Doppler effect in transparent media and the index of refraction

[member 743765]

If there is a light source that is moving relative to an observer and both are inside water for example for an observer at rest relative to the source the emitted light will have a certain frequency,however for the other observer in relative motion the frequency will change and hence the refractive index is that true? I want to mention that I know about the drag coefficient and I think that 1/n +-(1-1/n^2)v/c must equal 1/the index of refraction of the new frequency from doppler

 

[Ibix]

Are you thinking of Fizeau's experiments with light in flowing water? There the key point is that light travels in water at $c/n$ relative to the water, but if you are moving relative to the water you use relativistic velocity addition to compute the speed of the light with respect to you, which you can then interpret as a modified refractive index of moving water.

I'm not sure where frequency measurement would come in to this.

 

[member 743765]

Are you thinking of Fizeau's experiments with light in flowing water? There the key point is that light travels in water at $c/n$ relative to the water, but if you are moving relative to the water you use relativistic velocity addition to compute the speed of the light with respect to you, which you can then interpret as a modified refractive index of moving water.

I'm not sure where frequency measurement would come in to this.

From doppler effect when a source of light is in relative motion with respect to the observer its frequency shifts relative to the observer and this new frequency has its own index of refraction that I think must equal the modified index of refraction and yes consider fizau experiment

 

[Ibix]

So, you have a source that emits at $\nu_0$ in its rest frame, but it's moving at some speed with respect to the water so in the water rest frame it emits at $\nu_1$ then it is the refractive index at this frequency (say $n_1$) that you need to consider. The light travels at $v_1=c/n_1$ in this frame. Transform $v_1$ into your receiver's rest frame, call this $v_2$, and the apparent refractive index is $c/v_2$.

 

[member 743765]

So, you have a source that emits at $\nu_0$ in its rest frame, but it's moving at some speed with respect to the water so in the water rest frame it emits at $\nu_1$ then it is the refractive index at this frequency (say $n_1$) that you need to consider. The light travels at $v_1=c/n_1$ in this frame. Transform $v_1$ into your receiver's rest frame, call this $v_2$, and the apparent refractive index is $c/v_2$.

Yes and this n2 must equal the modified index of refraction ....am I right?

 

[Ibix]

Yes and this n2 must equal the modified index of refraction ....am I right?

I don't know what you mean here by "the modified index of refraction".

 

[member 743765]

I don't know what you mean here by "the modified index of refraction".

I mean n2 must equal (1/n1+-(1-1/n1^2)v/c)^-1

 

[Vanadium 50]

This question is swirling around "what does the OP mean?"

OP, please:

• Write a clear statement of the question.
• Define your terns. "n is the index of refraction" is not adequate. Instead try "n is the index of refraction of water in the frame where it is at rest".
• Please use LaTeX.

The first step in getting an answer is to clearly ask the question.

 

[Ibix]

I mean n2 must equal (1/n1+-(1-1/n1^2)v/c)^-1

That was what you said the drag coefficient was, which is not the same as the refractive index. The drag coefficient is $f=1-1/n^2$, and the $n$ in that expression would be $n_1$ (edit: not $c/v_2$, as I originally wrote) in the notation of my last post.

I believe that the expression you quote for $f$ is a low speed approximation. You would need to Taylor expand the expression derived from SR as I outlined and discard all except the leading order term to obtain it.

 

[member 743765]

That was what you said the drag coefficient was, which is not the same as the refractive index. The drag coefficient is $f=1-1/n^2$, and the $n$ in that expression would be $c/v_2$ in the notation of my last post.

I believe that the expression you quote for $f$ is a low speed approximation. You would need to Taylor expand the expression derived from SR as I outlined and discard all except the leading order term to obtain it.

I'm saying that n in the drag coefficient is n1 and the expression I wrote is the factor that modifies the speed and that this factor is 1/n2

 

[Ibix]

I'm saying that n in the drag coefficient is n1

Agreed - I've corrected my post above.

the expression I wrote is the factor that modifies the speed and that this factor is 1/n2

I don't think so. What you wrote is rather hard to parse because you haven't used LaTeX (please start doing so - it's straightforward and there is a LaTeX Guide linked immediately below the reply box), but it appears to be Fresnel's expression for the drag coefficient measured in the frame where the water is moving at $v$ parallel to the light.

 

[member 743765]

Agreed - I've corrected my post above.

I don't think so. What you wrote is rather hard to parse because you haven't used LaTeX (please start doing so - it's straightforward and there is a LaTeX Guide linked immediately below the reply box), but it appears to be Fresnel's expression for the drag coefficient measured in the frame where the water is moving at $v$ parallel to the light.

Sorry for being not so clear ....English is not my first language. What I meant is that when the observer and the source are inside the water and the source is moving relative to the observer there is a shift in frequency which must therefore changes the refractive index to that of the new frequency .....and this must be consistent with special relativity when you use the velocity transformations where you will find the drag coefficient part of the formula