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Lorentz invariance and volume

  1. Oct 14, 2007 #1
    Simple question.. How do you prove the volume element of momentum space (d3k/Ek) is Lorentz Invariant?

    I tried making it proportional to "velocity volume element" derived from the Lorentz transformations but didn't seem to get very far.
     
  2. jcsd
  3. Oct 14, 2007 #2
    Are you sure its supposed to be invariant?

    Pete
     
  4. Oct 14, 2007 #3
    Basically, what you need to prove is

    [tex] \frac{d^3k'}{E_{k'}} = \frac{d^3k}{E_{k}} [/tex]

    where 4-vectors [itex] (\mathbf{k'}, E_{k'}) [/itex] and [itex] (\mathbf{k}, E_{k}) [/itex] are related to each other by a Lorentz transformation. To do that you just need to calculate the Jacobian of the transformation [itex] \mathbf{k'} \to \mathbf{k} [/itex].

    Eugene.
     
  5. Oct 15, 2007 #4
    Yes, that's exactly what I was trying to show.
    I looked up Jacobian's and for that transformation I calculated the determinant to be: [tex]\gamma[/tex] (Lorentz factor). So I tried another approach and got a 3x4 matrix which I can't find how to solve.

    I know it should be quite trivial and only a few lines but I keep getting stuck. Please point me in the right direction to complete this. Do I even need the velocity transforms? Or does [tex]k_{x'}[/tex] transform like x' etc?
     
  6. Oct 15, 2007 #5
    I think you are on the right track here. The momentum-energy Lorentz transformations are (if the velocity v of the moving reference frame is along the x-axis)

    [tex] k'_x = k_x \cosh \theta - E_k/c \sinh \theta [/tex]
    [tex] k'_y = k_y[/tex]
    [tex] k'_z = k_z [/tex]
    [tex] E'_k = E_k \cosh \theta - k_x c \sinh \theta [/tex]

    where [itex] v = c \tanh \theta [/itex]; [itex] \gamma \equiv \cosh \theta [/itex]

    Eugene.
     
  7. Oct 15, 2007 #6
    Upon thinking this over more it occured to me that what you called "volume element of momentum space" and the quantity (d^3k/Ek) are not the same. If it was then it'd be

    momentum density = d^3p/dV

    That quantity is the compent of a 4-tensor and is not invariant. I don't see the connection between what you say and the expressions you write. And you never answered my question: Why do you think it should be invariant? Was this a homework assignment?

    Pete
     
  8. Oct 15, 2007 #7
    This was not homework but a practice question. To show (what they called a volume element) [tex]\frac{d^{3}k}{E_{k}}[/tex] is Lorentz invariant. The post above answers my question, thanks to both of you.
     
  9. Oct 15, 2007 #8
    You would be doing me a great service if you post the question exactly as it was stated in your source. I find this to be an interesting topic and I'd like to familiarize myself with it. Thanks.

    Best wishes

    Pete
     
  10. Oct 15, 2007 #9


    dans les transformation de lorentz V'=V/(1-v2/c2)^3/2
     
  11. Oct 15, 2007 #10


    Sorry I am French!
    In the transformation of the volume is lorentz V '= V / (1-v2/c2) ^ 3 / 2
     
  12. Oct 16, 2007 #11
    My question still remains to be answered, i.e. to Leeds student
    Best wishes

    Pete
     
  13. Oct 16, 2007 #12

    reilly

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    Science Advisor

    The easist way to show the invariance of d3k/E(k) is via a delta function,
    delta(E^^2 - P^^2 - M^^2) ==D, which is clearly invariant under LTs. Evaluate

    Integral( d4k D) and you will get the answer you want. (Note, this is a fairly standard approach, and can be found in many texts on E&M and QFT.)
    Regards,
    Reilly Atkinson
     
  14. Oct 16, 2007 #13
    I did, it was:
    Show that the volume element [tex]\frac{d^{3}k}{E_{k}}[/tex] is Lorentz invariant
     
  15. Oct 16, 2007 #14

    reilly

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    Science Advisor


    In fact, for those who are interested, the difference between old-fashioned QFT perturbation theory -- Weisskopf, Pauli and Heisenberg, Dirac, etc -- and modern covariant perturbation theory -- Feynman, Schwinger, Tomonaga -- is exactly the two versions of d3k/E(k) and d4k delta(E^^2- P^^2 - M^^2); The non-covariant approach uses the d3k approach, while the covariant approach uses the d4k approach. I'm quite sure that this difference is discussed by Zee in his field theory book.

    Regards,
    Reilly Atkinson
     
  16. Oct 16, 2007 #15
    Hmmm. That simple huh? Okay. Thank you.

    By the way, Welcome to the Forum!!! :smile:

    Best wishes

    Pete
     
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