# Can the Lorentz force expression be derived from Maxwell's equations?

The electromagnetic force on a charge $$e$$ is

$$\vec F = e(\vec E + \vec v\times \vec B),$$

the Lorentz force. But, is this a separate assumption added to the full Maxwell's equations? (the result of some empirical evidence?) Or is it somewhere hidden in Maxwell's equations?

• See David Z's answer to this Phys.SE question physics.stackexchange.com/q/15443/2451 – Qmechanic Feb 3 '12 at 14:11
• For anyone who imagines there's a lot of freedom in choosing the form of the force law, a nice exercise is to try to write down an alternative law that satisfies Lorentz invariance, i.e., that is a valid four-vector. (I'm ignoring constant factors.) The standard Lorentz force can be expressed as $F^{ab}v_b$, where $F$ is the e.m. tensor. $F^c_cF^{ab}v_b$? Vanishes. $F^{ab}v_bv^cv_c$? Reduces to the standard law. Derivatives aren't allowed, because then we won't have solutions to initial-value problems. – user4552 Sep 12 '13 at 3:53
• You assumed just few simplest possibilities. There is also the Landau-Lifshitz proposal for expression for four-force on a charged particle, $f^\mu = qF^{\mu\nu} u_\nu + C\frac{d}{d\tau}(qF^{\mu\nu} u_\nu)$. Very ugly, largely unnecessary and experimentally unconfirmed, but nevertheless still relativistic and Lorentz covariant. – Ján Lalinský Jan 22 '14 at 4:23

Maxwell's equations do not contain any information about the effect of fields on charges. One can imagine an alternate universe where electric and magnetic fields create no forces on any charges, yet Maxwell's equations still hold. ($$\vec{E}$$ and $$\vec{B}$$ would be unobservable and totally pointless to calculate in this universe, but you could still calculate them!) So you can't derive the Lorentz force law from Maxwell's equations alone. It is a separate law.

However...

• Some people count a broad version of "Faraday's law" as part of "Maxwell's equations". The broad version of Faraday's law is "EMF = derivative of flux" (as opposed to the narrow version $$\nabla\times\vec E = -\partial_t \vec B$$). EMF is defined as the energy gain of charges traveling through a circuit, so this law gives information about forces on charges, and I think you can derive the Lorentz force starting from here. (By comparison, $$\nabla\times\vec E = -\partial_t \vec B$$ talks about electric and magnetic fields, but doesn't explicitly say how or whether those fields affect charges.)

• Some people take the Lorentz force law to be essentially the definition of electric and magnetic fields, in which case it's part of the foundation on which Maxwell's equations are built.

• If you assume the electric force part of the Lorentz force law ($$\vec F = q \vec E$$), AND you assume special relativity, you can derive the magnetic force part ($$\vec F = q \vec v \times \vec B$$) from Maxwell's equations, because an electric force in one frame is magnetic in other frames. The reverse is also true: If you assume the magnetic force formula and you assume special relativity, then you can derive the electric force formula.

• If you assume the formulas for the energy and/or momentum of electromagnetic fields, then conservation of energy and/or momentum implies that the fields have to generate forces on charges, and presumably you can derive the exact Lorentz force law.

• So Faraday was playing a fool and Maxwell speculated without any idea how electric and magnetic fields affect charges? No, Faraday and Maxwell dealt with macroscopic charges (densities) and currents (current densities), not with a point-like charge as H. Lorentz did, that's the difference. – Vladimir Kalitvianski Feb 3 '12 at 17:29
• The expression of the Lorentz force in terms of electromagnetic fields can be derived from 4-divergence of EM field stress-energy tensor. But it is only half of the work. You can then derive the Lorentz force in terms of flux/charge densities if you assume that the action of magnetic field on flux 4-vector is rotation around direction of magnetic field, to the angle proportional to the magnitude of field. Similarly, the action of electric field on flux is Lorentz boost in the direction of electric field, proportional to the magnitude of the electric field. – Murod Abdukhakimov Feb 5 '12 at 13:14
• @Vladimir Kalitvianski, as you suspect, I am assuming the questioner is asking about "Maxwell's equations in modern form", i.e. the four equations as reformulated by Heaviside, in the form that appears in textbooks these days under the label "Maxwell's equations". I am not talking about what Maxwell and Faraday originally thought or wrote. If I'm remembering correctly, you're exactly right: Maxwell and Faraday did not have much or any logical separation between fields, forces, and charges. – Steve Byrnes Feb 5 '12 at 22:20
• @SteveB: I agree, if the Maxwell equations are understood as equations for the EMF sourced with a known four-current, then they do not describe the Lorentz force which is a part of "mechanical" equations. – Vladimir Kalitvianski Feb 6 '12 at 10:38
• @LarryHarson - "work done on charge" is the most common textbook definition of EMF; the line-integral of E is a less common definition (although I concede that you'll find it in some textbooks). Under your line-integral definition, "motional EMF" is not really EMF. Under your line-integral definition, a generator with a stationary wire and spinning magnet creates an EMF, but a generator with a spinning wire and stationary magnet does not create an EMF. Line-integral of E is a bad definition of EMF because it doesn't correspond to what a voltmeter measures or how experts actually use the term. – Steve Byrnes Aug 13 '13 at 13:22

