# Electromagnetic Tensor in Cylindrical Coordinates

I understand that the Electromagnetic Tensor is given by

$$F^{\mu\nu}\mapsto\begin{pmatrix}0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & -B_{z} & B_{y}\\ E_{y} & B_{z} & 0 & -B_{x}\\ E_{z} & -B_{y} & B_{x} & 0 \end{pmatrix}$$

where $\mu$, $\nu$ can take the values {0,1,2,3} or {$t,x,y,z$}.

So, for example $$F^{01}=F^{tx}=-E_x$$

My question is, what would the following expression be?

$$F^{t\rho}=?$$ or $$F^{z\rho}=?$$

where $\rho=\sqrt{x^{2}+y^{2}}$ is the radial coordinate in cylindrical coordinates?

And more generally, how can we construct the Electromagnetic Tensor in cylindrical coordinates? Where $\mu$, $\nu$ now take the values {$t,\rho,\varphi,z$}.

-
For more on E&M in curved space, see also Wikipedia. – Qmechanic Jun 4 '13 at 17:30

Just use the Jacobian of the coordinate system transformation. If your Cartesian coordinates are $\mu$ and $\nu$ and your cylindrical coordinates are $\mu', \nu'$, then there is a Jacobian ${f_\mu}^{\mu'}$ that allows you to write

$$F^{\mu' \nu'} = F^{\mu \nu} {f_\mu}^{\mu'} {f_\nu}^{\nu'}$$

where the Jacobian is given by

$${f_\mu}^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^\mu}$$

Now that's all well and good, but you might be thinking it's a bit abstract, and...it is. There's another way to do this instead, using what's called geometric algebra.

In geometric algebra, the EM tensor is called a bivector, taking on the form

$$F = F_{tx} e^t \wedge e^x + F_{ty} e^t \wedge e^y + \ldots = \frac{1}{2} F_{\mu \nu} e^\mu \wedge e^\nu$$

where $e^\mu$ represent basis covectors. What we've used here is called a wedge product, and orthogonal basis vectors will anticommute under it.

To extract the components in a new basis, you have a couple choices: (1) you can write the basis covectors in terms of the cylindrical basis and simplify. So that would entail writing $e^x$ and $e^y$ in terms of $e^\rho$ and $e^\phi$. This is equivalent to finding the inverse Jacobian.

However, there is another choice (2), which is to simply take the inner product of the basis vectors $e_\rho \wedge e_t, e_\phi \wedge e_t$ and so on with $F$. This requires a little more knowledge of geometric algebra, but you can write $e_\rho \wedge e_t$ in terms of $e_x \wedge e_t, e_y \wedge e_t$, and so on, which may be an easier computation.

I'll do the latter here to demonstrate the technique. See that $e^\rho = e^x \cos \phi + e^y \sin \phi$. We can then find $F^{t \rho}$ as:

\begin{align*}F^{t\rho} &= F \cdot (e^\rho \wedge e^t) \\ &= F \cdot (e^x \wedge e^t \cos \phi + e^y \wedge e^t \sin \phi) \\ &= F^{tx} \cos \phi + F^{ty} \sin \phi \end{align*}

This is no more exotic that finding the components of a vector in a new basis by finding the projection of the vector on each new basis vector.

-
Practically, the Jacobian Matrix of the transformation is ${f_\mu}^{\mu'} =\left[ \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & cos\left( \theta\right) & sin\left(\theta\right) & 0\\ 0 & -\frac{sin\left( \theta\right) }{R} & \frac{cos\left(\theta\right) }{R} & 0\\ 0 & 0 & 0 & 1 \end{array} \right]$ – Trimok Jun 4 '13 at 18:32