Covariant form of the electric field

Question

I'm a bit confused about some expressions about the relativity of the electric field.

The usual treatment ($c=1$) defines the four potential $A_\mu$ and $F = dA$ is the anti-symmetric 2-tensor. Then $F_{0i}$ is the electric field and the $F_{ij}$ is related to the magnetic field (say in the "laboratory" reference frame).

A different observer (primed) that is moving with three-velocity $v^i$ w.r.t. the laboratory frame measures

$F_{\mu'\nu'} = \Lambda^{\mu}_{\ \mu'}\Lambda^{\nu}_{\ \nu'}F_{\mu\nu}$

where $\Lambda^\mu_{\ \mu'}$ is the Lorentz transformation matrix. Then one can read off the electric field in the new boosted frame as $E'_i = F'_{0i}$ and so on.

However, I often see the electric field defined as a four-vector $E_\mu = F_{\mu\nu} u^\nu$ where $u^\mu$ is the four-velocity vector. If $u = (1,0,0,0)$, i.e. the object is at rest, then $E_\mu$ is just the normal electric field in the laboratory frame (with an extra temporal component that is zero).

What does this $E_\mu$ four-vector represent when the observer is moving? It's not the electric field in the frame of the moving observer, because that is given by $F'_{0i}$, so what is it?

References:

1. Wald page 64 (section 4.2 Special relativity).

2. https://arxiv.org/abs/1606.01226 page 6 (middle).

3. https://arxiv.org/abs/arXiv:1703.08757 page 4.

Answer 1

There are two different things you need to define the electric field 3-vector from the field strength tensor. First, you need a definition of your time coordinate. This is done by defining a 4-velocity $u^\mu$. Next, you need to define your spatial axes (i.e. the $x$, $y$ and $z$ coordinates). We do this by introducing 3 vectors $t_i^\mu$. In a static frame, these vectors are given by $$ u^\mu = (1,0,0,0), \qquad t^\mu_1 = ( 0,1,0,0) , \qquad t^\mu_2 = (0,0,1,0) , \qquad t^\mu_3 = ( 0,0,0,1) . $$ or more concisely, $t^\mu_i = \delta^\mu_i$. However, when I change frames, these vectors take on a new form. The electric field 3-vector is then defined as $$ E_i = u^\mu t_i^\nu F_{\mu\nu} \stackrel{\text{static frame}}{=}F_{0i} . $$ Now, let us move to a new frame described by a Lorentz matrix $\Lambda$ (w.r.t. the static frame). Then, the new vectors are $$ u'^\mu = \Lambda^\mu{}_\nu u^\nu = \Lambda^\mu{}_0 , \qquad t'^\mu_i = \Lambda^\mu{}_\nu t_i^\nu = \Lambda^\mu{}_i $$ The electric field 3-vector in the rotated frame is then $$ E'_i = u'^\mu t'^\nu_i F_{\mu\nu} = \Lambda^\mu{}_0 \Lambda^\nu{}_i F_{0i} = F'_{0i} $$

BTW, the two different viewpoints are the active and passive transformations, but the physics is the same. In this answer, I have presented the passive transformation which rotates the axes but keeps the fields invariant. The other definition of the electric field is the active one where you rotate the field strength but keep the axes fixed. Which type of transformation you use is simply a matter of convention. The physics is the same.

Answer 2

$E_\mu$ as you define it is the Lorentz force per unit charge. It is the sum of electric and magnetic forces. This is trivially equal to the electric field for a charge at rest.$^*$ Note that electric and magnetic force can strictly only be distinguished for electrostatic and magnetostatic cases.

To call this the electric field would be confusing.

The statements in Wald, GR, p64, strike me as odd.

$^*$ Edit in response to @Prahar's answer