Fresnel equations and special relativity

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So I'm thinking about Einstein's thought experiment in which an observer on a train has a clock that denotes time by bouncing light off a mirror and measuring the time for it to return. The setup is such that for the observer on the train, the light is traveling vertically - strictly perpendicular to the motion of the train. The light's path is just the line up to the mirror, back down to the detector.

For an observer on the train platform when the train is moving, the light's path makes an isosceles triangle whose height is the same, and whose base is determined by the velocity of the train (and the height and the speed of light).

That is all well established. What I am wondering about is replace the mirror at the top with partially silvered mirror that allows some transmission. The amount of light reflected versus transmitted depends on the angle of incidence. The observer on the train sees the light incident on the mirror at 0 degrees and maximum reflection. The observer on the platform sees the light incident on the mirror at some non-zero angle (the faster the train is moving, the higher the angle of incidence). The observer on the train then expects much higher intensity at his detector than the observer on the platform does.

To take it a step further, imagine that on the train the system is slightly angled so that the light from the source is not strictly perpendicular to the floor of the train, and the spot where it reflects back to is actually a hole in the floor of the train. Outside underneath the train, along the tracks, is a detector to measure the intensity of light. What intensity of light is measured at this external detector? If you could swap between closing the hole in the floor and measuring the intensity in the frame of reference of the train with opening the hole in the floor and measuring the intensity in the frame of reference of the platform, would they be the same?

See attached image for a crude diagram of the situations.
 

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  • #2
Bill_K
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Note that the mirror is moving along with the train. The atoms in the mirror are what's responsible for the reflection/transmission ratio, and they make this decision in their own rest frame. So the stationary observer should not be surprised to see that the moving mirror has an R/T ratio appropriate for normal incidence.
 
  • #3
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Good point. If it were atoms striking the surface and non-relativistic velocities, it would be obvious. Can light be treated the same way though?
 
  • #4
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So I'm thinking about Einstein's thought experiment in which an observer on a train has a clock that denotes time by bouncing light off a mirror and measuring the time for it to return. The setup is such that for the observer on the train, the light is traveling vertically - strictly perpendicular to the motion of the train. The light's path is just the line up to the mirror, back down to the detector.[..]
What I am wondering about is replace the mirror at the top with partially silvered mirror that allows some transmission. The amount of light reflected versus transmitted depends on the angle of incidence. The observer on the train sees the light incident on the mirror at 0 degrees and maximum reflection. The observer on the platform sees the light incident on the mirror at some non-zero angle (the faster the train is moving, the higher the angle of incidence). The observer on the train then expects much higher intensity at his detector than the observer on the platform does.
Surely the Fresnel equations are meant for a system in rest; for other cases one should include aberration and the Doppler effect.
To take it a step further, imagine that on the train the system is slightly angled so that the light from the source is not strictly perpendicular to the floor of the train, and the spot where it reflects back to is actually a hole in the floor of the train. Outside underneath the train, along the tracks, is a detector to measure the intensity of light. What intensity of light is measured at this external detector? If you could swap between closing the hole in the floor and measuring the intensity in the frame of reference of the train with opening the hole in the floor and measuring the intensity in the frame of reference of the platform, would they be the same? [..]
This is simpler: nothing to do with Fresnel anymore, just Doppler and aberration.
See sections 7 and in particular 8 of:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
 
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  • #5
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Thank you, I think section 8 does address it. I'll need dig deeper, but the conclusion that you should just study it in the frame where the optics are at rest is basically the answer. But I don't really see what the doppler effect has to do with it (it's not a frequency / wavelength question).
 
  • #6
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Thank you, I think section 8 does address it. I'll need dig deeper, but the conclusion that you should just study it in the frame where the optics are at rest is basically the answer. But I don't really see what the doppler effect has to do with it (it's not a frequency / wavelength question).
In fact it is particularly section 7 (Doppler and aberration) that is relevant. A of section 8 is the amplitude A of section 7; the intensity is given right at the end of section 7 as A2. :smile:
 
  • #7
Bill_K
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Sorry, this reference is for a perfect reflector, not a partial one.
 
  • #8
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In fact it is particularly section 7 (Doppler and aberration) that is relevant. A of section 8 is the amplitude A of section 7; the intensity is given right at the end of section 7 as A2. :smile:
Yeah, I don't think that has anything to do with reflection or partial reflection. Are you just throwing references and equations around?
 
