0
$\begingroup$

Lavoisier law states that for any isolated system, mass must be conserved over time.

Also, Einstein's equation $E = mc²$ shows energy-mass equivalency.

So, since mass must be conserved and energy and mass are equivalent, can I think of conservation of energy as a consequence of Lavoisier law?

$\endgroup$
  • $\begingroup$ I would say that energy conservation is the more fundamental law. When speaking of mass conservation in terms of chemistry (Lavoisier), that does not include the possibility to convert mass into energy ore vise verse. $\endgroup$ – Andréas Sundström Jul 17 at 9:15
3
$\begingroup$

Lavoisier was writing before relativity was known and his statement is wrong: mass is not conserved when relativistic effects are non-negligible. You cannot use his statement to imply that total energy is conserved.

The conservation of energy is due to a fundamental symmetry called time shift invariance. Noether's theorem tells us that this symmetry is linked a conserved quantity, and in the case of time shift invariance the conserved quantity is energy.

The equation $E=mc^2$ is a special case that applies only when the object with the mass $m$ is stationary. The full equation is:

$$ E^2 = p^2 c^2 + m^2 c^4 $$

where $p$ is the relativistic momentum. This reduces to $E=mc^2$ when $p=0$ i.e. when the mass $m$ is stationary.

$\endgroup$
  • $\begingroup$ i dont think it is wrong in relativistic setup. The conservation of energy holds for isolated systems and if isolated system is composed of several constituents moving with respect to each other (like gass in a box), then the kinetic energy also adds up to the overal mass of the whole system. Of course there is difference between mass - being defined by a rest frame of the system and energy that is frame dependent, but if you restrict yourself to rest frames of the system you are interested in, the two quantities are equal. $\endgroup$ – Umaxo Jul 17 at 10:40
  • 1
    $\begingroup$ If you consider two bodies of rest mass $m$ colliding and fusing in the COM frame then the rest mass of the final object is greater than $2m$. $\endgroup$ – John Rennie Jul 17 at 10:42
  • $\begingroup$ Of course it is, you need to add up the kinetic energy as i stated. But when you are measuring the two m´s and then both of them together, you are not in an isolated system. In relativity you cannot say that if mass of two isolated constituents is m, the mass of the whole system is 2m, but this has no influence on the fact that the mass in an ISOLATED system is conserved $\endgroup$ – Umaxo Jul 17 at 10:48
  • 1
    $\begingroup$ @Umaxo The two masses are an isolated system. There are no external forces acting on them. No energy is transferred in or out of the system. $\endgroup$ – John Rennie Jul 17 at 10:50
  • $\begingroup$ they are not. The rest mass m is measured only for one particle and that particle is not isolated system since other particle will soon hit it from outside. If you would measure the two particles mass as a whole- one single isolated system, then you would get the correct rest mass that contains the kinetic energy also, even if they didnt hit themselves yet. F.e. if you have an ideal gass in a jar, then the rest mass of the whole jar contains kinetic energy of the molecules also. But no molecule is isolated since they hit each other all the time. $\endgroup$ – Umaxo Jul 17 at 21:39

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.