Lecture 16 March 29, 2010
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1 Lecture 16 March 29, 2010 We know Maxwell s equations the Lorentz force. Why more theory? Newton = = Hamiltonian = Quantum Mechanics Elegance! Beauty! Gauge Fields = Non-Abelian Gauge Theory = Stard Model Anyway, let s look for s actions.
2 for a Motion is x(t). Action A = L( x, x, t) dt. Hamilton: actual path extremizes the action. Doesn t look Lorentz invariant, but all observers must agree (after suitable Lorentz transformation). So A should be a scalar. Start with a free particle. What could action be? Can t depend on x, for translation invariance. What property of path through space-time can we use? How about proper length? A = mc 2 dτ = mc dx µ dx µ = mc U α U α dτ = mc 2 1 u 2 c 2 dt.
3 So the is L( x, u, t) = mc 2/ γ( u) = mc 2 1 u 2 c 2. Note L is not an invariant, but Ldt γl are. Canonical Momentum (in 3-D language) ( P ) i = L u i = mu i 1 u 2 c 2 = ( p ) i, as we previously explored. Euler-Lagrange: d L L = 0, dt u i x i gives p i = constant, as x i is an ignorable coordinate. So this is correct for a free particle.
4 L for particle in a field What if charge q in an external field? Can depend on x µ, but only through the fields dependence on it. Can involve U α = dx α /dτ, but need it in combination as a scalar. Could use A α or F αβ, but U α U β F αβ 0, so only possibility linear in fields is γl int = q c U αa α, = L int = qφ + q u A, c with usual electrostatic vector potentials. Note first term looks like -PE as expected (as L = T V often). So the full lagrangian for the particle is L( x, u, t) = mc 2 1 u 2 c 2 + q c u A( x, t) qφ( x, t), the canonical momentum becomes m u q P = L/ u = + A( x, 1 u 2 c t) = p + q A, c c 2 not just the ordinary momentum p = mγ u.
5 The equations of motion are now d dt L u i }{{} P i L x i = dp i dt + q c d dt A i }{{} ( Ai t + u j j A i q c u j i A j + q i Φ = dp i dt + q A i c t + q iφ + q c (u j j A i u j i A j ) ( d p = 0 = dt + q d ) A c dt + q Φ q ( ) c u A d p dt = q E + q c u B so we see that this gives us the correct Lorentz force equation. ) i
6 The Hamiltonian What is the Hamiltonian? H = P u L, but reexpressed in terms of P rather than u. As u = p/mγ(u) = p m 1 u 2 /c 2 = u = mγ(u) = c p p 2 + m 2 c 2, p 2 + m 2 c 2 /c. Then we need to substitute p P qa/c. Thus ( ) P P qa/c + m 2 c 2 H = mγ(u) ( ) 2 P qa/c + m 2 c 2 ( ) q P qa/c A cmγ(u) + qφ = = + qφ mγ(u) (cp qa) 2 + m 2 c 4 + qφ. Note H is the total energy, the kinetic energy p 0 c + eφ, so this just verifies (p 0 ) 2 p 2 = m 2 c 2.
7 This L still doesn t have dynamical E&M fields - we will come to that later. First Recall from Classical Mechanics: Slowly varying perturbation on an integrable system with cyclic action-angle variables: action is adiabatic invariant. Apply this to motion transverse to uniform static magnetic field. Action J = P d r is an invariant. Need to use canonical momentum P = p + qa/c, not just p = mγ v. So J = mγ v d r + q A d r. c We have circular motion 1 with v = ω B r. 1 Note J12.38 says d v/dt = v ω B = ω B v, which explains the unexpected minus sign.
8 So the first term in J is mγ v d r = 2π 0 mγω B a 2 dθ = 2πmγω B a 2. As mγ ω B = qb/c, this is just 2qΦ B /c, where Φ B is the magnetic flux through the orbit. The second term in J, so q c A d r = q c S A = q c J = qφ B /c = q c Bπa2 = π c q S n B = q c Φ B, p 2 B is an adiabatic invariant, as are Ba 2 p2 B. These are conserved if B varies slowly compared to the gyroradius of the particle s motion.
9 So p 2 /B may be constant. In a purely magnetic field, speed γ are constant, but the transverse speed v B, while v 2 = v 2 + v2 is constant. So if particle drifts into a region of stronger B, v 2 may grow to use up all of v 2, v will vanish reverse. This is a magnetic mirror. Field lines converge where field gets strong, B so Lorentz force has v a component pushing F particle back into the weaker field region. This is called a magnetic mirror or magnetic bottle. Note that those particles with negligible v will not get confined.
