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Proving the Lorentz invariance of an integration measure? QFT related?

  1. Nov 1, 2011 #1
    qftproblem21.jpg

    So, first off, I'm thinking Lorentz invariant quantities are the same in any inertial frames S and S' regardless of their relative velocity.

    I'm thinking I need to show that
    [itex]\frac{d^3k}{(2\pi)^32E(\vec{k})} = \frac{d^3k'}{(2\pi)^32E'(\vec{k'})}[/itex] where the primed & unprimed quantities denote different frames.

    We also have [itex] E(\vec{k}) = \sqrt{\vec{k}^2 + m^2}[/itex] in the denominator. This isn't Lorentz invariant - the mass term is, but the 3 - momentum [itex]\vec{p} = \hbar \vec{k}[/itex] is not, therefore E is not.

    [itex]E \neq E'[/itex] where [itex] E'(\vec{k'}) = \sqrt{\vec{k}'^2 + m^2}[/itex].

    This is making me think that neither E nor d3k are Lorentz invariant, so that when they are used in this fraction, the Lorentz invariance from each one somehow cancels out overall.

    Then we come to the hint. I 'm really not sure what to make of this, I mean, I can write [itex]\delta(x^2 - x_0^2) = \frac{1}{2|x|}(\delta(x-x_0) + \delta(x+x_0))[/itex] with k's and m's: [itex]\delta(k^2 - m^2) = \frac{1}{2|k|}(\delta(k-m) + \delta(k+m))[/itex]
    to get an apparently Lorentz invariant expression: [itex] \frac{d^4k}{(2\pi)^3}\delta(k^2 - m^2)\theta(k_0) = \frac{d^4k}{(2\pi)^3}\frac{1}{2|k|}(\delta(k-m) + \delta(k+m))\theta(k_0) [/itex]
    but I'm not sure where this gets me, other than guessing that
    [itex] \delta(k^2 - m^2)\theta(k_0) [/itex] is also Lorentz invariant seeing as I am told d4k is.

    I have no idea how to proceed with this. I don't know what to make of the delta functions either.
     
  2. jcsd
  3. Nov 2, 2011 #2

    vela

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    You should recognize that [itex]d^4k\ \delta(k^2-m^2)\theta(k_0)[/itex] is manifestly Lorentz invariant because [itex]k^2 = k^\mu k_\mu[/itex] is invariant.

    The idea here is to note that [itex]k^2-m^2 = k_0^2-\mathbf{k}^2-m^2[/itex] and to do the integral over k0.

    Intuitively, you can think of the volume [itex]d^3k[/itex] being reduced by a factor [itex]\gamma[/itex] due to length contraction while E(k) is reduced by the same factor due to time dilation, so their ratio is invariant.
     
  4. Nov 2, 2011 #3
    right, so, if I'm not mistaken, [itex]k^\mu = (k_0, k_1, k_2, k_3)[/itex] and [itex]k_\mu = (k_0, -k_1, -k_2, -k_3)[/itex]. So, [itex]k^2 = k^\mu k_\mu = (k_0, k_1, k_2, k_3).(k_0, -k_1, -k_2, -k_3) = k_0^2 - k_1^2 -k_2^2 -k_3^2 = k_0^2 - \vec{k}^2[/itex].
    Is there some reason why I should instantly recognise this as an invariant quantity?
    Do you always get an invariant quantity if you take any 4-vector and do this with it?
    Also, what am I supposed to make of the [itex]\theta(k_0)[/itex] or the d4k? What about them says Lorentz invariant?

    I'm meant to be computing [itex]\frac{1}{(2\pi)^3}∫ \delta(k_0^2 - \vec{k}^2 - m^2)\theta(k_0)d^4k[/itex] right?
    I'm struggling to be change the [itex]d^4k[/itex] into something involving a dk0.
    Can I start off by saying [itex]dk^\mu = (dk_0, dk_1, dk_2, dk_3)[/itex]?

    Actually, can I say that [itex] d^4k = d^3k dk_0 [/itex]?
     
    Last edited: Nov 2, 2011
  5. Nov 2, 2011 #4

    vela

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    Yes, the product of any two four-vectors, including a four-vector with itself, is an invariant. It's analogous to the regular dot product being invariant under rotations.
    Think about the coordinate transformation [itex]k'^\mu = \Lambda^\mu{}_\nu k^\nu[/itex] where [itex]\Lambda^\mu{}_\nu[/itex] is a Lorentz transformation. How are the volume elements related? What about the signs of k0 of k'0?
    Yes, [itex] d^4k = d^3k\:dk_0 [/itex]
     
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