Electromagnetic Theory. 1 /56. Electromagnetic Theory. Summary: • Maxwell's
equations. • EM Potentials. • Equations of motion of particles in electromagnetic ...
Electromagnetic Theory Summary: • Maxwell’s equations • EM Potentials • Equations of motion of particles in electromagnetic fields • Green’s functions • Lienard-Weichert potentials • Spectral distribution of electromagnetic energy from an arbitrarily moving charge
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1 Maxwell’s equations curlE = –
∂B ∂t
Faraday’s law
1 ∂E curlB = µ 0 J + ----c2 ∂t ρ divE = ----ε0
Field diverges from electric charges
divB = 0
No magnetic monopoles
– 12
ε 0 = 8.854 ×10 1 ε 0 µ 0 = ----c2
Ampere’s law
Farads/metre
µ 0 = 4π × 10 – 7 Henrys/metre
8
c = 2.998 ×10 m/s ≈ 300, 000 km/s
Conservation of charge ∂ρ + div J = 0 ∂t
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Conservation of energy E×B 1 B 2 ∂ 1 2 --- ε 0 E + --- ------ + div -------------- = – ( J ⋅ E ) µ 2 µ 0 ∂t 2 0 Electromagnetic energy density
Poynting - Work done on flux particles by EM field
Poynting Flux This is defined by E×B S = -------------µ0
ε ijk E j B k S i = --------------------µ0
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Conservation of momentum Momentum density
Maxwell’s stress tensor
∂ S i ∂M ij ----- – = – ρE i – ( J × B ) i ∂ t c 2 ∂ x j
Rate of change of momentum due to EM field acting on matter
Electric part
Magnetic part Bi B j B 2
1 2 M ij = ε 0 E i E j – --- E δ ij + ----------- – --------- δ ij µ 2 2µ 0 0 = – Flux of i cpt. of EM momentum in j direction 2 Equations of motion
Charges move under the influence of an electromagnetic field according to the (relativistically correct) equation: dp p×B ------ = q ( E + v × B ) = q E + -------------- dt γm
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Momentum and energy of the particle are given by: p = γmv E = γmc 2
v 2 – 1 / 2 γ = 1 – ---- c 2 E 2 = p2c2 + m2c4
3 Electromagnetic potentials 3.1 Derivation divB = 0 ⇒ B = curl A curlE = – ⇒E+
∂B ∂A ⇒ curl E + = 0 ∂t ∂t
∂A = – gradφ ∂t
⇒ E = – gradφ –
∂A ∂t
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Summary: E = – gradφ –
∂A ∂t
B = curl A
3.2 Potential equations Equation for the vector potential A Substitute into Ampere’s law: 1 ∂ ∂A curl curl A = µ 0 J + ----- – gradφ – ∂t c2 ∂t 2
1 ∂ 1∂ A ----– ∇2 A + ----- gradφ + grad div A = µ 0 J c2 ∂t c2 ∂t 2 Equation for the scalar potential φ Exercise: Show that ∇2φ + div
∂A ρ = – ----∂t ε0
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3.3 Gauge transformations The vector and scalar potentials are not unique. One can see that the same equations are satisfied if one adds certain related terms to φ and A , specifically, the gauge transformations A′ = A – gradψ
φ′ = φ +
∂ψ ∂t
leaves the relationship between E and B and the potentials intact. We therefore have some freedom to specify the potentials. There are a number of gauges which are employed in electromagnetic theory. Coulomb gauge div A = 0 Lorentz gauge 1 ∂φ ----- + div A = 0 c2 ∂t 2
1∂ A ⇒ ----– ∇2 A = µ 0 J 2 2 c ∂t
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Temporal gauge φ = 0 ⇒
∂ ρ div A = – ----∂t ε0
2
1∂ A ----+ curl curl A = µ 0 J 2 2 c ∂t The temporal gauge is the one most used when Fourier transforming the electromagnetic equations. For other applications, the Lorentz gauge is often used. 4 Electromagnetic waves
For waves in free space, we take E = E 0 exp [ i ( k ⋅ x – ωt ) ] B = B 0 exp [ i ( k ⋅ x – ωt ) ]
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and substitute into the free-space form of Maxwell’s equations, viz., curlE = –
∂B ∂t
divE = 0
1 ∂E curlB = ----c2 ∂t div B = 0
This gives: k × E0 ik × E 0 = iωB 0 ⇒ B 0 = ---------------ω 1 ω ik × B 0 = – ----- iωE 0 ⇒ k × B 0 = – ----- E 0 c2 c2 ik ⋅ E 0 = 0 ⇒ k ⋅ E 0 = 0 ik ⋅ B 0 = 0 ⇒ k ⋅ B 0 = 0 We take the cross-product with k of the equation for B 0 : ( k × E0 ) ( k ⋅ E 0 )k – k 2 E 0 ω k × B 0 = k × --------------------- = ----------------------------------------- = – ----- E 0 ω ω c2
