Optics - Reflection of image from a moving mirror

[chaimc]

Optics -- Reflection of image from a moving mirror

I have a dufficult question which I couldn't find an answer to, yet:

If I am standing in front of a mirror - I can see my image, naturally.

Now, let's say that the mirror is actually a very long train and it moves parallel to me .

And the question: will my image be reflected back to me in ALL speeds whatsoever or will it shift in the dircetion of the mirror-train in "very high speeds"?

Thanks for your wise answers!

 

[ProTerran]

Well, i guess reflected image from the moving mirror will suffer some amount of shift - it depends on the velocity of the train. Reflection of a e-m wave requires interaction with the mirror matter (i.e. electrons).

 

[chrisbaird]

Your post is not very clear. Are you on the train or watching the train speed by?

If the mirror is at rest with respect to you (you are on the train, looking at a mirror), you will so no difference from if the train was at rest.

 

[chaimc]

Hi Chris.

I am at rest and the mirror is speeding by. Will I receive back the light that I have emitted?

 

[chrisbaird]

Yes. It's the induced change in the currents/fields in a material that give rise to reflections, not little balls bouncing off of atoms. So the change in current and therefore the reflection would be independent of the motion of the material's atoms.

 

[chaimc]

...and that includes very high (and theoratical) speeds as the speed of light or "very-near" of the speed of light?

(thx).

 

[chrisbaird]

Good question.

 

[chaimc]

...and I was hoping to get from you a good answer...?

 

[jtbell]

This thread has now been moved from the Classical Physics forum to the Relativity forum based on the clarification in post #6.

 

[ghwellsjr]

Your relative speed to the mirror will make no difference in the image that is reflected back to you.

 

[nitsuj]

ghwellsjr

the visible light leaving you reaches the mirror, some event happens, light is "reflected" back to you. Now imagine the color of light that is leaving you is blue, just guessing here but I think the light returning to you would be red(or some other lesser wave). I picture it as "stretching" that same bit of wave (energy) across a greater distance, c being c, light is seen as a lesser wave length.

Is that right?

 

[ghwellsjr]

ghwellsjr

the visible light leaving you reaches the mirror, some event happens, light is "reflected" back to you. Now imagine the color of light that is leaving you is blue, just guessing here but I think the light returning to you would be red(or some other lesser wave). I picture it as "stretching" that same bit of wave (energy) across a greater distance, c being c, light is seen as a lesser wave length.

Is that right?

No, assuming a perfect mirror.

 

[nitsuj]

why is there no Doppler effect there? It's not the perfect mirror is it? ()

EDIT: I just read what chrisbaird wrote in post #5.
EDIT AGAIN: just read post #7. So my question still stands.

 

[yuiop]

why is there no Doppler effect there? It's not the perfect mirror is it? ()

I think the answers are correct. If there was a blue light on the train it would be red shifted according to the track side observer due to relativistic Doppler shift. This shift is purely due to time dilation of the source. Oddly enough it seems that if the track side observer has a similar blue light he would not see any red shift in the reflected blue light while an observer on the train would see the track side light as red shifted.

This seems slightly odd at first and it might help to think in terms of tennis balls which is slightly easier to visualise. Imagine a person on the train fires balls at the track side at a rate of once per second according to his own clock. They arrive at the track side at a rate of once every ten seconds because the train clock is running slower than the track clock according to the the track side observer. Now if the track side observer fires balls at the passing train at rate of once per second, there is no good reason why they should not bounce back at a rate of once per second and by similar reasoning there is no reason why the reflected light should be red shifted.

<EDIT> However the reflected tennis balls would be deflected sideways by the passing train unlike the photons. It would be interesting to see if a paradox would be created if the reflected photons were deflected. Now that I think about it, I am not totally convinced reflected photons are not deflected and the best way to settle the issue would be to create a paradox for the false position.

 

[nitsuj]

Now if the track side observer fires balls at the passing train at rate of once per second, there is no good reason why they should not bounce back at a rate of once per second and by similar reasoning there is no reason why the reflected light should be red shifted.

Cool stuff, thanks yuiop. That example was clear. Maybe another way to word it is the observer is at rest with the light source (themself). I was thinking of the mirror as the light source once it is reflected.

 

[chaimc]

Thanks for the answers so far, but i just want to make sure I understood corectly:

I - As the mirror is moving sideways and in vertical to "me" it keeps a fixed distance from "me". Therfore, is the color shift still expected to happen?

