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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

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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
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(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
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1 Answer 1

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:

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This was not helpful... –  Abramo Oct 25 '10 at 16:39
    
@Abramo: How was it not? Are you having trouble with the notation used in the documents I linked to? If you can at least tell me your background and why you don't find those resources helpful, I may be able to tailor an answer more to your needs. –  Willie Wong Oct 25 '10 at 16:49
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The product of four dirac deltas is exactly the delta on 4-space. –  Willie Wong Oct 26 '10 at 11:37
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Hum, perhaps, if you are willing, it would be better for you to learn about distributions. ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/… has a decent and short introduction, and by the end of it it solves the linear wave equation. That would perhaps be more illustrative than whatever non-rigorous thing I can come up with. –  Willie Wong Oct 26 '10 at 12:38
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Now i see...... –  Abramo Oct 27 '10 at 10:38
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