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Electric and magnetic fields transform into each other under boosts; the invariants $\vec{E}\cdot\vec{B}$ and $E^2 - c^2B^2$, fields of fast charges, and Maxwell's equations in covariant form.
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By the end of this lesson you will be able to transform electric and magnetic fields between frames, find the field of a fast charge, use the invariants to find frames where one field vanishes, and write Maxwell's equations in covariant form.
You know Maxwell's equations, the potentials $V$ and $\vec{A}$, the Lorentz force and the fields of moving and radiating charges. From special relativity you know Lorentz transformations, $\gamma$, and four-vectors. This lesson brings them together: Maxwell's theory was relativistic all along, and electric and magnetic fields turn out to be two faces of one object.
| Term | What it means |
|---|---|
| Field transformation | The rule giving $\vec{E}'$ and $\vec{B}'$ in a frame moving at $\vec{v}$ from $\vec{E}$ and $\vec{B}$. |
| Lorentz invariant | A quantity all inertial observers agree on, such as $\vec{E}\cdot\vec{B}$ or $E^2 - c^2B^2$. |
| Field tensor | The antisymmetric four-by-four array $F^{\mu\nu}$ holding the six components of $\vec{E}$ and $\vec{B}$. |
| Four-current | $J^\mu = (c\rho, \vec{J})$, charge and current density combined into one four-vector. |
| Four-potential | $A^\mu = (V/c, \vec{A})$, the scalar and vector potentials combined. |
| Charge invariance | The total charge of an object is the same in every frame, unlike its length or energy. |
| Drift velocity | The velocity $\vec{E}\times\vec{B}/B^2$ of the frame in which crossed fields become purely magnetic. |
Suppose one observer sees fields $\vec{E}$ and $\vec{B}$, and a second moves past at velocity $\vec{v}$. Components along $\vec{v}$ are the same for both. The perpendicular components mix:
$$\vec{E}'_\perp = \gamma(\vec{E} + \vec{v}\times\vec{B})_\perp, \qquad \vec{B}'_\perp = \gamma\left(\vec{B} - \frac{\vec{v}\times\vec{E}}{c^2}\right)_\perp.$$
A pure electric field in one frame has a magnetic part in another, and the reverse. Two combinations never change:
$$\vec{E}\cdot\vec{B} \quad\text{and}\quad E^2 - c^2B^2.$$
If $\vec{E}\cdot\vec{B} = 0$ and $E < cB$, some frame sees a pure magnetic field; if $E > cB$, some frame sees a pure electric field. A fast point charge's field follows from boosting its Coulomb field: at the same distance it is $\gamma$ times stronger broadside and $\gamma^2$ times weaker ahead. In four-vector language the six field components form an antisymmetric tensor $F^{\mu\nu}$, and all four Maxwell equations collapse into two: $\partial_\mu F^{\mu\nu} = \mu_0J^\nu$ and its dual.
Another way: picture
Picture a charge at rest with its field lines spread evenly in all directions. Now watch it race past. Length contraction squeezes the pattern along the motion, so the lines crowd into a flat disk perpendicular to the velocity, like the spokes of a pancake. Ahead and behind the field is weak; to the side it is strong. And because the charge moves, it is a current, and the observer also sees a magnetic field circling the line of motion.
Another way: steps
1. Set up axes along the boost. Let $x$ lie along $\vec{v}$. Then $E_x$ and $B_x$ are unchanged, and $E'_y = \gamma(E_y - vB_z)$, $E'_z = \gamma(E_z + vB_y)$, $B'_y = \gamma(B_y + vE_z/c^2)$, $B'_z = \gamma(B_z - vE_y/c^2)$. 2. Substitute and simplify. 3. Check with the invariants. Compute $\vec{E}\cdot\vec{B}$ and $E^2 - c^2B^2$ in both frames; they must agree.
Other checks. At $v \ll c$, $\gamma \approx 1$ and the magnetic correction $\vec{v}\times\vec{E}/c^2$ is tiny; the electric correction $\vec{v}\times\vec{B}$ is the familiar motional term. A boost of $-\vec{v}$ applied after $\vec{v}$ must return the original fields. And no boost can turn a field with $\vec{E}\cdot\vec{B} \ne 0$ into a pure electric or pure magnetic field.
Take a neutral wire carrying a current: positive ions at rest, electrons drifting. A charge $q$ moving parallel to the wire at the electrons' speed feels a magnetic force toward or away from it. Now jump into the charge's frame. There the charge is at rest, so there can be no magnetic force on it. Yet the force must still be there, since a deflection is something all observers agree on.
