Electromagnetic radiation

Transcription

1 Eletromagneti radiation Solution of Maxwell equations with external urrent The eletromagneti field generated by an external (expliitly given) four-urrent J µ (x) is given as solution of Maxwell µ F µ = 4 J (8..) Under the Lorenz gauge µ A µ (x) =the equation takes the form of non-homogeneous d Alembert A µ (x) = 4 J µ (x). (8..) Due to linearity of equation (8..) its solution an be represented by the integral A µ (x) = 4 d 4 x D(x x )J µ (x ). (8..3) where D(x x ) is an unknown funtion whih must be determined. Note that suh solution is not unique - it is determined up to solution of homogeneous A µ =. The expression (8..3) must A µ (x) = 4 d 4 D(x x ) J µ (x )= 4 {z } J µ (x) (8..4) 4 (x x ) It means that D(x) is fundamental solution D is equal to the Dira delta in sense of generalized funtions (@ D(x),'(x)) = ( 4 (x),'(x)). (8..5) 47

2 8. ELECTOMAGNETIC ADIATION We managed to simplify the problem to question of solution of the d Alembert equation with -soure The fundamental solution of d Alembert equation The equation (8..5) an be solved by means of the Fourier transform. One an perform the Fourier transform with respet to all spae-time oordinates what leads to an algebrai equation. It turns out that solution of the algebrai equation has poles. As a result one annot simply alulate the integral representing the inverse Fourier transform. This problem an be solved by replaing the variable of integration by a omplex variable and performing integration with the help of Cauhy s theorem. In order to avoid this disussion we shall perform Fourier transform involving only spatial oordinates x. Plugging expression D(x, x) =F x [ D(x, k)] (8..6) 569 and 4 (x) = (x ) 3 (x) = (x )F x [] = F x [ (x )] (8..7) 56 to the equation (8..5) we get ((@ r )Fx [ D(x, k)],'(x, x)) = (Fx [ (x )],'(x, x)), (Fx [(@ + k ) D(x, k)],'(x, x)) = (Fx [ (x )],'(x, x)). 56 It an be ast in the form ((@ + k ) D(x, k), '(x, k)) = ( (x ), '(x, k)), (8..8) where '(x, k) Fx ['](x, k)) T is a test funtion. The solution of this equation is a fundamental solution of the harmoni osillator D ret/adv (x, k) =± (±x ) sin(kx ), (8..9) k 564 where k = k. The solution D ret/adv (x, x) reads D ret/adv (x, x) = ( ) 3 = ± (±x ) ( ) 3 d 3 ke ik x Dret/adv (x, k) d 3 ke ik x sin(kx ). k 48

3 8. Solution of Maxwell equations with external urrent D ret and D adv are alled retarded and advaned solutions. We hoose spherial oordinates suh that the axis ˆk 3 oinides with the vetor x we get d 3 k = k sin #dkd#d', k x = kr os #, r = x. 565 It is onvenient to define a new oordinate u = os #. It gives D ret/adv (x, x) = ± (±x ) ( ) 3 = ± (±x ) ( ) = ± (±x ) r( ) d' k dk sin(kx ) k dkk e ikx e ikx ik dk[e ik(x r) du e ikru! e ikr e ikr ikr e ik(x +r) ]+ dk[e ik(x r) e ik(x +r) ] C {z } A dk[eik(x r) e ik(x +r) ] 566 = ± (±x ) r( ) dk[e ik(x r) e ik(x +r) ] = ± (±x ) 4 r [ (x r) (x + r)]. (8..) Let us observe that for r> it holds (±x ) (x r) = (x r), (±x ) (x ± r) =. 567 It gives D ret/adv (x, x) = (±x ) 4 r [ (x r)+ (x + r)] = (±x ) (x ), (8..) where x =(x ) r. The last equality is obtained from (f(z)) = X k (z z k ) f (z k ), where f(z k)= Fundamental solution in lower dimensions Test funtion '(x) in general depends on x =(x, x), where x is position vetor in threedimensional spae. When physial system has symmetry suh that we do not expet dependene on one or two spatial dimensions fundamental solution has different form. Suh fundamental solutions an be obtained integrating out over all irrelevant oordinates. 49

