How do I pass a magnetic field to any coordinate system?

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Hello,
For example, an electric field vector, such as the gradient of the potential, passes in the following way to any coordinate system:$$E = -\triangledown{}V = - \frac{{\partial V}}{{\partial x^i}}g^{ij}e_j$$
But what about a vector of a magnetic field? How would it be expressed in any coordinate system?

Thanks!
 

Answers and Replies

  • #2
Orodruin
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In relativity, the electric and magnetic fields are part of the electromagnetic field tensor, which is expressed as
$$
F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu,
$$
which holds in any coordinate system.

If you are just interested in the expression of the curl in curvilinear coordinates in three dimensions, you have
$$
\nabla \times \vec A = \vec e_i \eta^{ijk} \partial_j A_k,
$$
where ##\vec e_i## are the tangent vector basis vectors, ##\eta^{ijk} = (1/\sqrt{g}) \epsilon^{ijk}##, and ##A_k## are the covariant components of ##\vec A##. Here ##g## is the metric determinant.
 
  • #3
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Thanks for the answer, but it's not exactly what I was finding for ..

I know from Maxwell's third equation that:$$\triangledown{} \times{} \vec{E}= -\frac{{\partial B}}{{\partial t}}$$Therefore$$(\triangledown{} \times{} \vec{E}) {\partial t}= -{\partial \vec{B}} \Longrightarrow{} \frac{{\partial t}}{{\partial r}}\times{}\frac{{\partial V}}{{\partial r}}={\partial \vec{B}}\Longrightarrow{}\vec{B}=\triangledown{}t\times{}\triangledown{}V$$ Is it correct?
And from here I do not know how to continue ..
 
  • #4
Orodruin
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Thanks for the answer, but it's not exactly what I was finding for ..

I know from Maxwell's third equation that:$$\triangledown{} \times{} \vec{E}= -\frac{{\partial B}}{{\partial t}}$$Therefore$$(\triangledown{} \times{} \vec{E}) {\partial t}= -{\partial \vec{B}} \Longrightarrow{} \frac{{\partial t}}{{\partial r}}\times{}\frac{{\partial V}}{{\partial r}}={\partial \vec{B}}\Longrightarrow{}\vec{B}=\triangledown{}t\times{}\triangledown{}V$$ Is it correct?
And from here I do not know how to continue ..
No. Sorry, but this makes no sense.
 
  • #5
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Yes, you're right..

But if I use the Biot-Savart law I have:$$B = \displaystyle\frac{1}{c^2}v\times{}E\Longrightarrow{}B c^2= v\times{}E\Longrightarrow{}-B c^2=\frac{dr}{dt}\times{}\triangledown{}V$$How would this be to any coordinate system? Could I pass the velocity and the gradient separately and then make a cross product?
 

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