Lecture 18 April 5, 2010

Transcription

1 Lecture 18 April 5, 2010 Darwin Particle dynamics: x j (t) evolves by F k j ( x j (t), x k (t)), depends on where other particles are at the same instant. Violates relativity! If the forces are given by a potential energy V ( x j (t), x k (t)), that also violates relativity, unless V δ( x j x k ). Not very useful. But we know how to treat charged particles interacting electromagnetically if they are not moving too fast. We learned as freshmen how to do the lowest order (c ): Darwin The V ( x j, x k ) = q jq k x j x k T = 1 2 mj v 2 j. This encapsulates the effect of the E one particle produces on the other.

2 To next order In our relativistic treatment L int = q j ( Φ( x j ) + 1 ) c u j A( x j ), j Φ( x j ) = k scalar potential. q k x j x k is the c limit for the Darwin The Magnetic forces require moving particles to be produced, moving particles to feel their effect. So these are v 2 /c 2 effects. To this order, Φ A depend on choice of gauge. Choose Coulomb ( A = 0), not Lorenz, because then 2 Φ = 4πρ, Φ is determined by instantaneous information: Φ( r, t) = d 3 r ρ( r, t) r r to all orders in v/c!

3 Magnetic interaction From σ F σj = 4πJ j /c we have 1 2 ( ) c 2 t 2 A 2 A + 1 c t Φ + A = 4π J/c. The A is zero in Coulomb gauge. Working accurate to order (v/c) 2 we may drop the 1 2 A term, as A is c 2 t 2 already order (v/c) 1. Thus we may take Darwin The Particle j contributes q j v j δ 3 ( x x j ) to J( x ) 2 A = 4π c J + 1 c t Φ. q j x x j to Φ( x ), v j ( x x j ) so it contributes q j x x j 3 to Φ t. The Green s function for Laplace s equation is 1/ x x, which we apply to the right h side:

4 A( x) = = = d 3 x ( 1 x x J( x c ) 1 4πc ) t Φ( x ) d 3 x [ qj v j x x c δ3 ( x x j ) q ( j 4πc vj ( x )] x j ) x x j 3 q j v j c x x j + q ( j d 3 x vj ( x ) x j ) 1 4πc x x j 3 x x Darwin The where we have integrated by parts thrown away the surface at infinity. The gradient action on a function of x x, so we can pull out of the integral. Let r = x x j y = x x j. Then A( x) = q j v j c r q j 4πc d 3 y v j y y 3 1 y r

5 The integral can be done by choosing z r v j in the xz plane: d 3 y v j y 1 y 3 y r = 0 y 2 dy π 0 dθ sin θ 2π 0 dφ y(cos θv jz + sin θ cos φv jx ) y 3 1 y 2 + r 2 2yr cos θ The φ integral kills the v jx term then what remains is 2πv jz times 1 u dy du 0 1 y 2 + r 2 2yru = 1, though this integral is not as straightforward as Jackson claims. Writing v jz = v j r/r, we have A( r) = q [ j vj c r 1 ( )] 2 vj r. r Darwin The

6 Applying the gradient, we get A j ( x k ) = q j 2c x j x k [ v j + ( x k x j ) v j ( x k x j ) x k x j Multiplying by q k v k /c to get the appropriate contribution to L int, correcting the free-particle, mc 2 γ 1 + mc mv mv4 /c 2, we get the Darwin L Darwin = 1 m j vj c 2 m j vj j + 1 4c 2 j k j j k q j q k r jk ]. q j q k r jk [ v j v k + ( v j ˆr jk )( v k ˆr jk )], Darwin The where of course r jk := x j x k, r jk := r jk, ˆr jk = r jk /r jk. This is used in atomic physics (with v α for Dirac) in plasma physics.

