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Electric and Magnetic in Lagrangian and Hamiltonian Formalism

Benjamin Hornberger

10/26/01

Phy 505, Classical Electrodynamics, Prof. Goldhaber Lecture notes from Oct. 26, 2001 (Lecture held by Prof. Weisberger)

1 Introduction

Conservative forces can be derived from a V (q, t). Then, as we know from , we can write the Lagrangian as

L(q, q,˙ t) = T − V, (1) where T is the kinetic of the system. The Euler-Lagrangian equa- tions of motion are then given by à ! d ∂L ∂L − = 0. (2) dt ∂q˙i ∂qi In three dimensions with cartesian Coordinates, this can be written as d ³ ´ ∇~ L − ∇~ L = 0. (3) dt ~v ~ Here, ∇~v means the with respect to the velocity coordinates. Now we generalize V (q, t) to U(q, q,˙ t) – this is possible as long as L = T − U gives the correct .

1 2 LORENTZ LAW 2

2 Law

The Lorentz force in is given by: Ã ! ~v F~ = Q E~ + × B~ , (4) c where Q is the , E~ (~x,t) is the electric field and B~ (~x,t) is the magnetic field. If the sources (charges or currents) are far away, E~ and B~ solve the homogeneous Maxwell equations. In Gaussian Units, they are given by

∇~ · B~ = 0 (5)

1 ∂B~ ∇~ × E~ + = 0 (6) c ∂t The magnetic field B~ can be derived from a A~:

B~ = ∇~ × A~ (7) If we plug this into Eq. (6), we get   1 ∂A~ ∇~ × E~ +  = 0 (8) c ∂t So the expression in square brackets is a vector field with no and can be written as the gradient of a ϕ:

1 ∂A~ E~ + = −∇~ ϕ (9) c ∂t or

1 ∂A~ E~ = −∇~ ϕ − (10) c ∂t This we plug into Eq. (4) for the Lorentz force law and we get    1 ∂A~ ³ ´ F~ = Q −∇~ ϕ −  − ~v × ∇~ × A~  . (11) c ∂t 3 LAGRANGIAN FORMALISM 3

If we apply the general general vector relation ³ ´ ³ ´ ~a × ~b × ~c = ~b (~a · ~c) − ~a ·~b ~c (12) to the triple vector in the square brackets, we get ³ ´ ³ ´ ³ ´ ~v × ∇~ × A~ = ∇~ ~v · A~ − ~v · ∇~ A.~ (13) So the equation for the Lorentz force law is now    1 ∂A~ ³ ´ ³ ´ F~ = Q −∇~ ϕ −  + ~v · ∇~ A~ − ∇~ ~v · A~  . (14) c ∂t

Now let’s look at the total of A~(~x,t):

X d ~ ∂ ~ ∂ ~ A (~x,t) = A (~x,t) + vj A (~x,t) (15) dt ∂t j ∂xj | {z } =(~v·∇~ )A~(~x,t) The right side of the equation corresponds to the first two terms in the square brackets of Eq. (14), and we can write   1 dA~ 1 ³ ´ F~ = Q −∇~ ϕ − + ∇~ ~v · A~  (16) c dt c

3 Lagrangian Formalism

3.1 The Lorentz Force Law in the Lagrangian Formal- ism Q ~ Let’s try to add a vector potential term UA~(~x,~v, t) = − c ~v · A to the La- grangian: 1 Q L = mv2 − Q ϕ(~x,t) + ~v · A~ (17) |2 {z } |c {z } I II If we apply the Euler-Lagrangian equation of motion (Eq. (3)) on part I of Eq. (17), we get 3 LAGRANGIAN FORMALISM 4

d~v m + Q ∇~ ϕ = 0, (18) dt and applying it to part II gives

d ³ ´ Q dA~ Q ³ ´ ∇~ U − ∇~ U = − ∇~ ~v · A~ = 0 (19) dt ~v A~ A~ c dt c Altogether, the Euler-Lagrangian equation of motion, applied on the La- grangian of Eq. (17), gives

d~v Q dA~ Q ³ ´ m + Q ∇~ ϕ + − ∇~ ~v · A~ = 0 (20) dt c dt c d~v ~ If we identify m dt with the force F , given by ’s Law, we can solve Eq. (20) for F~ :   1 dA~ 1 ³ ´ F~ = Q −∇~ ϕ − + ∇~ ~v · A~  (21) c dt c which is just the correct expression for the Lorentz Force Law, given by Eq. (16).

