# Speed of light constant in NON-vacuum?

## Main Question or Discussion Point

I know that the speed of light in a vacuum is always measured to be the same regardless of the observer's frame of reference. My question is, does the same apply to the speed of light as it travels through any non-vacuum medium? My gut feeling tells me yes, since its speed through the medium is simply an expression of the laws of physics, just as it is with no medium, which also remain the same no matter what your frame of reference relative to them is. Am I right?

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I couldn't find a solid answer there. Either that or I didn't understand the explanation. Is the speed of light only less in matter because it reflects back and forth, or are the fundamental characteristics of the wave itself changed to propogate more slowly? Or none of the above?

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PAllen
2019 Award
I know that the speed of light in a vacuum is always measured to be the same regardless of the observer's frame of reference. My question is, does the same apply to the speed of light as it travels through any non-vacuum medium? My gut feeling tells me yes, since its speed through the medium is simply an expression of the laws of physics, just as it is with no medium, which also remain the same no matter what your frame of reference relative to them is. Am I right?
This is wrong. You missed the example of Cerenkov radiation. Here, a massive particle goes faster than light through a medium. So, lab frame light is going one direction. Frame of this particle it is going the other direction. There can be particles which keep up with light in a medium. The aggregate speed of light in a medium is not fundamental at all, in the way speed of light in a vaccuum is.

OK.... Let me make this simple, since I still don't know what the answer is or if you have even explained it.

Say light travels at 0.9c through water, as measured from a stationary observer relative to the water. Will it travel at 0.9c through the water as measured by an observer moving 0.9c relative to the water?

PAllen
2019 Award
OK.... Let me make this simple, since I still don't know what the answer is or if you have even explained it.

Say light travels at 0.9c through water, as measured from a stationary observer relative to the water. Will it travel at 0.9c through the water as measured by an observer moving 0.9c relative to the water?
No. I thought I answered exactly this question in my last post.

Bill_K
The fundamental speed in nature is c, both inside the water and outside. Yes, light waves propagate more slowly in water, but as the Cerenkov example illustrates, the limiting speed is still c regardless. In water, particles can travel faster than light waves.
Say light travels at 0.9c through water, as measured from a stationary observer relative to the water. Will it travel at 0.9c through the water as measured by an observer moving 0.9c relative to the water?
Special relativity gives us a formula to combine velocities: v = (v1 + v2)/(1 + v1v2/c2). In your example, take v1 = v2 = 0.9 c.

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Special relativity gives us a formula to combine velocities: v = (v1 + v2)/(1 - v1v2/c2). In your example, take v1 = v2 = 0.9 c.
And I am familiar with this equation. I guess what I fundamentally don't understand is what is happening to light as it passes through a medium, and why it slows down.

Bill_K
The passage of light through a medium is a collective motion involving the atoms/ions/molecules that constitute the substance. An atom exposed to a constant uniform electric field will become polarized, i.e. develop an electric dipole moment. An oscillating field will induce an oscillating dipole moment, and this itself generates an oscillating field that combines with the incident one. The wave we see is a superposition of all these contributions.

If the index of refraction is greater than 1, the wave speed is less than c. But as the Wikipedia article on Refractive Index points out, the index of refraction is sometimes less than 1, and the wave speed is greater than c.

In either case, note that it is the phase velocity we're talking about, not the group velocity or velocity with which energy is transmitted.

PAllen
2019 Award
Note, though, for visible light in ordinary transparent media, the group velocity differs only slightly from the phase velocity.

I think jmvizanko is asking is $\frac{1}{\sqrt{\epsilon\mu}}$ an invariant

Interesting stuff, I would have thought in a vacuum there are simply no atoms in the way to absorb and re-emit photons but through a 'medium' it is interupted by them. Am I missing something here?

Interesting stuff, I would have thought in a vacuum there are simply no atoms in the way to absorb and re-emit photons but through a 'medium' it is interupted by them. Am I missing something here?
Refraction is not exactly absorption and subsequent re-emission, but it does imply interference of matter with light. Light waves can be considered as propagating at c, whereby all the time primary waves are being replaced by retarded secondary waves as a result of the interaction with matter. That results in an effectively slowed down resultant wave. This was discussed in several threads (and also in post #9 here above).

This was discussed... also in post #9 here above.
Okay, I was fairly sure (but not certain) that was what Bill K's post was getting at; just dumbing it down to my level lol.

the wavelength of the oscillating field is changed in a non vacuum medium.as wavelength in a medium=origional wavelength/n(n is refractive index).but as Bill K said the energy transfer rate is the same as c.if somebody is travelling at a speed of 0.9c in water then his clock will decelerate and acoordingly the relative velocities will be given by w=(v1+v2)/(1+v1V2/c^2) as per the frame of referrence.

the wavelength of the oscillating field is changed in a non vacuum medium.as wavelength in a medium=origional wavelength/n(n is refractive index).but as Bill K said the energy transfer rate is the same as c.if somebody is travelling at a speed of 0.9c in water then his clock will decelerate and acoordingly the relative velocities will be given by w=(v1+v2)/(1+v1V2/c^2) as per the frame of referrence.
So it's not because the photons are being absorbed by atoms and re-emitted? Or in other words you are saying (and Bill K) that the rate at which the photon "moves" (absorbs/emits) through the atom is still c for it's original path?

the atoms and molecules do come on the path.but the oscillating field that decides.it's a superposition principle of too many waves.but the energy transfer rate is a constant c.

PAllen
2019 Award
the atoms and molecules do come on the path.but the oscillating field that decides.it's a superposition principle of too many waves.but the energy transfer rate is a constant c.

1) Phase velocity of the wave is slower (for normal materials) by the index of refraction.
2) The constant for determining the physics of any microscopic interactions, and for velocity addition and time dilation, remains c.
3) The energy and signal transport rate is (for normal materials) the group velocity, which will differ from the phase velocity by an amount related to the dispersion of the material. For light, in normal transparent media, this will differ by a very small amount from the phase velocity, and will be slower than the phase velocity.

ZapperZ
Staff Emeritus
Wait till many people discover that in waveguides, the speed of light even in vacuum can also be less than "c"!

:)

Hopefully, this will make people sit up and take notice of the difference between the phase velocity and the group velocity that have been mentioned.

Zz.

Thank you.actually I didn't know the concept very clearly.Now, i am getting what you want to say.You made me clear.

I think jmvizanko is asking is $\frac{1}{\sqrt{\epsilon\mu}}$ an invariant
In Wave Mechanics $\ c^2 = \frac{1}{\epsilon\mu} \$ is an invariant, i.e., $\ c^2 = \mathbf{v}_{phase} \mathbf{v}_{group} \$ while $\ c = \frac{1}{\sqrt{\epsilon_0\mu_0}} \$ is not an invariant.