Conservation of angular momentum and RoS

[PeterDonis]

While the movement of the mirrors, and sound waves in the tubing, are certainly a nightmare to compute, I'm pretty sure that no such movement can add to the angular momentum of the structure.

If the structure were perfectly rigid, this might make sense. But it isn't.

As argued before, these sound waves are exactly the same when the photons move the other way.

That's not the point. The point is that the sound waves can carry angular momentum. Or, rather, you can't just help yourself to the assumption that they don't. It may be that this case is too complicated to analyze here if we have to include the sound waves; but that doesn't mean the sound waves have no effect. It means we can't compute their effect, so this problem is too complicated for us to solve. Which is why pervect is suggesting trying a simpler problem instead.

 

[PeterDonis]

Angular (3-or 4-)momentum = ∑\sum angular moments of parts
without explicitly stating "when".

If you use 4-momentum and 4-angular momentum, there is no "when". You don't have to split spacetime into space and time in order to use 4-vectors, 4-tensors, etc., which means you don't have to adopt any assumptions about "when". That's why 4-vectors, tensors, etc. are such good tools to learn to use.

 

[SlowThinker]

As argued before, these sound waves are exactly the same when the photons move the other way.

That's not the point. The point is that the sound waves can carry angular momentum. Or, rather, you can't just help yourself to the assumption that they don't. It may be that this case is too complicated to analyze here if we have to include the sound waves; but that doesn't mean the sound waves have no effect. It means we can't compute their effect, so this problem is too complicated for us to solve. Which is why pervect is suggesting trying a simpler problem instead.

OK so last try before I give up.
Do you agree that any infinitesimal movements, internal stresses, sound waves and anything else you can think of, in the front-bottom corner, are an exact mirror image of those in the front-top corner?
To me, the answer is "obviously yes". And I'm having trouble seeing how a perfectly symmetric movement could result in a net angular momentum.

 

[PeterDonis]

Do you agree that any infinitesimal movements, internal stresses, sound waves and anything else you can think of, in the front-bottom corner, are an exact mirror image of those in the front-top corner?

In which frame? You're bringing "when" into it again, and "when" is frame-dependent. It's impossible for all these movements to be "exact mirror images" in all frames. If you are content to work only in the center of mass frame, then you can reason more or less the way you have been, and just accept that your reasoning is only valid in that one frame; but if you were content with that, you wouldn't have needed to ask all these questions, would you?

If you're looking for a way to go beyond one frame and understand conservation laws in a more general way, the best approach I know of is to learn about 4-vectors and 4-tensors and how to express conservation laws in terms of them. Those expressions will automatically be valid in any frame, and in fact you often don't even need to pick a frame to understand what they're telling you physically. But you have to give up the urge to constantly try to view things from the viewpoint of a particular frame; the 4-vector 4-tensor approach is fundamentally about geometry and spacetime invariants, not about how things look from any particular frame.

 

[SlowThinker]

Do you agree that any infinitesimal movements, internal stresses, sound waves and anything else you can think of, in the front-bottom corner, are an exact mirror image of those in the front-top corner?

In which frame? You're bringing "when" into it again, and "when" is frame-dependent. It's impossible for all these movements to be "exact mirror images" in all frames.

In any reference frame that's moving horizontally.
Cancellation would also occur in a vertically moving frame, even though between different corners. I'm not sure if that means the cancellation occurs in all inertial reference frames, but it looks possible.
After all, zero angular momentum cannot transform into non-zero, can it? Even if it's angular 4-momentum.

So, are you saying, that angular 4-momentum of this system is always constant, no matter at which time I view it? I could even count some photons twice, if there is no limit on "when" the summation of parts is taking place.

Well I was hoping this system could help me understand how the angular 4-momentum works. Too bad I got stuck on such a silly issue.
Thanks for participation, everyone.

 

[jartsa]

SlowThinker, in your post #28 there is an error in photon energies.

In some frame the photons have equal energies. In that frame let's give everything a boost to the right. What happens is:
Boost to the right increases energy of photons moving to the right.
Boost to the right decreases energy of photons moving to the left.

Another way to say it: An observer of photons accelerates for a while, he sees some photons climbing up in a pseudo-gravity field and redshift, and he sees some photons traveling down in said field and blueshift.

 

[SlowThinker]

SlowThinker, in your post #28 there is an error in photon energies.

In some frame the photons have equal energies. In that frame let's give everything a boost to the right. What happens is:
Boost to the right increases energy of photons moving to the right.
Boost to the right decreases energy of photons moving to the left.