The resolution is length contraction. In the new frame the ions move and are contracted, the electrons are at rest and less contracted than before, and the wire is no longer neutral. The net charge makes an electric field, and its force on $q$ matches exactly the magnetic force seen in the lab. Magnetism is what the electric force looks like between moving charges, once relativity is taken seriously, and the effect is measurable even though the drift speeds are less than a millimeter per second, because the charges in a wire are so enormous in number.
Boosting a Coulomb field gives the field of a charge moving at constant velocity. At the same distance from the charge's present position, the field points radially from that present position, and its strength depends on the angle $\theta$ from the motion:
$$E = \frac{q}{4\pi\epsilon_0r^2}\,\frac{1 - \beta^2}{(1 - \beta^2\sin^2\theta)^{3/2}}.$$
Along the motion it is reduced by $\gamma^2$; perpendicular to it, enhanced by $\gamma$. A detector beside the path of a relativistic particle feels a sharp pulse lasting about $b/\gamma v$, where $b$ is the distance of closest approach. For $\gamma$ in the thousands, the field of a passing charge looks almost exactly like a flat pulse of light, which is why heavy ions colliding at the Relativistic Heavy Ion Collider can be treated as swarms of photons that collide with each other.
Because $\vec{E}\cdot\vec{B}$ and $E^2 - c^2B^2$ are the same for all observers, they classify fields. A plane light wave has $\vec{E}\cdot\vec{B} = 0$ and $E = cB$, so both vanish, and no boost can make light purely electric or purely magnetic: every observer sees a wave, only redder or bluer.
Crossed fields with $E < cB$ can be made purely magnetic by boosting at $v = E/B$ along $\vec{E}\times\vec{B}$. In that frame a charged particle simply circles, so in the lab it drifts at $E/B$ while gyrating, whatever its charge or mass. That $E\times B$ drift carries plasma across magnetic fields in fusion devices and in Earth's magnetosphere. If $E > cB$, no boost removes the electric field; instead some frame sees a pure electric field, which accelerates charges steadily instead of turning them.
The six components of $\vec{E}$ and $\vec{B}$ do not form a four-vector; they form an antisymmetric rank-two tensor,
$$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu,$$
built from the four-potential $A^\mu = (V/c, \vec{A})$. Its time-space components are $E/c$ and its space-space components are $B$. Transforming $F^{\mu\nu}$ with two Lorentz matrices reproduces the field rules above, and the invariants are its two scalar contractions, $F_{\mu\nu}F^{\mu\nu} \propto B^2 - E^2/c^2$ and one proportional to $\vec{E}\cdot\vec{B}$.
The gauge freedom of the potentials becomes $A^\mu \to A^\mu + \partial^\mu\lambda$, which leaves $F^{\mu\nu}$ untouched, and the Lorenz gauge condition becomes the single invariant statement $\partial_\mu A^\mu = 0$.
Charge density and current density combine into the four-current $J^\mu = (c\rho, \vec{J})$, and the continuity equation becomes $\partial_\mu J^\mu = 0$. Maxwell's equations with sources, Gauss's law and the Ampère–Maxwell law, become one equation,
$$\partial_\mu F^{\mu\nu} = \mu_0J^\nu,$$
and the two source-free equations, no magnetic monopoles and Faraday's law, become an identity satisfied automatically by any $F$ built from a potential. The Lorentz force becomes $dp^\mu/d\tau = qF^{\mu\nu}u_\nu$, whose time component is the power the field delivers.
Written this way the equations look the same in every inertial frame, which is what Einstein noticed in 1905: Maxwell's equations were already relativistic, and it was Newtonian mechanics that needed changing.
Einstein's paper on special relativity opens with an asymmetry. Move a magnet past a conducting loop and the textbook explanation is an electric field induced by the changing magnetic field. Move the loop past a stationary magnet and the explanation is the magnetic force $q\vec{v}\times\vec{B}$ on the moving charges. The two stories are different, yet the current is identical: only the relative motion matters.
The field transformations remove the puzzle. The loop's charges feel $\vec{E}'$ in their own frame, and $\vec{E}' = \vec{E} + \vec{v}\times\vec{B}$ at low speed. What one observer calls a magnetic force another calls an electric field; neither is more real. The motional emf of the Faraday's law lesson and the induced emf are one phenomenon seen from two frames.
This course has built electromagnetism from Coulomb's law to its relativistic form. The next steps go beyond it. Quantum electrodynamics replaces classical fields with photons and explains the pathologies of radiation reaction; its predictions, such as the electron's magnetic moment, agree with experiment to about one part in a trillion. The same tensor structure, generalized, underlies the weak and strong forces.
General relativity starts from the same place Einstein did, the demand that physical laws look the same to all observers, and extends it to accelerated frames and gravity. Gravitational waves, detected since 2015 by the LIGO observatories in Washington and Louisiana, are the gravitational analogue of the electromagnetic waves this course derived from Maxwell's equations.
