I have a small confusion in interpreting various forms of energy. Suppose two particles move towards each other with velocity $v$ and stick to each other as seen by an observer on the ground. The collision is not elastic. From Newtonian point of view, energy is lost in the collision and is liberated in the form of heat and sound (may be other forms as well).

But when we define $E$ as $\gamma mc^2$, we don't need to include any such energy losses. Why? Heat and sound are still produced, so where are those contributions in the expression of energy?

EDIT: When I say that losses are not included, I mean something like this; when writing down energy conservation using relativistic expression, we end up getting final mass greater than initial mass and we say that the energy missing is in the form of excess mass. So I asked that what about the energy in the form of heat and sound? Whatever expression you use, if heat and sound are produced, they are produced, regardless of the dynamics. So why do we account for energy losses as heat and sound for Newtonian case while we say that that energy is in excess mass while using relativistic expression when there is heat and sound in both cases?

EDIT-2: Replying to @enumaris

Applying energy conservation in the ground frame:

$$\frac{2m_{oi}c^2}{\sqrt{1-\frac{v^2}{c^2}}}=2m_{of}c^2$$ So the arguement that one needs is that final rest mass $m_{of}=m_{oi}/\sqrt{1-\frac{v^2}{c^2}}$. Here I didn't write down the energy losses due to other forms and used only rest mass arguement.

  • $\begingroup$ Don't understand why you're saying that when using the relativistic E (=${\gamma}mc^2$) that we don't need to include energy losses. Suggest expanding your thoughts and describing precisely what you mean. $\endgroup$ – Samuel Weir Apr 17 '18 at 16:54
  • $\begingroup$ Why do you say that losses to thermal energy are not included? $\endgroup$ – garyp Apr 17 '18 at 16:54
  • $\begingroup$ Just take a look, I made an edit. Is it fine now? $\endgroup$ – Ankur Singh Apr 17 '18 at 17:04
  • $\begingroup$ @ab. "So why do we account for energy losses as heat and sound for the Newtonian case while..." - If you're looking at the collision of, say, two identical masses in their center of mass frame then, yes, you have a combined mass of $2{\gamma}mc^2$ immediately after the collision. That excess energy will then typically go into other forms of energy such as heat, sound, light, x-rays, etc. In short, I think that that "excess mass" state is generally a temporary situation. $\endgroup$ – Samuel Weir Apr 17 '18 at 17:14
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    $\begingroup$ Can you show a source that stated that when you define $E=\gamma mc^2$ you don't need to include energy losses? It would appear that some context was missed when you were learning this since it's certainly untrue that you don't have to consider these losses in relativity. You can certainly still produce heat and sound in special relativity, it would be a very unphysical theory otherwise! $\endgroup$ – enumaris Apr 17 '18 at 17:16

The expression E=mc^2 tells the amount of binding energy in the nucleus of an atom caused by mass dissipation, m. When the nucleons rearrange or undergo decay of nucleus. The change in configuration always cause increase in magnitude of binding energy. An equivalent amount of energy is released in the form of radiation, mostly gamma rays. The heat and sound energy effects are caused by that radiation itself. Sound is merely vibration of air molecules which is caused during nuclear bomb explosion due to its interaction with intense gamma rays(shock waves). The whole energy is released purely in the form of radiation. Heat and sound are their after effects. Hope that helped.


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