Why does the tube contraction seem to change the speed of light?

[randyu]

"bgq
Posts: 59
YES!
I think I got it!
Let t1 be the time from E to F, and t2 from F to E as measured by A.
According to Lorentz transformation:
t1 = γ(T0/2 + vL0/c2)
t2 = γ(T0/2 - vL0/c2)
d = (L0/γ + vt1) + (L0/γ - vt2)
= 2L0/γ + v(t2 - t2)
= 2L0/γ + v(2γvL0/c2)
= 2L0/γ + 2γL0v2/c2
= 2L0(1/γ + γv2/c2)

so then solving, d = gamma * 2Lo, not divided by gamma as orginally posted,
this confuses me, any explanations would be helpful, thanks

 

[harrylin]

[..]
Let t1 be the time from E to F, and t2 from F to E as measured by A.
According to Lorentz transformation:
t1 = γ(T0/2 + vL0/c2)
t2 = γ(T0/2 - vL0/c2)
d = (L0/γ + vt1) + (L0/γ - vt2)
= 2L0/γ + v(t2 - t2)
= 2L0/γ + v(2γvL0/c2)
= 2L0/γ + 2γL0v2/c2
= 2L0(1/γ + γv2/c2)

so then solving, d = gamma * 2Lo, not divided by gamma as orginally posted,
this confuses me, any explanations would be helpful, thanks

Welcome to physicsforums.

You referred to post #28.

How did you get from
d= 2L0(1/γ + γv2/c2)
to
d = γ * 2Lo ?

Perhaps you did not look at it carefully.
(1/γ + γv²/c²) = 1/γ(1 + γ²v²/c²) ≠ γ
[EDIT: I was wrong, see next]

 

[randyu]

hi harrlylin,

Y=1/sqt(1-v2/c2) > v2/c2=(Y2-1)/Y2
then,
d=2Lo * 1/Y+Y(Y2-1)/Y2= 2Lo*Y
is this not right, thanks

 

[harrylin]

hi harrlylin,

Y=1/sqt(1-v2/c2) > v2/c2=(Y2-1)/Y2
then,
d=2Lo * 1/Y+Y(Y2-1)/Y2= 2Lo*Y
is this not right, thanks

Oops - I must admit that I did not check bqq's derivation.
And yes you are right. Regretfully I did not follow that conversation, [STRIKE]and I won't look into that now. Maybe someone else will, or I will later. [/STRIKE]
What I can say already, is that I see nowhere claimed that d= 2L0/Y

Oh OK I see it: the answer is in post #22. Did you read that?

d is the path length that the light ray travels, and as d>2Lo I suppose that it is the path length in the rest system, with respect to which the apparatus is moving. Do you follow that?

It's like the Michelson-Morley experiment. And it is very well explained in post #17.
So, to elaborate: You walk from the rear of the car to the front of the car and back. As seen from the train tracks, the distance is greater.
- http://en.wikipedia.org/wiki/Michelson–Morley_experiment

 

[randyu]

Hi, there is something I can't understand:

Consider a stationary observer at A. Now consider an observer B in a train that moves with constant velocity v with respect to A. In the train, B tries to measure the speed of light using an empty tube of length L₀ (proper length as measured by B). He sends a light signal at extremity E, the signal reaches extremity F (where a mirror exists) and return back to the extremity E. B measure this duration T₀ (proper period as measured by B).

Now B measure the speed of light as:

C_B = 2L₀/T₀

Now according to A the length of the tube is contracted and the time is dilated, so he measure the speed of light as:

C_A = 2L/T = (2L₀/γ)/(γT₀) = (L₀/T₀)/(γ^2) = C_B/(γ^2)

which is different from the value measure by B (divided by Gamma squared)

thanks harrylin, what I was thinking is from post #1, d=2Lo/Y and t=YTo which gave c as "divided by gamma squared".
This was wrong from the beginning, the length in the stationary frame is longer than the moving length Lo as you state, confusing because each is moving relative to the other. And also the time is YTo in the "stationary" frame, longer.
I guess this all makes sense, I just get confused about time dilation/expansion using t in different ways it seems to me, sometime an interval, sometimes ticks. Oh well, will come together sooner or later.
Thanks.

 

[harrylin]

thanks harrylin, what I was thinking is from post #1, d=2Lo/Y and t=YTo which gave c as "divided by gamma squared".
This was wrong from the beginning, the length in the stationary frame is longer than the moving length Lo as you state, confusing because each is moving relative to the other. And also the time is YTo in the "stationary" frame, longer.
I guess this all makes sense, I just get confused about time dilation/expansion using t in different ways it seems to me, sometime an interval, sometimes ticks. Oh well, will come together sooner or later.
Thanks.

That's why it is necessary to make sketches.
Cheers