Summary: Electromagnetic Theory 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 Electromagnetic Theory 1 /56
1 Maxwell s equations curle curlb B t µ 0 J ρ dive ---- ε 0 divb 0 + 1 E ---- t c 2 Faraday s law Ampere s law Field diverges from electric charges No magnetic monopoles Conservation of charge 12 ε 0 8.854 10 Farads/metre µ 0 4π 10 7 Henrys/metre 1 8 ε 0 µ 0 ---- c 2.998 10 m/s 300, 000 km/s c 2 ρ t + divj 0 Electromagnetic Theory 2 /56
Conservation of energy Poynting Flux This is defined by 1 --ε t 2 0 E 2 1 + -- ----- B2 2 µ + div ------------- E B ( J E) 0 µ 0 Electromagnetic energy density Poynting flux - Work done on particles by EM field S E B ε ijk E j B k ------------- S i -------------------- µ 0 µ 0 Electromagnetic Theory 3 /56
Conservation of momentum 2 Equations of motion Momentum density t S i ---- c 2 M ij Maxwell s stress tensor 1 M ij ε 0 E i E j --E 2 2 δ ij ρe x i ( J B) i j Electric part Rate of change of momentum due to EM field acting on matter Magnetic part B i B j + ---------- Charges move under the influence of an electromagnetic field according to the (relativistically correct) equation: µ 0 B 2 --------δ 2µ ij 0 Flux of i cpt. of EM momentum in j direction dp ----- q( E + v B) q E + ------------- p B dt γm Electromagnetic Theory 4 /56
Momentum and energy of the particle are given by: p v γmv γ 1 ---- 2 12 / c 2 E γmc 2 E 2 p 2 c 2 + m 2 c 4 3 Electromagnetic potentials 3.1 Derivation E divb 0 B curle A + t E curl A B A curl E + 0 t t gradφ gradφ A t Electromagnetic Theory 5 /56
Summary: 3.2 Potential equations E A gradφ B curl A t Equation for the vector potential Substitute into Ampere s law: A curl curl A µ 0 J + 1 ---- t c 2 gradφ A t 2 1 A ---- c 2 t 2 2 A 1 + ---- gradφ + grad div A µ0 J t c 2 Equation for the scalar potential Exercise: Show that φ 2 φ + A div t ρ ---- ε 0 Electromagnetic Theory 6 /56
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 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 Lorentz gauge A A gradψ φ φ 1 φ ---- t c 2 1 A ---- t 2 c 2 div A 0 2 + div A 0 2 A µ 0 J ψ + t Electromagnetic Theory 7 /56
Temporal gauge div A t φ 0 ρ ---- ε 0 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 1 A ---- t 2 c 2 2 + curl curl A µ0 J For waves in free space, we take E E 0 exp[ i( k x ωt) ] B B 0 exp[ i( k x ωt) ] Electromagnetic Theory 8 /56
and substitute into the free-space form of Maxwell s equations, viz., This gives: B 1 E curle curlb ---- t c 2 t dive 0 div B 0 ik E 0 iωb 0 B 0 ik B 0 k E 0 --------------- ω 1 ----iωe 0 k B 0 c 2 ik E 0 0 k E 0 0 ik B 0 0 k B 0 0 ω ----E c 2 0 We take the cross-product with k of the equation for B 0 : ( k E 0 ) ( k E k B 0 k -------------------- 0 )k k 2 E 0 ---------------------------------------- ω ω ω ----E c 2 0 Electromagnetic Theory 9 /56
and since k E 0 0 the well-known dispersion equation for electromagnetic waves in free space. The - sign relates to waves travelling in the opposite direction, i.e. We restrict ourselves here to the positive sign. The magnetic field is given by The Poynting flux is given by ω 2 c 2 ------ k 2 E0 0 ω ± ck E E 0 exp[ i( k x + ωt) ] k E 0 1 B 0 --------------- ω -- k c -- k E 0 S E B ------------- µ 0 κ E 0 --------------- c and we now take the real components of E and B: E E 0 cos( k x ωt) B κ E 0 --------------- cos( k x ωt) c Electromagnetic Theory 10 /56
E 0 where is now real, then S E 0 ( κ E 0 ) --------------------------------- cos k x ωt µ 0 c 2 ( ) cε 0 ( E 2 0 κ ( κ E 0 )E 0 ) cos 2 ( k x ωt) cε 0 E 2 0 κ cos 2 ( k x ωt) The average of cos 2 ( 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: S cε 0 -------E 2 2 0 κ 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: dp ----- q( v B) dt q ------ ( p B) γm Electromagnetic Theory 11 /56
Conservation of energy There are a number of constants of the motion. First, the energy: p dp ----- 0 dt and since then E 2 p 2 c 2 + m 2 c 4 There for E γmc 2 is conserved and γ is constant - our first constant of motion. Parallel component of momentum E de ------ c dt 2 p dp ----- c dt 2 p dp ----- 0 here. dt The component of momentum along the direction of B is also conserved: d dt ---- p --- B B dp B q ----- --- ------ B dt B γm --- ( B p B ) 0 p γmv is conserved Electromagnetic Theory 12 /56