My reasoning goes as follows: in the rest frame, imagine we set the photons to have a wavelength 1000 times shorter than the length of a tube.
This must hold in all frames. So the count of wavelengths in each tube stays constant.
Since the top tube moves against the photon, the photon has to wave much faster than before, to reach that 1000 waves very quickly.
On the other hand, the photons in the bottom tube have a lot of time to make those 1000 waves, so they appear red-shifted.
I agree this is very counter-intuitive, but it seems it must be true. Perhaps you need to boost "to the left" to agree with my scenario.

Another way to say it: An observer of photons accelerates for a while, he sees some photons climbing up in a pseudo-gravity field and redshift, and he sees some photons traveling down in said field and blueshift.

Or, are you saying the photon changes color in flight? That only happens in accelerated motion. My example is inertial.

 

[PeterDonis]

are you saying, that angular 4-momentum of this system is always constant, no matter at which time I view it?

Once more: with angular 4-momentum, there is no time. 4-vectors, 4-tensors, etc. are independent of any splitting of spacetime into space and time. They are geometric objects in 4-dimensional geometry, not "things" that change or don't change with "time".

I was hoping this system could help me understand how the angular 4-momentum works.

If by "this system", you mean 4-vectors, 4-tensors, etc., you haven't even tried to actually use it. You are still using your old system of concepts, and it isn't working.

 

[jartsa]

My reasoning goes as follows: in the rest frame, imagine we set the photons to have a wavelength 1000 times shorter than the length of a tube.
This must hold in all frames. So the count of wavelengths in each tube stays constant.
Since the top tube moves against the photon, the photon has to wave much faster than before, to reach that 1000 waves very quickly.
On the other hand, the photons in the bottom tube have a lot of time to make those 1000 waves, so they appear red-shifted.
I agree this is very counter-intuitive, but it seems it must be true. Perhaps you need to boost "to the left" to agree with my scenario.

Good reasoning, wrong result, therefore there must be an error in some premise.

My reasoning is this: Doppler shift. And I guarantee that there is no error.

The fronts of moving things emit (reflect) blueshifted light.
The rears of moving things emit (reflect) redshifted light.
Right?

 

[PeterDonis]

In the interest of following my own advice, I'm going to at least make a start at analyzing a "photon and mirror" scenario using the angular momentum 4-tensor. I'm going to use a simpler scenario than the one in the OP, along the lines pervect suggested in an earlier post.

The basic setup is as follows: we have four mirrors which, at some instant in the center of mass frame of the mirrors, are all at rest at the four corners of a square. I'll put the four corners at coordinates (in the CoM inertial frame) $(x, y) = (1, 1), (-1, 1), (-1, -1), (1, -1)$. We also have four photons, which, at the same instant in the CoM inertial frame, are just bouncing off the four mirrors; the photons all have energy $k$ in the CoM frame, and you can see that, in that frame just before the bounce, the four photons (in the same order as the mirrors they are just about to bounce off of) have momentum $k$ in, respectively, the positive $y$, negative $x$, negative $y$, and positive $x$ directions. (We are using units where $c = 1$.) Just after the bounce, the photons will have momentum $k$ in, respectively, the negative $x$, negative $y$, positive $x$, and positive $y$ directions.

Let's focus in on just one photon and mirror to start; we'll pick the mirror at $(x, y) = (1, 1)$ and the photon bouncing off of it, which has momentum $k$ in the positive $y$ direction before the bounce and the negative $x$ direction after it.

The angular momentum 4-tensor is $M^{ab} = X^a P^b - X^b P^a$, where $X^a$ is the 4-position and $P^a$ is the 4-momentum. What we want to do is evaluate this tensor just before and just after the bounce for the photon. We assume the bounce happens at $t = 0$ in the CoM frame, so in that frame, we have $X^a = (0, 1, 1)$ both before and after the bounce. (We are leaving out all $z$ components since nothing of interest is happening in the $z$ direction.) Before the bounce, we have $P^a = (k, 0, k)$, and after the bounce, $P^a = (k, -k, 0)$.

If we then calculate $M^{ab}$ before the bounce, we have, for example, $M^{01} = X^0 P^1 - X^1 P^0 = 0 - k = -k$; and after the bounce, we have, for example, $M^{01} = X^0 P^1 - X^1 P^0 = 0 - k = -k$. So that component does not change. Similar calculations show that none of the components change, so $M^{ab}$ is the same both before and after the bounce; its value is:

$$
M^{ab} = \left[ \begin{matrix}
0 & -k & -k \\
k & 0 & k \\
k & -k & 0
\end{matrix} \right]
$$

Similar calculations at the other three corners will show that the same is true there; none of the photon bounces transfer any angular momentum from the photons to the mirrors. So SlowThinker's intuition that, in the CoM frame, the bounces do not exchange angular momentum is correct.

