- #1

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Is it just change the Ex,Ey,Ez to Eρ,Eθ,Ez?

Is F

^{σρ}F

_{σρ}still equal to 2(B^2-E^2)

- Thread starter ngkamsengpeter
- Start date

- #1

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Is it just change the Ex,Ey,Ez to Eρ,Eθ,Ez?

Is F

- #2

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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.

[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

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- #3

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

[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

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- #4

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Use this to calculate F

[tex]

F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu

[/tex]

[tex]

F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu

[/tex]

- #5

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

[tex]

F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu

[/tex]

Thanks.

- #6

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When you transform from rectangular coordinates to cylindrical you must transform the differential operators also. This will remove the factor of rI am new to this, can you show me some steps and the final answer so that I can check with my one?

Thanks.

See here for the transformed operators

http://people.rit.edu/pnveme/pigf/Coords/Diff_operators_cyl.html [Broken]

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

Last edited by a moderator:

- #7

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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]

[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]

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- #8

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Let me see if I understand it correctly. the tensor transformation rule isThe 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]

[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

Many thanks for your help.

- #9

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(I haven't helped much so far ).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.

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 A

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 .

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- #10

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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/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.

[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].

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.

[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].

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- #11

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First of all, why replace AI 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/CylindricalCoordinates.html

Assume all your calculation correctly, then the final answer is just replaced the E

[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?

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- #12

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We have to raise the index of AFirst of all, why replace A_{θ}with A_{θ}/R. We must have a reason for that, cannot just replace it in order to get the answer.

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.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?

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

- #13

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Yes. It was interesting. I can understand the AWe have to raise the index of A_{μ}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.

Many thanks for your help.

- #14

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I think the contraction isYes. It was interesting. I can understand the A_{θ}/r now but I still dont quite understand with your calculation. What is the final answer of F in terms of E_{r},E_{θ},E_{z}?

Many thanks for your help.

[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

- #15

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Yes. I get this answer also but how can we simply define vectors? In cylindrical coordinate, vector should be defined asI 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 theEandBvectors.

[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?

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- #16

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Thanks.