II - (I think...) I did not receive an answer to my question (after Chris and the extension of my question to ultra high speeds) - will i receive my reflection back to where I am standing or will it shift sideways even by a fraction?

II-a - I believe that the process of the obsortion of the light and the rebouncing of it must take some time, though almost neglible. So there must be a shift of some degree. If it is so, there is a shift of the image. True or not?

Thanks for the answers.

 

[ghwellsjr]

Repeating:

Your relative speed to the mirror will make no difference in the image that is reflected back to you.

 

[yuiop]

Thanks for the answers so far, but i just want to make sure I understood corectly:

I - As the mirror is moving sideways and in vertical to "me" it keeps a fixed distance from "me". Therfore, is the color shift still expected to happen?

II - (I think...) I did not receive an answer to my question (after Chris and the extension of my question to ultra high speeds) - will i receive my reflection back to where I am standing or will it shift sideways even by a fraction?

I have given this some more thought and now I am convinced there will be no deflection of the reflection and it will not shift sideways by even by a fraction at relativistic speeds. This is more obvious when you consider the reflected light path from the rest frame of the train with the mirror mounted on it. The light source at the side of the track is moving in this reference frame and the light from it is subject to relativistic aberration and so the light path is diagonal. In the reference frame the mirror is stationary and so the light beam is reflected perfectly normally at the same angle as the angle of incidence and returns to the moving light source. When you transform back to the reference frame of the light source it is obvious there is no deflection of the reflected light ray.

II-a - I believe that the process of the obsortion of the light and the rebouncing of it must take some time, though almost neglible. So there must be a shift of some degree. If it is so, there is a shift of the image. True or not?

IF reflection is a process of absorbtion and re-emission (not sure about that) and IF that process takes a finite non zero interval of time then there might be a small shift of the image at relativistic speeds. It would need someone familiar with the intricate details of the process of reflection (not me) or a very sensitive experiment to settle that. Off hand I cannot think of a thought experiment that would refute a delay in the reflection process.

 

[nitsuj]

Hmmm, looks like there would be a doppler effect.


The Doppler Effect from a Uniformly Moving Mirror (what do you make of the paper yuiop?)

Very interestingly, the paper goes on to mention how Einstien discussed the moving mirror & the "special case" doppler in his 1905 paper.
But that it is often excluded from texts.

I'm gunna read into it some more.


so at the same time
ghwellsjr why is there no doppler shift?

 

[ghwellsjr]

Hmmm, looks like there would be a doppler effect.


The Doppler Effect from a Uniformly Moving Mirror (what do you make of the paper yuiop?)

Very interestingly, the paper goes on to mention how Einstien discussed the moving mirror & the "special case" doppler in his 1905 paper.
But that it is often excluded from texts.

I'm gunna read into it some more.


so at the same time
ghwellsjr why is there no doppler shift?

The mirror in the paper is moving away from the source of light. Chaimc's mirror is moving parallel to himself, the source of the light.

 

[nitsuj]

The mirror in the paper is moving away from the source of light. Chaimc's mirror is moving parallel to himself, the source of the light.

What does parallel to himself mean? I see he wrote it, but there is also relative motion between the two. So I am confused, is the image below "parallel to himself"? In that image I'd agree there is no shift

I was thinking of the scenario as;

 

[chaimc]

Here are my "original" drawings (I didn't notice I can do that...):

Am I wrong somewhere along this line of thought?

Thanks again.

 

[yuiop]

Here are my "original" drawings (I didn't notice I can do that...):

Am I wrong somewhere along this line of thought?

Thanks again.

Your last pic (y2) is not correct. When the train is stationary and the source at the side of the track is moving the light should follow a diagonal path (like an upside down 'V' pretty much like the one in the classic Einstein light clock ). In the rest frame of the train, where the signal is emitted is different to where it is received.

 

[chaimc]

...and this is...:'I have given this some more thought and now I am convinced there will be no deflection of the reflection and it will not shift sideways by even by a fraction at relativistic speeds' ??

 

[yuiop]

Your last pic (y2) is not correct. When the train is stationary and the source at the side of the track is moving the light should follow a diagonal path (like an upside down 'V' pretty much like the one in the classic Einstein light clock ). In the rest frame of the train, where the signal is emitted is different to where it is received.

