# Arbitrary Velocity C in Special Relativity

1. Oct 28, 2008

### Legion81

I am working through Einstein's "On the Electrodynamics of Moving Bodies", and I can't figure out why we are using c instead of a random constant velocity. It seems to me that c is an arbitrary value for the argument. I'm sure there is an explanation... I just can't think of it. Can anyone clear up my confusion?

Oh, in case you want to review the paper:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

2. Oct 28, 2008

### JesseM

You could construct a set of coordinates systems such that if an object was moving at some arbitrary speed v less than c in one of them, it would be moving at the same speed v in all of them. But in this case the postulate that the laws of physics (and in particular Maxwell's laws) work the same in all these coordinate systems would be empirically false. Of course, it is theoretically possible that the postulate that the laws of physics work the same in all the inertial coordinate systems given by the Lorentz transformation (which uses c) could turn out to be false too, but so far all the experimental evidence supports this postulate, and Einstein made the postulate because it was the only way that Maxwell's laws could apply in every inertial coordinate system (in any coordinate system where Maxwell's laws work, it must be true that all electromagnetic waves move at c, regardless of the velocity of the source of the waves).

3. Oct 28, 2008

### Legion81

That makes sense why c must be used for the Electrodynamical part, but not (to me at least) the Kinematical part. Why couldn't you use, say, a baseball moving at v0 instead of the ray of light for the argument in "Simultaneity" and "Relativity of Lengths and Times"?

4. Oct 28, 2008

### JesseM

Because if you constructed a set of coordinate systems using a different invariant speed, the laws of physics (including, but not limited to, electromagnetism) wouldn't obey the same the same equations in the different coordinate systems that make up the set. The fact that they do obey the same equations in the coordinate systems given by the Lorentz transformation is just a feature of how the laws of physics work in our universe--laws of physics with this characteristic are called "Lorentz invariant", and all the fundamental laws that physicists have found have this characteristic.

5. Oct 28, 2008

### bernhard.rothenstein

I think that a look at
Abraham A Ungar
"Formalism to deal with Reichenbach's special theory of relativity"
Foundation of Physics 21 691 1991
could be of help.

6. Oct 28, 2008

### rbj

i thought the answer to that (we've had a few "Why Light" threads with the same sorta question) is that it is in the nature of space and time (not particularly of light or of E&M) that all of these instantaneous fundamental interactions; gravitation, E&M, or nuclear, that all of these interactions propagate through space and time at this same finite speed which is, as we measure it with our meter sticks and clocks, c. since light is just the E&M interaction, it propagates at c, just as gravitational disturbances do. but baseball moving at v0 are not the same as any of these four fundamental interactions.

7. Oct 28, 2008

### Legion81

I am not too familiar with the "Lorentz invariant", so I will have to do some reading on that... sounds kind of important. I'll look up "Formalism to deal with Reichenbach's special theory of relativity" as well. Thanks.

8. Oct 29, 2008

### JesseM

Are you familiar with the Lorentz transformation? That's the coordinate transformation that Einstein derives in the 1905 paper. It says that if you have two inertial coordinate systems, one using coordinates (x,y,z,t) and the other using coordinates (x',y',z',t'), and the second coordinate system is moving at speed v along the x-axis of the first (with the spatial origins of each system coinciding at t=t'=0), then if you know the x,y,z,t coordinates of some event as measured in the first coordinate system, the x',y',z',t' coordinates of the same event as measured in the second coordinate system would be:

x'=gamma*(x - v*t)
y'=y
z'=z
t'=gamma*(t - vx/c^2)

with gamma = 1/squareroot(1 - v^2/c^2)

So to say the laws of physics are "Lorentz-invariant" means that if you write the equations representing the laws of physics the first coordinate system, the perform this transformation on the equations, the equations come out unchanged. Something I wrote about this a while ago in post #16 of this thread:

9. Nov 17, 2008

### karanmohan

Hello friends, i am also following the derivation of Einstein in OEMB paper. However, i have a few questions that you may be able to help me with. When einstein says let x'=x-vt, it seems he is referring to a point stationary in the moving frame k, and moving at speed v. Thus, x' is the coordinate of this point when measured from the origin of frame K (the stationary frame). However, if this is the case, should it not read x'=x+vt, as the location is increasing along the positive x-axis with time?

Also, I was curious as to how one can say the velocity of light in the y or z direction is sqrt(c^2-v^2). This seems rather arbitrary and troubling actually, as i assumed that the speed of light is c in all frames (towards the end, he also mentions a ray always travels at c+v in the x-direction) - is this a contradiction to his statement that the velocity of light is always c?

10. Nov 17, 2008

### JesseM

A while ago I tried to follow along with Einstein's derivation and write down my understanding of what he was doing--look at my post #4 on this thread and see if you find it helpful. I think what Einstein was doing in the steps you mentioned was to consider a coordinate system constructed by doing a Galilei transformation on an inertial coordinate system built out of physical rulers and clocks--in Newtonian physics different inertial coordinate systems are related by the Galilei transformation, but in relativity, this new coordinate system would not in itself be a valid inertial coordinate system and would not reflect measurements made by any set of physical rulers and clocks, because in relativity inertial coordinate systems are related by the Lorentz transformation rather than the Galilei transformation.

11. Nov 17, 2008

### karanmohan

hi Jesse, thank you for the response. I went through the other derivation, and while it works well, I dont understand why you would need to introduce a third frame, which as you mentioned is galilean, but wrong because the correct transform would have to be lorentz. So it appears from your explanation that Einstein first introduces an incorrect transformation, derives a transform between k and Kg, and then from that gets a relationship between Kg and K. Thus, is he getting the correct result by doing 2 wrong things?

if only this derivation was as straight forward as his light-sphere derivation in the 1907 paper *sigh*

12. Nov 17, 2008

### JesseM

There is nothing "wrong" with using a non-inertial coordinate system, SR does not forbid you from using one, it just says you can't assume that the laws of physics will look the same in this coordinate system as they would in an inertial one. And Einstein doesn't make such an assumption--if he did, then he would have assumed light still moved at c in this non-inertial coordinate system, but as you pointed out he correctly said it'd move at sqrt(c^2-v^2) in the y or z direction.

13. Nov 18, 2008

### Staff: Mentor

This is simply "proof by contradiction". Make an unjustified assumption (the laws of physics have the same form under a Galilean transformation) find a contradiction (the speed of light does not equal c in all frames) and therefore you can reject your unjustified assumption (conclude the laws of physics do not have the same form under a Galilean transformation).

14. Nov 18, 2008

### JesseM

I don't think he assumes that at all...what step are you referring to? I think he just uses the coordinate system obtained by doing a Galilei transform on the original relativistic frame for various intermediate steps, but never does he make any assumption that the laws of physics will work the same way in this coordinate system (as I pointed out earlier, he explicitly says that light will not move at c in this coordinate system, for example).

15. Nov 18, 2008

### Naty1

In addition to the coordinate transformations mentioned above, and that "c" works, Maxwell's equations on electromagnetism show c^2 = 1/(epsilon)(mu), permeability and permitivity...

As in Lorentz transformations lots seems to depend on "c".....to say it another way, "c" has been measured experimentally..