Energy, Momentum, and Lagrangians

Transcription

1 Energy, Momentum, and Lagrangians Lecture 10 1 The Poynting vector Previously an expression for the flow of energy in terms of the fields was developed. Note the following uses the MKS system of units in order to be consistent with the previous development. S = E H This expression represents the power crossing a unit area. For calculational purposes we are generally interested in the average value of the power flow. This is obtained by our previous conventions. S = (1/2)Re ( E H ) The Poynting vector was developed using the power loss in a resistive medium through the work done by the fields on charges, J E d 3 x. Note that the power in the fields flowing into a volume can be written as ; W t field = [ E H] da In the above W is the energy, and da is the area surrounding the volume containing the fileds. Use Gauss law to write this as; dw field = S d 3 x Now define W as the energy per unit volume, and collect all terms involving energy flow and loss. In differential form this equation is; S + J E = W t The above equation states that the power flowing out of a volume per unit volume plus the power put into moving the charges in the volume equals the time change of the energy density in the volume. This equation represents conservation of energy. If J E = 0 and we consider (cw, S) a 4-vector, the above covarient conservation equation is produced when (cw, vecs) is contracted with the 4-vector gradient. 1

2 2 Momentum flow In a similar way to the power flow, use Maxwell s equations and the Lorentz force to develop the time change of momentum. The Lorentz force is written; d P M = [ρ E + J B] d 3 x In the above expression, P M is the mechanical momentum in the moving charge. Now substitute E = ρ/ǫ and H = J + (1/cµ) E t into the Lorentz force equation. Use the remaining Maxwell equations to write the equation below; d P M + d [ǫ E B] d 3 x = ǫ [ E( E) E ( E) + c 2 B( B) c 2 ( B ( B))] d 3 x Note that P field is identified by the momentum per unit volume in the field, P field = (1/c 2 ) ( E H) d 3 x The above equation is interpreted as; Momentum/V olume = (Energy/c)/[(ct)Area] = (1/c 2 )(Energy/(area t)) Then S = E H is the power flow through a surface due to the fields. This power flows with a velocity, c, and the term (Energy/c) equals the momentum of the electromagnetic wave. Look at the remaining term which can be expressed in its vector components. In the 1 direction this expression for the E field only is; [ E( E) E ( E)] 1 = i x i [E 1 E i (1/2) E Eδ i1 ] The above is the divergence of a rank 2 tensor. Compiling all the components for the E as well as the B field, the tensor has the form; Tαβ = ǫ[e α E β + c 2 B α B β (1/2)( E E + c 2 B B)] This tensor is called the Maxwell stress tensor. The integral form of the equation developed above is; d [ P M + P Field ] = T αβ da β Interpretating this equation, the integral over da β is over the area surrounding the volume 2

3 t a b c E h g y f x E 2 2 E 1 e d d Figure 1: The surfaces on which we will evaluate the Maxwell stress tensor. which contains the momentum. Identify T αα as the pressure (f orce/area) on the surface α. This has the correct units of energy density. The tensor element Tαβ is the momentum density times the velocity or a shearing force on the area element α. In differential form the above equation is; T + (ρ E + J B) = G t Here, G = (1/c 2 )[ E H] is the momentum density, and the equation represents momentum conservation. The momentum density through a surface plus mechanical momentum density of the charge currents equals the time change of the momentum density in the enclosed volume. 3 Examples Look at the z component of the force applied by the fields. T zz = (ǫ/2)[e 2 z E2 y E2 x ] + (ǫc2 /2)[B 2 z B2 y B2 x ] Tzy = ǫ(e z E y + c 2 B z B y ) Tzx = ǫ(e z E x + c 2 B z B x ) Then calculate the force on a conducting plate inserted in a capacitor. Simple analysis shows that the field of the capacitor induces charge on the conducting plate, which causes a force on the plate pulling it into the capacitor. Consider Figure 1. Ignore dimensions perpendicular to the plane of the drawing and assume that the fields within the capacitor are uniform and perpendicular to the planar surfaces, ie the field of an infintite parallel plate capacitor. The dashed line indicates a surface (include dimensions in and out of the plane of the drawing) over which the stress tensor is evaluated. We need to only consider the E field. 3

