Electric and Magnetic Forces 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 Potential V (q; t). Then, as we know from classical mechanics, we can write the Lagrangian as L(q; q;˙ t) = T ¡ V; (1) where T is the kinetic energy 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 ´ r~ L ¡ r~ L = 0: (3) dt ~v ~ Here, r~v means the gradient 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 equations of motion. 1 2 LORENTZ FORCE LAW 2 2 Lorentz Force Law The Lorentz force in Gaussian Units is given by: à ! ~v F~ = Q E~ + £ B~ ; (4) c where Q is the electric charge, 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 r~ ¢ B~ = 0 (5) 1 @B~ r~ £ E~ + = 0 (6) c @t The magnetic field B~ can be derived from a vector potential A~: B~ = r~ £ A~ (7) If we plug this into Eq. (6), we get 2 3 1 @A~ r~ £ 4E~ + 5 = 0 (8) c @t So the expression in square brackets is a vector field with no curl and can be written as the gradient of a scalar potential ': 1 @A~ E~ + = ¡r~ ' (9) c @t or 1 @A~ E~ = ¡r~ ' ¡ (10) c @t This we plug into Eq. (4) for the Lorentz force law and we get 0 2 31 1 @A~ ³ ´ F~ = Q @¡r~ ' ¡ 4 ¡ ~v £ r~ £ A~ 5A : (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 cross product in the square brackets, we get ³ ´ ³ ´ ³ ´ ~v £ r~ £ A~ = r~ ~v ¢ A~ ¡ ~v ¢ r~ A:~ (13) So the equation for the Lorentz force law is now 0 2 31 1 @A~ ³ ´ ³ ´ F~ = Q @¡r~ ' ¡ 4 + ~v ¢ r~ A~ ¡ r~ ~v ¢ A~ 5A : (14) c @t Now let’s look at the total time derivative of A~(~x;t): X d ~ @ ~ @ ~ A (~x;t) = A (~x;t) + vj A (~x;t) (15) dt @t j @xj | {z } =(~v¢r~ )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 0 1 1 dA~ 1 ³ ´ F~ = Q @¡r~ ' ¡ + r~ ~v ¢ A~ 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'(~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 r~ ' = 0; (18) dt and applying it to part II gives d ³ ´ Q dA~ Q ³ ´ r~ U ¡ r~ U = ¡ r~ ~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 r~ ' + ¡ r~ ~v ¢ A~ = 0 (20) dt c dt c d~v ~ If we identify m dt with the force F , given by Newton’s Law, we can solve Eq. (20) for F~ : 0 1 1 dA~ 1 ³ ´ F~ = Q @¡r~ ' ¡ + r~ ~v ¢ A~ 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~ + r~ Λ(~x;t) (22) 1 ' (~x;t) ! '0 = ' ¡ Λ(˙ ~x;t) ; (23) c where Λ(~x;t) is an arbitrary scalar function. If we plug these new scalar and vector potentials into the Lagrangian (Eq. (17)), it changes to Q ³ ´ L ! L0 = L + Λ(˙ ~x;t)) + ~v ¢ r~ 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' + ~v ¢ A~ (26) 2 c 4 Hamiltonian Formalism 4.1 The Hamiltonian for the EM-Field We know the canonical momentum from classical mechanics: @L pi = (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 '; (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' (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 1 0 B C B~ = B ¢ zˆ = @ 0 A (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 j~pj2 Q Q2 ³ ´ ³ ´ H = + Q' ¡ ~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 angular momentum. So the Hamiltonian is j~pj2 Q H ' + Q' ¡ 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 0 1 1 X µ Q ¶ Q @A @' @ j A 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 1 0 B C B~ = B ¢ zˆ = @ 0 A (40) B For the vector potential, we choose 0 1 0 B C A~ = x ¢ B ¢ yˆ = @ x ¢ B A (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 } :=!2 2 L :=q1 QB where we define the Larmor Frequency !L := 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 !2 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¯ ! (46) n 2 L.