# Why must a pure force depend on the four-velocity?

TL;DR Summary
Why must a pure force depend on the four-velocity of the test-particle?
Disclaimer: Please read in the following formulas ##E/c^2## instead of ##m##, because W. Rindler used relativistic mass, what might be confusing with today's usage of the term "mass".

I am reading the chapter "38. The formal structure of Maxwell's theory" in Wolfgang Rinder's book "Introduction to Special Relativity", 2nd edition.

In chapter "35. Three-force and four-force" he defined a force ##\mathbf{F} = \frac{d}{d\tau}(m_0\mathbf{U}) ## as "pure", if it does not change the rest-mass ##m_0##. A counter-example would be a force during an elastic collision. With ##\mathbf{U} \cdot \mathbf{F} = \gamma^2 (c^2\frac{dm}{dt} - \mathbf{f} \cdot \mathbf{u}) = c^2\frac{dm_0}{d\tau} = \gamma c^2\frac{dm_0}{dt}## he derived by arguing in the rest-frame of the particle:

The necessary and sufficient condition for a force to be pure is $$\mathbf{U} \cdot \mathbf{F} = 0 \Leftrightarrow m_0 = \text{constant}\ \ \ \ \ (35.9)$$
In chapter "38. The formal structure of Maxwell's theory" he starts with:
W. Rindler said:
The only two assumptions that we shall specifically make about the electromagnetic force will be that it is a pure force (i.e. rest-mass preserving) and that it acts on particles in proportion to the charge q which they carry. Beyond that, only ‘simplicity’ and some analogies with Newton's gravitational theory will guide us.

We begin by considering — and rejecting — certain simple possibilities.

Consider a field of three-force ##\mathbf{f}## which, like the Newtonian gravitational force, acts on a particle independently of its velocity, in some frame S. From the transformation equations (35.6) for ##\mathbf{f}## we then see that in another frame S’ such a force will depend on the velocity ##\mathbf{u}## of the particle on which it acts. So velocity-independence is not a Lorentz-invariant condition we can impose on a field of three-force.

We could, however, suppose that there exists a field of four-force ##\mathbf{F}## which acts on any particle independently of its velocity [such as the gradient field (35.10)]. By (35.5), the corresponding three-force would be ##\mathbf{f}= \gamma^{-1}(u)(F^1, F^2, F^3)##, which, as expected, depends on the particle’s velocity. But it would not be a pure force [cf. (35.9)], since, for fixed velocity-independent ##\mathbf{F}##, ##\mathbf{U} \cdot \mathbf{F}## cannot vanish for arbitrary ##\mathbf{U}## unless ##\mathbf{F}## itself vanishes. So we must here reject this type of field.

The next simplest case, and the one that actually applies in Maxwell's theory, is that of a force which everywhere depends linearly on the velocity of the particles on which it acts.
...
In the last section, he refers - besides to equation (35.9), which a am showing above - to the following equations:
##(35.10)\ \ \ \ \ F_\mu = \partial \Phi / \partial x^\mu##
##(35.5)\ \ \ \ \ \ \mathbf{F} = ... = \gamma(u)(\frac{1}{c}\frac{dE}{dt},\mathbf{f} )##

My question relates to the text, which I marked bold. Why must a pure force depend on the four-velocity of the test-particle?

Last edited:

Mentor
Why must a pure force depend on the four-velocity of the test-particle?
The text says so, right after what you bolded. If ##F## is independent of ##U##, the only way to always have ##U \cdot F = 0## for any arbitrary ##U## is to have ##F = 0##. Or, in ordinary language, in order for ##F## to always be orthogonal to ##U## no matter what ##U## is (which is what ##U \cdot F = 0## says), ##F## has to depend on ##U##.

vanhees71 and Sagittarius A-Star
2022 Award
If ##F## does not depend on ##U##, how would they always be orthogonal?

vanhees71 and Sagittarius A-Star
Mentor
That is interesting that he makes the EM force being a pure force a foundational assumption. The EM force is not always "pure" in this sense.