Electrons in a beam repel each other, which should make any beam blow apart. But in the lab the electrons are also parallel currents, which attract. The magnetic attraction is $\beta^2$ times the electric repulsion, leaving a net force of $1/\gamma^2$ of the repulsion alone. In the beam's own frame there is only repulsion, but the lab clock sees it act slowly, by time dilation, and the two views agree.
This is why the hardest part of an electron accelerator is the first few meters. At the electron gun, $\gamma$ is close to one and space charge is fierce, so designers accelerate electrons to several MeV as fast as possible. By $50$ MeV, $\gamma \approx 100$ and the self-force has fallen ten thousandfold. The free-electron lasers at SLAC in California and elsewhere depend on bunches whose density is set in those first meters, before relativity mostly switches the repulsion off.
The solar wind is a plasma streaming from the Sun at about $400$ km/s, carrying a magnetic field of about $5$ nT. In the plasma's own frame there is no electric field, because charges move freely and cancel it. In Earth's frame, the field transformation gives $\vec{E} = -\vec{v}\times\vec{B}$, about $2$ mV/m.
That small field acts across the magnetosphere, roughly $30$ Earth radii wide, and amounts to hundreds of kilovolts. A fraction of it reaches the polar ionosphere as the polar cap potential, typically tens of kilovolts and more during storms, which drives plasma convection over the poles and powers the aurora. Space weather forecasters at the National Oceanic and Atmospheric Administration watch the solar wind's speed and magnetic field upstream of Earth precisely because their product sets this field.
It is natural to think a region either has a magnetic field or it does not, as a fact about the world. But a charge at rest has only an electric field, and the same charge seen by a moving observer has a magnetic field too. Which part of the electromagnetic field you call electric and which magnetic depends on how you move. What all observers agree on are the invariants and the physical outcomes, such as whether a particle is deflected.
A second misconception is that relativistic effects in electromagnetism need relativistic speeds. Magnetism itself is a relativistic effect of charges drifting at millimeters per second: the correction is tiny per charge, but the enormous numbers of electrons in a wire, whose electric forces would otherwise cancel exactly, make it the force that runs every electric motor.
A capacitor has a uniform field $E_0 = 1.0$ MV/m along $y$. Write the fields in its rest frame.
$\vec{E} = E_0\hat{y}, \qquad \vec{B} = 0$
A pure electric field.
Find $\gamma$ for an observer moving along $x$ at $0.6c$.
$\gamma = \dfrac{1}{\sqrt{1 - 0.36}} = 1.25$
The boost is across the field.
Transform the electric field.
$E' = \gamma E_0 = 1.25\ \text{MV/m}$
The plates contract toward each other along $x$, so their charge density rises.
Transform the magnetic field.
$B' = \dfrac{\gamma vE_0}{c^2} = \dfrac{1.25 \times 0.6 \times 10^6}{3.00 \times 10^8} = 2.5 \times 10^{-3}\ \text{T}$
The moving plates are sheets of current.
Check the invariant.
$E'^2 - c^2B'^2 = 1.5625 \times 10^{12} - 0.5625 \times 10^{12} = 1.0 \times 10^{12}\ \text{V}^2/\text{m}^2$
Equal to $E_0^2$, as it must be.
Find the Coulomb field of a proton at $1.0$ μm.
$E_0 = \dfrac{8.988 \times 10^9 \times 1.602 \times 10^{-19}}{(1.0 \times 10^{-6})^2} = 1440\ \text{V/m}$
The field if it were at rest.
Let it move with $\gamma = 5$. Find $\beta$.
$\beta = \sqrt{1 - \tfrac{1}{25}} = 0.980$
Close to light speed.
Find the field broadside at $1.0$ μm.
$E_\perp = \gamma E_0 = 7200\ \text{V/m}$
Enhanced by $\gamma$.
Find the field straight ahead at $1.0$ μm.
$E_\parallel = \dfrac{E_0}{\gamma^2} = 57.6\ \text{V/m}$
Reduced by $\gamma^2$.
Estimate the pulse duration at an impact parameter of $1.0$ μm.
$\Delta t \approx \dfrac{b}{\gamma v} = \dfrac{10^{-6}}{5 \times 0.980 \times 3.00 \times 10^8} = 6.8 \times 10^{-16}\ \text{s}$
A sharp kick lasting less than a femtosecond.
A region has $E = 90$ MV/m along $y$ and $B = 0.50$ T along $z$. Check the invariants.
$\vec{E}\cdot\vec{B} = 0, \qquad E^2 - c^2B^2 = 8.1 \times 10^{15} - 2.25 \times 10^{16} < 0$
Magnetic-type fields.
