Electromagnetic Field Transformation
A insight of special relativity is the recognition that electric and magnetic fields are not independent entities but rather intertwined components of a single, unified structure: the electromagnetic field.
The apparent character of this field—whether it is perceived as purely electric, purely magnetic, or a combination of both—depends on the inertial reference frame of the observer.
Albert Einstein, in his 1905 paper “On the Electrodynamics of Moving Bodies,” introduced this concept by questioning an apparent asymmetry in the standard formulation of Maxwell’s theory.
It is known that Maxwell’s electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take, for example, the reciprocal electrodynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet, whereas the customary view draws a sharp distinction between the two cases in which either the one or the other of these bodies is in motion.
This example is involving a magnet and an electric conductor such a wire and it can be simulated to the problem of a charged particle, contained in a wire, in a field of a magnet, with the wire that is moving in the rest frame of the magnet. The wire is oriented along the y axis and it is moving along the x axis, and an electron in the wire is dragged along. There is also a uniform magnetic field B_z acting on the system.
The electron feels a Lorentz force equal to the charge e times the cross product of its velocity v with the magnetic field B_z, so the force will be perpendicular to both of these vectors, and directed upward, with a current that, based on the convention of negative charge of an electron, flows downward.
That is the way how the laboratory observer describe the situation: the effective current is produced by the motion of the charge in a magnetic field.
Let’s now consider the same physical situation in the frame of the electron: in this frame, the electron is at rest, and the magnet moves to the left with a velocity v.
Since the electron is at rest, the force that acts on it should be due to an electric field. Since the two situation are describing the same phenomenon, one given by a pure magnetic field, the other given by a pure electric field, the only possible conclusion is that a moving magnet must create an electric field.
A magnet whose field is oriented in the z (or z^\prime) direction and moving to the left, it creates a downward electric field, which exerts an upward force eE_y on a stationary electron (the force on an electron is opposite to the electric field because it has a negative charge).
To explore this, we analyze the canonical problem Einstein described. Consider two inertial reference frames. The first is the laboratory frame, denoted as S. The second frame, S^\prime, moves with a constant velocity \mathbf{v} = v \hat{\mathbf{x}} relative to S.
Scenario 1: The Laboratory Frame (S)
In frame S, a uniform and static magnetic field is directed along the z-axis, \mathbf{B} = B_z \hat{\mathbf{k}}. An electron, representing a charge within a conducting wire, moves with velocity \mathbf{v} = v \hat{\mathbf{x}}. In this frame, there is no electric field, \mathbf{E} = \mathbf{0}. The electron experiences a force governed by the Lorentz force law:
\mathbf{F} = e(\mathbf{E} + \mathbf{v} \times \mathbf{B}) = e(\mathbf{v} \times \mathbf{B})
Substituting the given vectors, we find:
\mathbf{F} = e(v \mathbf{i} \times B_z \mathbf{k}) = e v B_z (\mathbf{i} \times \mathbf{k}) = -e v B_z \mathbf{j}
“The mathematical expression \mathbf{F} = -e v B\_z \mathbf{j} evaluates to an upward force. This is because the charge e is a negative number, which makes the scalar coefficient -e v B\_z positive, resulting in a force vector in the positive \mathbf{j} direction.
Scenario 2: The Electron’s Rest Frame (S^\prime)
Now, let us analyze the situation from the perspective of the frame S^\prime, which is co-moving with the electron. In this frame, the electron’s velocity is zero (\mathbf{v^\prime} = \mathbf{0}). The magnet, which was at rest in S, is now observed to move with velocity -\mathbf{v} = -v \hat{\mathbf{x}}.
Since the electron is stationary in S^\prime, it cannot experience a magnetic force, as the term \mathbf{v^\prime} \times \mathbf{B^\prime} is zero. However, the physical outcome—the force exerted on the electron—must be consistent across all inertial frames. Therefore, in frame S^\prime, the electron must experience a purely electric force:
\mathbf{F^\prime} = e(\mathbf{E^\prime} + \mathbf{v^\prime} \times \mathbf{B^\prime}) = e\mathbf{E^\prime}
This implies that the magnetic field present in frame S must manifest, at least in part, as an electric field \mathbf{E^\prime} in frame S^\prime.
