Sanity Check: Four-Force & Variable Mass

[SiennaTheGr8]

In classical mechanics, there's some debate about whether to define force as $\mathbf{f} = m\mathbf{a}$ or as $\mathbf{f} = \dot {\mathbf{p}} = d (m\mathbf{v})/dt$. The former is much more popular I think, and it has the virtue of always having a Galilean-invariant magnitude $f$, whereas the latter has the virtues of emphasizing a conserved quantity (momentum) and smoothing the transition to SR but leaves $f$ frame-dependent in a variable-mass situation ($\mathbf{f} = \dot{m} \mathbf{v} + m\mathbf{a}$).

In SR, we have the four-vector relation $\mathbf{F} = d\mathbf{P} / d\tau = d(m\mathbf{V})/d \tau$, which also equals $m \mathbf{A}$ if $m$ is constant. But here's my sanity check: even if $m$ isn't constant, the magnitude $F$ of the four-force is Lorentz-invariant, right? $\mathbf{F} = (dm / d \tau) \mathbf{V} + m \mathbf{A}$, but here $V$ is Lorentz-invariant, and $\mathbf{V} \cdot \mathbf{A} = 0$, so everyone should get the same value for $F$.

I'm 99.9% sure about this, especially since by other means I can calculate that $F$ must be invariant ($F^2$ equals the difference of the squares of the proper power and the proper force). Mostly just looking for confirmation that the logic above is sound.

There's something rather pleasing about this result, IMO.

 

[Sorcerer]

I am pretty sure that all four-vectors are Lorentz invariant.

 

[PeterDonis]

even if mmm isn't constant, the magnitude $F$ of the four-force is Lorentz-invariant, right?

Yes. The four-force is a 4-vector, and the magnitude of any 4-vector is Lorentz invariant.

I am prettyvsure that all four-vectors are Lorentz invsrient.

More precisely, the magnitudes of all 4-vectors are Lorentz invariant. In other words, given a 4-vector $F^\mu$, the number $g_{\mu \nu} F^\mu F^\nu$ is Lorentz invariant, where $g_{\mu \nu}$ is the metric tensor. (That number is actually the squared magnitude of the vector, which is more useful in most cases.)

 

[andrewkirk]

In SR, we have the four-vector relation $\mathbf{F} = d\mathbf{P} / d\tau = d(m\mathbf{V})/d \tau$, which also equals $m \mathbf{A}$ if $m$ is constant.

My understanding of the formula $\mathbf P = m\mathbf V$ is that the two bolded vectors are four-vectors and $m$ is the rest mass of the particle which, by definition, is constant. So the equation $\mathbf{F} = (dm / d \tau) \mathbf{V} + m \mathbf{A}$ collapses to $\mathbf{F} = m \mathbf{A}$ since $dm/d\tau=0$.

 

[PeterDonis]

$m$ is the rest mass of the particle which, by definition, is constant

No, it isn't. It is invariant--meaning its value at a given event is the same in all reference frames--but it is not constant; its value at different events in spacetime can be different. For a simple example, consider an object emitting radiation: its rest mass will decrease with proper time along its worldline.

 

[andrewkirk]

No, it isn't. It is invariant--meaning its value at a given event is the same in all reference frames--but it is not constant; its value at different events in spacetime can be different. For a simple example, consider an object emitting radiation: its rest mass will decrease with proper time along its worldline.

I was assuming the example was intended to apply only to theoretically pure point particles, where radiation would not apply.

If the object can lose rest-mass, either by particle emission (as in a rocket) or by radiation emission, the equation seems to me to require additional specification in order to avoid becoming ambiguous. What exactly is the meaning of the position vector $\mathbf X$ whose tau derivative is $\mathbf V$? If it relates to a centre of mass does it include ejected mass and radiation, and so on in the CoM calculation? It seems to me to become quite unclear what the items in the equation refer to.

 

[PeterDonis]

If the object can lose rest-mass, either by particle emission (as in a rocket) or by radiation emission, the equation seems to me to require additional specification in order to avoid becoming ambiguous. What exactly is the meaning of the position vector $\mathbf X$ whose tau derivative is $\mathbf V$?

In the example I gave, it would be the 4-position of the object that is radiating. I was idealizing the object as a point particle, which is what is usually implied when you use 4-vectors. If you're modeling an object with internal structure, either you would model it as multiple point particles, possibly connected by interactions idealized as springs or something like that, or you would pass to a continuous model, where the object is described by a stress-energy tensor, not a 4-vector. The continuous case introduces complications that I think are beyond the scope of this thread; that's why I was using the simple point particle model as an illustration.

 

[PeterDonis]

I was assuming the example was intended to apply only to theoretically pure point particles, where radiation would not apply.

There is no issue with having a point particle radiate, if you're not concerned with the details of the radiation but just want to have it carry energy and momentum. The radiation could be modeled as a series of "photons"--idealized massless particles carrying finite 4-momentum and moving on null worldlines. (Or you could model it as a continuous "null dust" with a traceless stress-energy tensor, but again, I think that's more complication than is necessary for this discussion.)

 

[SiennaTheGr8]

Thank you.

 

[andrewkirk]

There is no issue with having a point particle radiate, if you're not concerned with the details of the radiation but just want to have it carry energy and momentum.

There's a tricky nuance that crops up if the particle radiates, about whether $\mathbf F,\mathbf P$ are interpreted as the force on, and momentum of, the radiating particle, or on/of the system of particle plus its emissions.

If it is of the system then, by conservation of momentum, $d\mathbf P / d\tau = 0$, so the equation does not predict the worldline of the radiating particle.

On the other hand, if it is of the radiating particle, we need to note that $\mathbf F$ includes the 'recoil' force applied to the particle every time it emits/radiates a particle (be it massive or massless). So if the radiating particle is charged and is moving in an electric field, we need to take account of the fact that $\mathbf F$ is not just the electric force on the particle, but the combination of that force with the recoil force from its emissions.

 

[PeterDonis]

if the radiating particle is charged and is moving in an electric field, we need to take account of the fact that $\mathbf F$ is not just the electric force on the particle, but the combination of that force with the recoil force from its emissions.

Yes, agreed. I was glossing over that by pretending that the particle was uncharged, but of course in the real world uncharged particles don't radiate, at least not electromagnetically.

 

[vanhees71]

See

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

Sect. 2.1.

Please note that mass is mass and energy is energy. Since 1908 (Minkowski's famous talk on SR spacetime) we know that mass is a Lorentz scalar and should be treated as such. Why there's still some "relativistic mass" around, is a mystery to me. Einstein abandoned this idea very quickly, why not all textbookwriters since then? I'm puzzled, why misconceptions seem to be longer lived than their resolutions!

 

[SiennaTheGr8]

See

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

Sect. 2.1.

Please note that mass is mass and energy is energy. Since 1908 (Minkowski's famous talk on SR spacetime) we know that mass is a Lorentz scalar and should be treated as such. Why there's still some "relativistic mass" around, is a mystery to me. Einstein abandoned this idea very quickly, why not all textbookwriters since then? I'm puzzled, why misconceptions seem to be longer lived than their resolutions!

I always enjoy your contributions (and your SRT document! any plans to finish it?), but I don't see the relevance of this response—nobody in this thread is confused about the distinction between energy and mass, and "relativistic mass" hasn't come up at all. Did you mean to post this in a different thread?

 

[vanhees71]

No, in #4 was a suggestion that $m$ is dependent on proper time. I've overread that this misconception was corrected in the next step. Sorry for the confusion. I hope to find the time to write the SR script further. The next chapter is electromagnetism...