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Basic Lorentz Transformation

  1. Jul 8, 2008 #1
    Problem
    Write down the transformation from a frame S to a frame S' moving at +0.5 c in the x direction and then to another frame S'' moving at +0.5 c in the x direction relative to S'. What is the complete transformation from S to S''? What relative speed between frames S and S'' does your answer imply?

    Answer?
    Well, the Lorentz transformation matrix is just

    [tex]
    \hat{L}=

    \begin{pmatrix}
    \gamma & -\gamma\beta & 0 & 0\\
    -\gamma\beta & \gamma & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1\\
    \end{\pmatrix}
    [/tex]

    Now, it happens that Lorentz matrices are closed under multiplication (ie [tex]\hat{L} \cdot \hat{L} = \hat{L_1}[/tex]). If we let

    [tex]
    \hat{L_1} =

    \begin{pmatrix}
    G & -GB & 0 & 0\\
    -GB & G & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1\\
    \end{pmatrix}
    [/tex]

    we find that [tex]B = 2\beta/(1+\beta^2)[/tex], [tex]G = (1+\beta^2)\gamma^2[/tex]. Now, since [tex]\beta = 0.5[/tex], we have that [tex]B = 0.8[/tex], so the relative speed between S and S'' would be [tex]0.8 c[/tex]. I'm a bit confused here... why wouldn't the answer just be [tex]0.5c + 0.5c = c[/tex]?
     
  2. jcsd
  3. Jul 8, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's how velocities add under Galilean transformations, not Lorentz transformations. Review the relativistic addition of velocities. (What if the speeds were 0.6c instead of 0.5c? Would you expect the total speed to be greater than c?)
     
  4. Jul 8, 2008 #3
    Hmm okay, I haven't learned this yet. That's the next section in my book, heheh... but is my original answer of 0.8c correct?
     
  5. Jul 8, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Absolutely. :smile:
     
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