Trying to understand where the $\frac{1}{r sin(\theta)}$ and $1/r$ bits come in the definition of gradient.

I've derived the spherical unit vectors but now I don't understand how to transform cartesian del into spherical del at all. People keep saying use the chain rule, but I don't see it!

Any help?

  • $\begingroup$ I believe the way of deriving them from truly first principles should involve pulling back the metric from $\mathbb{R}^3$ when embedding $S^2$... A perhaps less fundamental but still satisfying way of doing things is defining $x,y,z$ in terms of $r,\theta,\phi$ and working from there. $\endgroup$
    – Danu
    Commented Sep 25, 2013 at 10:21
  • $\begingroup$ I mean how do you go about converting cartesian into spherical polars? $\endgroup$ Commented Sep 25, 2013 at 10:24
  • $\begingroup$ Would math.stackexchange.com be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Sep 25, 2013 at 11:46
  • $\begingroup$ @Qmechanic In Australia, we learn this identity in second year university Physics. I am just now messing about with the derivation myself as I already know how to do this using a general result from pure maths but finding a derivation without using that level of abstraction might be of interest to the general physics student. How do you draw the line between maths and physics? Not without a lot of blood on the carpet I would think. $\endgroup$ Commented Oct 11, 2014 at 9:32

3 Answers 3


You asked for a proof from "first principles". So let's do it. I'll highlight the most common sources of errors and I'll show an alternative proof later that doesn't require any knowledge of tensor calculus or Einstein notation.

The hard way

First, the coordinates convention:

$$(r,\theta,\phi) \rightarrow (x,y,z) = (r\sin\theta\cos\phi,\;r\sin\theta\sin\phi,\;r\cos\theta)$$

The same way we can express $(x,y,z)$ as $x\,\mathbf{\hat e}_x + y\,\mathbf{\hat e}_y + z\,\mathbf{\hat e}_z$, we can also express $(r,\theta,\phi)$ as $r'\,\mathbf{\hat e}_r + \theta'\,\mathbf{\hat e}_\theta + \phi'\,\mathbf{\hat e}_\phi$, but now the coefficients are not the same: $(r',\theta',\phi') \neq (r,\theta,\phi)$, in general. This is because spherical coordinates are curvilinear, so the basis vectors are not the same at all points. For small variations, however, they are very similar. More precisely, relative to a point $\vec{\mathbf p}_0 = (x,y,z)$, a neighbor point $\vec{\mathbf p}_1 = (x+\Delta x,\;y+\Delta y,\;z+\Delta z)$ can be described by $\Delta \vec{\mathbf p} = (\Delta x,\Delta y,\Delta z)$ and, in spherical coordinates, if this variation is "infinitesimal", then $d\vec{\mathbf p} = (dr, d\theta, d\phi) = dr\,\mathbf{\hat e}_r + d\theta\,\mathbf{\hat e}_\theta + d\phi\,\mathbf{\hat e}_\phi$. This is basically the motivation for defining the (unnormalized) basis as:

$$ \vec{\mathbf e}_r = \frac{\partial\vec{\mathbf p}}{\partial r}, \quad \vec{\mathbf e}_\theta = \frac{\partial\vec{\mathbf p}}{\partial \theta}, \quad \vec{\mathbf e}_\phi = \frac{\partial\vec{\mathbf p}}{\partial \phi} $$

But this is not normalized yet. Coincidentally, $||\partial\vec{\mathbf p}/\partial r||$ turns out to be $1$, but $||\partial\vec{\mathbf p}/\partial \theta|| = r$, as we'll see. So the actual basis should be defined as:

$$ \hat{\mathbf e}_r = \frac{\vec{\mathbf e}_r}{||\vec{\mathbf e}_r||}, \quad \hat{\mathbf e}_\theta = \frac{\vec{\mathbf e}_\theta}{||\vec{\mathbf e}_\theta||}, \quad \hat{\mathbf e}_\phi = \frac{\vec{\mathbf e}_\phi}{||\vec{\mathbf e}_\phi||} $$


$$\begin{align} \vec{\mathbf e}_r &=\;( & \sin\theta\cos\phi&, & \sin\theta\sin\phi&, & \cos\theta) \\ \vec{\mathbf e}_\theta &=\;( & r\cos\theta\cos\phi&, & r\cos\theta\sin\phi&, & -r\sin\theta) \\ \vec{\mathbf e}_\phi &=\;( & -r\sin\theta\sin\phi&, & r\sin\theta\cos\phi&, & 0) \end{align}$$

