I have the following problem: I know a unit vector in spherical polar coordinates, let's call it $\vec{e}_{s}$, in the form $\vec{e}_{s} = f(r,\theta)\vec{e}_{r} + g(r,\theta)\vec{e}_{\theta} + c\vec{e}_{\phi}$, where $f,g$ are known functions and $c$ is a constant. Furthermore I know that this vector is the tangent of some curve $C$ at every point $(r,\theta,\phi)$. My question is how can I compute the line integral of its divergence along $C$. I know that:$\begin{equation} \int_{C}\nabla\vec{e}_{s}ds=\int_{c}\nabla\vec{e}_{s}\vec{e}_{s}d\vec{r} \end{equation}$

where $d\vec{r}$ is the infinitesimal length element (in this case in spherical coordinates), but I can't get any further. I know that, in principle, given a parametric curve $C(t)$ one can easily compute (numerically) the line integral of a by expressing $(r,\theta) = (r(t),\theta(t))$ but I don't know how can I do this in my case.

I've seen the examples for NIntegrate here for Line integrals, but it seems to me I would need the curve to be parametrized by a single variable and the same to hold for my vector.

I know that this is not specifically a question for Mathematica, but the integral is to be evaluated numerically and I intend to do this in Mathematica.

Any help on this would be greatly appreciated!

  • 1
    $\begingroup$ Are you sure about: $$\int_{C} \nabla \cdot\vec{e}_{s} d s=\int_{C} \nabla \cdot\vec{e}_{s} \vec{e}_{s} d \vec{r}$$ ? Are the tangents $\mathbf{\alpha}^\prime(t)$ of your curve always unit vectors? If your curve has nonzero 'acceleration' $\mathbf{\alpha}^{\prime\prime}(t)$ i.e not a line or circle, then I don't think this is correct. $\endgroup$ – flinty May 18 at 16:59
  • 1
    $\begingroup$ The divergence in spherical polar is: $$\nabla \cdot \mathbf{A}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} A_{r}\right)+\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta A_{\theta}\right)+\frac{1}{r \sin \theta} \frac{\partial A_{\varphi}}{\partial \varphi}$$ In your case $A_r=f(r,\theta), A_\theta=g(r,\theta), A_\phi=c$ $\endgroup$ – flinty May 18 at 17:15
  • 1
    $\begingroup$ Hello! You might be right. Anyway, I do not know the curve parametrization. I integrate along some path that I require to be tangent to $\vec{e}_s$ at every point. I know how to write the divergence in polar coordinates. My problem is that I can't wrap my mind around how to parametrize the curve using a single parameter. The only information that I have is the expressions of the functions $f,g$ in terms of $r,\theta$. $\endgroup$ – Kvothe May 18 at 18:30
  • $\begingroup$ Just to be a little more explicit: i do not know the curve $C$. In principle I'm defining it to be the curve to which the vector $\vec{e}_s$ is always tangent. Maybe there is an error in this logic $\endgroup$ – Kvothe May 18 at 18:37

Update: technically it's ill-defined because you cannot take the divergence of $\vec{e}_s$ which is restricted to the curve. See here https://math.stackexchange.com/questions/3682769/line-integral-of-the-divergence-of-a-curves-unit-tangent-vector . So I've edited this to not confuse anyone passing by.

However 'ill-defined' is unsatisfying, so based on physical intuition I'm going to carry on saying it's zero. The reasoning is entirely based on the real-world meaning of divergence as average flux over a boundary and is not very mathematical.

Hand waving: Imagine the curve is actually a thin wire. The average amount of unit vector field flowing into a small box along the wire equals the amount flowing out - you have net zero flux. Also note if you had a closed curve in 2D, there's zero flux flowing in or out of the bounded region. In higher dimensions, there's zero flux going across the curve - it only flows along with it.

An example of such a curve could be a line, circle, or a helix like below:

a[t_] := {Cos[t]/2, Sin[t]/2, Sqrt[3/4] t}
f[x_, y_, z_] := {-y, x, Sqrt[3/4]}
a'[1] == f @@ a[1] (*test f[x(t),y(t),z(t)] gives unit vector a'[t]*)
Div[f[x, y, z], {x, y, z} ](* is zero *)

Here's how you can get the divergence (in spherical) of your $\vec{e}_s$ using Mathematica

 Div[{f[r, \[Theta]], g[r, \[Theta]], c}, {r, \[Theta], \[Phi]}, 

$$ \nabla\cdot\vec{e}_s=\left[\frac{2 f(r,\theta)}{r}+\frac{\partial f}{\partial r}\right]+\left[\frac{1}{r}\cot(\theta)g(r,\theta)+\frac{1}{r}\frac{\partial g}{\partial \theta}\right]+\left[0\right] $$

| improve this answer | |
  • $\begingroup$ I think you are right, but I followed a slightly different path. Consider a vectro $\vec{\alpha}(t) = \alpha_x(t)\vec{e}_{x} + \alpha_{y}(t)\vec{e}_y + \alpha_z(t)\vec{e}_z$. Now, the integral in the original post can be written as: $$ \int_{C}\nabla\vec{\alpha}\vec{\alpha}d\vec{r} = \int_{C}dt\left[\left(\frac{\partial \alpha_x}{\partial x}\alpha_x\frac{dx}{dt}\right) + (...)_y + (...)_z\right] $$ where $(...)_{y,z}$ contain the same expressions as the first brackets but with the indices $y,z$ instead of $x$. Comment continues bellow due to space limitations $\endgroup$ – Kvothe May 20 at 7:18
  • 1
    $\begingroup$ Then, using the chain rule, the integral gives the square of $\vec{\alpha}$: $$ \int_{C}\nabla\vec{\alpha}\vec{\alpha}d\vec{r} = \frac{1}{2}\vec{\alpha}^2(b)-\frac{1}{2}\vec{\alpha}^2(a) $$ In my case, in which $\vec{\alpha}$ is a unit vector, the two values at the ends of the curve ($a,b$) are equal both to $1$. So, indeed the line integral is 0. I'm not very confident in this reasoning because it gives a very general result which I did not find in any textbook. This might be because I'm actually wrong and have omitted something or because it is not very useful. $\endgroup$ – Kvothe May 20 at 7:22
  • 1
    $\begingroup$ I also asked about this elsewhere: math.stackexchange.com/questions/3682769/… $\endgroup$ – flinty May 20 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.