Electric and Magnetic Forces in Lagrangian and Hamiltonian Formalism Benjamin Hornberger 1/26/1 Phy 55, Classical Electrodynamics, Prof. Goldhaber Lecture notes from Oct. 26, 21 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 Lq, q, t = T V, 1 where T is the kinetic energy of the system. The Euler-Lagrangian equations of motion are then given by d dt L q i L q i =. 2 In three dimensions with cartesian Coordinates, this can be written as d v L L dt =. 3 Here, v means the gradient with respect to the velocity coordinates. Now we generalize V q, t to Uq, 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: F = Q E + v c B, 4 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 B = 5 E + 1 c B t = 6 The magnetic field B can be derived from a vector potential A: If we plug this into Eq. 6, we get B = A 7 E + 1 c A = 8 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 ϕ: or E + 1 c A t = ϕ 9 E = ϕ 1 A 1 c t This we plug into Eq. 4 for the Lorentz force law and we get F = Q ϕ 1 A c t v A. 11
3 LAGRANGIAN FORMALISM 3 If we apply the general general vector relation a b c = b a c a b c 12 to the triple vector cross product in the square brackets, we get v A = v A v A. 13 So the equation for the Lorentz force law is now F = Q ϕ 1 A c t + v A v A. 14 Now let s look at the total time derivative of A x, t: d A dt x, t = A t x, t + j v j x j A x, t } {{ } = v A x,t 15 The right side of the equation corresponds to the first two terms in the square brackets of Eq. 14, and we can write F = Q 3 Lagrangian Formalism ϕ 1 da c dt + 1 c v A 16 3.1 The Lorentz Force Law in the Lagrangian Formalism v A to the La- Let s try to add a vector potential term U A x, v, t = Q c grangian: L = 1 2 mv2 Q ϕ x, t + Q v A }{{}} c {{} I II 17 If we apply the Euler-Lagrangian equation of motion Eq. 3 on part I of Eq. 17, we get
3 LAGRANGIAN FORMALISM 4 m d v dt + Q ϕ =, 18 and applying it to part II gives d v U A U A = Q da dt c dt Q c v A = 19 Altogether, the Euler-Lagrangian equation of motion, applied on the Lagrangian of Eq. 17, gives m d v dt + Q ϕ + Q c d A dt Q c v A = 2 If we identify m d v dt with the force F, given by Newton s Law, we can solve Eq. 2 for F : F = Q ϕ 1 da c dt + 1 c v A 21 which is just the correct expression for the Lorentz Force Law, given by Eq. 16. 3.2 How does a gauge transformation affect this Lagrangian? As we know, E and B fields are invariant under gauge transformations A x, t A = A + Λ x, t 22 ϕ x, t ϕ = ϕ 1 c Λ x, t, 23 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 L L = L + Q c Λ x, t + v Λ x, t 24 The expression in brackets is just the total time derivative of Λ x, t, so we get
4 HAMILTONIAN FORMALISM 5 L = L + Q d 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 L = 1 2 mv2 Q ϕ + Q c v A 26 4 Hamiltonian Formalism 4.1 The Hamiltonian for the EM-Field We know the canonical momentum from classical mechanics: Using the Lagrangian from Eq. 26, we get The Hamiltonian is then given by p i = L ẋ i 27 p i = mv i + Q c A i 28 H = i p i ẋ i L = 1 2 mv2 + Q ϕ, 29 where v resp. ẋ must be replaced by p: Solving Eq. 28 for v i and plugging into Eq. 29 gives H = 1 2m p Q A c 2 + Q ϕ 3 So the kinetic momentum in is in this case given by P = m v = p Q c A 31
4 HAMILTONIAN FORMALISM 6 Example: Uniform constant magnetic field We assume B in z-direction: B = B ẑ = B The vector potential can then be written as 32 A = 1 2 B r 33 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. 3, we get H = p 2 2m + Q ϕ Q 2mc p B r + Q2 B r B r 34 8m 2 c 2 }{{} B 2 r 2 B r 2 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 H p 2 2m + Q ϕ Q L 2mc B 36 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 ẋ i = H p i and ṗ i = H x i. 37
4 HAMILTONIAN FORMALISM 7 If we apply these equations on the Hamiltonian Eq. 3, we get ẋ i = 1 m [ p i Q ] c A i 38 ṗ i = 1 m j p j Q Q c A j c A j x i Q ϕ x i 39 Example: Uniform constant magnetic field Again we look at a constant magnetic field in z-direction no other potential: B = B ẑ = For the vector potential, we choose A = x B ŷ = B x B 4 41 This is again an arbitrary choice which gives the correct result for B. We put this vector potential into the Hamiltonian and get H = 1 [ p 2 z + p 2 x + p y QB 2 ] 2m c x = H z + H. 42 The second part H of the Hamiltonian can be written as H = p2 x 2m + m 2 [ ] QB 2 [ x c ] 2 } mc {{} QB p y }{{} :=ωl 2 :=q1 2 43 where we define the Larmor Frequency ω L := QB and introduce a new mc coordinate q 1 := x c p QB y. Furthermore, we set p x = p 1, p y = p 2 and p z = p 3. H = p2 1 2m + 1 2 m ω2 L q 2 1 44
4 HAMILTONIAN FORMALISM 8 This is just the Hamiltonian for a harmonic oscillator. In Quantum Mechanics, we can use the commutator [q 1, p 1 ] = i h, 45 and for the harmonic oscillator, the energy eigenvalues are E n = n + 1 h ω L 46 2