10 treating x µ as dynamical The mc 2 1 u 2 /c 2 certainly doesn t look like a covariant formulation, we treated it as a functional to determine x(t), which is certainly not a covariant way of saying things. On the other h mc dx µ dx µ = mc η µν dx µ dx ν is a very covariant way of looking at the action, but what do we vary? All of m µ? or only the spatial part? Note that if we think of x µ (λ) as a parameterized path, we may write the action dx A = mc η µ µν dλ dx ν dλ dλ, think of varying the function x µ (λ) look for an extremum in the usual way. This gives d η µν dxν dλ = 0, dλ dx η µ dx ν µν dλ dλ
11 or dx µ dλ = dx Cµ η µ dx ν µν dλ dλ. Doesn t determine dxµ dλ! Though it looks like four equations, it is really only three, for contracting it with itself gives η µν dx µ dλ dx ν dλ = dx µ dx ν C2 η µν dλ dλ, which does nothing to determine dxν dλ but only that C 2 = 1. This should not be surprising. The path length doesn t depend on how it is parameterized, so any change x µ (λ) x µ (σ(λ)) will not change A, as long as σ(λ) is monotone.
12 Inability to predict the future is a sign of gauge invariance, though in this case it is not the gauge invariance we are used to for E&M. Here it is not a serious problem, because we can choose to use proper time as our parameter, providing the additional equation η µν dx µ dτ dx ν dτ = c2, = dxµ dτ = 1 m pµ = constant.
13 Action for particles in fields A int = q c U µ A µ dτ = q c dx µ dτ A µ dτ = q c A τ dx µ? The last expression is clearly covariant, the penultimate one gives the for the parameterized path x L = mc η α x β αβ λ λ q c A x α α λ with action Ldλ. P α = L xα λ = mc xα λ dx η µ dx ν µν dλ dλ + q c A α m x α λ τ τ + q c A α, Remember in Euler-Lagrange d/dλ is a stream derivative, so d dτ A α = U µ A α x µ.
14 The Euler-Lagrange equations are or m d dτ U α + q c x µ τ d dτ P α = L x α A α x µ = + q c A β x β x α τ, m d dτ U α = q ( c U β Aβ x α A ) α x β = q c F αβu β.
15 Canonical Momentum Canonical Momentum P α = L xα λ = mu α + q c A α, where we have required our parameter λ to be c times the proper time. Note that the canonical momentum is constrained: ( P α q c A α ) ( P α q c Aα) = m 2 U α U α = m 2 c 2. which we found before as P 0 = H/c. Minimum substitution principle: To introduce electromagnetism for a particle, take a free particle replace p α P α := p α q A/c.
16 Dynamics of fields requires a density, a function of the fields 2, say φ i ( x, t). Euler-Lagrange becomes L µ ( φ i / x µ ) L = 0. φ i What are our fundamental fields? L(φ i, m uφ i, x ν ) will give second order differential equations, not Maxwell in F. But we know F = da, so second order in A µ is what we want. We have already seen particle action requires (q/c)a µ dx µ for a single charge. That is, each charge q i at x i contributes to L q i Φ( x i ) + q i c u i A( x i ). 2 Never done dynamics of fields? Need to read up, e.g. shapiro/507/gettext.shtml look at chapter 8 (or get book9 2.pdf from the same location).
17 For many charges, L int = i = 1 c ( q i Φ( x i ) 1 ) c q i u i A( x i, t) ( d 3 x ρ( x)φ( x) 1 ) J( x) c A( x) d 3 xa α ( x)j α ( x). This will give us the J µ on the right h side of the Euler equation from varying A µ, but we need something to give the left h side of Maxwell s equation, which should be linear in F, so we need a quadratic piece in L, Lorentz invariant with a total of two derivatives on A µ s. Let s try L = 1 16π F µν F µν 1 c J µa µ, where it is understood that F µν sts for µ A ν ν A µ is not an independent field.
18 The only contribution to L/ A µ, (taken with ν A µ fixed) is the J µ /c from the interaction term. We have so F µν ( Aρ x σ ) = δ σ µδ ρ ν δ σ ν δ ρ µ, L ( ) = 1 Aρ 4π F ρσ, x σ the full Euler-Lagrange equation is or 1 4π σf σµ + 1 c J µ = 0, σ F σµ = 4π c J µ. Thus we have derived Maxwell s equations (as df = 0 is automatic as F := da).
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