How E and B Transform

The field-strength tensor transforms as a rank-two tensor, $F^{\prime \mu \nu }={\Lambda }^{\mu }{}_{\alpha }{\Lambda }^{\nu }{}_{\beta }{F}^{\alpha \beta }$. Because $\vec{E}$ and $\vec{B}$ are its components, this single rule fixes how both fields change between frames. Carrying it out turns the abstract statement the fields mix into explicit formulas, and those formulas explain two things at once: why a moving charge's field is squashed, and why magnetism exists at all.¹

Conventions as before: signature $(-,+,+,+)$, and the primed frame $S^{\prime }$ moves at velocity $\vec{v}=v\hat{x}$ relative to $S$, with $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$.

The transformation rules

Apply the boost matrix

to $F^{\alpha \beta }$. Each primed component is a sum of two unprimed ones. For the electric field,

and for the magnetic field,

The component along the boost is unchanged; the transverse components mix and carry a factor $\gamma$. Written in terms of parts parallel and perpendicular to $\vec{v}$, the rules are frame-direction independent:

At low speed, $\gamma \to 1$ and the rules become ${\vec{E}}^{\prime }\approx \vec{E}+\vec{v}\times \vec{B}$ and ${\vec{B}}^{\prime }\approx \vec{B}-c^{-2}\vec{v}\times \vec{E}$, the Galilean field transformations that underlie motional emf.

The field of a uniformly moving charge

Take a charge $q$ at rest at the origin of $S^{\prime }$. There it produces a pure Coulomb field, ${\vec{B}}^{\prime }=0$ and

isotropic and radial. View it from $S$, in which the charge moves at $v\hat{x}$. Transforming the fields and re-expressing them in terms of the charge's present position gives, at the moment the charge passes the origin,

where $\theta$ is the angle between $\vec{r}$ and the direction of motion. The field still points radially from where the charge is now, but its strength is no longer isotropic:

• Along the motion ($\theta =0$): the field is weakened, $E_{\parallel }=\frac{1}{{\gamma }^2}\frac{q}{4\pi {ϵ}_0{r}^2}$.
• Transverse to the motion ($\theta =\frac{\pi }{2}$): the field is strengthened, $E_{\perp }=\gamma \frac{q}{4\pi {ϵ}_0{r}^2}$.

The field lines, uniformly spaced for a charge at rest, are swept out of the direction of motion and compressed into a transverse pancake. As $v\to c$ the compression becomes extreme, and the field approaches a sheet concentrated in the plane perpendicular to the motion — the limiting field of a light-like charge. The moving charge also carries a magnetic field circling the line of motion,

which is the field a current element produces, here derived purely by boosting a static Coulomb field.

A pure field boosted

The transformation converts field types. A capacitor at rest sets up a uniform $\vec{E}$ between its plates and no magnetic field. An observer moving parallel to the plates sees the same ${\vec{E}}_{\parallel }$ but, moving perpendicular to the field, measures a magnetic field ${\vec{B}}_{\perp }^{\prime }=-\gamma c^{-2}\vec{v}\times \vec{E}$. The moving charges on the plates constitute surface currents in that frame, and those currents are the source of ${\vec{B}}^{\prime }$. Nothing changed about the physical setup; the magnetic field is the same electric field, sliced by a different observer.

Magnetism as relativistic electrostatics

The most direct demonstration that magnetism is a relativistic effect is the force between a current-carrying wire and a moving charge. Take a long neutral wire: a stationary lattice of positive ions with linear density $+{\lambda }_0$, and conduction electrons with density $-{\lambda }_0$ drifting at speed $v_d$ to produce a current $I={\lambda }_0{v}_d$. The wire is neutral, so in the lab frame a test charge $q$ outside it feels no electric force. Let the charge move parallel to the wire at speed $v$. In the lab, the only force is magnetic: the current's field $B={\mu }_{0}I/(2\pi d)$ at distance $d$ acts on the moving charge with force $F=qvB$.

Now view the situation from the test charge's rest frame. The positive ions were at rest in the lab and now stream backward at speed $v$, so their spacing contracts and their density rises in magnitude to ${\gamma }_v{\lambda }_0$. The electrons drifted at $v_d$ in the lab, so their speed in the new frame follows from velocity addition and differs from $v$; their spacing changes by a different factor. The two densities no longer cancel. The wire carries a net charge density in the test charge's frame,

producing an electric field that exerts a force on the stationary test charge. Working out the magnitude and transforming back gives exactly the magnetic force $F=qvB$ of the lab frame. What one observer calls a magnetic force, another calls the electric force of a wire that length contraction has left charged.

Summary

• Under a boost, field components along the velocity are unchanged; transverse components mix $\vec{E}$ and $\vec{B}$ and gain a factor $\gamma$: ${\vec{E}}_{\perp }^{\prime }=\gamma ({\vec{E}}_{\perp }+\vec{v}\times \vec{B})$ and ${\vec{B}}_{\perp }^{\prime }=\gamma ({\vec{B}}_{\perp }-c^{-2}\vec{v}\times \vec{E})$.
• Boosting a static Coulomb field gives the field of a uniformly moving charge: radial from the present position but compressed transverse to the motion, weakened by $1/{\gamma }^2$ ahead and strengthened by $\gamma$ to the side, with a circulating magnetic field $\vec{B}=c^{-2}\vec{v}\times \vec{E}$.
• A pure electric field acquires a magnetic part for a transversely moving observer; the moving source charges are the currents that produce it.
• The magnetic force between a current and a moving charge is the electric force of a differentially contracted, net-charged wire in the charge's frame. Magnetism is the relativistic consequence of electrostatics.