2
$\begingroup$

I have a vector field $\vec{A}$ that is given in spherical coordinates. $$\vec{A}=\frac{1}{r^2}\hat{e}_{r}$$ I need to calculate the flux integral over a unit sphere in origo (radius 1). I cannot use Gauss theorem since there exist a singularity in the volume. I have instead attempted to calculate it in the standard way with the integral below. $$\int_{\phi}^{ } \int_{\theta}^{ }\vec{A}(\vec{r}(\theta, \phi))\cdot(\frac{\partial{\vec{r}}}{\partial{\theta}}\times \frac{\partial{\vec{r}}}{\partial{\phi}})\space d\theta d\phi$$

So in this calculation i need a parameterization of the surface. I came up with this but have been informed that it is not a correct parametrization. $$\vec{r}(\theta, \phi)=1\hat{e}_{r}+\theta\hat{e}_{\theta}+\phi\hat{e}_{\phi}$$ $$\theta:0\rightarrow\pi$$ $$\phi:0\rightarrow2\pi$$

Why is that? And how should you describe a sphere in spherical coordinates? Or should I instead transform the vector field into cartesian coordinates? How is that done?

This is the rest of the calculation and my final answer. $$\frac{\partial{\vec{r}}}{\partial{\theta}}=1\hat{e}_{\theta}$$ $$\frac{\partial{\vec{r}}}{\partial{\phi}}=1\hat{e}_{\phi}$$ $$\frac{\partial{\vec{r}}}{\partial{\theta}}\times \frac{\partial{\vec{r}}}{\partial{\phi}}=1\hat{e}_{r}$$ $$\vec{A}(\vec{r}(\theta, \phi))=\frac{1} {(1)^2}\hat{e}_{r}=1\hat{e}_{r}$$ $$\vec{A}(\vec{r}(\theta, \phi))\cdot(\frac{\partial{\vec{r}}}{\partial{\theta}}\times \frac{\partial{\vec{r}}}{\partial{\phi}})=1\hat{e}_{r}\cdot1\hat{e}_{r}=1$$ $$\int_{0}^{2\pi} \int_{0}^{\pi}1\space d\theta d\phi=2\pi^2$$

$\endgroup$

3 Answers 3

2
$\begingroup$

The parameterization is not correct. The position vector has neither a $\theta$ component nor a $\phi$ component. Note that both of those compoents are normal to the position vector. Therefore, the sperical coordinate vector parameterization of a surface would be in general

$\vec r=\hat r(\theta,\phi)r(\theta,\phi)$.

For a spherical surface of unit radius, $r(\theta,\phi)=1$ and

$$\vec r=\hat r(\theta,\phi)$$

where the unit vector $\hat r(\theta,\phi)$ can be expressed on Cartesian coordinates as

$$\hat r(\theta,\phi)=\hat x\sin \theta \cos \phi+\hat y\sin \theta \sin \phi+\hat z\cos \theta$$

Now, we can show that the unit normal to the sphere is $\hat r(\theta,\phi)$ since $\frac{\partial \hat r}{\partial \theta}\times \frac{\partial \hat r}{\partial \phi}=\hat \theta \times \hat \phi=\hat r$.

Can you finish now?

$\endgroup$
1
  • $\begingroup$ Im still not sure how I should apply it to solve the whole problem, but I will give it some more thought. Not sure how $\vec{A}(\vec{r}(\theta, \phi))$ should be handled. Do you think it might be easier to transform the vector field into cartesian coordinates? $\endgroup$
    – user300219
    Commented Dec 21, 2015 at 21:51
0
$\begingroup$

I found a very similar problem and decided to solve my problem with the same method. I belive that this yields the correct answer.

The vector field: $\vec{A}=\frac{1}{r^2}\hat{e}_{r}$

The surface: $S = Unit \space sphere \space centered \space in \space origo$

The flux through the surface $S$ is given by: $\int_{S}\vec{A}\space\cdot\space d\vec{S}$

$$d\vec{S}=r^2sin\theta d\theta d\phi\hat{e}_{r}$$

$$\int_{S}\vec{A}\space\cdot\space d\vec{S}=\int_{s}(\frac{1}{r^2}\hat{e}_{r})\cdot(r^2sin\theta d\theta d\phi\hat{e}_{r})=\int_{S}sin\theta \space d\theta d\phi=\int_{0}^{2\pi}\int_{0}^{\pi}sin\theta \space d\theta d\phi=4\pi$$

$\endgroup$
0
$\begingroup$

Your parametrization is fine, the only thing is that when calculating the derivatives of the position:

$$\frac{\partial \mathbf r}{\partial u}, \frac{\partial \mathbf r}{\partial v} \tag{1}$$

(in your case $u := \phi$, $v :=\theta$) you either have to do this in Cartesian coordinates by transforming the coordinates (and afterwards transforming the vectors back*) or (as I've been told) you have to take the derivatives of the basis vectors ($\pmb{\hat r}, \pmb{\hat \phi}, \pmb{\hat \theta}$ in your case) by the chain rule, as their direction depends on position, but I'm not sure how to do that.

Now some useful shortcuts. In the case that $\pmb{\hat u}$ and $\pmb{\hat v}$ are orthogonal:

$$\left|\frac{\partial \mathbf r}{\partial u} \times \frac{\partial \mathbf r}{\partial v}\right| = \left|\frac{\partial \mathbf r}{\partial u}\right|\left| \frac{\partial \mathbf r}{\partial v}\right|=h_uh_v \tag{2}$$

where $h_u$ and $h_v$ are the scale factors, which are also found in the line element for orthogonal curvilinear coordinates:

$$\mathrm d\mathbf l = h_u \mathrm du\pmb{\hat u}+ h_v \mathrm dv\pmb{\hat v} + h_w \mathrm dw\pmb{\hat w} \tag{3}$$

(So in your case $h_\phi = r$ and $h_\phi = r\sin\theta$) Finally the direction of the area element can be reintroduced by geometrical interpretation, or by computing the cross product in (2) (without taking the norm) in the relevant coordinate system (this can be done in spherical coordinates), and normalizing the vector to a unit vector if neccessary.

*Confusion between points/positions and vectors is possible here, so be careful.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .