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Lorentz invariance of [itex] E_p \delta ({\bf p}- {\bf q}) [/itex]

  1. Sep 4, 2012 #1
    Can anybody help me with the proof that [itex] E_p \delta ({\bf p}- {\bf q}) [/itex] is a Lorentz invariant object?

    I did a boost along z axes and used the formula [tex] \delta (f(x)) = \frac{\delta(x-x_0)}{|f'(x_0)|} [/tex] and the factor in front of the delta function indeed is invariant but within the function I have something like this:

    [tex] E_p \delta (p_z -(v(E_q-E_p)+q_z)) [/tex]

    but not [itex] E_p \delta (p_z- q_z)[/itex]

    Thanks.
     
    Last edited: Sep 4, 2012
  2. jcsd
  3. Sep 5, 2012 #2
    Re: Lorentz invariance of [tex] E_p \delta ({\bf p}- {\bf q}) [/tex]

    Nobody? :(
     
  4. Sep 5, 2012 #3

    Bill_K

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    Re: Lorentz invariance of [tex] E_p \delta ({\bf p}- {\bf q}) [/tex]

    What makes you think it's Lorentz invariant?
     
  5. Sep 5, 2012 #4
    Because in QFT we specifically choose a normalization in such a way as to get scalar product invariant.
     
  6. Sep 5, 2012 #5

    Bill_K

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    Ok, now I understand! To distinguish between 3-dimensional and 4-dimensional quantities, let me call the energy-momentum 4-vector p, with components (E, k) and invariant norm p2 = m2.

    To begin with, the 4-dimensional delta function δ4(p - p0) is clearly Lorentz invariant. Write this as δ(E - E0) δ3(k - k0). And now change variables on the first factor. In place of the variable E, I want to use the variable p2. Well, one can show that δ(p2 - m2) = (1/2E0) δ(E - E0). [There's a second term with δ(E + E0) if you want to include the negative energies.]

    This gives us δ4(p - p0) = 2E0 δ(p2 - m2) δ3(k - k0), which is Lorentz invariant.
     
  7. Sep 5, 2012 #6
    Thank you, I got it. I made a mistake in calculation.
     
  8. Sep 5, 2012 #7

    Bill_K

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    That's good. By the way, the expression as you originally gave it is not Lorentz invariant. Only if you include the other delta function, δ(p2 - m2). And the variable you choose to replace E makes all the difference too. If you use some other f(E), you'll get something else instead of 2E0 as the factor in front.
     
  9. Sep 5, 2012 #8
    But [itex] \delta (p^2-m^2) [/itex] is Lorentz invariant.
     
  10. Sep 5, 2012 #9

    Bill_K

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    Consider the domain where these functions are nonzero. δ4(p - p0) is nonzero at just a single point - of course that's a Lorentz invariant domain. And δ(p2 - m2) is nonzero on a hyperboloid - that's Lorentz invariant too.

    But δ3(k - k0) is nonzero on an entire line - and that's not Lorentz invariant! It singles out a rest frame. And putting E0 in front of it doesn't change the fact.

    It's only when you multiply the two of them together, δ(p2 - m2) and δ3(k - k0) -- then the product is nonzero on the intersection of the line and the hyperboloid, namely a single point again -- that you get back to a Lorentz invariant domain.
     
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