would be $\frac{g}{3}$ in upward direction.
$2)$ Now making the forces for the second one:
We assume that the string is ideal, and as we know that an ideal string is massless and inextensible, which means that the tension at any point through out the string should be constant, because if it is not constant then the net force at that point would not be zero and thus the string would have infinite acceleration being massless, therefore the downward force that you apply at the free end of the string should be equal to the tension at that point, which eventually gets transferred to the block and lifts it up.
Thus,
$$|\vec{T}|=|\vec{F}| = 2mg$$
Let's say that the block moves upward with $a_2$ acceleration.
$$T-mg = ma_2$$
$$\implies 2mg - mg = ma_2$$
$$\implies mg = ma_2$$
$$\implies a_2 = g$$
So you see that in this case the block of mass m has an upward acceleration equal to $g$, i.e. 3 times the acceleration in first case (refer equation 3)
Thus $$\frac{a_1}{a_2} = \frac{\frac{g}{3}}{g} = \frac{1}{3}$$
$$\implies a_1 : a_2 = 1: 3$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do we know that the CMBR is the oldest light? How do we know that the CMBR is the oldest light which we can see? Is it based just on the facts 1.that waves redshift with expanding space, and 2.predictions of the big bang theory;
Or is there a way to know which light is older/younger?
How can You tell the difference between "normal" microwaves which could be emitted from anywhere, and Cosmic microwave background, when You are detecting it on somekind of light wave detector?
Or let me put it this way, whats the difference between a microwave generated in my oven, and the microwave with the same frequency as the one from the oven, but it was a lets say ultraviolet when it was generated but it redshifted to the frequency which is equal to the one generated in the oven?
| Distance is equivalent to time. The time at which the cosmic microwave background was emittied was the time when the universe made a phase transition from being a plasma to being atomic matter. During this time, the universe finally became transparent. Before this time, the universe was so hot that all matter was opaque. The CMB is the wall that separates these two times, so trying to see back farther would be like trying to see beneath the surface of the sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The pressure in a container of water is based on depth. So what happens if I remove the bottom of the container? So I understand that if we have a system that involves a container of water the pressure will equal atmospheric pressure at the top and as we go further down the container the pressure will increase with depth until it is at its maximum pressure at the bottom of the container. In terms of a 1-D fluid equation (neglecting viscosity) we have
$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} - \frac{\partial v}{\partial z} + g$$
$$0 = \frac{1}{\rho} \frac{\partial p}{\partial z} - 0 + g$$
$$\frac{\partial p}{\partial z} = \rho g$$
$$p = \rho g z + const$$
$$p = \rho g z $$
where we have taking the constant to be $0$.
Now say you removed the bottom of this container. Obviously the water starts to flow out the bottom. So the velocity changes with time. As the fluid is incompressible we still have $\frac{\partial v}{\partial z} = 0$.
So
$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} + g$$
But if pressure still balances with gravity we have no fluid flow! But the fluid does flow so the pressure must not balance with gravity anymore. So what is the pressure in the system now that the bottom of the container has been removed?
| If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why complexify in order to construct Dirac representation? Suppose we have a theory is covariant under the Spin group Spin(2n-1; 1). We consider the real vector space $V = R^{2n-1,1}$, which naturally comes with a Lorentzian inner product. On this vector space we introduce an orthonormal basis $e_0; e_1; ... ; e_{2n-1}$, where $e _0$ denotes the time direction.
To construct the Dirac representation of Spin(2n-1; 1) we take the complexified space $T = \mathbb{C} <e_1; ... ; e_n>$. My question is why is it that in order to construct the Dirac representation we complexify the space?
NOTE: Theory is even dimensional and of Lorentzian signature.
| We assume that OP asks apart from the facts that:
*
*Dirac representations by definition are complex;
*It is much easier to work with an algebraically closed field;
*Any real representation can be extended to a (possibly reducible) complex representation, so one is not missing anything by going complex.
In other words, OP is interested in why certain real Lie group representations cannot exist. Since it is well-known that every Lie group representation induces a corresponding Lie algebra representation, it will be enough for our purpose to show that certain real Lie algebra representations cannot exist.
So we are interested in whether there exists an $2^{[\frac{n}{2}]}$-dimensional$^1$ real spinor representation of $so(p,q)$, where $n=p+q\geq 2$?
A low dimension where this fails is $(p,q)=(3,0)$, i.e. 3D rotations, where we leave it as an exercise for the reader to check that the 1-dimensional pseudoreal/quaternionic spinor representation of the Lie algebra $so(3)\cong su(2)\cong u(1,\mathbb{H})$ has no real 2-dimensional irreducible subrepresentation.
OP only asks about even dimension $n$ with Minkowski signature. One may similarly show that $(p,q)=(5,1)$ fails, i.e that the direct sum of the 2-dimensional left and the 2-dimensional right pseudoreal/quaternionic Weyl spinor representations of the Lie algebra $so(5,1)\cong sl(2,\mathbb{H})$ has no real 8-dimensional irreducible subrepresentation.
Incidentally, Witten recently discussed real, pseudoreal and complex representations of fermions in arXiv:1508.04715.
--
$^1$ To understand where the dimension $2^{[\frac{n}{2}]}$ comes from, see e.g. this Phys.SE post.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Does sound show wave-particle duality? We know that light and electrons both show wave-particle duality. Or in other words we can say that they can be both seen as a wave and a particle. Can a similar theory be applicable for sound? Can sound also be explained as a particle as well as a wave?
| To plainly put what WetSavannaAnimal have said. Yes, sound waves can behave like a particle. When sound wave have enough energy to excite the particles that is use for traveling to their excited state, the sound wave becomes a Phonon.
Phonons act like particles that oscillates relative to each other and no longer function like a wave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do we see? Where do the photons disappear? I know that the light is reflected from a object to my eyes, but I don't understand exactly how. The photons appear from the light source and disappear in my eye! Can someone explain the phenomenon of where the photons go and do to allow us to see?
| From the wiki article on color vision as an illustration of how photons are absorbed:
Perception of color begins with specialized retinal cells containing pigments with different spectral sensitivities, known as cone cells. In humans, there are three types of cones sensitive to three different spectra, resulting in trichromatic color vision.
Each individual cone contains pigments composed of opsin apoprotein, which is covalently linked to either 11-cis-hydroretinal or more rarely 11-cis-dehydroretinal.
So it is molecules with different absorption spectra that absorb the optical photons and start the sequence of giving a signal to the brain. It is not a simple matter and belongs more to biology than to physics. The physics part is just that the photon hits a molecule and raises an electron to a higher level which generates a series of reactions that finally register in the brain.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 6,
"answer_id": 1
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.