4 Integration over ha is essentially zero as there is little field at these points. Integration over ab and gh cancel as does integration over cd and ef. Then integration over bc and fg are equal and only field components of E y are non-zero. E 1 = V/d E 2 = V/(d t) Let the wih of the conducting plate be w, and use the outward normal as the direction of the area vectors. The field tensor is then; (1/2)Ey T = 0 (1/2)Ey (1/2)Ey 2 The force on the plate is; F x = T xx da x V 2 F x = (ǫ/2) (d t) 2 w(d t) (ǫ/2) v2 d 2 wd F x = ǫv 2 wt 2d(d t) In this simple case, we can obtain the same result using the energy. Suppose the plate extends into the capacitor a distance x. Use the energy density in the fields to obtain; W = (1/2)[ǫE 2 + (1/µ)B 2 ] V olume W = (1/2)ǫ[ V 2 d 2 d(l x)w + V 2 (d t) 2 (d t)xw] F x = dw dx = (1/2)ǫ[ V 2 d w + V 2 (d t) w] This results in the same answer above when the stress tensor was used. Now consider another, more complicated example which better illustrates the power of the stress tensor. Find the radiation pressure on a spherical, metalic particle of radius, a, illuminated by a beam of plane-polarized light. The particle is assumed perfectly conducting. First develop an expresion for the radiation pressure using an approximation. The Poynting vector divided by c 2 is the momentum density in the EM fields, (Momentum c/(area time c 2 ). The force is the time change of the momentum density, G, so that (assume a plane EM wave in the z direction); dp z = F z = 2c Gz da z 4

5 In the above, da z is the surface area of the spherical particle struck by the wave. The factor of 2 comes from the change of momentum between the incoming to outgoing wave. The integral extends over the lower hemisphere and we want the time average. For a plane wave in MKS units, B 0 = E 0 /c. F z = 2/c r 2 dcos(θ) dφ cos(θ) ((1/2)Re[ E H ] z ) Then for the plane wave, H 0 = B 0 /µ = E 0 /cµ = ǫ c E 0. Substitution and integration yields; F z = πa 2 ǫe 2 0 However, to really work this problem correctly the stress tensor must be used. This means that we need all the fields surrounding the spherical particle. Therefore the scattering problem of a plane wave scattering from a conducting sphere must be obtained, and the total field, including both the incident and scattered waves, are used in the stress tensor At this point, assume we know these fields, ie both the incident and the scattered wave at the spherical surface of the particle in spherical coordinates are known. The force on the particle is obtained from; F z = (1/2)[ da Z T zz + da x T zx + da y T zy ] The factor of (1/2) comes from the time average, and use the time average stress tensor components T αβ. Put this in spherical coordinates and average over the azmuthal angle. F z = dσ cos(θ)[ǫ(e 2 θ +E2 φ E2 r ) + (1/µ)(B2 θ +B2 φ B2 r )] + sin(θ)[ǫe re θ +B r B θ /µ] Note that these fields are the total field (incident plus scattered) at the surface d σ. To obtain the fields look at Figure 2. The direction cosines of the axes in spherical coordinates are; l = sin(θ) cos(φ) m = sin(θ) sin(φ) n = cos(θ) and l dσ = dσ x, etc. Then; E x = E r sin(θ) cos(φ) + E θ cos(θ) cos(φ) E φ sin(φ) E y = E r sin(θ) sin(φ) + E θ cos(θ) sin(φ) + E φ cos(φ) 5

6 z θ E r Eφ π /2 + φ y x E θ Figure 2: The field directions in a spherical coordinate system which are then projected onto a Cartesian system using the direction cosines in order to find the stress tensor E z = E r cos(θ) E θ sin(θ) There is substantial algebra when substituting the fields into the stress tensor, collecting the terms, and averaging over the azmuthal angle. We will only need T zz to continue using the approximation below. Let ct = cos(θ) and st = sin(θ) T zz = (ǫ/2)[e 2 r (ct2 st 2 ) + E 2 θ (st2 cy 2 ) E 2 φ 4E re θ ct st] Ignore the scattered wave and choose a plane wave, incident along the z axis. For this case T zy = T zx = 0. T zz = (ǫ/2)e 2 x (1/2µ)B2 y Integration proceeds easily to obtain the previous result. Finally consider a long coaxial cable of length, L, with inner radius, a, and outer radius, b. There is a charge per unit length, λ, on the conductors and they carry a current, I, in opposite directions. The fields between the conductors are; E = 1 2πǫ λ ρ ˆρ ; B = µ 2π I ρ ˆφ The power flow down the cable is; Power = S da = (1/µ) [( E B) ẑ] daz Power = (1/µ) λµi 4π 2 ǫ dφ ρ ρ 2 dρ = I[ λ 2πǫ ln(b/a)] = IV 6

