Now it remains to be shown that A = B. I'll leave that to you.

AI Thread Summary
The discussion focuses on demonstrating the invariance of the spacetime interval, s^2 = x^2 + y^2 + z^2 - c^2t^2, using the Lorentz transformation. Participants explore how to express x' and t' in terms of x and t, emphasizing the algebraic manipulation needed to simplify the expressions. They discuss the importance of correctly applying the transformation equations and ensuring proper cancellation of terms during simplification. The conversation highlights the challenges of algebraic manipulation and the necessity of maintaining accuracy in calculations. Ultimately, the goal is to show that the transformed interval remains invariant under Lorentz transformations.
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1. Start with the expression x'^2 + y'^2 _ z'^2 -c^2t'^2 and show, with the aid of the Lorentz transofmration, that this quantity is equal to x^2 + y^2 + z^2 -c^2t^2. This result establishes the invariance of s^2 defined by s^2 = x^2 + y^2 + z^2 -c^2t^2



2. s^2 = x^2 + y^2 + z^2 -c^2t^2



3. So far I have y = y' and z = z' because of the transformation law. basically there is no lorentz contraction perpendicular to the motion of X so y = y' and z = z'

how do i really solve this problem? where should I start?
 
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So what does the Lorentz transformation look like for x' and t', in terms of x and t?
I mean, what is the correct full expression for

t' = ... t + ... x
x' = ... x + ... t
y' = y
z' = z
 
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Very nice. It's straightforward algebra then: plug them into
x^2 - c^2 t^2
and show that you can simplify it to
(x')^2 - c^2 (t')^2.
 
CompuChip said:
Very nice. It's straightforward algebra then: plug them into
x^2 - c^2 t^2
and show that you can simplify it to
(x')^2 - c^2 (t')^2.
so I would have
(x'+vt'/sqrt(1-v^2/c^2)^2 - c^2(t' + vx'/c^2 / sqrt(1-v^2/c^2) and simplify it algebraically to x'^2 - c^2t'^2 ?
 
Yes, the expression is a bit unclear, but it looks like what I got,
\frac{(x' + v t')^2 - (t' + v x' / c^2)^2}{(\sqrt{1 - v^2/c^2)^2}
Just work out the multiplication (don't forget the cross-term). Then you'll see some cancellation taking place and you can factor the rest into something\,{}\times (x'^2 - c^2t'^2).
 
thanks a lot, i will get on this now :D
 
I'm sort of stuck, my algebra isn't as fresh as I thought it would be. But am I on the right track though?

http://img264.imageshack.us/img264/1081/1233176495203dl2.jpg

^ My work
 
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You are on the right track, I see two terms cancelling.
Watch the squares however, I see c^2 t'^2 instead of (c^2 t')^2 and v x'^2 instead of (v x')^2...

Also it may help to write everything in one fraction.
 
  • #10
I need a common denominator to start canceling things out right, or can I do it before? This is the part where my algebra isn't as fresh as it should be. To get the common denominator, I have to multiply the left equation by c^4 (top and bottom) and multiple the right equation by 1 (top and bottom) correct?
 
  • #11
Okay I did what I said on the previous post and so far I have:

c^4[x'^2 + (vt')^2] - (c^2t')^2 - (vx'^2) / c^4(1-v^2/c^2)

What can I do now to move forward? I don't see anything else cancel out.
 
  • #12
Okay my way isn't right, still stuck on this :|
 
  • #13
It seems you're having a little problem with the algebra. Let me get you started.

Since y' = y and z' = z, it suffices to show that x'^2 - c^2 t'^2 = x^2 - c^2 t^2.
Plugging it in:

x^2 - c^2 t^2 = \left( \frac{ x' + v t' }{ \sqrt{1 - v^2 / c^2} } \right)^2 - c^2 \left( \frac{ t' + v x' / c^2 }{ \sqrt{1 - v^2 / c^2} } \right)^2.
Let me take the c^2 in the second term inside the brackets (c^2 a^2 = (c a)^2) and take the square root outside ( (a/b)^2 = a^2 / b^2, sqrt(a)^2 = a):
x^2 - c^2 t^2 = \frac{ \left( x' + v t' \right)^2 - \left( c t' + v x' / c \right)^2 }{ 1 - v^2 / c^2 }.
Open up the brackets:
x^2 - c^2 t^2 = \frac{ x'^2 + 2 x' v t' + v^2 t'^2 - c^2 t'^2 - 2 t' v x' - v^2 x'^2 / c^2 }{ 1 - v^2 / c^2 }.

Two terms cancel, after tracing the above steps carefully (make sure you understand which manipulations I did and that I didn't make any sign errors :-p) collect the terms in (x')^2 and (t')^2, i.e. write
x^2 - c^2 t^2 = \frac{ A x'^2 + B t'^2 }{ 1 - v^2 / c^2 }.
 
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