# Is there any way to justify or derive the form of the Lorentz force from relativity theory?

Lorentz force is in this form: $$\vec{F}=q[\vec{E}+\vec{u}\times\vec{B}]$$ As we know, it is Lorentz-invariant. Is there any way to justify or derive its form from relativity theory?

• I don't think the Lorentz force is Lorentz-invariant, because it depends in velocity. In fact, it's not even a four-vector. Can you explain further? Commented Jan 9, 2016 at 6:22
• @march it's form is invariant under Lorentz transformation. I mean if one transform both electric and magnetic, then the force is in this form again with new fields. Commented Jan 9, 2016 at 10:09
• well, when writen $f_\alpha=q F_{\alpha\beta}u^\beta$ it seems kinda obvious that this is the only possible covariant force linear in $\boldsymbol E,\boldsymbol B$. But the truth is, only experiments can justify this... Commented Jan 9, 2016 at 12:23

Although like any other physics, the Lorentz force law is experimentally measured, one could easily imagine an alternative universe where the discovery of relativity came before electromagnetic theory (suppose Michelson-Morely lived before Faraday, or look at the Ignatowskian approach to special relativity, where the form of the Lorentz transformation is argued without reference to light at all).

In this alternative history, two theoretical physicists might have been conversing:

Scene 1: A & B Talking About General Laws on "Electric Charge"

A: We know about this weird property recently discovered called 'electric charge'. What kind of laws would govern its motion?

B: Well, I've seen a charge sit pretty still in the laboratory or moving uniformly, so it's clearly possible to have an absence of electric effect. Let's postulate some electromagnetic field $F(\vec{X}, \vec{V},\,\cdots)$ ..."

A: Those of course all need to be four-vectors, or at least something to make the law Lorentz covariant ....

B: Of course, silly, I was just testing you. Anyhow, if $F$ to first order were linear, the law would have to be homogeneous ...

A: .... and you could get rid of $\vec{X}$ because, in the absence of other charges, the force doesn't depend on position ...

B: Of course: I was just testing you on that too ... anyhow, so we agree it's a linear, homogeneous function of velocity? ....

A: ... you mean four velocity ...

B: Of course: so a linear homogeneous function of four velocity is the simplest plausible thing to try. Hey quit interrupting my discovery of a new law will you ...

A: .. our law ...

B: OK. But you'll be second author, okay. Anyhow, here is what we have: our field has to be a rank two Lorentz covariant tensor to stand for a linear homogenous form of Newton's second law:

$$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu$$

where we'll write $q$ to measure the coupling strength between the 'charge' and the field.

A: Don't forget that a four velocity has a constant norm of unity .....

B:Oh DUUUH, I was just about to say that we can do better than this owing to obvious constraints like $\langle \vec{V},\,\vec{V}\rangle = 1$: so the acceleration has to be Minkowski-orthogonal to velocity, that gives us:

$$v^\mu\,\left(\eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu\right)v^\nu = 0;\forall \vec{V} \Rightarrow \eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu = 0$$

so the doubly covariant form of $F$ has to be skew symmetric to conserve the norm of the four velocity. Shall we publish now?

A: You do realize that that utter minging complete clot Heaviside is the editor of J. Modern Irreproducible Physics?

B: So ..?

A: We're never going to get that into print unless we write everything in his completely crap dotty crossy vectory notation! He's never going to go for that utterly indexxy $\eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu = 0$, he'll upend his cup of tea into his own lap and choke on his sticky bun as soon as he reads it ...

B: Oh, so how do we do that then?

A: Only if I'm first authoooor .....!

B: Oh, alright ....

Exeunt A & B to go off and eat sticky buns whilst A derives the dotty crossy version

Scene Two: After Sticky Buns

A: Here's the dotty crossy version of $\eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu = 0$. It can hold if and only if the force .....

B: the three-force .....

A: .... yes of course: we're talking dotty crossy here. Where was I? It can hold if and only if the three force acts according to:

$$\vec{F} = q\,\alpha\,(\vec{E} + \vec{V}\times \vec{B})$$

where $\alpha$ is an arbitrary scaling constant, which we can absorb into the definition of charge if we like, $\vec{E}$ is made of the off-diagonal zeroth row elements of $F$ and $\vec{B}$ is made of the three independent elements of the skew-symmetric $3\times 3$ lower right block of $F$.

The Lorentz force is not justified from relativity, but rather from experiments. In fact, the Lorentz force is used to define the fields $E$ and $B$. You are right that (the form) is Lorentz invariant.

EDIT: Most textbooks address the issue by saying that the relativistic (4-vector) form of the Lorentz force is the simplest that can be formed and so we use it as a working model a bit like how we use the standard expression for Poynting flux as a working model even though there are many ways to define it.