  • #9
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Yeah, I don't think that has anything to do with reflection or partial reflection. Are you just throwing references and equations around?
No, I did the effort of looking the answer up for you and giving you a link to it, despite you throwing "Fresnel equations" in the title which have nothing to do with your second, and main, question. Also reflection is irrelevant: in your sketches you can simplify your question by putting the light source at the place of the mirror.
 
  • #10
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OK then, rather than just point to the equation, can you explain why this has anything to do with the Doppler effect?

The fresnel equations are directly relevant because they explain the ratio of reflection and transmission as a function of angle of incidence, and my question is precisely about that.
 
  • #11
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OK then, rather than just point to the equation, can you explain why this has anything to do with the Doppler effect?

The fresnel equations are directly relevant because they explain the ratio of reflection and transmission as a function of angle of incidence, and my question is precisely about that.
OK, sorry I had forgotten that the top mirror is partly silvered. That is not Fresnel: Fresnel applies to the unsilvered glass fraction. But you correctly understood that the Fresnel equations relate to the rest frame of the optics.

Then about intensity: he derives there, without clearly showing how, that a higher frequency gives an increased amplitude and therefore an increased intensity. I think that this was discussed once on this forum... Maybe someone else can fill that in; I'm now going out to enjoy the sun. :smile:

Oh and note that relativistic Doppler also includes time dilation; as intensity is per time unit,
http://hyperphysics.phy-astr.gsu.edu/hbase/vision/photom.html#c1
time dilation necessarily affects the measured intensity.

ADDENDUM: I suddenly realise that it's very easy to explain. Suppose that you emit 1 MHz for 1 s: then you emitted 1 million cycles in 1 s, and that corresponds to a certain amount of emitted energy. If this is received at 2 MHz, then it is received in half a sec. The energy is received in half the time, and thus the received power (Watt=J/s) must be double. Double the power means double the intensity (see Hyperphysics). :tongue2:
 
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  • #12
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OK, sorry I had forgotten that the top mirror is partly silvered. That is not Fresnel: Fresnel applies to the unsilvered glass fraction. But you correctly understood that the Fresnel equations relate to the rest frame of the optics.

Then about intensity: he derives there, without clearly showing how, that a higher frequency gives an increased amplitude and therefore an increased intensity. I think that this was discussed once on this forum... Maybe someone else can fill that in; I'm now going out to enjoy the sun. :smile:

Oh and note that relativistic Doppler also includes time dilation; as intensity is per time unit,
http://hyperphysics.phy-astr.gsu.edu/hbase/vision/photom.html#c1
time dilation necessarily affects the measured intensity.

ADDENDUM: I suddenly realise that it's very easy to explain. Suppose that you emit 1 MHz for 1 s: then you emitted 1 million cycles in 1 s, and that corresponds to a certain amount of emitted energy. If this is received at 2 MHz, then it is received in half a sec. The energy is received in half the time, and thus the received power (Watt=J/s) must be double. Double the power means double the intensity (see Hyperphysics). :tongue2:
PS: My "easy explanation" above looks at the energy transfer to the track according to the rest frame of the optics; this is a big (first order) effect, and in classical physics that would be all. I hope that that clarifies why Doppler matters. However in SR there is, apart of time dilation, also the transformation of energy to the rest frame of the track to account for. That is a smaller effect which is derived in section 8.
 
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  • #13
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I understand the relativistic doppler effect. What I don't understand is how it applies to the situation I've described. It's not a change in frequency, and it's not a change in intensity; it's a change based on frame of reference of where the energy ends up.

In the train frame of reference, the angle of incidence is closer to normal, thus the should be more transmission (roughly speaking) than for that observed from platform frame of reference where there is a much more glancing angle of incidence. If transmission is different in these frames then the you have a photons ending up in radically different locations based on the observer's position.
 
  • #14
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If the intensity of the reflected light depends on the angle of incidence θ like Iref = Iinc sin(θ) then I transforms like Iref = Iinc/γ when seen from a moving frame. I thought it was interesting ...
 
  • #15
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I understand the relativistic doppler effect. What I don't understand is how it applies to the situation I've described. It's not a change in frequency, and it's not a change in intensity; it's a change based on frame of reference of where the energy ends up.
If you understand the relativistic Doppler effect - which concerns emitters and detectors in relative motion - then you can tell us how it applies to your train emitter and railway track detector which are in relative motion...
In the train frame of reference, the angle of incidence is closer to normal, thus the should be more transmission (roughly speaking) than for that observed from platform frame of reference where there is a much more glancing angle of incidence. If transmission is different in these frames then the you have a photons ending up in radically different locations based on the observer's position.
I already told you about your silvered mirror; your reply completely ignores that. Why?? :confused:
I don't grasp what you mean with your last sentence. But as you understand relativistic Doppler, please elaborate what you mean with "photons ending up in radically different locations based on the observer's position." Please give a calculation example. Likely more people will then get involved.
 
