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

--------------------------------------------------------------------------------

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]

--------------------------------------------------------------------------------

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

21. Jul 28, 2007

lightarrow

Maybe it's not c which varies, but the speed of light (which could be not exactly equal to c, depending on the frequency); c could be a limit for EM speed.
When we say that light's speed is equal to c, to which frequency do we refer? This is not specified, but, however, we have never used extremely high or extremely low frequencies, for example, so, how can we know it can't vary with frequency?

Last edited: Jul 28, 2007
22. Jul 28, 2007

ZapperZ

Staff Emeritus
But there's at least evidence that it doesn't change with frequency within the range that we know of already. There's zero evidence to the contrary. So we go by what we can verify, not what we can speculate. That's how physics works, and that's why you depend on things to work when you wake up every morning.

Zz.

23. Jul 28, 2007

lightarrow

I'm sure about that evidence and I'm not claiming a variation of light's speed with frequency, I'm just saying that the statement: "light's speed cannot vary with frequency because c is a dimensionful constant" is wrong. Do you agree?

24. Jul 28, 2007

ZapperZ

Staff Emeritus
I wouldn't know. All I know is that to speculate that c could vary with frequency is reaching way too far into the bottom of the barrel.

Zz.

25. Aug 13, 2007

Satie

Light propagates always at c in vacuum. A dielectric is atoms separated by vacuum. The velocity of the phase of a wave is the resultant of the radiation generated by the acceleration of the outermost electrons of the atom and thus the reradiation process creates a front that displaces the energy contained at c/n (n the refractive index). Everything is controlled by the delay in establishing a polarization in the medium, which manifest as a susceptibility with real and imaginary parts, taking account of propagation and absorption and related through the Kramers-Krönig relations. Isn´t that so? There is no contradiction whatsoever, because the radiated energy, between one atom and the next travels at c. Sorry if I misunderstood something in the original question.