Solutions to Jackson 12.3, 12.14, 12.16 PDF

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P507 Problem Set 21 Indiana Univeristy Ryan Michael Pennell April 24, 2018 1 Jackson 12.3 A particle with mass m and charge e moves in a uniform, static, electric field E0 . (a) Solve for the velocity and position of the particle as explicit functions of time, assuming that the initial velocity v0 w...

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P507 Problem Set 21 Indiana Univeristy Ryan Michael Pennell April 24, 2018

1

Jackson 12.3

A particle with mass m and charge e moves in a uniform, static, electric field E0 . (a) Solve for the velocity and position of the particle as explicit functions of time, assuming that the initial velocity v0 was perpendicular to the electric field. The equations of motion for a particle of charge e in external fields E and B can be written in the covariant from: dU α e αβ = F Uβ dτ mc where m is the mass, τ is the proper time, and the 4-velocity of the particle and where  0 −Ex −Ey  Ex 0 −Bz F αβ =   Ey B z 0 Ez −By Bx

U α = (γc, γu) = pα /m is  −Ez By   −Bx  0

For a constant electric field this equation can be broken into space and time components. e dU 0 = E0 · u, dτ mc Taking the time derivative of

dU 0 dτ

du e = E0 U 0 dτ mc

we find

1

d2 U 0 e du eE0 2 0 = E0 · =( ) U 2 dτ mc dτ mc This means that U 0 must be U 0 = Ae(eE0 /mc)τ + Be−(eE0 /mc)τ Now we can substitute this into

dU 0 dτ

to solve for u

u = u0 + Eˆ0 (Ae(eE0 /mc)τ + Be−(eE0 /mc)τ ) Plugging the U 0 and u terms back into the equation of motion we find (eE0 /mc)(Ae+(eE0 /mc)τ −Be−(eE0 /mc)τ ) =

e E0 ·(u0 +Eˆ0 (Ae(eE0 /mc)τ +Be−(eE0 /mc)τ ) mc

=⇒ E0 · u0 = 0 Note however, that the four-velocity satisfies the constraint U α Uα = c2 . This gives rise to a relation among the parameters 4AB − u0 = c2 If we take τ = 0 then u = u0 x,

U0 = A + B

This is satisfied by taking A = B, so that U 0 = Ae(eE0 /mc)τ +Ae−(eE0 /mc)τ = A(e(eE0 /mc)τ +e−(eE0 /mc)τ ) = 2A cosh( u = u0 + Eˆ0 (Ae(eE0 /mc)τ + Ae−(eE0 /mc)τ ) = u0 + 2AEˆ0 sinh( Where c A= 2 thus, since

now we know

1 γ0 = q 1− U 0 = 2A cosh(

r

1−

u20 c2

=⇒ u0 = γ0 v0

v02 c2

eE0 τ eE0 τ ) = cγ0 cosh( ) mc mc 2

eE0 τ ) mc

eE0 τ ) mc

and u = u0 + Eˆ0 (Ae(eE0 /mc)τ + Ae−(eE0 /mc)τ ) = γ0 v0 + cγ0 Eˆ0 sinh(

eE0 τ ) mc

Now find the four-vector by integrating U 0 Z τ mc2 γ0 eE0 τ U 0 dτ ′ = ( x0 = ) sinh( ) eE0 mc 0

and

x=

Z

τ

udτ ′ = γ0 v0 τ +

0

eE0 τ mc2 γ0 ˆ E(cosh( ) − 1) eE0 mc

These are the solutions in terms of proper time τ not of time t. To find the solutions in terms of time t we know that t=(

mc2 γ0 eE0 τ ) sinh( ) eE0 mc

τ=

eE0 t mc sinh−1 ( ) eE0 mcγ0

solving for τ we find

which means that u = γ0 (v0 + thus v(t) =

s

1+(

eE0 t ) mγ0

eE0 t eE0 t 2 ) (v0 + ) mcγ0 mγ0

and for position we have mcγ0 eE0 t mc2 γ0 ˆ x(t) = v0 sinh−1 ( )+ E0 ( eE0 mcγ0 eE0

s

1+(

eE0 t 2 ) − 1) mcγ0

(b) Eliminate the time to obtain the trajectory of the particle in space. Discuss the shape of the path for short and long times (define “short” and “long” times). xk =

mc2 γ0 eE0 τ [cosh( − 1], eE0 mc

=⇒ xk =

x⊥ = γ 0 v 0 τ

eE0 x⊥ mc2 γ0 [cosh( ) − 1] eE0 mcγ0 v0 3

For short times, t > mcγ0 /eE0 xk ≈

2

eE0 x⊥ mc2 γ0 ( mcγ ) 0 v0 e 2eE0

Jackson 12.14

An alternative Lagrangian density for the electromagnetic field is L=−

1 1 ∂ α Aβ ∂ α Aβ − J α Aα 8π c

(a) Derive the Euler-Lagrange equations of motion. Are they the Maxwell equations? Under what assumptions? The Lagrangian density: 1 ∂L = − Jν, ∂Aµ c

