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Do somebody knows anything about the Dirac's identity?

\begin{equation} \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) \end{equation}

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:

Following Willie's link

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

  • 0
    This solved everything! Thank you very very much!2010-10-27

1 Answers 1

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Your identity is in fact the expression for the fundamental solution of the linear wave equation in (1+3) dimensions. This should be in most textbooks on electrodynamics or intro to quantum field theory.

Google also tells me:

  • 0
    No, not multiplied by. Divided by. And outside. To make it even more clear $ \delta(x_\mu x^\mu) = \frac{\delta( x^0 - \sqrt{(x^1)^2 +(x^2)^2 + (x^3)^2}) }{ x^0 + \sqrt{(x^1)^2 + (x^2)^2 + (x^3)^2} } $. And only for x_0 > 0. If x_0 < 0, you have to swap the term inside the $\delta$ with the denominator. That you think what you wrote, which is incorrect, is obvious, already illustrates the subtlety of distributions.2010-10-27