Time dilation problem question

[Ibix]

Hmmm... Let me take a gander at making them see the same Doppler effect, in my obviously unqualified noobish ways... What about this scenario?Ibix shoots the light wave. Bartolomeo sees Ibix moving away at 0.9c. Since Ibix is moving away according to Bartolomeo's frame, Bartolomeo sees the following redshift:

$$f_O = f_E \sqrt{\frac{1-0.9}{1+0.9}}$$

Now, according to Ibix, when the light wave comes back, to him it is emitted from the mirror. As the mirror is moving away from him, he'll see the following Doppler shift:

$$f_O = f_E \sqrt{\frac{1-0.9}{1+0.9}}$$

Looks like it's exactly the same for Ibix if the reflected light can be treated as if it emitted from the mirror.

Unless... that assumption is not valid?

Not quite. Look at it from Bartolomeo’s frame. He receives a frequency $f_O$ and reflects back the same frequency. So what Ibix must see is the same as if Bartolomeo were emitting at $f_O$ - which means that the final frequency must be $$f_R=\sqrt {\frac {c-v}{c+v}}f_O=\frac {c-v}{c+v}f_E $$You can check this by letting Ibix be at rest at the origin and having Bartolomeo be at $x=vt$. At time $T$ and $T+1/f_E$ Ibix emits light pulses. Work out when they reflect and return back to the origin. The time difference between their return times ought to be $1/f_R$.

 

[Battlemage!]

Not quite. Look at it from Bartolomeo’s frame. He receives a frequency $f_O$ and reflects back the same frequency. So what Ibix must see is the same as if Bartolomeo were emitting at $f_O$ - which means that the final frequency must be $$f_R=\sqrt {\frac {c-v}{c+v}}f_O=\frac {c-v}{c+v}f_E $$You can check this by letting Ibix be at rest at the origin and having Bartolomeo be at $x=vt$. At time $T$ and $T+1/f_E$ Ibix emits light pulses. Work out when they reflect and return back to the origin. The time difference between their return times ought to be $1/f_R$.

So Ibix will see the light pulse even more strongly redshifted than Bartolomeo when it reflects back, correct?

Also, this should mean that each time the pulse reflects back it should continue to become even more red shifted until it's no longer visible, right?

 

[Bartolomeo]

@Ibix . You were right. Only blueshift when rotating and commoving coincide in one spatial position. A bit later regarding the test.

 

[Ibix]

So Ibix will see the light pulse even more strongly redshifted than Bartolomeo when it reflects back, correct?

Also, this should mean that each time the pulse reflects back it should continue to become even more red shifted until it's no longer visible, right?

Yes.

 

[Bartolomeo]

Yes.

And what had happened to the energy?
It seems that if the mirrors "run away" from each other, i.e photon travels longer and longer patch between mirrors with every oscillation, the photon makes more reddish at every turn. In case if distance between mirrors changes with time. It is like photon gas in a closed container. If the container expands, photons turns reddish.
But if distance between mirrors doesn't change with time, each reflection doesn't lead to change of color. It means if it was released green to the mirror, it will come back green.
Distance which photon travels between every reflection, to be exact.
In case if photon travels between "two parallel lines" distance he travels at every oscillation is the same, despite of how we see it - either up and down or zigzag.

 

[Battlemage!]

And what had happened to the energy?
It seems that if the mirrors "run away" from each other, i.e photon travels longer and longer patch between mirrors, the photon makes more reddish at every turn. In case if distance between mirrors changes with time. It is like photon gas in a closed container. If the container expands, photons turns reddish.
But if distance between mirrors doesn't change with time, each reflection doesn't lead to color. It means if it was released grin to a mirror, it will come back green.
Distance which photon travels between every reflection, to be exact.

Isn't there some energy change due to the interaction of the light and the mirrors?

I found this interesting article from the American Journal of Physics: On energy transfers in reflection of light by a moving mirror.Unfortunately I'm not at school right now so I don't have access to the full article, but the abstract is interesting, and it'd be interesting to see all the factors tied together in one treatment (but that's probably way above my pay grade).

