^{2}. 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:

A

_{0}+ K

_{A0}= A

_{1}+ K

_{A1}+ ½ E + ½E = A

_{1}+ K

_{A1}+ E

(2) In the moving frame:

B

_{0}+ K

_{B0}= B

_{1}+ K

_{B1}+ ½ E(1 – (v/c)cosΦ)(1 – v

^{2}/c

^{2})

^{-½}

+ ½ E(1 + (v/c)cosΦ)(1 – v

^{2}/c

^{2})

^{-½}

= B

_{1}+ K

_{B1}+ E(1 – v

^{2}/c

^{2})

^{-½}

Now taking a look at the energy difference of the particle in each of the frame:

(3) In the rest frame:

A

_{1}– A

_{0}= K

_{A0}– E – K

_{A1}= – ΔK

_{A}– E

(4) In the moving frame:

B

_{1}– B

_{0}= K

_{B0}– K

_{B1}– E(1 – v

^{2}/c

^{2})

^{-½}

= – ΔK

_{B}– E(1 – v

^{2}/c

^{2})

^{-½}

Whether the observer is at rest or moving with respect to the particle, the energy difference should be the same.

(5) – ΔK

_{A}– E = – ΔK

_{B}– E(1 – v

^{2}/c

^{2})

^{-½}

Calculating the difference in kinetic energy:

((6) ΔK = ΔK

_{A}– ΔK

_{B}= E(1 – v

^{2}/c

^{2})

^{-½}– E

= E ((1 – v

^{2}/c

^{2})

^{-½}– 1)

≈ E ((1 + ½v

^{2}/c

^{2}) – 1)

= ½ E (v

^{2}/c

^{2})

Einstein had reasoned that if the kinetic energy of the particle is smaller by ½ E (v

^{2}/c

^{2}), 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 mv

^{2}

If the particle was initially at rest,

(8) ΔK = K = ½ mv

^{2}

Equating (6) and (8)

½ mv

^{2}= ½ E (v

^{2}/c

^{2})

(9) Therefore E = mc

^{2}.

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