# Does c-invariance in SR postulate need inertial frames ?

1. Nov 12, 2009

### myspacetime

2 textbooks I have give postulate 2 of SR as:
"The speed of light in vacuum is constant in all inertial reference frames".
But "Classical Dynamics" by Jerry B. Marion has:
"The velocity of light in free space is a universal constant independent of any relative motion of the source and observer".

I think there is no need of inertial frame, i.e c would be the same for any observer and source, even if is in an accelerating spaceship doing the measuring. If it is so, then I think those texts are doing a great injustice to students as the notion of inertial frame, in itself, has deep and difficult implications.

2. Nov 12, 2009

### A.T.

3. Nov 12, 2009

### myspacetime

I only have a vague idea of SR, not GR. I read the wiki article and don't understand about the "measure ...average over finite distance" part.

I have a few questions:
1) If I simplify my question - If an experimenter on a space-lab (near some mass ), that can accelerate in any smooth manner, were to measure random sources of light, would the speed always be the constant c (everything relative to the space-lab or observer).
2) If answer to 1) is yes, than it seems postulate 2 of SR don't need the condition of inertial frames.
3) if measurements of c can differ, how strong is the experimental evidence.

Thanks

4. Nov 12, 2009

### A.T.

Average speed:
http://www.worsleyschool.net/science/files/average/velocity.html
That is simpler? Why bring "near some mass" into it? Stick with SR. If the space-lab accelerates the average speed of light traveling some distance in the lab can be different from c. But measured over an infinitesimally small distance with a local clock it is still c everywhere in lab.
The answer depends on how you measure the speed of light.
Experimental evidence for what? All above follows from constant c in inertial frames.

5. Nov 12, 2009

### curiousphoton

I'm confused. Are you suggesting the speed of light varies? How can the speed of light ever differ from c (besides in some medium)?

6. Nov 12, 2009

### A.T.

Measured by a local clock it doesn't, but clocks at different positions in an accelerated frame run a different rates. And so light propagates at different speeds at different positions in an accelerated frame, when measured by just one of the clocks. Follow the links in the wiki article:
http://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames

7. Nov 13, 2009

### bcrowell

Staff Emeritus
If you want to understand relativity, one of the things you run into is that SR is a lot easier than GR. It may be wise to learn SR first, with all the simplifications that it allows, before going back and doing everything in more generality with GR. In SR, there are inertial frames. In GR, there is in general no such thing as a global inertial frame; what you end up talking about are local free-falling frames, which have the same flat-space geometry as the global one in SR. Note that a free-falling frame is not inertial according to Newton.

A nicer way to look at SR, IMO, is that we simply take it as an empirical fact that clocks run at different rates depending on their state of motion. Given this, along with a few other assumptions, it is possible to derive the Lorentz transformations, prove that cause and effect has a certain maximum velocity (c), and then prove that massless particles such as photons move at exactly c. It's basically backwards to think of the speed of light as having some special status in relativity. If we learned tomorrow that photons have nonzero rest mass, and therefore move at somewhat less than c, it would have no effect on the basic logical structure of relativity. They would just get thrown into the same class as neutrinos: particles with very small mass.

8. Nov 13, 2009

### JesseM

It seems to me the quoted section can only be correct if we assume these "non-inertial frames" are still using inertial rulers and clocks to define speed. If we assume they define speed as just change in coordinate position divided by change in coordinate time, then since any arbitrary way of assigning events position and time coordinates qualifies as a "non-inertial frame", then pretty clearly the quoted statement would be wrong. For example, if x,t are the coordinates of an inertial frame, then this would be an example of a non-inertial frame:

x' = x + t*100c
t' = t

If a light ray passes the coordinates (x=0 light-seconds, t=0 seconds) and (x=10 l.s, t=10 s) in the inertial frame, in the non-inertial frame the ray has passed through the coordinates (x'=0, t'=0) and (x'=1010, t'=10), so the light's average speed measured over this finite distance would be 1010/10 = 101c in this frame.

9. Nov 14, 2009

### yuiop

Could you clarify what exactly qualifies as an inertial frame? Informally an inertial frame is taken to be an non accelerating frame, but even that raises complications because we have to define what is meant by non accelerating. If we define non accelerating to mean observers would not feel acceleration, then a free falling frame would qualify as an inertial frame. Assuming the example you give for an arbitary coordinate system is in flat space, then by the non-accelerating definition it is an inertial frame and using your coordinates we can say the speed of light is not necessarily c even in an inertial frame. If we define a non-inertial frame as one that has arbitrary coordinates then we need to define what qualifies as non-arbitrary coordinates, in order to define an inertial frame. I think a good start at defining an inertial frame would be a frame in which all clocks mantain a fixed spatial relationship to each other and all clocks run at the same rate. Are you simply defining an inertial frame as a frame in which the observers measure the speed of light to be c and constant over any arbitary distance within the frame?

Last edited: Nov 14, 2009
10. Nov 14, 2009

### JesseM

I'm defining an inertial frame the way Einstein did, a coordinate system which can be defined in terms of local readings on a set of non-accelerating rulers and clocks (moving slower than light of course), with the clocks having been synchronized using the Einstein synchronization convention. If you pick a spacetime origin, the set of all inertial frames with that origin will all be related to one another by the Lorentz transformation.

11. Nov 15, 2009

### myspacetime

I've got further clarification from sci.physics.relativity.

A.T's reply is the one I am looking for and it is resolved now for me. GR's stuff is no for me yet.

Thanks all.

12. Nov 15, 2009

### A.T.

How is that a counter example to the quoted wiki section?

Last edited by a moderator: May 4, 2017
13. Nov 15, 2009

### JesseM

Sorry, my mistake. I read the quote wrong, I thought it said that the average speed in a non-inertial frame would always be c while the "local speed" would not, but actually it was the reverse. But in this case "local speed" is kind of ambiguous, as one might take them to mean the instantaneous speed dx/dt in the coordinates of the non-inertial frame, when actually they are talking about the instantaneous speed in a locally inertial frame at that point, so the non-inertial frame isn't even involved and I don't think they should really say "in non-inertial frames the local speed of light is also c".

14. Nov 15, 2009

### A.T.

Why do you think the mean the later?
Is the above wrong if they actually mean the instantaneous speed dx/dt in the coordinates of the non-inertial frame?

15. Nov 15, 2009

### JesseM

Because it's obvious that the instantaneous velocity dx/dt of a photon need not be c if we are using the coordinates x and t of a non-inertial frame. For example, if we use the coordinate transformation I offered in post #8, then $$\Delta x' / \Delta t'$$ for the photon will be constant, it'll be 101c no matter which two points on its worldline you pick, so obviously in the limit as the distance between the two points goes to zero (which is what dx'/dt' means) it will be 101c as well.