p where is the component of momentum parallel to the magnetic field. We write the total magnitude of the velocity and since γ is constant, so is v and we put v cβ v vcosα 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: dp ----- dt q ------ γm e 1 e 2 e 3 p x p y p 0 0 B Electromagnetic Theory 13 /56
In component form: A quick way of solving these equations is to take the second plus i times the first: This has the solution φ 0 dp x -------- dt dp y -------- dt Ω B The parameter is an arbitrary phase. η q ------ p γm y B ηω B p y q ------ p γm x B ηω B p x qb --------- Gyrofrequency γm q ----- Sign of charge q d ---- ( p dt x + ip y ) iηω B ( p x + ip y ) p x + ip y Aexp( iφ 0 ) exp[ iηω B t] p x Acos( ηω B t + φ 0 ) p y Asin( ηω B t + φ 0 ) Electromagnetic Theory 14 /56
Positively charged particles: p x Acos( Ω B t + φ 0 ) p y Asin( Ω B t + φ 0 ) Negatively charged particles (in particular, electrons): p x Acos( Ω B t + φ 0 ) p y Asin( Ω B t + φ 0 ) We have another constant of the motion: p2 x + p2 y A 2 p 2 sin 2 α p2 where is the component of momentum perpendicular to the magnetic field. p Velocity The velocity components are given by: v x v y v z p x 1 ------ p γm y p z cβ sinαcos( Ω B t + φ 0 ) ηsinαsin( Ω B t + φ 0 ) cosα Electromagnetic Theory 15 /56
Position Integrate the above velocity components: x y z cβsinα ----------------- sin( Ω B t + φ 0 ) Ω B η----------------- cβsinα cos( Ω B t + φ 0 ) Ω B cβt cosα + x 0 y 0 z 0 This represents motion in a helix with cβ α Gyroradius r sin G ----------------- Ω B The motion is clockwise for η > 0 and anticlockwise for η< 0. Electromagnetic Theory 16 /56
y r G sin( Ω B t) y r G cos( Ω B t) Ω B t η > 0 η < 0 x r G cos( Ω B t) Ω B t x r G sin( Ω B t) In vector form, we write: x x 0 + r G sin( Ω B t + φ 0 ) ηcos( Ω B t + φ 0 ) 0 + cβt cosα 001 The parameter represents the location of the guiding centre of the motion. x 0 Electromagnetic Theory 17 /56
6 Green s functions Green s functions are widely used in electromagnetic and other field theories. Qualitatively, the idea behind Green s functions is that they provide the solution for a given differential equation corresponding to a point source. A solution corresponding to a given source distribution is then constructed by adding up a number of point sources, i.e. by integration of the point source response over the entire distribution. 6.1 Green s function for Poisson s equation A good example of the use of Green s functions comes from Poisson s equation, which appears in electrostatics and gravitational potential theory. For electrostatics: ρ e where is the electric charge density. In gravitational potential theory: 2 φ( x) ρ e ( x) ------------- ε 0 2 φ( x) 4πGρ m ( x) where ρ m ( x) is the mass density. Electromagnetic Theory 18 /56
The Green s function for the electrostatic case is prescribed by: 2 δ( x x ) G( x, x ) --------------------- where δ( x x ) is the three dimensional delta function. When there are no boundaries, this equation has the solution G( x, x ) G( x x ) G( x) 1 -------------- 4πε 0 r The general solution of the electrostatic Poisson equation is then For completeness, the solution of the potential for a gravitating mass distribution is: ε 0 φ( x) G( x x )ρ e ( x )d 3 x space 1 ----------- 4πε 0 φ( x) G space space ρ e ( x ) ----------------d x x 3 x ρ m ( x ) ----------------d x x 3 x Electromagnetic Theory 19 /56
6.2 Green s function for the wave equation In the Lorentz gauge the equation for the vector potential is: 1 A ---- t 2 and the equation for the electrostatic (scalar) potential is c 2 2 2 A µ 0 J 1 ---- φ( t, x) c 2 2 φ( t, x) t 2 These equations are both examples of the wave equation 2 ρ ---- ε 0 2 1 ---- ψ( t, x) c 2 2 ψ( t, x) Stx (, ) t 2 When time is involved a point source consists of a source which is concentrated at a point for an instant of time, i.e. S( x, t) Aδ( t t )δ 3 ( x x ) where A is the strength of the source, corresponds to a source at the point x x which is switched on at t t. Electromagnetic Theory 20 /56
