How E and B Fields Transform Between Frames
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You should be comfortable with the Lorentz transformation for coordinates and understand that electric and magnetic fields are components of a single antisymmetric tensor, \(F^{\mu\nu}\).
Why this matters
Electric and magnetic fields are not separate entities but different projections of a single electromagnetic field tensor, depending on the observer's motion.
In classical Galilean physics, electric and magnetic fields were thought to be distinct phenomena. However, Maxwell's equations are not invariant under Galilean transformations. Special relativity resolves this by showing that what one observer identifies as a purely electric field, another moving observer will perceive as a mixture of electric and magnetic fields. This unification is not a mathematical trick; it is a fundamental geometric property of spacetime. Understanding these transformations allows us to predict how electromagnetic forces appear to different observers and provides a deeper insight into the nature of electromagnetism as a relativistic phenomenon.
The Geometric View
Imagine a static electric charge at rest in frame \(S\). In this frame, there is only an electric field \(\mathbf{E}\) radiating outward, and the magnetic field \(\mathbf{B}\) is zero. Now consider an observer in frame \(S'\) moving with velocity \(\mathbf{v}\) relative to \(S\). To this observer, the charge is moving. A moving charge constitutes a current, and currents generate magnetic fields. Therefore, the observer in \(S'\) must detect a non-zero magnetic field \(\mathbf{B}'\). This is not a contradiction; it is a consequence of the relativity of simultaneity and the structure of spacetime. The "electric" and "magnetic" parts of the field are like the \(x\) and \(y\) components of a vector in Euclidean space: rotating your coordinate system (or changing your inertial frame) mixes these components. The total electromagnetic field is invariant, but its decomposition into \(\mathbf{E}\) and \(\mathbf{B}\) is frame-dependent.
Transformation of Transverse Components
Consider a boost along the \(x\)-axis with speed \(v\). Let \(\beta = v/c\) and \(\gamma = (1 - \beta^2)^{-1/2}\). The components of the fields parallel to the boost direction remain unchanged. This is because the Lorentz transformation does not mix time and space coordinates in the direction of motion in the same way it mixes transverse coordinates. Thus, the longitudinal components are invariant:
The transverse components, however, mix. The electric field in the new frame depends on both the electric and magnetic fields in the original frame. Specifically, the transverse electric field is given by:
Similarly, the transverse magnetic field transforms as:
These equations show that a pure electric field in \(S\) (where \(\mathbf{B} = 0\)) will appear in \(S'\) as a combination of an electric field \(\mathbf{E}'_{\perp} = \gamma \mathbf{E}_{\perp}\) and a magnetic field \(\mathbf{B}'_{\perp} = -\gamma \frac{1}{c^2} \mathbf{v} \times \mathbf{E}\). The magnetic field is perpendicular to both the velocity of the frame and the electric field.
Invariants and Field Strength
While the individual components of \(\mathbf{E}\) and \(\mathbf{B}\) change between frames, certain combinations remain invariant. These invariants are crucial for classifying electromagnetic fields. The first invariant is the difference between the squared magnitudes of the fields:
The second invariant is the dot product of the fields:
If \(I_1 > 0\), the field is "electric-like," meaning there exists a frame where the magnetic field vanishes. If \(I_1 < 0\), the field is "magnetic-like," and there exists a frame where the electric field vanishes. If \(I_1 = 0\) and \(I_2 = 0\), the field is a plane wave (like light), and no frame exists where either \(\mathbf{E}\) or \(\mathbf{B}\) is zero. These invariants provide a frame-independent characterization of the electromagnetic field, much like the length of a vector is invariant under rotation.
Physical Interpretation
The transformation laws can be understood physically through the concept of charge density and current density. In the rest frame of a charge distribution, there is a charge density \(\rho\) but no current density \(\mathbf{J}\). In a moving frame, the charge density transforms as \(\rho' = \gamma \rho\) due to length contraction, and a current density \(\mathbf{J}' = \rho' \mathbf{v}\) appears. Since currents generate magnetic fields, the appearance of \(\mathbf{B}'\) in the moving frame is a direct consequence of the relativistic transformation of charge and current. This perspective reinforces the idea that magnetism is a relativistic effect arising from the motion of charges, rather than a separate fundamental force.
The derivation
Transformation of Electric and Magnetic Fields
For a boost of speed v along the common x-axis relating inertial frames S and S′ (with β = v/c, γ = (1 − β2)−1/2), the electric and magnetic fields transform by mixing their transverse components. In particular a field that is purely electric in S is measured in S′ as an electric field and a magnetic field, with B′z = −γ v Ey/c2.
