Actually, the electric and magnetic fields from one *combined* tensor called the electromagnetic field tensor. This is a rank-2 tensor and takes the form*
$$
F^{\mu\nu}=\left(\begin{array}{cccc}
0 & -E_x & -E_y & -E_z \\
E_x & 0 & -B_z & B_y \\
E_y & B_z & 0 & -B_x \\
E_z & -B_y & B_x & 0
\end{array}\right)
$$
It has the following properties:

- It is anti-symmetric (so $F^{12}=-F^{21}$)
- It is traceless
- It has 16 elements, but only 6 distinct values
- When multiplied by its dual tensor ($G^{\mu\nu}$) it gives a Lorentz invariant value of $4\mathbf{B}\cdot\mathbf{E}$
- The inner product, $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2)$, is also a Lorentz invariant

You can also derive Maxwell's equations through the tensor by applying $\partial_\mu$ to it. Gauss's law and Ampere's law come from
$$
\partial_\mu F^{\mu\nu}=4\pi J^\nu
$$
where $J^\mu=\left(\rho,\,\mathbf{j}\right)$ is the four-current. The magnetic Gauss' law and Faraday's law come from applying the Bianchi identity to get
$$
\partial_\gamma F_{\mu\nu} + \partial_\mu F_{\nu\gamma} + \partial_\nu F_{\gamma\mu}=0
$$
Or more concisely,
$$\partial_{[\mu}F_{\nu\gamma]}=0$$

_{I am an astrophysicist, so I use cgs units; in SI, all electric fields have a factor of $1/c$.}

Physically, it tells you the direction in spacetime that the object is going. Because it is a unit vector and it points the direction in spacetime the object is going.

Recall how in 3d you can represent the direction of a velocity vector with a unit vector? You could choose any orthonormal basis you like and then the components were $$\left(\frac{v_x}{\sqrt{v_x^2+v_y^2+v_y^2}},
\frac{v_y}{\sqrt{v_x^2+v_y^2+v_y^2}},
\frac{v_z}{\sqrt{v_x^2+v_y^2+v_y^2}}\right).$$

And you could get it by taking $(\Delta x, \Delta y,\Delta z)$ for two points where the particle is at at really close moments in time and then making a unit vector version of it. It just tells you the direction in space it is going. And as a vector it's the same for anyone that measures distance like you do. No matter what basis they use, though with a different 3d basis they get different components.

The four velocity is like that, it's merely a direction in spacetime. And you get it by considering two events $(ct_1,x_1,y_1,z_1)$ and $(ct_2,x_2,y_2,z_2)$ where the particle is at that are really close together in time and subtracting them to get the vector between them $(c\Delta t,\Delta x, \Delta y,\Delta z)$ and then making a unit vector out of that.

And what's nice here is that every frame agrees on this vector, even if they give different components in a different 4d basis.

So **physically** all it does it tell you the direction in spacetime that an object is going, just like the unit 3 velocity tells you the direction in space something is going.

As an aside if you call $w=ct$ then it is like time but measured in meters and then $(w,x,y,z)$ are the components so having it as a 0th component (rather than a 4th) seems more natural.

Now there is an interesting relationship between energy and momentum and the unit tangent.

$$(E,c\vec p)=mc^2u,$$

Where $u$ is the unit tangent, $\vec p$ is the 3 momentum, $m$ is the rest mass, $E$ is the energy (rest energy and kinetic energy together) and $c$ is the speed of light.

Which means the fourth component is the specific energy (energy per unit mass) except with come factors of $c$ to make it dimension less. And it should be dimensionless since it is a component of a unit vector.

So $E/m$ or $E/mc^2.$

## Best Answer

The zeroth component of the 4-velocity $u^a=(\gamma ,\gamma \vec v/c)c$ is essentially the time-dilation factor $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ (multiplied by $c$ for dimensional purposes). Using the rapidity $\theta$ (the Minkowski angle between two timelike vectors), that zeroth component is essentially $\cosh\theta$.

In practice, [in geometric units] the 4-velocity is a unit-timelike vector. If drawn on a spacetime diagram, the tip would represent "one tick" of that object's clock. Thus, the time-component of the unit 4-velocity would be the apparent duration of that object's tick, namely the time-dilation factor multiplied by "one tick".

The zeroth component of the 4-force is the relativistic power.