7 I E I E I I E decelerates charge Figure 3: The movement of charges in the power flow in the coaxial cable to show the hidden momentum. The momentum in the field is ; P field = (1/c 2 ) S d 3 x = µλil 2π ln(a/b) ẑ Nothing appears to move, at least the CM of the system is constant, so there must be some other momentum which cancels the field momentum in order for momentum to be conserved. Look at the current loop in Figure 3. Charges on the left are accelerated toward the top and charges on the right decelerated toward the bottom. The current I = λu is the same in all the segments, but the number of charges, N t, at the top must move faster than number of charges, N b, at the bottom. Thus to keep the current constant there must be less charge at the top. Note the insert to the right of the figure showing the distance between the charges they are accelerated/decelerated. I = QN t L U t = QN b L U b Classically the momentum of each charge carrier is mu for U the velocity. Therefore; P classical = mn t U t mn b U b Substitute for the number of charges; P classical = mil Q mil Q = 0 However, relativistically the momentum is mγu. When we substitute for NU, we obtain; 7

8 P rel = γ t mn t U t γ b mn b U b = mil Q [γ t γ b ] Therefore must be work done, and the field momentum provides the work to move the charges. This momentum is cancelled by the resistance to this flow. There is also angular momentum in an EM field, and is obtained from; L field = (1/c 2 ) d 3 x x ( E H) In terms of a tensor formulation; M αβγ = T αβ x γ T αγ x β Conservation of angular momentum requires the divergence equation; α M αβγ = 0 4 Kinematics In the following, return to the use of Gausian units when using the field equations. In a collision between particles both energy and momentum must be conserved. Suppress the constant, c, in the following. Assume a process; A + B = C + D The conservation equations are; E A + E B = E c + E D Let; p A + p B = p C + p D E 0 = E A + E B p 0 = p C + p D The particles are placed on their mass shells so that E 2 p 2 = m 2. Now solve for the energy/momentum of particle C, and choose to let particle B be at rest, so p B = 0. and 8

9 CM Velocity CM CM in CM in Lab CM Velocity Figure 4: A cartoon of a transformation between the CM and Lab frames E B = m B. E 2 D p2 D = m D = (E 0 E C ) 2 ( p 0 p C ) 2 Solve this equation for p C. To simplify the equations define; α = (m 2 A + m2 B + m2 C m2 D + 2 E A m B )/(2E 0 ) β = p A cos(θ) Here θ is the angle between p A and p 0. The solution has the form; p = αβ ± α 2 m 2 C (1 β2 ) 1 β 2 If the masses, incident energies, momenta, and scattering angle are known, then the scattered energy and momenta are uniquely determined. There is the possibility of two solutions for appropriate masses and energies. This is because a boost may move a solution from the CM system so that both particles recoil into forward angles as shown in Figure 4. For a lorentz transformation between 2 reference frames with the boost in the 1 direction; E = γ(e + βp 1 ) p 1 = γ(p 1 + βe ) p 2 = p 2 p 3 = p 3 In polar coordinates; 9

10 p 1 = p cos(θ) p 2 = p sin(θ) cos(φ) p 3 = p sin(θ) sin(φ) One obtains; p cos(θ) = γ(p cos(θ ) + βe ) p sin(θ) = p sin(θ ) φ = φ E = γ(e + βp cos(θ )) 5 Phase space Suppose there are more than 2 particles in the exit channel of a scattering problem. In this case, there is no unique energy for any of the scattered particles, but a continuium of possible energies and momenta for the exit particles is obtained. Of course the total energy and momentum of the system of particles is conserved. Thus determine the probability of a given distribution of scattering momenta using phase space (density of states). Suppose there are n particles in the final state created by 2 particles in the initial state p 1, p 2. The transition probability to a specific final state is written; S(p 1, p 2 p 1, p 2 ) Sum all possible probabilities for un-measured particles (p i are incident particles and p j all final state particles). Write this as; P n n i=1 d 4 p j δ(p 2 j m 2 i ) δ( p j p 1 p 2 ) S(p 1, p 2 p 1, p 2 ) The 1 st delta function puts each particle on its mass shell. The 2 nd requires conservation of both energy and momentum. Now factor the transition probability from the integral (The assumption is that it changes little over the integration). P n S n i=1 d 4 p i δ(p 2 i m 2 i)δ( p j p 1 p 2 ) = S R n 10

11 The integral, R n, is the phase space (density of states) factor. It represents the number of possible states for specific set of measured kinematics. R n = n i=1 d 4 p i δ(p 2 i m 2 i ) δ( p j p 1 p 2 ) 6 Lagrangians A particle takes the path of least action for conservative motion. The action is defined as; A = L(x i (t), ẋ i (t), t) In the above, (q, q) are the conjugate variables of the Lagrangian, L, and the integral is over time. As written, this is a non-relativistic relation. To extend the relation into covarient form we integrate over proper time τ and because = γdτ, write; A = dτ [γl] Then [γl] must be a Lorentz invarient so that the total integral is invarient. Variation of the action leads to the Euler-Lagrange equations which can be solved to find the path of the particle as a function of time. For the moment, revert to a non-relativistic form for the motion of a charge in an EM field. Here V is the particle velocity. L = (1/2)mV 2 qφ + (q/c) V A The Euler-Lagrange equation results in; d V L L = 0 i x i L V i = mv i + (q/c)a i d(mv i ) = qφ x i + (q/c) x i[ V A] (qm/c) da i This is the non-relativistic expression for the Lorentz force with φ the scalar and A the vector potentials. Note that; E i = i φ (1/c) A i t da i = A i t + A i x j V j 11

12 [ V B] i = i ( A) ( V )A i Upon substitution of these equations into the Euler relation above, the Lorentz force is obtained. F i = qe i + (q/c)[ V B] i Now the form, φ (1/c)( V A), is almost a Lorentz invarient. It should have the invarient form; γ(φ (1/c) V A) = U β A β where U = (γc, γ V ) and A = (φ, A). Next the invarient Lagrangian for a freely moving relativistic particle is; L free = mc 2 1 (V/c) 2 = mc 2 /γ Note that γl is a Lorentz invarient. Insert this into the Euler-Lagrange equations; d (γmv i) = 0 Combine this Lagrangian with a Lagrangian for the EM interaction; L rel = mc 2 /γ (q/γ)[γ(φ (1/c) V A)] Then insertion into the action integrand using = γdτ, gives the Euler-Lagrange equations. We obtain, j = L V j = γmv j + (q/c)a j Here j is the momentum conjugate to V j, and it includes both the particle momentum and a field component. 7 Hamiltonian Use the identity ; dl = j L q j q j + L q j q j Substitute into the 1 st term in the above sum, the Euler equation; 12

13 To obtain; L qj = d L q j dl = j d [ q L j qj ] q j and finally; d [L L q qj j ] = 0 Thus define the Hamiltonian as; H = L qj q j L Then use the conjugte momentum derived earlier to write the Hamiltonian; H = V L After re-arrangement of terms, the following expression is obtained. (W qφ) 2 = (c q A) 2 + (mc 2 ) 2 Here we have replaced H by W the total energy. Note that it includes the field energy. U α U α = γ 2 (c 2 v 2 ) = c 2 The covarient Lagrangian is; L rel = [mc(u α U α ) 1/2 + (q/c)u α A α 8 Example Start with the relativistic Lagrangian; L = (1/2)mU α U α (q/c)u α A α Then note that U α U α = c 2 and; du α dτ = x0 τ Uα x 0 xi τ Uα x i 13

14 τ = Uα x α Apply this to the Lagrangian; Then use ; to get; L U α = g αβ [mu β + (q/c)a β ] d dτ L U α = [mu α Uβ x α + (q/c)u α Aβ x α ]g βγ F αβ U β = U β A β x α U A β α x α d dτ U L α x L α = d[muβ ] dτ (q/c)u α F αβ For α = 1, 2, 3 we obtain the force equation. Note that U α and x α are conjugate variables. 9 Gauge invarience In addition to being Lorentz invarient, the Euler equations are also gauge invarient. A gauge transformation has the form; A A + Λ φ φ Λ t The Lorentz condition expresses this symmetry as; A + (1/c) φ t = 0 In covarient form this is; α A α = 0 Apply the gauge transformation to the Lagrange components; U α A α U α A α + U α α Λ Put this into the Euler equations, this is; 14