In the case of no boundaries, the Green s function for the wave equation satisfies: 1 ---- c 2 t 2 2 Gt ( t, x x ) δ( t t )δ 3 ( x x ) The relevant solution (the retarded Green s function) is: 2 Gtx (, ) 1 4πr --------δ t r - c so that Gt ( t, x x ) ----------------------δ 1 4π x x t t x x --------------- c The significance of the delta function in this expression is that a point source at ( t, x ) will only contribute to the field at the point ( t, x) when x x t t + --------------- c Electromagnetic Theory 21 /56
x x i.e. at a later time corresponding to the finite travel time --------------- of a pulse from the point x. Equivalently, a disturbance which arrives at the point t, x had to have been emitted at a c time t t x x The time t --------------- is known as the retarded time. c The general solution of the wave equation is x x --------------- c ψ( t, x) dt Gt ( t, x x )St (, x )d 3 x space 1 ----- dt 4π space ------------------- St (, x ) x x δ t x x --------------- d c 3 x Electromagnetic Theory 22 /56
6.3 The vector and scalar potential Using the above Green s function, the vector and scalar potential for an arbitrary charge and current distribution are: A( t, x) φ( t, x) µ 0 ----- dt 4π 1 ----------- dt 4πε 0 space space δ ---------------------------------------------------------------------d ( t t x x c)j( t, x ) x x 3x δ( t t x x c)ρ e ( t, x ) -----------------------------------------------------------------------d x x 3 x 7 Radiation from a moving charge the Lienard-Weichert potentials 7.1 Deduction from the potential of an arbitrary charge distribution The current and charge distributions for a moving charge are: ρ( t, x) qδ 3 ( x X() t ) J( t, x) qvδ 3 ( x X() t ) where v is the velocity of the charge, q, and X() t is the position of the charge at time t.the charge q is the relevant parameter in front of the delta function since ρ ( t, x )d3x q δ 3 ( x X() t )d 3 x space space q Electromagnetic Theory 23 /56
Also, the velocity of the charge so that v() t dx() t ------------- Ẋ() t dt ρ( t, x) qδ 3 ( x X() t ) J( t, x) qx () t δ 3 ( x X () t ) With the current and charge expressed in terms of spatial delta functions it is best to do the space integration first. We have space δ( t t x x c)ρ e ( t, x ) -----------------------------------------------------------------------d x x 3 x space ------------------------------------------------------------------------------------ δ( t t x x c)δ 3 ( x X( t )) x x d3x qδ x X( t ) t t ------------------------ c ------------------------------------------------------ x X( t ) Electromagnetic Theory 24 /56
One important consequence of the motion of the charge is that the delta function resulting from the space integration is now a more complicated function of t, because it depends directly upon t and indirectly though the dependence on X( t ). The delta-function will now only contribute to the time integral when t t x X( t ) ------------------------ c The retarded time is now an implicit function of ( t, x), through X( t ). However, the interpretation of t is still the same, it represents the time at which a pulse leaves the source point, X( t ) to arrive at the field point ( t, x). We can now complete the solution for φ( t, x) by performing the integration over time: φ( t, x) 1 ----------- 4πε 0 qδ x X( t ) t t ------------------------ ------------------------------------------------------ c d x X( t ) t This equation is not as easy to integrate as might appear because of the complicated dependence of the delta-function on t. Electromagnetic Theory 25 /56
7.2 Aside on the properties of the delta function The following lemma is required. We define the delta-function by f()δt t ( a) dt f( a) Some care is required in calculating f()δgt t ( a) dt. Consider g 1 a where ( ) is the value of t satisfying g() t a. f()δgt t ( a) dt f()δgt t ( a) ----- dt dg dg ---------δ f() t ( ġ() t g a) dg f( g 1 ( a) ) ------------------------ ġ( g 1 ( a) ) Electromagnetic Theory 26 /56
