Here’s a quick rundown on one of the most famous equation in physics, E = mc2. Einstein knew from experiments previously done that a particle could decay and release gamma rays. He reasoned that when this happened, the particle would lose kinetic energy, and this could only be accounted by a loss of mass. So how did he come to that conclusion? He analysed the situation both in a rest frame and in a moving frame.
The value of γ was already known, but we can derive it from Relativistic Doppler Effect (to be the topic for another blog A).
By the law of conservation of energy:
The energy before decay = the energy after decay
(1) In the rest frame:
A0 + KA0 = A1 + KA1 + ½ E + ½E = A1 + KA1 + E
(2) In the moving frame:
B0 + KB0 = B1 + KB1 + ½ E(1 – (v/c)cosΦ)(1 – v2/c2)-½
+ ½ E(1 + (v/c)cosΦ)(1 – v2/c2)-½
= B1 + KB1 + E(1 – v2 /c2)-½
Now taking a look at the energy difference of the particle in each of the frame:
(3) In the rest frame:
A1 – A0 = KA0 – E – KA1 = – ΔKA – E
(4) In the moving frame:
B1 – B0 = KB0 – KB1 – E(1 – v2 /c2)-½
= – ΔKB – E(1 – v2 /c2)-½
Whether the observer is at rest or moving with respect to the particle, the energy difference should be the same.
(5) – ΔKA – E = – ΔKB – E(1 – v2 /c2)-½
Calculating the difference in kinetic energy:
((6) ΔK = ΔKA – ΔKB = E(1 – v2 /c2)-½ – E
= E ((1 – v2/c2)-½ – 1)
≈ E ((1 + ½v2/c2) – 1)
= ½ E (v2/c2)
Einstein had reasoned that if the kinetic energy of the particle is smaller by ½ E (v2/c2), the only way this can happen is that the particle must lose mass when emitting radiation.
By definition the kinetic energy is,
(7) K = ½ mv2
If the particle was initially at rest,
(8) ΔK = K = ½ mv2
Equating (6) and (8)
½ mv2 = ½ E (v2/c2)
(9) Therefore E = mc2.