0
$\begingroup$

If two mirrors were fixed normal to each other in a vacuumed cubicle travelling at some velocity V (see image below) and a single photon of light was emitted from point A at time t1. At time t2 would an observer (or some measurement device) see the photon returning back to point A or point C?

I am assuming that there is no horizontal component of velocity V that can be added to the photon of light emitted normally from mirror A, otherwise that would contravene the universal constant speed of light.

enter image description here

$\endgroup$
2
  • $\begingroup$ Well, what happens in you lab (or house)? Remember you are whizzing around in space as the Earth rotates... $\endgroup$
    – Jon Custer
    Nov 4, 2022 at 15:23
  • 1
    $\begingroup$ Say the mirrors are normal to each other on the roof and floor of my house. But that my house is moving horizontally at 0.9 c . If I was able to watch and measure the photon leaving A (in the frame of reference of my house), would it still return exactly to point A? $\endgroup$
    – Dubious
    Nov 4, 2022 at 23:55

4 Answers 4

3
$\begingroup$

The return point of the photon depends upon the direction in which it was travelling. If the photon was travelling normally to the mirrors in the frame of the mirrors, then it will return to point A. If it was travelling normal to the mirrors in the frame of the stationary observer, it will return to point C.

You must remember that directions of motion, as well as speeds of motion, are frame dependent. If you drop a ball on a moving train it will fall vertically. However to an observer on the platform, the ball will follow a sloping path. Likewise if the observer drops a ball vertically on the platform, it will drop at an angle in the frame of the train. So if you want to talk unambiguously about a ball dropping vertically, you have to say whether you mean in the frame of the train or the platform. The same applies to your thought experiment- you have to say whether the light is moving vertically down from the upper mirror in the frame of the mirror or the stationary observer.

Finally, as other answers have pointed out, it is simply wrong to say that 'there is no horizontal component that can be added to the photon'. Adding the horizontal component changes the direction of the photon, but not its speed, owing to the rule for the relativistic addition of velocities.

$\endgroup$
3
  • $\begingroup$ You said "If it was travelling normal to the mirrors in the frame of the stationary observer, it will return to point C". When I look at similar examples, it seems that the path of light to a stationary observer would not be C but as portrayed in this you-tube video from 3:09-3:52 youtube.com/… $\endgroup$
    – Dubious
    Nov 6, 2022 at 2:08
  • $\begingroup$ One of the key things to remember when trying to understand relativity is that all motion is relative. If you look at your reflection in a mirror, you may be standing in your frame of reference, but you and the mirror are travelling at near c in the frame of a passing muon. Does that stop the light bouncing directly back to you from the mirror in your frame? No. But the light takes an angled path in the frame of the muon. $\endgroup$ Nov 6, 2022 at 8:01
  • $\begingroup$ Your comment shows you have misunderstood my answer. The YouTube video shows light moving normal to the ground in the frame of the spaceship. If the light was moving normal to the ground in the frame of the earth, it would go straight up and down in the Earth frame, but appear to be taking an angled path in the frame of the spaceship- ie the opposite of what is shown in the video clip. $\endgroup$ Nov 6, 2022 at 8:05
2
$\begingroup$

I am assuming that there is no horizontal component of velocity V that can be added to the photon of light emitted normally from mirror A, otherwise that would contravene the universal constant speed of light.

Then you are failing to understand and really grapple with special relativity.

So you are saying, “I am assuming that you can't X because that would contradict Y” but we would like to say, “No, X and Y are both true, and your argument is actually ‘if X (and W, obviously), then not Y’, and the problem is that W that you think is obvious.

In this case, your W is the belief that if you add some horizontal component to the velocity, for instance by using a different reference frame, then the vertical component of the velocity of the light stays constant. And it actually does, to first order in $v_x/c$, which is why you can essentially ignore this effect. But then to second order you have $$v_y = \sqrt{c^2-v_x^2}\approx c -\frac12 {v_x^2\over c}$$ and the vertical component has been very slightly attenuated.

$\endgroup$
2
  • $\begingroup$ So if there is an attenuation of the vertical component then are you saying there must be a horizontal component so that the net speed is 'c' ? But that would mean that the photon has been emitted at an angle which isn't the case because the photon is emitted normal from the top mirror. $\endgroup$
    – Dubious
    Nov 5, 2022 at 0:05
  • 1
    $\begingroup$ @Dubious if you're sitting on a vehicle being pulled along with some speed, and you throw a ball straight upward, will an observer on the ground see the ball has some horizontal velocity? $\endgroup$
    – BowlOfRed
    Nov 5, 2022 at 0:27
1
$\begingroup$

That depends on how you aim it. I presume you are thinking of a case where the beam is oriented to produce photons normal to the mirror, and where with the mirrors at rest the photon returns to point A. Is that correct?