A photon can change the mass of an atom. And macroscopically if a EM wave is absorbed by a material it can get warmer which increases its mass.

vanhees71 and dextercioby
Mentor
A photon can change the mass of an atom. And macroscopically if a EM wave is absorbed by a material it can get warmer which increases its mass.
These processes can't be modeled with an electromagnetic force, at least not with the standard one. The standard EM force is ##f_\mu = q F_{\mu \nu} u^\nu##, where ##F_{\mu \nu}## is the EM field tensor. It is pure because of the antisymmetry of ##F_{\mu \nu}##, which makes it obvious that ##f_\mu u^\mu = 0##.

To model processes such as absorption or emission of EM waves (or photons) affecting the temperature (and hence the rest mass) of an object, you need a different model. There's no way to do it using the standard EM force given above. The best model we have of such processes, namely QED, does not model them using an EM force at all (you can of course derive the classical EM force given above as an appropriate limit of QED, but that limit excludes those absorption/emission processes).

dsaun777 and vanhees71
It is pure because of the antisymmetry of ##F_{\mu \nu}##, which makes it obvious that ##f_\mu u^\mu = 0##.
After the text I cited, W. Rindler says the same, but argues in the opposite direction. From ##\mathbf{U} \cdot \mathbf{F} = 0 ## he argues, that the electromagnetic field tensor, which he calls only "coefficient" at this stage, must be antisymmetric.

vanhees71
Mentor
After the text I cited, W. Rindler says the same, but argues in the opposite direction. From ##\mathbf{U} \cdot \mathbf{F} = 0 ## he argues, that the electromagnetic field tensor, which he calls only "coefficient" at this stage, must be antisymmetric.
Yes, that's because he's trying to derive the form of the Lorentz force from other assumptions. I was simply pointing out that the standard form of the Lorentz force is pure by his definition; I wasn't claiming that that is a derivation of anything.

vanhees71
Mentor
These processes can't be modeled with an electromagnetic force, at least not with the standard one. The standard EM force is ##f_\mu = q F_{\mu \nu} u^\nu##, where ##F_{\mu \nu}## is the EM field tensor. It is pure because of the antisymmetry of ##F_{\mu \nu}##, which makes it obvious that ##f_\mu u^\mu = 0##.
I would certainly consider the force density on a continuum to be standard also: $$f_\mu=F_{\mu \nu}J^\nu$$ That is a perfectly standard part of classical EM and need not preserve the mass.

To me, it seems like the assumption of a pure force is more of a statement about the matter being acted on than it is about electromagnetism itself. I would be immensely skeptical of any conclusions about EM obtained through this assumption.

Last edited:
vanhees71
Mentor
That is a perfectly standard part of classical EM and need not preserve the mass.
It will preserve the invariant mass of the 4-momentum stream described by ##J^\mu##, since ##f_\mu## will be orthogonal to ##J^\mu## because of the antisymmetry of ##F##. So you still can't use this model to describe something like absorption or emission of EM waves by matter that changes its temperature and hence its rest mass.

To me, it seems like the assumption of a pure force is more of a statement about the matter being acted on than it is about electromagnetism itself. I would be immensely skeptical of any conclusions about EM obtained through this assumption.
I don't think it's an assumption about either the nature of matter or the nature of EM by themselves, but an assumption (or a limitation) about the kind of interaction that one is modeling between them. The pure force assumption limits the kinds of interactions you can model; you can model things like a charged particle moving in an EM field, but you can't model things like absorption or emission of radiation by matter (except in some sort of limiting case where the interaction is perfectly elastic, so to speak--something like a solar sail that absorbs/emits no radiation but just responds to the radiation pressure).

Mentor
It will preserve the invariant mass of the 4-momentum stream described by ##J^\mu##, since ##f_\mu## will be orthogonal to ##J^\mu## because of the antisymmetry of ##F##. So you still can't use this model to describe something like absorption or emission of EM waves by matter that changes its temperature and hence its rest mass.
##J^\mu## is the four-current density, not a four-momentum stream (not sure what you mean by “stream” there). You seem to be implicitly assuming that ##J^\mu## is parallel to ##u^\mu##, but that is not generally the case.

