Can the Rest Mass of an Object Change Over Time?

ChrisVer
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I was wondering... can the rest mass of an object be time dependent? Like in a scenario where the body is losing mass?

(Sorry I meant for a title "rest mass time dependent?")
 
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Only if the object is losing mass by somehow radiating it away.
 
Or if you chop pieces off with a knife!
 
I think the usual application of this scenario is the rocket problem. :)
 
Well, this is the relativity subforum, where rest mass is synonymous with energy. What process doesn't change the energy?

The energy of a tree changes if I start walking to the left.
 
stedwards said:
this is the relativity subforum, where rest mass is synonymous with energy

No, it isn't. Energy is one component of the 4-momentum vector, and is frame-dependent. Rest mass is the invariant length of the 4-momentum vector, and is not frame-dependent.
 
stedwards said:
Well, this is the relativity subforum, where rest mass is synonymous with energy.
Using older terminology than Peter, you're talking about relativistic mass (a concept that has largely fallen out of favour, by the way), not rest mass.
 
ChrisVer said:
I was wondering... can the rest mass of an object be time dependent? Like in a scenario where the body is losing mass?
My $0.02 is that one should consider the dynamical group for the entire system. E.g., in a rocket, matter is ejected in one direction to change the rocket's velocity in the opposite direction. The rocket thus feels acceleration, so one must be cautious about naively applying the Poincare group to the rocket in isolation, and hence should also be cautious about the "invariant mass" Casimir of the Poincare group. Rather, one should probably decompose into com and relative coordinates, and find the full dynamical group applicable to the system with com dofs factored out. Then look at the Casimirs of that group applicable to the relative motion...
 
  • #10
PeterDonis said:
No, it isn't. Energy is one component of the 4-momentum vector, and is frame-dependent. Rest mass is the invariant length of the 4-momentum vector, and is not frame-dependent.

You love the word "no", Peter. No, no, no, no no. Intrinsic mass is normally associated with the norm of P.
 
  • #11
stedwards said:
[..] The energy of a tree changes if I start walking to the left.
- The energy of a tree is just as little affected by your change of choice of reference system as the amount of gasoline in a car is affected by your choice of using gallons or liters. Changing the observation (assuming no interaction) can not affect what is observed.
- Your choice of reference system is free, it does not depend on your state of motion (compare GPS)
 
  • #12
stedwards said:
You love the word "no", Peter. No, no, no, no no.
What's your problem, mate? Peter is not the one who made an incorrect statement. You did, in post #6, i.e.,
stedwards said:
[...] rest mass is synonymous with energy
That's just plain wrong. The word "no" is entirely appropriate in an answer to correct an error.

Intrinsic mass is normally associated with the norm of P.
If by "P" you mean 4-momentum, then,... PeterD said essentially the same thing. But that's different from your incorrect statement that "rest mass is synonymous with energy".
In modern parlance, "Intrinsic mass", "invariant mass", and "rest mass" are synonymous (though I prefer "invariant mass"). Cf. this Wiki page.
 
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  • #13
strangerep said:
applying the Poincare group to the rocket in isolation, and hence should also be cautious about the "invariant mass" Casimir of the Poincare group. Rather, one should probably decompose into com and relative coordinates, and find the full dynamical group applicable to the system with com dofs factored out. Then look at the Casimirs of that group applicable to the relative motion...

Yes I had the moving rocket in mind.
However I don't understand what you meant in the quoted message
 
  • #14
ChrisVer said:
Yes I had the moving rocket in mind.
However I don't understand what you meant in the quoted message
Uh oh. To answer that properly needs a lot more than a mere "$0.02 worth", but I don't have much spare time at the moment. Sorry. :oldfrown:
 

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