To formalize the transformation between \mathbf{E} and \mathbf{B}, we employ the tensor formalism of special relativity. The electric and magnetic fields are unified into the second-rank, antisymmetric electromagnetic field tensor, F^{\mu\nu}. Its components are given by:
F^{\mu\nu} = \begin{bmatrix} 0 & +E_x & +E_y & +E_z \\ -E_x & 0 & +B_z & -B_y \\ -E_y & -B_z & 0 & +B_x \\ -E_z & +B_y & -B_x & 0 \end{bmatrix}
The transformation of this tensor from frame S to S^\prime is governed by the Lorentz transformation matrix, \Lambda. For a boost with velocity v along the x-axis, the matrix and its inverse are:
\begin{aligned} & {\Lambda^\mu}_\nu = \begin{bmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\ & {(\Lambda^{-1})^\mu}_\nu = \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{aligned}
where \beta = v/c and \gamma = (1 - \beta^2)^{-1/2}. The components of the tensor in the new frame, \left(F^\prime\right)^{\mu\nu}, are found using the tensor transformation law:
\left(F^\prime\right)^{\mu\nu} = {\Lambda^\mu}_\sigma {\Lambda^\nu}_\tau F^{\sigma\tau}
Moving charge
In our specific problem, the fields in the laboratory frame S are \mathbf{E} = \mathbf{0} and \mathbf{B} = (0, 0, B_z).
The tensor in frame S is:
F^{\mu\nu} = \left[ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ & & ct & x & y & z \\ \hline 0 & ct & 0 & 0 & 0 & 0 \\ 1 & x & 0 & 0 & +B_z & 0 \\ 2 & y & 0 & -B_z & 0 & 0 \\ 3 & z & 0 & 0 & 0 & 0 \end{array} \right]
The electromagnetic tensor F^{\mu\nu} in this frame has only two non-zero components:
\begin{aligned} & F^{12} = B_z \\ & F^{21} = -B_z \end{aligned}
We now apply the transformation \left(F^\prime\right)^{\mu\nu} = {\Lambda^\mu}_\sigma {\Lambda^\nu}_\tau F^{\sigma\tau} component by component to find the fields in frame S^\prime, and we first compute the Electric Field Components in S^\prime. To simplify the summation all the 14 zero component of F^{\mu\nu} will not be added to each formula, but leaving only the two non-zeros.
For the x-component:
\begin{aligned} \left(E^\prime \right)^x & = \left(F^\prime \right)^{01} = {\Lambda^0}_\sigma {\Lambda^1}_\tau F^{\sigma \tau} \\ & = {\Lambda^0}_1 {\Lambda^1}_2 F^{1 2} + {\Lambda^0}_2 {\Lambda^1}_1 F^{21}\\ & = (-\gamma \beta) (0) F^{1 2} + (0) (\gamma) F^{21} = 0 \end{aligned}
For the y-component:
\begin{aligned} \left(E^\prime \right)^y & = \left(F^\prime \right)^{02} = {\Lambda^0}_\sigma {\Lambda^2}_\tau F^{\sigma \tau} \\ & = {\Lambda^0}_1 {\Lambda^2}_2 F^{1 2} + {\Lambda^0}_2 {\Lambda^2}_1 F^{21}\\ & = (-\gamma \beta) (1) F^{1 2} + (0) (0) F^{21} \\ & = -\beta\gamma F^{12} = \frac{-\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} B^z \end{aligned}
For the z-component:
\begin{aligned} \left(E^\prime \right)^z & = \left(F^\prime \right)^{03} = {\Lambda^0}_\sigma {\Lambda^3}_\tau F^{\sigma \tau} \\ & = {\Lambda^0}_1 {\Lambda^3}_2 F^{1 2} + {\Lambda^0}_2 {\Lambda^3}_1 F^{21}\\ & = (-\gamma \beta) (0) F^{1 2} + (0)(0) F^{21} = 0 \end{aligned}
We can compute the similar transformation for the magnetic field.
