Multivariable Calculus proof for Optics

[Blanchdog]

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.

 

[Orodruin]

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 $\oint_{S} F \cdot \hat n~dA = \int_V \nabla \cdot F~dV$

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.

What does this tell you about how the integral depends on the volume V?

 

[kuruman]

It will be beneficial for you to be able to decipher the symbols. It's really easy when you get used to the notation. You write
$\vec r=x~\hat x+y~\hat y+z~\hat z$
$\vec r'=x'~\hat x+y'~\hat y+z'~\hat z$
Then
$\vec r-\vec r'=(x-x')~\hat x+(y-y')~\hat y+(z-z')~\hat z$
the magnitude of the difference is the square root of the sum of the squares:
$|\vec r-\vec r'|=\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{1/2}$
The denominator is simply the magnitude cubed:
$|\vec r-\vec r'|^3=\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}.$

 

[Blanchdog]

What does this tell you about how the integral depends on the volume V?

That the volume V is arbitrary, the only thing that matters is whether it contains the point r' = r, and so the result of the integral is really just the value of the function at the point r' = r? But that seems weird since it looks like that would set the left side of the equation to zero. Am I on the right track at all?

 

[Orodruin]

That the volume V is arbitrary, the only thing that matters is whether it contains the point r' = r, and so the result of the integral is really just the value of the function at the point r' = r? But that seems weird since it looks like that would set the left side of the equation to zero. Am I on the right track at all?

Not really, no. You are assuming that the volume integrand is regular, but you know you are asked to show that it is proportional to a delta function.