Transformations of E and B

Electric and magnetic fields transform according to relativity too, although their exact behavior under a Lorentz transformation is more complicated than your average quantity. The transformation rules for an electromagnetic field are

\begin{align} &E'_{x}=E_{x},&E'_{y}&=\gamma(E_{y}-vB_{z}),&E'_{z}&=\gamma(E_{z}+vB_{y}) \\ &B'_{x}=B_{x},&B'_{y}&=\gamma\left( B_{y}+ \frac{v}{c^{2}}E_{z} \right),&B'_{z}&=\gamma\left( B_{z}- \frac{v}{c^{2}}E_{y} \right) \end{align}

where $v$ is the velocity of the new frame of reference with respect to the old one.

Derivation#

Within the context of relativity, through the knowledge that the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ are interrelated and of magnetism as a relativistic phenomenon, we want to take the final step and explain $\mathbf{E}$ and $\mathbf{B}$ themselves in terms of relativity. Like most things in relativity, $\mathbf{E}$ and $\mathbf{B}$ are just the spatial parts of their respective four-vectors — or, that's what you'd hope. In truth, that is not the case and the relativistic generalizations of the fields is quite a bit more complex.

To tackle the situation, we are to check how these two transform under a Lorentz transformation. To start, we need to lay the groundwork, which is to say we need to make some assumptions about how this might work and then check if it's true (experimentally or theoretically). We need to claim two things:

the electric charge is a relativistic invariant. It does not change under Lorentz transformation. A charge is a charge everywhere and anywhere.
the transformation rules are the same regardless of the origin of the fields. It does not matter how a field was created: stationary charge, straight moving charge, random motion in the core of the Sun, spinning particles in a cyclotron; it makes no difference. All fields should transform the same. If they don't, the field theory formulation must be dispensed with, as the fields are supposed to carry all information about electromagnetism on their own, without additional information (such as their origin).

We'll use a capacitor with surface charge densities $\pm \sigma_{0}$ as a test bench since it makes for rather simple electric fields (we'll assume that it is lying on its side so that its length $l$ lies on the $x$ axis). Assume we are observing the situation while at rest from a frame $\mathcal{S}_{0}$ in which the capacitor is also at rest. In this frame, its electric field is

E_{y}=\frac{\sigma_{0}}{\varepsilon_{0}}

The Lorentz transformation moves the capacitor to frame of reference $\mathcal{S}$ moving on the $x$ axis at a constant speed $v_{0}$ (relative to $\mathcal{S}_{0}$ ).

Diagram Moving capacitor relativity.svg|80%

Note how the velocity $v_{0}$ moves the frame of reference to the left, which means that the capacitor effectively moves to the right.

In fact, in $\mathcal{S}$ , it makes a field

E_{y}=\frac{\sigma}{\varepsilon_{0}}

where $\sigma$ is the surface charge density on the plates of the capacitor. Actually, there's two densities, $\pm \sigma$ , depending on whether you're looking at the top or bottom plate. Since $\mathcal{S}$ is moving, the length $l$ will be contracted by a factor $\gamma_{0}=1/\sqrt{ 1-v_{0}^{2}/c^{2} }$ compared to $\mathcal{S}_{0}$ , in which the density (at rest) is $\sigma_{0}$ , so $\sigma=\gamma_{0}\sigma_{0}$ . Also due to the frame's motion, the charges are moving with speed $v_{0}$ , so they'll create a surface current density $\mathbf{K}_{\pm}$ and produce a magnetic field

\mathbf{K}_{\pm}=\mp \sigma v_{0}\hat{\mathbf{x}}

The field points in the negative $z$ direction (use the right hand rule!) and has magnitude

B_{z}=-\mu_{0}\sigma v_{0}

as per Ampere's law. This is the situation in $\mathcal{S}$ . In another frame $\mathcal{S}'$ , moving at a constant velocity $v$ relative to $\mathcal{S}$ , the observed fields will be

E_{y}'=\frac{\sigma'}{\varepsilon_{0}},\qquad B_{z}'=-\mu_{0}\sigma'v'

where $v'$ is the velocity of $\mathcal{S}'$ relative to $\mathcal{S}_{0}$ , as per the Einstein velocity addition rule:

v'=\frac{v+v_{0}}{1+vv_{0}/c^{2}}

The quantity $\sigma'=\gamma'\sigma_{0}$ is weighed by $\gamma'=1/\sqrt{ 1-v'^{2}/c^{2} }$ . To find the general transformation rule, we want to express the quantities of $\mathcal{S}'$ in terms of those of $\mathcal{S}$ (i.e. from a general moving system to another general moving system). Using

