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- 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

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:

##(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?

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:

In the last section, he refers - besides to equation (35.9), which a am showing above - to the following equations:W. Rindler said:The only two assumptions that we shall specifically make about the electromagnetic force will be that it is apureforce (i.e. rest-mass preserving) and that it acts on particles in proportion to the chargeqwhich 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[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.pureforce

The next simplest case, and the one that actually applies in Maxwell's theory, is that of a force which everywhere dependslinearlyon the velocity of the particles on which it acts.

...

##(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?

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