Physics 504, Physics 504, Lecture 17 April 1, 2010 Spring 2010 For electromagnetism we have not a scalar but four fields Spring 2010 Electricity ν Electricity and A ,sowehavefour 4-vector fields and Canonical Momentum Density Magnetism Magnetism ∂ Shapiro P µ := L . Shapiro We have seen that in field theory the Lagrangian is an α ∂Aα(~x, t) integral of the Lagrangian density Canonical ∂ µ Canonical Momentum,  ∂x  Momentum, Tmunu Tmunu ν ξ Canonical Canonical (φi,∂φi/∂x ,x ) Momentum for Momentum for L E&M Last time we saw that the lagrangian density for the E&M The Stress The Stress (Energy- electromagnetic fields is (Energy- and the equations of motion come from the functional Momentum) Momentum) Tensor Tensor derivatives of L with respect to the local values of the Stress-Energy Stress-Energy for E&M 1 µν 1 µ for E&M fields, but because the Lagrangian density is local, these Ambiguities in = F Fµν JµA , Ambiguities in Lagrangian L −16π − c Lagrangian ∂ ∂ density density µν with µν with are given by L ν and L ν , which are currents currents ∂φi(x ) ∂(∂ρφi(x )) so the canonical momentum densities are µ Equations of Equations of functions of x . The Euler-Lagrange equations involved Motion for Motion for Aµ ∂ 1 Aµ density µ µ not a total momentum but a momentum . For a Green’s Pα := α L µ = F α, Green’s function for ∂(∂A (~x, t)/∂x ) −4π function for scalar field φj this would be wave equation wave equation F 2 µ ∂ because, as we saw last time, only the term depends P (xρ)= L . α µ j ∂∂ φ on ∂A /∂x . µ j xρ

Physics 504, µ Physics 504, The Stress (Energy-Momentum) Tensor Spring 2010 T ν for electromagnetism Spring 2010 Electricity Electricity and λ and Magnetism For electromagnetism, φi is replaced by A , Magnetism Discrete mechanics: H = Piq˙i L,˙qi Pi i − → Shapiro ∂ ∂Aλ Shapiro Field theory: HamiltonianP density µ µ T ν = λL µ ν δν . Canonical ∂(∂A /∂x ) ∂x − L Canonical Momentum, Momentum, ∂ ∂φi Tmunu Tmunu (~x):= Pi(~x)φ˙i(~x) (~x)= L . 0 0 Canonical The first factor in the first term is Canonical H −L ∂(∂φi/∂x ) ∂x −L Momentum for Momentum for Xi Xi E&M E&M The Stress The Stress (Energy- ∂ 1 µ (Energy- Momentum) L = F λ, Momentum) Time has been picked out. More generally, let Tensor ∂ ∂Aλ/∂xµ − π Tensor Stress-Energy ( ) 4 Stress-Energy for E&M for E&M ∂ ∂φi Ambiguities in Ambiguities in µ µ Lagrangian so our first (tentative) expression for the energy Lagrangian T ν = L µ ν δν . density density ∂(∂φi/∂x ) ∂x − L µν with µν with Xi currents momentum tensor is currents Equations of Equations of Motion for λ Motion for This object goes by the names energy-momentum µ 1 ∂A 1 αβ µ A Tµν = Fµλ ηµνF Fαβ . A Green’s ν Green’s tensor or stress-energy tensor or canonical stress function for −4π  ∂x − 4  function for tensor, and we see the hamiltonian density is the 00 wave equation wave equation component of this tensor. This expression has good and bad properties! We have seen T 00 is the Hamiltonian density.