4 8. ELECTOMAGNETIC ADIATION Let us onsider lass of test funtions onstant in diretion given by x 3 i.e. funtions of the form '(x,x,x ). It follows that (@ D (3) ret (x),'(x)) = (D(3) ret (x, 3' ) {z} apple = dx dx dx dx 3 D (3) ret (x, )'. (8..) 563 We shall denote D () ret (x,x,x ):= dx 3 D (3) ret (x, x) (8..3) 564 then HS of (8..) an be ast in the form (D () ret,@ '(x,x,x )) = (@ D () ret,'(x,x,x )) (8..4) is redued to + dimensions and (f,') = 3 dx dx dx f'. Sine test funtion ' dos not depend on x 3 we get ( 4 (x),'(x,x,x )) = '(,, ) = ( (x ) (x ) (x ),'(x,x,x )), (8..5) 567 so D () ret obeys distributional equation (@ D () ret (x),'(x)) = ( 3 (x),'(x)), x =(x,x,x ). (8..6) 568 It allows to interpret D () ret as a fundamental solution of +dimensional d Alembert equation Fundamental solution in +dimensions We still have to find expliit form of distribution D () ret. We define y := (x,x ) and y := y. Let us onsider expression dx d y'(x, y) dx d y'(x, y) dx 3 (x ) 4 x dx 3 4 (x x ) (x p y +(x 3 ) ) p y +(x 3 ) = where f(x 3 ):=x p y +(x 3 ) ). Equation f(x 3 )=has two solutions x 3 = p (x ) y, x 3 =+ p (x ) y (8..7) 4

5 8. Solution of Maxwell equations with external urrent whih holds for y applex. Last ondition restrits integration area to interior of dis with radius y = x. Applying formula (f(x 3 )) = X k (x 3 x 3 k ) f (x 3 ) x 3 =x 3 k = (x3 x 3 ) + x p 3 y +(x 3 ) (x3 x3 ) x 3 p y +(x 3 ) = x p (x ) y (x 3 x 3 )+ (x 3 x 3 ) (8..8) we get...= = dx y applex d y'(x, y) dx 3 4 dx y applex d y'(x, y) 4 p x (x 3 x 3 (x ) y )+ (x3 x 3 ) p y +(x 3 ) x p (x ) y x = One an extend integration range on by inlusion a Heaviside funtion (x gives...= It follows that D () ret is of the form dx d y'(x, y) y ) what (x y ) p (8..9) (x ) y D () ret (x) = (x x ) p (x ) x, (8..) 5634 where now x =(x,x ) Fundamental solution in +dimensions In order to find a fundamental solution of d Alembet equation in +dimensions we onsider a lass of test funtions whih depends only on x and x. Expression (D () ret,') an be integrated out on variable x (D () ret,')= dx dx = dx dx " D () ret (x,x ) z } { apple dx D () (x,x,x ) '(x,x ) dx # (x x ) p '(x,x )=... (x ) x 5636 where (x x ) restrit integration to interior of a irle x applex. It means that x, x apple x apple x, p (x ) (x ) apple x apple p (x ) (x ). 4

6 8. ELECTOMAGNETIC ADIATION Figure 8.: egion of integration. and then...= = " x p # (x dx dx ) (x ) p dx p '(x,x ) x (x ) (x ) (x ) (x ) (x ) x p 3 dx dx (x ) (x ) dx 4 p q 5 '(x,x )=... x (x ) (x ) (x ) (x ) (x ) 5637 With new variable u := x p (x ) (x ), du = dx p (8..) (x ) (x ) we get...= = = x dx dx 6 x 4 dx x x dx '(x,x ) du p u {z } arsin() arsin()= '(x,x ) apple dx dx (x x ) '(x,x ). (8..) 5638 It follows that fundamental solution of d Alembert equation in +dimensions is of the form D () ret (x) = (x x ). (8..3) 4

