# Electromagnetic tensor in cylindrical coordinates

1. Mar 12, 2012

### 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)

2. Mar 12, 2012

### Mentz114

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.

Last edited: Mar 12, 2012
3. Mar 12, 2012

### ngkamsengpeter

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

Last edited: Mar 12, 2012
4. Mar 13, 2012

### Mentz114

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

5. Mar 14, 2012

### ngkamsengpeter

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

Thanks.

6. Mar 14, 2012

### Mentz114

Last edited: Mar 14, 2012
7. Mar 14, 2012

### Mentz114

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
$$g_{mn}=\left[ \begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & {R}^{2} & 0\\ 0 & 0 & 0 & 1 \end{array} \right]$$

Last edited: Mar 14, 2012
8. Mar 15, 2012

### ngkamsengpeter

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 dont 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?

9. Mar 15, 2012

### Mentz114

(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.

Last edited: Mar 15, 2012
10. Mar 15, 2012

### Mentz114

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 $g_{mn}=diag(-1,1,R^2,1)$.

Last edited: Mar 15, 2012
11. Mar 16, 2012

### ngkamsengpeter

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?

Last edited: Mar 16, 2012
12. Mar 16, 2012

### Mentz114

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.

13. Mar 16, 2012

### ngkamsengpeter

Yes. It was interesting. I can understand the Aθ/r now but I still dont quite understand with your calculation. What is the final answer of F in terms of Er,Eθ,Ez?

14. Mar 16, 2012

### Mentz114

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.

15. Mar 17, 2012

### ngkamsengpeter

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?

Last edited: Mar 17, 2012
16. Mar 18, 2012

### ngkamsengpeter

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