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Can the change in energy change the mass?

  1. Oct 23, 2011 #1
    If we give an object kinetic energy, will its mass increase (because of E=mc^2)?
    If not, why shouldn't it, because we are raising the energy of the object, right?
  2. jcsd
  3. Oct 23, 2011 #2
    Etotal = Kinetic energy + rest energy = γmc2. Making the velocity of an object larger increases the total energy, but it does nothing for the mass that is constant. E=mc2 refers only to the rest energy, that is also constant and independent of velocity.
  4. Oct 23, 2011 #3
    oh, but what about that equation Einstein had stated, using that you could prove that no object can travel faster than light because it would have gained a lot of mass by then.
  5. Oct 23, 2011 #4
    No, you misunderstand - mass is invariant, though people sometimes talk about "relativistic mass", which is γm. Nothing having mass can reach the speed of light because it would take an infinite amount of energy to get it going that fast, not because it's mass has changed.
  6. Oct 24, 2011 #5
    If we apply any force (non-zero) to an object of any mass (finite) it will accelerate. The speed of light is finite, so the object can attain it with its acceleration, even if it takes millenniums to do so. I agree it will take a lot of energy to go that fast (W=Fd), but why infinite?
  7. Oct 24, 2011 #6
    Total relativistic energy is [itex]\gamma m c^{2}[/itex] where [itex]\gamma = \frac{1}{\sqrt{1-(\frac{v}{c})^{2}}}[/itex]. If v=c then this quantity goes to infinity.
  8. Oct 24, 2011 #7
    Maybe AudioFlux refered to the last two equations on page 919 in Einstein A "Zur Elektrodynamik bewegter Körper" Ann. Phys. (Leipzig) 17 [322] 891 – 921 (1905) [ http://www.christoph.mettenheim.de/einstein-dynamik.pdf [Broken] ]. This would be not even the "relativistic mass" but special cases for "m" in F=m·a (used in an inadmissible manner for variable mass plus arithmetic error).
    Last edited by a moderator: May 5, 2017
  9. Oct 24, 2011 #8
    I do not see any of those 'grand objections' to relativistic mass, call it something else, a rose by any other name.

    The term rest mass is well defined but if you look closer is it actually so far removed from relativistic mass? Typically there is nothing but rest in matter, and in some cases the relativistic mass is a large component to the rest mass. Even a substance cooled as close as possible to 0K still has motion thus, basically relativistic mass.

    Something to ponder on: if all elementary particles are points then can they even have mass in GR? Think about it: What is the stress-energy tensor for a point? Perhaps everything is relativistic mass, there is really no rest mass at all.
  10. Oct 24, 2011 #9
    Neither do I, but I see that the relativistic mass is not necessary.

    As the rest mass is not additive it is not quite clear what this means. This argumentation might work for a body at rest because its rest mass is equal to its relativistic mass but it fails for a moving body.
  11. Oct 24, 2011 #10
    If you accelerate a particle with so much energy that additional matter-antimatter particle pairs form upon the accelerated particle's collision with a nucleus, the extra particles, which are ejected from that collision, have their own rest mass derived from the energy provided by the work done during the acceleration. The more relativistic energy embodied by that particle, then clearly the more mass of particles that could be generated by that impact. This method is, after all, how the top quark was proven to exist in experiment.
  12. Oct 24, 2011 #11
    It matters to know whether the acceleration occurs along a geodesic or not. The former (coordinate acceleration) does not change the mass, whereas the latter (proper acceleration) does.
  13. Oct 24, 2011 #12
    The average rest energy per charge is less when the charges are bounded in an atom than when they are free. This is because some of that rest energy was converted into EM radiation. That EM radiation accounts for the missing m_0. Including this radiation as a part of the physical model is needed so that the tally of m_0 remains constant, in accord with the first law of thermodynamics. Modelling the m_0 of an incomplete system of particles as a constant will lead to problems if the m_0 of input and output of that system are unequal. In non-inertial situations, such as that described by the opening post, they are *not* equal. As far as proper accelerations go, conservation of momentum requires either an "action" mass or photon whose loss of m_0 is to the accelerated mass and losses, but this action mass or photon has been ignored, and it has been assumed that objects cannot absorb additional m_0 in any amount at all. Such has been, is, and remains a mistake. Relativistic mass that arises non-inertially involves the transfer of m_0's from one body to another. Relativitisic mass that arises due to coordinate acceleration has nothing at all to do with such m_0 transfer. The often cited relativistic energy equation *CANNOT* make this distinction.
    Last edited: Oct 25, 2011
  14. Oct 25, 2011 #13


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    Well, we can say that rest mass of object does not increase.
    Say we have reference mass and object under question. We measure mass of the object using reference mass.
    Then we accelerate our object and reference mass so that they remain at rest in respect to each other. If we now measure mass of our object in respect to reference mass the measurement is the same as before acceleration. So we can say that rest mass of our object has not changed.
    But to me this seems rather trivial and it does not demonstrate anything worth discussing.
  15. Oct 25, 2011 #14
    I think I understand now, thanks everyone.
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