7 8. Cauhy problem for an eletromagneti field Huygen s priniple Comparing solutions (8..), (8..) and (8..3) we an note that solutions in + and + dimensions depend on x applex so ontribution to solution at of wave equation at (x, x) omes from all events inside the past light one. In ontrary, solution in 3+ dimensions depends only on events whih are loated at the surfae of the past light one of the event (x, x). It means solution of wave equation propagates exatly with the speed of light in 3+ dimensions. This fat is known as Huygen s priniple and it is satisfied in all n +dimensions with n being an odd number n 3. Figure 8.: Fundamental solutions and their supports in +, + and 3+ dimensions Cauhy problem for an eletromagneti field Solution D ret D adv Sine retarded and advaned fundamental solution obey equation (@ D ret/adv,')=( 4,') then their differene D(z) :=D (3) ret (z) D(3) adv (z) is a very speial solution of homogeneous wave equation. Suh solution is very useful to express a general solution of wave equation in dependene on initial data. Plugging expression for solutions D (3) ret/adv in form of the Fourier 43

8 8. ELECTOMAGNETIC ADIATION transform with respet to spatial oordinates we get D(z) = ( ) 3 d 3 ke ik z (3) [ D ret (z, k) 3 = ( ) 3 d 3 ke ik z (z )+ ( z ) 3 {z } We have and It follows that D(z) := ( ) 3 d 3 ke ik z sin(kz ) 3 k D(3) adv (z, k)] sin(kz ). k (8..4) D = ( ) 3 d 3 ke ik z os(kz ). (8..5) 3 D(, z) =, D (, z) = (z, z) =. (8..6) z = Funtions D(z, z) and D (z, z) form two linearly independent solutions of wave equation Cauhy problem We shall onsider an initial problem for wave equation in 3+ dimensions. Let be an (3 dimensional) hypersurfae at given instant given by fixed number y = onst. An initial problem is given by wave A µ (x) = (8..7) and initial onditions a µ (x) :=A µ (x, x) x =y, (8..8) b µ Aµ (x, x) x =y. (8..9) We want to obtain solution whih satisfy the wave equation and the initial onditions at x = y. In order to solve this problem we shall perform a Fourier transform with respet to spatial oordinates. Substituting A µ (x, x) =F [õ (x, k)] = ( ) 3 3 d 3 ke ik x à µ (x, k) (8..3) into equation (@ A µ,')=we get (@ + k )õ (x, k), '(x, k) = (8..3) 44

9 8. Cauhy problem for an eletromagneti field Figure 8.3: Cauhy problem for wave equation. i.e. à µ (x, k) is a solution of equation of harmoni osillator. Fourier transforms of initial onditions read ã µ (k) :=F [a µ (x)] = õ (x, k) x =y, (8..3) bµ (k) :=F [b µ (x, k) x =y. (8..33) form Two linearly independent solutions of equation (@ + k ) D(z, k) =an be hosen in the D(z, k) = sin(kz ), D (z, k) D (z, k) = os(kz ) where z := x y They satisfy initial onditions D(, k) =and D (, k) =. A general solution is given by linear ombination à µ (x, k) = µ D (x y, k)+ µ D(x y, k) Sine õ (x, k) x =y = µ à µ (x, k) x =y = µ we an identify free oeffiients as µ =ã µ (k) and µ = b µ (k). It gives à µ (x, k) =ã µ (k) D (x y, k)+ b µ (k) D(x y, k). (8..34) 45

10 8. ELECTOMAGNETIC ADIATION Solution A µ (x, x) is given by A µ (x, x) = ( ) 3 = ( ) 3 d 3 ke ik x à µ (x, k) d 3 ke ik x d 3 ye ik y d 3 ze ik z a µ (y)d (x y, z)+b µ (y)d(x y, z) (8..35) Considering that ( ) 3 and integrating over z we get expression A µ (x, x) = (y ) d 3 ke ik (x y z) = 3 (x y z) d 3 y a µ (y)d (x y, x y)+b µ (y)d(x y, x y), where integration is performed over hypersurfae of initial onditions. (8..36) We shall express this formula in alternative form. Note, that D(z) is anti-symmetri i.e. D( z, z) = D(z, z). It is lear from the expression D( z, z) = ( ) 3 d 3 ke ik z sin( k ( z )) k whih after hanging a variable of integration k! k is equal to D(z, z). Considering y as a variable we an express derivative with respet to x by derivative with respet to y D (x D(x y, x y) D(y x, D(x y, x y) where the minus sign an be absorbed due to antisymmetry of D(z). Sine y is now a variable then integration on the surfae of initial data is realized by a onsidering expressions D(y x) (y x) at speifi value of this variable y =ȳ. We also a µ (y) =A µ (y) y =ȳ, bµ (y) y =ȳ, (8..37) 5675 then A µ (x) = (ȳ ) d 3 y apple A µ D(y x). (8..38) 46