7 The For Maxwell s electromagnetism: L EM = 1 16π F µν F µν 1 c J µa µ. Does not give complete equations of motion A µ. Consider adding a term proportional to A 2 : L = 1 16π F µν F µν + µ2 8π A µa µ 1 c J µa µ, known as the. Still have F µν := µ A ν ν A µ, not an independent field. homogeneous Maxwell equations still hold (as F = da). Extra term doesn t change P α µ (no α A β dependence), so change in equations of motion is just from L/ A µ = (µ 2 /4π)A µ, Darwin The β F βα + µ 2 A α = 4π c J α. One consequence comes from taking the 4-divergence of this equation:

8 Equations of Motion α β F βα +µ 2 α A α = 4π }{{} c α J α, }{{} 0 0 where the first vanishing is by symmetry the second assumes charge is still conserved, to the continuity equation α J α = 0 still holds. Thus α A α = 0 is an equation of motion, not a gauge condition! Then β F βα = A α, ( + µ 2) A α = 4π c J α. Darwin The In the absence of sources, this has solutions as before, ( ) A µ e i k x iω k t + A µ k + e i k x+iω k t, k k but with ω 2 = c 2 ( k 2 + µ 2 ).

9 Particle content Quantum mechanically we know p = i k E = i / t = ± ω, so ω 2 = c 2 ( k 2 + µ 2 ) tells us we have particles for which E 2 = P 2 c 2 + µ 2 2 c 2. Of course quantum field theoriests take = 1 c = 1, so this represents a massive photon with mass µ. Static solution: Darwin The If we consider a point charge at rest look for the static field it would generate, we need to solve 2 Φ + µ 2 Φ = 4πqδ 3 ( r) or ( r 2 Φ ) + r 2 µ 2 Φ = qδ(r). r r Away from r = 0 this clearly requires rφ(r) = Ce µr.

10 So Φ(r) = C e µr, Gauss s law tells us r 4πq = 4πR 2 dφ/dr R + µ 2 Φ 4πC, R 0 so C = q r<r Φ( x) = q e µr, with r = x. r This is the well-known Yukawa potential, which nuclear physicists had found was a good fit to the binding of nucleons in a nucleus, leading Yukawa to propose the existance of a massive carrier of the nuclear force, which we now know to be the π meson. Darwin The

11 A in s In the BCS theory of superconductivity, electrons form pairs, each pair acts like a boson. So the quantum mechanical state that each pair is in can be multiply occupied, superconductivity occurs when states develop macroscopic occupation numbers, 1. The wave function ψ( x) describing these particles is a complex function, with the density of particles n( x) = ψ ψ, so ψ = n( x)e iθ( x). We may approximate n( x) as being roughly constant. The velocity of these particles is related to the canonical momentum by v = 1 ( P q A m c ) which can be viewed as an operator acting between ψ ψ. It is the canonical momentum P which acts like i. Thus the current density is J = qψ vψ = nq ( θ q A m c ). Darwin The

12 If we take the curl of both sides of J = nq ( θ q A m c ), we get J = nq2 mc A = nq2 B, mc (1) as θ = 0. This equation doesn t quite say J = nq2 A, mc (2) but it does say, in a simply connected region, that the difference is the gradient of something, as such a gradient could be added to A by a gauge transformation, we might as well assume (2), which is known as the London equation. This gauge is still compatible with Lorenz (which can be viewed as determining A 0 ), so we have 2 A 1 2 A c 2 t 2 = 4π J c = 4πnq2 A, mc 2 which is the equation with µ 2 = 4πnq 2/ mc 2. Darwin The

13 London Penetration Depth At the boundary of the superconductor, if no current is crossing the boundary, we must have n A = 0. If we look for a static solution for a planar boundary z, uniform along the boundary, we have A e µz. The London penetration depth is λ L := 1 µ = mc 2 4πnq 2. Darwin The With q = 2e m = 2m e for the electron pair, taking n as the density of valence electrons, the penetration depth is of the order of tens of nanometers. As the A field is not penetrating further than that into the medium, any external magnetic field has been excluded.

14 Vortex Lines But magnetic field lines can enter the medium if our assumption of being able to do away with A by a gauge transformation is not correct. That could happen if the region of the superconductor is not simply connected that is, a flux line could enter destroy the superconducting region around which θ is incremented by a multiple of 2π. This is called a vortex line, corresponds to a quantized amount of flux, as A dl = 2πN c/q = A = Φ B, with N Z. S Darwin The With q = 2e, the quantum of flux is hc/2e.