7.3 Derivation of the Lienard-Wierchert potentials In the above integral we have the delta function δ( t t x X( t ) c) δ( t + x X( t ) c t) so that Differentiating this with respect to t : To do the partial derivative on the right, express x X( t ) in tensor notation: Now, gt ( ) t + x X( t ) c t dg( t ) x X( t ) -------------- ġ( t ) 1+ ------------------------ dt c 2 x X( t ) 2 x i x i 2x i X i ( t ) + X i ( t )X i ( t ) x X( t ) 2 2 x X( t ) Differentiating the tensor expression for x X( t ) gives: 2 x X ( t ) t x X( t ) 2 2x i X i ( t ) + 2X i ( t )X i ( t ) 2X i ( t )( x i X i ( t )) Electromagnetic Theory 27 /56
Hence, 2 x X( t ) The derivative of gt ( ) is therefore: x X ( t ) t 2X i ( t )( x i X i ( t )) X i ( t )( x i X i ( t )) x X( t ) ------------------------------------------- x X( t ) Ẋ( t ) ( x X( t )) ------------------------------------------- x X( t ) x X( t ) Ẋ( t ) ( x X( t )) c ġ( t ) 1 + ------------------------ 1 -------------------------------------------------- c x X( t ) x X( t ) Ẋ( t ) ( x X( t )) c --------------------------------------------------------------------------------- x X( t ) Hence the quantity 1 ( ġ( t )) which appears in the value of the integral is 1 x X( t ) ---------- --------------------------------------------------------------------------------- ġ( t ) x X( t ) Ẋ( t ) ( x X( t )) c Electromagnetic Theory 28 /56
Thus, our integral for the scalar potential: φ( t, x) where, it needs to be understood that the value of t involved in this solution satisfies, the equation for retarded time: We also often use this equation in the form: qδ x X( t ) t t ------------------------ 1 c ----------- 4πε 0 ------------------------------------------------------ d x X( t ) t q x X( t ) ----------- --------------------------------------------------------------------------- 4πε 0 Ẋ( t ) ( x X( t )) x X( t ) ------------------------------------------- c q 1 ----------- --------------------------------------------------------------------------- 4πε 0 Ẋ( t ) ( x X( t )) x X( t ) ------------------------------------------- c t t x X( t ) ------------------------ c x X( t ) t + ------------------------ t c ------------------------ 1 x X( t ) Electromagnetic Theory 29 /56
7.4 Nomenclature and symbols We define the retarded position vector: and the retarded distance The unit vector in the direction of the retarded position vector is: The relativistic β of the particle is r x X( t ) r x X( t ) n ( t ) r --- r. β( t ) Ẋ( t ) ------------ c Electromagnetic Theory 30 /56
7.5 Scalar potential In terms of these quantities, therefore, the scalar potential is: φ( t, x) q 1 ----------- --------------------------------------------------------------------------- 4πε 0 Ẋ( t ) ( x X( t )) x X( t ) ------------------------------------------- c q 1 ----------- ------------------------------ 4πε 0 r β( t ) r q ---------------- 1 ---------------------------------- 4πε 0 r [ 1 β( t ) n ] This potential shows a Coulomb-like factor times a factor ( 1 β( t ) n ) 1 which becomes extremely important in the case of relativistic motion. Electromagnetic Theory 31 /56
7.6 Vector potential The evaluation of the integral for the vector potential proceeds in an analogous way. The major difference is the velocity Ẋ( t ) in the numerator. Hence we can write A( t, x) µ 0 q Ẋ( t ) -------- --------------------------------------------------------------------------- 4π Ẋ( t ) ( x X( t )) x X( t ) ------------------------------------------- c µ 0 q Ẋ( t ) ---------- ---------------------------------- 4πr [ 1 β( t ) n ] q Ẋ( t ) A( t, x) µ 0 ε 0 ---------------- ---------------------------------- 4πε 0 r [ 1 β( t ) n ] c 1 β( t )φ( t, x) 1 q Ẋ( t ) ---- ---------------- ---------------------------------- 4πε 0 r [ 1 β( t ) n ] This is useful when for expressing the magnetic field in terms of the electric field. c 2 Electromagnetic Theory 32 /56
7.7 Determination of the electromagnetic field from the Lienard-Wierchert potentials To determine the electric and magnetic fields we need to determine The potentials depend directly upon x and indirectly upon x, t through the dependence upon t. Hence we need to work out the derivatives of t with respect to both t and x. Expression for Since, ----- t E B gradφ curl A A t x X( t ) t + ------------------------ t c Electromagnetic Theory 33 /56