0i i Physics 504, Physics 504, T = P Spring 2010 Conservation of Momentum Spring 2010 Electricity µ Electricity R and If has no explicit dependence on x , we can show and Magnetism Lµ Magnetism 00 0j ∂µT ν = 0, where ∂µ is the stream derivative. For If T is the energy density, is T the density of Shapiro Shapiro momentum? µ ∂ ∂ Canonical ∂µT ν = ∂µ L ∂νφi+ L ∂µ∂νφi ∂ν . Canonical Momentum,  ∂(∂µφi) ∂(∂µφi) − L Momentum, 0i 1 λ 1 1 Xi Xi T = F0λ∂iA = Ej∂iAj = E~ B~ + E~ ~ A~ . Tmunu Tmunu i Canonical Canonical 4π 4π 4π  ×  · ∇  Momentum for Momentum for E&M The derivative in the last term is given by the chain rule E&M The Stress The Stress Poynting tells us the first term is the correct expression. (Energy- (Energy- Momentum) Momentum) Tensor ∂ ∂ Tensor The second term is unwanted, and also not gauge Stress-Energy ∂ ∂ φ ∂ ∂ φ Stress-Energy for E&M ν = L ν i L ν µ i for E&M invariant. Ambiguities in − L − ∂φi − ∂ (∂µφi) Ambiguities in Lagrangian Xi Xi Lagrangian density density But we haven’t included charges in the momentum, and if µν with µν with currents so currents no charges, ~ E~ =0, E~ ~ A~i = ~ (AiE~ ) Ai ~ E~ is ∂ ∂ ∇· · ∇ ∇· − ∇· Equations of µ Equations of   Motion for ∂µT ν = ∂µ L L ∂νφi Motion for a total derivative, and won’t affect the total momentum. µ  ∂(∂ φ ) − ∂φ  µ A Xi µ i i A 3 0µ µ Green’s Green’s So we do have the good property d xT = P , the function for function for Z wave equation and the parenthesis vanishes by the equations of motion. wave equation total momentum. But we don’t have the right density. Thus we have µ ∂µT ν =0. (1) This is a good thing. Physics 504, µν Physics 504, Non-symmetry is a bad thing Spring 2010 Improving T Spring 2010 µν νµ Electricity Electricity Note: this T = T . This is a problem. We expect the and ρµν ν µρ and 6 0k Magnetism So consider ψ = A F /4π, and adding Magnetism angular momentum density to be given by ijkxjT , but µν Shapiro Shapiro that requires T to be symmetric. 1 ν µρ 1 ν µρ T µν ∂ρ (A F )= (∂ρA ) F So our is good at giving the total momentum and Canonical 4π 4π Canonical being conserved, but bad in not being symmetric or Momentum, Momentum, Tmunu µρ µ Tmunu because ∂ρF = 0 in the absence of a source J . But this gauge-invariant. Need modification keeping good Canonical Canonical Momentum for µν Momentum for E&M is just what we need to add to T to make E&M properties but changing bad ones. The Stress The Stress ρµν µρν (Energy- (Energy- If we have a tensor ψ = ψ (antisymmetric in first Momentum) Momentum) − Tensor µν µν 1 µρ ν 1 µρ ν 1 µν αβ Tensor ρµν µν Stress-Energy T F ∂ A F F η F F . Stress-Energy two indices) and we add ∂ρψ to T , for E&M Θ = + ρ = ρ αβ for E&M Ambiguities in 4π −4π  − 4  Ambiguities in Lagrangian Lagrangian µν ρµν density density ∆(∂µT )=∂µ∂ρψ =0, µν with µν with currents This expression has all the good properties and is also currents µν Equations of gauge invariant and symmetric. Furthermore, Equations of so the new T is also conserved. Furthermore, Motion for Motion for Aµ Aµ Green’s 1 1 1 Green’s function for 0i 0j i ~ ~ function for 3 0ν 3 ρ0ν 3 j0ν j0ν wave equation Θ = F F j = EjijkBk = (E B)i, wave equation d x∆T = d x∂ρψ = d x∂jψ = njψ 0 −4π 4π 4π × Z Z Z ZS → the correct momentum density or energy flux, as given by for surface S . So adding ∂ ψρµν keeps all the good →∞ ρ Poynting. properties.

