Lorentz Covariance of Maxwell's Equations
April 2025
It is (famously!) possible to show that Maxwell’s equations imply the electric & magnetic fields follow a wave equation . Additionally, Maxwell’s equations are Lorentz covariant . If Maxwell’s equations are Lorentz covariant, then the wave equation must be too; let’s also show that this is the case, and that it violates Galilean relativity . Lastly let’s verify the Lorentz covariance of Maxwell’s equations.
This article will make more sense after reading the previous article: Equation Covariance under Coordinate Transforms.
Contents
Relativity & Inertial Frames
An inertial reference frame is a frame which is not experiencing any acceleration, and so moves with constant velocity. The two relevant transforms for this are the Galilean transform and the Lorentz transform . They can be derived from first principles (without reference to the speed of light) as per “Nothing But Relativity” - P. Pal .
The laws of physics have no preferred inertial reference frame (as far as we know), which means they should be covariant under boosts.
Notation & Basic Properties
Say we are in an inertial reference frame \(S\) with coordinates \(x\) and \(t\). Now consider a second inertial reference frame \(S^\prime\) with coordinates \(x^\prime\) and \(t^\prime\) which moves with velocity \(u\) relative to \(S\). Denote the operator which transforms between \(S\) coordinates and \(S^\prime\) coordinates as \(\Lambda_u\), or in vector notation \(\mathbf x^\prime = \Lambda_u \mathbf x.\)
Some basic properties that we know the Galilean & Lorentz transforms must obey are
| Property | Meaning | |
|---|---|---|
| Identity | \(\Lambda_0 = \mathbf I\) | If \(u=0\) the transform must have no effect. |
| Inverse | \(\Lambda^{-1}_u = \Lambda_{-u}.\) | The inverse of the transform exists and must be given by the forward transform with \(-u.\) |
It feels, perhaps, like the identity property should follow from the inverse property applied to \(u=0\). However that only tells us that \(\Lambda_0\) is involutory , not that it is the identity.
Gradient Transformations
As per the previous article the gradients are transformed from \(S\) to \(S^\prime\) by using the chain-rule:
$$\partial_\mu \phi(\mathbf x) = \sum_\nu \partial_\nu^\prime \phi^\prime(\mathbf x^\prime) \partial_\mu x^\prime_\nu$$That is
$$\begin{align} \partial_x \phi &= \partial_x x^\prime \partial_{x^\prime} \phi^\prime + \partial_x t^\prime \partial_{t^\prime} \phi^\prime \\ \partial_t \phi &= \partial_t x^\prime \partial_{x^\prime} \phi^\prime + \partial_t t^\prime \partial_{t^\prime} \phi^\prime \end{align}$$We will calculate these transformations for the Galilean & Lorentz transformations below.
Galilean Transformation
Galilean relativity has an absolute time but varying spatial coordinates. All observers agree on when events took place; the order & simultaneity of events is agreed upon.
$$\begin{align} x^\prime &= x - ut &\qquad x &= x^\prime + ut\\ t^\prime &= t &\qquad t &= t^\prime \end{align}$$It should be trivially clear that both the identity and inverse properties are obeyed by the Galilean transform.
The gradient transforms are given by
$$\begin{align} \partial_x\phi &= \partial_{x^\prime}\phi^\prime \\ \partial_t\phi &= \partial_{t^\prime}\phi^\prime - u \partial_{x^\prime}\phi^\prime \end{align}$$This result can be intuitively understood by considering a function which is stationary in \(S^\prime\), i.e. \(\phi^\prime(x^\prime, t^\prime) = f(x^\prime)\). This clearly means that \(\partial_{t^\prime} \phi^\prime = 0.\) In \(S\) the function will, however, appear to be moving and so have non zero time derivative, caused purely by the change of coordinates. Further, how fast the function changes in \(S\) is clearly related to it’s spatial gradient at that point and the velocity \(u\).