For example ##u^\mu## is always timelike, but ##J^\mu## is spacelike in an ordinary circuit. You can use ##f_\mu=F_{\mu\nu}J^\nu## to model an ordinary current carrying resistor, which will give ##f^\mu## is actually parallel to ##u^\mu## and thus changing the mass.

The pure force assumption limits the kinds of interactions you can model;
I, for one, wouldn’t accept a formulation of classical electromagnetism that cannot even model a resistor, nor would I trust a derivation that rests on an assumption that is contradicted by an ordinary resistor.

dextercioby
Mentor
##J^\mu## is the four-current density
If you mean the charge-current density 4-vector, then I don't understand where the equation you wrote down comes from. The corresponding Maxwell equation is ##\partial_\nu F^{\mu \nu} = 4 \pi J^\mu##. You can't write a four-force equation with just the charge-current density; you have to include the mass of the matter somewhere.

If you mean something else by "four-current density", then I'm not sure what you mean by it.

Mentor
I, for one, wouldn’t accept a formulation of classical electromagnetism that cannot even model a resistor, nor would I trust a derivation that rests on an assumption that is contradicted by an ordinary resistor.
The "pure force" Rindler is talking about, and which I wrote down an equation for, is the standard Lorentz force. There's nothing wrong with it in the domain for which it is intended. That domain just doesn't cover all of electromagnetism, since, as already noted, it makes a restrictive assumption about the interaction of EM fields and matter.

I don't think Rindler claims that his "pure force" formulation does cover all of electromagnetism, although I'm not familiar enough with the book in question to say for sure. If he does make such a claim, I would agree that it can't be correct; but I would expect him to have been knowledgeable enough to realize that.

Mentor
If you mean the charge-current density 4-vector, then I don't understand where the equation you wrote down comes from.
You can see it here: https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism#Charge_continuum but it is just the covariant way of writing the standard Lorentz force density: $$\vec f=\rho \vec E + \vec J \times \vec B$$

You can't write a four-force equation with just the charge-current density; you have to include the mass of the matter
Why? The four force density does not depend on the mass. It depends on the charge and current densities.

You can use ##f_\mu=F_{\mu\nu}J^\nu## to model an ordinary current carrying resistor, which will give ##f^\mu## is actually parallel to ##u^\mu## and thus changing the mass.
I found this equation at Wikipedia under "Charge continuum". I'm not sure if it applies also to a resistor, were positive charges are at rest and electrons are moving.

EDIT: Maybe it does apply also there. Rindler writes something like ##\partial_\mu F^{\mu \nu} = 4 \pi \rho_0u^\nu= 4 \pi J^\nu## with the statement, that if the elementary particles move incoherently, the "rest frame" is that for which, in the limit of small volumes, there is no net current.

I assume, that it is derived from the following equation, by simply setting setting ##u^\nu = J^\nu / \rho_0## ...
##f_\mu = q F_{\mu \nu} u^\nu##

The electromagnetic field tensor would then still to be calculated from the divergence
##\partial_\nu F^{\mu \nu} = 4 \pi J^\mu##.

Last edited:
Mentor
it is just the covariant way of writing the standard Lorentz force density
Hm, I see a relation given in the Wikipedia article you referenced to the standard EM stress-energy tensor. I'll take a look at that.

In terms of that same Wikipedia article, Rindler appears to be talking about the "charged particle" force, which is the one I wrote down in an earlier post. That force is pure, but as I've said, it is obviously only of restricted application.

The four force density does not depend on the mass. It depends on the charge and current densities.
Yes, sorry, I misspoke. I meant the 4-velocity, but you addressed that elsewhere in your post.

Dale
Mentor
I'm not sure if it applies also to a resistor, were positive charges are at rest and electrons are moving.
Suppose that we have a pure E field in the x direction. Then the electromagnetic field tensor is
$$F^{\mu \nu } =\left( \begin{array}{cccc} 0 & -{E_x} & -{E_y} & -{E_z} \\ {E_x} & 0 & -{B_z} & {B_y} \\ {E_y} & {B_z} & 0 & -{B_x} \\ {E_z} & -{B_y} & {B_x} & 0 \\ \end{array}\right) =\left( \begin{array}{cccc} 0 & -{E_x} & {0} & {0} \\ {E_x} & 0 & {0} & {0} \\ {0} & {0} & 0 & {0} \\ {0} & {0} & {0} & 0 \\ \end{array} \right)$$ And if that field is applied to a resistive material then we have $$J^{\mu }=\left(\rho,{j_x},{j_y},{j_z}\right) =\left(0,{\sigma E_x},{0},{0}\right)$$ Then $$f_\mu = F_{\mu\nu}J^\nu =\left(\sigma {E_x}^2 ,0 ,0,0\right)$$