I haven't seen this mentioned in the answers so I thought I should at least mention it. If you take the perspective that Maxwell's equations are the equations describing a $U(1)$ gauge field, then minimal coupling (which is, in a sense, the only gauge invariant way of coupling matter to a gauge field) ensures than any charged particle obeys the Lorentz force law, with the only freedom being the value $e$ of its charge. So while Maxwell's equations themselves, without some additional assumptions, may not necessarily imply the Lorentz force law, $U(1)$ gauge invariance does imply the Lorentz force law. In fact, if you take $U(1)$ gauge invariance as being the fundamental starting point, then it implies both Maxwell's equations and the Lorentz force law. Again, this is a matter of perspective, so I am not disagreeing with the other answers, but I think that this is the modern point of view.

• With a better success one can proceed from the Lorentz force directly, why invoke U(1)? U(1) cannot be a fundamental starting point because it implies already the mechanical and the wave equations in place with their strict physical meaning. Gauge invariance is like invariance with respect to the potential energy constant shift $V_0$; the mechanical equations with forces do not include the absolute value of the potential energy at all, there is nothing to speak about. $V_0$-invariance cannot be a fundamental starting point to derive physics because it follows from physics, not vice versa. – Vladimir Kalitvianski Feb 5 '12 at 12:45
• Minimal coupling is not the only way of coupling matter to electromagnetic field. Consider, for instance, Pauli coupling. – Murod Abdukhakimov Feb 5 '12 at 13:08
• Minimal coupling does not work alone, it needs renormalizations, i.e., additional counter-terms to produce physical results. – Vladimir Kalitvianski Feb 5 '12 at 16:17

Yes, the Lorentz force law can be derived from Maxwell's equations (up to a multiplicative constant), with only a few assumptions about what it means to talk about a field theory.

If we start from Maxwell's equations in a vacuum, we observe that they are Lorentz invariant. Therefore we expect that any force law had better be Lorentz invariant. If you like, you can add this as an explicit assumption.

Applying Noether's theorem for time-translation symmetry, we get an energy conservation law for an energy whose density is $u=(1/8\pi)(\textbf{E}^2+\textbf{B}^2)$. The factor of $1/8\pi$ is arbitrary and not specified by Noether's theorem. There is also nonuniqueness in the sense that you can add certain kinds of terms to this expression involving things like second derivatives of the fields, but I don't think those terms have any effect on the following argument, because the argument will depend only on the integral of $u$, not on its local density, and the added terms only give surface terms in the integral, and those vanish. This ambiguity is discussed in the Feynman lectures, section II-27-4.

Now add the source terms to Maxwell's equations. Consider two sheets of charge $\pm Q$ in the form of a parallel-plate capacitor with a small enough gap so that the interior field is nearly uniform. The energy $U=\int u dV$ is finite and calculable from the geometry. If we move one sheet closer to the other by $dx$, the energy in the electric field changes by $dU$. The total force between the sheets is $F_{total}=dU/dx$, which we can also calculate.

Now when we talk about a field theory, we assume that it's local in some sense. For this reason, the force acting on a small piece of charge $q$ in our capacitor can only depend on the field at that point, not on the field elsewhere. But the field has no transverse variation, so given $\textbf{F}_{total}$, we can infer the contribution $\textbf{F}$ from the force acting on $q$. The field is actually discontinuous in our example, but one can deal with that issue, which produces a factor of 2. The result of this example is $\textbf{F}=q\textbf{E}$, and the only possible wiggle room is that we could have chosen a different constant of proportionality in our definition of $u$. In other words, we could have changed the conversion factor between electromagnetic energies and mechanical energies, but we had no other freedom here. We could have chosen this conversion factor such that $\textbf{F}$ would vanish identically, but then electromagnetic fields would be undetectable with material devices, so this possibility is not very interesting.

Once the electrical part of the Lorentz force law is established, the full Lorentz force law follows from Lorentz invariance.