Find the boost that removes the electric field.
$v = \dfrac{E}{B} = \dfrac{9.0 \times 10^7}{0.50} = 1.8 \times 10^8\ \text{m/s}$
Along $\vec{E}\times\vec{B}$, the $x$ direction.
Find $\beta$ and $\gamma$.
$\beta = 0.60, \qquad \gamma = 1.25$
A familiar pair.
Transform the magnetic field.
$B' = \gamma\left(B - \dfrac{vE}{c^2}\right) = 1.25(0.50 - 0.18) = 0.40\ \text{T}$
Since $vE/c^2 = 1.8 \times 10^8 \times 9.0 \times 10^7/9.0 \times 10^{16} = 0.18$ T.
Check with the invariant.
$-c^2B'^2 = -1.44 \times 10^{16} = 8.1 \times 10^{15} - 2.25 \times 10^{16}$
The two frames agree.
Describe a proton's motion in the lab.
$\text{circles in the moving frame, so drifts at } 1.8 \times 10^8\ \text{m/s in the lab}$
The $E\times B$ drift, the same for every charge.
Find the Lorentz factor.
$\gamma = 1.25$
At $\beta = 0.6$.
Transform the electric field.
$E' = 1.25 \times 2.0 = 2.5\ \text{MV/m}$
Enhanced by $\gamma$.
Transform the magnetic field.
A proton at rest makes a field of $4$ V/m at a certain distance. The proton now moves with $\gamma = 3$. What field does it make at the same distance, straight out to the side of its motion?
Complete the worked solution: a charge at rest makes a field of $900$ V/m at some distance. Moving with $\gamma = 2$, find its field at that distance straight ahead, then straight to the side, and how much larger the side field is, in V/m.
Divide by $\gamma^2 = 4$ ahead.
$E_\parallel = \dfrac{900}{4} =$ a
The field thins along the line of motion.
Multiply by $\gamma = 2$ to the side.
$E_\perp = 2 \times 900 =$ b
The field crowds into the transverse plane.
Subtract the two fields.
$E_\perp - E_\parallel =$ c
Their ratio is $\gamma^3 = 8$.
Match each statement about fields under a boost $\vec{v}$ to its expression.
| unchanged by the boost | $\gamma(\vec{E} + \vec{v}\times\vec{B})_\perp$ | $\vec{E}\cdot\vec{B}$ | $E^2 - c^2B^2$ | |
|---|---|---|---|---|
| parallel components | ||||
| perpendicular electric field | ||||
| dot-product invariant | ||||
| difference-of-squares invariant |
A capacitor makes a pure electric field $E_0$ in its rest frame. An observer moves parallel to the plates, across the field, at $\beta = 0.6$, $0.8$, $0.96$ and $0.28$. Fill in $E'/E_0$ and $cB'/E_0$ for each.
| $E'/E_0$ | $cB'/E_0$ | |
|---|---|---|
| $\beta = 0.6$ | ||
| $\beta = 0.8$ | ||
| $\beta = 0.96$ | ||
| $\beta = 0.28$ |
In the lab there is only a magnetic field of $2$ mT. An observer flies across it at $6 \times 10^5$ m/s. What electric field does the observer measure, in V/m?
Answer: V/m
In the lab, a uniform electric field of $90$ MV/m points along $y$ and a uniform magnetic field of $0.5$ T points along $z$. In the frame moving along $x$ in which the electric field vanishes, how strong is the magnetic field, in teslas?
Answer: T
Electrons in an accelerator beam repel each other electrically but, moving together, attract each other magnetically. At a total energy of $1.022$ MeV per electron, what is the net sideways force as a percentage of the electric repulsion alone?
Answer: %
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the worked solution: a charge at rest makes a field of $400$ V/m at some distance. Moving with $\gamma = 2$, find its field at that distance straight ahead, then straight to the side, and how much larger the side field is, in V/m.
Divide by $\gamma^2 = 4$ ahead.
$E_\parallel = \dfrac{400}{4} =$ a
The field thins along the line of motion.
Multiply by $\gamma = 2$ to the side.
$E_\perp = 2 \times 400 =$ b
The field crowds into the transverse plane.
Subtract the two fields.
$E_\perp - E_\parallel =$ c
Their ratio is $\gamma^3 = 8$.
You can transform fields between frames. Explain to someone why magnetism is a relativistic effect.
19. Your turn: a capacitor's field is $E_0 = 2.0$ MV/m. An observer moves across it at $0.6c$. Find the electric and magnetic fields the observer measures., step 3
$B' = \dfrac{1.25 \times 0.6 \times 2.0 \times 10^6}{3.00 \times 10^8} = 5.0 \times 10^{-3}\ \text{T}$
Five millitesla from a purely electric source.