Lorentz Transformation
Special relativity uses the Lorentz transformation which has no absolute time. Observers can disagree when events took place; the simultaneity & even order of events is not agreed upon.
$$\begin{align} x^\prime &= \gamma (x - ut) &\qquad x &= \gamma(x^\prime + ut^\prime)\\ t^\prime &= \gamma (t - ux / c^2) &\qquad t &= \gamma(t^\prime + ux^\prime / c^2) \end{align}$$where
$$\gamma = \frac{1}{\sqrt{1 - \tfrac{u^2}{c^2}}}$$When \(u=0\) we have \(\gamma=1\) so clearly the identity rule is obeyed. It is also simple to show the inverse rule holds by summing the equations for \(x^\prime\) and \(t^\prime\):
$$\begin{align} x^\prime + ut^\prime &= \gamma(x-ut) + u\gamma \Big(t - \frac{ux}{c^2}\Big) \\ &= \gamma x \Big(1 - \frac{u^2}{c^2}\Big) \\ &= \gamma^{-1} x \\ \Rightarrow x &= \gamma (x^\prime + ut^\prime) \end{align}$$and similarly
$$\begin{align} \frac{u}{c^2} x^\prime + t^\prime &= \frac{u}{c^2}\gamma(x - ut) + \gamma\Big(t - \frac{ux}{c^2}\Big) \\ &= \gamma t \Big(1 - \frac{u^2}{c^2}\Big) \\ &= \gamma^{-1} t \\ \Rightarrow t &= \gamma(t^\prime + ux^\prime / c^2) \end{align}$$Notice that if \(u\ll c\) we have \(\gamma \approx 1\), \(u/c^2\approx 0\) and we recover the Galilean transform. This is why our intuition tells us reality obeys Galilean relativity.
The gradient transformation for the Lorentz transform is a tad more complex than for the Galilean, but the concepts are identical. We have
$$\begin{align} \partial_x x^\prime &= \gamma &\quad \partial_t x^\prime &= -\gamma u \\ \partial_x t^\prime &= -\gamma u/c^2 &\quad \partial_t t^\prime &= \gamma \end{align}$$yielding
$$\begin{align} \partial_t \phi &= \gamma \big( \partial_{t^\prime} \phi^\prime -u\partial_{x^\prime} \phi^\prime \big)\\ \partial_x \phi &= \gamma \big( \partial_{x^\prime} \phi^\prime - \frac{u}{c^2} \partial_{t^\prime} \phi^\prime \big) \end{align}$$The Wave Equation
The wave equation in one dimension is
$$\frac{\partial^2 \psi}{\partial t^2} = c^2 \frac{\partial^2 \psi}{\partial x^2}$$It has solutions which look like
$$\psi = A f(x - ct) + B g(x + ct)$$which is a combination of a right-moving part with profile \(f\) and a left-moving with profile \(g.\) Both ‘waves’ travel with velocity \(c.\)
Let’s just dive in a see how it transforms.
Galilean Boost
To keep things manageable lets consider each side separately. Transforming the time derivative first
$$\begin{align} \partial_{tt}\psi &= (\partial_{t^\prime} - u \partial_{x^\prime})(\partial_{t^\prime} - u \partial_{x^\prime})\psi^\prime \\ &= (\partial_{t^\prime t^\prime} - 2 u \partial_{x^\prime t^\prime} + u^2 \partial_{x^\prime x^\prime})\psi^\prime \\ &= \partial_{t^\prime t^\prime} \psi^\prime - 2 u \partial_{x^\prime t^\prime} \psi^\prime + u^2 \partial_{x^\prime x^\prime}\psi^\prime \end{align}$$The spatial derivative is trivial
$$\begin{align} \partial_{xx}\psi &= \partial_{x^\prime x^\prime}\psi^\prime \end{align}$$Which means the transformed wave equation is given by
$$\begin{align} \partial_{t^\prime t^\prime}\psi^\prime &= (c^2 - u^2)\partial_{x^\prime x^\prime}\psi^\prime + 2u \partial_{x^\prime t^\prime} \psi^\prime \\ &\neq c^2 \partial_{x^\prime x^\prime} \psi^\prime \end{align}$$If \(u = 0\) this does reduce back to the original wave equation; a useful sanity check. But in general this is not the same equation. The wave equation is not invariant under Galilean transformations!