$$\begin{align} ||\vec{\mathbf e}_r||^2 &= \sin^2\theta(\cos^2\phi + \sin^2\phi) + \cos^2\theta & &= 1 \\ ||\vec{\mathbf e}_\theta||^2 &= r^2\cos^2\theta(\cos^2\phi + \sin^2\phi) + r^2\sin\theta & &= r^2 \\ ||\vec{\mathbf e}_\phi||^2 &= r^2\sin^2\theta(\sin^2\phi + \cos^2\phi) & &= r^2\sin^2\theta \end{align}$$

$$\begin{align} \hat{\mathbf e}_r &= \vec{\mathbf e}_r & &= & &( & \sin\theta\cos\phi&, & \sin\theta\sin\phi&, & \cos\theta) \\ \hat{\mathbf e}_\theta &= \vec{\mathbf e}_\theta/r & &= & &( & \cos\theta\cos\phi&, & \cos\theta\sin\phi&, & -\sin\theta) \\ \hat{\mathbf e}_\phi &= \vec{\mathbf e}_\phi/(r\sin\theta) & &= & &( & -\sin\phi&, & \cos\phi&, & 0) \end{align}$$

You may verify that this also forms an orthogonal basis (hence orthonormal). For example:

$$\begin{align} \hat{\mathbf e}_r \cdot \hat{\mathbf e}_\theta &= \sin\theta\cos\theta\cos^2\phi + \sin\theta\cos\theta\sin^2\phi - \sin\theta\cos\theta \\ &= 0 \end{align}$$

That does not need to happen in general.

To go from one set of coordinates to the other using the basis vectors, we solve:

$$ \begin{bmatrix}\hat{\mathbf e}_r \\ \hat{\mathbf e}_\theta \\ \hat{\mathbf e}_\phi\end{bmatrix} = \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \begin{bmatrix}\hat{\mathbf e}_x \\ \hat{\mathbf e}_y \\ \hat{\mathbf e}_z\end{bmatrix} $$

for $\hat{\mathbf e}_x$, $\hat{\mathbf e}_y$, and $\hat{\mathbf e}_z$ in terms of $\hat{\mathbf e}_r$, $\hat{\mathbf e}_\theta$, and $\hat{\mathbf e}_\phi$. Then any vector $\vec{\mathbf p} = x\,\mathbf{\hat e}_x + y\,\mathbf{\hat e}_y + z\,\mathbf{\hat e}_z$ can be written in the form $r'\,\mathbf{\hat e}_r + \theta'\,\mathbf{\hat e}_\theta + \phi'\,\mathbf{\hat e}_\phi$ by simple substitution. Since this particular basis is orthonormal, there's an alternative way: simply use the dot product. For example, to get $r'$:

$$\begin{align} \vec{\mathbf p}\cdot\mathbf{\hat e}_r &= r'\,\mathbf{\hat e}_r\cdot\mathbf{\hat e}_r + \theta'\,\mathbf{\hat e}_\theta\cdot\mathbf{\hat e}_r + \phi'\,\mathbf{\hat e}_\phi\cdot\mathbf{\hat e}_r \\ &= r' \end{align}$$

Now to the gradient. Using matrix notation, we can write the gradient as a row vector and the formula for the chain rule becomes:

$$\begin{align} \vec\nabla f &= \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{bmatrix} \\ &= \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} & \frac{\partial r}{\partial z} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} & \frac{\partial \theta}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{bmatrix} \end{align}$$

Call the matrix on the right $J$ (it's the Jacobian matrix). Note that this also works the other way around too:

$$ \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} \\ = \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{bmatrix} \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} & \frac{\partial z}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \theta} \\ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \end{bmatrix} $$

And call this other matrix $J'$. We can invert the first equation to get $\vec\nabla f \,J^{-1} = \vec\nabla f \,J'$ $\Rightarrow$ $\vec\nabla f \,\left(J^{-1}-J'\right) = 0$. Since this works for an arbitrary $f$, we have $J^{-1} - J' = 0$ $\Rightarrow$ $J' = J^{-1}$. An important consequence is that, in general:

$$ \frac{\partial a}{\partial b} \neq \left(\frac{\partial b}{\partial a}\right)^{-1} $$

It appears that the OP made this mistake in a comment, confusing $\partial r/\partial x$ with $(\partial x/\partial r)^{-1} = 1/(\sin\theta\cos\phi)$, as would be the case if we were using regular (instead of partial) derivatives.