For the x component:
\begin{aligned} (B^\prime)^x & = (F^\prime)^{23} = {\Lambda^2}_\sigma {\Lambda^3}_\tau F^{\sigma\tau} \\ & = {\Lambda^2}_1 {\Lambda^3}_2 F^{12} + {\Lambda^2}_2 {\Lambda^3}_1 F^{21} \\ & =(0)(0)F^{12} + (1)(0)F^{21} = 0 \end{aligned}
For the y component:
\begin{aligned} (B^\prime)^y & = (F^\prime)^{31} = {\Lambda^3}_\sigma {\Lambda^1}_\tau F^{\sigma\tau} \\ & = {\Lambda^3}_1 {\Lambda^1}_2 F^{12} + {\Lambda^3}_2 {\Lambda^1}_1 F^{21} \\ & = (0)(0)F^{12} + (0)(\gamma)F^{21} = 0 \end{aligned} For the z component:
\begin{aligned} (B^\prime)^z & = (F^\prime)^{12} = {\Lambda^1}_\sigma {\Lambda^2}_\tau F^{\sigma\tau} \\ & = {\Lambda^1}_1 {\Lambda^2}_2 F^{12} + {\Lambda^1}_2 {\Lambda^2}_1 F^{21} \\ & = (\gamma)(1)F^{12} + (0)(0)F^{21} \\ &= \gamma F^{12} = \gamma B_z = \frac{1}{\sqrt{1 - v^2/c^2}} B_z \end{aligned}
the fields in the moving frame (the primed frame) for this specific scenario are:
\begin{aligned} \mathbf{E^\prime} & = \begin{pmatrix} 0, & -\gamma v B_z, & 0 \end{pmatrix} \\ \mathbf{B^\prime} & = \begin{pmatrix} 0, & 0, & \gamma B_z \end{pmatrix} \end{aligned}
This demonstrates that for the observer in the primed frame, the magnetic field in the z-direction is stronger, and an electric field has appeared in the negative y-direction.
The result confirms our initial physical intuition.
The purely magnetic field in the lab frame S has transformed into a combination of a magnetic field and an electric field in the electron’s rest frame S^\prime.
Let us verify the consistency of the forces. The force on the electron in S^\prime is:
\mathbf{F^\prime} = e\mathbf{E^\prime} = e(0, -\gamma v B_z, 0) = -\gamma e v B_z \hat{\mathbf{j}} This should relate to the force calculated in the original frame, \mathbf{F} = -e v B_z \hat{\mathbf{j}}.
The forces are related by the factor \gamma: F^\prime_y = \gamma F_y. This is the correct relativistic transformation for a force component that is perpendicular to the direction of the boost.
The framework is internally consistent: the appearance of an electric field in the new frame produces exactly the force required by the principles of relativity. The simple observation of a magnet and a conductor in relative motion necessitates the unified and observer-dependent nature of the electromagnetic field.
The procedure of applying (F^\prime)^{\mu\nu} = \Lambda F \Lambda^T can be generalized for arbitrary electric and magnetic fields.
For a Lorentz boost with velocity v along the positive x-axis, the general transformation equations for the components of \mathbf{E} and \mathbf{B} are:
\begin{aligned} E^\prime_x &= E_x \\ E^\prime_y &= \gamma (E_y - v B_z) \\ E^\prime_z &= \gamma (E_z + v B_y) \\ B^\prime_x &= B_x \\ B^\prime_y &= \gamma \left(B_y + \frac{v}{c^2} E_z\right) \\ B^\prime_z &= \gamma \left(B_z - \frac{v}{c^2} E_y\right) \end{aligned}
We can verify that for the initial conditions (\mathbf{E} = \mathbf{0}, \mathbf{B} = (0, 0, B_z)), these general formulas yield the exact results we derived from the tensor transformation.
Moving magnet
Let’s consider now the same experiment from in the frame of the electron at rest, which is equivalent to assume a magnet moving to left with a velocity -v in the x direction.
We have:
\begin{aligned} & \mathbf E = \begin{bmatrix} 0 & E_y & 0 \end{bmatrix} \\ & \mathbf B = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \end{aligned} The electromagnetic tensor in this frame at is:
F^{\mu\nu} = \left[ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ & & ct & x & y & z \\ \hline 0 & ct & 0 & 0 & +E_y & 0 \\ 1 & x & 0 & 0 & 0 & 0 \\ 2 & y & -E_y & 0 & 0 & 0 \\ 3 & z & 0 & 0 & 0 & 0 \end{array} \right] There are only 2 non-zero components to this tensor:
F^{02} = -F^{20} = E_y
… Coming soon second part…
References
EINSTEIN, Albert, 1905. On the Electrodynamics of Moving Bodies. English translation from "Zur elektrodynamik bewegter Korper", prepared by John Walker (1999). Annalen der Physik, 17, 891-921.
SUSSKIND, Leonard and FRIEDMAN, Art, 2018. Special Relativity and Classical Field Theory - The Theoretical Minimum. London: Penguin Books Ltd. ISBN 978-0-14-198501-5.