\sigma'=\gamma' \sigma_{0},\quad\sigma=\gamma_{0}\sigma_{0},\quad\Rightarrow \quad \sigma'=\frac{\gamma'}{\gamma_{0}}\sigma

we can write

E'_{y}=\frac{\gamma'}{\gamma_{0}} \frac{\sigma}{\varepsilon_{0}},\quad B'_{z}=- \frac{\gamma'}{\gamma_{0}}\mu_{0}\sigma v'

The ratio of gammas is

\frac{\gamma'}{\gamma_{0}}=\frac{\sqrt{ 1-v_{0}^{2}/c^{2} }}{\sqrt{ 1-v'^{2}/c^{2} }}=\frac{1+vv_{0}/c^{2}}{\sqrt{ 1-v^{2}/c^{2} }}=\gamma\left( 1+ \frac{vv_{0}}{c^{2}} \right)

where, intuitively, $\gamma=1/\sqrt{ 1-v^{2}/c^{2} }$ . Thus, inverting the field relations to express one field in terms of the other:

B_{z}=-\mu_{0}\sigma v_{0}=- \frac{\sigma v_{0}}{c^{2}\varepsilon_{0}}\quad\Rightarrow \quad \frac{\sigma}{\varepsilon_{0}}=- \frac{c^{2}B_{z}}{v_{0}}=E_{y}

to express the fields in $\mathcal{S}'$ in terms of the fields in $\mathcal{S}$ :

E'_{y}=\gamma\left( 1+ \frac{vv_{0}}{c^{2}} \right) \frac{\sigma}{\varepsilon_{0}}=\gamma\left( E_{y}- vB_{z} \right)

and

B'_{z}=-\gamma\left( 1+ \frac{vv_{0}}{c^{2}} \right)\mu_{0}\sigma\left( \frac{v+v_{0}}{1+vv_{0}/c^{2}} \right)=\gamma\left( B_{z}- \frac{v}{c^{2}}E_{y} \right)

These are the transformation rules for $E_{y}$ and $B_{z}$ . To find the other components, we just align the capacitors on different axes. Say the capacitor is now standing up vertical on the $xy$ plane instead of laying flat on the $xz$ plane like in the image, the fields in $\mathcal{S}$ would be

E_{z}=\frac{\sigma}{\varepsilon_{0}},\qquad B_{y}=\mu_{0}\sigma v_{0}

(the only practical change is the direction of the magnetic field and that the contracted length is $w$ instead of $l$ ). The rest of the formulas run the same and we get

E_{z}'=\gamma(E_{z}+vB_{y}),\qquad B_{y}'=\gamma\left( B_{y}+ \frac{v}{c^{2}}E_{z} \right)

Finally, if we make the capacitor stand up on the $yz$ axis, then little happens. This is because the contracted length here is the distance between plates, but since the electric field doesn't depend on it, nothing changes:

E_{x}'=E_{x}

In this case, there is no magnetic field, so we can't figure out the transformation of $B_{x}$ from this system. What we can do is use a different electrical component altogether: a long solenoid centered on the $x$ axis at rest in $\mathcal{S}$ .

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The field of a resting solenoid with $n$ turns per unit length is

B_{x}=\mu_{0}nI

When moving at speed $v$ in $\mathcal{S}'$ , the turns per unit length become denser as per $n'=\gamma n$ . However, we also need to care about time dilation on the electric current, since it is charge per unit time:

I'=\frac{I}{\gamma}

All in all, these two factors cancel and we get

B'_{x}=B_{x}

Putting all of these transformations together finally gives us

\boxed{\begin{align} &E'_{x}=E_{x},&E'_{y}&=\gamma(E_{y}-vB_{z}),&E'_{z}&=\gamma(E_{z}+vB_{y}) \\ &B'_{x}=B_{x},&B'_{y}&=\gamma\left( B_{y}+ \frac{v}{c^{2}}E_{z} \right),&B'_{z}&=\gamma\left( B_{z}- \frac{v}{c^{2}}E_{y} \right) \end{align}}\tag{1}

The fields are only transformed in directions perpendicular to motion (the opposite of distance contraction).

Interestingly, even we observe no fields at a given point in $\mathcal{S}$ , $\mathbf{E}=0$ and $\mathbf{B}=0$ in $\mathcal{S}$ , we still observe something at that same point in $\mathcal{S}'$ :

\boxed{\mathbf{E}'=\mathbf{v}\times \mathbf{B}',\quad \mathbf{B}'=- \frac{1}{c^{2}}(\mathbf{v}\times \mathbf{E}')}

The fields "exist" for one observer but not for another, and in the new system they are related by these vector products.