I know that in special relativity Electric and Magnetic fields mix together in different reference frames, but my question is about classical mechanics.

It seems weird to me is that the Lorentz Force law has velocity in it, it doesn't make much sense in classical mechanics, and I assume the Lorentz force law was derived before Special relativity.

In classical mechanics acceleration should be the same in all reference frames, so let's take two examples:

  1. There's a constant magnetic field and a particle moving, if I'm moving with the particle it looks to me as if it is stationary so it should not have any magnetic field acting on it. so why is it accelerating?
    I assume the problem here is that it is impossible to have a constant magnetic field in both frames, but I'm not sure why.

  2. There are two charged particles moving in parallel to each other, meaning they have the same velocity, again if I'm moving together with the particles it looks to me as if they are stationary and should not have any magnetic field acting on them.

I realize the real answer is to use special relativity, but my question is how did Lorentz think about this before special relativity was discovered and how is this problem "solved" in classical mechanics? I assume this also has some connection with what the definitions of Electric Field and Magnetic Field actually are.

  • $\begingroup$ In case 1 there is a Lorentz force, because the charge crosses magnetic field lines. In case 2 there is no Lorentz force because charge does not cross magnetic field lines. If magnet moves, magnetic field lines move with the magnet. $\endgroup$
    – kartsa
    Commented Mar 2, 2012 at 3:35

2 Answers 2


It's in Lorentz' 1895 paper:

"Versuch einer theorie der electrischen und optischen erscheinungen bewegten kΓ΅rpern"

which you can find copies of on the net. It's buried under notation alien to current eyes, and is in German. I can read German, but it's a drag for me to do so.

Here's an English translation:

Attempt of a Theory of Electrical and Optical Phenomena in Moving Bodies

I can't vouch for its accuracy.

The "moving" - as in Einstein's 1905 paper - refers specifically to the condition $𝐆 β‰  𝟎$ (or, in his notation: $𝔭 β‰  0$) of being in a "non-stationary" frame. That's discussed in further detail below.

He mentions the force in Β§12 of Abschnitt I; lays out some equations - for empty space - in that section, and the equations for moving media in Abschnitt II. You need a Rosetta Stone to penetrate the text. So, for that, let's lay out a few points of reference.

He apparently expressed the Lorentz force, there and later, as a force $𝐅_1 = 𝐄 + 𝐯×𝐁$ per unit charge, rather than as a force outright $𝐅 = e(𝐄 + 𝐯×𝐁)$. Part of this was already there with the velocity-correction $𝐄 + 𝐆×𝐁$ that is already present in the constitutive law for the electric field - see below. The velocity $𝐆$ is relative to the "stationary" frame. Again: see below. So adding on the extra velocity term $(𝐯 - 𝐆)×𝐁$ for the velocity $𝐯$ of the moving charge is a natural step - and a step required to make the force law independent of any specific frame.

Our current understanding includes the Maxwell equations: $$βˆ‡Β·π, \quad βˆ‡Γ—π„ + \frac{βˆ‚π}{βˆ‚t} = 𝟎, \quad βˆ‡Β·πƒ = ρ, \quad βˆ‡Γ—π‡ - \frac{βˆ‚πƒ}{βˆ‚t} = 𝐉.$$ This holds across the board, relativistically and non-relativistically as is, because they live at a deeper layer of geometry in which such distinctions (even distinctions between space-like and time-like) are not present.

It was also common then - and now - to define the "total current": $$𝐂 = 𝐉 + \frac{βˆ‚πƒ}{βˆ‚t},$$ so that one could just as well write $βˆ‡Γ—π‡ = 𝐂$. The current $𝐂$ is the one used in circuit analysis, not $𝐉$.

Our understanding, at present, also includes the following constitutive relations: $$𝐃 + α𝐆×𝐇 = Ξ΅(𝐄 + 𝐆×𝐁), \quad 𝐁 - α𝐆×𝐄 = ΞΌ(𝐇 - 𝐆×𝐃).$$ For Relativity, $Ξ± = (1/c)^2$ and these are the Maxwell-Minkowskii relations. For non-relativistic theory $Ξ± = 0$, and they are the relations posed by Maxwell with the correction $-𝐆×𝐃$ later added by Thomson.

The velocity $𝐆$ is relative to the frame in which the constitutive laws assume isotropic form $𝐃 = Ρ𝐄$ and $𝐁 = μ𝐇$. In that frame, $𝐆 = 𝟎$, and may be referred to as the "stationary" frame.

