2
$\begingroup$

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

  • 1
    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?2010-10-27
  • 1
    (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. )2010-10-27
  • 0
    I'll try it soon, thanks!2010-10-27
  • 0
    This solved everything! Thank you very very much!2010-10-27

1 Answers 1

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:

  • 0
    This was not helpful...2010-10-25
  • 0
    @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.2010-10-25
  • 0
    Yeah, the problem is that i don't know anything about distributions and their notations. I had a look at your links, but i simply can't find a formula equivalent to mine, with the product of 4 Dirac's Delta on the right.2010-10-26
  • 0
    If you could tailor an answer as simple as you can i will appreciate it a lot, and i will tell you if there's something i can't understand. Even an intuitive idea would be great! Thanks2010-10-26
  • 1
    The product of four dirac deltas is exactly the delta on 4-space.2010-10-26
  • 0
    You know, I racked my brain for almost half and hour, but I must apologise: I cannot right now think of a way of explaining an expression whose left and right hand sides only make sense as distributions without using the language or at least the philosophy of distributions. I may try again later. (And I just want to note that the equation in, say Proposition 1.24 of the third link is in fact completely equivalent to your identity.)2010-10-26
  • 1
    Hum, perhaps, if you are willing, it would be better for you to learn about distributions. http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19940029887_1994029887.pdf 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.2010-10-26
  • 0
    Thank you very very much for your time. I will study the material and then come back and tell you! ;-)2010-10-26
  • 0
    I tried to understand your third link, but proposition 1.24 is still misterious for me. There are too many recursions in the definition of E.2010-10-26
  • 0
    And in the NASA's pdf i really can't find a relation equivalent to my identity... maybe i'm not ready for these thing yet2010-10-26
  • 0
    Look, I hate to say this, but there's no royal way to learning. If you are serious about wanting to find out about this identity, you should try to understand distributions/generalised functions. The actual meaning of the term on the left hand side of your expression is actually quite subtle. When $x^0$ is positive, the term $$\delta(x_\mu x^\mu) = \delta( x^0 - \sqrt{(x^1)^2 +(x^2)^2 + (x^3)^2}) / (x^0 + \sqrt{(x^1)^2 + (x^2)^2 + (x^3)^2})$$ which is not entirely obvious without knowing something about distributions.2010-10-26
  • 0
    The problem is that most derivations for the Green's function to the wave equation (e.g. http://farside.ph.utexas.edu/teaching/jk1/lectures/node19.html ) leave the Green's function in the form of the right hand side of the expression I wrote in my previous comment. To deduce that that form is equal to the left hand side (and hence to the left hand side of your identity) is the step requiring distribution theory. If you are willing to take that step on faith, maybe you can find another easier to read presentation of the derivation for the **Green's function** to the wave equation.2010-10-26
  • 0
    I feel quite serious, but i'm working on a bachelor thesis, so i can't spend too much time on this subject. However maybe you were saying $$\delta(x_\mu x^\mu) = \delta( x^0 - \sqrt{(x^1)^2 +(x^2)^2 + (x^3)^2})(x^0 + \sqrt{(x^1)^2 + (x^2)^2 + (x^3)^2})$$, which is obvious, without knowing anything about distributions. Or were you saying something else?2010-10-27
  • 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