For mechanical waves this should not be too surprising; they are traveling in a medium (e.g. sound waves travel in air) and the rest frame of the medium picks out a preferred internal frame. Indeed if we look at the transformed solution we see it becomes
$$\psi^\prime = A f(x^\prime - (c-u)t) + B g(x^\prime + (c+u)t)$$As expected it is now a combination of a right-moving part with speed \(c - u\) and a left-moving part with speed \(c + u.\)
Lorentz Boost
Exactly as above, plug in the gradient transforms
$$\begin{align} \partial_{tt}\psi &= \gamma^2(\partial_{t^\prime} - u \partial_{x^\prime})(\partial_{t^\prime} - u \partial_{x^\prime})\psi^\prime \\ &= \gamma^2(\partial_{t^\prime t^\prime} + u^2 \partial_{x^\prime x^\prime} - 2 u \partial_{t^\prime x^\prime})\psi^\prime \\ \end{align}$$This time the spatial derivative looks similar to the time derivative
$$\begin{align} c^2\partial_{xx}\psi &= c^2\gamma^2\Big(\partial_{x^\prime} - \frac{u}{c^2} \partial_{t^\prime}\Big)\Big(\partial_{x^\prime} - \frac{u}{c^2} \partial_{t^\prime}\Big)\psi^\prime \\ &= c^2\gamma^2\Big(\partial_{x^\prime x^\prime} + \frac{u^2}{c^4} \partial_{t^\prime t^\prime} - 2 \frac{u}{c^2} \partial_{t^\prime x^\prime}\Big)\psi^\prime \\ \end{align}$$Which means
$$\begin{align} \gamma^2(\partial_{t^\prime t^\prime} + u^2 \partial_{x^\prime x^\prime})\psi^\prime &= c^2\gamma^2\Big(\partial_{x^\prime x^\prime} + \frac{u^2}{c^4} \partial_{t^\prime t^\prime}\Big)\psi^\prime \\ \gamma^2\Big(1 - \frac{u^2}{c^2}\Big)\partial_{t^\prime t^\prime}\psi^\prime &= c^2\gamma^2\Big(1 - \frac{u^2}{c^2}\Big) \partial_{x^\prime x^\prime}\psi^\prime \\ \partial_{t^\prime t^\prime}\psi^\prime &= c^2 \partial_{x^\prime x^\prime}\psi^\prime \quad\square \end{align}$$The wave equation is invariant under Lorentz boosts.
This is (presumably?) related to the fact that the two vectors lying along the 1d light-cone are eigenvectors of the Lorentz transformation. If we visualize the transformation this is clear
The eigenvalues are different but that does not matter because the speed of the wave is equal to the gradient of the light-cone vectors. It is unrelated to their magnitudes.
Maxwell’s Equations
Maxwell’s equations read
$$\begin{align} \nabla\cdot\mathbf E &= \frac{\rho}{\varepsilon_0} &\quad \nabla\cdot\mathbf B &= 0 \tag{1}\label{eq:gauss} \\ \nabla\times\mathbf E &= -\partial_t\mathbf B &\quad \nabla\times\mathbf B &= \mu_0\big(\mathbf J + \varepsilon_0 \partial_t \mathbf E\big) \tag{2}\label{eq:faraday_ampere} \\ \end{align}$$First, let’s derive the wave equation from Maxwell’s equations.
Electromagnetic Waves
The speed of light is \(c\) only in a vacuum, so we can consider free space. In this case both \(\rho\) (the charge density ) and \(\mathbf J\) (the current density ) are zero, simplifying things to
$$\begin{align} \nabla\cdot\mathbf E &= 0 &\quad \nabla\cdot\mathbf B &= 0 \tag{3}\label{eq:gauss_freespace} \\ \nabla\times\mathbf E &= -\partial_t\mathbf B &\quad \nabla\times\mathbf B &= \mu_0\varepsilon_0 \partial_t \mathbf E \tag{4}\label{eq:faraday_ampere_freespace} \\ \end{align}$$The wave equation in 3D reads
$$\partial_{tt}\phi = c^2\nabla^2\phi $$so let’s be guided by this. We need the second time derivative to appear. There’s already a first time derivative in equations \eqref{eq:faraday_ampere_freespace}, so let’s simply take the derivative of these and go from there.