The only other argument I know (which I think is better) is essentially to show that the relativistic form is identical to the electric force in the rest frame of a particle and that therefore the form holds generally in all frames. The argument goes something like this:

In frame S in which a particle with charge q is at rest (momentarily), the force is:

$F_i=qE_i$

(from the definition of the electric field $E$). We are free to form a 4-vector of the form:

$F_{ik}U_k$

where $U_k$ is the 4-velocity. In the rest frame this 4-velocity is:

$U_k=(c,0)$

and so it is true that

$qF_{ik}U_k/c=(0,qE_i)$

(note that we divided by c to make things match). So we can see that the components of $qF_{ik}U_k/c$ and the Lorentz force are equal in the rest frame. And since a tensor equation that is valid in one frame must be valid in all frames, the form $qF_{ik}U_k/c$ for the force must be true in all frames.

• Ok but you know what the OP meant, right? They want to know if that form is an inevitable consequence of a small set of "reasonable" assumptions about Nature. Yes, those assumptions must be verified by experiment, but that's not the question. Commented Jun 1, 2016 at 0:57
• You're right, I've extended my answer to give the usual explanation :) Commented Jun 1, 2016 at 9:58

It's a matter of dimensional analysis. Assume the force and power are sums of monomial terms, each being a positive or zero power of any of the various quantities that might be involved.

Let $$Q$$ and $$P$$ respectively be the dimensions of electric and magnetic charge. Then $$PQ$$ is the dimension of action $$ML^2/T$$ where $$M$$ denotes mass, $$L$$ length and $$T$$ time duration. Let $$e$$, $$g$$, $$𝐯$$, $$𝐁$$, $$𝐃$$, $$𝐄$$ and $$𝐇$$, respectively denote electric charge, magnetic charge, velocity, the magnetic induction, the displacement, the electric field and magnetic field. Their dimensions are: $$[e] = Q, \hspace 1em [g] = P, \hspace 1em [𝐯] = \frac{L}{T}, \hspace 1em [𝐁] = \frac{P}{L^2}, \hspace 1em [𝐃] = \frac{Q}{L^2}, \hspace 1em [𝐄] = \frac{P}{LT}, \hspace 1em [𝐇] = \frac{Q}{LT}.$$

The only combinations that produce vectors with the dimension of force $$ML/T^2 = PQ/(LT)$$ are: $$e𝐄, \hspace 1em e𝐯×𝐁, \hspace 1em g𝐇, \hspace 1em g𝐯×𝐃.$$ The only combinations that produce scalars with the dimension of power $$ML^2/T^3 = PQ/T^2$$ are: $$e𝐯·𝐄, \hspace 1em g𝐯·𝐇.$$

Check their transform behavior under boosts, and you'll see that the combinations of these that transform together are: $$(e(𝐄 + 𝐯×𝐁), e𝐯·𝐄), \hspace 1em (g(𝐇 - 𝐯×𝐃), g𝐯·𝐇).$$ For stationary charges, i.e. when $$𝐯 = 𝟎$$, it's already given that the force on an electric charge $$e$$ would be $$e𝐄$$ and the force on a magnetic charge $$g$$ would be $$g𝐇$$. So, these are the respective 4-vectors for force and power in moving frames of reference.

Relativity actually has no bearing in the matter. The same arguments and conclusions apply non-relativistically, as well.

• Mathematically, force need not be linear in fields, see the expression of effective force in the Landau-Lifshitz equation. Commented Sep 7, 2023 at 7:42
• I think what you meant to say is that they need not be polynomial in the fields, since the discussion covers all polynomial cases, not just linear cases. There is no assumption of linearity anywhere, but only of polynomiality. You can also resort to the condition that at 𝐯 = 𝟎 the force reduce to the familiar forms of e𝐄 or g𝐇 and the power to 0. Under transform to 𝐯 ≠ 𝟎, (e𝐄,0) and (g𝐇,0) respectively become (e(𝐄 + 𝐯×𝐁), e𝐯·𝐄) and (g(𝐇 - 𝐯×𝐃), g𝐯·𝐇). Commented Sep 7, 2023 at 14:45
• I mean the Landau-Lifshitz equation for point particle in external field, see the equation (1) in here (they put $c=1$, but you can recover it to get the idea how force depends on both $m,c$). researchgate.net/publication/… Commented Sep 7, 2023 at 15:16
• Your resulting force/power pair is linear in fields, but there is a counterexample - the LL force has terms quadratic in Fμν. Thus your analysis makes some restrictive assumption - it seems the restrictive assumption is that the force does not depend on $m$ or $c$. Commented Sep 7, 2023 at 15:19
• There is no counterexample. The only polynomials that can be formed, vector in the force, scalar for the power from e,g,𝐁,𝐃,𝐄,𝐇 and 𝐯 are those cited. This is very easy to see. In each monomial, call their respective powers e,g,B,D,E,H,v. Then e+D+H=1=g+B+E, 2(B+D)+(E+H)-v=1(or2), E+H+v=1(or0) for force (or power). Thus e+g=1, (e,g)=(1,0) or (0,1). e=1 means D=0=H,B+E=1, g=1 means B=0=E,D+H=1. All powers are 0 or 1, except the cases v=2,B+D=1 for power; and no scalars arise there. The only way out is either (1) go non-polynomial or (2) bring something else in. Commented Sep 8, 2023 at 12:03