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

[raistlin]

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!

 

[Orodruin]

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.

 

[raistlin]

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 ..

 

[Orodruin]

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.

 

[raistlin]

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?