- #1

Blanchdog

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- Homework Statement
- Prove that the identity ##\nabla_{\textbf{r}} \frac{1}{|\textbf{r} - \textbf{r}'|} = -\frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3}## where ##\nabla_{r}## operates only on r, treating r' as a constant vector.

Next, prove that ##\nabla_{\textbf{r}} \frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3} = 0## except at r = r' where a singularity situation occurs.

Finally, use the divergence theorem to show that the function ##\nabla_{\textbf{r}} \frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3}## is ## 4\pi## times the threee-dimensional delta function: ##\delta^3(\textbf{r'} - \textbf{r}) \equiv \delta(x' - x)\delta(y' - y)\delta(z' - z)##

- Relevant Equations
- The Dirac delta function is defined indirectly as ##f(t) = \int_{-\infty}^{\infty} f(t') \delta (t' - t) dt' ##

The divergence theorem for a vector function F is ##\oint_{S} F \cdot \hat n~dA = \int_V \nabla \cdot F~dV##

Part A)

For part A I forgo breaking down the identity into it's component x, y, and z parts, and just take the r derivative treating r' as a constant vector. This seems to give the right answer, but to be entirely honest I'm not sure how I'd go about doing this component by component. I figure I'd replace the numerator with x - x' (or whatever direction I'm working with) but what would I do with the denominator? Would I just take the components there too, would I need to use the distance equation?

Part B) Really not sure what to do with this part.

Part C)

By the divergence theorem, I get $$\oint_{S} \frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3} \cdot \hat n~dA = \int_V \cdot \nabla_{\textbf{r}} \frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3}~dV$$

From part B, the argument in the integral on the right side must be zero except at ##r' = r##, which makes both sides of the relation zero except at that point. I'm sure there's a way to relate that to a delta function somehow but I'm not sure what it is.

For part A I forgo breaking down the identity into it's component x, y, and z parts, and just take the r derivative treating r' as a constant vector. This seems to give the right answer, but to be entirely honest I'm not sure how I'd go about doing this component by component. I figure I'd replace the numerator with x - x' (or whatever direction I'm working with) but what would I do with the denominator? Would I just take the components there too, would I need to use the distance equation?

Part B) Really not sure what to do with this part.

Part C)

By the divergence theorem, I get $$\oint_{S} \frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3} \cdot \hat n~dA = \int_V \cdot \nabla_{\textbf{r}} \frac{(\textbf{r} - \textbf{r}')}{|\textbf{r} - \textbf{r}'|^3}~dV$$

From part B, the argument in the integral on the right side must be zero except at ##r' = r##, which makes both sides of the relation zero except at that point. I'm sure there's a way to relate that to a delta function somehow but I'm not sure what it is.