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If you understand the relativistic Doppler effect - which concerns emitters and detectors in relative motion - then you can tell us how it applies to your train emitter and railway track detector which are in relative motion...

I already told you about your silvered mirror; your reply completely ignores that. Why?? :confused:
I don't grasp what you mean with your last sentence. But as you understand relativistic Doppler, please elaborate what you mean with "photons ending up in radically different locations based on the observer's position." Please give a calculation example. Likely more people will then get involved.

The doppler effect concerns the change in frequency for different frames of reference. What does frequency have to do with it?

Partially silvered mirrors just have higher reflectivity than other materials. If it really is a problem, replace the silvered mirrors with regular glass or some transparent material with a higher index of refraction. It doesn't change fundamental problem.
 
  • #17
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The doppler effect concerns the change in frequency for different frames of reference. What does frequency have to do with it?
That looks, at first sight, inaccurately phrased; but in view of your question it must be worse.
As I tried to explain in posts #11 the Doppler effect first of all accounts for the effect of receiving a number of cycles over a different duration than the duration of emission, so that the received power (and thus also the intensity) differs from the emitted power - that is in a single reference system. I even gave a simple calculation example. What is not clear about that? :confused:
Partially silvered mirrors just have higher reflectivity than other materials. If it really is a problem, replace the silvered mirrors with regular glass or some transparent material with a higher index of refraction. It doesn't change fundamental problem.
Yes, for your example it is better to take regular glass. Now, if you give a numerical example with your calculation of "photons ending up in radically different locations based on the observer's position", then there will be something more precise to discuss.
 
  • #18
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The doppler effect concerns the change in frequency for different frames of reference. What does frequency have to do with it?

Partially silvered mirrors just have higher reflectivity than other materials. If it really is a problem, replace the silvered mirrors with regular glass or some transparent material with a higher index of refraction. It doesn't change fundamental problem.
What problem ? You asked a question in your original post. Please explain the problem you see.
 
  • #19
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That looks, at first sight, inaccurately phrased; but in view of your question it must be worse.
As I tried to explain in posts #11 the Doppler effect first of all accounts for the effect of receiving a number of cycles over a different duration than the duration of emission, so that the received power (and thus also the intensity) differs from the emitted power - that is in a single reference system. I even gave a simple calculation example. What is not clear about that? :confused:

Yes, for your example it is better to take regular glass. Now, if you give a numerical example with your calculation of "photons ending up in radically different locations based on the observer's position", then there will be something more precise to discuss.
What is not clear, as I have stated repeatedly, is what this has to do with the original question. If you want to continue, can you please look at the diagram I provided?

To restate: the train frame of reference sees the light incident on the glass at near normal incidence, meaning high transmittance. More photons, more intensity end up going through the glass than being reflected. The station frame of reference sees the light incident at a much more glancing angle of incidence, meaning much lower transmittance. Fewer photons, less intensity end up going through the glass than being reflected.

So in one frame of reference there are more photons, more energy above the glass. In the other frame of reference there are more photons below (reflected from) the glass.

So what does frequency have to do with any part of this? Yes, the frequency of the light observed in the 2 frames is different. What does that have to do with the above observations?
 
  • #20
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What problem ? You asked a question in your original post. Please explain the problem you see.
harrylin has a problem with the use of silver mirrors saying the fresnel equations don't apply. That is the problem. I said if that is a problem, replace the silvered mirrors in the example with glass.
 
  • #21
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assume / given:
* the train is moving at 0.5*c
* the light source is stationary in the train frame of reference
* in the train frame of reference the light is incident perpendicular to the plane of the glass
* the light is s-polarized with respect to the direction of motion of the train
* glass index of refraction is ~1.5
* air index of refraction ~1

fresnel equations:
http://en.wikipedia.org/wiki/Fresnel_equations

In the train frame of reference based on the fresnel equations, the reflection coefficient is ~5%

from the station the light is incident on the glass at 45 degrees. The reflection coefficient is ~10%

The station observer sees / expects twice as many photons to be reflected as the train based observer.

Now, as I've said before, I understand from the section on power that the above is wrong - I just don't understand why. How does the doppler effect tie into this?
 
  • #22
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What is not clear, as I have stated repeatedly, is what this has to do with the original question. If you want to continue, can you please look at the diagram I provided?

