Prove that s^2=(s')^2 using the Lorentz Transformation

[castrodisastro]

Homework Statement

I am learning special relativity and we came across the invariant quantity s = x² - (ct)². Our professor wants us to prove it. I admit that this is a proof and belongs in the mathematics section but I didn't see an Algebra section and this is most easily identified by those learning special relativity.

The assignment simply states

"Prove s² = s'²"

Homework Equations

s²= x²-(ct)²

$\gamma$=[1-($\frac{v}{c}$)²]^-1/2

x' = $\gamma$(x-vt)

t' = $\gamma$(t-(vx/c²)

The Attempt at a Solution

My textbook is telling me in one sentence that if we apply the lorentz transformation to x and t then s² = s'²...so I did that...

I choose to start with s'² = x'²-(ct')²

Applying the lorentz transformation to x' and t' our equation becomes...

s'² = ($\gamma$(x-vt))²-(c($\gamma$(t-(vx/c²))²

Expanding what we have takes us to...

s'² = ($\gamma$²(x²-2vt+(vt)²)-(c²$\gamma$²(t²)-2(v/c²)x+(v²/c⁴)x²))

If I combine some terms...

s'² = $\gamma$²[x²(1-(v²/c⁴)+t²(v-1)+2v((x/c²)-t)]

From here I tried a couple of different things on scratch paper but I couldn't see particular direction that would simplify it all down. Am I just not being patient enough and not seeing that it gets worse before it gets better?

Thanks in advance.

 

[Chestermiller]

Homework Statement

I am learning special relativity and we came across the invariant quantity s = x² - (ct)². Our professor wants us to prove it. I admit that this is a proof and belongs in the mathematics section but I didn't see an Algebra section and this is most easily identified by those learning special relativity.

The assignment simply states

"Prove s² = s'²"

Homework Equations

s²= x²-(ct)²

$\gamma$=[1-($\frac{v}{c}$)²]^-1/2

x' = $\gamma$(x-vt)

t' = $\gamma$(t-(vx/c²)

The Attempt at a Solution

My textbook is telling me in one sentence that if we apply the lorentz transformation to x and t then s² = s'²...so I did that...

I choose to start with s'² = x'²-(ct')²

Applying the lorentz transformation to x' and t' our equation becomes...

s'² = ($\gamma$(x-vt))²-(c($\gamma$(t-(vx/c²))²

Expanding what we have takes us to...

s'² = ($\gamma$²(x²-2vt+(vt)²)-(c²$\gamma$²(t²)-2(v/c²)x+(v²/c⁴)x²))

If I combine some terms...

s'² = $\gamma$²[x²(1-(v²/c⁴)+t²(v-1)+2v((x/c²)-t)]

From here I tried a couple of different things on scratch paper but I couldn't see particular direction that would simplify it all down. Am I just not being patient enough and not seeing that it gets worse before it gets better?

Thanks in advance.

You made a bunch of algebra errors. The last equation you had that was correct was: s'² = x'²-(ct')²

Chet