Electromagnetic tensor in cylindrical coordinatesby ngkamsengpeter Tags: coordinates, cylindrical, electromagnetic, tensor 

#1
Mar1212, 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 F^{σρ}F_{σρ} still equal to 2(B^2E^2) 



#2
Mar1212, 03:38 PM

PF Gold
P: 4,081

The field tensor will give the same contraction whatever the coordinates. This is what defines tensors. Changing the coordinate basis from x^{a} > x^{A}
[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 ds^{2}= dt^{2}  dz^{2}  dr^{2}  r^{2}dθ^{2}, I think. 



#3
Mar1212, 11:16 PM

P: 194

[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^2E^2/c^2) also in cylindrical or any coordinates right? Can anyone tell me which part I didt it incorrectly? Thanks 



#4
Mar1312, 06:05 AM

PF Gold
P: 4,081

Electromagnetic tensor in cylindrical coordinates
Use this to calculate F
[tex] F_{\mu\nu}=\partial_\mu A_\nu\partial_\nu A_\mu [/tex] 



#5
Mar1412, 01:22 AM

P: 194

Thanks. 



#6
Mar1412, 07:41 AM

PF Gold
P: 4,081

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 



#7
Mar1412, 11:57 AM

PF Gold
P: 4,081

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) \,Eysin\left( \theta\right) \,Ex}{R} & Ez\\ sin\left( \theta\right) \,Eycos\left( \theta\right) \,Ex & 0 & \frac{Bz}{R} & cos\left( \theta\right) \,By+sin\left( \theta\right) \,Bx\\ \frac{cos\left( \theta\right) \,Eysin\left( \theta\right) \,Ex}{R} & \frac{Bz}{R} & 0 & \frac{sin\left( \theta\right) \,Bycos\left( \theta\right) \,Bx}{R}\\ Ez & cos\left( \theta\right) \,Bysin\left( \theta\right) \,Bx & \frac{sin\left( \theta\right) \,Bycos\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] 



#8
Mar1512, 03:10 AM

P: 194

[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 E_{R},E_{θ},E_{z} instead of Ex,Ey,Ez, how should I do that? Many thanks for your help. 



#9
Mar1512, 12:17 PM

PF Gold
P: 4,081

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 4potential A_{μ} 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 . 



#10
Mar1512, 05:49 PM

PF Gold
P: 4,081

I think I've solved this. By replacing A_{θ} with A_{θ}/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]. 



#11
Mar1612, 03:19 PM

P: 194

Assume all your calculation correctly, then the final answer is just replaced the E_{x},E_{y},E_{z} by E_{r},E_{θ},E_{z} 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? 



#12
Mar1612, 09:41 PM

PF Gold
P: 4,081

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



#13
Mar1612, 10:23 PM

P: 194

Many thanks for your help. 



#14
Mar1612, 11:21 PM

PF Gold
P: 4,081

[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. 



#15
Mar1712, 01:43 AM

P: 194

[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? 



#16
Mar1812, 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. 


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