---

Fernando Garcia

Home     Research     Blog     Other     About me

---

PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.63

(a) A direct computation gives

$$\begin{array}{rl}\mathrm{\nabla }\cdot \mathbf{v}& =\mathrm{\nabla }\cdot \left(\frac{1}{r}\hat{\mathbf{r}}\right)\\ & =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{1}{r}\right)\\ & =\frac{1}{r^2}\frac{\partial }{\partial r}(r)\\ & =\frac{1}{r^2}\end{array}$$

Using the divergence theorem

$$\int \mathbf{v}\cdot d\mathbf{a}=\int \mathrm{\nabla }\cdot \mathbf{v}d\tau$$

let's first take a surface integral over a sphere of radius $R$:

$$\begin{array}{rl}\oint \mathbf{v}\cdot d\mathbf{a}& =\int \left(\frac{1}{R}\hat{\mathbf{r}}\right)\cdot (R^2\sin \theta d\theta d\varphi \hat{\mathbf{r}})\\ & =R{\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi \\ & =4\pi R\end{array}$$

And now a volume integral of the divergence

$$\begin{array}{rl}\int \mathrm{\nabla }\cdot \mathbf{v}d\tau & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }{\int }_{r=0}^R\left(\frac{1}{r^2}\right)(r^2\sin \left(\theta \right)drd\theta d\varphi )\\ & =4\pi R\end{array}$$

In this case (contrary to $\mathbf{v}=\hat{\mathbf{r}}/r^2$), the divergence theorem works perfectly.

In general, we have:

$$\begin{array}{rl}\mathrm{\nabla }\cdot (r^n\hat{\mathbf{r}})& =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2{r}^n\right)\\ & =\frac{1}{r^2}\frac{\partial }{\partial r}(r^{n+2})\\ & =\frac{1}{r^2}(n+2)r^{n+1}\\ & =(n+2)r^{n+1-2}\\ \text{(For }n\ne -2\text{)}& & =(n+2)r^{n-1}\end{array}$$

If $n=-2$, then we have the delta Dirac $\mathrm{\nabla }\cdot (\hat{\mathbf{r}}/r^2)=4\pi {\delta }^3(\mathbf{r})$ (see Eq. 1.99, page 48).

(b) A quick inspection at the curl formula in spherical coordinates tells us that

$$\mathrm{\nabla }\times (r^n\hat{\mathbf{r}})=\mathbf{0}$$

Using problem 1.61b, we see that the integrand on right hand side of

$${\int }_{\mathcal{V}}(\mathrm{\nabla }\times \mathbf{v})d\tau =-{\oint }_{\mathcal{S}}\mathbf{v}\times d\mathbf{a}$$

Is a cross product of parallel vectors and thus vanishes. This means that the left hand side is also zero, as expected from the explicit computation of the curl.

Back to PHYS405 site