As a brief aside, we can also add up the four angular momentum tensors we compute for the four photons to get the total angular momentum of the system; it turns out to be, in the CoM frame:

$$
M^{ab} = \left[ \begin{matrix}
0 & 0 & 0 \\
0 & 0 & 4k \\
0 & -4k & 0
\end{matrix} \right]
$$

So, in the CoM frame, the angular momentum tensor is indeed purely spatial, as I said in an earlier post.

But what happens if we boost to a different frame? First of all, as I said before, tensor equations are valid in all frames, so if we have shown that a bounce transfers zero angular momentum in one frame, it must transfer zero angular momentum in every frame. That is why the tensor formalism is so powerful; we can pick a frame that makes the component calculation easy, and then, as long as we can write the result in tensor form, we can invoke Lorentz invariance to conclude that the equation is valid in all frames.

But let's go ahead and work through the gory details for this example to see how it goes. We'll boost in the $x$ direction with velocity $u$; that will transform all 4-vectors from $V^a = (v^0, v^1, v^2)$ to $V'^a = \left[ \gamma \left( v^0 - u v^1 \right), \gamma \left( - u v^0 + v^1 \right), v^2 \right]$. So, taking the first photon again, its new 4-position will be $X'^a = ( - \gamma u, \gamma, 1 )$, and its new 4-momentum will be $P'^a = ( \gamma k, - \gamma u k, k )$ before the bounce, and $P'^a = \left[ \gamma \left( 1 + u \right) k, - \gamma \left( 1 + u \right) k, 0 \right]$ after the bounce. Going through the same calculation as before, and remembering that $\gamma^2 \left( 1 - u^2 \right) = 1$, we again see that $M'^{ab}$ is the same before and after the bounce; in this new frame, it is given by

$$
M'^{ab} = \left[ \begin{matrix}
0 & -k & - \gamma \left( 1 + u \right) k \\
k & 0 & \gamma \left( 1 + u \right) k \\
\gamma \left( 1 + u \right) k & - \gamma \left( 1 + u \right) k & 0
\end{matrix} \right]
$$

So each bounce transfers zero angular momentum in the new frame as well. Hence, it doesn't matter that in the new frame, the bounces are happening at different times; each one individually transfers no angular momentum, so there is no need to view them in "pairs" for things to cancel out.

So far, we have not looked at the mirrors at all; but it should be evident, first, that in the CoM frame, before the bounces, each mirror has a vanishing angular momentum tensor (since they have zero momentum, being at rest). After the bounces, each mirror must still have a vanishing angular momentum tensor, since no angular momentum is transferred; and if you compute $M^{ab}$ for each mirror after the bounce, using conservation of linear momentum through the bounce to find $P^a$ for each mirror after the bounce, you will see that $M^{ab}$ still vanishes. And, since $M^{ab}$ is a tensor, if it vanishes in one frame, it vanishes in all frames, so the mirrors must have zero angular momentum in all frames.

So SlowThinker's original intuition was actually correct: the photons cannot transfer any angular momentum to the mirrors at all! But now we can see that this is true individually, at each bounce at each mirror, so there is no problem with relativity of simultaneity, since there is no need to "pair up" bounces at opposite mirrors and do any cancellation to zero out the angular momentum transfer.

 

[jartsa]

Good reasoning, wrong result, therefore there must be an error in some premise.

My reasoning is this: Doppler shift. And I guarantee that there is no error.

The fronts of moving things emit (reflect) blueshifted light.
The rears of moving things emit (reflect) redshifted light.
Right?

Seriously, wavelengths must contract in the tube into which the photons gather, so that all the photons can fit in there.
A water tube analogy: A water tubing moving very fast is length contracted, the water that moves faster than the tubing is even more length contracted, that's how the water fits in the part of tubing where the water accumulates.

 

[SlowThinker]

Seriously, wavelengths must contract in the tube into which the photons gather, so that all the photons can fit in there.
A water tube analogy: A water tubing moving very fast is length contracted, the water that moves faster than the tubing is even more length contracted, that's how the water fits in the part of tubing where the water accumulates.

Length contraction always works in team with time dilation and relativity of simultaneity. You need to take all 3 into account.
I'm pretty sure that my reasoning does that.
You can imagine the standard light clock: the number of waves stays constant between bounces as well, in all frames. Faster moving clock have longer wave, that is, bigger redshift.