 

[stevendaryl]

And what had happened to the energy?

When a pulse of light bounces off a mirror, it imparts energy and momentum to the mirror.

 

[Bartolomeo]

Isn't there some energy change due t

o the interaction of the light and the mirrors?

I found this interesting article from the American Journal of Physics: On energy transfers in reflection of light by a moving mirror.

Unfortunately I'm not at school right now so I don't have access to the full article, but the abstract is interesting, and it'd be interesting to see all the factors tied together in one treatment (but that's probably way above my pay grade).

When a pulse of light bounces off a mirror, it imparts energy and momentum to the mirror.

There are two mirrors and photon gas between them. Someone wants to bring the mirrors closer to each other. He has to make some work and transfer some energy into system, because photons tend to push the mirrors away from each other. That is, if the mirrors are not rigidly fixed photons push them apart and transmit some energy to mirrors. Thus, if mirrors approach each other, photons will gain energy and will be more blueish. If mirrors recede, even by inertia, photons will transfer some energy to them.
But if mirrors are rigidly fixed, color doesn't change after reflection.
Of course, we assume that the mirrors are "ideal".

 

[stevendaryl]

But if mirrors are rigidly fixed, color doesn't change after reflection.
Of course, we assume that the mirrors are "ideal".

That's not 100% true. Every time a light pulse bounces off a mirror, then it imparts momentum and energy to the mirror. Of course, the energy loss is negligible in the limit of a very massive mirror or very low-energy light pulse.

 

[Bartolomeo]

That's not 100% true. Every time a light pulse bounces off a mirror, then it imparts momentum and energy to the mirror. Of course, the energy loss is negligible in the limit of a very massive mirror or very low-energy light pulse.

Obviously in real life. It "heats up" mirror. I mentioned that mirror is ideal, so us to omit "heating up"
If "ideal" mirrors move parallel to each other. One mirror "on the floor" is in motion and sends green photon. Photon travels together with this mirror "up", while mirror moves to the "right" in positive direction. When mirrors at the points of closest approach, the photon hits the mirror "on the top". Mirror "on the top" sees blue photon, because source approaches and immediately reflects it. Photon approached this mirror at oblique angle and leaves at oblique angle again. Photon comes back to "floor mirror". Since this mirror recedes from the mirror "on the top" photon redshifts, i.e. blue photon turns green again. Photon left the "floor mirror" at right angle and came back at right angle again.
This is consideration from longitudinal perspective. We can describe the same situation as the Transverse effect. In this case mirror "in the bottom" will be at rest, but mirror "on the top" in motion. Mirror in the bottom (at origin) sends GREEN photon straight up. Mirror "on the top" at that moment was at point (-X, Y) . When "top" mirror comes into point (Y,0) the mirrors at the closest approach. Due to ABERRATION photon blueshifts, i..e. turns BLUE. It means. that "top" mirror sees acceleration of clock, which is at rest. "Top" mirror reflects blue photon. Photon comes back to bottom mirror "straight from the top", from the point (Y,0)
Photon redshifts, since was released by moving source. "Bottom" mirror "sees" green photon which was BLUE at the source. It is in accordance with time dilation of a moving clock.
By the way, do you see reciprocity of observations in both cases. I don't.

 

[stevendaryl]

That's not 100% true. Every time a light pulse bounces off a mirror, then it imparts momentum and energy to the mirror. Of course, the energy loss is negligible in the limit of a very massive mirror or very low-energy light pulse.

If you work out the energy/momentum for an elastic collision between a photon and a perfect mirror, what you find is that if the momentum of the photon is $p$ before the collision, then after the collision, it will be:

$p' = \frac{-p (E-Pc)}{E+Pc}$

where $E$ is the total energy (mirror plus photon) and $P$ is the total momentum. In the case where the mirror is very massive, so that the energy due to the photon is negligible, $P \approx \gamma m v$ where $m$ is the mass of the mirror, and $v$ is its speed, and $E \approx \gamma mc^2$. So you get:

$p' = -p \frac{c-v}{c+v}$

The energy of the photon changes from $pc$ to $|p'|c = pc \frac{c-v}{c+v}$

 

[stevendaryl]

Obviously in real life. It "heats up" mirror.