...and this is...:'I have given this some more thought and now I am convinced there will be no deflection of the reflection and it will not shift sideways by even by a fraction at relativistic speeds' ??

The above two statements of mine do not contradict each other, if that is what you are suggesting.

In the rest frame of the train, the light leaves the moving track side observer, travels diagonally, reflects off the train and returns diagonally to the observer who is now in a different position according to an observer on the train. In either rest frame (train or track) the signal always returns to the source. There is no shift from the point of view of the observer at the side of the track.

 

[chaimc]

Maybe I wasn't too clear and I missed an important factor - I am pretending in the darwings that it is a very short burst of light, a single photon even. Does it make a difference now?

Thanks.

 

[yuiop]

Maybe I wasn't too clear and I missed an important factor - I am pretending in the darwings that it is a very short burst of light, a single photon even. Does it make a difference now?

Thanks.

What I said before is true even with a single photon. Check out this youtube video http://www.youtube.com/watch?v=XhJitbhsKvI&feature=related of the Einstein light clock. Note the diagonal path of the photon when the source is moving. Even thought the source is moving, the diagonal light path ensures the photon always returns to the source. If you switch to the point of view where the source is stationary, the photon just goes straight out to the mirror and back along the same path and it makes no difference whether the mirror is stationary or moving in your scenario.

 

[Subplotsville]

A related problem is where the test surface is a type of wire mesh that reflects radio waves if the regular spacing of the mesh is narrow enough given the wavelength of the wave. In this case, the wire mesh and radio wave are such that the radio wave would pass through the wire mesh if the mesh was at rest with respect to the transmitter, but the Lorenz contraction due to the motion of the wire mesh reduces the spacing of its grid to where it should reflect the (appropriately polarized) radio wave. Does it indeed reflect? If so, how would an observer riding with the wire mesh account for the reflection?

 

[yuiop]

A related problem is where the test surface is a type of wire mesh that reflects radio waves if the regular spacing of the mesh is narrow enough given the wavelength of the wave. In this case, the wire mesh and radio wave are such that the radio wave would pass through the wire mesh if the mesh was at rest with respect to the transmitter, but the Lorenz contraction due to the motion of the wire mesh reduces the spacing of its grid to where it should reflect the (appropriately polarized) radio wave. Does it indeed reflect? If so, how would an observer riding with the wire mesh account for the reflection?

Good question. The observer sees the the radio waves as redshifted (therefore longer wavelength) and I am guessing that would be his explanation for why the radio waves do not pass through the mesh.

 

[Subplotsville]

Good question. The observer sees the the radio waves as redshifted (therefore longer wavelength) and I am guessing that would be his explanation for why the radio waves do not pass through the mesh.

Good answer. Thanks for the reply.

Let's tweak the setup a bit then. The radio transmitter is now on the moving platform along with the wire mesh surface. Someone in motion with these should of course always observe the signal to pass through the surface, as when the platform is at rest. However, at a certain speed, an observer on the ground will see both a contracted mesh and a lengthened wave to the extent that the signal should reflect. Which actually happens?

 

[Saw]

Good answer. Thanks for the reply.

Let's tweak the setup a bit then. The radio transmitter is now on the moving platform along with the wire mesh surface. Someone in motion with these should of course always observe the signal to pass through the surface, as when the platform is at rest. However, at a certain speed, an observer on the ground will see both a contracted mesh and a lengthened wave to the extent that the signal should reflect. Which actually happens?

Is this subject the same / analogous to the one treated here?

 

[chaimc]

(I, myself, am not a phisicist.)

1- Yuiop, on mesage #27 sent me to learn about time dilation. Apart from the interpataion of the famous Michelson Morley experiment, what other evidence is there to Time dilation?
2- Historicaly, where there any other interpretaions of that experiment?


3- I got lost somehwere along the line when "you people" started talking abput radio waves - is their behaviour identical to light waves?

 

[Austin0]

Good answer. Thanks for the reply.

Let's tweak the setup a bit then. The radio transmitter is now on the moving platform along with the wire mesh surface. Someone in motion with these should of course always observe the signal to pass through the surface, as when the platform is at rest. However, at a certain speed, an observer on the ground will see both a contracted mesh and a lengthened wave to the extent that the signal should reflect. Which actually happens?