Physics 504, µν Physics 504, Ambiguities in Spring 2010 Θ with currents Spring 2010 L Electricity Electricity 1 µν and The energy-momentum tensor of the electromagnetic field and F Fµν is gauge invariant, but the interaction term −16π Magnetism is Magnetism 1 d4xJ xρ Aµ xρ Shapiro µν 1 µρ ν 1 µν αβ Shapiro in the action, µ( ) ( ) is not, so is not a ΘEM = F F η F Fαβ , − c Z L −4π  ρ − 4  µ Canonical Canonical E~ B~ J µν unique function of the physical state ( and and ). Momentum, ∂ Momentum, Tmunu and is conserved ( µΘEM = 0 if there are no sources. Tmunu Is there an ambiguity in the action under a gauge Canonical Canonical Momentum for What if there are? Momentum for transformation Aµ Aµ0 = Aµ + ∂µΛ? This adds a piece E&M E&M → 4 µ The Stress 1 The Stress A /c d xJ ∂ (Energy- µν µρ ν µν αβ (Energy- to the action ∆ = (1 ) µΛ. But Momentum) 4π∂µΘ = ∂µ F Fρ + η F Fαβ Momentum) − Tensor  4  Tensor R Stress-Energy Stress-Energy for E&M for E&M 4 µ µ 4 µ Ambiguities in µρ ν µρ ν 1 αβ ν Ambiguities in d xJ ∂µΛ= nµ J Λ d xΛ ∂µJ , Lagrangian =(∂µF ) Fρ + F ∂µFρ + F ∂ Fαβ Lagrangian Z ZS − Z density 2 density µν with µν with 0 currents π currents 0 4 ρ ν 1 αβ ν ν ν S−→ Equations of = J Fρ + F ∂αFβ ∂βFα + ∂ Fαβ Equations of → | {z } Motion for c 2  −  Motion for | {z } µ µ A 4π ρ ν 1 αβ νρ A so this will not affect the action. Green’s = J Fρ + F η (∂αFβρ + ∂βFρα + ∂ρFαβ) Green’s function for c 2 function for More generally, adding a total divergence to the wave equation wave equation =0 as dF =ddA=0 lagrangian density in a field theory, like adding a total 4π | {z } = J ρF ν =0. time derivative in a particle theory, does not affect the c ρ 6 equations of motion, and is irrelevant to the physics. Not conserved!

ν Physics 504, Physics 504, PEM is not conserved Spring 2010 Total Momentum is conserved Spring 2010 Electricity Electricity and and Magnetism Consider a charged particle of mass mi, charge qi at point Magnetism ~xi(t). Its mechanical 4-momentum changes by Thus the total 4-momentum of the electromagnetic field Shapiro Shapiro dP ν dP ν ν 1 3 0ν Canonical (i) 1 (i) 1 qi ν ρ Canonical PEM = d xΘ (~x), Momentum, = = F ρ(~xi)Ui . Momentum, c Z Tmunu dt γi dτ γi c Tmunu Canonical Canonical Momentum for Momentum for E&M E&M is not conserved, but rather The Stress This particle corresponds to a 4-current The Stress (Energy- (Energy- Momentum) Momentum) ν Tensor ρ ~ 3 3 Tensor dPEM 1 d 3 0ν 3 0ν Stress-Energy J =(cρ, J)=(cqiδ (~x ~xi),qiuiδ (~x ~xi) Stress-Energy = d x Θ (~x)= d x∂0Θ (~x) for E&M − − for E&M dt c dt Z Z Ambiguities in 1 ρ 3 Ambiguities in Lagrangian = qiγi− Ui δ (~x ~xi). Lagrangian density − density 1 3 ρ ν 1 3 iν µν with µν with = d xJ (~x)F (~x) d x∂ Θ currents currents c ρ − c i Z Z Equations of Plugging this into our expression for the change in the Equations of Motion for Motion for 1 µ momentum of the electromagnetic field, we have µ = d3xJ ρ(~x)F ν(~x), A A ρ Green’s Green’s c Z function for ν function for wave equation dPEM qi 3 ν 1 ρ 3 qi ν ρ wave equation = d xFρ (~x)γi− Ui δ (~x ~xi)= F ρ(~xi)Ui , as the second term is the integral of a divergence. dt c Z − −cγi