15 η η η i 1 i i+1 a a Figure 5: A set of coupled oscillators used to develop the Lagrangian for a continuous system d β Λ β (U α α Λ) = 0 10 Lagrangian for continuious systems Previously we discussed the motion for a few degrees of freedom, ie motion of a few particles acted on by a force. If we wish to include the fields, there will be an infinite number of degrees of freedom as each field point acts as an oscillator. Look at a system of coupled 1-D oscillators connected by springs as shown in Figure 5. The masses oscillate linearly with displacements η i. Assume a spring constant k and mass m to obtain the non-relativistic kinetic and potential energies; T = (1/2) m η i ; V = (1/2) k(η i+1 η i ) 2 The force on particle i is ; F i = [k(η i+1 η i ) k(η i η i 1 )]/a The non-relativistic Lagrangian is ; L = T V = (1/2) [m η 2 i k(η i+1 η i ) 2 For each i the Euler Lagrange equations are; (m/a) η i ka[(η i+1 η i ) (η i η i 1 )]/a 2 = 0 Take the continuium limit as a 0. Let m/a µ and ka Y where Y is Young s modulus. The index i corresponds to the position coordinate x. Change the sum to an integral. The system Lagrangian is then; L = (1/2) dx [µ η 2 Y ( η x )2 ] 15

16 and the Euler equation is then; µ d2 η 2 Y d2 η dx 2 = 0 Note that x is not a generalized coordinate. Each value of x indentifies a generalized coordinate, η(x, t). It is now obvious how to go to a 3-D system, since in a continuum limit the number of oscillators, and the Lagrangian becomes a Lagrangian density; L dx dy dz [Lagrange density] In a relativistic formulation, spatial and time coordinates are treated symmetrically, so the Lagrange density has the form; L = L(η, η x i,, x, y, z) where i runs over all dimensions including time. From the action is; A = dx dy dz [Lagrange density] Again the η i and not the coordinates, are varied. Using the calculus of variations, the Euler Lagrange equations are obtained. In covarient form these are; 4 d L dx µ L µ=1 ( ηj x µ) η j = 0 This result shows that the conjugate variables are the amplitudes, η i, and their derivatives, η j x µ. 11 Lagrangian for the field Now develop the field contributions to a Lagrangian density through symmetry considerations. This development is similar to the way one might explore possible mathematical expressions to express physical processes and insert symmetries. Use the above section to develop a Lagrange density from 4-vector potentials, A α, and their corresponding velocities, β A α. Possible invarient forms for the fields are; F αβ F αβ, F αβ F αβ, J α A α, A α A α. The form, F αβ F αβ, representing the field tensor contracted with its dual, is a pseudo-scalar, and is not by itself, acceptable. Other possible combinations involve the combination of field components resulting in powers of the field to orders higher than quadratic. Thus, first consider interaction terms linear in A. 16

17 Thus assume a Lagrange density of the form; L = (1/16π)F αβ F αβ (1/c)J α A α The Euler equations then reduce to ; (1/4π) β F βα = (1/c)J α These are the inhomogeneous field equations of Maxwell. As expected, they are produced from a linear interaction term in the Lagrangian. 12 Generalization Return to the Lagrangian for a particle interacting with the EM field. We previously found this Lagrangian density could be written as; γl = κ 1 U α U α + κ 2 U α A α The 2 nd term is the interaction term in the last section. The 1 st term is due to the motion of the charge. If we add a constant term to this Lagrangian it could have been combined with the velocity U α U α = c 2, and of course renormalized to c 2. We should combine the Lagrangian for the field energy F αβ F αβ with the above lagrangian to include the field equations as well as those for the charged particle. Now suppose we allow a non-linear interaction by introducing in the Lagrangian the form κ 3 A α A α. A non-limear term means that the field can self-interact, and this causes problems with infinities in the energies. We do not explore this further, but note that the constant κ 3 is equivalent to a mass, and helps to understand how such infinities can be removed through renormalization of the mass. In any event, the Lagrangian density for the field would have the form; L = (1/16π)F αβ F αβ + κ 3 A α A α (1/c)J α A α The Euler equation is; α F αβ + 8πκ 3 A α = (4π/c)J α The additional term 8πκ 3 A α no longer allows a gauge symmetry because the resulting equations depend on the potentials as well as the fields. Recognize that this term adds a mass to the wave equation. This resuts in the covarient form below. 17

18 (1/c 2 ) 2 A α t 2 In the static limit this reduces to; 2 A α + m 2 γ A α = (4π/c)J α 2 A α m 2 γ A α = (4π/c)J α with solution for the scalar potential; φ e mγr r The above is the Proca equation. Of course m γ is in units of inverse distance. This shows that if photon has no mass, the EM potentials will be proportional to 1/r, and is the basic reason for Gauss law and gauge invarience. 18