It is clear that the kinetic energy can be derived as $(m-m_0)c^2$. However, why do we say that $m_0c^2$ is the rest mass energy? It seems that this mass-energy equivalence for rest mass is just a easy convention but cannot be proven, i.e. we can allocate any value to the rest mass energy. Is that true?
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ Word of warning: never, never, never use the "relativistic mass" as a single symbol. Rest mass is $m$, "relativistic mass" is $\gamma m$. Hiding the $\gamma$ factor in with the $m$ is something physicists did about a century ago, before they knew any better, and it only leads to confusion. $\endgroup$– user10851Commented Jan 15, 2015 at 11:05
-
2$\begingroup$ I'd go further and suggest that you purge any mention of "relativistic mass" from your brain. The concept is unnecessary and leads to confusion. It was abandoned many decades ago, but appears to persist. I'm not sure why. People must be reading old books. And perhaps writing new books based on old outdated books? $\endgroup$– garypCommented Jan 15, 2015 at 15:06
-
$\begingroup$ How is $(m-m_0)c^2$ equal to kinetic energy? it's equal to kinetic energy plus some terms with higher orders in v $\endgroup$– Shuheng ZhengCommented Apr 6, 2022 at 20:46
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
The mass energy equivalence is proved every day in nuclear reactors around the world. When a $^{235}$U nucleus fissions the amount of energy given off is exactly equivalent to the mass change multiplied by $c^2$.
The most precise test I know of was done in 2005 by a combined team from MIT, NIST and ILL in 2005. They found that $E$ differs from $mc^2$ by at most 0.0000004, or four-tenths of 1 part in 1 million.
answered Jan 15, 2015 at 10:29
$\begingroup$
$\endgroup$
0
No, it is not true and it can be checked in nuclear reactions. When a nucleus disintegrates, the difference in mass between the parent nucleus and all the decay products (nuclei and particles), goes to the kinetic energies of the decay products.
$\begingroup$
$\endgroup$
3
$m_{0}c^{2}$ is the energy of the particle in the frame where the particle is at rest, that is the frame where its momentum is zero. Thats why its called rest mass energy. If you look for the expression for the kinetic energy of the photon you will find that there is no such analog "rest mass energy" term in this case because you cant find any inertial frame where the photon is at rest because you will need to find a frame that moves at speed of light and that frame does not exist
-
$\begingroup$ The question wasn't about the rest mass, but about why it stores an energy of $m_0 c^2$. $\endgroup$– SofiaCommented Jan 15, 2015 at 10:36
-
$\begingroup$ Ok, sorry I didnt get the question. $\endgroup$ Commented Jan 15, 2015 at 10:45
-
$\begingroup$ you still can give an interesting answer and get points, if you can offer an example different from what was already said. Good luck! $\endgroup$– SofiaCommented Jan 15, 2015 at 10:49
$\begingroup$
$\endgroup$
Antimatter annihilations convert one hundred percent of rest mass into energy. So this means there is mass energy equivalence with rest mass. For example in positron electron annihalation, 511 Kev from the electron and 511 Kev from the positron are combined to give a total of 1022 Kev or 1.022 Mev of energy in antimatter annihilations.
