# Electromagnetic tensor in cylindrical coordinates

• ngkamsengpeter
In summary, the electromagnetic field tensor in cylindrical coordinates can be found using the transformation matrix provided. However, transforming the tensor directly using the tensor transformation rule does not yield the correct result. The tensor must be written in terms of ER, Eθ, and Ez instead of Ex, Ey, and Ez. Finding the field tensor directly in cylindrical coordinates may require a different approach.
ngkamsengpeter
I can find the metric tensor in cylindrical coordinates to be [1,-1,-1/r^2,-1] but how about the electromagnetic field tensor and thus the energy stress tensor?
Is it just change the Ex,Ey,Ez to Eρ,Eθ,Ez?
Is FσρFσρ still equal to 2(B^2-E^2)

The field tensor will give the same contraction whatever the coordinates. This is what defines tensors. Changing the coordinate basis from xa -> xA
$$F^{AB}F_{AB} = \Lambda^A_a\Lambda^B_bF^{ab}\Lambda^a_A\Lambda^b_B F_{ab}=F^{ab}F_{ab},\ \ \Lambda^a_A=dx^a/dx^A$$The flat space cylindrical metric is ds2= dt2 - dz2 - dr2 - r22, I think.

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Mentz114 said:
The field tensor will give the same contraction whatever the coordinates. This is what defines tensors. Changing the coordinate basis from xa -> xA
$$F^{AB}F_{AB} = \Lambda^A_a\Lambda^B_bF^{ab}\Lambda^a_A\Lambda^b_B F_{ab}=F^{ab}F_{ab},\ \ \Lambda^a_A=dx^a/dx^A$$The flat space cylindrical metric is ds2= dt2 - dz2 - dr2 - r22, I think.

Yes. I guess it should be the same in cylindcal coordinates but I try to work it by using

$$n_{\mu \nu }= \left[ \begin{array}{cccc}c^2& 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right]$$
$$n^{\mu \nu }= \left[ \begin{array}{cccc} \frac{1}{c^2} & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{-1}{r^2} & 0 \\ 0 & 0 & 0 & -1 \end{array} \right]$$

$$F_{\mu \nu }= \left[ \begin{array}{cccc} 0 & -E_{\rho} & -E_{\phi} & -E_{z} \\ E_{\rho} & 0 & B_{z} &-B_{\phi} \\E_{\phi} & -B_{z} & 0 & B_{\rho} \\E_{z} & B_{\phi} & -B_{\rho} & 0 \end{array} \right]$$

and I get
$$F^{\mu \nu }= \left[ \begin{array}{cccc} 0 & \frac{E_{\rho}}{c^2} & \frac{E_{\phi}}{c^2 r^2} & \frac{E_{z}}{c^2} \\ \frac{-E_{\rho}}{c^2} & 0 & \frac{B_{z}}{r^2} &-B_{\phi} \\\frac{-E_{\phi}}{c^2 r^2} & \frac{-B_{z}}{r^2} & 0 & \frac{B_{\rho}}{r^2} \\\frac{-E_{z}}{c^2} & B_{\phi} & \frac{-B_{\rho}}{r^2} & 0 \end{array} \right]$$

And thus
$$F^{\mu \nu }F_{\mu \nu }≠2(B^2-\frac{E^2}{c^2})$$

But since it is tensor, so I guess it should be equal to 2(B^2-E^2/c^2) also in cylindrical or any coordinates right? Can anyone tell me which part I didt it incorrectly?

Thanks

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Use this to calculate F
$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$

Mentz114 said:
Use this to calculate F
$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$

I am new to this, can you show me some steps and the final answer so that I can check with my one?

Thanks.

ngkamsengpeter said:
I am new to this, can you show me some steps and the final answer so that I can check with my one?

Thanks.

When you transform from rectangular coordinates to cylindrical you must transform the differential operators also. This will remove the factor of r2 that is giving you the wrong answer.

