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My Own website regarding physics and math.

  1. Feb 13, 2005 #1
    I've set up my own website,


    Please critique the articles on math and physics, especially the more mathematical ones. I fear that there could be problems in my presentation or derivations of certain formulas (in physics) or that my understanding of the concept is just wrong !!!

    I have a little question as well:

    In deriving the lorentz transform for the x-coordinate, [tex]x'=\gamma(x-vt)[/tex] as compared to the galilean transform [tex] x'=x-vt[/tex]. By taking into account length contraction [tex] x'=\frac{1}{\gamma} x[/tex], shouldn't the lorentz transform be [tex] x=\gamma(x-vt)[/tex] instead of [tex]x'=\gamma(x-vt)[/tex], since if we substitute the length contraction formula, x' is already replaced by x divided by one over gamma?
  2. jcsd
  3. Feb 13, 2005 #2


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    Please,set up the problem.Which system is moving wrt which,in what direction,...blah,blah,...

  4. Feb 13, 2005 #3
    ^hmmm, apologies if i wasn't specific enough.

    Frame S' is moving with velocity v relative to a rest frame S in the positive x direction (i.e. towards the right).
  5. Feb 13, 2005 #4


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    I'm afraid the bolded part is incorrect.It should be [itex]x'=\gamma (x-vt) [/tex] or else [tex] x=\gamma (x'+vt') [/tex].You can find the rigurous proof (which is pretty digestable) in any SR (electrodynamics) book.

  6. Feb 14, 2005 #5
    ^ hmm ok, understood, I'll dig up the modern physics textbook again...thanks for the help....

    : )
  7. Feb 14, 2005 #6


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    Then inverse of the Lorentz transform

    x' = gamma*(x-v*t)
    t' = gamma*(t-v*x)

    is just

    x = gamma*(x'+v*t')
    t = gamma*(t' + v*x')

    You should be able to simply solve the linear equations to verify this - also note that the inverse of the Lorentz transform is a Lorentz transform with v=-v, which should be obvious if you think about it for a bit.
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