Register to reply

Electromagnetic tensor in cylindrical coordinates

by ngkamsengpeter
Tags: coordinates, cylindrical, electromagnetic, tensor
Share this thread:
ngkamsengpeter
#1
Mar12-12, 12:49 PM
P: 194
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)
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
Mentz114
#2
Mar12-12, 03:38 PM
PF Gold
P: 4,087
The field tensor will give the same contraction whatever the coordinates. This is what defines tensors. Changing the coordinate basis from xa -> xA
[tex]
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
[/tex]


The flat space cylindrical metric is ds2= dt2 - dz2 - dr2 - r22, I think.
ngkamsengpeter
#3
Mar12-12, 11:16 PM
P: 194
Quote Quote by Mentz114 View Post
The field tensor will give the same contraction whatever the coordinates. This is what defines tensors. Changing the coordinate basis from xa -> xA
[tex]
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
[/tex]


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

[tex]
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]
[/tex]
[tex]
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]
[/tex]

[tex]
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]
[/tex]

and I get
[tex]
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]
[/tex]

And thus
[tex]
F^{\mu \nu }F_{\mu \nu }≠2(B^2-\frac{E^2}{c^2})
[/tex]

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

Mentz114
#4
Mar13-12, 06:05 AM
PF Gold
P: 4,087
Electromagnetic tensor in cylindrical coordinates

Use this to calculate F
[tex]
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu
[/tex]
ngkamsengpeter
#5
Mar14-12, 01:22 AM
P: 194
Quote Quote by Mentz114 View Post
Use this to calculate F
[tex]
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu
[/tex]
I am new to this, can you show me some steps and the final answer so that I can check with my one?

Thanks.
Mentz114
#6
Mar14-12, 07:41 AM
PF Gold
P: 4,087
Quote Quote by ngkamsengpeter View Post
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/Co...ators_cyl.html

http://en.wikipedia.org/wiki/Del_in_...al_coordinates
Mentz114
#7
Mar14-12, 11:57 AM
PF Gold
P: 4,087
The transformation (t,x,y,z) -> (T,R,θ,Z)
[tex]
t \rightarrow T,\ \ x \rightarrow R \cos(\theta),\ \ y \rightarrow R \sin(\theta),\ \ z \rightarrow Z
[/tex]
is done by this matrix
[tex]
\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]
[/tex]

The transformed field tensor is
[tex]
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]
[/tex]

and the contraction [itex]g_{an}g_{bm}F^{ab}F^{mn} = -2 E^2 + 2B^2[/itex], where
[tex]
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]
[/tex]
ngkamsengpeter
#8
Mar15-12, 03:10 AM
P: 194
Quote Quote by Mentz114 View Post
The transformation (t,x,y,z) -> (T,R,θ,Z)
[tex]
t \rightarrow T,\ \ x \rightarrow R \cos(\theta),\ \ y \rightarrow R \sin(\theta),\ \ z \rightarrow Z...
[/tex]
Let me see if I understand it correctly. the tensor transformation rule is
[itex]\widetilde{A^\mu}=\frac{∂\widetilde{x^\mu}}{∂x^\nu}A^\nu[/itex]
So the transformation matrix you given is [tex]\frac{∂\widetilde{x^\mu}}{∂x^\nu} [/tex]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 [tex]\frac{∂\widetilde{x^1}}{∂x^1} =\frac{∂R}{∂x}=\frac{1}{Cos(\theta)}[/tex]

And I dont quite understand how the Field tensor transform. I would think of it transform with the same matrix with the following formula:
[tex]
\widetilde{F^{\mu\nu}}=\frac{∂\widetilde{x^\mu}}{∂x^ρ}\frac{∂\widetilde {x^\nu}}{∂x^σ}=F^{ρσ}
[/tex]

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?

Many thanks for your help.
Mentz114
#9
Mar15-12, 12:17 PM
PF Gold
P: 4,087
Quote Quote by ngkamsengpeter View Post
If I want to write the tensor in terms of ER,Eθ,Ez instead of Ex,Ey,Ez, how should I do that?

Many thanks for your help.
(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 [itex]F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu[/itex].

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

I'll think about this. Maybe I've made another mistake. My calculation above is horribly wrong .
Mentz114
#10
Mar15-12, 05:49 PM
PF Gold
P: 4,087
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 [itex]F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu[/itex].

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

[added later]
Using equation (117) the magnetic fields are
[tex]
\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}
[/tex]
and the electric fields are
[tex]
\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}
[/tex]
now we can write
[tex]
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]
[/tex]
and (finally ?) [itex]g_{ma}g_{bn}F^{ab}F^{mn}= 2(-E^2+B^2)[/itex] where [itex]g_{mn}=diag(-1,1,R^2,1)[/itex].
ngkamsengpeter
#11
Mar16-12, 03:19 PM
P: 194
Quote Quote by Mentz114 View Post
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 [itex]F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu[/itex].

The curl operator in cylindrical coords is equation (117) on this page
http://mathworld.wolfram.com/Cylindr...ordinates.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
[tex]
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]
[/tex]
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?
Mentz114
#12
Mar16-12, 09:41 PM
PF Gold
P: 4,087
Quote Quote by ngkamsengpeter View Post
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
[tex]
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]
[/tex]
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.
ngkamsengpeter
#13
Mar16-12, 10:23 PM
P: 194
Quote Quote by Mentz114 View Post
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 dont quite understand with your calculation. What is the final answer of F in terms of Er,Eθ,Ez?
Many thanks for your help.
Mentz114
#14
Mar16-12, 11:21 PM
PF Gold
P: 4,087
Quote Quote by ngkamsengpeter View Post
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?
Many thanks for your help.
I think the contraction is
[tex]-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) ) [/tex]

[tex]-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) ) [/tex]

depending how you define the E and B vectors.
ngkamsengpeter
#15
Mar17-12, 01:43 AM
P: 194
Quote Quote by Mentz114 View Post
I think the contraction is
[tex]-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) ) [/tex]

[tex]-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) ) [/tex]

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
[tex]
\vec{E}=E_r \hat{r}+ E_\theta \hat{\theta}+E_z \hat{z}
[/tex]
Is that mathematically correct if we simply defined the vector just to fits into the answer?
ngkamsengpeter
#16
Mar18-12, 11:40 PM
P: 194
Can anyone verified if what Mentz114 did is correct or not? Can we defined the E and B vector like that?
Thanks.


Register to reply

Related Discussions
Fluid stress tensor in cylindrical coordinates? Classical Physics 2
Gradient of a tensor in cylindrical coordinates Special & General Relativity 5
Cylindrical coordinates to cartesian coordinates Calculus & Beyond Homework 2
Differential Cartesian Coordinates Into Cylindrical Coordinates Calculus & Beyond Homework 3
Rank 3 tensor created by taking the derivative of electromagnetic field tensor Advanced Physics Homework 1