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

$ \boldsymbol{u} \cdot \boldsymbol{f} = \boldsymbol{u} \cdot {d\boldsymbol{p} \over d\tau} = \boldsymbol{u} \cdot ({dm_0 \over d\tau}\boldsymbol{u} + m_0{d\boldsymbol{u} \over d\tau}) = c^2 {dm_0 \over d\tau} + m_0 \boldsymbol{u} \cdot {d\boldsymbol{u} \over d\tau} = c^2 {dm_0 \over d\tau} $

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 $dm_0 \over d\tau$ 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? :-)

  • $\begingroup$ Read the wikipedia article on rest mass... it is not by definition constant -- you're thinking of the fact that it's invariant across all inertial frames. and the invariant will only be changed by escaping (ex: via light or heat). $\endgroup$ Jan 6, 2012 at 0:20

1 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.

  • $\begingroup$ I see. So your answer basically answers all my questions :-) Thanks! $\endgroup$
    – Rafid
    Jan 6, 2012 at 13:33
  • $\begingroup$ Hmm, the last sentence sounds strange. Isn't the invariant mass of photos always zero in vacuum? If not, is their velocity timelike? Also, since the interactions propagate with the speed of light, they will never catch up with a photon, so a system of two photons cannot interact except when they collide. $\endgroup$
    – bkocsis
    Nov 17, 2020 at 22:37
  • $\begingroup$ @bkocsis It is a mathematical fact of four vector algebra. The individual photons always have c the velocity of light, their added four vector has an invariant mass if the angle between them is non zero. The two photon interaction is very very improbable, but the decay of a massive particle is not, the first example is the pi0, the recent example is the Higgs to two photons cms.cern/news/cms-precisely-measures-mass-higgs-boson $\endgroup$
    – anna v
    Nov 18, 2020 at 5:39
  • $\begingroup$ Agreed, the center of mass velocity of two photons is timelike, so in the center-of-mass frame they are literally at rest and their invariant mass is their total energy. But imagine a spherical shell expanding isotropically with the speed of light. The enclosed mass is zero once inside of the expanding shell of radiation. So while the invariant mass is finite for a system of photons, the gravity they generate at infinite time must always be exactly zero. $\endgroup$
    – bkocsis
    Nov 19, 2020 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.