 

[jartsa]

Length contraction always works in team with time dilation and relativity of simultaneity. You need to take all 3 into account.
I'm pretty sure that my reasoning does that.
You can imagine the standard light clock: the number of waves stays constant between bounces as well, in all frames. Faster moving clock have longer wave, that is, bigger redshift.

So let's say I have a light clock on my table, where light bounces vertically.

If I boost that light clock to the left, energy of the light increases.
If I boost that light clock to the right, energy of the light increases.

Now our problem is: How do we convince SlowThinker about that.

Hmmm ... $$ E'= \gamma E $$ where E = energy in energy's rest frame, isn't it like that how we transform energy between frames?

 

[SlowThinker]

If I boost that light clock to the left, energy of the light increases.
If I boost that light clock to the right, energy of the light increases.

Now our problem is: How do we convince SlowThinker about that.

You may want to consider the possibility that you are wrong
I'm still digesting Peter's post but

So, taking the first photon again, its new 4-position will be $X'^a = ( - \gamma u, \gamma, 1 )$, and its new 4-momentum will be $P'^a = ( \gamma k, - \gamma u k, k )$ before the bounce, and $P'^a = \left[ \gamma \left( 1 + u \right) k, - \gamma \left( 1 + u \right) k, 0 \right]$ after the bounce.

seems to suggest that the vertical photon's energy is $\gamma k$ (green in my picture), while the returning photon's energy is $\gamma(1+u)k$ (violet in my picture). This agrees, both qualitatively and quantitatively, with what I said.

 

[PeterDonis]

Hmmm ..
$$
E'= \gamma E
$$
where E = energy in energy's rest frame, isn't it like that how we transform energy between frames?

No. Energy is the "time" component of the 4-momentum vector; you have to transform the whole vector, and then take the "time" component of the transformed vector to get the energy in the new frame.

 

[SlowThinker]

So I'm starting to see what's going on: the tubes lead through time as well as space, and the photons rotate in a square that is not purely spatial. However, trying to read Peter's math, I found 2 strange things and I can't quite see where it gets wrong.
1.

So, taking the first photon again, its new 4-position will be $X'^a = ( - \gamma u, \gamma, 1 )$, and its new 4-momentum will be $P'^a = ( \gamma k, - \gamma u k, k )$ before the bounce

I find it strange that a photon, that is clearly moving to the right (and up), has a negative momentum along the x-axis.
All the math leading to this seems correct. When I tried to replace $u$ with $-u$, it broke the position of the corner (moved it into the future rather than past).
So I'm not quite sure what's wrong here. (I am?)

2. Out of curiosity, I wanted to find all 8 momenta (before and after each corner) but got stuck pretty fast. Let's follow the photon to the next corner.
Again the front-top corner is ${X'}_{FT}^a = ( - \gamma u, \gamma, 1 )$ and, unless I'm mistaken, the rear-top corner is ${X'}_{RT}^a=(\gamma u, -\gamma, 1)$. The momentum along the top tube is $P'^a = \left[ \gamma \left( 1 + u \right) k, - \gamma \left( 1 + u \right) k, 0 \right]$ . Now let's look at the angular momentum 4-tensors.
The ${M'}_{FT}^{01}$ component is
$${M'}_{FT}^{01}=X_{FT}^0 P^1 - X_{FT}^1 P^0=-\gamma u(-\gamma (1+u) k)\,-\,\gamma \gamma (1+u)k=\gamma^2(1+u)k(u-1)=-k$$
but
$${M'}_{RT}^{01}=X_{RT}^0 P^1 - X_{RT}^1 P^0=\gamma u (-\gamma(1+u)k)\,-\,-\gamma \gamma (1+u)k=\gamma^2 k(1+u)(-u+1)=k$$
That looks like this component of the angular 4-momentum changes during flight
This looks like a pretty basic mistake but again, I can't find it. Other components seem to match.

 

[PeterDonis]

I find it strange that a photon, that is clearly moving to the right (and up), has a negative momentum along the x-axis.

No, in the primed frame, it's moving to the left and up. The frame itself is moving to the right (positive x direction) relative to the CoM frame, so a photon that is moving purely up (positive y direction) in the CoM frame will be moving to the left (negative x' direction) and up (positive y' direction) in the primed frame.

the rear-top corner

It really helps if you don't use words but math; words are ambiguous, math is not. I don't know which corner you think is the "rear-top corner".

