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Deriving the lorentz transforms

  1. Apr 18, 2006 #1
    It's a really easy question, I know, but I must be doing something stupid. Can someone please spell out how to get the right hand side matrix form out of the individual equations?

  2. jcsd
  3. Apr 18, 2006 #2

    Andrew Mason

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    If you multiply the matrix by the 4-vector (t', x', y', z') it should result in the 4-vector (t, x, y, z) as set out on the left side of the arrow. I think the matrix is wrong, though. The numerator of the second term in the top row should be v/c^2 and the numerator of the first term in the second row should be v.

  4. Apr 18, 2006 #3
    See I thought that as well, but they have the inverse of that matrix in the book too & it matches up with the one on wikipedia. Apparently it's gotten by switching V for (-V). Does it work out if you use the 4 vector (ct', x', y', z') & (ct, x, y, z)?

    Edit: Right.. it does.. I knew I was doing something stupid sorry.
  5. Apr 18, 2006 #4
    Wait don't go. I have more foolish questions once I figure out how to post in that .. latex is it?
  6. Apr 18, 2006 #5
    \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

    g_{\it ij} \pd{}{V^k}{} (V^i V^j )= 2 g_{\it kj} V^j [/tex]
    Last edited: Apr 18, 2006
  7. Apr 18, 2006 #6
    K got it. Why does that work? Something to do with the symmetry of the metric....
  8. Apr 18, 2006 #7


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    No need for any special symmetry. It follows from [itex] {\partial V^i \over \partial V^k} = \delta^i_k [/itex] and similarly if i is replaced by j. Then you just need to rename a dummy index in one of the terms and you get the answer provided.
  9. Apr 18, 2006 #8
    Thanks both of you. I don't know where i'd go when my brain isn't working. :)
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