# Basic doubt in SR regarding speed of light in different frames.

1. May 31, 2012

suppose we have a moving frame and a rest frame..we know time dilution and length contraction had occured in moving frame wrt rest frame,this is to make sure that speed of light is same in any frame..but this is not the case..consider measurement of speed of light in moving frame wrt rest frame we get...speed=distance/time
=x/t(gamma)(gamma)
Hence it is c/(gamma)^2
But we need it to be c right.?
So having a additional (gamma)^2 should be rectified and how's it done.?

Last edited: May 31, 2012
2. May 31, 2012

### Staff: Mentor

How are you arriving at this formula?

3. May 31, 2012

well length of displacement of light has measure in moving frame is x,then same displacement wrt stationary frame is x/(gamma)...similarly time taken for light beam to move x as per moving frame is t,same time t as measured by stationary frame is (gamma)t......so speed of light wrt to stationaru frame happens to be x/t(gamma)(gamma)..which is what I got

PS-i ve written wrongly in my first post but now it's rectified..

4. Jun 1, 2012

### Staff: Mentor

You've written either the length or the time in the "stationary" frame incorrectly. I can't tell which because your nomenclature for the frames is rather confusing when combined with your discussion.

If the stationary frame is truly "stationary", then the time in it should be larger, which t(gamma) is; but then length in it should also be *larger* (since the length should be contracted, i.e., smaller, in the *moving* frame), so the length in the "stationary" frame should be x(gamma), not x/(gamma).

OTOH, if you really meant for what you are calling the "moving" frame to be at rest and the "stationary" frame to be in motion, then the length in the "stationary" frame should indeed be shorter, which x/(gamma) is; but then the time in the "stationary" frame should also be shorter (since what you are calling the "moving" frame should have the *larger* time), so it should be t/(gamma), not t(gamma).

In either case, the gammas cancel when you take the ratio length/time to get the speed of light, so you get the same result, c.

5. Jun 1, 2012

### stevendaryl

Staff Emeritus
If you look at the Lorentz transforms, there are actually 3 effects, and they are all three involved in the invariance of the speed of light:
1. Length contraction
2. Time dilation
3. Relativity of simultaneity

The meaning of the latter is that clocks that are synchronized according to one coordinate system are NOT synchronized according to another. You have to take this into account when computing how things look in different frames--if you only use length contraction and time dilation, you'll get inconsistent results.

Here's the derivation using Lorentz transformations:

x' = γ (x - vt)
t' = γ (t - vx/c2)

So dividing them gives:
x'/t' = (x-vt)/(t - vx/c2)

If we let u = x/t and u' = x'/t', we find:
u' = (ut - vt)/(t - uvt/c2)
= (u-v)/(1-uv/c2)

So what that means is that if an object has speed u in the unprimed frame, then it has speed u' = (u-v)/(1-uv/c2) in the primed frame. In the special case in which u=c (the object is a light signal), we find:

u' = (c-v)/(1-cv/c2) = c

Now, the derivation in terms of length contraction, time dilation and relativity is a little complicated. But suppose that we measure the speed of light in the primed frame this way:

1. Take two clocks at rest in the primed frame.
2. Bring them together at x'=0 and set them to the same time, say t1'.
3. Now, slowly move one of the clocks to the point x' = L. Let t2' be the time on that clock when it reaches its destination.
4. At time t3' according to the first clock at x'=0, send a light signal from the first clock to the second.
5. Let t4' be the time on the second clock when the light signal is received.
6. Then the measured speed of light will be c = L/(t4' - t3')

To convince yourself of the last point from the point of view of the unprimed observer requires considering a bunch of different factors:
1. The moving clocks are running slow, by a factor of γ.
2. The second clock must travel even faster than the first clock in order to get to the point x' = L. So while it is traveling, its time dilation is even greater than the other clock. So afterward, it is no longer in synch with the first clock. Working it out, you'll find that it is running slow, compared with the first clock by an amount vL/c2.
3. The distance between the clocks isn't L, but is L/γ, according to the unprimed observer.
4. Light takes longer than time L/(c γ) to go from the first clock to the second, because the second clock is moving.

If you take all these into account, you'll find that the measured speed of light in the two frames is the same.

6. Jun 1, 2012

really thanks for helping me out
But I find that
Uprime===
[u/(gamma)+{[1-1/(gamma)](u.v)/v^2 -1}v]/[1-(v.u)/c^2]

Here when u=speed of light and which is perpendicular to v,ie u.v=0
Is not ending with u prime to be 'c'
Am i wrong anywhere..I found that derivation to be right ..

7. Jun 1, 2012

### Staff: Mentor

Once again, where are you getting this formula for velocity composition? It doesn't look like any of the ones I'm familiar with. See, for example, this page:

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

8. Jun 1, 2012

### TheEtherWind

Using length contraction and time dilation to disprove the invariance of the speed of light is like assuming the invariance of the speed of light (postulate 2), from which length contraction and time dilation are derived from, to disprove the invariance of the speed of light. In short, you assumed the invariance of the speed of light to disprove it.

9. Jun 3, 2012

Ha..thats true..but that formulae is given in -
THEORY OF RELATIVITY--c.MOLLER

10. Jun 3, 2012

### Staff: Mentor

That formula from Moller is a vector equation. (I didn't check if you wrote it correctly.) You can't just plug in magnitudes and get an answer.

Right after he gives that complicated general formula, he states that for perpendicular velocities (where v.u = 0) it boils down to this:
$$\vec{u} = \vec{v} + \vec{u'}/\gamma$$
Let's take the case where u' = c and is in the y direction; v is in the x direction.

ux = v
uy = c/γ

If you compute the magnitude of that velocity you'll find that it equals c. (As it must, of course!)