11 8. Cauhy problem for an eletromagneti field Let us hek if this formula has all properties. Ating (with respet to x) we get that A µ (x) is a solution of wave A µ (x) = (ȳ ) d 3 y 4A @ D(y x) {z } D(y x) 5 =. {z } At x =ȳ (also y =ȳ ) we get D(y x, y x) x =y =ȳ = ( D(y x, y x) = x =y =ȳ ( ) 3 = 3 (y x), d 3 ke ik (y x) sin( k (y x )) k x =y =ȳ =, d 3 ke ik (y x) os( k (y D(y x, y x) = 3 (y x), x @y D(y x, y x) = x =y =ȳ ( ) 3 d 3 ke ik (y x) k sin( k (y x )) x =y =ȳ x =y =ȳ = Now we an hek initial onditions A µ (x) x =ȳ = d 3 ya µ (y, y) 3 (y (ȳ ) x) =A µ (y, x) =a µ (x) µ = d 3 3 x =ȳ (ȳ (y (y, x) = b µ (x). x =ȳ In the last step we shall generalize formula (8..38) on expression whih depends on initial data given at Cauhy surfae whih is a spae like 3-dim surfae. Note, that not always there exist a surfae of simultaneity (ȳ ) so generalization to (whih loally looks like a piee of ) is very useful. A volume integral over an be interpreted as integral over hypersurfae where the in- tegration element d 3 y an be interpreted as ds, where ds µ = 3! "µ ds. Clearly, a natural generalization is! and d 3 y! ds µ. We still have to generalize = 3 (y x) whih results in x =y x =y d x) y f(y) @y we have to generalize 3. This an be done by substituting 3 by simple layer 3 µ x =y = n µ 3 (y x) 3 µ(y x) (8..39) 47

12 8. ELECTOMAGNETIC ADIATION 5689 suh that ds µ n 3 µ (y x) f(y) =f(x), (8..4) 569 where n µ is a time-like four-vetor orthogonal to the surfae. It follows that for x 569 ds (y x)f(y) =f(x). µ We get the following expression for A µ (x) apple A µ (x) = ds A µ D(y x) D(y x)@aµ An important fat about (8..4) is that it does not depend on partiular hoie of Cauhy s surfae. To show this we hoose two Cauhy surfaes and. We take a 4-dim volume Figure 8.4: egion and its border delimited by those surfaes and a time-like at spatial infinity and assume that A µ! suffiiently quikly for x!. A µ (y) D(y with respet to y), then we an hoose expression whih is identially zero = d 4 y D(y x)@ A µ (y) A µ (y)@ D(y x) = d 4 y@ [D(y x)@ A µ (y) A µ (y)@ D(y x)] d 4 y [@ D(y x)@ A µ A µ (y)@ D(y x)] {z } I = ds [D(y x)@ A µ (y) A µ (y)@ D(y apple = + ds [D(y x)@ A µ (y) A µ (y)@ D(y x) =(derivatives where the integral at spatial infinity vanishes. It follows that integrals on surfaes and are equal. 48

13 8. Cauhy problem for an eletromagneti field Kirhhoff s formula Substituting a ausal solution x >y D(x y, x y) = 4 (x y x y ) x y into formula Kirhhoff s formula apple A µ (x, x) = d 3 y a D(x y, x y)+b µ (y)d(x y, x y) we get where b µ Aµ t=t. A µ (x, x) = apple d 3 y (x y x y ) x y d 3 y (x y x y ) x y b µ (y) (8..43) a µ (y), (8..44) Cauhy problem in (+) dimensions Before going bak to desription of eletromagnetism we shall disuss a solution of initial problem for non-homogeneous (fored) wave equation in + dimensions. Sine all omponents of A µ obeys the same equation then we shall onsider only one omponent i.e. u tt u xx = f(t, x), u(,x)='(x), u t (,x)= (x). (8..45) We shall solve the problem integrating equation (8..45) over a region of dependene of event (x, t) (x is hosen as first oordinate and t as a seond one) whih is a triangle delimited by segments C, C and C 3, see Fig.8.5. A segment C is loated at t =and its ends orrespond with events (x t,) and (x + t,). The initial onditions for wave equation are given on this segment. Segments C and C 3 are light-one surfaes given by x = x + (t t ) and x = x (t t ) respetively. We hoose anti-lokwise orientation of the whih onords with ordering of oordinates. Integrating equation (8..45) we get d (u t t u x x ) = d f(t,x ) {z } {z } I left I right where I right is given expliitly by expression (8..46) I right = t x+(t t dt ) dx f(t,x ). (8..47) x (t t ) 49