Interesting question.
My thought is that, no matter the velocity, the signal will pass through the mesh and be detected by the ground observer but it will, in all cases, be blue shifted rather than red.
Due to the aberration from the relative velocity the signal path in the ground frame would not be orthogonal but would be at some forward angle and therefore blue shifted.

Or not ;-) In which case you have come up with a thorny one.

 

[Saw]

the signal will (...) be blue shifted rather than red.

I have a problem with any explanation resorting to blue or red shift: isn't it a position dependent question? When the moving source is approaching, the observer sees the signal as blue-shifted. Instead another observer in the same frame, for which the source is receding, sees the signal as redshifted. However, it seems the answer to the question "what happens" should be the same in all frames and from all positions of those frames...

 

[Austin0]

I have a problem with any explanation resorting to blue or red shift: isn't it a position dependent question? When the moving source is approaching, the observer sees the signal as blue-shifted. Instead another observer in the same frame, for which the source is receding, sees the signal as redshifted. However, it seems the answer to the question "what happens" should be the same in all frames and from all positions of those frames...

Hi Saw ,,,long time. In general you are quite correct regarding position dependence but in this setup there are limitations.
Assuming a flat mesh, aligned parallel to travel, then the only signals that would pass through in the train frame would be those traveling from the emitter on an orthogonal path [with some limited deviation] relative to the screen. Those outside this narrow angle would be reflected.Clearly this path, which is orthogonal in the train frame, is at some forward angle in the ground frame.
So if the observers were lined up along the track , only a limited number of observers, falling within this narrow sector at any given time, would receive the signal. But this is expected because the other observers ,further up and down the line would be at too acute an angle [wrt the mesh ] to receive a signal in any case.
It seems unlikely there could be a red shift. For this to occur the path angle would have to be toward the rear in the ground frame. Since aberration shifts emited path angles in the train frame, forward in the ground frame, for a signal to have an angle toward the rear in the ground frame would require the emitted signal to have an even greater angle toward the rear in the train F and also wrt the mesh.
Having thought it over it would also depend on boundary conditions. Distance of emitter from mesh, wavelength etc. I guess if the conditions were loose enough and a wide enough angle of signals could pass through the mesh then some red shifted signals would be observed on the ground.
In this case it seems like it would still only be observed by a limited number of observers at a time but then each would receive a changing signal, shifting from blue through the spectrum to red.
Interestin scenario. Thoughts?

 

[Saw]

Hmm. Yes, you are right in that, to determine what happens, what matters is how the signal arrives (red or blue shifted) in the position where it arrives. You are also right in holding that the signal is blue shifted in the sense that, in the ground frame, the signal is projected with a direction that has two components, one in the Y axis and another in the +X axis. Peak n, since it is emitted later than n-1, is emitted from a place that is farther away from the origin in the +X axis of the ground frame. However, the peculiarity here is that no single observer receives all peaks. If you make observers small enough, you would get one observer for each peak...

Is this what is called transverse Doppler shift? I read the wikipedia entry on Relativistic Doppler Shift and the section on the Transverse one, but did not understand what it says about the latter.

 

[Austin0]

Hmm. Yes, you are right in that, to determine what happens, what matters is how the signal arrives (red or blue shifted) in the position where it arrives. You are also right in holding that the signal is blue shifted in the sense that, in the ground frame, the signal is projected with a direction that has two components, one in the Y axis and another in the +X axis. Peak n, since it is emitted later than n-1, is emitted from a place that is farther away from the origin in the +X axis of the ground frame. However, the peculiarity here is that no single observer receives all peaks. If you make observers small enough, you would get one observer for each peak...

Is this what is called transverse Doppler shift? I read the wikipedia entry on Relativistic Doppler Shift and the section on the Transverse one, but did not understand what it says about the latter.

If by peaks you are referring to wave peaks then of course every observer gets a full set [i.e.at least one photon] otherwise you must explain what you mean.
On the two components. If a signal has a positive x component in direction it is blue shifted if negative , red shifted.[assuming +x travel]
I'm not sure but I think this is transverse Doppler in a nutshell. I believe the math is simply the R Doppler with the inclusion of the trig to calculate the variation with angle.But may be wrong
My assumption is that you have maximal blue shift on distant approach, maximal red shift on distant recession and then shift through the spectrum in between in passing, as the angle changes. At some point having no shift, probably at the closest point. If this is incorrect hopefully someone will supply the right picture.ciao