In the CoM frame, here are the four mirrors with their 4-positions $X^a = (t, x, y)$ at the instant of the bounce, and the 4-momenta $P^a = (E, p^x, p^y)$ of the photons bouncing off of them at time $t = 0$ in that frame, just before and just after the bounce:

Mirror A: $X^a = (0, 1, 1)$. Photon bouncing off mirror A: $P^a = (k, 0, k)$ before bounce; $P^a = (k, -k, 0)$ after bounce.

Mirror B: $X^a = (0, -1, 1)$. Photon bouncing off mirror B: $P^a = (k, -k, 0)$ before bounce; $P^a = (k, 0, -k)$ after bounce.

Mirror C: $X^a = (0, -1, -1)$. Photon bouncing off mirror C: $P^a = (k, 0, - k)$ before bounce; $P^a = (k, k, 0)$ after bounce.

Mirror D : $X^a = (0, 1, -1)$. Photon bouncing off mirror D: $P^a = (k, k, 0)$ before bounce; $P^a = (k, 0, k)$ after bounce.

In the moving frame, which is moving in the positive $x$ direction at speed $u$ relative to the CoM frame, the above 4-positions and 4-momenta transform to:

Mirror A: $X^a = ( - \gamma u, \gamma, 1)$. Photon bouncing off mirror A: $P^a = ( \gamma k, - \gamma u k, k )$ before bounce; $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0)$ after bounce.

Mirror B: $X^a = ( \gamma u, - \gamma, 1)$. Photon bouncing off mirror B: $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0 )$ before bounce; $P^a = ( \gamma k, - \gamma u k, -k)$ after bounce.

Mirror C: $X^a = ( \gamma u, - \gamma, -1)$. Photon bouncing off mirror C: $P^a = ( \gamma k, - \gamma u k, - k)$ before bounce; $P^a = ( \gamma (1 - u) k, \gamma ( 1 - u ) k, 0 )$ after bounce.

Mirror D : $X^a = ( - \gamma u, \gamma, -1)$. Photon bouncing off mirror D: $P^a = ( \gamma (1 - u) k, \gamma ( 1 - u ) k, 0)$ before bounce; $P^a = ( \gamma k, - \gamma u k, k )$ after bounce.

That looks like this component of the angular 4-momentum changes during flight

Your intuition that that shouldn't happen is correct. Hopefully the above will help you to re-check your computations.

 

[PeterDonis]

Let's follow the photon to the next corner.

If by this you mean, let's look at what the angular momentum tensor looks like when the photon has reached its next corner, you have to account for the fact that time has passed, so the 4-position vectors of the mirrors in the CoM frame will now have nonzero $t$ components. Let's look at just mirror B, since that's the next mirror that the photon that bounces off mirror A at $t = 0$ in the CoM frame will hit.

It will take 2 units of time for the photon--call it photon a--to travel from mirror A to mirror B, since the mirrors are two units of distance apart (in this case, two units in the $x$ direction). So when photon a reaches mirror B, the 4-position of mirror B in the CoM frame will be $X^a = (2, -1, 1)$. The 4-momentum of photon a before the bounce will be the same as when it left mirror A, i.e., $P^a = (k, -k, 0)$. If you calculate the 4-angular momentum tensor using these values, you should get the same answer as when you calculated it for photon a just leaving mirror A at time $t =0$ in the CoM frame. And if you then transform the new 4-position vector into the primed frame, and use it to calculate the 4-angular momentum of photon a in that frame just before it hits mirror B, you should get the same answer as what you calculated in the primed frame for photon a just leaving mirror A.

 

[SlowThinker]

It will take 2 units of time for the photon--call it photon a--to travel from mirror A to mirror B, since the mirrors are two units of distance apart (in this case, two units in the $x$ direction). So when photon a reaches mirror B, the 4-position of mirror B in the CoM frame will be $X^a = (2, -1, 1)$. The 4-momentum of photon a before the bounce will be the same as when it left mirror A, i.e., $P^a = (k, -k, 0)$. If you calculate the 4-angular momentum tensor using these values, you should get the same answer as when you calculated it for photon a just leaving mirror A at time $t =0$ in the CoM frame. And if you then transform the new 4-position vector into the primed frame, and use it to calculate the 4-angular momentum of photon a in that frame just before it hits mirror B, you should get the same answer as what you calculated in the primed frame for photon a just leaving mirror A.

I still can't quite make it to add up. Just a quick question:
The speed of the Center of mass is to be ignored, or should I somehow subtract its speed from that of the photon?
I'll check my spreadsheet again and again and hopefully make it work eventually.