• -1 because you've replaced the need for the Lorentz force with the need for a definition of energy in terms of $E$, independently of from Maxwell's equations. – Larry Harson Sep 12 '13 at 0:35
• @LarryHarson: No, the definition of $E$ follows from Noether's theorem, as stated in the answer. – user4552 Sep 12 '13 at 0:40
• @LarryHarson You and I probably disagree on when to downvote, but here's my perspective so please consider it (you don't have to agree): Even if you were right and the definition of E didn't come from Noether's theorem as Ben says (I'm with Ben on this one) this isn't reason to downvote. A shift in perspective afforded by a "replacement" is not trivial and helps me understand subtle subjects. I certainly want to see as many "trivial perspective shifts" as I can: I'm only one person and I learn enormously from other peoples' perspective. Physics would be dreadfully the poorer otherwise. – Selene Routley Sep 12 '13 at 1:46
• @LarryHarson I agree that Ben's answer takes on more information to get to the result, so it strictly isn't answering the question EXACTLY as the OP thought of it, but he says that at the beginning ("a few assumptions") and, hey, sometimes questions just don't have a simple answer: there is quite a spectrum of meanings for "independent" unless you're talking true mathematical axiom systems. – Selene Routley Sep 12 '13 at 1:51
• Clever! Question: To apply Noether's theorem, my understanding is you need an EM Lagrangian. If so, can't you just add a free particle Lagrangian to the EM portion and deduce the particle equations of motion directly? – Art Brown Sep 12 '13 at 3:25

Steve B gives a very, very good answer, but I have one thing to add to his third point. He says if you assume the electric part of the force, you can derive the magnetic part from relativity. I have a different derivation for the magnetic part that doesn't exactly use relativity in an obvious way. I take a freely propagating e-m wave travelling between two metal plates. From Maxwell's equations we can get the induced charges in the plates, and also the induced currents. If we know the electrostatic force due to the charges, then the two plates must be attracted to each other. It turns out that the magnetic force is exactly equal and opposite to the electric force, so there is no net force between the plates. It's a nice calculation, and I'd like to say it allows me to derive the magnetic force, but I was never able to think of a physical reason why I would be entitled to assume that the total force between the plates must be zero.

• Can you check the link to your blog? – Greg P Feb 3 '12 at 17:31
• Thanks, Greg. I had an extra backslash at the end. It's OK now. – Marty Green Feb 3 '12 at 18:05

# Retraction

I have removed my own answer (which however can still be found in the revision record), because it has a counterfactual implication.

Consider a velocity field $${\bf u}({\bf x},t)$$, where $${\bf x}$$ is the position vector and $$t$$ is time. Let us say that $${\bf u}$$ preserves magnetic flux if and only if the magnetic flux through every closed curve, every part of which moves at velocity $${\bf u}$$, is constant — as if the flux were moving at that velocity. Then (as I said) Faraday's law for a fixed loop $${\cal C}$$ reduces to $$$$\label{2}\tag{2} \oint_{\cal C}{\bf E}\cdot d{\bf x} = -\oint_{\cal C}{\bf u}\times{\bf B}\cdot d{\bf x} \,.$$$$ So far: so good. But then I claimed that in so far as the $${\bf E}$$ field was due to the moving flux, we could localize the influence and interpret the above equality as element-by-element, obtaining $$$$\label{4}\tag{4} {\bf E} = -{\bf u}\times{\bf B}$$$$ as Faraday's law for a velocity field $${\bf u}$$ that preserved magnetic flux. Similarly, for the Maxwell-Ampère law (with no conduction current) for a velocity $${\bf u}$$ that preserved the electric displacement flux, I claimed $$$$\label{6}\tag{6} {\bf H} = {\bf u}\times{\bf D} \,.$$$$ Together, (4) and (6) would imply that if the velocity $${\bf u}$$ is flux-preserving (in both senses), then both $${\bf E}$$ and $${\bf H}$$ are perpendicular to $${\bf u}$$. This in turn would imply that a wave traveling at a flux-preserving velocity in an isotropic medium is TEM.

That implication is wrong. Counterexamples include:

• The TE and TM modes of a straight lossless rectangular waveguide; and

• the evanescent wave due to total internal reflection of a plane sinusoidal wave by a plane interface, and the superposition of the incident and reflected waves; both the evanescent wave and the superposition are TE for the s polarization and TM for the p polarization, but not both at once.

In both cases, a waveform travels at an obvious fixed velocity (that of the evanescent wave in the latter case), with no other change, so that the velocity is flux-preserving.

So the existence of a flux-preserving velocity does not give us a license to interpret the integral forms of the Faraday and Maxwell-Ampere laws in a localized manner.

Philosophically, the problem seems to be this: Because a flux-preserving velocity does not exist except in special cases, the flux per se is not some sort of "stuff" that moves, and does not become such in those cases in which, per accidens, a flux-preserving velocity happens to exist. And even if we accept the premise that all instantaneous influence is local, we cannot construct a valid physical argument by localizing the influence of "stuff that moves" if we don't physically have "stuff that moves"!

I hasten to add that equations (4) and (6) are still correct if we take $${\bf u}$$ as the ray velocity, the determination of which was the original reason for my interest in this matter.

So, in terms of my original purpose, the problem is this: Flux preservation doth not a ray velocity make.