ν ν and the total momentum, PEM + P(i) is conserved. µ Physics 504, But we assumed the Lorenz condition, which constrains Physics 504, Equations of Motion for A Spring 2010 0 Spring 2010 Electricity ωA ~k A~ Electricity the coefficients ~k ~k = 0. These are the Euler-Lagrange tell us and ± ∓ · ± and Magnetism solutions for an electromagnetic wave in empty space. Magnetism σµ σ µ µ σ 4π µ ∂σF = ∂σ∂ A ∂ ∂σA = J . Shapiro Shapiro − c Without imposing the Lorenz condition, σ Canonical Canonical If we knew ∂σA = 0 (the Lorenz condition), we could Momentum, σ µ µ σ Momentum, Tmunu ∂σ∂ A ∂ ∂σA =0, Tmunu discard second term, and have Canonical − Canonical Momentum for Momentum for E&M µ E&M σ µ 4π µ The Stress which is inadequate to determine the evolution of A (~x, t) The Stress (Energy- (Energy- ∂σ∂ A = J , Momentum) Momentum) c Tensor in time. Fourier transform: Tensor Stress-Energy σ µ ν µ σ ν Stress-Energy for E&M kσk A˜ (k ) k kσA˜ (k ) = 0, which is not four for E&M which has solutions given by Ambiguities in − Ambiguities in Lagrangian k Lagrangian 1) a particular solution, given in terms of the Green’s density independent equations, because dotting with ρ gives density µν with µν with currents currents function on J, and σ µ ν µ σ ν 2 2 ρ ν Equations of kσk kµA˜ (k ) kµk kσA˜ (k )=(k k )kρA˜ (k )=0, Equations of 2) an arbitrary solution of the homogeneous wave Motion for − − Motion for σ µ Aµ Aµ equation ∂σ∂ A = 0. The homogeneous solution is Green’s ˜ρ ν Green’s function for telling us nothing about A (k ). Euler-Lagrange only function for wave equation wave equation µ i~k ~x iω t µ i~k ~x+iω t determine the components transverse to k. A e · − ~k + A e · ~k , ~k + ~k  −  This is gauge invariance again. No physics constrains the X~k gauge transformation Λ(~x, t) in the future, so Aµ is where ω = c ~k , where ω = c ~k . underdetermined. | | | |

Physics 504, Physics 504, Solving the inhomogeneous equation Spring 2010 Fourier transformed equation Spring 2010 µ Electricity 4 µ 1 4 ikµz 2 ˜ µ Electricity But we are free to impose the Lorenz condition. Let’s do and As δ (z )= (2π)4 d ke− , we have k D(k )= 1, so and Magnetism − Magnetism so. the solution for theR Green’s function is Shapiro Shapiro µ e ikµz Now we turn to the inhomogeneous equation ˜ µ 1 µ 1 4 − Canonical D(k )= 2 , and D(z )= 4 d k 2 . Canonical µ β µ 4π µ Momentum, −k −(2π) Z k Momentum, A = ∂β∂ A = J , Tmunu Tmunu c Canonical µ Canonical Momentum for 4 µ 1 4 ikµz Momentum for As δ (z )= 4 d ke , the solution for the with the solution E&M (2π) − E&M The Stress The Stress (Energy- Green’s function isR (Energy- µ 4π 4 µ Momentum) Momentum) Tensor Tensor A (x)= d x0D(x, x0)J (x0), µ c Z Stress-Energy ikµz Stress-Energy for E&M µ 1 µ 1 4 e− for E&M Ambiguities in D˜(k )= , and D(z )= d k . Ambiguities in Lagrangian 2 4 2 Lagrangian where D(x, x0) is a Green’s function for D’Alembert’s density −k −(2π) Z k density µν with µν with equation currents currents 4 2 xD(x, x0)=δ (x x0). Equations of Looks like what we did for Poisson, but here k =0is Equations of − Motion for 2 ~ 2 ~ Motion for µ more difficult, as it requires only k = k , not k =0. µ We are interested in solving this in all of spacetime. No A 0 A Green’s ~ Green’s function for For Poisson, trouble from k = 0 gives ambiguity of function for boundaries, translation invariance, so wave equation wave equation ψ = V0 + ~r C~ , a uniform E~ and ambiguous constant in Φ. D(x, x0)=D(x x0)=D(z). Solve by Fourier transform: · − For wave equation: arbitrary waves satisfying free wave 1 µ 4 µ ˜ µ ikµz equation. write D(z)= 4 d k D(k )e− . (2π) Z Deform the integration path to resolve the singularities.

Physics 504, Physics 504, Disambiguate with Contour choice Spring 2010 Retarded Green’s function Spring 2010 Electricity Electricity and Consider D for contour r. If source at x0 = 0, and we look and Clarify ambiguity by specifying how to avoid the Magnetism for A later, z0 > 0. Close contour in lower half plane, Magnetism 0 0 Shapiro ik0z Im k0 z Shapiro singularities and writing where e− = e−| | 0, so this semicircle k−→ 0 | |→∞ ik0z Canonical D πi Canonical 1 3 i~k ~z e− Momentum, adds nothing. So = 2 times the sum of the Momentum, D(z)= d ke · dk0 . Tmunu − Tmunu − π 4 2 ~ 2 residues. minus because clockwise. The residues are (2 ) Z ZΓ k0 k Canonical Canonical −| | Momentum for ik z0 ik z0 Momentum for E&M e 0 e 0 E&M The Stress − − The Stress (Energy- Res + Res (Energy- Specifying contour Momentum) ~ ~ ~ ~ ~ ~ Momentum) k0= k (k0 + k )(k0 k ) k0= k (k0 + k )(k0 k ) Tensor | | −| | Tensor Γ’s avoidance of the k 0 Stress-Energy | | −| | | | −| | Stress-Energy ~ r for E&M i ~k z0 i ~k z0 ~ 0 for E&M poles at k0 = k . Ambiguities in e− | | e | | sin( k z ) Ambiguities in ±| | Lagrangian = + = i | | . Lagrangian density ~ ~ ~ density Three such con- F µν with 2 k 2 k − k µν with a currents | | − | | | | currents tours are shown. 0 Equations of But if z < 0 close in upper half plane, no residues, Equations of Integrand analytic The retarded (r), advanced Motion for Motion for Aµ D = 0, so all together Aµ except at poles so (a), and Feynman (F ) con- Green’s Green’s function for 0 ~ 0 function for the contours may tours for defining the Green’s wave equation Θ(z ) 3 i~k ~z sin( k z ) wave equation Dr(z)= d ke · | | . be deformed while function. (2π)3 Z ~k | | avoiding the poles. D is rotationally invariant, so we may choose the North pole along ~z using spherical coordinates. Physics 504, Physics 504, We get Spring 2010 Simplifying D Spring 2010 Electricity Electricity and We may simplify Dr by noting and 0 0 Magnetism Magnetism Θ(z ) ∞ 2 ikR cos θ sin(kz ) 1 Dr(z)= k dk dθ sin θe 0 0 2 Shapiro sin(kR) sin(kz )= cos(k(R z )) cos(k(R + z0)) Shapiro (2π) Z0 k 2 − − 0   Θ(z ) ∞ 0 Canonical 1 i(z0 R)k i(z0+R)k i(z0 R)( k) Canonical = e − e + e − − = 2 dk sin(kR) sin(kz ), Momentum, Momentum, 2π R Z0 Tmunu 4h − Tmunu Canonical i(z0 R)( k) Canonical Momentum for +e − − Momentum for R ~z E&M E&M where = . The Stress i The Stress | | (Energy- (Energy- This is the retarded Green’s function aka causal, as the Momentum) 0 Momentum) Tensor Tensor µ Θ(z ) ∞ i(z0 R)k i(z0+R)k A Stress-Energy D z dk e