Is it only the speed of light in a vacuum that is constant for all observers?

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Assume that both light and an observer are travelling through water with a refractive index of 1.33. Without getting my calculator out, I think this means that the light would be travelling at about 0.75c.
The observer is travelling in the opposite direction at about 0.25c (it is only a thought experiment) does she measure the speed of the light as it passes her as “c”, or “0.75c”?
 

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  • #2
Drakkith
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The light would propagate through the water at .75c or so in the observers frame, but the frequency would be higher just like it would in space.
 
  • #3
Doc Al
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Assume that both light and an observer are travelling through water with a refractive index of 1.33. Without getting my calculator out, I think this means that the light would be travelling at about 0.75c.
The observer is travelling in the opposite direction at about 0.25c (it is only a thought experiment) does she measure the speed of the light as it passes her as “c”, or “0.75c”?
Neither. To calculate the speed of the light with respect to the moving observer, you must apply the 'addition of velocity' formula. Turns out to be about 0.84c.
 
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Now I am really confused! Drakkith says 0.75c; Doc Al says 0.84c.

Doc Al, perhaps if you could explain the process by which you arrived at your figure it might help me. The sad fact is that I am neither a scientist, nor a mathematician, just a dabbler with a reluctance to take anything on "faith".

Thanks for your replies.
 
  • #5
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Assume that both light and an observer are travelling through water with a refractive index of 1.33. Without getting my calculator out, I think this means that the light would be travelling at about 0.75c.
The observer is travelling in the opposite direction at about 0.25c (it is only a thought experiment) does she measure the speed of the light as it passes her as “c”, or “0.75c”?

it shud be c,as max speed in present dimension goes till "c",as in vaccum c+c also gives c
the increase in speed of light exists,hence even in water max speed limit for c.
cheers!!!
 
  • #6
Doc Al
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Now I am really confused! Drakkith says 0.75c; Doc Al says 0.84c.
Well, at least one of us is wrong. :wink:
Doc Al, perhaps if you could explain the process by which you arrived at your figure it might help me. The sad fact is that I am neither a scientist, nor a mathematician, just a dabbler with a reluctance to take anything on "faith".
Sure. (Of course, unless you know something about relativity, you'll be taking my explanation on 'faith'. Oh well.)

Lets say the water is at rest on earth. So, with respect to the earth, the beam of light moves at a speed of 0.75c to the east while the 'observer' moves at 0.25c to the west.

Classically, to find the relative velocity of the light pulse with respect to the observer you'd just add the velocities: w = u + v = 0.75c + 0.25c = c. But according to special relativity, velocities don't add like that. Instead that equation must be modified like so:
w = (u + v)/(1 + u*v/c^2)

Plug in the numbers and you'll see how I got 0.84c.

Note that for ordinary, everyday speeds that are much less than c, w = u + v works just fine.

For more on how velocities add in special relativity, see:
http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html" [Broken]
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html#c1"

To answer your title question directly, Is it only the speed of light in a vacuum that is constant for all observers? Yes! That's exactly it. It's the speed c that's invariant, not speeds less than c.
 
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  • #7
Well, at least one of us is wrong. :wink:

Sure. (Of course, unless you know something about relativity, you'll be taking my explanation on 'faith'. Oh well.)

Lets say the water is at rest on earth. So, with respect to the earth, the beam of light moves at a speed of 0.75c to the east while the 'observer' moves at 0.25c to the west.

Classically, to find the relative velocity of the light pulse with respect to the observer you'd just add the velocities: w = u + v = 0.75c + 0.25c = c. But according to special relativity, velocities don't add like that. Instead that equation must be modified like so:
w = (u + v)/(1 + u*v/c^2)

Plug in the numbers and you'll see how I got 0.84c.

I think I agree with Drakkith here, although I may be wrong. Like in a vacuum, shouldn't the speed of light depend on one over the square root of epsilon and mu? And shouldn't the values for epsilon and mu in a certain medium be independent of whatever speeds you're traveling at? If so, then shouldn't light propagate at the same speed in a certain medium regardless of what speed an observer is travelling at? (I'm just paraphrasing the arguments my book made for the constant speed of light in a vacuum :smile: )
 
  • #8
Vanadium 50
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No, because epsilon and mu are observer-dependent. (Indeed, they are not even scalars)

Consider the limit of "optical flypaper" - an index of refraction of a zillion, so any light that falls on it travels through it with an arbitrarily slow speed. In this case, the speed of the flypapered light in any frame is the same as the flypaper.

