OK, I have to show the following:

$$ \nabla \times \left( \frac{1}{r^2} \hat r \right) = 0$$

This should be pretty easy, but I wanted to be sure I was doing this correctly.

I set up the matrix:

$$ \begin{bmatrix} \hat r & \hat \theta & \hat \phi \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ \frac{1}{r^2} & 0 & 0 \\ \end{bmatrix} =\left(\frac{\partial}{\partial \theta}(0)-\frac{\partial}{\partial \phi}(0)\right)\hat r-\left(\frac{\partial}{\partial r}(0)-\frac{\partial}{\partial \phi}(\frac{1}{r^2})\right)\hat \theta-\left(\frac{\partial}{\partial r}(0)-\frac{\partial}{\partial \theta}(\frac{1}{r^2})\right)\hat \phi$$ which leaves me with 0 because $\frac{\partial}{\partial \theta}(\frac{1}{r^2})$ and $\frac{\partial}{\partial \phi}(\frac{1}{r^2})$ are both zero.

This is correct, yes? I know this is ridiculously simple a problem but I want to make sure I did not forget everything I learned last semester. (Also, I was curious if there is a more rigorous proof, tho this is for a phys and not a math class).

Edit: BTW this is in spherical (I think -- the assignment uses $\hat r$ so I am going with that).

  • $\begingroup$ That determinant formula for the curl is only valid in cartesian coordinates! It would also give you zero for the curl of $\hat\theta$, which is clearly wrong. $\endgroup$ – Anthony Carapetis Sep 14 '13 at 0:05
  • $\begingroup$ that wold mean I have to use the other formula for curl I found -- $$ \nabla \times u = \hat r \frac{1}{sin \theta}\left[\frac{\partial}{\partial \theta} (u_{\phi} sin \theta)-\frac{\partial u_{\theta}}{\partial \phi} \right] + ... $$ yes? $\endgroup$ – Jesse Sep 14 '13 at 0:13
  • $\begingroup$ This can be handled as a special case of my answer to [this question][1]. [1]: math.stackexchange.com/questions/497293/… $\endgroup$ – Robert Lewis Sep 18 '13 at 18:19
  • $\begingroup$ The answer is definitely not trivial. It uses a citing, which is accepted literary, academic, scientific practice. Not too put too fine a point on it, but where did such policy come from? $\endgroup$ – Robert Lewis Sep 18 '13 at 18:20

Any vector field that can be expressed in the form $f(r)\mathbf{\hat r}$ must necessarily have zero curl (where the function is smooth, at least).

This can be seen by noting that, if you have a scalar field $g(r)$, and you take its gradient, you get $g'(r)\mathbf{\hat r}$. As such, with $f(r)=g'(r)$, you get the vector field. Now, you have $$\nabla \times f(r)\mathbf{\hat r} = \nabla \times \nabla g(r) = 0$$

In particular, for $f(r)=\frac1{r^2}$, you have $g(r)=-\frac1r$.

Note that this only applies if the function is independent of $\theta$ and $\phi$.


Actually we can check this is true using the following two facts:

  • curl of a gradient field is zero.

  • cross product of two parallel vector fields is zero.

I am assuming your $\hat{r} = (x,y,z)$, and $r= \sqrt{x^2+y^2+z^2}$, then it is not hard to check that $$ \frac{1}{r^2} \hat{r} = \frac{\nabla r}{r}. $$ Now using the product rule for curl, a scalar function $f$ and a vector field $\hat{g}$: $$ \nabla \times (f\hat{g}) = \nabla f \times \hat{g} + f\nabla\times \hat{g}. $$ We have $$ \nabla \times \left(\frac{1}{r^2} \hat{r}\right)= \nabla \times \left(\frac{\nabla r}{r}\right) = \nabla \times (\nabla r) \frac{1}{r} + \nabla \left(\frac{1}{r}\right)\times \nabla r = 0 - \frac{1}{r^2}\nabla r\times \nabla r=0. $$


Set $U(r) = \frac{1}{r^2}$; then this problem is a special case of the following, which also appears here.