A common misconception is that the magnetic field is "caused" by the electric field in the sense of a causal mechanism. In reality, both are aspects of the same tensor \(F^{\mu\nu}\). The transformation is a change of basis in spacetime, not a dynamic process. The fields do not "generate" each other; they are different components of a single geometric object that appear differently to different observers. C
Worked example
Problem. A point charge \(Q = 1.0 \times 10^{-6}\ \mathrm{C}\) is at rest at the origin of frame \(S\). An observer in frame \(S'\) moves with speed \(v = 0.6c\) along the \(x\)-axis. Calculate the electric and magnetic fields at a point \(P\) with coordinates \((0, 1.0\ \mathrm{m}, 0)\) in frame \(S\).
- Determine the fields in frame \(S\). Since the charge is at rest, \(\mathbf{B} = 0\). The electric field at \(P\) is radial. The distance \(r = 1.0\ \mathrm{m}\). Using Coulomb's law, \(E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}\). With \(\frac{1}{4\pi\epsilon_0} \approx 9.0 \times 10^9\ \mathrm{N\,m^2/C^2}\), \(E = 9.0 \times 10^9 \times \frac{1.0 \times 10^{-6}}{1.0^2} = 9.0 \times 10^3\ \mathrm{N/C}\). The direction is along the \(y\)-axis, so \(\mathbf{E} = 9.0 \times 10^3\ \hat{\mathbf{j}}\ \mathrm{N/C}\).
- Identify the components relative to the boost direction (\(x\)-axis). \(E_x = 0\), \(E_y = 9.0 \times 10^3\ \mathrm{N/C}\), \(E_z = 0\). \(B_x = B_y = B_z = 0\).
- Calculate \(\gamma\). \(\beta = 0.6\), so \(\gamma = \frac{1}{\sqrt{1 - 0.6^2}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25\).
- Apply the transformation for \(E'_y\). Since \(B = 0\), \(E'_y = \gamma E_y = 1.25 \times 9.0 \times 10^3 = 1.125 \times 10^4\ \mathrm{N/C}\).
- Apply the transformation for \(B'_z\). Using \(\mathbf{B}'_{\perp} = -\gamma \frac{1}{c^2} \mathbf{v} \times \mathbf{E}\). Here \(\mathbf{v} = v \hat{\mathbf{i}}\) and \(\mathbf{E} = E_y \hat{\mathbf{j}}\). \(\mathbf{v} \times \mathbf{E} = v E_y (\hat{\mathbf{i}} \times \hat{\mathbf{j}}) = v E_y \hat{\mathbf{k}}\). Thus, \(B'_z = -\gamma \frac{v E_y}{c^2}\). Note that \(v/c = \beta\), so \(v/c^2 = \beta/c\). \(B'_z = -\gamma \beta \frac{E_y}{c}\). Substituting values: \(B'_z = -1.25 \times 0.6 \times \frac{9.0 \times 10^3}{3.0 \times 10^8} = -0.75 \times 3.0 \times 10^{-5} = -2.25 \times 10^{-5}\ \mathrm{T}\).
Units check. \(E\) is in \(\mathrm{N/C}\). \(B\) is in \(\mathrm{T}\) (\(\mathrm{N/(A\,m)}\)). The term \(\frac{v E}{c^2}\) has units \(\frac{(\mathrm{m/s})(\mathrm{N/C})}{(\mathrm{m/s})^2} = \frac{\mathrm{N}}{\mathrm{C} \cdot (\mathrm{m/s})} = \frac{\mathrm{N}}{\mathrm{A\,m}} = \mathrm{T}\). The units are consistent.
Three quick tests
- If a frame moves such that \(\mathbf{E} \cdot \mathbf{B} \neq 0\), can you find a frame where \(\mathbf{B} = 0\)?
Answer
No. If \(\mathbf{E} \cdot \mathbf{B} \neq 0\), the field is neither purely electric nor purely magnetic in any frame. Both fields are non-zero in all inertial frames. - What is the value of the invariant \(E^2 - c^2 B^2\) for a plane electromagnetic wave?
Answer
Zero. For a plane wave, \(E = cB\), so \(E^2 - c^2 B^2 = 0\). This is a characteristic of radiation fields. - Why does a static charge produce a magnetic field in a moving frame?
Answer
Because the moving observer sees the charge as moving, creating a current. Currents generate magnetic fields. This is a consequence of the relativistic transformation of charge density and current density.
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.