The Magnetic Force as a Relativistic Effect
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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 electromagnetic field tensor, not separate entities.
Why this matters
Magnetism is not a fundamental force distinct from electricity; it is the relativistic shadow of the electric force.
Historically, electric and magnetic forces were treated as separate phenomena until Maxwell unified them. However, the deeper unification comes from Special Relativity. When we observe a charged particle moving relative to a source charge, the "magnetic" force we calculate in our frame is actually a consequence of the length contraction and time dilation affecting the electric field distribution. This lecture demonstrates that the magnetic force on a moving charge is a relativistic correction to the electric force, emerging naturally from the transformation of the electromagnetic field tensor.
The Geometric Origin of Magnetism
In classical Newtonian mechanics, forces are absolute. If you push a box, everyone agrees on the force. In Special Relativity, forces are frame-dependent. Consider a simple scenario: a line of positive charges at rest in frame \( S \). To an observer in \( S \), these charges exert a purely electric force on a test charge. There is no magnetic field because the source charges are stationary.
Now, imagine an observer in frame \( S' \) moving parallel to the line of charges. In \( S' \), the source charges are moving. Moving charges constitute a current. We know from classical electromagnetism that currents generate magnetic fields. Therefore, the observer in \( S' \) detects both an electric field (due to the charge density) and a magnetic field (due to the current). The "magnetic" part of the force experienced by a test charge in \( S' \) is not a new fundamental interaction; it is the manifestation of the electric field in a different inertial frame, distorted by the Lorentz contraction of the charge distribution.
Field Transformation and the Emergence of B
The precise relationship between the electric field \( \mathbf{E} \) and magnetic field \( \mathbf{B} \) in different frames is given by the field transformation laws. For a boost with velocity \( \mathbf{v} \) along the \( x \)-axis, the transverse components of the fields mix. If a frame \( S \) contains only an electric field \( \mathbf{E} \) and no magnetic field (\( \mathbf{B} = 0 \)), an observer in \( S' \) moving with velocity \( \mathbf{v} \) relative to \( S \) will measure a magnetic field.
Using the transformation law for the magnetic field perpendicular to the boost direction, we have:
Since \( \mathbf{B} = 0 \) in the rest frame \( S \), this simplifies to:
This equation is profound. It shows that a magnetic field in one frame is simply the electric field in another frame, viewed through the lens of relativistic kinematics. The factor \( 1/c^2 \) ensures the correct units, and the cross product \( \mathbf{v} \times \mathbf{E} \) indicates that the magnetic field is perpendicular to both the velocity of the frame and the electric field.
The Magnetic Force as a Relativistic Correction
Let us consider the force on a test charge \( q \) moving with velocity \( \mathbf{u} \) in frame \( S' \). The Lorentz force law in \( S' \) is:
We substitute the transformed fields. For low velocities (\( v \ll c \)), we can approximate \( \gamma \approx 1 \). The electric field transforms as \( \mathbf{E}' \approx \mathbf{E} \) (for components perpendicular to the boost, to first order). The magnetic field is \( \mathbf{B}' \approx -\frac{\mathbf{v} \times \mathbf{E}}{c^2} \).
The magnetic part of the force is:
Using the vector triple product identity \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B}) \), and assuming \( \mathbf{u} \) is parallel to \( \mathbf{v} \) for simplicity, we find that the magnetic force is proportional to \( v u / c^2 \). This term is of order \( v^2/c^2 \) relative to the electric force. Thus, magnetism is a small relativistic correction to the electric force, becoming significant only when velocities are a substantial fraction of the speed of light.
Consistency with the Field Tensor
The consistency of this picture is guaranteed by the structure of the electromagnetic field tensor \( F^{\mu\nu} \). The electric and magnetic fields are not independent vectors but components of a single antisymmetric tensor. The transformation of this tensor under Lorentz boosts automatically ensures that the laws of physics (Maxwell's equations) are covariant.
If we write the Lorentz force in covariant form as \( \frac{dp^\mu}{d\tau} = q F^{\mu\nu} u_\nu \), the separation into electric and magnetic parts is purely a matter of the observer's frame. The "magnetic force" is the spatial component of the four-force arising from the time-space components of the tensor \( F^{\mu\nu} \) (which correspond to the electric field) when the particle has spatial velocity. This geometric view eliminates the paradox of "why" a magnetic field exists; it is simply the electric field seen from a moving frame.
