I've become confused about spherical coordinates when dealing with electric fields.

The way I always understood spherical coordinates is something like the below picture. To define a vector, you give it a distance outwards (r), and two angles to get a final position. Below, the $\theta$ and $\phi$ components are measured in radians.

Spherical coordinates

(Courtesy Wikipedia.org)

However, you can also have, say, an electric field in spherical coordinates. In this case, the unit vectors $\theta$ and $\phi$ don't define angles but rather values of the vector fields. So, in the case of electric fields, we might have $E_\theta = 10\text{ Vm}$. That is, at every point there will be this electric field component in the theta direction.

So, it seems there are two different ways of dealing with spherical coordinates. One, where the $\theta$ and $\phi$ components represent angles, and one where they represent values of the components in those directions.

This would then give you two different measures of lengths of the vectors. In the first case, the length of the vector is always given by the r component. In the second case, you take $|\vec{E}|=\sqrt{E_r^2+E_\theta^2+E_\phi^2}$.

What am I mixing up here?


2 Answers 2


You can't take the root mean square of the spherical coordinate parameters because they aren't all the same units (one is a length measurement while the other two are angle values). Well you can, but the output is meaningless.

To convert spherical parameters to Cartesian coordinates, you use simple trig:

\begin{align}E_z & = r \cos(\theta) \\ E_x & = r \sin(\theta)\cos(\phi) \\ E_y &= r \sin(\theta)\sin(\phi) \end{align}

Now if you take the root mean square of $E_x$, $E_y$ and $E_z$ you will get the correct $r$.

  • $\begingroup$ can you change the $sin$ to $\sin$ etc using backslash please. $\endgroup$ May 28, 2015 at 6:55
  • $\begingroup$ Note that defining a local orthogonal coordinate system at $(r,\theta,\phi)$ as $\vec e_r$,$\vec e_\theta$,$\vec e_\phi$, one might find cases where the projection of $\vec E$ on ,e.g. $\vec e_\theta$ is called $E_\theta$. This are Cartesian coordinates then and root would make sense. $\endgroup$ May 28, 2015 at 6:59
  • $\begingroup$ @mikuszefski you should edit the answer yourself if you would like to see notation corrected. $\endgroup$
    – Danu
    May 28, 2015 at 10:45
  • $\begingroup$ @Danu Well, in principle that is a good idea. An edit, however, requires a certain amount of changed characters (you probably know that) and five backslashes are not sufficient. As I don't want to mess in an otherwise correct answer only to get the right amount of characters, I asked the poster to do it. $\endgroup$ May 28, 2015 at 13:01
  • $\begingroup$ Hi @mikuszefski, I think you've just described what's been confusing me. In the problem I was working on, E had three components in spherical coordinates, and each was the strength of the field in each of those orthogonal directions, not an angle from an axis. So, in that case, the coordinates are defined differently? Could you clarify that a bit? $\endgroup$
    – Rupert
    May 28, 2015 at 23:39

For an electric field usually you have three components like $(E_x, E_y, E_z)$ in Cartesian coordinate system. Now you want to rewrite the same vector in a spherical coordination, what you should do is as follows: first you write the vector like the electric field as $\mathbf{E}=|E|\mathbf{e_r}$ where $|E|$ is given by the $|E|=\sqrt{E_x^2+E_y^2+E_z^2}$ and $\mathbf{e_r}$ is a function of $\theta$ and $\phi$. Then One should have $\theta=\text{acos}(E_z/|E|)$ and $\phi=\text{atan}(E_y/E_x)$. If you like to write in a three component form, it is always (|E|, 0, 0).

  • $\begingroup$ But what if you have a field which has no radial direction? For example, picture the magnetic field around an infinitely long wire. I could write that as $\textbf{B} = |B|\textbf{e}_\theta$ in cylindrical coordinates, couldn't I? $\endgroup$
    – Rupert
    May 28, 2015 at 23:42
  • $\begingroup$ @Rupert yes, of course you can have the magnetic field in the $\theta$ direction. Actually this is the reason we are using a spherical coordination. For an electric field, it always has an origin (for example an electron) which means its force direction is along the radial direction $e_r$ always, and then for a magnetic field associated the electric field, based on the Maxwell equation $i\omega\mathbf{B}=\nabla \times \mathbf{E}$ you know that the magnetic field is along the $\theta$ direction. $\endgroup$
    – Liu Feng
    May 29, 2015 at 1:16

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