14 8. ELECTOMAGNETIC ADIATION Figure 8.5: Dependene region of solution of wave equation in + dimension. 57 Expression I left an be omputed using Stokes theorem. A Stokes theorem da (r A) A dl in two dimensions takes the form 57 I dy dy (@ {z } A )= (A dy + A dy ). (8..48) We shall identify dy = dx and dy = dt. LHS of wave equation an be ast in the following t x u x {z} xu) {z It allows to express I left as integral on the I I left I (A dy + A dy {z} t tu). (8..49) {z A [ (@ t u)dx (@ x u)dt ]=I + I + I 3, (8..5) where integrals I,I,I 3 are omputed on C,C,C 3 respetively Integral I. On a line C we have dt =so the integral reads I = x+t x t dx t u(t,x )] t = = x+t x t dx (x ) (8..5) Integral I. 43

15 8. Cauhy problem for an eletromagneti field The light-one C is suh that dx = dt, what gives I = [ (@ t u)dx (@ x u)dt ] C apple dx = (@ t u)( dt ) (@ x u) C = [@ t udt x udx ]= du C C = [u(t, x) u(,x+ t)] = [u(t, x) '(x + t)]. (8..5) Integral I 3. Similarly, on C 3 we have dx =+dt, what gives I 3 = [ (@ t u)dx (@ x u)dt ] C 3 apple dx = (@ t u)(dt ) (@ x u) C 3 = [@ t udt x udx ]= du C 3 C 3 = [u(,x t) u(t, x)] = [u(t, x) '(x t)]. (8..53) 578 Plugging this result to equation I + I + I 3 = I right we get u(t, x) ['(x + t)+'(x t)] x+t x t dx (x )= t x+(t t dt ) dx f(t,x ) x (t t ) whih gives u(t, x) = ['(x + t)+'(x t)] + x+t Note that last term an be also ast in the form t where we define x+(t t dt ) dx f(t,x )= dt dx x (t t ) x t dx D () ret (t, x) := (t (x )+ t x+(t t dt ) dx f(t,x ). x (t t ) apple ((t t ) x x ) {z } D () ret (t t,x x ) x ). f(t,x ), (8..54) (8..55) 43

16 8. ELECTOMAGNETIC ADIATION The eletromagneti field of a single eletri harge 8.3. The Lienard-Wihert potentials In order to understand the phenomenon of eletromagneti radiation we shall study an arbitrary motion of a point-like harge q. We will assume that trajetory of the partile is a given funtion. Suh physial system is a fored system and it requires an external power soure to maintain a desired motion of the partile. In further part we shall analyze a partile under external fore. In suh a ase a trajetory of the partile is determined by two fators: an external fore and an eletromagneti radiation whih ats as a dissipative fore. In urrent analysis the worldline of the harge is a given funtion z µ = z µ ( ). The eletri urrent density assoiated with the partile reads J µ (x) =q d dzµ d 4 (x z( )). (8.3.56) The four-urrent density does not vanish only at the worldline of the partile. The retarded/advaned potential is obtained as A µ (x) = 4 d 4 x D ret/adv (x x )J µ (x ) = 4 q d 4 x d (±(x x )) [(x x ) ] dzµ 4 (x z( )) d = q d (±(x z )) dzµ d [(x z( )) ]. (8.3.57) The equation (x z( )) =has two solution ret and adv. There are two lightlike fourvetors whih point out from events z( ret ) and z( adv ) at the worldline of the partile to the event x. We shall denote them by µ ( ret ):=x µ z µ ( ret ) and µ ( adv ):=x µ z µ ( adv ). They satisfy µ ( ret/adv ) µ ( ret/adv )= It follows that [(x z( )) ]= ( ret ) ż µ µ ret + ( adv) ż µ µ adv (8.3.58) A µ (x) =q ż µ (±(x apple ( z ret ) ( ))) ż µ + ( adv) µ ret ż µ µ adv = q (±(x z ( ret )))ż µ ( ret ) ż µ µ ret + q (±(x z ( adv )))ż µ ( adv ) ż µ µ adv (8.3.59) 43