∂L 1 = − ∂ ν Aµ ∂(∂ν Aµ ) 4π

The Euler-Lagrange equations 1 1 − Jµ + ∂ ν ∂ ν Aµ = 0 c 4π or

4π Jν c = ∂µ Aν − ∂ν Aµ the Maxwell equation

∂ ν ∂ ν Aµ = In the Lorenz gauge ∂ µ Aµ = 0, Fµν ∂ µ Fµν = ( 4π c )Aν becomes

∂ µ ∂µ Aν − ∂ν (∂ µ Aµ ) =

4π Jν c

This makes Fµν = ∂µ Aν − ∂ν Aµ since ∂ µ Aµ = 0. (b) show explicitly, and with what assumptions, that this Lagrangian density differs from (12.85) by a 4-divergence. does this added 4-divergence affect the action or the equations of motion? L0 = −

1 1 Fµν F µν − J µ Am u 16π c 4

can be expressed as ∆L =

1 1 Fµν F µν − ∂µ Aν Aν 16π 8π =−

=−

3

1 ∂µ Aν ∂ ν Aµ 8π

1 µ 1 ∂µ (Aν ∂ ν Aµ − Aµ ∂ ν Aν ) − (∂ Aµ )2 8π 8π

Jackson 12.16

(a) Starting with the Proca Lagrangian density (12.91) and following the same procedure as for the electromagnetic fields, show that the symmetric stress-energy-momentum tensor for the Proca fields is Θαβ =

1 αγ 1 1 [g Fγλ F λβ + g αβ Fλv F λv + µ2 (Aα Aβ − g αβ Aλ Aλ )] 4π 4 2

(12.91)

Lproca = −

L=−

1 µ2 1 Fαβ F αβ + Aα Aα − J α Aα 16π 8π c

µ2 1 Fµν F µν + Aµ Aµ 16π 8π

The covarient generalization T αβ =

X ∂Lem 1 ∂ β Aλ − g αβ Lem = − g αµ Fµλ ∂ β Aλ − g αβ Lem λ 4π ∂(∂α A ) k

Where F µν = ∂ µ Aν − ∂ ν Aµ 1 1 ∂L Fαβ = − Fβα = λ 4π 4π ∂(∂α A )

T αβ = −

=−

1 αλ β 1 αβ 1 1 αλ [F Fλ − g Fαβ F β + µ2 g αβ A2 ] − F ∂λ Aβ 4π 4 2 4π

1 1 αλ β 1 αβ αλ β 1 2 αβ 2 [F Fλ − g F Fλ + µ g A − (∂λ F αλ )Aβ ] − ∂λ (F αλ Aβ ) 4π 4 2 4π

5

1 T αβ = Θαβ + ∂λ ( F λα Aβ ) 4 Θαβ = T αβ − TDαβ =

1 αµ 1 (g Fµλ F λβ + g αβ Fµλ F µλ ) 4π 4

1 αγ 1 1 [g Fγλ F λβ + g αβ Fλv F λv + µ2 (Aα Aβ − g αβ Aλ Aλ )] 4π 4 2 (b) For these fields in interaction with the external source J β , as in (12.91), show that the differential conservation laws take the same from as for the electromagnetic fields, namely, Θαβ =

∂α Θαβ = ∂α Θαβ = −

Jλ F λβ c

1 1 [∂α F αλ Fλβ +F αλ ∂α Fλβ − Fσλ ∂ β F σλ −µ2 (∂α Aα Aβ +Aα ∂α Aβ −Aλ ∂ β Aλ )] 4π 2

1 1 [(∂α F αλ + µ2 Aλ )Fλβ + Fσλ (∂ σ F βλ + ∂ λ F σβ + ∂ β F λσ )] 4π 2 1 λ β = − J Fλ c 1 = Jλ F λβ c (c) Show explicicitly that the time-time and space-time components of Θαβ are =−

1 2 [E + B 2 + µ(A0 A0 + A · A)] 8π 1 Θi0 = [(E × B)i + µ2 Ai A0 ] 4π The time-time component is 1 1 1 Θ00 = − [F 0i Fi0 + (E 2 − B 2 ) − µ2 ((A0 )2 − ((A0 )2 − A2 ))] 4π 2 2 1 2 [E + B 2 + µ2 ((A0 )2 + A2 )] = 8π The time-space components are 1 Θ0i = − [Fj0 F ij − µ2 A0 Ai ] 4π 1 1 1 [(E×B)i +µ2 A0 Ai ] = − [E j (−ǫijk B k )−µ2 A0 Ai ] = − [−ǫijk E j B k −µ2 A0 Ai ] = 4π 4π 4π Θ00 =

6...