Divergence in spherical coordinates

[wifi]

Problem:

For the vector function $$\vec{F}(\vec{r})=\frac{r\hat{r}}{(r^2+{\epsilon}^2)^{3/2}}$$

a. Calculate the divergence of $\vec{F}(\vec{r})$, and sketch a plot of the divergence as a function $r$, for $\epsilon$<<1, $\epsilon$≈1 , and $\epsilon$>>1.

b. Calculate the flux of $\vec{F}$ outward through a sphere of radius R centered at the origin, and verify that it is equal to the integral of the divergence inside the sphere.

c. Show that the flux is $4 \pi$ (independent of R) in the limit $\epsilon \rightarrow 0$.

Solution:

a. Calculating the divergence is simply a matter of plugging the function correctly into the spherical coordinates divergence formula. I get $$F_r=\frac{r}{(r^2+{\epsilon}^2)^{3/2}} \Rightarrow \frac{\partial F}{\partial r}=3r^2(r^2+{\epsilon}^2)^{-3/2}-3r^4(r^2+{\epsilon}^2)^{-5/2}$$

Therefore,

$$\nabla \cdot \vec{F}= 3(r^2+{\epsilon}^2)^{-3/2}-3r^2(r^2+{\epsilon}^2)^{-5/2}$$

So for the different values of $\epsilon$ do I just substitute?

 

[wifi]

Also can someone point me in the right direction for parts b. and c.? Thanks.

 

[wifi]

Anyone?

 

[ehild]

Check your ∂F/∂r.

Anyway, the end result is correct.

As for b), what is flux?

ehild

 

[paranormal]

Yes, you can substitute different values of $\epsilon$ to get a better understanding of how the divergence changes. When $\epsilon$ is very small, the divergence is close to 3/r^2, which means that the vector field is diverging strongly away from the origin. As $\epsilon$ increases, the divergence decreases and approaches 0 as $\epsilon$ approaches infinity. This means that the vector field becomes more uniform and less divergent as $\epsilon$ increases.

b. To calculate the flux, we use the divergence theorem, which states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence over the enclosed volume. In this case, the closed surface is a sphere of radius R centered at the origin. Therefore, the flux is given by:

\Phi=\int_V \nabla \cdot \vec{F} dV

Since the vector field is spherically symmetric, we can use the divergence formula in spherical coordinates to simplify the integral. This gives us:

\Phi=\int_0^{2\pi} \int_0^{\pi} \int_0^R \left(3(r^2+{\epsilon}^2)^{-3/2}-3r^2(r^2+{\epsilon}^2)^{-5/2}\right) r^2 \sin{\phi} dr d\phi d\theta

Solving this integral gives us:

\Phi=\frac{4\pi}{3}\left(1-\frac{R^3}{(R^2+{\epsilon}^2)^{3/2}}\right)

c. To show that the flux is independent of R in the limit $\epsilon \rightarrow 0$, we simply take the limit as $\epsilon$ approaches 0 in the above equation. This gives us:

\lim_{\epsilon \rightarrow 0} \Phi=\frac{4\pi}{3}

Therefore, the flux is independent of R in the limit $\epsilon \rightarrow 0$, which means that the vector field is uniform and non-divergent in this limit. This is consistent with our previous understanding that as $\epsilon$ approaches 0, the vector field becomes more uniform and less divergent.