The coefficients $Ρ$ and $μ$ are properties of the medium, with $Ρμ > 0$; the speed $V = 1/\sqrt{Ρμ}$ is a characteristic of the medium and determines the speed of wave propagation, including that of light - relative to the stationary frame.

In the case where $Ρμ = Ξ±$, provided that $|𝐆| < V$, all frames are equivalent to the stationary frame, the Maxwell-Minkowski relations are equivalent to their $𝐆 = 𝟎$ version, for all $𝐆$, and $𝐆$ drops out of the picture. In that case, $V = c$, the invariant speed mandated by Special Relativity. These are the relations for the vacuum. Hence, $c$ is referred to as the speed of light in a vacuum.

The equations laid out by Lorentz in Abschnitt II are: $$\begin{align} β… _b\ & \text{Div}\ 𝔑 = ρ \\ β…‘_b\ & \text{Div}\ β„Œ = 0 \\ β…’_b\ & \text{Rot}\ β„Œ' = 4πρ𝔳 + 4Ο€\dot{𝔑} \\ β…£_b\ & \text{Rot}\ 𝔉 = -\dot{β„Œ} \\ β…€_b\ & 𝔉 = 4Ο€V^2𝔑 + [π”­Β·β„Œ] \\ β…₯_b\ & β„Œ' = β„Œ - 4Ο€[𝔭·𝔑] \\ β…¦_b\ & π”ˆ = 𝔉 + [π”³Β·β„Œ] \end{align}$$ In other contexts, in Abschnitt I, he also makes mention of the convection current $β„­ = ρ𝔳$ and the total current $𝔖 = β„­ + \dot{𝔑}$. He also uses the notation $$Ξ” = \left(\frac{βˆ‚}{βˆ‚x}\right)^2 + \left(\frac{βˆ‚}{βˆ‚y}\right)^2 + \left(\frac{βˆ‚}{βˆ‚z}\right)^2$$ and $Ξ”'$ for $Ξ”$, with the effect of the velocity $𝔭$ subtracted off, and $(βˆ‚/βˆ‚t)_1$ being $βˆ‚/βˆ‚t$ with the effect of the velocity $𝔭$ subtracted off. His $\dot{(\_)}$, going by context, is equal to our (and his) $βˆ‚/βˆ‚t$, while his "Div" is equivalent to our $βˆ‡Β·(\_)$ and his "Rot" to our $βˆ‡Γ—(\_)$, where $$βˆ‡ = \left(\frac{βˆ‚}{βˆ‚x}, \frac{βˆ‚}{βˆ‚y}, \frac{βˆ‚}{βˆ‚z}\right).$$ His vector product $[\_Β·\_]$ is what we today write as the vector cross product $(\_)Γ—(\_)$.

The Rosetta Stone, with Lorentz on top, and our notation on the bottom, is: $$ \left(β„Œ, 𝔉, π”ˆ, β„Œ', 𝔑, ρ, β„­, 𝔖, 𝔳, 𝔭\right) \\ β‡’ \\ \left(\frac{𝐁}m, \frac{𝐄}m, \frac{𝐅_1}m, 4Ο€m𝐇, m𝐃, mρ, m𝐉, m𝐂, 𝐯, -𝐆\right) $$ for the objects, where $m = \sqrt{ΞΌ/(4Ο€)}$; and $$ \left(Ξ”, Ξ”', \text{Div}, \text{Rot}, \dot{\left(\_\right)}, \left(\frac{βˆ‚}{βˆ‚t}\right)_1, [\_Β·\_]\right) \\ β‡’ \\ \left(βˆ‡Β², βˆ‡Β² - \frac{(π†Β·βˆ‡)^2}{V^2}, βˆ‡Β·(\_), βˆ‡Γ—(\_), \frac{βˆ‚}{βˆ‚t}, \frac{βˆ‚}{βˆ‚t} + π†Β·βˆ‡, (\_)Γ—(\_)\right) $$ for the operators. The specific factors used are those required to also reconcile the energy integrals, which were cited elsewhere in the article.

With these correspondences, his equations become: $$\begin{align} β… _b\ & βˆ‡Β·πƒ = ρ \\ β…‘_b\ & βˆ‡Β·π = 0 \\ β…’_b\ & βˆ‡Γ—π‡ = 𝐉 + \frac{βˆ‚πƒ}{βˆ‚t} \\ β…£_b\ & βˆ‡Γ—π„ = -\frac{βˆ‚π}{βˆ‚t} \\ β…€_b\ & 𝐄 = 𝐃/Ξ΅ - 𝐆×𝐁 \\ β…₯_b\ & 𝐇 = 𝐁/ΞΌ + 𝐆×𝐃 \\ β…¦_b\ & 𝐅₁ = 𝐄 + 𝐯×𝐁 \end{align}$$ with the constitutive law $𝐉 = ρ𝐯$ for the current.

