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I have been looking around for a satisfactory answer to prove that

$$\frac{d^3\vec{p}}{2E_{\vec{p}}}$$

where $E_{\vec{p}}=+\sqrt{(|\vec{p}|c)^2+(mc^2)^2}$, is Lorentz invariant. The standard answer seems to be that the above measure is equal to

$$d^4p \, \delta (p^2-m^2)\theta(p_0)$$

where $p^2=p_0^2-|\vec{p}|^2$, and $\theta(x)$ is the step function. I understand these two are equivalent, but I don't understand why the second has to be Lorentz invariant, in particular why the Dirac delta has to be Lorentz invariant. I have found a document (section 2.1) that proves that $\delta^{(4)}(p-p')$ is Lorentz invariant, but I can't find a way to extend their method successfully here. In fact, all I can seem to get tells me the above isn't Lorentz invariant, and that in fact it should transform to

$$\frac{d^3\vec{p}'}{2\gamma E_{\vec{p}'}}$$

which makes sense from $d^4p$ being Lorentz invaraint and $dp_0$ transforming proportional to $\gamma$, but it isn't what everyone else says. What's the problem here?


An alternative way to "derive" the factor of $1/\gamma$:

$$\delta (p^2-m^2)\theta(p_0)=\frac{1}{E_{\vec p}}\delta(p_0-E_{\vec{p}})$$

Now, as $dp_0$ transforms to $\gamma dp_0$, the $\delta(p_0-E_{\vec{p}})$ should transform to $\gamma^{-1}\delta(p_0-E_{\vec{p}})$, as can be shown with an analogous argument to the one shown in the document. The $E_{\vec{p}}$ also transforms to $\gamma(E_{\vec{p}}-v p_x)$ for an $x$-boost, but this doesn't seem to solve it.

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  • $\begingroup$ Well, the argument of this Dirac delta is rather obviously itself Lorentz invariant, isn't it? $\endgroup$ – ACuriousMind Mar 1 '15 at 21:17
  • $\begingroup$ Yes, but that isn't enough to show it is Lorentz invariant. The way distributions transforms is defined here: mathworld.wolfram.com/GeneralizedFunction.html But more simply, you should show it in an analogous way to the document I linked. $\endgroup$ – guillefix Mar 1 '15 at 21:19
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    $\begingroup$ For a linear transformation though, the transformation law reduces to a factor of the absolute value of the determinant. In one dimension: $\delta(ax) = |a|^{-1}\delta(x)$, the generalization is naturally $a \to \det A$. The determinant of a Lorentz transformation is $\pm 1$. $\endgroup$ – Robin Ekman Mar 1 '15 at 22:25
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/83260/2451 and links therein. $\endgroup$ – Qmechanic Jun 28 '15 at 12:58
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I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the $p_3$ direction so that $p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$

$$ \delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3}$$ $$ = \delta^3(p'-q')\gamma(1+\beta\frac{dE}{dp_3})= \delta^3(p'-q')\frac{\gamma}{E}(E + \beta E\frac{dE}{dp_3})$$ $$ = \delta^3(p'-q')\frac{\gamma}{E}(E+\beta p_3)= \delta^3(p'-q')\frac{E'}{E}.$$

So for a 3-momentum delta function, the quantity $E\delta^3(p-q)$ is Lorentz invariant. Combined with the fact that $\delta^4(p-q)$ is invariant, we conclude $\frac{1}{E}\delta(p_0-q_0)$ is invariant.

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  • $\begingroup$ Thank you, I think this solves it! I will look at it more carefully tomorrow, and give you the accepted answer then. $\endgroup$ – guillefix Mar 2 '15 at 1:37
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    $\begingroup$ By the way, for the record: the derivatives you show should be partial derivatives. This is in fact what made me get the wrong result. $\endgroup$ – guillefix Mar 20 '15 at 20:41
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Any Lorentz transformation will leave $p^2$ unchanged, hence the mass of the Dirac $\delta$ will remain at $m^2$. Observe that, for physical reasons, one usually only considers the component of the (full) Lorentz group that is connected to the identity. Any transformation in this proper subgroup leaves the sign of $p_0$ (which by the spectral condition is assumed to be strictly positive) invariant, hence the Heaviside function is also relativistically invariant.

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