If what you really want to do is show that $\nabla \times U(r) \hat r = 0$ for spherically stmmetric functions $U(r)$, then perhaps the easiest thing to do is use the identity

$\nabla \times (f \hat V) = \nabla f \times \hat V + f \nabla \times \hat V, \tag{1}$

a standard result from vector calculus, which can easily be derived from the well-known "determinant formula" for $\nabla \times$:

$\nabla \times \hat Y = \begin{bmatrix} \hat {\mathbf i} & \hat {\mathbf j} & \hat {\mathbf k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat Y_x & \hat Y_y & \hat Y_z \end{bmatrix}. \tag{2}$

If we set $\hat Y = f \hat V$ in (2) and grind, we get

$\nabla \times f \hat V = (\frac{\partial}{\partial y} (f \hat V_z) - (\frac{\partial}{\partial z} (f \hat V_y))\hat {\mathbf i} + (\frac{\partial}{\partial z} (f \hat V_x) - (\frac{\partial}{\partial x} (f \hat V_z))\hat {\mathbf j}$ $+ (\frac{\partial}{\partial x} (f \hat V_y) - (\frac{\partial}{\partial y} (f \hat V_x))\hat {\mathbf k}, \tag{3}$


$\nabla \times f \hat V = ((f_y \hat V_z + f \hat V_{z, y}) - (f_z \hat V_y + f \hat V_{y, z}))\hat {\mathbf i} + ((f_z \hat V_x + f \hat V_{x, z}) - (f_x \hat V_z + f \hat V_{z, x}))\hat {\mathbf j}$ $+ ((f_x \hat V_y + f \hat V_{y, x}) - (f_y \hat V_x + f \hat V_{x, y}))\hat {\mathbf k}, \tag{4}$

after which some simple algebraic re-arrangement yields (1). For example, the coefficient of $\hat{\mathbf i}$ is

$(f_y \hat V_z - f_z \hat V_y) + f(\hat V_{z, y} - \hat V_{y, z}). \tag{5}$

In (3), (4), and (5), I have used subscript notation for partials, e.g. $f_x = \frac{\partial f}{\partial x}$ and $\hat V_{x,y} = \frac{\partial V_x}{\partial y}$ etc. This formula may also be found in this wikipedia article.

Applying (1) to $U(r) \hat r$, we have

$\nabla \times U(r) \hat r = \nabla U(r) \times \hat r + U(r) \nabla \times \hat r. \tag{6}$


$\nabla \times \hat r = 0, \tag{7}$

which is easy to see by direct calculation, using (2) if you like, so we are left with

$\nabla \times U(r) \hat r = \nabla U(r) \times \hat r; \tag{8}$

but I claim that for any function $U(r)$ which only depends on $r$, $\nabla U(r)$ is in fact collinear with $\hat r$; to see this, write

$U(r) = U((x^2 + y^2 + z^2)^{\frac{1}{2}}), \tag{9}$

so that for example

$\frac{\partial U}{\partial x}(r) = \frac{dU}{dr}(r) \frac{x}{r}, \tag{10}$

by the chain rule, since $\frac{\partial r}{\partial x} = \frac{x}{r}$. Similar expressions hold for the $y$ and $z$ derivatives as well. Now (10), when combined with its $y$ and $z$ counterparts, gives

$\nabla U(r) = r^{-1}\frac{dU}{dr}(r) \hat r, \tag{11}$,

and substituting this in (8) yields

$\nabla \times U(r) \hat r = r^{-1}\frac{dU}{dr}(r) \hat r \times \hat r = 0, \tag{12}$

since $\hat r \times \hat r$ vanishes identically. Thus we see that

$\nabla \times U(r) \hat r = 0, \tag{13}$

as was required. QED.

CAVEAT: One should of course take care applying these or any derivative formula to a function or vector field at the origin $(0, 0, 0)$ where $r = 0$, since $\sqrt{x^2 + y^2 + z^2}$ is continuous but not differentiable there. But if $U(r)$ is sufficiently smooth everywhere, then I do b'lieve she'll fly, Wilbur!

Hope this helps. Cheers!

and as always

Fiat Lux!

  • $\begingroup$ wow, this is so unnecessarily complicated! $\endgroup$ – tom Sep 18 '13 at 19:24
  • $\begingroup$ @tom: how so? It basically uses a similar technique to Shuhao Cao's answer, but explains it in more detail. Seriously, not trying to be argumentative, but since you brought up the point . . . $\endgroup$ – Robert Lewis Sep 18 '13 at 19:46

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