physics2u
Depth
PU-204 / L26 · lecture

How E and B Transform Between Frames

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You must be comfortable with the Lorentz transformation for spacetime coordinates and the definition of the electromagnetic field tensor \( F^{\mu\nu} \) as established in the previous lecture.

Why this matters

Electric and magnetic fields are not separate entities but different aspects of a single physical object viewed from different states of motion.

In classical pre-relativistic physics, electricity and magnetism were treated as distinct phenomena governed by separate laws. However, special relativity reveals that what one observer perceives as a purely electric field, another observer moving relative to the first will perceive as a mixture of electric and magnetic fields. This unification is not merely a mathematical curiosity; it explains why magnetic forces exist at all. Magnetism can be understood as the relativistic partner of electrostatics. Understanding how these fields transform allows us to predict the behaviour of charged particles in accelerators, analyse the fields of moving currents, and appreciate the deep symmetry underlying Maxwell’s equations.

The Geometric View

Imagine the electromagnetic field as a single geometric object, much like a vector in space. Just as a vector’s components change when you rotate your coordinate axes, the components of the electromagnetic field change when you change your frame of reference via a Lorentz boost. In one frame, you might see only an electric field. In another frame moving relative to the first, that same physical situation will appear to have both electric and magnetic components. There is no absolute distinction between "electric" and "magnetic"; the split depends entirely on the observer’s velocity. This perspective dissolves the historical separation between the two forces and shows that they are two faces of the same coin.

Transformation of Parallel Components

Consider a Lorentz boost along the \( x \)-axis with velocity \( v \). The components of the fields parallel to the direction of motion remain invariant. In tensor language, \( E_x \) and \( B_x \) sit in the entries \( F^{01} \) and \( F^{23} \). A boost along \( x \) mixes only the indices 0 and 1, and an entry carrying both of them (or neither) is unchanged.

\[ E'_x = E_x, \quad B'_x = B_x \]

This invariance is physically intuitive. If you move along the line of a uniform electric field generated by a parallel plate capacitor, the charge density and the geometry of the plates do not change in a way that alters the field strength in the direction of motion. Similarly, a solenoid moving along its own axis keeps its axial field: length contraction packs the turns more closely by a factor \( \gamma \), while time dilation lowers the current in each turn by the same factor, so \( nI \) and hence \( B_x \) are unchanged. The parallel components are "decoupled" from the boost.

Mixing of Perpendicular Components

The components perpendicular to the boost direction mix. This is where the relativistic nature of the fields becomes most apparent. The electric field in the new frame depends on both the electric and magnetic fields in the original frame, and vice versa. The mixing is governed by the Lorentz factor \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \).

\[ \begin{aligned} E'_y &= \gamma\left(E_y - vB_z\right) \\ E'_z &= \gamma\left(E_z + vB_y\right) \end{aligned} \]
\[ \begin{aligned} B'_y &= \gamma\left(B_y + \frac{v}{c^2}E_z\right) \\ B'_z &= \gamma\left(B_z - \frac{v}{c^2}E_y\right) \end{aligned} \]

Notice the symmetry: the electric field transforms into a magnetic field and vice versa, scaled by \( v/c^2 \) for the magnetic part. This reciprocal relationship ensures that the speed of light remains invariant and that the Lorentz force law \( \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \) is consistent across all inertial frames. The term \( \vec{v} \times \vec{B} \) in the electric transformation is the familiar motional EMF, while the term \( \vec{v} \times \vec{E}/c^2 \) in the magnetic transformation is less commonly encountered but equally fundamental.

Physical Interpretation: Magnetism as Relativity

A powerful way to see the unity of electricity and magnetism is to consider a pure electric field in one frame. Suppose a point charge is at rest in frame \( S \). It produces a radial electric field \( \vec{E} \) and no magnetic field \( \vec{B} = 0 \). Now consider an observer in frame \( S' \) moving with velocity \( \vec{v} \) relative to the charge. According to the transformation laws, this observer will measure a magnetic field:

\[ \vec{B}' = -\gamma \frac{\vec{v} \times \vec{E}}{c^2} \]

Since \( \vec{E} \) is radial and \( \vec{v} \) points along a fixed direction, the cross product \( \vec{v} \times \vec{E} \) is perpendicular to both, resulting in a magnetic field that circulates around the direction of motion. This is precisely the magnetic field of a moving charge. In this sense magnetism is the relativistic companion of electrostatics, an effect arising from the transformation of the electric field when the source charge is in motion relative to the observer. This insight resolves the historical puzzle of why magnetic forces exist: they are the "shadow" of electric forces cast by the geometry of spacetime.