A somewhat unrelated question, can we say that a photon with 4-momentum (p,q,r,s) has wavelength hc/p?

 

[jartsa]

You may want to consider the possibility that you are wrong
I'm still digesting Peter's post but
seems to suggest that the vertical photon's energy is $\gamma k$

Vertical photon's energy is proportional to gamma. So energy is Increased not decreased. I don't know what the k includes though.
I'll explain simple Doppler shift in a simple way. Pay attention. Or ignore if you dislike simple stuff:

A moving thing waves, it creates a start of a crest at position p, then the thing either chases or runs away from the start of the crest for a time: (1/frequency of the thing) / 2. Then the thing creates the end of the crest, near the start of the crest if the thing was chasing the wave, far away from the start of the crest if the thing was running away from the wave.

Crest length is: velocity of the wave times the time we calculated, minus closing or separation speed between the thing and the wave multiplied by the time we calculated.

Oh yes, the frequency of the moving thing is: the frequency in the thing's rest frame divided by gamma.

 

[PeterDonis]

The speed of the Center of mass is to be ignored, or should I somehow subtract its speed from that of the photon?

The way I derived quantities in the moving frame was to first derive them in the CoM frame and then Lorentz transform. If you do that, you only have to worry about the 4-vectors that describe each photon (4-position and 4-momentum); you don't have to worry about what the CoM is doing.

can we say that a photon with 4-momentum (p,q,r,s) has wavelength hc/p?

If p has units of energy, yes, the photon's frequency will be p / h (strictly speaking, $\hbar$, but we'll ignore that here), and its wavelength will be hc / p, using conventional units. However, often physicists will use units in which $\hbar = c = 1$, so energy, frequency, momentum, and inverse wavelength all have the same units; in those units, the frequency and wavelength would be p and 1 / p, respectively.

 

[PeterDonis]

Vertical photon's energy is proportional to gamma.

Yes, that' what $\gamma k$ means.

I don't know what the k includes though.

$k$ is just the photon's energy in the CoM frame, as should be obvious from the 4-momentum vectors that I wrote down in that frame. I am using units in which $\hbar = c = 1$, as I described in my response to SlowThinker just now.

the frequency of the moving thing is: the frequency in the thing's rest frame divided by gamma.

You're ignoring 4-vectors again. Frequency and inverse wavelength are components of a 4-vector; in fact, if we take into account quantum mechanics, it is just the 4-momentum vector divided by $\hbar$. So you have to Lorentz transform frequency and wavelength the same way you transform energy and momentum, as components of a 4-vector.

Also, what do you mean by "the frequency of the moving thing", as opposed to the frequency of the light it's emitting? The frequency of the light does not get transformed by $\gamma$; it gets transformed by the Doppler shift factor, which is not $\gamma$. (If you look at the math, you will see that this is because, as I said just now, the frequency-wave number 4-vector has to be transformed as a 4-vector.)

 

[jartsa]

You're ignoring 4-vectors again.

Maybe I have to use 4-vectors then. But I must use them in my own special way. So if I boost 4-positions of two consecutive wave crests, find 3-positions of those boosted 4-positions, subtract those 3-positions, do I get a boosted wave-length?

 

[PeterDonis]

if I boost 4-positions of two consecutive wave crests, find 3-positions of those boosted 4-positions, subtract those 3-positions, do I get a boosted wave-length?

Have you tried it?

You might also want to try boosting the 4-momentum of the light, since that is really what defines the light's frequency and wavelength--see post #56.

 

[pervect]

I still can't quite make it to add up. Just a quick question:
The speed of the Center of mass is to be ignored, or should I somehow subtract its speed from that of the photon?
I'll check my spreadsheet again and again and hopefully make it work eventually.

A somewhat unrelated question, can we say that a photon with 4-momentum (p,q,r,s) has wavelength hc/p?

This is the first time the "center of mass" has come up in this thread. A few remarks were made earlier that I think were helpful.

https://en.wikipedia.org/wiki/Relativistic_angular_momentum
In any frame, the "space-space" components of this 4-tensor will be the ordinary angular momentum in that frame (but as an antisymmetric 3-tensor instead of a pseudovector), and the "time-space" components will be the "mass moment".

[the total angular momentum 4-tensor for all 4 photons before the boost, which is constant in coordinate time t]

$$
M^{ab} = \left[ \begin{matrix}
0 & 0 & 0 \\
0 & 0 & 4k \\
0 & -4k & 0
\end{matrix} \right]
$$

So we see in the original frame, there is no mass-moment, i.e. the origin of the system is at the center of mass-energy. I'll add that I didn't double check all the work here personally, due to laziness.