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and since k ⋅ E 0 = 0 ω -----– k 2 E 0 = 0 ⇒ ω = ± ck c2 2
the well-known dispersion equation for electromagnetic waves in free space. The - sign relates to waves travelling in the opposite direction, i.e. E = E 0 exp [ i ( k ⋅ x + ωt ) ] We restrict ourselves here to the positive sign. The magnetic field is given by κ × E0 k × E0 1 k B 0 = ---------------- = --- --- × E 0 = --------------- ck ω c The Poynting flux is given by E×B S = -------------µ0 and we now take the real components of E and B : E = E 0 cos ( k ⋅ x – ωt )
κ × E0 B = ---------------- cos ( k ⋅ x – ωt ) c
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where E 0 is now real, then E0 × ( κ × E0 ) S = ---------------------------------- cos2 ( k ⋅ x – ωt ) µ0 c = cε 0 ( E 02 κ – ( κ ⋅ E 0 )E 0 ) cos2 ( k ⋅ x – ωt ) = cε 0 E 02 κ cos2 ( k ⋅ x – ωt ) The average of cos2 ( k ⋅ x – ωt ) over a period ( T = 2π ⁄ ω ) is 1 ⁄ 2 so that the time-averaged value of the Poynting flux is given by: cε 0 ------- E 02 κ 〈 S〉 = 2 5 Equations of motion of particles in a uniform magnetic field
An important special case of particle motion in electromagnetic fields occurs for E = 0 and B = constant . This is the basic configuration for the calculation of cyclotron and synchrotron emission. In this case the motion of a relativistic particle is given by: q dp ------ = q ( v × B ) = ------- ( p × B ) γm dt
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Conservation of energy There are a number of constants of the motion. First, the energy: dp p ⋅ ------ = 0 dt and since E 2 = p2c2 + m2c4 then dp dp dE E ------- = c 2 p ------ = c 2 p ⋅ ------ = 0 here. dt dt dt There for E = γmc 2 is conserved and γ is constant - our first constant of motion. Parallel component of momentum The component of momentum along the direction of B is also conserved: q B B dp B d ----- p ⋅ ---- = ------ ⋅ ---- = ------- ---- ⋅ ( p × B ) = 0 γm B B dt B dt ⇒ p || = γmv || is conserved
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where p || is the component of momentum parallel to the magnetic field. We write the total magnitude of the velocity v = cβ and since γ is constant, so is v and we put v || = v cos α where α is the pitch angle of the motion, which we ultimately show is a helix. Perpendicular components Take the z –axis along the direction of the field, then the equations of motion are: e1 e2 e3 q dp ------ = ------- p p p γm x y || dt 0 0 B
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In component form: d px q --------- = ------- p y B = ηΩ B p y γm dt d py q --------- = – ------- p x B = – ηΩ B p x γm dt qB Ω B = ---------- = Gyrofrequency γm q η = ----- = Sign of charge q A quick way of solving these equations is to take the second plus i times the first: d ----- ( p x + i p y ) = – iηΩ B ( p x + i p y ) dt This has the solution p x + i p y = A exp ( iφ 0 ) exp [ – iηΩ B t ] ⇒ p x = A cos ( ηΩ B t + φ 0 )
p y = – A sin ( ηΩ B t + φ 0 )
The parameter φ 0 is an arbitrary phase.
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Positively charged particles: p x = A cos ( Ω B t + φ 0 )
p y = – A sin ( Ω B t + φ 0 )
Negatively charged particles (in particular, electrons): p x = A cos ( Ω B t + φ 0 )
p y = A sin ( Ω B t + φ 0 )
We have another constant of the motion: p x2 + p y2 = A 2 = p 2 sin2 α = p ⊥2 where p ⊥ is the component of momentum perpendicular to the magnetic field. Velocity The velocity components are given by: vx
px
sin α cos ( Ω B t + φ 0 )
1 v y = ------- p y = cβ – η sin α sin ( Ω t + φ ) B 0 γm vz pz cos α
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Position Integrate the above velocity components: cβ sin α ------------------ sin ( Ω B t + φ 0 ) ΩB
x0 x + y0 cβ sin α y = η ------------------ cos ( Ω B t + φ 0 ) ΩB z z0 cβt cos α This represents motion in a helix with cβ sin α Gyroradius = r G = -----------------ΩB The motion is clockwise for η > 0 and anticlockwise for η < 0 .