No, I'm not talking about that. I'm talking about the case of an ideal mirror. When a light pulse bounces off an ideal mirror, it imparts energy to the mirror.

 

[Bartolomeo]

No, I'm not talking about that. I'm talking about the case of an ideal mirror. When a light pulse bounces off an ideal mirror, it imparts energy to the mirror.

What that energy does with that "perfect mirror"? What that energy turns into?
Which amount of energy the "perfect" mirror has to consume? What that amount depends on?
Is it perfect, if it consumes energy?

 

[stevendaryl]

What that energy does with that "perfect mirror"? What that energy turns into?
Which amount of energy the "perfect" mirror has to consume? What that amount depends on?
Is it perfect, if it consumes energy?

Saying it's a perfect mirror says that all the energy of the mirror is in the form of kinetic energy (no energy due to heating up the mirror).

Let $E$ be the total energy of mirror plus photon. Let $P$ be the total momentum of mirror plus photon. Both of these quantities are conserved in a collision. Let $M$ be the mass of the mirror. It is unchanged by a collision in the case of a perfect mirror. (If the mirror heats up, then $M$ increases.) Let $p$ be the momentum of the photon, and let $p_m$ be the momentum of the mirror. Then we have:

Then what we have is:
$p + p_m = P$
$|p|c + \sqrt{p_m^2 c^2 + M^2 c^4} = E$

Rearrange and square the second equation, to get:
$p_m^2 c^2 + M^2 c^4 = E^2 - 2 E |p| c + p^2 c^2$

Now, we use the equation for momentum to rewrite $p_m$:
$p^2 c^2 - 2 P p c^2 + P^2 c^2 + M^2 c^4 = E^2 - 2E|p|c + p^2 c^2$

So: $E^2 - P^2 c^2 - M^2 c^4 = 2(-Pc^2 \pm Ec) p$
(where $\pm$ is the sign of $p$)

So: $p = \pm \frac{E^2 - P^2 c^2 - M^2 c^4}{2c (E \mp Pc)}$

So for a fixed $P$ and $E$, there are two solutions for the momentum of the photon:

$p_{+} = \frac{E^2 - P^2 c^2 - M^2 c^4}{2c (E - Pc)}$

$p_{-} = -\frac{E^2 - P^2 c^2 - M^2 c^4}{2c (E + Pc)}$

$\frac{p_{-}}{p_{+}} = -\frac{E-Pc}{E+Pc}$

If the energy of the photon is negligible compared to the rest energy of the mirror, then $\frac{E-Pc}{E+Pc} \approx \frac{c-v}{c+v}$ where $v$ is the original velocity of the mirror.

$p_{+}$ is the momentum before the collision, and $p_{-}$ is the momentum after the collision. The change in momentum of the photon goes into making the mirror go a tiny bit faster. It's a negligible change to the speed of the mirror, but just enough to change the total energy of the mirror by exactly the amount that is lost by the photon.

 

[Bartolomeo]

Saying it's a perfect mirror says that all the energy of the mirror is in the form of kinetic energy (no energy due to heating up the mirror).

Let $E$ be the total energy of mirror plus photon. Let $P$ be the total momentum of mirror plus photon. Both of these quantities are conserved in a collision. Let $M$ be the mass of the mirror. It is unchanged by a collision in the case of a perfect mirror. (If the mirror heats up, then $M$ increases.) Let $p$ be the momentum of the photon, and let $p_m$ be the momentum of the mirror. Then we have:

Then what we have is:
$p + p_m = P$
$|p|c + \sqrt{p_m^2 c^2 + M^2 c^4} = E$

Rearrange and square the second equation, to get:
$p_m^2 c^2 + M^2 c^4 = E^2 - 2 E |p| c + p^2 c^2$