Now we have two ways of calculating the matrix $J$. Directly or by calculating $J'$ first and then inverting it. Let's do it directly. We're gonna need the expressions for $r$, $\theta$, and $\phi$ in terms of $x$, $y$, and $z$ (for other coordinate systems this might be very difficult to obtain):

$$\begin{align} r &= \sqrt{x^2+y^2+z^2} \\ \theta &= \arctan\left(\frac{\sqrt{x^2+y^2}}{z}\right) \\ \phi &= \arctan\left(\frac{y}{x}\right) \end{align}$$

The partial derivatives are:

$$\begin{align} \frac{\partial r}{\partial x} &= \frac{x}{\sqrt{x^2+y^2+z^2}} & &= \sin\theta\cos\phi \\ \frac{\partial r}{\partial y} &= \frac{y}{\sqrt{x^2+y^2+z^2}} & &= \sin\theta\sin\phi \\ \frac{\partial r}{\partial z} &= \frac{z}{\sqrt{x^2+y^2+z^2}} & &= \cos\theta \end{align}$$

$$\begin{align} \frac{\partial \theta}{\partial x} &= \frac{zx}{\sqrt{x^2+y^2}\left(x^2+y^2+z^2\right)} & &= \frac{\cos\theta\cos\phi}{r} \\ \frac{\partial \theta}{\partial y} &= \frac{zy}{\sqrt{x^2+y^2}\left(x^2+y^2+z^2\right)} & &= \frac{\cos\theta\sin\phi}{r} \\ \frac{\partial \theta}{\partial z} &= -\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2} & &= -\frac{\sin\phi}{r} \end{align}$$

$$\begin{align} \frac{\partial \phi}{\partial x} &= -\frac{y}{x^2+y^2} & &= -\frac{\sin\phi}{r\sin\theta} \\ \frac{\partial \phi}{\partial y} &= \frac{x}{x^2+y^2} & &= \frac{\cos\phi}{r\sin\theta} \\ \frac{\partial \phi}{\partial z} &= 0 & &= 0 \end{align}$$

Our Jacobian is then:

$$ J = \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \frac{\cos\theta\cos\phi}{r} & \frac{\cos\theta\sin\phi}{r} & -\frac{\sin\phi}{r} \\ -\frac{\sin\phi}{r\sin\theta} & \frac{\cos\phi}{r\sin\theta} & 0 \end{bmatrix} $$

Alternatively, we could have calculated the inverse Jacobian (which is straightforward) and then inverted it (which is a nightmare). We can use Wolfram Alpha to confirm that it gives the same result:

Wolfram Alpha input and result

Finally, we use the dot product to find the coefficients $r'$, $\theta'$, and $\phi'$:

$$ r' = \vec\nabla f\cdot\mathbf{\hat e}_r = \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} J \begin{bmatrix}\sin\theta\cos\phi \\ \sin\theta\sin\phi \\ \cos\theta\end{bmatrix} = \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} = \frac{\partial f}{\partial r} $$

$$ \theta' = \vec\nabla f\cdot\mathbf{\hat e}_\theta = \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} J \begin{bmatrix}\cos\theta\cos\phi \\ \cos\theta\sin\phi \\ -\sin\theta\end{bmatrix} = \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} \begin{bmatrix}0 \\ 1/r \\ 0\end{bmatrix} = \frac{1}{r}\frac{\partial f}{\partial \theta} $$

$$ \phi' = \vec\nabla f\cdot\mathbf{\hat e}_\phi = \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} J \begin{bmatrix}-\sin\phi \\ \cos\phi \\ 0\end{bmatrix} = \begin{bmatrix}\frac{\partial f}{\partial r} & \frac{\partial f}{\partial \theta} & \frac{\partial f}{\partial \phi}\end{bmatrix} \begin{bmatrix}0 \\ 0 \\ 1/(r\sin\theta)\end{bmatrix} = \frac{1}{r\sin\theta}\frac{\partial f}{\partial \phi} $$


$$ \vec\nabla f = \frac{\partial f}{\partial r}\mathbf{\hat e}_r + \frac{1}{r}\frac{\partial f}{\partial \theta}\mathbf{\hat e}_\theta + \frac{1}{r\sin\theta}\frac{\partial f}{\partial \phi}\mathbf{\hat e}_\phi $$

A much better way

We're gonna need a new notation to avoid having to use different letters for $x$, $y$, and $z$, for example. Instead let's use indices from $1$ to $3$. For Cartesian coordinates we'll use the letter $x$, and for spherical coordinates we'll use the letter $r$. The following should be self explanatory:

$$ \vec{\mathbf p} = \sum_i x_i\mathbf{\hat x}^i = \sum_k r_k\mathbf{\hat r}^k $$