We can determine ----- by differentiation of this implicit equation. t ----- t 1 + -- ( x X( t )) 1 c t ( x X( t )) ----- -------------------------- Ẋ( t ) ------------ + ----- 1 t x X( t ) c t Ẋ( t ) ( x X( t )) 1 ------------------------------------------- c x X( t ) ----- 1 t Solving for t : ----- t ----- t x X( t ) -------------------------------------------------------------------------------------- x X( t ) ( Ẋ( t ) c) ( x X( t )) r ------------------------------------- r Ẋ( t ) r c 1 ----------------------------- 1 β( t ) n Electromagnetic Theory 34 /56
Expression for ------ t x i Again differentiate the implicit function for t : ------ x i ------ x i ( x i X i ( t )) ( x j X j ( t ))X j ( t ) c + ---------------------------- ---------------------------------------------------- ------ 0 c x X( t ) x X( t ) x i ------ x i ------ 1 x i β j ( x j X j ( t )) ------------------------------------ x X( t ) x i x X( t ) β( t ) ( x X( t )) -------------------------------------------------------------------------- x X( t ) ( x i X i ( t )) c -------------------------------------------------------------------------- x X( t ) β( t ) ( x X( t )) X i + --------------------------- 0 c x X( t ) x i X i --------------------------- c x X( t ) i.e. ------ x i or t x i ------------------------------------- r Ẋ( t ) r c c 1 n ----------------------------- 1 β( t ) n c 1 n i ----------------------------- 1 β( t ) n Electromagnetic Theory 35 /56
The potentials include explicit dependencies upon the spatial coordinates of the field point and implicit dependencies on ( t, x) via the dependence on t The derivatives of the potentials can be determined from: φ x i t φ x i t A ------- i A i t ------- ----- t xi t + φ ------ x i x i In dyadic form: ε ijk A i ------- x j A k -------- x j t t A i ------- x j ε ijk t A k -------- x j A i + ------- ------- x j t ε ijk x ------- + -------- Ak j φ t φ φ t + t A ------ A x t ------ ----- t x t x curl A t curl A t + t A ------ x Electromagnetic Theory 36 /56
Electric field The calculation of the electric field goes as follows. Some qualifiers on the partial derivatives are omitted since they should be fairly obvious E A φ ------ φ t t φ [ c 1 β( t )φ] t ------------------------------ ----- t φ φ t t + β -- ----- t c t φ --β t c ( ) ----- t t The terms t + β -- ----- t c t 1 ( n β) -- --------------------- c 1 β n ----- t 1 -------------------------- ( 1 β n ) Other useful formulae to derive beforehand are: r ------ c 1 β r ------ c 1 β n 1 1 In differentiating ------------------------------ it is best to express it in the form ---------------------. r ( 1 β n ) r β r Electromagnetic Theory 37 /56
With a little bit of algebra, it can be shown that q n β φ t ------------------ 4πε 0 r 2 ---------------------------- ( 1 β n ) 2 φ qc [ ------------------ β n β2 + c 1 r β n ] 4πε 0 r 2 ------------------------------------------------------------- ( 1 β n) 2 Combining all terms: E q [( n β) ( 1 β ------------------ 2 + c 1 r β n ) c 1r β ( 1 β n )] 4πε 0 r 2 ------------------------------------------------------------------------------------------------------------------------------- ( 1 β n ) 3 The immediate point to note here is that many of the terms in this expression decrease as. However, the terms proportional to the acceleration only decrease as. These are the radiation terms: r 1 r 2 E rad q [( n β)β n β ( 1 β n )] ----------------- ------------------------------------------------------------------------- 4πcε 0 r ( 1 β n ) 3 Electromagnetic Theory 38 /56
Magnetic field We can evaluate the magnetic field without going through more tedious algebra. The magnetic field is given by: where B curl A curl A t + t curl c 1 φβ ( ) t + t A ------ A ------ Now the first term is given by: t c 1 n --------------------- 1 β n c 1 n ----- t curl ( c 1 φβ) c 1 φ t β and we know from calculating the electric field that q n β φ t ------------------ 4πε 0 r 2 ---------------------------- ( 1 β n ) 2 ξ( n β) Electromagnetic Theory 39 /56
