# Q & A: relativistic dynamics

Q:
Dear Professor, I am an Italian that, after focussing all his studies on humanities only, has decided to remedy his complete ignorance in the field of exact and natural sciences. I have studied a university textbook of mathematics (calculus, linear algebra, geometry ...) and am teaching myself some physics. My textbook summarily talks about the conservation of relativistic momentum and then proposes many exercises, but I am not totally sure that my assumption are correct, and I take the liberty to bother you with some question, since I know no expert in physics and don't attend physics classes (though I am considering the possibility of studying physics at university, notwithstanding my three children :-)). 1)An isolated system doesn't exchange energy with the outside. Is it correct to say that since the variation of the total energy of an isolated system (for ex. an isolated system where a collision happens) is DeltaE=0 [I write Delta because I am not sure that the Greek letter is correctly displayed] and the variation of the total relativistic mass of the system is Deltam=DeltaE/c^2 [I use the symbol ^ to precede the exponent] then the variation of the total relativistic mass of the system is Deltam=DeltaE/c^2=0? 2) I know from classical mechanics that collisions systems acted upon by a null resultant force conserves momentum; if they are elastic collisions, they also conserve kinetic energy, if they are inelastic collisions, they conserve only momentum, not kinetic energy. I think the same applies in relativistic mechanics: am I right? 3) Since the total energy of a system is E=m_0c^2/sqrt(1-(v/c)^2) [where m_0 is the total mass and where I use the Python notation sqrt to mean "square root"] and its total kinetic energy is K=mc^2-m_0c^2, I think that the non-kinetic energy of the system is E-K=m_0c^2. So, if non-kinetic (E-K=m_0c^2) energy changes (ex.: when a spring is compressed), is it correct to say that the total rest mass m_0=(E-K)/c^2 of the system change? I think that Delta(E-K)=c^2Deltam_0 ... Is gravitational potential energy included in E=m_0c^2/sqrt(1-(v/c)^2) and does the rest mass change also apply when gravitational potential energy changes? Hoping that I am not bothering you too much and that you can help me, I thank you with all my heart and, if you are Christian, I wish you a happy Easter. Yours sincerely, Davide
- Davide (age 34)
Genova, Italy
A:
Davide- We're very pleased to help out.

1. You are correct regarding E. You are also correct regarding m if it is simply defined to be E/c2, i.e. the inertial mass, the quantity that appears in p=mv. Some people prefer to use m to describe the rest mass, sqrt(E2-p2c2)/c2. For an isolated system, it too is conserved since both E and p are conserved, so delta_m is zero either way. If someone decides to describe the system as the sum of a number of subsystems (e.g. elementary particles) then the sum of their separate energies is conserved so long as the interaction energies are not important. Then the sum of their inertial masses is conserved. The sum of their rest masses need not be conserved, since the sub-systems can exchange momentum.

2. yes

3. Yes, the rest mass changes when the non-kinetic part of the energy changes. Note that due to the conservation laws, that cannot happen due to any purely internal processes in an isolated system, but requires interactions with the outside. When people say that some nuclear reaction changes the rest mass, what they are really doing is taking the sum of the rest masses of some parts, not the rest mass of the total system. Those two quantities can be very different, as you can easily work out in two-particle examples.

As far as the gravitational potential energy goes, I hesitate to give too confident an answer since my knowledge of General Relativity is sketchy. However, to lowest order  you do include the gravitational potential energy in the total energy and thus in the mass calculations. You can see where this leads to complications, since the gravitational energy is a source of the mass which enters into the gravitational fields, leading to the famous non-linearity of General Relativity. For our purposes here, consider the following situation: two stars in a highly elliptical mutual orbit. Some of the energy goes back and forth between kinetic and gravitational potential forms. If they contributed differently to the inertial mass, then momentum conservation would require that the entire binary speed up and slow down from the point of view of someone who says it has net momentum. From the center of mass point of view it wouldn't accelerate at all. That would make even approximate inertial frames impossible. So the gravitational potential energy counts as part of the total energy.

These were great questions, very clear.

Mike W.

(published on 04/23/2011)

## Follow-Up #1: you're welcome

Q:
Thank you with all my heart! Enlightening answer!
- Davide (age 34)
Genoa, Italy
A:
mw

(published on 04/26/2011)