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Lorentz Transformation

  1. Jan 27, 2009 #1
    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?
     
  2. jcsd
  3. Jan 28, 2009 #2

    CompuChip

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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
     
  4. Jan 28, 2009 #3
    [​IMG]
     
  5. Jan 28, 2009 #4

    CompuChip

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    Very nice. It's straightforward algebra then: plug them into
    [itex]x^2 - c^2 t^2[/itex]
    and show that you can simplify it to
    [itex](x')^2 - c^2 (t')^2[/itex].
     
  6. Jan 28, 2009 #5
    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 ?
     
  7. Jan 28, 2009 #6

    CompuChip

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    Yes, the expression is a bit unclear, but it looks like what I got,
    [tex]\frac{(x' + v t')^2 - (t' + v x' / c^2)^2}{(\sqrt{1 - v^2/c^2)^2}[/tex]
    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[itex]\,{}\times (x'^2 - c^2t'^2)[/itex].
     
  8. Jan 28, 2009 #7
    thanks a lot, i will get on this now :D
     
  9. Jan 28, 2009 #8
  10. Jan 28, 2009 #9

    CompuChip

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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.
     
  11. Jan 28, 2009 #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?
     
  12. Jan 28, 2009 #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.
     
  13. Jan 28, 2009 #12
    Okay my way isn't right, still stuck on this :|
     
  14. Jan 29, 2009 #13

    CompuChip

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    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:

    [tex]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.[/tex]
    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):
    [tex]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 }.[/tex]
    Open up the brackets:
    [tex]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 }.[/tex]

    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 :tongue:) collect the terms in (x')^2 and (t')^2, i.e. write
    [tex]x^2 - c^2 t^2 = \frac{ A x'^2 + B t'^2 }{ 1 - v^2 / c^2 }.[/tex]
     
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