Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am teaching myself about vector fields and came across the following question:

Is the following force field $\vec{F}$ conservative, where $\vec{F}(r,\theta,\varphi)$ is defined by: $$F_{r}=2ar\sin(\theta)\sin(\varphi),\: F_{\theta}=ar\cos(\theta)\sin(\varphi),\: F_{\varphi}=ar\cos(\varphi)$$

A simple test to determine whether a force field is conservative is to see if the following is true: $$\nabla\times \vec{F}=\vec{0}$$

Where by abuse of notation we have: $$\nabla=\frac{\partial \hat{\boldsymbol{\imath}}}{\partial x}+\frac{\partial \hat{\boldsymbol{\jmath}}}{\partial y}+\frac{\partial \hat{\boldsymbol{k}}}{\partial z}$$

However, as we are using a curvilinear co-ordinate basis I'm not sure how $\nabla$ should be defined?

Further to JohnD's answer, I have tried to derive his expression for $\nabla$, however, I have not managed to come to the right answer.

Taking partial derivatives of $r$ with respect to $x,y$ and $z$ we get:

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

Our partial derivatives of $\theta$ with respect to our cartesian co-ordinates are:

\begin{align}\frac{\partial \theta}{\partial x}&= \frac{z\frac{\partial r}{\partial x}}{\sqrt{r^{2}-z^{2}}}=\frac{r\cos(\theta)\sin(\theta)\cos(\varphi)}{r\sqrt{1-\cos^{2}(\theta)}}=\cos(\theta)\cos(\varphi) \\ \frac{\partial \theta}{\partial y}&=\frac{z\frac{\partial r}{\partial y}}{\sqrt{r^{2}-z^{2}}}=\frac{r\cos(\theta)\sin(\theta)\sin(\varphi)}{r\sqrt{1-\cos^{2}(\theta)}}=\cos(\theta)\sin(\varphi) \\ \frac{\partial \theta}{\partial z} &= \frac{z\frac{\partial r}{\partial z}-r}{r\sqrt{r^{2}-z^{2}}} = \frac{r\cos^{2}(\theta)-r}{r\sqrt{r^{2}-r^{2}\cos^{2}(\theta)}}=-\frac{\sin(\theta)}{r}\end{align}

And finally partial derivatives of $\varphi$:

\begin{align}\frac{\partial \varphi}{\partial x}&=-\frac{y}{x^{2}(1+\frac{y^{2}}{x^{2}})}=-\frac{r\sin(\theta)\sin(\varphi)}{r^{2}\sin^{2}(\theta)\cos^{2}(\theta)(1+\tan^{2}(\varphi))}=-\frac{\sin(\varphi)}{r} \\ \frac{\partial \varphi}{\partial y}&=\frac{1}{x(1+\frac{y^2}{x^{2}})}=\frac{1}{r\sin(\theta)\cos(\varphi)(1+\tan^{2}(\varphi))}=\frac{\cos(\varphi)}{r\sin(\theta)} \\ \frac{\partial \varphi}{\partial z}&=0\end{align}

We therefore get:

\begin{align}\frac{\partial}{\partial x}&\mapsto \sin(\theta)\cos(\varphi)\frac{\partial}{\partial r}+\cos(\theta)\cos(\varphi)\frac{\partial}{\partial \theta} - \frac{\sin(\varphi)}{r}\frac{\partial}{\partial \varphi} \\ \frac{\partial}{\partial y} &\mapsto \sin(\theta)\sin(\varphi)\frac{\partial}{\partial r} + \cos(\theta)\sin(\varphi)\frac{\partial}{\partial \theta} + \frac{\cos(\varphi)}{r\sin(\theta)}\frac{\partial}{\partial \varphi} \\ \frac{\partial}{\partial z} &\mapsto \cos(\theta)\frac{\partial}{\partial r} - \frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\end{align}

However summing coefficients of $\frac{\partial}{\partial r}$, $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \varphi}$ doesn't give what is expected, so what have I done wrong?

share|cite|improve this question
See Del symbol in curlinear coordinate – Shuchang Dec 25 '13 at 14:22
up vote 1 down vote accepted

You've written $\nabla$ a bit confusingly. It may be clearer to write it (in Cartesian) as

$$\nabla = \hat i \frac{\partial}{\partial x} + \hat j \frac{\partial}{\partial y} + \hat k \frac{\partial}{\partial z}$$

This makes clear that the partial derivatives aren't of the basis vectors.

In general, $\nabla$ uses a basis of dual (or "cotangent") vectors. For a Cartesian basis, the dual basis vectors are identical to the ordinary basis vectors, so this property is somewhat less apparent.

The natural basis vectors associated with a spherical coordinate system are

$$\begin{align*} e_r &\equiv \frac{\partial}{\partial r} \vec r = \hat r \\ e_\theta &\equiv \frac{\partial}{\partial \theta} \vec r = r \hat \theta \\ e_\varphi &\equiv \frac{\partial}{\partial \varphi} \vec r = r \hat \varphi \sin \theta \end{align*}$$

The dual basis vectors--$e^r, e^\theta, e^\varphi$--can be directly computed using a triple product formula:

$$e^r = \frac{e_\theta \times e_\varphi}{e_r \cdot (e_\theta \times e_\varphi)}$$

...and similarly for $e^\theta, e^\varphi$ by permutation of variables, but since the coordinate system is orthogonal, all you really have to do is divide by the squared magnitude of each basis vector. Hence,

$$\begin{align*} e^r &= e_r \\ e^\theta &= \frac{e_\theta}{r^2} = \frac{\hat \theta}{r} \\ e^\varphi &= \frac{e_\varphi}{r^2 \sin^2 \theta} = \frac{\hat \varphi}{r \sin \theta} \end{align*}$$

Once the dual basis vectors are known, the expression for $\nabla$ follows:

$$\begin{align*} \nabla &= e^r \frac{\partial}{\partial r} + e^\theta \frac{\partial}{\partial \theta} + e^\varphi \frac{\partial}{\partial \varphi} \\ &= \hat r \frac{\partial}{\partial r} + \frac{1}{r} \hat \theta \frac{\partial}{\partial \theta} + \frac{1}{r \sin \theta} \hat \varphi \frac{\partial}{\partial \varphi}\end{align*}$$

share|cite|improve this answer

In spherical coordinates, $x=r\sin\theta\cos\varphi$, $y=r\sin\theta\sin\varphi$, $z=r\cos\theta$. Use this change of variables in conjunction with the multivariable chain rule to express ${\partial \over \partial x}$, ${\partial \over \partial y}$, ${\partial \over \partial z}$ in terms of $r,\theta,\varphi$ to obtain $$ \nabla_\text{spherical}=\left\langle {\partial\over \partial r},{1\over r}{\partial\over \partial\theta},{1\over r\sin\theta}{\partial\over \partial \varphi}\right\rangle. $$

The chart linked in the comment above is a very helpful reference.

share|cite|improve this answer
Thank you for your helpful answer, I tried to perform a change of variable (as per the update in my question) however I didn't get the right answer; would you mind having a look through it? Thanks! – Shaktal Dec 25 '13 at 17:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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