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Space-time interval Lorentz co/invariant

  1. Sep 2, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hi guys!
    I must show by brute force that ##\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2## is invariant under Lorentz transforms.


    2. Relevant equations
    Lorentz transforms:
    ##\Delta t' = \gamma \left ( \Delta t -\frac{\vec v \cdot \Delta \vec x}{c^2} \right )## and ##\vec \Delta x'=\gamma (\Delta \vec x - \vec v t)##.
    I will use the convention ##\Delta \vec x= (\Delta x, \Delta y , \Delta z )## and ##\vec v = (v_x ,v_y, v_z)##

    3. The attempt at a solution
    Basically I must show that ##\Delta s'^2=\Delta s^2##. Where ##\Delta s'^2=-c^2\Delta t'^2+\Delta x'^2+\Delta y'^2+ \Delta z'^2##.
    I have that ##\Delta x'^2+\Delta y' ^2+\Delta z'^2=\gamma ^2[\Delta x^2+\Delta y^2+\Delta z^2 +\Delta t^2 \underbrace{(v_x^2+v_y^2+v_z^2)}_{=\vec v^2}-2\Delta t (\Delta x v_x+\Delta yv_y +\Delta z v_z)]##
    While ##\Delta t'^2=\gamma ^2 \{ \Delta t^2-\frac{2\Delta t}{c^2}(v_x\Delta x+v_y\Delta y +v_z \Delta z)+\frac{1}{c^4}[v_x^2 \Delta x^2+v_y^2\Delta y^2+v_z^2 \Delta z^2+2(v_x\Delta xv_y\Delta y+ v_y\Delta y v_z\Delta z+v_x\Delta x v_z\Delta z)] \}##.
    So then I performed ##\Delta s'^2## and after all the simplifications I could do, I reached that ##\Delta s'^2=\gamma ^2 \left [ \Delta t^2 (\vec v^2-c^2) -\frac{\vec v \cdot \vec x}{c^2} +\Delta \vec x +\frac{2}{c^2} (v_x\Delta xv_y\Delta y+ v_y\Delta y v_z\Delta z+v_x\Delta x v_z\Delta z) \right ]## which does not seem equal to ##\Delta s##. I've redone the algebra 4 times (lost hours on this), I reach the same answer every time. So either the expression I found is right but I must simplify it even more and it indeed equals ##\Delta s##, or I'm wrong somewhere.
    I'd appreciate some help... thank you!
     
  2. jcsd
  3. Sep 2, 2013 #2

    TSny

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    For a boost in an arbitrary direction, I think your expression for ##\Delta t'## is ok. But the correct expression for ##\vec \Delta x'## is more complex than what you wrote. See for example

    http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_any_direction

    Are you sure you can't orient your axes so that the x-axis is parallel to the boost?
     
  4. Sep 2, 2013 #3

    fluidistic

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    I see, thank you very much.
    Due to the complexity of the expression for a boost in any direction, I guess I can orient the axes so that it results to be in the x-direction. I will consider this case only. :biggrin:
     
  5. Sep 2, 2013 #4

    fluidistic

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    At first glance I get ##\Delta s'^2=\Delta t^2 (-c^2\gamma^2+v^2)+\gamma ^2 \left ( \Delta x^2+v^2 \Delta t^2 - \frac{v^2\Delta x^2}{c^2} \right )##.
    I guess I'll have to redo the math.
    Edit:Yeah I forgot 2 terms. I reach now ##\Delta s'^2=\Delta t^2[v^2+\gamma(v^2-c^2)]+\Delta x^2+\Delta y^2+\Delta z^2##. Almost the desired result but apperently I still have some job to do :)

    Edit 2:This reduces to ##\Delta s'^2=\Delta t^2 (v^2-c^2)+\Delta \vec x##. I'm almost there. But I would reach the result only if v=0... which is not correct.
     
    Last edited: Sep 2, 2013
  6. Sep 2, 2013 #5

    vela

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    Everything multiplying ##\Delta t^2## should be proportional to ##\gamma^2##, so your expression in your first edit looks kinda odd.
     
  7. Sep 2, 2013 #6

    TSny

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    Trace back to see how you got the first term of ##v^2## in the first brackets. I don't think that should be there.
     
  8. Sep 2, 2013 #7

    fluidistic

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    I've tracked it down (already on my own), at ##\Delta x'=\gamma (\Delta x - v \Delta t)##. When I must take the square of this quantity, there's a term ##\gamma ^2 v^2\Delta t^2##. I couldn't cancel this term out though I must re-recheck the algebra.


    Edit: Nevermind, I made a mistake. I now reach the desired result! Thank you guys.
     
    Last edited: Sep 2, 2013
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