$$\begin{align} \partial_{tt}\mathbf B &= -\partial_t \nabla\times\mathbf E & \partial_{tt}\mathbf E &= \tfrac{1}{\mu_0\varepsilon_0}\partial_t\nabla\times\mathbf B &\quad \text{Differentiate}\\ &= -\nabla\times\partial_t \mathbf E & &= \tfrac{1}{\mu_0\varepsilon_0}\nabla\times\partial_t\mathbf B &\quad\text{Linearity} \\ &= -\tfrac{1}{\mu_0\varepsilon_0} \nabla\times\nabla\times\mathbf B & &= -\tfrac{1}{\mu_0\varepsilon_0}\nabla\times\nabla\times\mathbf E &\text{Substitute \eqref{eq:faraday_ampere_freespace}} \\ &= -\tfrac{1}{\mu_0\varepsilon_0} \big(\nabla \cdot \mathbf B - \nabla^2 \mathbf B\big) & &= -\tfrac{1}{\mu_0\varepsilon_0}\big(\nabla \cdot \mathbf E - \nabla^2 \mathbf E\big) &\quad\text{Identity} \\ &= \tfrac{1}{\mu_0\varepsilon_0} \nabla^2 \mathbf B & &= \tfrac{1}{\mu_0\varepsilon_0}\nabla^2 \mathbf E &\quad\text{Substitute \eqref{eq:gauss_freespace}} \\ \end{align}$$The identity in the penultimate step can be shown easily through direct calculation.
Consider the x-component and churn through the direct calculation:
$$\begin{align} (\nabla\times\nabla\times\mathbf A)_x &= \partial_y (\partial_x A_y - \partial_y A_x) - \partial_z (\partial_z A_x - \partial_x A_z) \\ &= \partial_{yx} A_y - \partial_{yy} A_x - \partial_{zz} A_x + \partial_{zx} A_z \\ &= \partial_{yx} A_y - \partial_{yy} A_x - \partial_{zz} A_x + \partial_{zx} A_z + \partial_{xx} A_x - \partial_{xx} A_x \\ &= \partial_x(\partial_x A_x + \partial_y A_y + \partial_z A_z)-(\partial_{xx} + \partial_{yy} + \partial_{zz}) A_x \\ &= (\nabla (\nabla\cdot\mathbf A))_x - (\nabla^2 \mathbf A)_x \end{align}$$The y and z components end up being the same and so we get the result
$$\nabla\times\nabla\times\mathbf A = \nabla (\nabla \cdot\mathbf A) - \nabla^2 \mathbf A$$Thus all components of both \(\mathbf E\) and \(\mathbf B\) are governed by wave equations with velocity
$$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$$Since Maxwell’s equations imply the wave equation they cannot be Galilean covariant. The wave equation is, however, Lorentz covariant. The fact the wave equation pops out of Maxwell’s equations does not tell us that they are Lorentz covariant, but it is at least not inconsistent with them being Lorentz covariant.
One interpretation of the wave equation breaking Galilean covariance is that the electric and magnetic fields travel in some medium, named the luminiferous aether . This would pick out a special reference frame where the aether is at rest. There are numerous issues with such a hypothesis; the aether would have to have very strange properties and (more importantly) experiments failed to detect any motion relative to the aether. The Wikipedia article has a nice discussion of the issues with such theories.
One objection which I don’t really see mentioned anywhere is that Maxwell’s equations are basically phenomenological, based on the results of experiments carried out on Earth. If motion relative to the aether caused effects (as Galilean relativity says it must) then we would have presumably measured these while formulating Maxwell’s equations. That is to say, if the aether did exist how come we ended up with Maxwell’s equations being in the rest frame of the aether from measurements taken on Earth?
