Derivation of Length Contraction (special relativity)

[Chewy0087]

Homework Statement

Show the derivation for legnth contraction, given that from the example given;

The example shows a moving light clock - two walls which are moving in a direction at speed u and a beam of light from wall A to wall B then back to A and so forth.

From Ao (starting point) to B1 (distance moved after the beam has got there)

=t1 = "Moving" legnth / c - u

And from B1 - A2

t1 (2) = "Moving" Length / c + u

Homework Equations

t subscript 1 = 2L / c * sqroot (1 - u²/c²)

Basically the time dilation equation is given (and i managed to work out in an earlier question)

The Attempt at a Solution

Well I added up t1 and t1 (2), to get the time taken by the light to do one back and forth getting;

t1 + t1(2) = gamma*time = 2 * Moving Length / c*(1- u²/c²)

Making Moving length the subject;

Moving Length = (c*(gamma*time))*(1-u²/c²)

Put gamma into the equation;

Moving Length = (c/2)*(1-u²/c²)*2L/(c*sqrt(1-u²/c²))

And from here I'm totally stuck, I do know that the final answer is "Normal" length / gamma but really stuck taking the final steps... :/

Any help would be appreciated

 

[Gokul43201]

t1 + t1(2) = gamma*time = 2 * Moving Length / c*(1- u²/c²)

Making Moving length the subject;

Moving Length = (c*(gamma*time))*(1-u²/c²)

You're missing a factor of 2 here, but you seem to have got it right on the next step.

Put gamma into the equation;

Moving Length = (c/2)*(1-u²/c²)*2L/(c*sqrt(1-u²/c²))

And from here I'm totally stuck, I do know that the final answer is "Normal" length / gamma but really stuck taking the final steps... :/

Any help would be appreciated

By "moving length", do you mean the wall separation measured in the lab frame or the moving (light clock) frame? Think about this.

As far as the math goes, you're nearly done. It's just a matter of canceling out terms from the numerator and denominator.