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I want to calculate the electromagnetic tensor components in cylindrical coordinates. Suppose I did not know that those components are given in Cartesian coordinates by $$(F^{\mu \nu})= \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0 \end{pmatrix}.$$

I want to derive the result in the same manner I did in the Cartesian coordinates case, i.e., using that $F^{ \mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$, where $A^\alpha=(V,\vec{A})$, $\vec{B} = \nabla \times \vec{A}$ and $\vec{E} = -\nabla V - \partial \vec{A} / \partial t$. Using the formulas for curl and gradient in cylindrical coordinates, we find $$ \vec{E} = - \left( \frac{\partial V}{\partial r} + \frac{\partial A_r}{\partial t} \right)\hat{r} \ - \left( \frac{1}{r}\frac{\partial V}{\partial \phi} + \frac{\partial A_\phi}{\partial t} \right)\hat{\phi} - \left( \frac{\partial V}{\partial z} + \frac{\partial A_z}{\partial t} \right)\hat{z} $$ and $$ \vec{B} = \left( \frac{1}{r}\frac{\partial A_z}{\partial \phi} - \frac{\partial A_\phi}{\partial z} \right)\hat{r} \ +\left(\frac{\partial A_r}{\partial z} - \frac{\partial A_z}{\partial r} \right)\hat{\phi} \ +\frac{1}{r}\left(\frac{\partial (r A_\phi)}{\partial r} - \frac{\partial A_z}{\partial r} \right)\hat{z}. \ $$ The invariant interval is given by $ds^2 = -dt^2 + dr^2 + r^2 d\phi^2 + dz^2$, (with $c=1$). Therefore, the metric tensor reads $$(g_{\mu \nu})= \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & r^2 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix},$$ and its inverse is $$(g^{\mu \nu})= \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1/r^2 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}.$$

Which implies $\partial^0 = -\partial_0$, $\partial^1 = \partial_1$, $\partial^2 = \frac{1}{r^2}\partial_2$ and $\partial^3 = \partial_3$.

So, for example, $$ F^{ 01} = \partial^0 A^1 - \partial^1 A^0 = -\partial_0 A^1 - \partial_1 A^0 = -\frac{\partial A_r}{\partial t}-\frac{\partial V}{\partial r} = E_r, $$ which is reassuring. Now, $$ F^{02} = \partial^0 A^2 - \partial^2 A^0 = -\partial_0 A^2 - \frac{1}{r^2}\partial_2 A^0 = -\frac{\partial A_\phi}{\partial t}-\frac{1}{r^2}\frac{\partial V}{\partial \phi}. $$ However, I cannot identify this quantity with any component of the electric field. This last expression looks almost like $E_\phi$, except for an extra $\frac{1}{r}$ multiplying $\partial V / \partial \phi$. What went wrong here?

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The problem is that there is a mismatch between the vector basis that you are using to write the 4-vector potential and the ones you are considering for your metric, which are not unitary. The standard cylindrical coordinate basis should have a Minkwoski metric, since we don't really have curvature in this case. The only difference is then a $1/r$ factor in the $\phi$ component.

Therefore, in order to be consistent, you need to replace $A^2 \rightarrow \frac{A_{\phi}}{r}$. Then, you will see that $F^{02}=\frac{E_{\phi}}{r}$, which, again, is consistent with your metric.

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