Lorentz Transformation
It’s possible to formulate electromagnetism in a “manifestly covariant” fashion; see Wikipedia . However for me there’s something visceral about seeing it pop out of the equations directly, so here I apply a very direct (“zero-thought”) approach. Also, just directly using the covariant formulation feels like jumping ahead to the answer!
Maxwell’s equations are 8 coupled PDEs which makes directly transforming the equations a bit verbose, but it’s simple enough. Above, we defined the Lorentz transform in the x direction. This is fine to consider without loss of generality because we could always rotate first such that this is true. In this case the gradients for the y and z directions remain unchanged.
For reasons that will become clear, we will denote the transformed physical quantities with calligraphic symbols. By direct substitution we get
Gauss’ Laws:
$$\begin{align} %-- symbols for directly transformed \def\Ep{\mathcal E^\prime} \def\Bp{\mathcal B^\prime} \def\rhop{\varrho^\prime} \def\Jp{\mathcal J^\prime} \def\pp{\partial^\prime} %-- \gamma(\pp_x - \tfrac{u}{c^2}\pp_t) \Ep_x + \pp_y \Ep_y + \pp_z \Ep_z &= \rhop / \varepsilon_0 \tag{5}\label{eq:xformed_gauss_e}\\ \gamma(\pp_x - \tfrac{u}{c^2}\pp_t) \Bp_x + \pp_y \Bp_y + \pp_z \Bp_z &= 0 \tag{6}\label{eq:xformed_gauss_b} \end{align}$$Faraday’s Law:
$$\begin{align} \pp_y \Ep_z - \pp_z \Ep_y &= - \gamma(\pp_t - u\pp_x) \Bp_x \tag{7}\label{eq:xformed_faraday_x} \\ \pp_z \Ep_x - \gamma(\pp_x - \tfrac{u}{c^2}\pp_t) \Ep_z &= -\gamma(\pp_t - u\pp_x) \Bp_y \\ \gamma(\pp_x - \tfrac{u}{c^2}\pp_t) \Ep_y - \pp_y \Ep_x &= -\gamma(\pp_t - u\pp_x) \Bp_z \end{align}$$Ampere’s Law:
$$\begin{align} \pp_y \Bp_z - \pp_z \Bp_y &= \mu_0 \big(\Jp_x + \varepsilon_0 \gamma(\pp_t - u\pp_x) \Ep_x \big) \tag{8}\label{eq:xformed_ampere_x} \\ \pp_z \Bp_x - \gamma(\pp_x - \tfrac{u}{c^2}\pp_t) \Bp_z &= \mu_0 \big(\Jp_y + \varepsilon_0 \gamma(\pp_t - u\pp_x) \Ep_y \big) \\ \gamma(\pp_x - \tfrac{u}{c^2}\pp_t) \Bp_y - \pp_y \Bp_x &= \mu_0 \big(\Jp_z + \varepsilon_0 \gamma(\pp_t - u\pp_x) \Ep_z \big) \end{align}$$These certainly don’t look exactly like Maxwell’s equations. Despair not, it’s possible to rearrange them to a form which does look like Maxwell’s equations in \(S^\prime.\)
Gauss’ Laws
Looking at Gauss’ laws we can see some time derivatives have appeared. We can eliminate these by rearranging the transformed Faraday’s and Ampere’s laws:
$$\begin{align} \mu_0\varepsilon_0 \gamma \pp_t \Ep_x &= \pp_y \Bp_z - \pp_z \Bp_y + \mu_0 (\gamma u \varepsilon_0 \pp_x \Ep_x - \Jp_x) \\ \gamma \pp_t \Bp_x &= \pp_z \Ep_y - \pp_y \Ep_z + \gamma u \pp_x \Bp_x \\ \end{align}$$We can simplify a bit by remembering \(1/\mu_0\varepsilon_0 = c^2\)
$$ \gamma \pp_t \Ep_x = c^2\pp_y \Bp_z - c^2\pp_z \Bp_y + \gamma u\pp_x \Ep_x - \Jp_x / \varepsilon_0 \\ $$then substituting these two
$$\begin{align} \gamma\pp_x \Ep_x - \tfrac{u}{c^2}(c^2\pp_y \Bp_z - c^2\pp_z \Bp_y + \gamma u \pp_x \Ep_x - \Jp_x / \varepsilon_0) + \pp_y \Ep_y + \pp_z \Ep_z &= \rhop / \varepsilon_0 \\ \gamma \pp_x \Bp_x - \tfrac{u}{c^2}(\pp_z \Ep_y - \pp_y \Ep_z + \gamma u \pp_x \Bp_x) + \pp_y \Bp_y + \pp_z \Bp_z &= 0 \end{align}$$Now gather like terms involving gradients in the same axes together
$$\begin{align} \gamma(1 - \tfrac{u^2}{c^2})\pp_x \Ep_x + \pp_y \Ep_y -u \,\pp_y \Bp_z + \pp_z \Ep_z + u \,\pp_z \Bp_y &= \rhop / \varepsilon_0 - \tfrac{u}{c^2} \Jp_x / \varepsilon_0\\ \gamma(1 - \tfrac{u^2}{c^2}) \pp_x \Bp_x + \pp_y \Bp_y + \tfrac{u}{c^2}\pp_y \Ep_z + \pp_z \Bp_z - \tfrac{u}{c^2}\pp_z \Ep_y &= 0 \end{align}$$Multiply by \(\gamma\) and cancel the x-axis prefactor
$$\begin{align} \pp_x \Ep_x + \gamma(\pp_y \Ep_y -u \,\pp_y \Bp_z + \pp_z \Ep_z + u \,\pp_z \Bp_y) &= \gamma(\rhop - \tfrac{u}{c^2} \Jp_x) / \varepsilon_0 \\ \pp_x \Bp_x + \gamma(\pp_y \Bp_y + \tfrac{u}{c^2}\pp_y \Ep_z + \pp_z \Bp_z - \tfrac{u}{c^2}\pp_z \Ep_y) &= 0 \end{align}$$Pull things inside the gradients
$$\begin{align} \pp_x \Ep_x + \pp_y (\gamma ( \Ep_y -u \Bp_z)) + \pp_z(\gamma (\Ep_z + u \Bp_y)) &= \gamma(\rhop - \tfrac{u}{c^2} \Jp_x)/ \varepsilon_0 \\ \pp_x \Bp_x + \pp_y (\gamma ( \Bp_y + \tfrac{u}{c^2} \Ep_z)) + \pp_z (\gamma (\Bp_z - \tfrac{u}{c^2} \Ep_y)) &= 0 \end{align}$$We are at an interesting point here. The equations we have here look kind of like Gauss’ laws but only if we make the following identifications:
$$\begin{align} E^\prime_x &= \Ep_x &\quad B^\prime_x &= \Bp_x \\ E^\prime_y &= \gamma(\Ep_y - u\Bp_z) &\quad B^\prime_y &= \gamma(\Bp_y + \tfrac{u}{c^2}\Ep_z) \\ E^\prime_z &= \gamma(\Ep_z + u\Bp_y) &\quad B^\prime_z &= \gamma(\Bp_z - \tfrac{u}{c^2}\Ep_y) \\ \end{align}$$and
$$\rho^\prime = \gamma(\rhop - \tfrac{u}{c^2}\Jp_x)$$yielding the transformed versions of Gauss’ laws
$$\begin{align} \nabla^\prime\cdot\mathbf E^\prime = \rho^\prime / \varepsilon_0 & \quad & \nabla^\prime\cdot\mathbf B^\prime = 0 \end{align}$$Faraday’s Law
Now let’s tackle Faraday’s law. I found this more natural to do “in reverse”, by which I mean we start with the equations in terms of \(\mathbf E^\prime\) and \(\mathbf B^\prime\) then use their definitions to show these are equivalent to the directly transformed equations involving \(\mathcal{E}^\prime\) and \(\Bp\).
The method is simple; start with the x-component of equation \eqref{eq:faraday_ampere} in frame \(S^\prime\), substitute in the mapping above, then show this is equivalent to the directly transformed equation \eqref{eq:xformed_faraday_x}.
$$\begin{align} \pp_y E^\prime_z - \pp_z E^\prime_y &= -\pp_t B^\prime_x \\ \gamma\pp_y (\Ep_z + u\Bp_y) - \gamma\pp_z (\Ep_y - u\Bp_z) &= -\pp_t \Bp_x \\ \pp_y \Ep_z - \pp_z \Ep_y + u(\pp_y \Bp_y + \pp_z \Bp_z) &= -\gamma^{-1}\pp_t \Bp_x \\ \pp_y \Ep_z - \pp_z \Ep_y - u(\gamma(\pp_x - \tfrac{u}{c^2}\pp_t)\Bp_x) &= -\gamma^{-1}\pp_t \Bp_x \\ \pp_y \Ep_z - \pp_z \Ep_y &= -\gamma((\gamma^{-2} + \tfrac{u^2}{c^2})\pp_t - u \pp_x )\Bp_x \\ \pp_y \Ep_z - \pp_z \Ep_y &= -\gamma(\pp_t - u \pp_x )\Bp_x \quad\square \end{align}$$where we have used
$$\pp_y \Bp_y + \pp_z \Bp_z = -\gamma(\pp_x - \tfrac{u}{c^2}\pp_t)\Bp_x$$from equation \eqref{eq:xformed_gauss_b}, and \((\frac{1}{\gamma^2} + \frac{u^2}{c^2})\equiv 1.\)
The important thing to note here is that the same combinations of the raw transformed components identified above also obey Faraday’s law.
I’ve only shown the x-component here, but the y- and z-component derivations are analogous without even the minor “trick” of having to use the transformed Gauss’ law \eqref{eq:xformed_gauss_b}!
Ampere’s Law
Of course we now turn to Ampere’s law. Since it’s the magnetic equivalent of Faraday’s law the approach is exactly the same. However, in contrast to Faraday’s law, Ampere’s law is inhomogeneous; thus we should expect to have to make a further identification to handle \(\mathbf J^\prime.\) Indeed, this is exactly what happens.
Proceed as above
$$\begin{align} \pp_y B^\prime_z - \pp_z B^\prime_y &= \mu_0(J^\prime_x + \varepsilon_0\pp_t E^\prime_x) \\ \gamma\pp_y (\Bp_z - \tfrac{u}{c^2}\Ep_y) - \gamma\pp_z (\Bp_y - \tfrac{u}{c^2}\Ep_z) &= \mu_0(J^\prime_x + \varepsilon_0\pp_t \Ep_x) \\ \pp_y \Bp_z - \pp_z\Bp_y - \tfrac{u}{c^2}(\pp_y\Ep_y + \pp_z\Ep_z) &= \gamma^{-1}\mu_0(J^\prime_x + \varepsilon_0\pp_t \Ep_x) \\ \pp_y \Bp_z - \pp_z\Bp_y - \tfrac{u}{c^2}(\tfrac{\rhop}{\varepsilon_0} - \gamma(\pp_x - \tfrac{u}{c^2}\pp_t)\Ep_x) &= \gamma^{-1}\mu_0(J^\prime_x + \varepsilon_0\pp_t \Ep_x) \end{align}$$Where this time we have used
$$\pp_y\Ep_y + \pp_z\Ep_z = \tfrac{\rhop}{\varepsilon_0} - \gamma(\pp_x - \tfrac{u}{c^2}\pp_t)\Ep_x$$from equation \eqref{eq:xformed_gauss_b}. Collect like terms
$$\begin{align} \pp_y \Bp_z - \pp_z\Bp_y &= \mu_0\tfrac{J^\prime_x}{\gamma} + \tfrac{u}{c^2}\tfrac{\rhop}{\varepsilon_0} + (\tfrac{\mu_0\varepsilon_0}{\gamma} + \tfrac{u^2}{c^4})\pp_t \Ep_x - \tfrac{u}{c^2}\gamma\pp_x\Ep_x \\ &= \mu_0\tfrac{J^\prime_x}{\gamma} + \mu_0 u \rhop + \mu_0\varepsilon_0\gamma((\tfrac{1}{\gamma^2} + \tfrac{u^2}{c^2})\pp_t - u\pp_x)\Ep_x \\ &= \mu_0(\tfrac{J^\prime_x}{\gamma} + u \rhop + \varepsilon_0\gamma(\pp_t - u\pp_x)\Ep_x) \\ \end{align}$$This is equivalent to equation \eqref{eq:xformed_ampere_x} if we make the identification
$$J^\prime_x = \gamma(\Jp_x - u\rhop)$$We can perform the same procedure easily for the y and z-components, for which we find
$$\begin{align} J^\prime_y &= \Jp_y \\ J^\prime_z &= \Jp_z \\ \end{align}$$Thus we find that certain combinations of the transformed quantities obey all of Maxwell’s equations. Thus proving the Lorentz covariance of the equations.