So, if he originated this, then his contribution was to take the force law for the "moving" electric force $𝐄 + 𝐆×𝐁$ and to then make it frame-independent, by generalizing the $𝐆$ in the $𝐆×𝐁$ term to $𝐯$. Maxwell's $𝐄$, in fact, was actually already defined as the "moving" electric force, not as our $𝐄$. It muddied up his equations and formalism.

As you can see ... by the way ... Lorentz' equations are a combination of Maxwell's equations plus the non-relativistic version of the constitutive law, along with the per-unit-charge force law and the decomposition law $𝐉 = ρ𝐯$ for the current. Lorentz' theory was firmly couched in non-relativistic physics because of the absence of the relativistic corrections $+ α𝐆×𝐇$ and $- α𝐆×𝐄$ with $Ξ± = (1/c)^2$ present in the Maxwell-Minkowski relations.


I really don't know the exact historical chain of events, so I might even give a result that popped up after SR came into existence. In fact, I figured out two different ways of explaining this, which I've given here. The first one is IMHO classical, but the fact that Coulomb's law is only for static cases may be incorrect in classical physics. I doubt it, though; Maxwell knew the relation of EM fields with EM waves. The second explanation makes sense from a pure classical POV, before Maxwell. They conflict each other, though they both explain it. So I'm giving both here. Comments appreciated on which is more correct.

I'm referring to electric field as E and magnetic field as B here, with standard notations.

Answer #1

In classical mechanics, you can solve this by changing the definitions of electric and magnetic fields. They're the same thing. Moving with a velocity makes an E field into a B field or vice versa. Aside from that, Coulomb's law is only applicable for electro-static situations. When the particle is moving, the E field is different.

In the end, only force has to be the same in inertial frames. If E became B on a change of velocity, the formulae will be such that the force stays the same. An observer travelling along with the moving particles will see an E, no B, whereas an observer at "rest" (basically moving relative to the particles) will see a B field and a smaller E field. But, both observers will feel the same force, and they will see the particles being attracted/repelled by the same amount.

One way to look at this is from the fact that EM fields are transmitted/mediated by EM radiation. So, going at a velocity changes the behaviour of the waves in your frame.

Actually, I had this confusion a few years ago (for two parallel particles) I assumed electrostatic force in both cases, and got some strange results. Knowing that SR had its origin somewhere in electromagnetism, I assumed an unknown length contraction and resolved it. Surprisingly, the lorentz factor popped up in my equations (since $\mu_0\epsilon_0=1/c^2$ except that the length contraction was in the perpendicular direction. This is all the result of using electrostatic force in both cases.

Note that the shift in fiels is pretty tiny for nonrelativistic cases, owing to the $c^2$.

For your first case, the problem becomes trivial after this. In your frame, a bit of the magnetic field lines are electric field lines. Problem solved.

In fact, one can look at a magnetic field as a sort of 'reserve E field'. When a particle moves through a magnetic field, in its frame, it sees itself at rest. So the force it feels is an electric field, which was 'drawn from' the 'reserve E field' (i.e., the magnetic field). One can look at it the other way around as well, though not

Answer #2

Remember, in classical mechanics, we do have the lumineferous aether, which acts as an absolute reference frame for light. So even in classical mechanics, wierd things do happen when going near-lightspeeds. Your reference frames are no longer equivalent, anyways. Since EM fields are transmitted by em waves, the aether playe a crucial part. At near lightspeeds, $\mu_0\epsilon_0=1/c^2$ becomes significant in your equations.

  • $\begingroup$ Your second answer is more likely to be what people historically thought. Since Maxwell's equations are not invariant under Galilean relativity, they introduced a special frame in which ether is static. In other words, in electrodynamics before relativity, it is wrong to change reference frame. $\endgroup$
    – Siyuan Ren
    Commented Mar 2, 2012 at 2:48
  • $\begingroup$ @KarsusRen Good point. I remember something about Maxwell's laws behaving crazily on using various reference frames. I guess Answer#1 is the just-before-relativity explanation. $\endgroup$ Commented Mar 2, 2012 at 13:05

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