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The rod and the observer moves with velocity and w.r.t

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c23

c2

V = =c−2

3c

2

1− c. .23

c

21

c2

c4

= = 0.968l′ l0 1 − V 2

c2

− −−−−−√ l0

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Here

at

So ,

Now

Extracting u

we get,

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= Fdp

dt

p = F t + kp = 0

t = 0k = 0⇒ p = F t = mu

1− u2

c2√

⇒ F (1 − ) =t2 u2

c2 m2u2

u = Ftc

+m2 c2 F 2 t2√

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M + Δ

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M can be excited to a state of mass by photon

capture.

From momentum conservation,

From energy conservation

Squaring and rearranging, we get,

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M + Δ

0 + = phνc

M + hν =c2 (M + Δ +)2c4 p2C 4

− −−−−−−−−−−−−−−−√

ν =Δ(Δ+2M)c2

2Mh

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