# Question about Dirac Delta function

1. Aug 9, 2013

### yungman

In page 555, Appendix B of Intro to electrodynamics by D Griffiths:
$$\nabla\cdot \vec F=-\nabla^2U=-\frac{1}{4\pi}\int D\nabla^2\left(\frac{1}{\vec{\vartheta}}\right)d\tau'=\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r)$$
where $\;\vec{\vartheta}=\vec r-\vec r'$.

Is it supposed to be equal to $D(\vec r')$ as equation (1.98) in page 50 stated $\int f(\vec r)\delta(\vec r-\vec a)d\tau'=f(\vec a)$?

Thanks

2. Aug 9, 2013

### vanhees71

I don't precisely understand, what's your question. The most non-trivial statement is that
$$\Delta_{\vec{x}} \frac{1}{4 \pi |\vec{x}-\vec{x}'|}=\delta^{(3)}(\vec{x}-\vec{x}').$$
To prove this, it is most simple to solve for the Green's function of the Laplace operator, i.e.,
$$\Delta G(\vec{x})=-\delta(\vec{x}).$$
One way is to use the method of Fourier transformation, i.e., write
$$G(\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \tilde{G}(\vec{k})\exp(\mathrm{i} \vec{k} \cdot \vec{x}).$$
Then you find
$$\Delta G(\vec{x})=-\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3}\vec{k}^2 \tilde{G}(\vec{k})\exp(\mathrm{i} \vec{k} \cdot \vec{x}).$$
On the other hand, the Dirac $\delta$ distribution (it's not a function but a distribution!) has the Fourier representation
$$\delta^{(3)}(\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3}\exp(\mathrm{i} \vec{k} \cdot \vec{x}).$$
This implies
$$\tilde{G}(\vec{k})=\frac{1}{\vec{k}^2}.$$
Now you have to do the Fourier transform. Unfortunately the integral is not absolutely convergent. This we cure by restricting the integration in momentum space to the ball of radius $\Lambda$,
$$G(\vec{x})=\int_{B_{\Lambda}} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \frac{\exp(\mathrm{i} \vec{k} \cdot \vec{x})}{\vec{k}^2}.$$
We introduce spherical coordinates with polar axis in direction of $\vec{x}$. Then we can do the angular integrals immediately (substituting $u=\cos \vartheta$)
$$G(\vec{x})=\frac{1}{4 \pi^2} \int_0^{\Lambda} \mathrm{d} K \int_{-1}^1 \mathrm{d} u \exp(\mathrm{i} K r u)=\frac{1}{2 \pi^2 r} \int_0^{\Lambda} \frac{\sin(k r)}{k}.$$
Now we can let $\Lambda \rightarrow \infty$ and use the standard integral
$$\int_0^{\infty} \mathrm{d} k \frac{\sin(k r)}{k}=\frac{\pi}{2}$$
to finally obtain
$$G(\vec{x})=\frac{1}{4\pi |\vec{x}|}.$$

3. Aug 9, 2013

### HallsofIvy

No. In that last equation, the integration is with respect to the "primed" values and the result is f of the unprimed a. In the equation you are asking about, the integration is again with respect to the "primed" variables and the result is again a function of the unprimed variable.

4. Aug 9, 2013

### yungman

Thanks for your response. That's exactly my question. I don't understand why using $D(\vec r)$ instead of $D(\vec r')$?

I don't understand why $\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r)$ where integration with respect to the "primed" variables gives a function of the unprimed variable? Is it like this:
$$\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=-\int D(\vec r')\delta^3(\vec r'-\vec r)d\tau'=D(\vec r)$$
as $\delta(-x)=-\delta(x)\Rightarrow\;\delta^3(\vec r-\vec r')=-\delta^3(\vec r'-\vec r)$

Last edited: Aug 9, 2013
5. Aug 9, 2013

### klawlor419

Its because the primed r inside of the D is just a tool for you to note that you are integrating that function over all space, while weighting it with the 3-D point function.

Once you do the integral, it takes on the value of the function D evaluated at the point r (unprimed). The function at all other points outside of that the integral is zero because its weight with the Dirac function

6. Aug 10, 2013

### andrien

how did you got that?Delta function is an even function.What you have written holds for δ'(-x).

7. Aug 10, 2013

### yungman

I double check my notes, I was wrong.
$$\delta(-x)=\delta(x)\; and \; \delta(\vec r-\vec r')=\delta(\vec r'-\vec r)$$

Does this work better?