e Stress-Energy effects on of the source are felt only after the source for E&M so r( )= 2 − for E&M Ambiguities in 8π R Z h − i Ambiguities in acts. Lagrangian −∞ Lagrangian density 0 density µν with Θ(z ) µν with currents = [δ(z R) δ(z + R)] currents πR 0 − − 0 The contour a gives the advanced Green’s function useful Equations of 4 Equations of Motion for 0 Motion for only if you want to configure an incoming field which µ Θ(z ) µ A = δ(z0 R), A Green’s Green’s would magically be totally dissolved by a given source. function for 4πR − function for wave equation wave equation where the second δ was dropped because both z0 and R Finally the contour F gives the Feynman propagator, are positive. So the Green’s function only contributes which is used in quantum field theory. when the source and effect are separated by a lightlike path, with ∆z0 = ∆~z . | |

µ Physics 504, Physics 504, Full solution for A Spring 2010 Radiation Field. Spring 2010 Electricity Electricity and and So how do we describe the field when we know what the Magnetism Magnetism sources are throughout space-time? We can use any of the Shapiro Shapiro Green’s functions to get the inhomogeneous contribution, Canonical Canonical and then allow for an arbitrary solution of the Momentum, Of course the source may be persistent, for example if Momentum, Tmunu Tmunu homogeneous equation. Thus we can write Canonical there is a net charge, but we may often consider that the Canonical Momentum for µ Momentum for E&M effect of the source is confined to the change from A (x) E&M 4π The Stress µ in The Stress µ µ 4 µ (Energy- A x (Energy- A = A (x)+ d x0Dr(x x0)J (x0) Momentum) to out( ). Then we define the radiation field to be Momentum) in c Z − Tensor Tensor Stress-Energy Stress-Energy π for E&M for E&M µ 4 4 µ Ambiguities in µ µ µ 4π 4 µ Ambiguities in = A (x)+ d x0Da(x x0)J (x0). Lagrangian A (x)=A (x) A (x)= d x0D(x x0)J (x0), Lagrangian out c Z − density rad out in density µν with − c Z − µν with currents currents If the sources are confined to some finite region of Equations of D z D z D z . Equations of Motion for where ( ):= r( ) a( ) Motion for µ − µ space-time, there will be no contribution from Dr at times A A µ Green’s Green’s earlier than the first source, and A (x) describes the function for function for in wave equation wave equation fields before that time. Also after the last time that the µ source influences things, the field will be given by Aout(x) alone.

µ Physics 504, Better expression for J from charges Spring 2010 Electricity and Magnetism The expression we wrote earlier for the current density of Shapiro

a point charge, Canonical Momentum, ρ 1 ρ 3 Tmunu J = qiγi− Ui δ (~x ~xi) Canonical − Momentum for E&M The Stress (Energy- can be written in this four-dimensional language as Momentum) Tensor Stress-Energy ρ for E&M ρ µ 0 1 3 Ambiguities in J (x )=qi dtδ(t x /c)γi− Ui δ (~x ~xi(t)) Lagrangian Z − − density µν with 4 µ µ ρ currents = qic dτδ (x xi (τ))Ui , Equations of Z − Motion for Aµ Green’s where τ measures proper time along the path of the function for wave equation particle.