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

• A
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
Staff Emeritus
Homework Helper
Gold Member
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.

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
Staff Emeritus
Homework Helper
Gold Member
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.

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?