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

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Homework Help Overview

The discussion revolves around proving the equation s² = s'² using the Lorentz transformation in the context of special relativity. Participants are exploring the relationships between spacetime intervals and the transformations between different inertial frames.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the expressions for s'² using the Lorentz transformation equations. There is a focus on expanding and simplifying the terms involving gamma and beta, and questions arise about how to eliminate certain factors in the equations.

Discussion Status

Some participants have provided hints and guidance on factoring and manipulating the equations. There is an ongoing exploration of how to relate gamma to the terms in the equation, and while some progress has been made, not all participants are in agreement on the next steps or the final form of the proof.

Contextual Notes

Participants express frustration with the complexity of the problem and the need to reconcile various terms, particularly the presence of the factor (1 - beta²). There is an acknowledgment of the cognitive load involved in working through the physics concepts after extended study.

beecher
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Homework Statement



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


Homework Equations



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

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 can't tell how to progress and make it equal to the original s2 = x2 - (c2t2).

Any insight would be greatly appreciated.

Thanks!
 
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x^2-\beta^2x^2-c^2t^2+\beta^2c^2t^2=x^2(1-\beta^2)-c^2t^2(1-\beta^2)=______?:wink:
 
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.
 
Hint: \gamma=___?
 
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 don't 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
 
Are you forgetting about the gamma^2 you had here:

beecher said:
Cancel like terms
s'2 = gamma2 [x2 + Beta2*c2t2 - c2t2 - Beta2*x2

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

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