# Speed of light

1. Jan 8, 2007

### anantchowdhary

Does speed of light Change with respect to accelerating frames

It shouldn't be as gravitational time dilation exists

2. Jan 8, 2007

### Hootenanny

Staff Emeritus
No, the speed of light is constant when viewed from any reference frame.

3. Jan 8, 2007

### quantum123

No, the speed of light slows down to = 0 at the Schwarzschild radius of the black hole, to the frame of the external observer.

Last edited: Jan 8, 2007
4. Jan 8, 2007

### Hootenanny

Staff Emeritus
This is not true, the photon is simply red shifted to zero frequency.

5. Jan 8, 2007

### quantum123

If you look at the Schwarzschild solution, you just let ds=0 (light-like), you will obtain
dr/dt = 1 - 2GM/r , which proves what I said.

6. Jan 8, 2007

### Chris Hillman

Oh dear oh dear...

Depends upon what you mean by "speed".

I am guessing that you mean velocity of a particle very near some observer, with respect the frame field (local Lorentz frame) of that observer. If so, the speed of light "at the level of tangent spaces" is indeed unity (in geometric units) in any Lorentzian manifold.

But note that "coordinate speed" is not in general geometrically or physically meaningful.

Also, there are many distinct operationally significant notions of "distance in the large" and thus "velocity in the large", even in flat spacetime (for accelerating observers, such as Rindler or Bell observers).

Here, in a Lorentzian manifold, Hootenanny's "reference frame" is what I called a frame above. A frame field is a set of vector fields which forms a basis for the tangent space at eacn event. Just a vector is one bit of a vector field, a frame is one bit of a frame field. Frames are also called vierbeins, among various other names.

Note that quantum123 is referring to a coordinate speed (slope) computed for radial null geodesics in a particular chart (the exterior Schwarzschild chart) for a particular exact solution of the EFE, the Schwarzschild vacuum. But the word "frame" is inappropriate here since frames are associated with operationally significant quantities, but coordinate speeds are not in general physically meaningful.

This is an easy way of finding the radial null geodesics in the above mentioned chart, and of verifying the property of the coordinate speeds which you mentioned.

But of course light never "slows down" in any Lorentzian manifold (that wouldn't even make sense!), which is an indication that this refers to a spurious property of a particular chart. That is, it is a (bad) property of a particular representation of (part of) this spacetime, not a propery of the spacetime itself. That is why it is not geometrically or physically meaningful.

7. Jan 9, 2007

### quantum123

Well it is bad to you but good to me. It is relative.
To the observer, the light photon will approach the event horizon, both its speed(dr/dt) and vibration slowing down, virtually stopping at the horizon. The same goes for any mass>0 object.

8. Jan 9, 2007

### pervect

Staff Emeritus
As Chris Hillman points out at some length, the word "speed" is ambiguous. The speed of light is always constant when measured with local clocks and local rulers. In fact, we have to go back to a very old defintion of the meter to even talk about measuring the speed of light - using the modern SI defintion of the meter, the speed of light (defined in the frame-field sense that Chris Hillman talked about at length) is a constant by defintion.

Therfore you need to be somewhat careful about assuming that people share the same coordinate-based defintion of speed that you are using. Your notion of "speed" is not compatible with the SI defintion of the speed of light, for instance, while Chris Hillman's defintion is.

9. Jan 9, 2007

### Thrice

SR vs GR again?

10. Jan 9, 2007

### quantum123

Speed |v| = sqrt(v1^2+v2^2+v3^2)
v1=d/dt(x1)
v2=d/dt(x2)
v3=d/dt(x3)
d/dt(x1)=d/dt(r)sin(theta)cos(phi)
d/dt(x2)=d/dt(r)sin(theta)sin(phi)
d/dt(x3)=d/dt(r)cos(theta)
Hence
d/dt(r)=0
=> d/dt(x1)=d/dt(x2)=d/dt(x3)=0
=> v1=v2=v3=0
=> |v|=0
=> speed =0

Last edited: Jan 9, 2007
11. Jan 9, 2007

### anantchowdhary

please gimme a definite answer.yes or no?And how do u visulaize curvature of space-time when there is nothing to curve

12. Jan 9, 2007

### JesseM

In terms of local inertial coordinate systems (defined in terms of measurements made by freely-moving observers in a small region of space and time) the speed of light is always c. But in terms of nonlocal coordinate systems in curved spacetime it doesn't have to be, and likewise, if you use a non-inertial (accelerating) coordinate system in flat spacetime it also may not be c.

13. Jan 9, 2007

### Chris Hillman

Hi, Thrice

If you are referring to the existence of multiple distinct notions of "distance in the large", and thus of "velocity in the large", these arise even in flat spacetime, when you study accelerated observers such as the Bell or Rindler observers (IOW, arises even in scenarios you would study using str). This is a (fundamental) issue in Lorentzian manifolds, it doesn't really have anything to do with gtr versus str.

14. Jan 27, 2007

### quantum123

http://www.geocities.com/physics_world/gr/uniform_light.htm

Speed of Light in a Uniform Field

Home

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The metric in a uniform gravitational field is, by the equivalence principle, identical to the metric in a accelerated frame

[Eq. 1] ds2 = -c2(1 + gz/c2)2dt2 + dx2 + dy2 + dz2

where c is the speed of like in a vacuum in a Minkowski frame of reference.

Since light moves on null geodesice, i.e. we set ds2 = 0 in [Eq. 1]

[Eq. 2] -c2(1 + gz/c2)2dt2 + dx2 + dy2 + dz2 = 0

Divide through by dt2, and substitute vx = dx/dt, vy = dy/dt, vz = dz/dt, v2 = vx2 + vy2 + vz2

-c2(1 + gz/c2)2 + dx2/dt2 + dy2/dt2+ dz2/dt2 = -c2(1 + gz/c2)2dt2 + (dx/dt)2 + (dy/dt)2+ (dz/dt)2 = 0

-c2(1 + gz/c2)2 + vx2 + vy2 + vz2 = -c2(1 + gz/c2)2 + v2 = 0

v2 = c2(1 + gz/c2)2

[Eq. 3] v = (1 + gz/c2)c

Let F = gz. Note that we choose the arbitrary constant, that is usually associated with a Newtonian potential, to be zero.

[Eq. 4] v = (1 + F/c2)c

This is exactly the result obtained by Einstein in 1907 [1]

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References

[1] On the Relativity Principle and the Conclusions Drawn from It, Albert Einstein, Jahrbuch der Radioaktivitat und Electronik 4 (1907)

15. Jul 25, 2007

### Ruian

16. Jul 25, 2007

### rbj

17. Jul 25, 2007

### Ruian

maybe they are wrong or right...

tnx for the reply maybe the studies are wrong or right... but speed of light might change then it will be a small speed diff. but we are not yet sure of this...

18. Jul 27, 2007

### rbj

the question you have to ask yourself is: how would we ever know that c has changed in a context where absolutely nothing else did (particularly the dimensionless constants of nature)? we measure (or perceive) only dimensionless quantities, and if you think you measured a change in c, you measured a change in some ratio of c against some other like-dimensioned quantity (perhaps $e^2/(4 \pi \epsilon_0 \hbar)$) and that changing dimensionless quantity is the only salient measure.

19. Jul 27, 2007

### Ruian

I am not saying that I FOUND c to be changing. What i am only saying is that we still must consider what some studies are presented. like speed of light might change. It is maybe its right or wrong. We still dont know. Currently we are considering c to be constant because
it was supported by experiments and used by some theories... Remember that Nicolaus Copernicus was once not beleived that the sun is the center of the universe...

20. Jul 27, 2007

### JesseM

I think rbj's point is that the notion of a dimensionful constant changing is inherently meaningless, because the only constants that actually have physical meaning are dimensionless ones...this article explains the idea pretty well:

http://www.phys.unsw.edu.au/~dzuba/varyc.html