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Now, let's just start with what I do know and what I don't know:

I know that $$KE=m_0c^2(γ-1)=Δmc^2,$$ which is not too hard to derive using the Lorentz Transformations if you know basic calculus (correct me if, in the derivations, I have made some assumptions I shouldn't have). From this follows, then, assuming a stationary body possesses some energy $E_0$ and, consequently, that $$E=E_0+KE,$$ that $$ΔE=ΔKE=Δmc^2.$$

I do NOT know that the rest mass energy is, in fact, $E_0=m_0c^2$.

Assume that I also know everything that came before Einstein - i.e., Maxwell's equations, the De Broglie principle etc - and that I am not a very advanced physicist, to say the least.

Using this information, can you derive that $E=γm_0c^2$?

I know an article that is kind of trying to do just that, but I get a feeling it is deliberately over-complicating things to make everything as valid as possible; here is the article. That said, maybe, you could simply go through the same method as in the article but in a layman's terms?

Either way, any valid derivation which wholly proves the mass-energy equivalence identity for ALL scenarios without going any deeper than special relativity and empirical or intuitively explicable principles would do the trick all day long.

EDIT: this provides a very, very basic but, otherwise, a valid derivation for $E=m_0c^2$; however, in that derivation, one question arises: how can mass just disappear? I do realise that the mass must have been emitted as light in order to conserve the total momentum of the box, but how does that work in terms of changes in the electric field? You see, say, the electrons in the object emitting the light vibrate, thereby creating an 'ever-changing' (always non-zero second derivative) change in the electric field, which then propagates by subsequent alternating changes in the electric and electromagnetic fields, this propagation capable of 'the amount of energy spent on vibration of the electron'-worth of work - so where does the mass-to-energy transformation step in???

Max
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  • Possible duplicates: http://physics.stackexchange.com/q/43813/2451 , https://physics.stackexchange.com/q/178960/2451 and links therein. – Qmechanic May 26 '17 at 21:07
  • Try minutephysics video on the subject, youtube it – Ismasou May 26 '17 at 21:09
  • @Ismasou Well, it's invalid - as it uses a changing reference frame to compare with a constant frame, which breaks one of the postulates of special relativity – Max May 26 '17 at 22:17
  • @Max what do you mean by "changing"? Do you mean non-inertial? – probably_someone May 26 '17 at 23:12
  • @probably_someone Well, the video uses a complex frame which jumps from inertial to non-inertial and requires the total energy observed in that frame to be the same as the total energy observed in a constantly non-inertial frame... if that makes sense? – Max May 26 '17 at 23:22
  • @Max Just to make sure we're on the same page, "inertial" means "not accelerating," so both stationary frames and frames moving at a constant velocity are inertial. Therefore, I don't see this "constantly non-inertial frame" that you're talking about. – probably_someone May 26 '17 at 23:28
  • @probably_someone damn that, I meant not moving when I was talking about 'inertial', and I meant inertial when I was talking about 'constantly inertial' – Max May 26 '17 at 23:39

1 Answers1

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Using postulates of relativity, one can derive that when body is subject to external force and is not receiving any other energy:

$$ \text{work done by the force on the body} = \text{increase of }\gamma m_0c^2. $$

One can also derive that when body at rest is receiving energy in the form of radiation while not external force acts on it: $$ \text{energy transferred to the body via the radiation} = \text{increase of }m_0c^2. $$

Based on these results, relativistic energy of any body is defined to be $\gamma m_0 c^2$.

  • Agreed on the former; however, how the latter can be shown would completely have answered my original question. Luckily, I have already got an insight here: https://physics.stackexchange.com/questions/178960/how-to-derive-e-mc2?rq=1 – Max May 26 '17 at 23:17