Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 {$t,\rho,\varphi,z$}.

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

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
2  
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

Your Answer

 
discard

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.