Now, we use the equation for momentum to rewrite $p_m$:
$p^2 c^2 - 2 P p c^2 + P^2 c^2 + M^2 c^4 = E^2 - 2E|p|c + p^2 c^2$

So: $E^2 - P^2 c^2 - M^2 c^4 = 2(-Pc^2 \pm Ec) p$
(where $\pm$ is the sign of $p$)

So: $p = \pm \frac{E^2 - P^2 c^2 - M^2 c^4}{2c (E \mp Pc)}$

So for a fixed $P$ and $E$, there are two solutions for the momentum of the photon:

$p_{+} = \frac{E^2 - P^2 c^2 - M^2 c^4}{2c (E - Pc)}$

$p_{-} = -\frac{E^2 - P^2 c^2 - M^2 c^4}{2c (E + Pc)}$

$\frac{p_{-}}{p_{+}} = -\frac{E-Pc}{E+Pc}$

If the energy of the photon is negligible compared to the rest energy of the mirror, then $\frac{E-Pc}{E+Pc} \approx \frac{c-v}{c+v}$ where $v$ is the original velocity of the mirror.

$p_{+}$ is the momentum before the collision, and $p_{-}$ is the momentum after the collision. The change in momentum of the photon goes into making the mirror go a tiny bit faster. It's a negligible change to the speed of the mirror, but just enough to change the total energy of the mirror by exactly the amount that is lost by the photon.

If mirror gains kinetic energy, it will start moving and recede from another mirror. But, if the mirrors are rigidly fixed (on the opposite sides of a room) the same amount on energy it had swallowed on reception of a photon it will transfer to the photon when spitting it out.
Otherwise it is not a mirror, but a well-polished boots, for example.

 

[Bartolomeo]

@Ibix. I'm thinking. You say that - something is wrong with setup of experiment. I break my head for quite a long time and I cannot figure out how to conduct such an experiment, so us to get redshift for the both observes in the same experiment. I mean observation of frequency. I would be most grateful for your advice.

 

[stevendaryl]

If mirror gains kinetic energy, it will start moving and recede from another mirror. But, if the mirrors are rigidly fixed (on the opposite sides of a room) the same amount on energy it had swallowed on reception of a photon it will transfer to the photon when spitting it out.
Otherwise it is not a mirror, but a well-polished boots, for example.

No, it is not possible for a photon to bounce off a mirror without imparting momentum to the mirror. It doesn't matter whether the mirror is perfect, or not. I just went through the math of this.

The correct conclusion is that, relativistically, it is impossible for something to be "rigidly fixed". Any object will compress slightly if you crash something into it. Any object will stretch slightly if you jerk on one end.

 

[Battlemage!]

If mirror gains kinetic energy, it will start moving and recede from another mirror. But, if the mirrors are rigidly fixed (on the opposite sides of a room) the same amount on energy it had swallowed on reception of a photon it will transfer to the photon when spitting it out.
Otherwise it is not a mirror, but a well-polished boots, for example.

You can't have it perfectly rigidly fixed, however, since there can be no perfectly rigid rods, if that matters.

 

[Ibix]

@Ibix. I'm thinking. You say that - something is wrong with setup of experiment. I break my head for quite a long time and I cannot figure out how to conduct such an experiment, so us to get redshift for the both observes in the same experiment. I mean observation of frequency. I would be most grateful for your advice.

I don't think it's possible with a single light pulse. I was only commenting that your description of the experiment was incomplete - you said "along the y axis" without saying according to whom.

I think the best you can do is require both parties to follow the same experimental procedure. Both fire a light pulse along their own y-axis and both reflect the other's pulse back. Assuming the pulses have the same proper frequency at emission both observers will agree on the frequency of the pulse they receive from the other and on the frequency of their own pulse reflected back. (Also they will agree that the reflection times were simultaneous and that the other ship emitted the pulse first and received the reflected pulse last.)

This is the reciprocity we expect.

 

[Bartolomeo]

I don't think it's possible with a single light pulse. I was only commenting that your description of the experiment was incomplete - you said "along the y axis" without saying according to whom.