The derivation

◆ DERIVATION EMBED · D-176 · not re-derived here

Relativistic Transformation of E and B

Starting from the covariant transformation law of the electromagnetic field tensor, \( F'^{\mu\nu} = \Lambda^{\mu}{}_{\alpha}\Lambda^{\nu}{}_{\beta}F^{\alpha\beta} \), we derive the explicit mixing of the electric and magnetic fields under a Lorentz boost of speed \( v \) along the \( x \)-axis: the field components parallel to the boost are unchanged, \( E'_{\parallel} = E_{\parallel} \), \( B'_{\parallel} = B_{\parallel} \), while the perpendicular components mix as \( \vec{E}'_{\perp} = \gamma\left(\vec{E} + \vec{v}\times\vec{B}\right)_{\perp} \) and \( \vec{B}'_{\perp} = \gamma\left(\vec{B} - \dfrac{\vec{v}\times\vec{E}}{c^{2}}\right)_{\perp} \). A pure electric field in one frame is therefore an electric and magnetic field in another: magnetism is the relativistic residue of moving charge.

\[ \begin{aligned} E'_x &= E_x & E'_y &= \gamma\left(E_y - vB_z\right) & E'_z &= \gamma\left(E_z + vB_y\right) \\ B'_x &= B_x & B'_y &= \gamma\left(B_y + \tfrac{v}{c^{2}}E_z\right) & B'_z &= \gamma\left(B_z - \tfrac{v}{c^{2}}E_y\right) \end{aligned} \]
Open full derivation →
Tier 3 · watch out

A common misconception is that the transformation laws imply that electric and magnetic fields are "real" only in their own frames. In fact, the invariants \( E^2 - c^2B^2 \) and \( \vec{E} \cdot \vec{B} \) are the same in all frames. If \( \vec{E} \cdot \vec{B} = 0 \) in one frame, it is zero in all frames. This means that if a field configuration is purely electric in one frame (where \( \vec{B} = 0 \)), it cannot be transformed into a purely magnetic field in another frame; it will always have an electric component. The distinction between "purely electric" and "purely magnetic" is frame-dependent, but the existence of a non-zero electric component is invariant if \( E^2 > c^2B^2 \). C

Worked example

Problem. Crossed fields: in frame \(S\) there is a uniform \(\mathbf{E} = 3.0 \times 10^{4}\ \hat{\mathbf{y}}\ \mathrm{V\,m^{-1}}\) and \(\mathbf{B} = 2.0 \times 10^{-4}\ \hat{\mathbf{z}}\ \mathrm{T}\). Find the boost along \(x\) that removes the electric field, the magnetic field in that frame, and check the invariant \(E^2 - c^2B^2\). Take \(c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}\).

  1. Invariants first: \(\mathbf{E}\cdot\mathbf{B} = 0\) and \(E^2 - c^2B^2 = 9.0 \times 10^{8} - 3.6 \times 10^{9} = -2.7 \times 10^{9}\ \mathrm{V^2\,m^{-2}} \lt 0\), so a frame with a purely magnetic field exists.
  2. Set \(E'_y = \gamma(E_y - vB_z) = 0\): \(v = E/B = 3.0 \times 10^{4}/(2.0 \times 10^{-4}) = 1.5 \times 10^{8}\ \mathrm{m\,s^{-1}} = 0.5c\), so \(\gamma = 1/\sqrt{0.75} = 1.155\).
  3. Magnetic field: \(B'_z = \gamma(B_z - vE_y/c^2) = 1.155\,(2.0 \times 10^{-4} - 0.5 \times 10^{-4}) = 1.73 \times 10^{-4}\ \mathrm{T}\).
  4. Check: \(-c^2B'^2 = -9.0 \times 10^{16} \times 3.0 \times 10^{-8} = -2.7 \times 10^{9}\ \mathrm{V^2\,m^{-2}}\), the same invariant.
  5. Physically, \(v = E/B\) is the \(\mathbf{E}\times\mathbf{B}\) drift speed: a charge released at rest drifts at this velocity, because in the drifting frame it sees only a magnetic field and simply circles.
\[ v = \frac{E}{B} = 0.5c, \qquad \mathbf{E}' = 0, \qquad \mathbf{B}' = 1.73 \times 10^{-4}\ \hat{\mathbf{z}}\ \mathrm{T} \]

Units check. \(E/B\) has units \(\mathrm{V\,m^{-1}}/(\mathrm{V\,s\,m^{-2}}) = \mathrm{m\,s^{-1}}\), a speed; \(vE/c^2\) has units \(\mathrm{V\,s\,m^{-2}} = \mathrm{T}\).

CHECK YOURSELF

Three quick tests

  1. If \( \vec{E} \) and \( \vec{B} \) are perpendicular in one frame, are they necessarily perpendicular in another?
    AnswerYes, because \( \vec{E} \cdot \vec{B} \) is a Lorentz invariant. If it is zero in one frame, it is zero in all frames.
  2. Can a purely magnetic field in one frame be transformed into a purely electric field in another?
    AnswerNo. If \( B^2 > E^2/c^2 \) in one frame, the field is "magnetic-like" in all frames. It will always have a magnetic component, though the electric component may change.
  3. Why does the parallel component of the field not change under a boost?
    Answer\( E_x \) and \( B_x \) live in the tensor entries \( F^{01} \) and \( F^{23} \); a boost along \( x \) mixes only indices 0 and 1, which leaves both entries unchanged.
PROBLEMS

Problem set for PU-204 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-176 · Threads: force, energy, light