Can non-free forces change the rest mass?

Question

While reading Hobsen et al.'s "General Relativity: An Introduction for Physicists", I came across a bit confusing derivation. Multiplying the 4-force and 4-velocity, the following derivation can be made

\begin{align*} \boldsymbol{u} \cdot \boldsymbol{f} &= \boldsymbol{u} \cdot {d\boldsymbol{p} \over d\tau} \\[4pt]&= \boldsymbol{u} \cdot \left({dm_0 \over d\tau}\boldsymbol{u} + m_0{d\boldsymbol{u} \over d\tau}\right) \\[4pt]&= c^2 {dm_0 \over d\tau} + m_0 \boldsymbol{u} \cdot {d\boldsymbol{u} \over d\tau} \\[4pt]&= c^2 {dm_0 \over d\tau} \end{align*}

After this derivation, the authors make the following conclusion:

where we have (twice) used the fact that $\boldsymbol{u} \cdot \boldsymbol{u} = c^2$. Thus, we see that in special relativity the action of a force can alter the rest mass fo a particle! A force that preserves the rest mass is called a pure force and must satisfy $\boldsymbol{u} \cdot \boldsymbol{f} = 0$

But I have the following objections and questions about this derivation:

1. The rest mass is by definition a constant, so it should have been considered a constant while differentiating.

2. If we go back to Newton's second law, which is still valid under the special theory of relativity (though with some correction), the mass is the resistance of a body to changes in velocity, i.e. the larger the mass is, the stronger the force we need to change its velocity. But a non-free force seems to contradict this basic concept when $\large dm_0 \over\large d\tau$ is negative, because this means that the force is reducing the resistance of the body towards the force. As a funny comparison, imagine that the harder you push a heavy box, the lighter it becomes (which is obviously not the case even in Newtonian mechanics, not to mention that special relativity predicts the opposite, i.e. the faster the body is, the harder it becomes to increase its velocity)!!

3. Unless the mass is being converted to energy or transferred somewhere else (which is not inferred from the derivation, as the derivation comes straightforward from the force equation without depending on any other equation), where is the mass going?! Isn't this contradictory to the conservation of mass an energy law?

4. If we assumed in this derivation that the rest mass is variable, why didn't we do so in many other derivations in the special theory of relativity?

5. Do we have examples of such forces anyway? :-)

Answer 1

I second the suggestion of @ChrisGerig in the comment above about reading the wiki article.

This is the relevant paragraph:

If the system consists of more than one particle, the particles may be moving relative to each other in the center of momentum frame, and they will generally interact through one or more of the fundamental forces. The kinetic energy of the particles and the potential energy of the force fields increase the total energy above the sum of the particle rest masses, and contribute to the invariant mass of the system. The sum of the particle kinetic energies as calculated by an observer is smallest in the center of momentum frame (or rest frame if the system is bound).

What they call a particle is not an elementary particle, i.e. one which is point like and whose invariant mass is constant on all frames. Once there is a system of particles, even two photons, their invariant mass is variable.

Answer 2

Thus, we see that in special relativity the action of a force can alter the rest mass fo a particle!

This demonstration of force changing the rest mass is nonsense, based on a misconception of what net force is. Only if we define net force $f^\mu$ as $\frac{d(m u^\mu)}{d\tau}$, or if we assume net force obeys this relation (as a physical law), then $u_\mu f^\mu$ is giving rate of change of mass $m$. But this a wrong idea of net force, since any reasonable concept of net force that can be used to formulate and describe equations of motion for variable mass systems contradicts it. More arguments on this below.

But before that, notice that applying the conclusion (force altering mass) to a non-relativistic motion of a falling body losing mass equally in all directions (e.g. a falling water droplet that evaporates in all directions), we would be lead to say that net force alters mass of the falling droplet. Of course, that is nonsense, as the only contribution to net force is gravity. The mass in this example changes for different reasons, and net gravity force acting on the body has nothing to do with it. One can't find how fast the mass is changing just from net gravity 3-force. And if there was no external force, the droplet would sit in place, losing mass, and no actual force could be blamed for it losing mass.

The whole argument is most probably based on the misconception that net 4-force is always, even if rest mass of the body $m$ changes in time, equal to $$ \frac{d(mu^\mu)}{d\tau}. $$

This statement isn't true, because in non-relativistic limit, it doesn't connect to value of net force in non-relativistic mechanics when describing variable mass systems. In non-relativistic mechanics, the correct relation between net force and change of velocity, for both constant and variable mass systems, is given by Newton's 2nd law :

$$ \mathbf F_{net} = m\mathbf a. $$ It is not true that variable mass systems obey $\mathbf F_{net} = \frac{d\mathbf p}{dt}$; this is a common idea but it is clearly incorrect, and easily disproven e.g. by the fact the rocket equation is correct.

