# Exercises in Electrodynamics

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Exercises in Electrodynamics
```Exercises in Electrodynamics
Based on course by Yuri Lyubarsky and Edited By Avry Shirakov
Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel
This exercise pool is intended for an undergraduate course in “Electrodynamics 1”.
EM wave perpendicularly incidents a moving mirror
In this solution we will use the photon picture and look through the collision of the photon with the moving mirror
using the conservation laws of the elastic collision.
Consider a photon with ~ω energy and ~ω/c momentum. We start form finding the frequency shift (Doppler effect).
So, starting with the momentum vector of the photon moving in the ẑ direction, p = (~ω/c, 0, 0, ~ω/c).
Transforming the photon into the moving mirror frame (v = vẑ).





γ − γβ
γ 0 0 −γβ
~ω/c
0

 0 1 0 0   0  ~ω 
p0 = 
=


0
0 0 1 0  0 
c
− (γ − γβ)
−γβ 0 0 γ
~ω/c
In the mirror frame the incident of the photon with the mirror occurs, after the elastic incident the momentum vector
is p00 = ~ω/c (γ − γβ, 0, 0, γ − γβ). Now after the incident, its time to move from the mirror frame back to the photon
frame.






γ 0 0 −γβ
~ω/c (γ − γβ)
γ − γβ
1
0
0
 0 1 0 0 
 ~ω

 ~ω
2 0 
p000 = 
(γ − γβ) 
(γ − γβ) 
=
=
0 0 1 0 
0
0
0 
c
c
−γβ 0 0 γ
~ω/c (γ − γβ)
− (γ − γβ)
−1
Comparing the frequency of the incident photon to the original
1+β
2
2
ω
ω 000 = (γ − γβ) ω = γ 2 (1 − β) ω =
1−β
000
or in other words ω ≥ ω for 0 ≤ β ≤ 1.
Finding the frequency of the incident photon, we are left to transform the amplitudes of the incident fields. Generally
the field at the left side of the mirror are (the field is ŷ polarized):
E = Ei ei(kz−ωt) ŷ + Er ei(−kz−ωt) ŷ
Ei i(kz−ωt)
Er i(−kz−ωt)
B=
e
x̂ −
e
x̂
c
c
Transforming the fields from the lab frame to the
mirror frame:
Ei
0
ŷ = γ (1 + β) Ei ŷ = −E0r
Ei = γ (Ei + v × Bi ) = γ (Ei ŷ + vẑ × Bi x̂) = γ Ei + v
c
v
v v
B0i = γ Bi − 2 × Ei = γ Bi x̂ − 2 ẑ × Ei ŷ = γ Bi + 2 Bi = γ (1 + β) Bi x̂ = −B0r
c
c
c
here we used a previously found relation that at the incident plane z = 0 the electric field is zero, following Er = −Ei .
(we assume that the velocity of the mirror is not changed after the incident).
Now we transform the reflected field back to the lab frame:
v
2
E00r = γ (E0r + v × B0r ) = γ (Er ŷ + vẑ × Bi x̂) = γ −γ (1 + β) Ei − γ (1 + β) Ei ŷ = −γ 2 (1 + β) Ei ŷ
c
v
v
v
2 Ei
B00r = γ B0r − 2 × E0r = γ Br x̂ − 2 ẑ × Er ŷ = γ −γ (1 + β) Bi − γ (1 + β) Bi x̂ = −γ 2 (1 + β)
x̂
c
c
c
c
So that Er > Ei and Br > Bi .
Taken from Unknown
```
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