Magnetism is Just Electricity in Another Frame

 Recently, I came across something that shook my whole understanding of electrodynamics. I had studied the different electric and magnetic fields their interactions and how one (quite mysteriously I might add) transforms to the other. A moving charge creates a magnetic field. A change in the magnetic field causes a current. Why do these different fields just change into each other based on the relative motion of their sources? The answer lies in special relativity. Rigorously of course, one could prove that the magnetic field is just the electric field in another reference frame using the relativistic transformation of the fields however, the setup that gave me intuition and helped build the idea is deriving the Biot Savart Law from special relativity.
\[B =  \frac{\mu_0I}{2\pi r}
\]
The setup

Take an infinitely long wire, we aim to find the force on a moving charged particle adjacent to such a wire. Of course, the force is\[F = qv \times B\]
But consider this, what causes the particle to experience such a force in the first place. Special Relativity says that the force is due to the electric field of the wire. Looking at the setup we see that it is obvious that the wire doesn't have any electric field, it has equal number of protons and the electrons. The electrons move in the direction of the charge but they maintain the charge distribution. Thus, the system is neutral. How can it have an electric field. Well, I have said that according to Special relativity there is an electric field of the wire, so there must be some loopy relativity action going on; think about it, both the electrons and the adjacent charged particle are undergoing velocity (for simplicities sake we take both these velocities to be equal v), thus there must be some adjustments in the reference frame to be made. In the initial setup, we account for the charge density of the positive and negative particles in the wire.\[\lambda_+ +\lambda_- = \frac{dq_+}{dl} +\frac{dq_-}{dl} = 0\]This accounts for the neutral charge we assumed for the wire. Now we look at the system from the point of view of the adjacent moving particle of velocity v. Since the electrons in the wire are assumed to have the same velocity as the charged particle, they appear to be at rest whereas the positive metal cations appear to be moving backwards at speed v. Due to this relative motion, an effect of special relativity takes place called lenght contraction. Due to this, the positively charged cations appear to be closer together as the lenght shrinks.
\[\lambda_+' = \frac{dq_+}{dl}\gamma = \lambda_+\gamma\]
Where gamma is the Lorentz factor\[\gamma = \frac1{\sqrt{1-v^2/c^2}}\]
Additionally, we had measured \(\lambda_-\) in the lab frame of reference where it was moving, whereas in the reference of the charged particle it is stationary. Thus we use a similar but opposite analysis\[\lambda_- =\lambda_-'\gamma\]\[\Rightarrow \lambda_-/\gamma = \lambda_-'\]
Thus the new total charge density of the wire is\[\lambda_+'+ \lambda_-' = \lambda_+\gamma + \frac{\lambda_-}{\gamma} = \lambda_+(\gamma-1/\gamma),\;\;\;\;\;\textit{Since \(\lambda_+ = -\lambda_-\)}\]

Next, we find the electric field. The electric field of an infinitely long wire is taken as\[E = \frac{\lambda}{2\pi\epsilon_0r}\]
Thus the electric field of our wire in the moving reference frame is \[E' = \frac{\lambda_+(\gamma-1/\gamma)}{2\pi\epsilon_0r}\]
\[\Rightarrow F' = E'q = q\frac{\lambda_+(\gamma-1/\gamma)}{2\pi\epsilon_0r}\]
This force must translate to the one we see in the lab frame of reference. \[F = F'/\gamma = qv \times B\]
\[\Rightarrow q\frac{\lambda_+(\gamma-1/\gamma)}{2\pi\epsilon_0r\gamma} = qv \times B\]
Additionally, the velocity of the charged particle will be perpendicular to the circular magnetic field of the wire thus\(qv \times B = qvB\)
\[\Rightarrow q\frac{\lambda_+(\gamma-1/\gamma)}{2\pi\epsilon_0r\gamma} = qvB\]
Additionally, we simplify the gamma terms:
\[\gamma- 1/\gamma = \frac{1}{\sqrt{1-v^2/c^2}}-\sqrt{1-v^2/c^2} = \frac{v^2/c^2}{\sqrt{1-v^2/c^2}} = (v^2/c^2)\gamma\]
\[\Rightarrow qvB = q\frac{\lambda_+(v^2/c^2)\gamma}{2\pi\epsilon_0r\gamma}\]
Canceling the gammas, the qv's on both sides and taking \(\lambda_+v = I\), we get
\[B = \frac{I/c^2}{2\pi\epsilon_0r}\]
Additionally, \[c^2 = \frac{1}{\mu_0 \epsilon_0}\]
\[\Rightarrow B = \frac{\mu_0 I}{2\pi r}\]
Hence the law is proved.

With this, we have shown an example of how the effect that we consider to be the magnetic field is merely the electric field under relativistic transformation. While proper transformation of the electrodynamic fields would give a clearer picture, this example gives intuition as to how the electric and magnetic fields are more closely related.