The reason for this is a moving magnetization results in a polarization and vice versa, and that's what causes epsilon and mu to deviate from their vacuum values.
 
  • #9
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Thanks folks, I think I am beginning to see through the "haze".
BTW, could someone please explain epsilon and mu - Greek letters, obviously - but what is their significance here?
 
  • #10
Vanadium 50
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Permittivity and permeability.
 
  • #11
No, because epsilon and mu are observer-dependent. (Indeed, they are not even scalars)

Consider the limit of "optical flypaper" - an index of refraction of a zillion, so any light that falls on it travels through it with an arbitrarily slow speed. In this case, the speed of the flypapered light in any frame is the same as the flypaper.

The reason for this is a moving magnetization results in a polarization and vice versa, and that's what causes epsilon and mu to deviate from their vacuum values.

Thanks for the clarification! So in the original scenario, would we then use the relativistic velocity addition formula, like what Doc Al did?
 
  • #12
Vanadium 50
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That's probably the easiest way to do it.
 
  • #13
K^2
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Refraction index is not invariant* under Lorentz transformation, so might actually be able to measure a difference in speed of light propagation in water from different reference frames.

* Consider what happens to energy levels of an atom under Lorentz transformations.
 
  • #14
Doc Al
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so might actually be able to measure a difference in speed of light propagation in water from different reference frames.
Sure you can. Recall Fizeau's classic experiment of 1851 which measured the speed of light in moving water. (And which is most easily explained using the relativistic addition of velocity approach that I described above.)
 
  • #15
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Its important to distinguish between phase velocity and group velocity. The (phase) refractive index, which is most commonly used, may be less than one, and even negative in some exotic conditions. This does not contradict relativity, since the phase velocity is the wavelength times the frequency, which is not necessarily the velocity of the propagating wave front.
 
  • #16
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The speed of light should be the same for every observer in all reference frames, I've gotta believe that if the speed of light in water is 0.75c, then it must be this speed in water for all reference frames.
 
  • #17
Vanadium 50
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The speed of light should be the same for every observer in all reference frames

That's the speed of light in vacuum.

I've gotta believe that if the speed of light in water is 0.75c, then it must be this speed in water for all reference frames.

See above.
 
  • #18
Rap
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The speed of light should be the same for every observer in all reference frames, I've gotta believe that if the speed of light in water is 0.75c, then it must be this speed in water for all reference frames.

You have to look at whats happening microscopically. Light hits an atom, it gets absorbed, then a little later it gets re-emitted, with some phase change that might not be purely a function of the time between emission and absorption. If you are looking at a beam thru water, what is the speed of light? The original photons have all been absorbed and re-emitted many times. The light you see is light (re)emitted from atoms near your eye. Is the speed of light the distance from where the beam began to your eye divided by the time from when it entered to the time you first saw it? That will be less than or equal to c (c=speed of light in vacuum). Or will it be the wavelength times the frequency of the light hitting your eye? That can be greater, equal, or less than c. For someone moving with respect to the water, the atoms will not only be absorbing, waiting, re-emitting, but moving as well. c is the universal constant, not the speeds calculated as the result of many absorptions and emissions from water that may or may not be moving.
 
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  • #19
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Huh, ok, thanks for the clarification.
 
  • #20
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The speed of light should be the same for every observer in all reference frames, I've gotta believe that if the speed of light in water is 0.75c, then it must be this speed in water for all reference frames.

Does this not take us back to the original question? Earlier responses suggest that it is only in a vacuum that the speed of light is the same for all observers, which fits well with the idea that the constancy is not a unique feature of light ,but would apply to anything that was able to attain a speed of "c".
 
  • #21
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soothsayer said:
The speed of light should be the same for every observer in all reference frames, I've gotta believe that if the speed of light in water is 0.75c, then it must be this speed in water for all reference frames.

Looking again at your post, would I be right in thinking that what you are saying here is that if you plug the speed of light into the equation w = (u + v)/(1 + u*v/c^2), as 0.75, then the speed of light in water will be constant for all observers. This is a point that has just been raised by a poster in another forum, and the maths seem to work.
 

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