I think the best you can do is require both parties to follow the same experimental procedure. Both fire a light pulse along their own y-axis and both reflect the other's pulse back. Assuming the pulses have the same proper frequency at emission both observers will agree on the frequency of the pulse they receive from the other and on the frequency of their own pulse reflected back. (Also they will agree that the reflection times were simultaneous and that the other ship emitted the pulse first and received the reflected pulse last.)

This is the reciprocity we expect.

Hmmm... IMHO there is one minor problem. If Bartolomeo and Ibix want to release photon along his own Y axis each, we both have to keep our tubes a right angle to direction of travel. One straight up and another straight down. This is the case of Joe and Aliens we started from. We simply will not see each other, I am afraid to say.

 

[Ibix]

So use two tubes - one for the emission and return of your own pulse and obe for the reflection of the other guy's pulse. Or rotate one tube after you emit and before you reflect, then rotate back in time to receive your own pulse reflected back.

 

[Ibix]

Actually you can't use a single tube for reflection because you don't send it back the way it came. You need a v-shaped tube with a mirror at the point, and a single tube for your emission and reception of returned pulse.

 

[Bartolomeo]

Actually you can't use a single tube for reflection because you don't send it back the way it came. You need a v-shaped tube with a mirror at the point, and a single tube for your emission and reception of returned pulse.

It is exactly moving mirror case. Mirror receives photon through one tube which looks into front and spits out through another which tilted back. This tube back will not release photon at right angle in mirrors frame
I think that you understand everything very well now. You will put all pieces together and will see the full picture very clearly.

 

[Bartolomeo]

So use two tubes - one for the emission and return of your own pulse and obe for the reflection of the other guy's pulse. Or rotate one tube after you emit and before you reflect, then rotate back in time to receive your own pulse reflected back.

It is not physics already. It is rather psyhics. It is the same as to conduct one experiment at 10 AM and another at 2 PM.

 

[Bartolomeo]

I will post another trick. If you are not bored yet. Would you mind?
The idea behind all this communication is very simple. Reciprocity of observations occurs, when you change reference frame. I introduce my rest frame first, you introduce your rest frame then. If we conduct observations on one chosen frame, there is no recoprocity of observations. Transverse Doppler effect makes troubles to that (change of frames I mean). It shows that reference frame is mutual property but not private.

 

[Ibix]

I already knew all of this. It doesn’t invalidate any point I am making. Both parties have a v tube to reflect the other's light pulse and a straight tube to emit and receive their own. This is the reciprocity that is expected.

I'm slightly at a loss to understand what it is that you expect.

 

[Ibix]

It is not physics already. It is rather psyhics. It is the same as to conduct one experiment at 10 AM and another at 2 PM.

Of course it's the same, if you do the same experiment. As I noted earlier, you keep running antisymmetric experiments, swapping participants, then acting surprised when the results swap. That is the reason for the difference between your 10am and 2pm experiments.

 

[Ibix]

Reciprocity of observations occurs, when you change reference frame.

Not true in general, as we've discussed at length. The fact that the laws of physics are the same in any inertial frame does not mean that all physical situations must be described the same way in all inertial frames. If it did there would be no need to consider frames.

It shows that reference frame is mutual property but not private.

I have no idea what this is meant to mean.

This conversation isn't going anywhere, so I'm out. Nothing you are saying invalidates anything I understand about relativity. If it causes problems for your understanding, then it would seem to be your understanding at fault.

 

[Battlemage!]

Not true in general, as we've discussed at length. The fact that the laws of physics are the same in any inertial frame does not mean that all physical situations must be described the same way in all inertial frames. If it did there would be no need to consider frames.

I have no idea what this is meant to mean.

This conversation isn't going anywhere, so I'm out. Nothing you are saying invalidates anything I understand about relativity. If it causes problems for your understanding, then it would seem to be your understanding at fault.

I wish you wouldn't as this conversation is helping my own understanding, lol.

 

[stevendaryl]

The original poster, @IvicaPhysics, seems to be satisfied that his questions have been answered, so I'm going to use that as a justification for closing this thread.