# C / speed of light

1. Dec 16, 2007

### p4h

Quick, maybe stupid, question:

If we say the speed of light is a constant, how come when we talk about c, we talk about the speed of light in vacuum? I mean it shouldn't mean anything if we see c as a constant? Or maybe I'm missing something?

2. Dec 16, 2007

### Feldoh

You have to take into account optical density of a material. When light travels through a vacuum then it has the expected value of c. Light can appear slower than it actually is in certain mediums.

3. Dec 16, 2007

### p4h

So it's purely optical?

4. Dec 16, 2007

### Feldoh

Err, I said that very badly. c the velocity of the speed of light in a vacuum will never ever change. However the observed speed of the speed of light will be different when not in a vacuum.

The speed of light through a medium is v = c/n where n is the refraction index of a certain material. The refraction index in a vacuum is 1 hence in a vacuum v = c, but everywhere else the value of n will be greater than 1 which will slow down light when is passes through particular media

5. Dec 17, 2007

### p4h

Ah okay :) Thanks

6. Dec 18, 2007

### country boy

Don't forget about the speed of light in a gravitational field in a vacuum.

7. Dec 19, 2007

### Feldoh

Something to do with light being bent as it travels through space or something?

8. Dec 19, 2007

### HallsofIvy

Staff Emeritus
What exactly is your point? The speed of light in a gravitational field, in a vacuum, is c. A gravitational field changes the direction of light, it does not change its speed.

9. Dec 19, 2007

### pervect

Staff Emeritus
If you use local clocks and rulers, the speed of light in a vacuum is always 'c', regardless of any gravitational field. This is the modern interpretation.

See for instance the sci.physics.faq Is the speed of light constant?

Thus while some people, possibly including Einstein (the meaning of his original statement is debatable) have stated that the speed of light varies with proximity to a large mass, this is the result of NOT using local clocks and rulers. The physics is really simpler when one always uses local clocks and rulers to measure speed. And using this modern interpretation, the speed of light in a vacuum, c, is always constant.

Last edited: Dec 19, 2007
10. Dec 22, 2007

### country boy

See pervect's response. The modern meaning of "constant c" is based on the concept of local measurement, which is a useful idealization. An accurate measurement of the speed of light over a distance in a gravitational field will not give c. But such a measurement can never be strictly local.

11. Dec 22, 2007

### JesseM

But there is no single procedure for measuring the speed of an object over a nonlocal region in curved spacetime--that depends on your choice of coordinate system (for example, the speed of a light beam passing near a black hole might be different in Schwarzschild coordinates than it is in Eddington-Finkelstein coordinates). And as long as we're talking about non-inertial coordinate systems, there's no need to refer to gravity, since even in flat spacetime the speed of light can be different than c in a non-inertial coordinate system.

Last edited: Dec 23, 2007
12. Dec 22, 2007

### country boy

What do you mean by "flat spacetime," and "non-inertial?"

13. Dec 23, 2007

### JesseM

Flat spacetime is where the curvature is everywhere zero--the spacetime of special relativity. An inertial coordinate system is one where the laws of SR hold exactly, and an observer at rest in these coordinates experiences no G-forces. In a curved spacetime you can only have an inertial coordinate system in the limit as the size of the spacetime region covered by the coordinate system approaches zero--the infinitesimal neighborhood of a single point in spacetime, essentially. In flat spacetime, Rindler coordinates would be an example of a non-inertial coordinate system.

14. Dec 23, 2007

### country boy

Thanks for the introduction to Rindler coordinates. Interesting. If I understand correctly, it is a way to map a special accelerating frame onto a Minkowski space.

But my reply pertained to c being always the same in the modern interpretation and how this depends on the concept of local measurements. Your response seems consistent with my suggestion that purely local measurements cannot be achieved in practice.

15. Dec 23, 2007

### pervect

Staff Emeritus
For modest accelerations, the affect of acceleration on measurements will be very minor. Errors will be on the order of $\approx g d/ c^2$. So for instance if g is 10 m/s^2 (1 gravity) and d is 10 km, the error is about 1 part in 10^13.

The effect of acceleration on measurements can be thought of as the effect of "the" gravitational field on measurements, via the equivalence principle - at least if one uses the usual notion of "gravitational field" that is imported from Newtonian theory.

16. Dec 24, 2007

### HallsofIvy

Staff Emeritus
I think we've gotten way off the orginal simple question:
What you are missing is that the speed of light (in vacuum) is constant relative to any frame of reference even if it is moving. Imagine standing on a flatbed truck moving at 40 mph with another person. Let us say the other person throws a ball to you with speed, relative to the truck, of 60 mph. Since you are moving with the truck, the speed of the ball, relative to you, is 60 mph. The speed of the ball, relative to a person standing on the roadside, would be (approximately) 60+ 40= 100 mph. Light doesn't work that way. The same beam of light would have the same speed, c, relative to a person on the truck, a person on the roadway, or a person in a fast moving space ship.

I said "approximately" above because one of the results of relativity is that "normal" velocities" don't quite add that way: If object A is moving toward you with speed (relative to you) of u and object B is moving toward you from the other side (and so directly toward object A) with speed (relative to you) v, then they are moving, relative to one another, with speed
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
where c is the speed of light. Notice that, as long as u and v separately are less than c, that "sum" is also less than c. Notice also that, if either u is equal to c, that will be equal to c no matter what v is.

17. Dec 26, 2007

### RandallB

What you are describing is most simply thought of and physically observed in the Shapiro effect.
Where radar signals passing close to the sun were seen to take longer than expected, more so than could be accounted for by and minor bend causing a longer distance to be traveled. Real observations showed that the distances coved by photons very near the sun were shorter than indicated by a speed of “c”. However when close to the sun the photons are in a locale using a slower time reference and using that shorter distance traveled and shorter measure of time (both as observed form the distant reference frame) gives us the same constant speed of “c” for photons in all local frames.

By defining time and distance in those fields to be slower and shorter, GR had predicted the relative speed of light in strong gravitational fields would be slow long before the effect was actually measured.