How do you calculate the time component of a 4-Vector

[Philosophaie]

A stationary Monopole exist at the Origin. I am trying to get an understanding of the time derivative of a Four-Vector of $\vec{B}$ and $\vec{E}$

$\vec{B} = B_r \hat r + B_\theta \hat \theta + B_\phi \hat \phi + \frac{1}{c}B_t \hat t$

$\vec{E} = E_r \hat r + E_\theta \hat \theta + E_\phi \hat \phi + \frac{1}{c}E_t \hat t$

How do you calculate $B_t$ and $E_t$?

 

[Dale]

There is no such thing as a four vector of B and E. They are components of an antisymmetric rank 2 tensor

https://en.m.wikipedia.org/wiki/Electromagnetic_tensor

 

[Philosophaie]

Even thought it is not a 4-Vector can it have a $B_t$ component?

 

[Ibix]

No. The tensor Dale mentioned has six independent components which correspond to the three components of the electric field three-vector and the three components of the magnetic field three-vector.

The electric potential and the magnetic vector potential together form a four vector.

 

[Philosophaie]

Antisymmetric rank 2 tensor has six independent components which correspond to the three components of the electric field three-vector and the three components of the magnetic field three-vector?

 

[Dale]

Yes, see: https://en.m.wikipedia.org/wiki/Electromagnetic_tensor

 

[vanhees71]

The 6 components of the electromagnetic field are in fact components of an antisymmetric Minkowski tensor. It's easier to remember the relations in terms of the four-potential:
$$F_{\mu \nu} = \partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
The time-space components are (latin indices run from $1$ to $3$, and I use the (+---) convention of the metric)
$$F_{0n}=\partial_0 A_n-\partial_n A_0=-\frac{1}{c} \partial_t A^n-\partial_n A^0=E^n=-E_n,$$
and the space-space components
$$F_{mn}=\partial_m A_n-\partial_n A_m=-\partial_m A^n + \partial_n A^m=-\epsilon^{lmn} B^l=-B^{mn}=-B_{mn}.$$
This is in a fixed inertial reference frame.

For the following we also need the Hodge dual
$$(\dagger F)^{\mu \nu}=\frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F^{\rho \sigma}.$$
After some algebra one gets
$$(\dagger F)^{0 n}-B^n.$$

You can rewrite this in a coordinate-free way. Take a reference frame with the time-like unit vector called $u^{\mu}$. For the original frame, used above we have $(u^{\mu})=(1,0,0,0)$. So you can define four-vector electric-field components by
$$E^{\mu}=F^{\mu \nu} u_{\nu}, \quad B^{\mu}=(\dagger F)^{\mu \nu} u_{\nu}.$$
In the reference frame, where $u^{\mu}=(1,0,0,0)$ a single monopole sitting at rest in the origin of the spatial coordinate system, there is no time component. You have
$$\vec{E}=0, \quad \vec{B}=\frac{g}{4 \pi r^3} \vec{x}$$
where $g$ is the magnetic charge of the point monopole. It's completely analogous to electrostatics.