Therefore, c 1 φ t β c 1 ξ( n β) β c 1 ξ( n β) ( β n + n) c 1 ξ( n β) n c 1 φ t n Hence, we can write the magnetic field as: B c 1 φ t n c 1 A n ------ ----- c t 1n φ t A ------ t Compare the term in brackets with E φ t + φ x t A ------ t Since t n, then B c 1 ( n E) This equation holds for both radiative and non-radiative terms. Electromagnetic Theory 40 /56
Poynting flux The Poynting flux is given by: E B E ( n E) S ------------- ------------------------------ cε µ 0 cµ 0 [ E 2 n ( E n )E] 0 We restrict attention to the radiative terms in which E rad r 1 For the radiative terms, E rad n so that the Poynting flux, q [( 1 β n )( n β ) n β ( 1 β n )] ----------------- ----------------------------------------------------------------------------------------------- 4πcε 0 r ( 1 β n ) 3 0 S cε 0 E 2 n This can be understood in terms of equal amounts of electric and magnetic energy density (( ε 0 2)E 2 ) moving at the speed of light in the direction of n. This is a very important expression when it comes to calculating the spectrum of radiation emitted by an accelerating charge. Electromagnetic Theory 41 /56
8 Radiation from relativistically moving charges Trajectory of particle v 2 γ Illustration of the beaming of radiation from a relativistically moving particle. Note the factor ( 1 β n ) 3 in the expression for the electric field. When 1 β n 0 the contribution to the electric field is large; this occurs when β n 1, i.e. when the angle between the velocity and the unit vector from the retarded point to the field point is approximately zero. We can quantify this as follows: Let θ be the angle between β( t ) and n, then 1 β n 1 β θ 1 1 1 -------- 1 cos 1 2γ 2 --θ 2 2 1 1 1 -------- θ2 2γ 2 ----- 2 1 -------- 2γ 2 θ + ----- 2 2 1 -------- 2γ 2 ( 1 + γ 2 θ 2 ) Electromagnetic Theory 42 /56
So you can see that the minimum value of 1 β n is 1 ( 2γ 2 ) and that the value of this quantity only remains near this for θ 1 γ. This means that the radiation from a moving charge is beamed into a narrow cone of angular extent 1 γ. This is particularly important in the case of synchrotron radiation for which γ 10 4 (and higher) is often the case. 9 The spectrum of a moving charge 9.1 Fourier representation of the field Consider the transverse electric field, E() t, resulting from a moving charge, at a point in space and represent it in the form: E() t E 1 ()e t 1 + E 2 ()e t 2 where e 1 and e 2 are appropriate axes in the plane of the wave. (Note that in general we are not dealing with a monochromatic wave, here.) The Fourier transforms of the electric components are: E α ( ω) e iωt E α ()t t d E α () t 1 ----- e 2π iωt E α ( ω) dω Electromagnetic Theory 43 /56
The condition that E α () t be real is that E α ( ω) E * α ( ω) Note: We do not use a different symbol for the Fourier transform, e.g. is indicated by its argument. Ẽ α ( ω). The transformed variable 9.2 Spectral power in a pulse Outline of the following calculation E α () t t Diagrammatic representation of a pulse of radiation with a duration t. t Consider a pulse of radiation Calculate total energy per unit area in the radiation. Use Fourier transform theory to calculate the spectral distribution of energy. Show this can be used to calculate the spectral power of the radiation. The energy per unit time per unit area of a pulse of radiation is given by: dw ----------- Poynting Flux ( cε dtda 0 )E 2 () t ( cε 0 )[ E 2 1 () t + E 2 2 () t ] Electromagnetic Theory 44 /56
where E 1 and E 2 are the components of the electric field wrt (so far arbitrary) unit vectors e 1 and e 2 in the plane of the wave. The total energy per unit area in the From Parseval s theorem, α component of the pulse is The integral from to can be converted into an integral from 0 to using the reality condition. For the negative frequency components, we have so that E α dw αα -------------- ( cε da 0 ) E 2 α ()t t d E 2 α ()t t d 1 ----- E 2π α ( ω) 2 dω ( ω) E * α ( ω) E * α ( ω) E α ( ω) E α ( ω) 2 E 2 α ()t t d 1 -- E π α ( ω) 2 dω 0 Electromagnetic Theory 45 /56