The derivations
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.
The Electromagnetic Field Tensor
Assembles the electric and magnetic fields into the antisymmetric rank-2 field tensor whose transformation unifies them as one geometric object.
Open full derivation →A common misconception is that the magnetic force does no work. While \( \mathbf{F}_{\text{mag}} \cdot \mathbf{v} = 0 \) in a single inertial frame, the total power delivered by the electromagnetic field is frame-dependent. In a frame where both \( \mathbf{E} \) and \( \mathbf{B} \) are present, the electric field may do work, and the "magnetic" part of the force in another frame may appear to change the kinetic energy due to the transformation of the velocity vector itself. The invariant quantity is the change in the particle's proper time and four-momentum, not the instantaneous power in a specific frame. C
Worked example
Problem. An infinite line of charge with \( \lambda = 5.0 \times 10^{-6} \ \mathrm{C/m} \) is at rest in frame \( S \). A test charge \( q = 2.0 \times 10^{-9} \ \mathrm{C} \) moves parallel to the line at \( v = 0.6c \), a distance \( r = 0.10 \ \mathrm{m} \) away. Find the force on the test charge in \( S \), and then in the test charge's own rest frame \( S' \). Use \( 1/(4\pi\epsilon_0) = 9.0 \times 10^9 \ \mathrm{N \cdot m^2/C^2} \) and \( c = 3.0 \times 10^8 \ \mathrm{m/s} \).
- Field in \( S \): \( E = \dfrac{\lambda}{2\pi\epsilon_0 r} = \dfrac{2(9.0 \times 10^9)(5.0 \times 10^{-6})}{0.10} = 9.0 \times 10^5 \ \mathrm{N/C} \), pointing radially away from the line. The line is at rest, so \( \mathbf{B} = 0 \) in \( S \).
- Force in \( S \): purely electric, \( F = qE = (2.0 \times 10^{-9})(9.0 \times 10^5) = 1.8 \times 10^{-3} \ \mathrm{N} \). The magnetic force is zero because there is no magnetic field in this frame.
- In \( S' \) the line moves at \( 0.6c \) (\( \gamma = 1.25 \)). Its charge density is \( \lambda' = \gamma\lambda = 6.25 \times 10^{-6} \ \mathrm{C/m} \), so \( E' = \gamma E = 1.125 \times 10^6 \ \mathrm{N/C} \). The moving line also carries a current and so produces \( B' = \gamma v E/c^2 = 2.25 \times 10^{-3} \ \mathrm{T} \), but the test charge is at rest in \( S' \) and feels no magnetic force.
- Force in \( S' \): \( F' = qE' = 2.25 \times 10^{-3} \ \mathrm{N} = \gamma F \), exactly the transformation rule for a force perpendicular to the velocity. The two frames describe one interaction, split differently into electric and magnetic parts.
Units check. \( \lambda/(\epsilon_0 r) \) is \( (\mathrm{C/m})/((\mathrm{C^2/(N\,m^2)})\,\mathrm{m}) = \mathrm{N/C} \); \( qE \) is in newtons; \( vE/c^2 \) is \( (\mathrm{m/s})(\mathrm{V/m})/(\mathrm{m^2/s^2}) = \mathrm{V\,s/m^2} = \mathrm{T} \).
Three quick tests
- Can a frame exist where the magnetic field is zero but the electric field is non-zero?
Answer
Yes, provided \( \mathbf{E} \cdot \mathbf{B} = 0 \) and \( E^2 - c^2 B^2 > 0 \) (both are invariants). In that frame the field is purely electric. - If \( \mathbf{E} \) and \( \mathbf{B} \) are perpendicular and \( E = cB \), what is the nature of the field?
Answer
It is a radiation field (null field). No frame exists where either \( \mathbf{E} \) or \( \mathbf{B} \) is zero. - Why is the magnetic force on a charge moving parallel to a current-carrying wire a relativistic effect?
Answer
It arises from the length contraction of the charge densities in the wire, which creates a net electric field in the charge's rest frame, equivalent to a magnetic force in the lab frame.
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.
Derivations cited: D-105 · D-104 · Threads: force, energy, light