Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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? :-)

share|improve this question
    
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). –  Chris Gerig Jan 6 '12 at 0:20
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
I see. So your answer basically answers all my questions :-) Thanks! –  Promather Jan 6 '12 at 13:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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