This is the usual power density in a resistor, and it is the rate at which the resistor is gaining thermal energy. Since the resistor is gaining energy but is not gaining any momentum this will change the mass of the resistor.

dextercioby, vanhees71 and Sagittarius A-Star
This is the usual power density in a resistor, and it is the rate at which the resistor is gaining thermal energy. Since the resistor is gaining energy but is not gaining any momentum this will change the mass of the resistor.

The four-current of individual electrons in the resistor is still time-like. The electrons gain kinetic energy and momentum in (x) direction by a pure force, but loose them regularly by collisions, which create heat.
E.M. Purcell - Electricity and Magnetism said:
4.8 Energy dissipation in current flow
...
If an electric field ##\mathbf{E}## is driving the ion of charge q, then ##\mathbf{F} = q\mathbf{E}##, and the rate at which work is done is ##q\mathbf{E} \cdot \mathbf{v}##. The energy thus expended shows up eventually as heat.
...
The ion acquires some extra kinetic energy, as well as momentum, between collisions.

Last edited:
Mentor
The four-current of individual electrons in the resistor is still time-like. The electrons gain kinetic energy and momentum in (x) direction by a pure force, but loose them regularly by collisions, which create heat.
Don’t you think classical EM should be enough to handle a simple macroscopic resistor without requiring a quantum mechanical analysis?

Mentor
Don’t you think classical EM should be enough to handle a simple macroscopic resistor without requiring a quantum mechanical analysis?
I don't think the description given was necessarily quantum mechanical, but it was statistical (but classical Maxwell-Boltzmann statistics should be sufficient). Such an analysis might not be necessary to describe a resistor, but it should be consistent with a continuum description, since ultimately the continuum description has to emerge from the behavior of individual electrons and atoms.

Sagittarius A-Star
Mentor
I don't think the description given was necessarily quantum mechanical, but it was statistical
If you are actually talking about electrons then you are either talking about QM or chemistry. If you are not actually talking about electrons then you should use a different word (e.g. “classical point charge”). Bringing electrons into classical EM or circuits is a “pet peeve” of mine. They are generally neither necessary nor beneficial.

Classical EM is formulated in terms of fields, charge densities, and current densities. Classical point charges (which electrons are not) actually cause several paradoxes, so they are to be avoided in any definitions and whenever they are not directly useful. When needed their behavior can be derived from the continuum as ##\rho=q\delta## and ##\vec j=q\delta \vec v##, so starting with the “statistical approach” is thinking about classical EM the wrong way around, IMO.

Such an analysis might not be necessary to describe a resistor, but it should be consistent with a continuum description, since ultimately the continuum description has to emerge from the behavior of individual electrons and atoms
They are certainly consistent. QM simply justifies the classical constitutive relations, like Ohm’s law.

Last edited:
vanhees71
I don't think Rindler claims that his "pure force" formulation does cover all of electromagnetism, although I'm not familiar enough with the book in question to say for sure.
Here is the content of the chapters I cited in a shorter form:

Dale
Gold Member
2022 Award
For a treatment of point-particle mechanics in special relativity, as far as it can be formulated, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

One should, however, be aware that classical point-particle mechanics cannot be consistently formulated. Even in the most simple case of charged particles in electrodynamics there's no completely consistent dynamics for such particles and the em. field due to the infamous "radiation-reaction problem". The best approximation one can get is the Landau-Lifshitz approximation of the Lorentz-Abraham-Dirac equation.

The situation is much better in the context of quantum theory. It turns out that what you really get is a non-Markovian dissipative equation of motion. See, e.g.,

https://doi.org/10.1016/0375-9601(91)90054-C

Dale