# Prove s^2 = s'^2 using the Lorentz Transformation

1. Sep 19, 2009

### beecher

1. The problem statement, all variables and given/known data

Using the Lorentz transformation, prove that s2 = s'2

2. Relevant equations

s2 = x2 - (c2t2)
x' = gamma * (x-Beta*ct)
t' = gamma * (t - Beta*x/c)

3. The attempt at a solution
s2 = x2 - (c2t2)
Therefore
s'2= x'2 - (c2t'2)

and x' = gamma * (x-Beta*ct)
So x'2 = [gamma * (x-Beta*ct)]2
Using FOIL x'2 = gamma2 (x2 - 2xBeta*ct + Beta2*c2t2)

and t' = gamma * (t - Beta*x/c)
So t'2 = [gamma * (t - Beta*x/c)]2
Using FOIL t'2 = gamma2 (t2 - 2Beta*tx/c + Beta2*x2/c2)

combining

s'2 = gamma2 [x2 - 2Beta*cxt + Beta2*c2t2 - c2t2 + 2Beta*cxt - Beta2*x2
Cancel like terms
s'2 = gamma2 [x2 + Beta2*c2t2 - c2t2 - Beta2*x2

This is where I get stuck and cant tell how to progress and make it equal to the original s2 = x2 - (c2t2).

Any insight would be greatly appreciated.

Thanks!

2. Sep 19, 2009

### gabbagabbahey

$x^2-\beta^2x^2-c^2t^2+\beta^2c^2t^2=x^2(1-\beta^2)-c^2t^2(1-\beta^2)=$______?

3. Sep 19, 2009

### beecher

Thanks, factoring it out like that made things more clear.
That gives 1-beta2(x2 - c2t2)
This is what we want except for the factor of 1 - beta2 in the front which I'm still not sure how to get rid of, or account for.

4. Sep 19, 2009

### gabbagabbahey

Hint: $\gamma=$___?

5. Sep 19, 2009

### beecher

gamma = (1-Beta2)-1/2.
But I'm still confused. Possibly because I've been busting my brains on physics for many hours now. Where will the gamma get me? i need to completely remove the 1-beta2 factor dont I?
I can change it into terms of gamma but then its still there in front.
gamma = (1-Beta2)-1/2
thus 1-Beta2) can be called gamma-2 since it is the inverse of gamma2.
Im still stuck though

6. Sep 19, 2009

### gabbagabbahey

???

7. Sep 19, 2009

### beecher

s'2 = gamma2[1-Beta2(x2-c2t2)
s'2 = gamma2[Gamma-2(x2-c2t2)
s'2 = x2-c2t2

Finally, finished! I feel exhilaration and also relief, I can go to bed now lol. Thank you so much for your help tonight. I can't really pay you back other then by trying my best to help others in the way you have helped me. Thanks a million

8. Sep 19, 2009

### gabbagabbahey

You're welcome!...And get some sleep:zzz: