If Maxwell's equations were around long before special relativity, and yet they don't make sense without special relativity (like if you think of a moving charge next to an electrically neutral wire carrying current, then in the lab frame it experiences magnetic force and in its rest frame, there's no force - unless you take length contraction into account), how were they accepted at the time?

  • 9
    $\begingroup$ Because they do make sense without special relativity. $\endgroup$
    – Kyle Kanos
    Feb 25, 2019 at 0:04
  • $\begingroup$ This problem was resolved by Lorentz in his Theory Of Electrons that contained the Lorentz Transformation. He received a Nobel Prize for it in 1902. $\endgroup$
    – safesphere
    Feb 25, 2019 at 9:24

4 Answers 4


At the time of Maxwell, it was believed that the electric and magnetic fields were separate fields, though related through the latter two of Maxwell's equations (Ampere's laws & the induction eq'n). So there really was no conflict that you somehow perceive there is/was/should be, it all made verifiable sense (cf. the Wikipedia entry on the History of Maxwell's equations, for instance).

What relativity did was enable us to view the electric and magnetic fields as a single entity, the electromagnetic tensor, $$ F^{\mu\nu}=\left(\begin{array}{cccc} 0&-E_x&-E_y&-E_z\\ E_x&0&-B_z&B_y\\ E_u&B_z&0&-B_x\\ E_z&-B_y&B_x&0\\ \end{array} \right) $$ The connection between the two fields is discussed in many other places on this site:

All of which would be good reading for you.

  • $\begingroup$ But, ultimately, they don't really make sense without special relativity because of the fact that they spontaneously determine the speed of light in all the frames that they are true in, right? And people realized this and that led to the whole Lorentz-Poincare-Michealson-Morley thingy. $\endgroup$
    – user87745
    Feb 26, 2019 at 1:37
  • $\begingroup$ @DvijMankad the individual laws were experimentally determined, so to say they don't make sense is ridiculous. SR added more knowledge about the E&M fields, but it doesn't invalidate any knowledge about them separately. $\endgroup$
    – Kyle Kanos
    Feb 26, 2019 at 1:46

At the time it was a mystery how they could work. One leading theory was the "luminiferous ether". This was a medium which pervaded all things and through which maxwell waves propagated. This ether provided (in theory) the rest frame for the equations.

As the theory was tested by, e.g. the Michaelson Morley, experiment and by the observation of photons it was modified. So, for example, the idea was advanced that the ether could be dragged by bodies (akin to general relativistic frame dragging) and that it was composed of filaments which kept packets of wave energy intact.


The history of Maxwell's equations is a long and hardy story. 1865, James Clerk Maxwell published "A Dynamical Theory of the Electromagnetic Field", unifying all then available results on electricity and magnetism. In this article - which is worthwhile reading - he also postulated the existence of electromagnetic waves, as well as their propagation speed. His theory is presented in the form of 20 equations in 20 unknowns; vector analysis was not yet known at that time.

What we know today as the four "Maxwell's equations" is the work of Oliver Heaviside, who published them in "Electromagnetic Theory", 1893 onwards, 4 volumes. These four equations can be further condensed in 4-space into a single equation, the "Fundamental equation of Electrodyanmics":

((1/c^2)∙∂^2/(∂t^2) - ∂^2/(∂x1^2) - ∂^2/(∂x2^2) - ∂^2/(∂x3^2) )A = μ0∙J

wherein A=(φ/c, (A1, A2, A3)), and J=(ρc, (J1, J2, J3))

(This is the equivalent of Poisson's equation which governs flow processes in 3-space)

James Clerk Maxewll based his theory on the hypothesis of the luminiferous ether, which made its way from there into telecommunication jargon ("sending through the ether" was used synonymous to "wireless"). Heinrich Hertz and Oliver Lodge (the true inventor of radiocommunication; see "The work of Hertz and some of his successors", lectures given in 1894) were firmly adhering to the ether theory.

Michelson and Moorley found no evidence for the existence of a luminiferous ether; however their experiment was interpreted in a 3-dimensional frame; in the 4-dimensional space-time of Special Relativity, the Michelson-Moorley-setup is an inertial system which cannot be used to prove or to disprove the existence of the luminiferous ether. But that our Universe is 4-dimensional was not known at that time.

In 4-dimensional space, physics becomes much simpler, indeed:

Albert Einstein's formulas E=m∙c^2 and the Relativity Invariant E^2/c^2 - p⃗^2= m0^2∙c^2 can be combined (measuring distance in light-seconds; c=1) to E^2 = m^2 = m0^2 + p⃗^2 = m0^2 + p1^2 + p2^2 + p3^2. That means that the rest mass is simply the fourth component of the momentum vector, and the energy is the total momentum.

According to Leonhard Euler's 4-square-identity, any sum of 4 squares can be written as a product of two sums of four squares each. Hence (M0^2 + P1^2 + P2^2 + P3^2) = (r0^2 + r1^2 + r2^2 + r3^2)∙(m0^2 + p1^2 + p2^2 + p3^2).

Herein are

M0 = (r0∙m0 - r1∙p1 - r2∙p2 - r3∙p3)

P1 =(r0∙p1 + r1∙m0 + r2∙p3 - r3∙p2)

P2 =(r0∙p2 - r1∙p3 + r2∙m0 + r3∙p1)

P3 =(r0∙p3 + r1∙p2 - r2∙p1 + r3∙m0)

which represents the quaternion multiplication rule (proof by simple algebraic evaluation)

The squares sums may be interpreted as scalar (inner) products of a vector with itself, hence as a the square of a metric length.

Lets take the vectors as follows:

A⃗ = (m0, p1, p2, p3), a phyical system

R⃗ = (r0, r1, r2, r3), an interaction operator

with (r0^2 + r1^2 + r2^2 + r3^2) = 1; (for energy conservation)

P⃗ = (M0, P1, P2, P3), the resulting physical system

then we can write: P⃗ = R ⃗ * A ⃗ , or also P⃗ = P1⃗ + P2⃗ = R⃗ * (A1⃗ + A2⃗) = R⃗ * A⃗ This is a formula describing the energy conservation in a physical interaction. Energy can only be transmitted, shared, or cumulated between the starting system A⃗ or parts of it (A1⃗ + A2⃗) and the resulting system P⃗ or parts of it P1⃗ + P2⃗, but never be created nor annihilated. The multiplication * sign herein designates the quaternion multiplication rule as given above.

By the way, Leonhard Euler's 4-square-identity in some way prefigures Maxwell's equations, but it was found in a completely different, unrelated context, more than 100 years before. See: https://www.e-periodica.ch/cntmng?pid=fng-001:2017:106::158


Maxwell's equations, as they are known today, were not around before Special Relativity and differ from what Maxwell wrote - which were covariant under the Galilei transforms; which Maxwell went to pains to show, in his treatise.

I've made the comments, pertaining to all the myths underlying your remarks, elsewhere in greater detail,

How did Maxwell's theory of electrodynamics contradict the Galilean principle of relativity? (Pre-special relativity)
How did Maxwell's theory of electrodynamics contradict the Galilean principle of relativity? (Pre-special relativity)

but I'll elaborate on it in here; particularly with some further remarks on Lorentz' equations - which are also non-relativistic.

First, your comments are based on a widespread folklore myth that Maxwell's theory was somehow relativistic and its "discrepancy" was what led to Special Relativity. In fact, the equations today known as "Maxwell's equations" are not what Maxwell originally posed; and what he posed - particularly when including the correction later added by Thomson - is Galilean covariant.

This myth is widely propagated - even within the Physics community; even by programs, like PBS's "Einstein's Big Idea". It's totally wrong and is based on simple ignorance: the 19th century literature is virtually inaccessible to contemporary eyes, because its notations are totally different and things that look similar (like Maxwell's $𝐄$ field) are different; and most people don't properly contextualize what they're reading. So, it goes unread and is forgotten, or misunderstood and out of context.

Second, is the equally false folklore myth that Lorentz somehow made the Maxwell's equations relativistic. In fact, the equations posed by Lorentz are equivalent to those posed by Maxwell, with Thomson's correction, and are non-relativistic. This is, in particular, in contrast with the equations written by Einstein and Laub, as well as by Minkowski, in 1908, that now go under the name: the Maxwell-Minkowski relations.

The Maxwell-Minkowski relations include a relativistic correction that is absent in the Maxwell-Thomson relations.

Third, is the notion that Maxwell's equations call out any space-time geometry at all! They don't. The equations - when written correctly - are covariant with respect to all coordinate transforms.

The equations, themselves, are non-metrical. They reside at a deeper layer of geometry where such distinctions between relativistic versus non-relativistic, between space-like and time-like, where notions such as parallelism, orthogonality, angle, distance, etc. do not exist.

Instead, what is not covariant, what breaks general covariance, where a specific space-time structure is called out, is the constitutive relations that connect the displacement field and magnetic field with the electric field and magnetic induction.

This is the one and only place where a difference between the relativistic versus non-relativistic paradigm occurs. It is the only place where any dependence on an underlying space-time metric or geometry at all arises. That's where the distinction between the Maxwell-Thomson(-Lorentz) relations and the Maxwell-Minkowski(-Einstein-Laub) relations arises.

This is the part of Maxwell's equations that has general covariance: $$ 𝐁 = ∇×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t} \hspace 1em⇐\hspace 1em ∇·𝐁 = 0, \hspace 1em ∇×𝐄 + \frac{∂𝐁}{∂t} = 𝟬, \\ ∇·𝐃 = ρ, \hspace 1em ∇×𝐇 - \frac{∂𝐃}{∂t} = 𝐉 \hspace 1em⇒\hspace 1em ∇·𝐉 + \frac{∂ρ}{∂t} = 0. $$ where $$∇ = \left(\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}\right).$$

I can't begin to overemphasize that I literally mean that these are covariant with respect to all coordinate transforms. The components $$ φ, \hspace 1em 𝐀 = \left(A_x, A_y, A_z\right), \hspace 1em 𝐄 = \left(E_x, E_y, E_z\right), \hspace 1em 𝐁 = \left(B^x, B^y, B^z\right), \\ ρ, \hspace 1em 𝐉 = \left(J^x, J^y, J^z\right), \hspace 1em 𝐃 = \left(D^x, D^y, D^z\right), \hspace 1em 𝐇 = \left(H_x, H_y, H_z\right) $$ that the coordinates $(t,x,y,z)$ could be anything - not just Cartesian, but any four independent functions of the coordinates. The very same equations will hold intact, in the very same form down to the last component, regardless of what you call $t$, $x$, $y$ and $z$.

That's because the quantities can be grouped together into a set of differential forms: $$ A = 𝐀·d𝐫 - φ dt, \hspace 1em F = 𝐁·d𝐒 + 𝐄·d𝐫∧dt, \\ J = ρdV - 𝐉·d𝐒∧dt, \hspace 1em G = 𝐃·d𝐒 - 𝐇·d𝐫∧dt, $$ where $$d𝐫 = (dx, dy, dz), \hspace 1em d𝐒 = (dy∧dz, dz∧dx, dx∧dy), \hspace 1em dV = dx∧dy∧dz.$$ The equations previously laid out can then be written as: $$dA = F, \hspace 1em dF = 0, \hspace 1em dG = J, \hspace 1em dJ = 0.$$

The differential forms $A$, $F$, $G$, $J$ are all invariant under arbitrary coordinate transforms. Using that invariance, for instance, you can determine the transforms of the components. Thus, for instance, if $(t',x',y',z') = (x + vt, x - vt, y, z)$, for some speed $v ≠ 0$ (and we'll even make it variable $v = v(t)$ as a function of $t$, then the coordinate differentials transform as $$(dt', dx', dy', dz') = (dx + (v'(t) t + v(t)) dt, dx - (v'(t) t + v(t)) dt, dy, dz).$$ Then, from the invariance of $A$, we may write: $$\begin{align} A &= A_{x'} dx' + A_{y'} dy' + A_{z'} dz' - φ' dt' \\ &= A_{x'} (dx - (v' t + v) dt) + A_{y'} dy + A_{z'} dz - φ' (dx + (v' t + v) dt) \\ &= \left(A_{x'} - φ'\right) dx + A_{y'} dy + A_{z'} dz - \left(φ' - A_{x'}\right) (v' t + v) dt, \end{align}$$ or matching components: $$φ = \left(φ' - A_{x'}\right) (v' t + v), \hspace 1em A_x = A_{x'} - φ', \hspace 1em A_y = A_{y'}, \hspace 1em A_z = A_{z'}.$$ Similarly, for the differential operators, using the invariance of $$dt \frac{∂}{∂t} + d𝐫·∇ = dt' \frac{∂}{∂t'} + d𝐫'·∇',$$ we get a similar set of transforms: $$\frac{∂}{∂t} = \left(\frac{∂}{∂t'} + \frac{∂}{∂x'}\right) (v' t + v), \hspace 1em ∇ = ∇' + \frac{∂}{∂t'}.$$ You'd also have to substitute to get these fully in terms of the primed coordinates: $$v' t + v = v' \frac{t' - x'}{2v} + v = \frac{t' - x'}2 + v,$$ which is not as easy as it looks, since $v$ is a function of $t$: $$v = v(t) = v\left(\frac{t' - x'}{2v}\right),$$ which means it becomes a function $v = V(t' - x')$ of $u = t' - x'$ given implicitly by: $$V(u) = v\left(\frac{u}{2V(u)}\right).$$

You can carry out the analysis on $F$, $G$ and $J$ to find the transforms for their components and substitute into the Maxwell equations, with the transformed differential operators to find that the very same equations arise - but with the primes on the components and coordinates. The coordinates $t'$ and $x'$ are not even time and space coordinates anymore. They have units of length and are mixtures of $t$ and $x$.

The part that is not covariant calls out a specific geometry. These are the constitutive relations. It can be written in the following general form: $$𝐃 + α𝐆×𝐇 = ε(𝐄 + β𝐆×𝐁), \hspace 1em 𝐁 - α𝐆×𝐄 = μ(𝐇 - β𝐆×𝐃),$$ where I assume $(α, β) ≠ (0, 0)$.

These go with the geometry that contains the following as its invariants: $$β dt^2 - α |d𝐫|^2, \hspace 1em dt \frac{∂}{∂t} - d𝐫·∇, \hspace 1em β ∇^2 - α \left(\frac{∂}{∂t}\right)^2.$$

In all cases, where $α ≠ βεμ$, the version of these relations where $𝐆 ≠ 𝟬$ (the "moving" versions) differs from the version $𝐆 = 𝟬$ (the "stationary" version). Only in the stationary version are the constitutive laws isotropic, with $𝐃$ and $𝐄$ aligned with $𝐃 = ε𝐄$, and $𝐁$ and $𝐇$ aligned with $𝐁 = μ𝐇$. For lack of a better name, that's the (unique) frame of isotropy.

For this to be consistent, the vector $𝐆$ must have a transform that is given, in infinitesimal form, for infinitesimal boosts $𝞄$ as: $$Δ𝐆 = -β 𝞄 + α 𝞄·𝐆 𝐆.$$

We can call out the cases here.

If $εμ > 0$, then this calls out a speed $$V = \sqrt{\frac{1}{εμ}},$$ that is associated with the wave motion predicted by the Maxwell equations coupled with these constitutive relations. The speed called out by the wave motion is relative to the frame of isotropy.

In cases where $α = βεμ$, provided $|𝐆|$ is not too large, the moving and stationary versions are mathematically equivalent and coincide. In that case, one can take them to be constitutive relations for a "vacuum" - i.e. a background that is "boost-invariant" or independent of the motion of an inertial observer. This really only has meaning if $αβ ≥ 0$, with $β ≠ 0$.

Carrollian Geometry - $c = 0$:
This is the case $α ≠ 0$ and $β = 0$. A Carrollian transform to a "moving" frame going in the $x$ direction, i.e. a Carrollian boost in the $x$ direction, is $$(t,x,y,z) → (t',x',y',z') = (t - αvx,x,y,z).$$ The coefficient $αv$ has the dimension of "slowness" or inverse speed; i.e. "miles per minute" or "marathon speed", because if you've ever run a marathon you end up going slow after 20 miles unless you're in shape. Normalizing $α = 1$ makes $v$ inverse speed.

It's named after Carroll because the speed zero is absolute, and under the transform to a moving frame things that were standing still continue to stand still - so they "move" without going anywhere.

The constitutive relations reduce to: $$𝐃 + 𝐆×𝐇 = ε𝐄, \hspace 1em 𝐁 - 𝐆×𝐄 = μ𝐇.$$

Galilean Geometry (the non-relativistic version) - $c = ∞$:
This is the case $α = 0$ and $β ≠ 0$. Here, the transforms may be written as $$(t,x,y,z) → (t',x',y',z') = (t,x - βvt,y,z).$$ Now, $βv$ has the the dimension of "speed" - meters per second and you can normalize $β = 1$, to make $v$, itself, a speed.

In that case, the constitutive relations reduce to a form: $$𝐃 = ε(𝐄 + 𝐆×𝐁), \hspace 1em 𝐁 = μ(𝐇 - 𝐆×𝐃),$$ that is equivalent to what Maxwell wrote - with the Thomson correction $-𝐆×𝐃$ - and is also equivalent to what Lorentz wrote.

Lorentz' equations and theory were non-relativistic. Though it is true he wrote down what we call the "Lorentz transforms", he only used it as to stage the Galilean transforms! There is an extra "corrective" transform that is tacked onto the "Lorentz transform" to recover the Galilean transform. The end result is that the equations preserved are those given by the above constitutive relations, not by the Maxwell-Minkowski(-Einstein-Laub) relations that has the relativistic corrections in it.

The absolute speed is infinity, which is the speed at which simultaneous happens. So, the back-door way of describing that is that "simultaneity" is absolute.

The only way $α = βεμ$ can happen is if $εμ = 0$. That would correspond to a "wave" speed of $V = ∞$. Only there can a "vacuum" form of the constitutive laws be asserted.

Timeless Space: $αβ < 0$:
Without further elaboration, this is just 4-dimensional Euclidean spatial geometry, where $t$ is just another spatial coordinate and there are no coordinates of time. It's not a space-time, but a timeless space.

There's no absolute speed, because there's no notion of "speed" at all, since there's no time dimension. Coordinate transforms are just rotations in 4D. I won't elaborate on it any further here.

Lorentzian Geometry: $αβ > 0$:
This is the case that has a finite, non-zero absolute speed $$c = \sqrt{\frac{β}{α}}.$$ Normalizing $β = 1$ and $α = (1/c)^2$, the constitutive relations can be written as: $$𝐃 + \frac{1}{c^2}𝐆×𝐇 = ε(𝐄 + 𝐆×𝐁), \hspace 1em 𝐁 - \frac{1}{c^2}𝐆×𝐄 = μ(𝐇 - 𝐆×𝐃).$$

That's the Maxwell-Minkowski relations.

The case $α = βεμ$ for the "vacuum" case happens if $V = c$: if the wave speed matches the invariant speed. Consequently, $c$ is called the speed of light in a vacuum.

The corresponding boost transform can be cast in a form that includes the other cases of $αβ ≥ 0$: $$(γ(t - αvx), γ(x - βvt), y, z), \hspace 1em γ = \frac{1}{\sqrt{1 - αβv^2}},$$ and includes the restriction $αβv^2 < 1$. The restriction is tautological, except for the Lorentzian case, where it reduces to the form $|v| < c$: that boost velocities must be sub-light.

Lorentz' Equations
I'll repeat what I posted elsewhere; here, in my reply to:

Lorentz force law in Newtonian relativity
Lorentz force law in Newtonian relativity

This comes out of Lorentz' 1895 paper (cited in the link) but is used throughout his later papers. The equations laid out in his 1895 paper were: \begin{align} Ⅰ_b\ & \text{Div}\ 𝔡 = ρ \\ Ⅱ_b\ & \text{Div}\ ℌ = 0 \\ Ⅲ_b\ & \text{Rot}\ ℌ' = 4πρ𝔳 + 4π\dot{𝔡} \\ Ⅳ_b\ & \text{Rot}\ 𝔉 = -\dot{ℌ} \\ Ⅴ_b\ & 𝔉 = 4πV^2𝔡 + [𝔭·ℌ] \\ Ⅵ_b\ & ℌ' = ℌ - 4π[𝔭·𝔡] \\ Ⅶ_b\ & 𝔈 = 𝔉 + [𝔳·ℌ] \end{align}

This correspondence to the form given above for the Maxwell-Thomson relations is: $$\left(ℌ, 𝔉, 𝔈, ℌ', 𝔡, ρ, ℭ, 𝔖, 𝔳, 𝔭\right) \\ ⇒ \\ \left(\frac{𝐁}m, \frac{𝐄}m, \frac{𝐅_1}m, 4πm𝐇, m𝐃, mρ, m𝐉, m𝐂, 𝐯, -𝐆\right)$$ where $m = \sqrt{μ/(4π)}$; and for the operators: $$\left(Δ, Δ', \text{Div}, \text{Rot}, \dot{\left(\_\right)}, \left(\frac{∂}{∂t}\right)_1, [\_·\_]\right) \\ ⇒ \\ \left(∇², ∇² - \frac{(𝐆·∇)^2}{V^2}, ∇·(\_), ∇×(\_), \frac{∂}{∂t}, \frac{∂}{∂t} + 𝐆·∇, (\_)×(\_)\right).$$ This yields: $$\begin{align} Ⅰ_b\ & ∇·𝐃 = ρ \\ Ⅱ_b\ & ∇·𝐁 = 0 \\ Ⅲ_b\ & ∇×𝐇 = 𝐉 + \frac{∂𝐃}{∂t} \\ Ⅳ_b\ & ∇×𝐄 = -\frac{∂𝐁}{∂t} \\ Ⅴ_b\ & 𝐄 = 𝐃/ε - 𝐆×𝐁 \\ Ⅵ_b\ & 𝐇 = 𝐁/μ + 𝐆×𝐃 \\ Ⅶ_b\ & 𝐅₁ = 𝐄 + 𝐯×𝐁 \end{align}$$ with the constitutive law $𝐉 = ρ𝐯$, cited elsewhere in the paper, for the current. His $𝐅₁$ is what we today call the Lorentz force, pro-rated to unit charge. Today, we write $𝐅 = e𝐅₁ = e(𝐄 + 𝐯×𝐁)$, for an electric charge $e$.

That's the Maxwell equations with the Lorentz force law and ... the Maxwell-Thomson(-Lorentz) relations - non-relativistic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.