3
$\begingroup$

Is the mass of a object at rest defined by $$E=mc^2$$ where $m$ is the rest mass. I.e. does the rest mass include every thing from thermal to gravitational potential energy and every other possible energy that it could have at rest. And thus if we write the following: $$total\ energy=mc^2+potential\ energy+thermal\ energy $$ are we double counting the potential energy and the thermal energy?

$\endgroup$
  • $\begingroup$ If you heat an object its mass increases because kinetic energy of particles,those that make the object increases.yes ofcourse thermal energy has mass. $\endgroup$ – Paul Jan 16 '15 at 13:36
  • 1
    $\begingroup$ Before anyone can give you a definitive answer to this question you would have to give definitive definitions for "potential energy" and "thermal energy" as you are using them. If you include microscopic degrees of freedom in the potential term than you can't have those same energies in the thermal term without double counting (and you should ditch the thermal term and include microscopic kinetic energies). Conventionally, though, you would leave the microscopic DOFs out of the potential terms and include them in the thermal term. $\endgroup$ – dmckee --- ex-moderator kitten Jan 16 '15 at 15:14
  • $\begingroup$ Related: physics.stackexchange.com/q/48490/2451 , physics.stackexchange.com/q/69080/2451 and links therein. $\endgroup$ – Qmechanic Aug 14 '16 at 17:41
7
$\begingroup$

The mass term includes all internal "energies". Heating up a body increases the internal kinetic energy. Binding energies also contribute to the mass (when nuclear fission occur, energy is freed and the products of the reaction are lighter than the original element), and this include any bond due to the fundamental forces of nature (which include, e.g., gravitational interaction, but only for parts within the body). Any extra energy coming from the interaction of the body as a whole with an external field doesn't contribute to the rest mass, which is a relativistic invariant.

$\endgroup$
  • $\begingroup$ Falling object has constant rest mass? Rest mass does not depend on altitude? So lifted object has constant rest mass? $\endgroup$ – stuffu Jan 21 '15 at 13:28
4
$\begingroup$

No, $E=mc^2$ covers only the mass energy of the object. Requirement of gravitational potential or thermal energy correctly require additional terms to the energy equation.

It's worth noting they are usually neglected because they are so small compared with the mass energy. Consider a rock of mass 1kg a few metres above the surface of the Earth:

  • Mass energy = $mc^2$ $\approx 10^{17}$J
  • GPE $\approx$ $mgh$ $\approx$ $50$J
  • Thermal energy = $mTq$ $\approx$ $300$kJ .
$\endgroup$
  • $\begingroup$ '''thermal energy correctly require additional terms to the energy equation.''' please elaborate a little on this. $\endgroup$ – Paul Jan 16 '15 at 14:25
  • $\begingroup$ If you really have a reason for needing the precise total energy, then you must consider contributions from sources other than mass energy. Thermal energy is one such source. $\endgroup$ – Kieran Hunt Jan 16 '15 at 14:30
  • 1
    $\begingroup$ You have to be careful here. In nuclear or particle physics, the mass of a composite system is usually defined as including the mass of the constituents and the energy of internal degrees of freedom. At macroscopic scales we call the energy in those internal degrees of freedom "thermal energy" and hold them to be separate from the mass (for mainly historical reasons that no one is trying to "correct" because the classical approach is simply more useful at human scales). So this answer is correct for usual classroom physics and macroscopic practice but leaves room for pedantry. $\endgroup$ – dmckee --- ex-moderator kitten Jan 16 '15 at 15:21
1
$\begingroup$

No, this equation only contains the mass and not the other forms of energy. In order for them to be added into the the equation and into the total energy (E), they have to be added separately but since they are negligible as compared to the mass energy, they are usually ignored. This means that their addition will not make much impact so generally they are not there in the formula.

$\endgroup$
0
$\begingroup$

There's no doubt that the energy of an object increases when the object is heated. Therefore there can't be any doubt that the mass of an object increases when the object is heated.

There's no doubt that the energy of an object increases when the object is heated by first lifting it up and then dumping it down on the hard ground. Therefore there can't be any doubt that the mass of an object increases when the object is heated by first lifting it up and then dumping it down.

Notice the pattern there: Undisputed increase of energy -> undisputed increase of mass.

When exactly does the energy of an object increase when the object is heated by first lifting it up and then dumping it down? One possible answer is: When the moving ground hits the still standing object. So therefore we can say that lifting an object does not change its energy or mass.

$\endgroup$
  • $\begingroup$ So you're saying that gravitational potential energy of an object does not cause an increase in mass? I'm not sure this is true. $\endgroup$ – Time4Tea Jan 17 '15 at 14:58
  • $\begingroup$ Potential energy of an object increases -> mass of the object increases. Lift a rock -> potential energy of the rock does not increase -> mass of rock stays same. Don't believe me? -> Read carefully the last sentence of Phoenix87's answer. $\endgroup$ – stuffu Jan 18 '15 at 10:40
  • $\begingroup$ I don't quite understand your comment. If you lift a rock then it gains gravitational potential energy. Phoenix87 said that energy due to external fields doesn't affect rest mass; that doesn't mean it doesn't affect mass. $\endgroup$ – Time4Tea Jan 18 '15 at 13:37
  • $\begingroup$ If you are in a tall accelerating rocket, carrying a rock "upwards", you feel the inertia of the rock staying constant. If you lift the rock over your head, you feel a decrease of inertia. Special relativity says so. General relativity agrees, and adds that in an uniform gravity field the same things happen. $\endgroup$ – stuffu Jan 18 '15 at 14:47
  • $\begingroup$ I'm not aware that special relativity says anything about non-inertial reference frames. $\endgroup$ – Time4Tea Jan 18 '15 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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