17 8.3 The eletromagneti field of a single eletri harge Figure 8.6: The worldline z µ ( ); the retarded and the advaned proper time Taking into aount that (±(x z ( ret/adv ))) =, (±(x z ( adv/ret ))) =, and ż µ µ ( ret ) >, ż µ µ ( adv ) < we get A µ (x) =±q ż µ ż ret/adv (x) = q ż µ ret/adv (8.3.6) 5736 where we have denoted ret/adv := (x z ( ret/adv ))ż ( ret/adv ) ± ż ret/adv (8.3.6) The expression (x) has interpretation of the spatial distane between an event x and events z( ret/adv ) taken in the instantaneous rest frame of the partile. The retarded Lienard Wihert potentials are given by (8.3.6) for = ret (x). Notie that a solution of the equation µ ( ret ) µ ( ret )=is a funtion of x so x enters to (8.3.6) expliitly and and through the funtion ret (x). The four-vetor µ ( ret ) has omponents µ ( ret )! (, ) =(, )=(, ˆn) (8.3.6) where =. We shall express omponents of the four vetor A µ (x) in terms of the vetors ˆn and = V. Considering that żµ = (, ) and µ ( ret )=(, ˆn) we get ż µ ( ret ) µ ( ret )= ( ˆn ) ret The zero omponent ( ret ) leads to relation 433

18 8. ELECTOMAGNETIC ADIATION Figure 8.7: The relations between vetors and t ( ret )=t whih represent relation between ret and t. The retarded eletromagneti potentials an be expressed in the form. A (t, x) = q ˆn t A(t, x) = q ˆn t (8.3.63) The eletromagneti field tensor The eletromagneti field tensor an be obtained diretly from the potentials (8.3.6) apple apple qż qż µ F µ (x) =@ (8.3.64) (x) (x) 5747 In order to simplify notation we define a symbol (ab) :=a b. The first term µ A (x) = q µ ż (8.3.65) 434

19 8.3 The eletromagneti field of a single eletri harge µ = ± [@ µ(x z )ż µ ż ] ± ( µ µ (x))ż + µ (x) ± żµ +(( z) )@ µ (x) (8.3.66) where (x) ret/adv (x). The µ (x) is obtained taking µ ( )= whih gives ( µ µ (x)) =and µ (x) = µ (ż) = ± µ =: ±k µ ) µ = k µ (8.3.67) where k µ ż µ = ±. Notie that a simplified notation has been used µ µ ret/adv, ret/adv and k µ k µ ret/adv. The µ takes the µ = ± żµ +[ (k z) ] k µ (8.3.68) Plugging this result to (8.3.65) we µ A = q apple (±k µż ) = qk µ appleż + ± z (±ż µż ) (k z)ż ( (k z) ) k µż q ż µ ż (8.3.69) The term ż µ ż is symmetri and therefore it does not ontribute to the eletromagneti field tensor where F ret/adv µ (x) =k µ k µ (8.3.7) appleż := q + ± z (k z)ż (8.3.7) When a harged partile is moving without aeleration z µ =then the eletromagneti field is ontains term. Choosing a spae-like four-vetor w µ whih satisfies w µ ż µ =, w µ w µ = we an express k µ as follows k µ ret/adv = (±żµ + w µ ). (8.3.7) 435