Four-Current & The Faraday Tensor
We ploughed right ahead and transformed all the quantities and their gradients from \(S\) to \(S^\prime\), and the combinations popped out naturally. What’s going on? Well, had we thought about it, we should have expected something like this to happen.
When transforming between reference frames vectors also transform, for example under rotation the Cartesian components of the velocity vector get “mixed” together (cf. the previous article.) Under rotation of the 3 spatial dimensions we would naturally expect the components of \(\mathbf E\) to combine together, and similarly for \(\mathbf B\) and \(\mathbf J\). Clearly something similar is happening here, but things look a little different we actually end up mixing \(\mathbf E\) with \(\mathbf B\) and \(\mathbf J\) with \(\rho\) and vice-versa.
The charge- and current-densities can be combined into a Lorentz covariant four-vector called the four-current . The mixing of the four components here can be understood intuitively as coming from two effects; the Lorentz transform causes length contraction which affects the charge density, and different relative motion of the charges causes causes different current density.
Charge density and current density are obviously related quantities so the fact they get mixed is expected. The electric and magnetic fields are perhaps a bit more surprising; they are coupled via Maxwell’s equations but if viewed as two vector fields why would we expect them to mix together under a boost? The answer is they are not really two separate 3-vector fields but together form the six degrees of freedom of the Faraday tensor (aka electromagnetic tensor).
Maxwell’s equations can be formulated directly using the Faraday tensor & the four-current, both of which we know transform properly under Lorentz transformations, making the Lorentz covariance manifest.
Aside: Duality Rotation
This mixing actually goes deeper than simply appearing under Lorentz boosts. Maxwell’s equations are suspiciously asymmetric between the electric and magnetic fields, and yet they are ultimately part of the same mathematical object. What gives?
The answer is that the equations are symmetric under a so-called duality rotation, which mixes the electric and magnetic fields together. Each component of the electric and magnetic fields get rotated together according to
$$\begin{align} E_i &\rightarrow E_i^\prime = E_i \cos\theta - B_i \sin\theta \\ B_i &\rightarrow B_i^\prime = B_i \cos\theta + E_i \sin\theta \\ \end{align}$$where \(i = x, y, z\) represents the component.
If you want to visualize this rotation note that it suffices to consider the 2d plane in which both \(\mathbf E\) and \(\mathbf B\) lie. This is because \(\mathbf E^\prime\) and \(\mathbf B^\prime\) are linear combinations of \(\mathbf E\) and \(\mathbf B,\) and so must lie in the same plane. In other words \(\mathbf E^\prime \in \text{span}(\{\mathbf E, \mathbf B\}),\) and similarly for \(\mathbf B^\prime.\)
The source (inhomogeneous) term also get appropriately rotated and now appears with non-zero value in both Faraday’s & Ampere’s laws. However, as experimenters we can never tell the difference between (say) an electron having fully electric charge and no magnetic charge, or \(1/\sqrt 2\) electric charge and \(1/\sqrt 2\) magnetic charge. The choice of saying we have no magnetic monopoles is mostly arbitrary, but does simplify electrostatics. (Assuming magnetic monopoles are not forbidden for some other reason. In which case the choice definitely not arbitrary, rather enforced.)