I believe we do see a mass-moment in the boosted frame, at least that's what I got via a Lorentz boost on the above tensor. It didn't quite match Peter's boosted version in the quoted post. Because Peter's posted version didn't have a factor of 4, I didn't quite trust it.

What I get via the direct boost (using a computer algebra package) is:

$$
\left[ \begin {array}{ccc} 0&0&4\,k{\gamma}\,v\\0&0&4\,k{\gamma}\\-4\,k{{\gamma}}\,v&-4\,k{\gamma}&0
\end {array} \right]
$$

It's not quite what I expected, but - assuming I didn't make an error in my rather hasty calculation, it confirms the idea that the mass-energy moment does change as a result of the boost. In other words, the 3-angular-momentum around the given point increases by a factor of gamma, which we can ascribe as being due to time dilation. Additionally the mass-energy moment changes, which we can ascribe as being due to the relativity of simultaneity. So the angular momentum around the given point doesn't change, but in the boosted frame this point is no longer the "center of mass-energy". I believe this is also what the various diagrams (which I haven't really checked either) are also telling us.

 

[PeterDonis]

I believe we do see a mass-moment in the boosted frame

We certainly should on physical grounds, since the CoM is moving in the boosted frame.

It didn't quite match Peter's boosted version in the quoted post.

My notation in post #45 was rather confusing, as I can see on re-reading. The boosted $M'_{ab}$ tensor I give there is only for the first photon, not for the total of all four. In other words, it corresponds to this:

its value is:
$$
M^{ab} = \left[ \begin{matrix} 0 & -k & -k \\ k & 0 & k \\ k & -k & 0 \end{matrix} \right]
$$

If you do the same computation of the boosted tensor individually for the other three photons, and add them all up to get the counterpart to this:

it turns out to be, in the CoM frame:
$$
M^{ab} = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 4k \\ 0 & -4k & 0 \end{matrix} \right]
$$

Then I think you get what you posted, i.e., this:

What I get via the direct boost (using a computer algebra package) is:
$$
\left[ \begin {array}{ccc} 0&0&4\,k{\gamma}\,v\\0&0&4\,k{\gamma}\\-4\,k{{\gamma}}\,v&-4\,k{\gamma}&0 \end {array} \right]
$$

 

[SlowThinker]

Surely I must be stupid or something. I can't figure out the 4-momentum even in the rest frame.

In the CoM frame, ...

Mirror A: $X^a = (0, 1, 1)$. Photon bouncing off mirror A: $P^a = (k, 0, k)$ before bounce; $P^a = (k, -k, 0)$ after bounce.

Mirror B: $X^a = (0, -1, 1)$. Photon bouncing off mirror B: $P^a = (k, -k, 0)$ before bounce; $P^a = (k, 0, -k)$ after bounce.

The 4-momentum of photon leaving mirror A: X x P = (0, 1, 1) x (k, -k, 0):
$M_{01}=0 (-k) - k = -k\text{ ,}\ \ \ \ \ \ \ \ \ \ M_{02}=-k\text{ , }M_{12}=+k$
This same photon arriving at B: X x P = (0, -1, 1) x (k, -k, 0):
$M_{01}=0 (-k) - (-1)k = +k\text{ , }M_{02}=-k\text{ , }M_{12}=+k$
So again, it looks like the angular 4-momentum (component 01) changes in flight.

Your intuition that that shouldn't happen is correct. Hopefully the above will help you to re-check your computations.

What am I doing wrong? This is pretty basic stuff, right?
I've tried to fix it in various ways but none made sense.

The funny thing is that the 4-angular momentum conserves in all 4 corners, even in the moving frame.

 

[PeterDonis]

This same photon arriving at B: X x P = (0, -1, 1) x (k, -k, 0):

No. The same photon arriving at B: X x P = (2, -1, 1) x (k, -k, 0). As I posted before, you have to update the 4-position to account for the travel time of the photon; it isn't at mirror B at time t = 0, it's there at time t = 2, and the time is part of the 4-position.

 

[SlowThinker]

No. The same photon arriving at B: X x P = (2, -1, 1) x (k, -k, 0). As I posted before, you have to update the 4-position to account for the travel time of the photon; it isn't at mirror B at time t = 0, it's there at time t = 2, and the time is part of the 4-position.