See here for the transformed operators
http://people.rit.edu/pnveme/pigf/Coords/Diff_operators_cyl.html

http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

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The transformation (t,x,y,z) -> (T,R,θ,Z)
$$t \rightarrow T,\ \ x \rightarrow R \cos(\theta),\ \ y \rightarrow R \sin(\theta),\ \ z \rightarrow Z$$
is done by this matrix
$$\left[ \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & cos\left( \theta\right) & sin\left(\theta\right) & 0\\ 0 & -\frac{sin\left( \theta\right) }{R} & \frac{cos\left(\theta\right) }{R} & 0\\ 0 & 0 & 0 & 1 \end{array} \right]$$

The transformed field tensor is
$$F^{mn}=\left[ \begin{array}{cccc} 0 & sin\left( \theta\right) \,Ey+cos\left( \theta\right) \,Ex & \frac{cos\left( \theta\right) \,Ey-sin\left( \theta\right) \,Ex}{R} & Ez\\ -sin\left( \theta\right) \,Ey-cos\left( \theta\right) \,Ex & 0 & -\frac{Bz}{R} & cos\left( \theta\right) \,By+sin\left( \theta\right) \,Bx\\ -\frac{cos\left( \theta\right) \,Ey-sin\left( \theta\right) \,Ex}{R} & \frac{Bz}{R} & 0 & -\frac{sin\left( \theta\right) \,By-cos\left( \theta\right) \,Bx}{R}\\ -Ez & -cos\left( \theta\right) \,By-sin\left( \theta\right) \,Bx & \frac{sin\left( \theta\right) \,By-cos\left( \theta\right) \,Bx}{R} & 0 \end{array} \right]$$

and the contraction $g_{an}g_{bm}F^{ab}F^{mn} = -2 E^2 + 2B^2$, where
$$g_{mn}=\left[ \begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & {R}^{2} & 0\\ 0 & 0 & 0 & 1 \end{array} \right]$$

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Mentz114 said:
The transformation (t,x,y,z) -> (T,R,θ,Z)
$$t \rightarrow T,\ \ x \rightarrow R \cos(\theta),\ \ y \rightarrow R \sin(\theta),\ \ z \rightarrow Z...$$
Let me see if I understand it correctly. the tensor transformation rule is
$\widetilde{A^\mu}=\frac{∂\widetilde{x^\mu}}{∂x^\nu}A^\nu$
So the transformation matrix you given is $$\frac{∂\widetilde{x^\mu}}{∂x^\nu}$$right?
However, I try to work it out the matrix, I have the one different from your one. for example the 1,1 component, is $$\frac{∂\widetilde{x^1}}{∂x^1} =\frac{∂R}{∂x}=\frac{1}{Cos(\theta)}$$

And I don't quite understand how the Field tensor transform. I would think of it transform with the same matrix with the following formula:
$$\widetilde{F^{\mu\nu}}=\frac{∂\widetilde{x^\mu}}{∂x^ρ}\frac{∂\widetilde{x^\nu}}{∂x^σ}=F^{ρσ}$$

However, when I do the matrix multiplication using the matrix you given, I cannot get the answer same as your one. Is it transform with another different matrix?

If I want to write the tensor in terms of ER,Eθ,Ez instead of Ex,Ey,Ez, how should I do that?

ngkamsengpeter said:
If I want to write the tensor in terms of ER,Eθ,Ez instead of Ex,Ey,Ez, how should I do that?

(I haven't helped much so far ).
What you've written is correct, but the matrix I've given transforms the inverse of the cylindrical metric to the inverse of the Minkowski metric, so it is not used correctly in my calculation. Anyhow, it wouldn't answer your question.

To find F directly in cylindrical coordinates is not as simple as I thought. Starting the 4-potential Aμ the field tensor is $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$.

If
$$\frac{\partial}{\partial {x^2}}=\frac{1}{R}\frac{\partial}{\partial \theta}$$
and $g^{\mu\nu}=diag(-1,1,1/R^2,1)$ then $g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}F_\mu\nu$ has factors of R and is not what is wanted. Maybe the potential A needs to be written differently.

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I think I've solved this. By replacing Aθ with Aθ/R and using the spatial curl operator defined below for the magnetic fields it works out correctly. The curl is the spatial ( and temporal ) parts of $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$.

The curl operator in cylindrical coords is equation (117) on this page
http://mathworld.wolfram.com/CylindricalCoordinates.html

So far I've done the calculation on a piece of paper and it gives the right contraction but I'll set up a script when I can to check it. As I said earlier, the differential operators must be correct for the coordinates used.

