Übungen zur Elektrodynamik (T3)

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1 Arnold Sommerfeld Center Ludwig Maximilians Universität München Prof. Dr. Ivo Sachs SoSe 08 Übungen zur Elektrodynamik (T3) Lösungen zum Übungsblatt 7 Lorentz Force Calculate dx µ and ds explicitly in terms of v and dt. In an inertial system (x µ ) = (ct, x) and hence if the worldline is parametrized w.r.t. the coordinate time: dx µ = (cdt, dx, dx, dx 3 ) = (cdt, cβdt). We may assume this since the Lagrangian is parametrisation invariant. = ds = dx µ dx µ = c dt c β dt = cdt β = c dt () γ = dx µ = ( cdt, cβdt ) () (ii) Calculate the Euler-Lagrange equation x i (t) for the action S to find the Lorentz force. Since from which we can read off the Lagrangian S = mc dt β e dt ( φ cβ A ) = dt (mc β + e ( φ cβ A )) (3) L = mc β + e ( φ cβ A ) (4) Now we need to compute the Euler-Lagrange equations x i d ( ) dt ẋ i = 0 (5) We find and x i = e ( i φ cβ i A ) = e i φ + eẋ j i A j (6) ẋ i = c β i = mc β i c + eca β i = γmẋ i + ea i (7) Taking the time derivative of (7) we get ( ) d dt ẋ i = d dt (γmẋi ) + e A i t + eẋj j A i (8)

2 where the last term appears due to the dependence A i (t, x i (t)). Inserting (6) and (8) in the Euler-Lagrange equations this yields e i φ + eẋ j i A j d dt (γmẋ i) e A i t eẋj j A i = 0 d ) dt (γmẋi ) = e ( i φ Ai + e ( ẋ j i A j ẋ j j A i) (9) t The left hand side is our momentum and the right hand side contains all terms of our field strength tensor F µν. (iii) Finally, express the covariant Lorentz force in terms of E and B. Since (ẋ ( A)) i = ɛ ijk ɛ klm ẋ j l A m = (δ il δ jm δ im δ jl )ẋ j l A m = j,k,l,m j,k,l,m = (0) ẋ j i A j ẋ j j A i = ẋ j i A j ẋ j j A i j where two indices where switched in ẋ j i A j and thus no sign was gained. By definition B = A and E i = i φ t A i, thus we arrive at the equation of motion with the r.h.s describing the Lorentz-force. d (γmẋ) = e (E + ẋ B) () dt Principle of Least Action Introduce the variation δϕ(x µ ) and show that the action can be rewritten as S = S + δs. In addition, show explicitly that δ µ ϕ can be rewritten as µ δϕ. For the variation of ϕ, we can write: ϕ = ϕ + δϕ () Regarding δϕ, we can think of it as a difference: δ µ ϕ = µ ϕ µ ϕ = µ (ϕ ϕ ) = µ δϕ (3) So the action integral takes the form S[ϕ + δϕ] = d 4 xl = d 4 x( µ ϕ + µ δϕ)( µ ϕ + µ δϕ) = d 4 x( µ ϕ µ ϕ + µ δϕ µ ϕ + µ ϕ µ δϕ + µ δϕ µ δϕ) = S[ϕ] d 4 x( µ δϕ µ ϕ + µ ϕ µ δϕ) = S[ϕ] d 4 xδ( µ ϕ µ ϕ) (4) δϕ is small, so we can safely assume that the square (δϕ) will not contribute.