To restate: the train frame of reference sees the light incident on the glass at near normal incidence, meaning high transmittance. More photons, more intensity end up going through the glass than being reflected. The station frame of reference sees the light incident at a much more glancing angle of incidence, meaning much lower transmittance. Fewer photons, less intensity end up going through the glass than being reflected.

So in one frame of reference there are more photons, more energy above the glass. In the other frame of reference there are more photons below (reflected from) the glass.

So what does frequency have to do with any part of this? Yes, the frequency of the light observed in the 2 frames is different. What does that have to do with the above observations?
The answer is that in both frames there are more photons passing through than reflected.

That the percentage is dependent on the angle in the source frame.

A comparable situation is a flat screen parallel to motion with a mesh only sufficiently wide to admit a certain wavelength. AN isotropic emitter behind the screen will only pass those photons which reach the screen on an orthogonal path from the moving emitter to screen moving with it. In a frame where this is in motion the path of photons which reach them through the screen will be angled in the forward direction in that frame.
So in the observation frame those photons shouldn't make it through the screen because their angle of incidence should prevent their passage. But they do.
 
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  • #23
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What is not clear, as I have stated repeatedly, is what this has to do with the original question. If you want to continue, can you please look at the diagram I provided?
Perhaps I concentrated too much on your diagram...
To restate: the train frame of reference sees the light incident on the glass at near normal incidence, meaning high transmittance. More photons, more intensity end up going through the glass than being reflected.
"More photons" complicates a supposedly SR description (QM is not SR). Let's stick to light waves.
The station frame of reference sees the light incident at a much more glancing angle of incidence, meaning much lower transmittance. [..] less intensity end up going through the glass than being reflected.
Ehm no, please check the units. Intensity is proportional to power -> J/s.
Power doesn't "end up going through the glass", it is what goes through the glass per second. That's what I have been telling you for the last three or four posts.
[..] So in one frame of reference there [is...] more energy above the glass. In the other frame of reference there [is.. more energy] below (reflected from) the glass. So what does frequency have to do with any part of this? Yes, the frequency of the light observed in the 2 frames is different. What does that have to do with the above observations?
It's good that you reduced it back to the simpler case. For that simpler case you won't need Doppler and you already had in principle the answer from my post #4: the Fresnel equations are meant for a system in rest. That were the top two sketches in your picture; and your question about that scenario was apparently settled, in principle, in your post #5. It may be good if you elaborate on your conclusion there by applying that conclusion to your calculation example. :tongue2:

Concerning your second scenario (bottom two sketches), you originally asked " What intensity of light is measured at this external detector" (on the track). I explained why the frequency of the light (and thus also the power and intensity) according to any 1 reference system is different between what is reflected and what is received; energy must be conserved.
 
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  • #24
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Perhaps I concentrated too much on your diagram...

"More photons" complicates a supposedly SR description (QM is not SR). Let's stick to light waves.
# photons is directly proportional to intensity, no?

Ehm no, please check the units. Intensity is proportional to power -> J/s.
Power doesn't "end up going through the glass", it is what goes through the glass per second. That's what I have been telling you for the last three or four posts.
This is minor semantics at best. Is it really critical to be re-phrased on a discussion forum to say "the intensity observed on the other side of the glass" instead of the intensity transmitted? Really? really?

It's good that you reduced it back to the simpler case. For that simpler case you won't need Doppler and you already had in principle the answer from my post #4: the Fresnel equations are meant for a system in rest. That were the top two sketches in your picture; and your question about that scenario was apparently settled, in principle, in your post #5. It may be good if you elaborate on your conclusion there by applying that conclusion to your calculation example. :tongue2:
The "more complicated" case was just an example of how the difference could be measured. It's not really that more complicated.

Concerning your second scenario (bottom two sketches), you originally asked " What intensity of light is measured at this external detector" (on the track). I explained why the frequency of the light (and thus also the power and intensity) according to any 1 reference system is different between what is reflected and what is received; energy must be conserved.
I think you need to look at the photoelectric effect experiments...
 
  • #25
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[..] This is minor semantics at best. Is it really critical to be re-phrased on a discussion forum to say "the intensity observed on the other side of the glass" instead of the intensity transmitted? Really? really?
Not at all - and I give up: evidently I am unable to explain to you that energy≠power, etc. :uhh:
The "more complicated" case was just an example of how the difference could be measured. It's not really that more complicated.
I explained what makes it more complicated and why; but none of my explanations arrived at their destiny. I hope for you that someone else can and is willing...
I think you need to look at the photoelectric effect experiments...
I think that this is hopeless... Sorry mate, and good luck!

PS: v=0.5c does not result in 45 degrees angle...
 
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