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

In summary, the homework statement is that using the Lorentz transformation, prove that s2 = s'2. Using FOIL, x'2 = gamma2 (x2-2xBeta*ct+Beta2*c2t2) and t'2 = gamma2 (t2-2Beta*tx/c+Beta2*x2/c2). Combining these two equations, s'2 = gamma2 [x2-c2t2] which gives 1-beta2(x2-c2t2). However, I'm still confused because I'm unable to remove the 1-beta2 factor. Any insight would be greatly appreciated.Thanks!
  • #1
beecher
15
0

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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  • #2
[itex]x^2-\beta^2x^2-c^2t^2+\beta^2c^2t^2=x^2(1-\beta^2)-c^2t^2(1-\beta^2)=[/itex]______?:wink:
 
  • #3
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
Hint: [itex]\gamma=[/itex]___?
 
  • #5
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
 
  • #6
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:
 
  • #7
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
You're welcome!...And get some sleep:zzz::smile:
 

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

What is the Lorentz Transformation?

The Lorentz Transformation is a set of equations used in special relativity to describe how measurements of space and time are affected by the relative motion between two observers.

Why is it important to prove s^2 = s'^2 using the Lorentz Transformation?

This proof is important because it demonstrates the validity of the Lorentz Transformation equations, which are the foundation of special relativity and have been extensively tested and confirmed by experiments.

What is s^2 and s'^2 in this context?

s^2 and s'^2 are terms used in the Lorentz Transformation equations to represent the spacetime interval, which is the difference between the squared distances in space and time between two events as measured by two different observers.

How do you prove s^2 = s'^2 using the Lorentz Transformation?

The proof involves substituting the Lorentz Transformation equations into the definition of the spacetime interval and simplifying the resulting equations to show that s^2 = s'^2.

What are the implications of proving s^2 = s'^2 using the Lorentz Transformation?

Proving this equation using the Lorentz Transformation confirms the principles of special relativity, such as the constancy of the speed of light, time dilation, and length contraction. It also provides a framework for understanding how the laws of physics are the same for all inertial observers, regardless of their relative motion.

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