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Hw6, 12.1, 12.5, 12.6, 12.9, 12.
The Euler-Lagrange equations of motion are
d dτ
∂L
∂uβ
∂L
∂xβ^
From L = −
m 2
uαuα −
q c
uαAα
we can calculate ∂L ∂uβ^
= −muβ −
q c
Aβ
and ∂L ∂xβ^
q c
uα
∂Aα ∂xβ
Now plugging these results into the Euler-Lagrange equations yields
d dτ
−muβ −
q c
Aβ
q c
uα
∂Aα ∂xβ
−m
d^2 xβ dτ 2
q c
dAβ dτ
− uα
∂Aα ∂xβ
q c
∂β^ Aα^ − ∂αAβ^
) (^) dxα dτ
a) E < B Following section 12.3, we boost into K’ frame moving relative to the lab frame at velocity
u = c
E × B
B^2
The E-field vanishes in K’ frame, leaving only the uniform B-field
B′^ =
B^2 − E^2
B^2
B yˆ
The trajectory in this frame is, as done in section 12.2, a helix winding around the B-field
~x(t) = v‖t yˆ′^ + Re
ia( ˆx′^ − i zˆ′)exp(−iωB′ t)
We now boost back into the lab frame using equation 11.18 to obtain
x(t′) = a cos(ωB′ t′) y(t′) = v‖t′ z(t′) = γ (a sin(ωB′ t′) + ut′)
where
γ =
1 − (u/c)^2
and
a =
cp′⊥ eB′
Note that t′^ is not the time in the lab frame. It is acceptable nonetheless because we just want a parametric expression for the trajectory. So, in the lab frame the particle makes a helix with the same circular frequency while uniformly translating along a line in the yz plane. b) E > B We boost into frame K’ at velocity
u = c
E × B
E^2
We have only a uniform E-field in this frame. We have encountered this situation in our last homework. We showed that the particle moves with increasing speed in the direction of the E-field. The mass gets heavier, and the velocity component perpendicular to the field becomes smaller over time. Taking our results from 12.3, we can then boost back into the lab frame.
x(t′) =
mc^2 γ 0 eE′
eE′t′ mcγ 0
y(t′) =
mcγ 0 eE′^
v 0 sin φ 0 sinh−^1
eE′t′ mcγ 0
z(t′) = γ
mcγ 0 eE′^
v 0 sin φ 0 sinh−^1
eE′t′ mcγ 0
From our last homework, we know we can boost into a frame K’ in which E′^ and B′^ are parallel, thus reducing to part b of the question. Choosing E′^ and B′^ to be in the z-direction, we have a system of 2nd order ODEs
duα dτ
e mc
F αβ^ uβ
Expanding these equations leads to
du^0 dτ
eE mc
u^3 du^3 dτ
eE mc
u^0
du^1 dτ
eB mc
u^2
du^2 dτ
eB mc
u^1
We see that the 0 and 3 components will behave like sinh(ρφ) and cosh(ρφ) whereas 1 and 2 like sin(φ) and cos(φ). This is typical of these problems: the E-field generates an asymptotic
and we do not have curvature drift. On the other hand, we will have the gradient drift due to |B| changing in the r and φ directions. From
▽⊥|B| = ˆr
∂B
∂r
r sin θ
∂B
∂φ
= −
3 M
r^4
ˆr
we have for the drift velocity
vG = −
3 a^2 ωB 2 r
φˆ
Because a ≪ R,
vG = R φ˙ ≈ −
3 a^2 ωB 2 R
c) From equation 12.
(v‖)^2 = (R θ˙)^2 = v 02 − v^2 ‖ 0
B(z) B 0
Let θ = π/2 + ǫ. From expanding B about θ = π/2 and differentiating the equation above with respect to time, we obtain
ǫ¨ = −
v^2 ‖ 0 2 R^5
9 M
B 0
ǫ
Hence, the frequency at which θ oscillates about π/2 is
ωB
a R
The Euler-Lagrange equation is
∂μ^
∂L
∂(∂μAν^ )
∂L
∂Aν
where
L = −
8 π
∂αAβ ∂αAβ^ −
c
JαAα
Straightforward differentiation yields
∂μ∂μAν =
4 π c
Jν
Recall from equation 11.141 that Maxwell’s equations are
∂μ(∂μAν^ − ∂ν^ Aμ) =
4 π c
Jν
Hence, our Euler-Lagrange equation becomes Maxwell’s equations if
∂μAμ = 0
b) Straightforward calculation shows the difference between the Lagrangian in part a and that given in 12.85 is
L 12. 85 − La =
8 π
∂ν (Aμ∂μAν^ )
which clearly is a 4-divergence. Hence, by Stokes theorem the difference in the action is a boundary term, which is kept constant. Therefore, the equations of motion remain the same.