3.2 How does a gauge transformation affect this La- grangian? As we know, E~ and B~ fields are invariant under gauge transformations

A~ (~x,t) → A~0 = A~ + ∇~ Λ(~x,t) (22)

1 ϕ (~x,t) → ϕ0 = ϕ − Λ(˙ ~x,t) , (23) c where Λ(~x,t) is an arbitrary scalar function. If we plug these new scalar and vector into the Lagrangian (Eq. (17)), it changes to Q ³ ´ L → L0 = L + Λ(˙ ~x,t)) + ~v · ∇~ Λ(~x,t) (24) c The expression in brackets is just the total time derivative of Λ(~x,t), so we get 4 HAMILTONIAN FORMALISM 5

Q d L0 = L + Λ(~x,t) (25) c dt . But as we know, adding to the Lagrangian a total time derivative of a function of ~x and t does not change the equations of motion.

So, the Lagrangian for a particle in an electromagnetic field is given by 1 Q L = mv2 − Q ϕ + ~v · A~ (26) 2 c 4 Hamiltonian Formalism

4.1 The Hamiltonian for the EM- We know the canonical from classical mechanics: ∂L = (27) ∂x˙ i Using the Lagrangian from Eq. (26), we get Q p = mv + A (28) i i c i The Hamiltonian is then given by

X 1 2 H = pix˙ i − L = mv + Q ϕ, (29) i 2

where v resp.x ˙ must be replaced by p: Solving Eq. (28) for vi and plugging into Eq. (29) gives ¯ ¯ 1 ¯ Q ¯2 H = ¯~p − A~¯ + Q ϕ (30) 2m ¯ c ¯ So the kinetic momentum in is in this case given by Q P~ = m~v = ~p − A~ (31) c 4 HAMILTONIAN FORMALISM 6

Example: Uniform constant magnetic field We assume B~ in z-direction:   0   B~ = B · zˆ =  0  (32) B

The vector potential can then be written as 1 A~ = B~ × ~r (33) 2 This is an arbitrary choice, but it is easy to prove that it gives the correct result for B~ . Now suppose the particle is bound in a strong central potential and B~ is relatively weak. If we plug the vector potential (Eq. (33)) into the Hamiltonian (Eq. (30)), we get

|~p|2 Q Q2 ³ ´ ³ ´ H = + Q ϕ − ~p · B~ × ~r + B~ × ~r · B~ × ~r (34) 2m 2mc 8m2c2 | {z } 2 B~ 2~r2−(B~ ·~r)

The last term in this equation can be neglected for a bound particle in a weak field. For the mixed scalar / cross product in the second term, we can write

~p · B~ × ~r = ~r × ~p · B~ = L~ · B,~ (35) where L~ is the . So the Hamiltonian is

|~p|2 Q H ' + Q ϕ − L~ · B~ (36) 2m 2mc The last term is this Hamiltonian causes the ordinary Zeeman Effect.

4.2 Hamiltonian Equations of Motion The Hamiltonian equations of motion are given by ∂H ∂H x˙ i = andp ˙i = − . (37) ∂pi ∂xi 4 HAMILTONIAN FORMALISM 7

If we apply these equations on the Hamiltonian (Eq. (30)), we get 1 · Q ¸ x˙ = p − A (38) i m i c i   1 X µ Q ¶ Q ∂A ∂ϕ  j  p˙i = pj − Aj − Q (39) m j c c ∂xi ∂xi

Example: Uniform constant magnetic field Again we look at a constant magnetic field in z-direction (no other potential):   0   B~ = B · zˆ =  0  (40) B For the vector potential, we choose   0   A~ = x · B · yˆ =  x · B  (41) 0 This is again an arbitrary choice which gives the correct result for B~ . We put this vector potential into the Hamiltonian and get " # 1 µ QB ¶2 H = p2 + p2 + p − x = H + H . (42) 2m z x y c z ⊥

The second part H⊥ of the Hamiltonian can be written as " # p2 m ·QB ¸2 c 2 H = x + x − p (43) ⊥ 2m 2 mc QB y | {z } | {z } :=ω2 2 L :=q1 QB where we define the Larmor ωL := mc and introduce a new c coordinate q1 := (x − QB py). Furthermore, we set px = p1, py = p2 and pz = p3.

p2 1 H = 1 + m ω2 q2 (44) ⊥ 2m 2 L 1 4 HAMILTONIAN FORMALISM 8

This is just the Hamiltonian for a harmonic oscillator. In Quantum Me- chanics, we can use the commutator

[q1, p1] = ih,¯ (45) and for the harmonic oscillator, the energy eigenvalues are µ 1¶ E = n + h¯ ω (46) n 2 L