Using equation (117) the magnetic fields are
\begin{align} B_R &= \frac{1}{R}\frac{\partial A_z}{\partial \theta}-\frac{1}{R}\frac{\partial A_\theta}{ \partial Z}\\ B_\theta &= \frac{\partial A_R}{\partial z}-\frac{\partial A_z}{ \partial R}\\ B_Z &= \frac{1}{R} \left[ \frac{\partial A_\theta}{\partial r}-\frac{\partial A_Z}{ \partial \theta} \right] \end{align}
and the electric fields are
\begin{align} E_\theta &= \frac{1}{R} \left[\frac{\partial A_\theta}{\partial t}-\frac{\partial A_t}{ \partial \theta} \right]\\ E_R &=\frac{\partial A_R}{\partial t}-\frac{\partial A_t}{ \partial R}\\ E_Z &=\frac{\partial A_Z}{\partial t}-\frac{\partial A_t}{ \partial Z} \end{align}
now we can write
$$F^{\mu\nu}= \left[ \begin{array}{cccc} 0 & \frac{\partial}{\partial\,t}\,A_R-\frac{\partial}{\partial\,R}\,A_t & \frac{\frac{\partial}{\partial\,t}\,A_\theta-\frac{\partial}{\partial\,\theta}\,A_t}{R} & \frac{\partial}{\partial\,t}\,A_z-\frac{\partial}{\partial\,z}\,A_t\\ \frac{\partial}{\partial\,R}\,A_t-\frac{\partial}{\partial\,t}\,A_R & 0 & \frac{\frac{\partial}{\partial\,R}\,A_\theta-\frac{\partial}{\partial\,\theta}\,A_R}{R} & \frac{\partial}{\partial\,R}\,A_z-\frac{\partial}{\partial\,z}\,A_R\\ \frac{\frac{\partial}{\partial\,\theta}\,A_t-\frac{\partial}{\partial\,t}\,A_\theta}{R} & \frac{\frac{\partial}{\partial\,\theta}\,A_R-\frac{\partial}{\partial\,R}\,A_\theta}{R} & 0 & \frac{\frac{\partial}{\partial\,\theta}\,A_z-\frac{\partial}{\partial\,z}\,A_\theta}{R}\\ \frac{\partial}{\partial\,z}\,A_t-\frac{\partial}{\partial\,t}\,A_z & \frac{\partial}{\partial\,z}\,A_R-\frac{\partial}{\partial\,R}\,A_z & \frac{\frac{\partial}{\partial\,z}\,A_\theta-\frac{\partial}{\partial\,\theta}\,A_z}{R} & 0 \end{array} \right]$$
and (finally ?) $g_{ma}g_{bn}F^{ab}F^{mn}= 2(-E^2+B^2)$ where $g_{mn}=diag(-1,1,R^2,1)$.

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Mentz114 said:
I think I've solved this. By replacing Aθ with Aθ/R and using the spatial curl operator defined below for the magnetic fields it works out correctly. The curl is the spatial ( and temporal ) parts of $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$.

The curl operator in cylindrical coords is equation (117) on this page
http://mathworld.wolfram.com/CylindricalCoordinates.html
First of all, why replace Aθ with Aθ/R. We must have a reason for that, cannot just replace it in order to get the answer.
Assume all your calculation correctly, then the final answer is just replaced the Ex,Ey,Ez by Er,Eθ,Ez only right? That is your final answer should be
$$F^{\mu \nu }= \left[ \begin{array}{cccc} 0 & E_{\rho} & E_{\phi} & E_{z} \\ -E_{\rho} & 0 & -B_{z} &B_{\phi} \\-E_{\phi} & B_{z} & 0 & -B_{\rho} \\-E_{z} & -B_{\phi} & B_{\rho} & 0 \end{array} \right]$$
Isn't that same as the first attempt I made?

And I am sure that it does not contract correctly. What is your final answer of F that give the correct contraction as you mentioned?

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ngkamsengpeter said:
First of all, why replace Aθ with Aθ/R. We must have a reason for that, cannot just replace it in order to get the answer.
We have to raise the index of Aμ with the inverse of the cylindrical metric because the curl operators I'm using are contravariant. That's why the curls are components of a contravariant rank-2 tensor.