The trouble is, at that time, the center of mass is at position (2,0,0) as well, so the "4-distance" from COM to B is (0, -1, 1).
Or, at which point to I advance the COM to new time? If I don't, then after a full cycle, leaving mirror A, I end up with $(8, 1, 1) \times (k, -k, 0) = -9k$ for the 01 component.
It gets even worse in the moving frame.

 

[PeterDonis]

The trouble is, at that time, the center of mass is at position (2,0,0) as well, so the "4-distance" from COM to B is (0, -1, 1).

The 4-position vector is not the "4-distance" from the point to the CoM. (There is no such thing anyway; 4-distance is something that goes between events, not objects or worldlines.) It is the 4-position of the event relative to the spacetime origin of the frame. The very fact that you are getting the wrong answer by taking the "4-distance" to the CoM should be a clue that you are doing it wrong.

at which point to I advance the COM to new time?

You don't. The spacetime origin of the frame is an event, not an object. Remember, this is geometry: the spacetime origin of coordinates does not move. It's a particular point in the geometry, and the 4-position of any other point is taken relative to that particular point. The same goes for 4-angular momentum; it is 4-angular momentum about a particular point in spacetime, not about a particular point in space.

 

[jartsa]

I have now boosted an energy-momentum vector, witch had some momentum to the right, to left. IOW I changed to a frame that moves to the left.

There was more energy in the boosted frame than in the original frame. (Just as I expected)

So right moving light should have extra energy in the picture in post #28.

 

[PeterDonis]

I changed to a frame that moves to the left.

I'm not sure about post #28, but in my analysis in post #45, my "moving" frame was moving to the right (positive x direction) relative to the CoM frame. In this frame, a right-moving photon has less energy, and a left-moving one has more (which is consistent with the analysis I posted).

 

[SlowThinker]

I'm not sure about post #28, but in my analysis in post #45, my "moving" frame was moving to the right (positive x direction) relative to the CoM frame. In this frame, a right-moving photon has less energy, and a left-moving one has more (which is consistent with the analysis I posted).

You seem to love to use unusual coordinates to confuse people
Now that I realized your movement is the opposite of the one in my pictures, your numbers do support Jartsa.
But that, unfortunately, means that my reasoning is wrong in this point as well, although again, I can't see where.

Surely the number of waves between bounces must be invariant in all frames?
It is true for the vertical photons, but not for the horizontal ones, if your 4-momenta are correct. A left-moving photon in a left-moving square has shorter wavelength, and spends more time+distance in flight, so the number of waves it makes is $\gamma^2(1+u)^2$-times higher.

 

[PeterDonis]

You seem to love to use unusual coordinates to confuse people

That wasn't my intent; to be honest, I wasn't sure in which direction things were supposed to be moving in previous scenarios in this thread, so I picked a frame moving to the right because that's the first version of the Lorentz transformation that I learned.

A left-moving photon in a left-moving square has shorter wavelength

No, it doesn't, it has longer wavelength. A left moving photon in a right-moving frame has shorter wavelength.

 

[SlowThinker]

No, it doesn't, it has longer wavelength. A left moving photon in a right-moving frame has shorter wavelength.

Now I'm thoroughly confused.

Mirror A: $X^a = (0, 1, 1)$. Photon bouncing off mirror A: $P^a = (k, 0, k)$ before bounce; $P^a = (k, -k, 0)$ after bounce.

Mirror B: $X^a = (0, -1, 1)$. Photon bouncing off mirror B: $P^a = (k, -k, 0)$ before bounce; $P^a = (k, 0, -k)$ after bounce.

Moving Frame:

Mirror A: $X^a = ( - \gamma u, \gamma, 1)$. Photon bouncing off mirror A: $P^a = ( \gamma k, - \gamma u k, k )$ before bounce; $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0)$ after bounce.

Mirror B: $X^a = ( \gamma u, - \gamma, 1)$. Photon bouncing off mirror B: $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0 )$ before bounce; $P^a = ( \gamma k, - \gamma u k, -k)$ after bounce.

So we have, for positive $u$, the whole square moving to the left (-x axis) (cross-check: negative x-component of P before bounce off A).
Also the photon from A to B is moving to the left.
The photon's 4-momentum is $( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0)$. So the A-B photon's wavelength, as agreed before, is $$\lambda_{AB}=\frac{h}{\gamma(1+u)k}$$
with friends
$$\lambda_{BC}=\frac{h}{\gamma k}$$
$$\lambda_{CD}=\frac{h}{\gamma(1-u)k}$$
Clearly, $\lambda_{AB}$, the wavelength of a left moving photon in left moving square, is the shortest, contradicting the first quote (which I think is true, BTW).