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

Staff Emeritus
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?