3 (ii) Discuss under which assumption the boundary terms add no contribution and can be neglected. Using Gauß s theorem in four dimensions we find that for a ball of radius R centred at zero, denoted B R and corresponding sphere S R µ φ µ φd 4 x = ( µ (φ µ φ) (φ µ µ φ))d 4 x = φ µ φ d 3 x µ B R B R S R B R (φ µ µ φ)d 4 x To integrate over R 4 we have to let R, thus we may integrate by parts if the first term on the right hand side vanishes in that limit. A sufficient condition for this to occur would be that φ is compactly supported or that φ µ φ vanishes faster than R 3. (iii) Finally, derive the Euler Lagrange equation by imposing that the variation of the action does not contribute to the total action. Per the principle of least of action, a small variation should not have any effect on the action, thus 0 =! δs = d 4 x( µ δϕ µ ϕ + µ ϕ µ δϕ) = d 4 x µ δϕ µ ϕ (5) IBP = d 4 x( µ µ ϕ)δϕ where we assumed that the conditions for integration by parts are met. Thus the Euler Lagrange equation of this system is 0 = µ µ φ = ϕ (6) 3 Covariant Maxwell Equations Compute the eom by either applying the principle of least action or using the Euler Lagrange equation. Using principle of least action: ( ) δs = d 4 xδ F µν F µν + A µ j µ ( ) = d 4 x (F µν δf µν + δf µν F µν ) + δa µ j µ ( ) = d 4 x F µν δ( µ A ν µ A µ ) + δa µ j µ µ 0 ( ) = d 4 x (F µν µ δa ν F µν ν δa µ ) + δa µ j µ µ 0 ( ) IBP = d 4 x ( µ F µν δa ν ν F µν δa µ ) δa µ j µ µ 0 ( ) = d 4 x ν F νµ + j µ δa µ! = 0 µ 0 = µ F µν = µ 0 j ν (7) We used the antisymmetric property of the field strength tensor ( F µν = F νµ ) in the last equality.

4 (ii) First derive the inhomogeneous Maxwell equations. Set the index ν = 0 and inserting the definition of the electric field: µ F µ0 = 0 F 00 + a F a0 = c E = µ 0j 0 = µ 0 cρ = ρ cɛ 0 (8) which yields Gauß s law. Now setting ν = i and using the definition of the electric and magnetic field F ij = ε ijk B k : ( µ 0 j i = µ F µi = 0 F 0i + j F ji = i c te + B) (9) which is Maxwell s fourth equation. (iii) Second, derive the homogeneous Maxwell equations from the Bianchi Identity. The Bianchi identity is α F βγ + β F γα + γ F αβ = 0 (0) If all three indices are space indices then we may summarise this identity as ɛ ijk i F jk = 0. In this case using we find F ij = ε ijk B k B i = ε ijkf jk () For the other equation let α = i, β = j γ = 0, and apply ε ijk to get i B i = ε ijk i F jk = 0 () B = 0 (3) ( 0 = ε ijk i F j0 + j F 0i + 0 F ij ) ( = ε ijk j F 0i + j F 0i + 0 F ij ) = ε ijk j F }{{} 0i + ε ijk 0 F ij = E i/c (4) = c ε ijk i E j + ε ijk 0 F ij = c ( E) k + c t B k = E + t B = 0 (iv) Third, derive the continuity equation. Taking the divergence of the covariant inhomogeneous Maxwell equation: 0 = ν µ F µν + ν j ν = ν j ν (5) By decomposing the index ν into a temporal and spatial part, we get the ordinary expression of the continuity equation: ν j ν = 0 j 0 + i j i = c tcρ + i j i = t ρ = j (6)

5 (v) Use the Legendre transformation and find an expression in terms of E and B. a) F µν = µ A ν ν A µ which yields ( 0 A κ ) = η µρ η νσ ( 0 A κ ) (F ρσf µν ) = ( (η µ0 η νκ η µκ η ν0 )F µν + (η 0ρ η κσ η κρ η 0σ )F ρσ ) = (F 0κ F κ0 + F 0κ F κ0 ) = µ 0 F 0κ (7) We have F 0i = Ei c and F ij = ɛ ijk B k. Thus using ) F µν F µν = (B E c (8) Furthermore ( 0 A µ ) 0A µ = µ 0 F 0µ 0 A µ = µ 0 c Ei t A i = ɛ 0 E t A (9) Since E i = i φ t A i we have ɛ 0 E i t A i = ɛ 0 E i i φ E i E i = ɛ 0 E i i φ + E Inserting this into (9) the Hamiltonian becomes ( H ɛ 0 E + ɛ 0 E i i φ + ( B ɛ 0 µ 0 E ) ) + A µ j µ µ 0 ( ) ) (ɛ 0 E + µ0 B ɛ 0 φ E + A µ j µ ( ) ) (ɛ 0 E + µ0 B φρ + A µ j µ ( ) ) (ɛ 0 E + µ0 B + A i j i (30) where we used (φe) = i (φe i ) = E i i φ + φ i E i = E i i φ + φ E as well as the Maxwell-equation E = ɛ 0 ρ and from the first to the second line we integrated by parts using Gauß s theorem and dropped the boundary term due to the fall-off conditions of the Maxwell-fields. In the last line we used that A 0 j 0 = φ c cρ = φρ.