I am assuming that there is no horizontal component of velocity V that can be added to the photon of light emitted normally from mirror A, otherwise that would contravene the universal constant speed of light.

But the emitter in your scenario already has a horizontal component. In fact, we can calculate the total momentum of this object in the horizontal direction.

When the photon leaves, the mass of the mirror must be reduced. Therefore either the speed is the same as before and the momentum is less or the momentum is the same and the speed has increased. The second is not possible on energy conservation. So the first is true.

The total momentum of the system must be constant, and the momentum of the mirror has reduced. Therefore the photon must be carrying some of the horizontal momentum of the system. To an observer where the mirrors are stationary and to one where the mirrors are moving, the photon returns to point A.

$\endgroup$
3
  • $\begingroup$ If you read my previous comment in reply to Jon Custer , do you think the photon will return back to point A? $\endgroup$
    – Dubious
    Nov 5, 2022 at 0:08
  • $\begingroup$ I've done a search and there is a similar type of question raised on the link physics.stackexchange.com/questions/166448/…. But one of the comments made said the following " If you move along with the mirrors the photon will still appear to move perpendicular to the mirrors." which doesn't make any sense to me. $\endgroup$
    – Dubious
    Nov 5, 2022 at 0:23
  • 1
    $\begingroup$ @Dubious If the mirrors and the emitter are aligned in such a way that the photon returns to A when the mirrors are at rest, then it will also return to A when the mirrors are moving sideways. $\endgroup$
    – BowlOfRed
    Nov 5, 2022 at 0:28
0
$\begingroup$

First, I am calling $T1$ frame $T$ and $T2$ is now $T'$, because: tradition makes SR problems clearer.

It's unclear how the light is being transmitted. Suppose it's a laser with its phase fronts parallel to the mirror in frame $T$ (that is, the laser is stationary w.r.t. to the mirrors, and emits parallel to their normal).

Then the bounce is perpendicular and it returns to the laser. (If this were not true, then the internal mirrors in the laser would only work in the Universe's absolute rest frame, which doesn't exist, ergo: there is no absolute rest frame ....so much easier than the Michelson-Morley experiment...but I digress).

In the moving frame, the hyperplane of simultaneity is tilted w.r.t to the direction of motion, hence the phase fronts are tilted. Note: phase is a Lorentz scalar, everyone agrees what it is at an event in spacetime:

$$ \phi(x, t) = kx-\omega t = k_{\mu}x^{\mu} = k'_{\mu}x'^{\mu} = k'x'-\omega' t' = \phi(x', t') $$

But the laser aperture spans many events.

For instance, at $t=0$ lets say $\phi(x, 0)=0$. But:

$$t' = \gamma\big(t-\frac{vx}{c^2}\big) = -\gamma\frac{vx}{c^2}\ne 0$$

The phase front is tilted in $T'$ so the laser is not perpendicular to the mirror.

If you want to calculate the angle, then just start with:

$$ k_{\mu} = (\omega/c, k_x, k_y, k_z) = (k,0,0,-k) $$

and boost along $-x$.

$\endgroup$
7
  • $\begingroup$ Many thanks for your explanation but I'm afraid I'm not qualified enough to understand hyperplane of simultaneity. I've done a google search and will try to read up on it, but I suspect they will be mathematical explanations and way over my head. So if I'm an observer inside the cubicle, the photon of light will just move between A and B irrespective of the magnitude of the cubicle velocity V. And the reason is explained by a mathematical formula that proves that the photon is being emitted at an angle and not normal to the mirrors? $\endgroup$
    – Dubious
    Nov 6, 2022 at 1:34
  • $\begingroup$ Isn't my question similar to aberration effect of light from distant stars where the telescope has to be tilted at an angle to capture the photons of light? This is where one has to take into the account the velocity of the orbiting earth and the speed of light? Or is this a different situation entirely because the photon was from a source with a different frame of reference to the telescope? $\endgroup$
    – Dubious
    Nov 6, 2022 at 1:48
  • $\begingroup$ @Dubious hyperplane of simultaneity is a big word for "now". Moving frames, even with the same origin, have different definitions of now everywhere but the origin. $\endgroup$
    – JEB
    Nov 6, 2022 at 4:07
  • $\begingroup$ I've done a search in Physics Stack Exchange and someone has actually raised the same question. So are the answers incorrect? physics.stackexchange.com/questions/544526/… $\endgroup$
    – Dubious
    Nov 7, 2022 at 2:55
  • $\begingroup$ @Dubious that question is different. It states the photon has wave vector $k_{\mu}=(k,0,k,0)$, and a box moves along $u_{\mu}=\gamma(c, v, 0, 0)$, then in the box's rest frame, $k'_x \ne 0$. Basic Lorentz transform of a 4-vector. $\endgroup$
    – JEB
    Nov 8, 2022 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.