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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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2 Answers 2

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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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  • $\begingroup$ For people curious why this is, the natural basis vector for the angular coordinate has length $\langle e_\theta, e_\theta\rangle = r^2$, which is the definition of the metric components. However physicists love to work with unit vectors and so almost always pull out the size of the basis explicitly $e_\theta = r \hat{\theta}$. This leaves the usual basis employed by physicists as neither the naturally covariant or contra-variant basis, and as such the components of vectors, both raised and lowered, will be off by an $r$ somewhere. $\endgroup$
    – Craig
    May 8, 2023 at 1:03
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If you were like me who thought for two minutes before understanding Ben's answer, here is a longer but maybe more detailed explanation, and hopefully, more straightforward to understand.

Let us first rephrase the answer in a more detailed fashion.

The unit bases in cylindrical coordinates, denoted by $(\hat{r},\hat{\phi},\hat{z})$ above, are the non-coordinate basis. This is discussed clearly in, for instance, the textbook of Schutz A First Course in General Relativity, by the end of Chapter 5. To be specific, these bases (as well as the components) do not transform as tensor (eg. vector or 1-form). By employing the same convention, let us denote the correct coordinate basis one-forms by ${\vec{e}}^1, {\vec{e}}^2, {\vec{e}}^3$. These bases can be obtained by the transformation rule of tensors. For instance, explicit calculations give $${\vec{e}}^2=\frac{\partial x^2}{\partial {x'}^\mu}{\vec{e}'}^\mu =\frac{\partial \phi}{\partial x}{\vec{e}}^x+\frac{\partial \phi}{\partial y}{\vec{e}}^y=-\frac1r\sin\phi {\vec{e}}^x+\frac1r\cos\phi {\vec{e}}^y=-\frac1r\sin\phi \hat{x}+\frac1r\cos\phi \hat{y}=\frac1r \hat{\phi}$$ for basis one-form, and similarly, one finds $${\vec{e}}_2=-r\sin\phi \hat{x}+r\cos\phi \hat{y}=r \hat{\phi}$$ for basis vector, where we have made use of the explicit form of coordinate transformation $x=r\cos\phi, y=r\sin\phi$ and the fact ${\vec{e}}^x={\vec{e}}_x=\hat{x}, {\vec{e}}^y={\vec{e}}_y=\hat{y}$, namely, the coordinate bases in Cartesian coordiante are unit vectors.

Now, a vector can be written down in both bases, coordinate one and non-coordinate one as $$\vec{A}=A^1 {\vec{e}}_1+A^2{\vec{e}}_2+A^3{\vec{e}}_3=A_r\hat{r}+A_\phi\hat{\phi}+A_z\hat{z}$$ Therefore we have $A^2=\frac{A_\phi}{r}$. We note that the question at hand is about looking for a relation between $F^{02}$ and $E_\phi$, the latter is defined as the component in the front of the relevant unit basis vector, namely, $\hat{\phi}$. Before proceed further, we emphasize that $A_\phi$ or $E_\phi$ is NOT a component of any real vector in the context of tensorial algebra.

By putting all the pieces together, in co-moving frame (see comments below) one finds $$E^{2}= F^{02} = \partial^0 A^2 - \partial^2 A^0 = -\partial_0 A^2 - \frac{1}{r^2}\partial_2 A^0 = -\frac1r\frac{\partial A_\phi}{\partial t}-\frac{1}{r^2}\frac{\partial V}{\partial \phi}=\frac{E_\phi}{r},$$ namely, everything is indeed consistent. As mentioned, the about part has been pointed out in Ben's answer.

However, when I explored a bit further regarding the original question proposed by Thiago, the following question was raised.

In terms of the scalar and vector potentials, the electric field and magnetic induction are defined as rank-two tensors. These quantities, for instance $F^{02}\equiv \partial^0A^2-\partial^2A^0$ are eventually compared against vectors, for instance, $E^2$. Even though these components are shown to be identical for a particular case, the equalities are not expected to hold in any coordinate system. Therefore, one might wonder if there is a more serious problem: why such a comparison is meaningful at all?

To solve the second issue, let us note that a more general definition (see this paper for instance) of electromagnetic fields in terms of $F^{\mu\nu}\equiv \partial^\mu A^\nu-\partial^\nu A^\mu$ are $$E^i=F^{ i \alpha}u_\alpha,$$ $$B^i= {^*F}^{i\beta}u_\beta$$ where ${^*F}^{\alpha\beta}\equiv \frac12\varepsilon^{\alpha\beta\gamma\delta}F_{\beta\delta}$ and $\varepsilon^{\alpha\beta\gamma\delta}$ is the Levi-Civita tensor. The mathematical aspect is obvious, one has to contract a rank-two tensor in order to obtain a vector. The physical interpretation of this choice is the following. It implies that one has to choose an observer whose world line is described by the four-velocity, the definition of the electromagnetic fields has to be implemented from his point of view. Of course, the four-velocity $u_\alpha$ falls back to $(1,0,0,0)$ in the co-moving frame of the observer and therefore from his viewpoint, the traditional definition is restored.

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