# Dirac's identity

Do somebody knows anything about the Dirac's identity?

$$\label{Dirac} \frac{\partial^2}{\partial x_{\mu}\partial x^{\mu}} \delta(xb_{\mu}xb^{\mu}) = -4\pi \delta(xb_0)\delta(xb_1)\delta(xb_2)\delta(xb_3)$$

here

• $xb$, is the 4-vector $x-b$ in Minkowsky spacetime

• $\delta$ is the Dirac delta function

• $x_0 = -x^0, \quad x_1 = x^1, \quad x_2 = x^2, \quad x_3, = x^3$.

Do you know where can i find some material about it?

Thanks!

UPDATE:

i've understood that a solution for the linear wave equation $$\square \psi(\mathbf{r},t) = g(\mathbf{r},t)$$ for a given $g(\mathbf{r},t)$ is $$\psi = \int \int g(r',t')G(r,r',t,t')dV'dt'$$ where $$G(r,r',t,t') = AG^+(r,r',t,t') + BG^-(r,r',t,t') , \qquad A + B = 1$$ and $$G^{\pm}(r,r',t,t') = \frac{\delta(t' - (t \mp | \mathbf{r} - \mathbf{r'} | / c))}{4\pi | \mathbf{r} - \mathbf{r'} | }$$

I think Dirac's follow from the solution of

$$\square \psi(r,t) = \delta(\mathbf{r},t)$$

But i'm not sure of the details. Can you Willie help me?

Thanks

-
Purely formally, you can plug in the four dimensional dirac delta $\delta(r,t) = \delta(r)\delta(t)$ into the integral for $\psi$. Then by the property of the dirac delta, the integral tells you that $\psi(r,t) = G(r,0,t,0)$. Which using the non-intuitive change of variable formula I mentioned in my earlier comment below, is the same as $\delta( t^2 - |r|^2/c^2)$. Does this help? –  Willie Wong Oct 27 '10 at 15:44
(The above is not at all rigorous mathematically; that the integral makes sense when you just "plug-in" the Dirac delta, and gives you the right answer, is a happy coincidence. ) –  Willie Wong Oct 27 '10 at 15:47
I'll try it soon, thanks! –  Abramo Oct 27 '10 at 16:04
This solved everything! Thank you very very much! –  Abramo Oct 27 '10 at 17:14