The total energy per unit area in the pulse, associated with the α component, is dw αα -------------- cε da 0 E 2 α ()t t d (The reason for the αα subscript is evident below.) cε 0 ------- E π α ( ω) 2 dω 0 [Note that there is a difference here from the Poynting flux for a pure monochromatic plane wave in which we pick up a factor of 1 2. That factor results from the time integration of cos 2 ωt which comes from, in effect, E α ( ω) 2 dω. This factor, of course, is not evaluated here since the pulse has an arbitrary spectrum.] 0 We identify the spectral components of the contributors to the Poynting flux by: dw αα -------------- dωda cε 0 ------- E π α ( ω) 2 dw αα The quantity -------------- represents the energy per unit area per unit circular frequency in the entire pulse, dωda i.e. we have accomplished our aim and determined the spectrum of the pulse. Electromagnetic Theory 46 /56
We can use this expression to evaluate the power associated with the pulse. Suppose the pulse repeats with period T, then we define the power associated with component α by: This is equivalent to integrating the pulse over, say several periods and then dividing by the length of time involved. 9.3 Emissivity dw αα ------------------- dadωdt 1 --- -------------- dw T dadω cε 0 ------- E πt α ( ω) 2 r dω da r 2 dω Variables used to define the emissivity in terms of emitted power. Region in which particle moves during pulse Consider the surface da to be located a long distance from the distance over which the particle moves when emitting the pulse of radiation. Then da r 2 dω and dw αα ------------------- dadωdt 1 ---- dw αα -------------------- dωdωdt r 2 dw αα -------------------- dωdωdt r 2 dw αα ------------------- dadωdt Electromagnetic Theory 47 /56
The quantity dw αα -------------------- dωdωdt cε 0 r 2 ------------- E πt α ( ω) 2 cε 0 r 2 ------------- E πt α ( ω)e * α ( ω) (Summation not implied) is the emissivity corresponding to the component of the pulse. e α 9.4 Relationship to the Stokes parameters We generalise our earlier definition of the Stokes parameters for a plane wave to the following: cε 0 I ω ------- [ E πt 1 ( ω)e * 1 ( ω) + E 2 ( ω)e * 2 ( ω) ] cε 0 Q ω ------- [ E πt 1 ( ω)e * 1 ( ω) E 2 ( ω)e * 2 ( ω) ] cε 0 U ω ------- [ E πt * 1 ( ω)e 2 ( ω) + E 1 ( ω)e * 2 ( ω) ] 1 V ω -- cε 0 ------- [ E i πt * 1 ( ω)e 2 ( ω) E 1 ( ω)e * 2 ( ω) ] The definition of is equivalent to the definition of specific intensity in the Radiation Field chapter. I ω Also note the appearance of circular frequency resulting from the use of the Fourier transform. Electromagnetic Theory 48 /56
We define a polarisation tensor by: 1 I αβ, ω -- I ω + Q ω U ω iv ω 2 U ω + iv ω I ω Q ω cε 0 -------E πt α ( ω)e * β ( ω) We have calculated above the emissivities, dw αα -------------------- dωdωdt cε 0 r 2 ------------- E πt α ( ω)e * α ( ω) (Summation not implied) corresponding to E α E * α. More generally, we define: dw αβ -------------------- dωdωdt and these are the emissivities related to the components of the polarisation tensor. cε 0 -------r πt 2 E α ( ω)e * β ( ω) I αβ Electromagnetic Theory 49 /56
In general, therefore, we have dw 11 -------------------- Emissivity for 1 dωdωdt 2 -- ( I ω + Q ω ) dw 22 -------------------- Emissivity for 1 dωdωdt 2 -- ( I ω Q ω ) dw 12 -------------------- Emissivity for 1 dωdωdt 2 -- ( U ω iv ω ) dw 21 -------------------- dωdωdt dw* 12 -------------------- Emissivity for 1 dωdωdt 2 -- ( U ω + iv ω ) Consistent with what we have derived above, the total emissivity is and the emissivity into the Stokes Q is ε ω I ε ω Q dw 11 -------------------- dωdωdt dw 11 -------------------- dωdωdt dw 22 + -------------------- dωdωdt dw 22 -------------------- dωdωdt Electromagnetic Theory 50 /56
Also, for Stokes U and V: ε ω U ε ω V dw 12 -------------------- dωdωdt dw 12 i -------------------- dωdωdt dw* 12 + -------------------- dωdωdt dw* 12 -------------------- dωdωdt Note the factor of r 2 in the expression for dw αβ dωdωdt. In the expression for the E vector of the radiation field E q [( n β) ( n β ) β ( 1 β n )] ----------------- ------------------------------------------------------------------------------ 4πcε 0 r ( 1 β n ) 3 E 1 r. Hence re and consequently r 2 E α E * β are independent of r, consistent with the above expressions for emissivity. The emissivity is determined by the solution for the electric field. Electromagnetic Theory 51 /56