From the definition of the basis vectors:

$$ \mathbf{\vec r}^k = \frac{\partial \vec{\mathbf p}}{\partial r_k}, \quad \mathbf{\hat r}^k = \frac{\mathbf{\vec r}^k}{||\mathbf{\vec r}^k||} = \frac{1}{h_k}\frac{\partial \vec{\mathbf p}}{\partial r_k} $$

Where $h_k \triangleq ||\mathbf{\vec r}^k||$. Expanding in the $x$ basis:

$$ \mathbf{\hat r}^k = \sum_j\frac{1}{h_k}\frac{\partial x_j}{\partial r_k}\mathbf{\hat x}^j $$

Now the gradient is just:

$$ \vec\nabla f = \mathbf{\vec F} = \sum_i F_i\mathbf{\hat x}^i = \sum_i \frac{\partial f}{\partial x_i}\mathbf{\hat x}^i $$

To get the $k$'th component in spherical coordinates ($F'_k$), use the dot product:

$$\begin{align} F'_k &= \mathbf{\vec F} \cdot \mathbf{\hat r}^k \\ &= \left(\sum_i \frac{\partial f}{\partial x_i}\mathbf{\hat x}^i\right) \cdot \left(\sum_j\frac{1}{h_k}\frac{\partial x_j}{\partial r_k}\mathbf{\hat x}^j\right) \\ &= \frac{1}{h_k}\sum_i \frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial r_k} \\ &= \frac{1}{h_k}\frac{\partial f}{\partial r_k} \end{align}$$

and we're done.

  • $\begingroup$ Thank you so much! It cleard a lot of confusion in my head, espacially the distinction between the jacobian and the change of basis matrix, which are similar in shape and idea and pretty confusing. Also, thank you for opening my eyes on the difference between coordinates and component. $\endgroup$ Commented Jul 31, 2020 at 2:15
  • $\begingroup$ By the way, would the last formula be true also for non orthogonal bases? $\endgroup$ Commented Jul 31, 2020 at 2:15
  • 1
    $\begingroup$ @JonasDaverio No, because dot products only give orthogonal projections. If you look at the coefficients $\frac{1}{h_k}\frac{\partial x_j}{\partial r_k}$, you could write that as a matrix. In the general case you'd have to use the inverse transpose of that. In the orthogonal case, the inverse transpose happens to be the own matrix. $\endgroup$
    – Wood
    Commented Aug 1, 2020 at 3:30
  • $\begingroup$ Oh yes, you're talking about the metric tensor, is that write? As it is diagonal for orthogonal bases, the inverse stays diagonal. $\endgroup$ Commented Aug 1, 2020 at 17:30
  • $\begingroup$ Very well answered. I don't know why this answer isn't on top $\endgroup$ Commented Oct 2, 2020 at 15:50

We take: $$x=r\sin\theta\cos\phi$$ $$y=r\sin\theta\sin\phi$$ $$z=r\cos\theta$$

Now, you know the definition of the gradient in Cartesian coordinates: $\vec{\nabla}=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$

Now, we use the chain rule or each component. For instance, $$\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}+\frac{\partial \phi}{\partial x}\frac{\partial}{\partial \phi}$$

After lots of cumbersome algebra, this will give you the correct form.

  • $\begingroup$ What part goes wrong? $\endgroup$
    – Danu
    Commented Sep 25, 2013 at 10:50
  • $\begingroup$ I get $\frac{\partial}{\partial x} = \frac{1}{sin(\theta)cos(\phi)} \frac{\partial}{\partial r} + \frac{1}{cos(\theta)cos(\phi)} \frac{\partial}{\partial \theta} - \frac{1}{sin(\theta)sin(\phi)} \frac{\partial}{\partial \phi}$! $\endgroup$ Commented Sep 25, 2013 at 11:05
  • $\begingroup$ And then how do the hat vectors get replaced? $\endgroup$ Commented Sep 25, 2013 at 11:06
  • $\begingroup$ you should express $r,\theta,\phi$ in terms of only $x,y,z$ $\endgroup$
    – Danu
    Commented Sep 25, 2013 at 11:07
  • $\begingroup$ How do you do that? I can't seem to separate them in the partials! $\endgroup$ Commented Sep 25, 2013 at 11:16

It follows from the general definition of the gradient as $$ \langle\nabla f(p)|v\rangle=d_pf(v)=\sum_i\left.\frac{\partial f}{\partial x^i}\right|_pdx^i(v) $$ where p is a point in space and v a vector in the tangent space. The summation is over the basis vectors of the tangent space. You can try to expand this expression to get the final result for the component $i$ $$ (\nabla f)_i=\frac{1}{h_i}\frac{\partial f}{\partial x^i} $$ This is the most useful formula. The quantity $h_i$ is the modulus of the $i$th tangent vector.