In relativistic mechanics, we have the relativistic, more accurate counterpart of 2nd law: $$ \mathbf F_{net} = m\frac{d(\gamma \mathbf v)}{dt}. $$

The 4-vector expression of this law is $$ f_{net}^\mu = m\frac{du^\mu}{d\tau}. $$

Again, the relation $\mathbf F_{net} = \frac{d\mathbf p}{dt}$ or $f_{net}^\mu = \frac{dp^\mu}{d\tau}$ does not hold for variable mass systems. There is no differentiating mass by time in 2nd law relating net force and change of velocity, whether non-relativistic or relativistic. The mass is never differentiated in 2nd law; time derivative of mass can come into the equation only through additional term in $\mathbf F_{net}$.

The theoretical reason is this. 2nd law in non-relativistic mechanics is based on experience with systems of constant mass. It can be generalized to systems of variable mass, like rocket, or evaporating droplet, but this generalization is not replacing $m\mathbf a$ with $d\mathbf p/dt$; instead, $m\mathbf a$ stays, and the meaning of $\mathbf F_{ext}$ is altered to contain additional terms, which we can interpret as additional forces due to the leaving/incoming mass. These forces (their value and direction) depend on details of how the mass is leaving/incoming. In case of the droplet, due to isotropic loss, there is no additional force; in case of a rocket, there is, but it depends on which direction the nozzle oriented into. These details can't be expressed by the bogus terms $\frac{dm}{dt}\gamma \mathbf v$ or $\frac{dm}{d\tau}u^\mu$ coming from the incorrect notion of net force.

All this can be seen on the rocket equation

$$ m\mathbf a = \mathbf F_{ext} + \frac{dm}{dt} \mathbf u' $$ where $\mathbf F_{ext}$ is sum of external forces (gravity and air friction), and $\mathbf u '$ is final velocity of combustion products in the frame of the rocket.

One can never derive this equation from the incorrect idea that net force is equal to $\frac{d(m\mathbf v)}{dt}$. This is easy to see, as one would get bogus term $\frac{dm}{dt}\mathbf v$ that is not present in the rocket equation. Instead, the rocket equation can be derived from 2nd law as defined above, applied to all particles making up the rocket and the combustion products, each of which has constant mass. The term $\frac{dm}{dt}\mathbf u'$ comes in as expression of force of the combustion products back on the compact rocket.

So since net force on a body with varying mass isn't given by $d(m\mathbf v)/dt$ in non-relativistic mechanics, it can't be given by $d(m\gamma\mathbf v)/dt$ in relativistic mechanics; and net 4-force can't be $d(mu^\mu)/d\tau$, as its spatial components would not approach the correct values of net force in the non-relativistic regime.

Instead, we should stick to the notion of net force that is sum of individual forces, and it is equal to sum of terms $m_a \frac{d\mathbf v_a}{dt}$, or $m_a \frac{d(\gamma_a \mathbf v_a)}{dt}$. Change of mass in mechanics happens only via loss/income of particles of constant mass, or in EM theory, it is due to change of internal energy of the body; it it is not connected in any universal way with the actual net force on the body.

If we want $f_{net}^\mu$ to be a 4-vector, zeroth component should be defined in analogous way as the 3D components: $$ f_{net}^i = m\frac{d u^i}{d\tau} = m\frac{d( \gamma v_i)}{d\tau}, $$ i.e. by $$ f_{net}^0 = m\frac{du^0}{d\tau} = mc\frac{d\gamma}{d\tau}. $$ So we have

$$ f_{net}^\mu = m\frac{du^\mu}{d\tau}, $$ with mass in front of time derivative. With this definition, even while mass can be changing in time, rate of change of mass is not determined by the "dot product" $u_\mu f_{net}^\mu$. Instead, we have

$$ u_\mu f_{net}^\mu = u_\mu m \frac{du^\mu}{d\tau} = 0, $$ whether mass changes or not.

This is not a problem, it is just a good definition connecting consistently to those in non-relativistic mechanics. Change of mass is a phenomenon that can't be explained as effect of net force alone, and it is nonsensical to try to express rate of change of mass in terms of net force in that way.