10 Fourier transform of the Lienard-Wierchert radiation field The emissivities for the Stokes parameters obviously depend upon the Fourier transform of re() t q n [ n β -------------- ( ) β ] ------------------------------------------------ 4πcε 0 ( 1 β n ) 3 where the prime means evaluation at the retarded time t given by r t t --- r x X( t ) c The Fourier transform involves an integration wrt t. We transform this to an integral over t as follows: t 1 dt -----dt --------------dt ( t t 1 β n )dt using the results we derived earlier for differentiation of the retarded time. Hence, re( ω) q -------------- 4πcε 0 q -------------- 4πcε 0 n [( n β ) β ] ------------------------------------------------e iωt ( 1 β n ) 3 ( 1 β n ) dt n [( n β ) β ] ------------------------------------------------e iωt ( 1 β n ) 2 dt Electromagnetic Theory 52 /56
The next part is e iωt exp iω t r + --- c Since then we expand r to first order in X. Thus, r x j X j ( t ) r x X( t ) x when x» X( t ) r ------- X i ( x i X i ( t )) -------------------------------- --- at X r r i 0 r r + r ------- X i ( t ) r ---X X i X j 0 r i ( t ) r n i X i r n X( t ) Note that it is the unit vector n r - which enters here, rather than the retarded unit vector n r Hence, exp( iωt) iω r n X( t ) t + - -------------------- n X( t ) exp exp iω t -------------------- c c c exp iωr -------- c x i x i Electromagnetic Theory 53 /56
iωr The factor exp -------- is common to all Fourier transforms re and when one multiplies by the complex conjugate it gives unity. This also shows why we expand the argument of the exponential to first c α ( ω) order in X( t ) since the leading term is eventually unimportant. The remaining term to receive attention in the Fourier Transform is n [( n β ) β ] ------------------------------------------------ ( 1 β n ) 2 We first show that we can replace n by n by also expanding in powers of X i ( t ). n i x i X i ( t ) ---------------------------- x j X j ( t ) x i --- r + x i X i ---------------------------- X j x j X j ( t ) X j 0 X j x i --- + δ r ij ( x i X i ) x j X j x j X j ( t ) 3 ( ) + ------------------------------------------ X j 0 X j n i + δ ij x i x j + --------- r 2 X j ----- r Electromagnetic Theory 54 /56
So the difference between n and n is of order X r, i.e. of order the ratio the dimensions of the distance the particle moves when emitting a pulse to the distance to the source. This time however, the leading term does not cancel out and we can safely neglect the terms of order X r. Hence we put, n [( n β ) β ] ------------------------------------------------ ( 1 β n ) 2 n [( n β ) β ] -------------------------------------------- ( 1 β n) 2 It is straightforward (exercise) to show that d ------ ---------------------------- n ( n β ) dt 1 β n n [( n β ) β ] -------------------------------------------- ( 1 β n) 2 Hence, re( ω) q --------------e iωr c 4πcε 0 d ------ n ( n β ) dt ---------------------------- 1 β n iω t n X( t ) exp -------------------- dt c One can integrate this by parts. First note that n ( n β ) ---------------------------- 1 β n 0 Electromagnetic Theory 55 /56
since we are dealing with a pulse. Second, note that, d dt ------ iω t n X( t ) -------------------- n X( t ) exp exp iω t -------------------- iω[ 1 β n] c c and that the factor of [ 1 β n ] cancels the remaining one in the denominator. Hence, re( ω) iωq --------------e iωr c n ( n β ) iω n X( t ) t -------------------- 4πcε 0 exp dt c In order to calculate the Stokes parameters, one selects a coordinate system ( e 1 and e 2 ) in which this is as straightforward as possible. The motion of the charge enters through the terms involving β( t ) and X( t ) in the integrand. Remark The feature associated with radiation from a relativistic particle, namely that the radiation is very strongly peaked in the direction of motion, shows up in the previous form of this integral via the factor ( 1 β n ) 3. This dependence is not evident here. However, when we proceed to evaluate the integral in specific cases, this dependence resurfaces. Electromagnetic Theory 56 /56