Assume all your calculation correctly, then the final answer is just replaced the Ex,Ey,Ez by Er,Eθ,Ez only right? That is your final answer should be
$$F^{\mu \nu }= \left[ \begin{array}{cccc} 0 & E_{\rho} & E_{\phi} & E_{z} \\ -E_{\rho} & 0 & -B_{z} &B_{\phi} \\-E_{\phi} & B_{z} & 0 & -B_{\rho} \\-E_{z} & -B_{\phi} & B_{\rho} & 0 \end{array} \right]$$
Isn't that same as the first attempt I made?

And I am sure that it does not contract correctly. What is your final answer of F that give the correct contraction as you mentioned?

My calculation is spelt out. F and g are given and the results are correct because there are no coordinates in the contraction. It just depends on how you define Ex, Ey etc.

I've had enough and I'm satisfied, it was interesting.

Mentz114 said:
We have to raise the index of Aμ with the inverse of the cylindrical metric because the curl operators I'm using are contravariant. That's why the curls are components of a contravariant rank-2 tensor.

My calculation is spelt out. F and g are given and the results are correct because there are no coordinates in the contraction. It just depends on how you define Ex, Ey etc.

I've had enough and I'm satisfied, it was interesting.

Yes. It was interesting. I can understand the Aθ/r now but I still don't quite understand with your calculation. What is the final answer of F in terms of Er,Eθ,Ez?

ngkamsengpeter said:
Yes. It was interesting. I can understand the Aθ/r now but I still don't quite understand with your calculation. What is the final answer of F in terms of Er,Eθ,Ez?

I think the contraction is
$$-2 ( -R^{-2}{E_\theta}^2-{E_r}^2-{E_z}^2+{B_\theta}^2 +R^{-2}{B_r}^2+R^{-2}{B_z}^2)=-2( -\vec{E}^2+\vec{B}^2) )$$

$$-2 ( -{E_\theta}^2-{E_r}^2-{E_z}^2+{B_\theta}^2 +{B_r}^2+{B_z}^2)=-2( -\vec{E}^2+\vec{B}^2) )$$

depending how you define the E and B vectors.

Mentz114 said:
I think the contraction is
$$-2 ( -R^{-2}{E_\theta}^2-{E_r}^2-{E_z}^2+{B_\theta}^2 +R^{-2}{B_r}^2+R^{-2}{B_z}^2)=-2( -\vec{E}^2+\vec{B}^2) )$$

$$-2 ( -{E_\theta}^2-{E_r}^2-{E_z}^2+{B_\theta}^2 +{B_r}^2+{B_z}^2)=-2( -\vec{E}^2+\vec{B}^2) )$$

depending how you define the E and B vectors.
Yes. I get this answer also but how can we simply define vectors? In cylindrical coordinate, vector should be defined as
$$\vec{E}=E_r \hat{r}+ E_\theta \hat{\theta}+E_z \hat{z}$$
Is that mathematically correct if we simply defined the vector just to fits into the answer?

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Can anyone verified if what Mentz114 did is correct or not? Can we defined the E and B vector like that?
Thanks.

## 1. What is the electromagnetic tensor in cylindrical coordinates?

The electromagnetic tensor in cylindrical coordinates is a mathematical representation of the electromagnetic field in a cylindrical coordinate system. It describes the electric and magnetic field components at any point in space and time.

## 2. How is the electromagnetic tensor expressed in cylindrical coordinates?

The electromagnetic tensor in cylindrical coordinates is typically expressed as a 4x4 matrix with 16 components. The first three components correspond to the electric field in the radial, azimuthal, and axial directions, while the last three components correspond to the magnetic field in the same directions.

## 3. What are the advantages of using cylindrical coordinates to represent the electromagnetic tensor?

Cylindrical coordinates are useful for describing systems with cylindrical symmetry, such as a cylindrical wire or a rotating cylinder. They also simplify the equations for calculating the electromagnetic field in these systems, making them easier to solve.

## 4. How is the electromagnetic tensor related to Maxwell's equations?

The electromagnetic tensor is closely related to Maxwell's equations, which describe the behavior of electric and magnetic fields. In fact, the components of the electromagnetic tensor can be used to calculate the electric and magnetic fields at any point in space and time.

## 5. Can the electromagnetic tensor be used to describe all types of electromagnetic fields?

Yes, the electromagnetic tensor can be used to describe all types of electromagnetic fields, including static fields, time-varying fields, and even non-electromagnetic fields such as gravitational and quantum fields. It is a powerful tool for understanding the behavior of the electromagnetic field in various systems.

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