Example: you want to compute the gradient in spherical coordinates. The basis of the tangent space is $\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta},\frac{\partial}{\partial\phi}\}$. Since $$ \begin{split} \left\|\frac{\partial}{\partial \theta}\right\|^2&= \left\| \frac{\partial x}{\partial\theta}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial\theta}\frac{\partial}{\partial y}+ \frac{\partial z}{\partial\theta}\frac{\partial}{\partial z} \right\|^2\\ &= r^2\cos^2\theta\cos^2\phi\underbrace{\left\|\frac{\partial}{\partial x}\right\|^2}_{=1}+ r^2\cos^2\theta\sin^2\phi\underbrace{\left\|\frac{\partial}{\partial y}\right\|^2}_{=1}+ r^2\sin^2\theta\underbrace{\left\|\frac{\partial}{\partial z}\right\|^2}_{=1}\\ &=r^2 \end{split} $$ Thus we get $$h_\theta=\left\|\frac{\partial}{\partial \theta}\right\|=r$$ In the same spirit you can calculate that $$ h_r=1\quad\text{and}\quad h_\phi=r\sin\theta $$ giving us the gradient in spherical coordinates $$ \nabla f=\frac{\partial f}{\partial r}\hat e_r+\frac{1}{r}\frac{\partial f}{\partial\theta}\hat e_\theta+\frac{1}{r\sin\theta}\frac{\partial f}{\partial\phi}\hat e_\phi $$

Proof for the first step

Expand the vector $|\nabla f\rangle$ in terms of basis vectors $$ |\nabla f\rangle =\sum_i(\nabla f)_i|e_i\rangle =\sum_i(\nabla f)_i\frac{1}{h_i}|\frac{\partial}{\partial x^i}\rangle $$ This is basically where the factor $h_i$ comes from. Now take $v=|\frac{\partial}{\partial x^k}\rangle$ and insert it in the first expression given above. Note that by definition of a dual vector we get $dx^i(|\frac{\partial}{\partial x^k}\rangle)=\delta_k^i$. The left-hand side is $$ \begin{split} \langle f|\frac{\partial}{\partial x^k}\rangle &=\sum_i(\nabla f)_i\frac{1}{h_i}\langle\frac{\partial}{\partial x^i}|\frac{\partial}{\partial x^k}\rangle\\ &=\sum_i(\nabla f)_i\frac{1}{h_i}h_i^2\delta_{ik}\\ &=(\nabla f)_kh_k \end{split} $$ Whreas the right-hand side $$ \sum_i\left.\frac{\partial f}{\partial x^i}\right|_pdx^i\left(|\frac{\partial}{\partial x^k}\rangle\right) =\sum_i\left.\frac{\partial f}{\partial x^i}\right|_p\delta^i_k =\frac{\partial f}{\partial x^k} $$ By comparing both expressions you obtain the claim.

  • $\begingroup$ Great answer, much better than mine. However, I'm not sure the asker is comfortable with this level of abstraction (tangent space, for instance, might be an unknown concept). $\endgroup$
    – Danu
    Commented Sep 25, 2013 at 10:35
  • $\begingroup$ Good answer, I've read about this method before, but you've made it simpler and I understand nearly all of it. I actually worked through this when derving spherical unit vectors. Just one thing: how do you get from the general definition at the top to your second expression? Where does this $h$ come from? $\endgroup$ Commented Sep 25, 2013 at 10:50
  • $\begingroup$ And why does $h_{\theta} = ||\frac{\partial}{\partial \theta}||$? $\endgroup$ Commented Sep 25, 2013 at 10:53
  • $\begingroup$ I just added a proof for the first step. $h_\theta=\left\|\frac{\partial}{\partial\theta}\right\|$ by definition, it's simply the modulus of the tangent vector $\frac{\partial}{\partial\theta}$ $\endgroup$
    – Stan
    Commented Sep 25, 2013 at 11:39
  • $\begingroup$ @Stan I am following a GR course, and can't seem to grasp the following part of the solution: As far is I seem to understand so far, your second formula gets this factor of $\frac{1}{h}$, due to a factor of $\frac{1}{h^2}$ first, coming from the metric, and a factor of h after that, in order to switch from tangent vectors to an orthonormal basis. The modulus of the tangent vector will always be positive wont it? Then a different orthonormal basis could have been found where $\phi$ is such that the theta component of the divergence reads $\frac{1}{r \,|\sin\theta|}$ How can i verify that the so $\endgroup$
    – Fictional
    Commented Nov 25, 2013 at 1:08

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