Saturday, 22 October 2016

homework and exercises - 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 {$t,\rho,\varphi,z$}.



Answer




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.


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