20 8. ELECTOMAGNETIC ADIATION Sine ż µ = onst then ret = adv. The symmetri parts ±ż µ ż anels out and we get F ret µ = F adv µ = q (ż µw ż w µ ) (8.3.73) In the instantaneous rest frame of the harged partile ż µ =(, ), w µ =(,ˆn) so the tensor F µ ontains a Coulomb field F i = qni F ij = The part of the eletromagneti field tensor proportional to is a proper field of the partile. It is present even though the partile moves without aeleration. On the other side the radiation field is proportional to. It is a free eletromagneti field emitted by the partile in a nonuniform motion The eletri and magneti omponents In this setion we derive expressions desribing retarded eletri and magneti fields. As a first step we write down the eletromagneti field tensor in terms of and n µ. Substituting k µ ret = µ (ż) = nµ (nż) and ret = (ż) = (nż) we get ż (n z) (nż)ż = q 4 (nż) + z (nż) apple ż = q (nż) + (nż) z 3 5 (n z)ż (nż) (8.3.74) 576 Then the eletromagneti field tensor has the form F µ = F oul µ + F rad µ where Fµ oul := q n µ ż n ż µ (nż) 3 ret (8.3.75) 576 F rad µ := q (nż)(n µ z n z µ ) (n z)(n µ ż n ż µ ) (nż) 3. (8.3.76) ret The part of eletromagneti field tensor (8.3.75) behaves as therefore it is alled a Coulomb part of eletromagneti field. Let us note that this part is always present even though the eletri harge moves uniformly i.e. without aeleration. On the other side the part (8.3.76) depend on four-aeleration z µ. Moreover, it behaves as so it dominates far from the aelerating harge. This field is a radiation field. 436

21 8.3 The eletromagneti field of a single eletri harge The eletri and magneti field an be derived diretly from the form of the eletromagneti field tensor. We shall denote the Coulomb part by E(ż), B(ż) and the radiation part by E( z), B( z). The four veloity and four aeleration of the harge read ż µ! (, ), z µ! 4 ( a)(, )+ (, a) what gives (nż) = ( ˆn ), (n z)! 4 ( a)( ˆn ) ˆn a. The eletri field reads E i (ż) =F i (ż) = q n i ż n ż i (nż) 3 and magneti field has the form n i = q ( ) ( ˆn ) 3 t i t = q n i v i 3 3 ( ˆn ) 3 t (8.3.77) B i (ż) = ijkf jk (ż) = q n j ż k n k ż j ijk (nż) 3 = q ( ) ijkn j ( k ) ( ˆn ) 3 t =(ˆn E(ż)) i t t = ijk q n j ż k (nż) 3 t = q ( ) ijkn j (n k k ) ( ˆn ) 3 t. (8.3.78) The magneti field is perpendiular to the eletri field and its magnitude behaves with distane exatly as magnitude of the eletri field. The the eletri field omponent that depends on aeleration reads Plugging E i ( z) =F i ( z) = q (nż)(n i z n z i ) (n z)(n i ż n ż i ) (nż) 3 = q [(nż) z (n z)ż ]n i [(nż) z i (n z)ż i ] (nż) 3 t t (8.3.79) (nż) z (n z)ż = ( ˆn )[ 4 ( a)] [ 4 ( a)( ˆn ) ˆn a] = 3 ˆn a (8.3.8) 437

22 8. ELECTOMAGNETIC ADIATION and (nż) z i (n z)ż i = ( ˆn )[ 4 ( a) i + a i ] [ 4 ( ((((((((( ( a)( ˆn ) (ˆn a)] i = 3 [(ˆn a) i +( ˆn )a i ] (8.3.8) to the formula (8.3.79) we get E i ( z) = q The magneti field reads 3 (ˆn a)n i 3 [(ˆn a) i +( ˆn )a i ] 3 3 ( ˆn ) 3 t = q (ˆn a)(n i i ) ˆn (ˆn )a i ( ˆn ) 3 t = q (ˆn [(ˆn ) a]) i ( ˆn ) 3 t. (8.3.8) B i ( z) = ijkf jk ( z) 5769 then Let us denote = q ijk = q ijk = q (nż)[n j z k n k z j ] (n z)[n j ż k n k ż j ] (nż) 3 (nż)[n j z k ] (n z)[n j ż k ] (nż) 3 ijk n j [(n z)ż k (nż) z k ] (nż) 3 = q ijkn j 3 [(ˆn a)( k ) ( ˆn )a k ] 3 3 ( ˆn ) 3 t = q ijkn j (ˆn a)(nk k ) ˆn (ˆn )a k ( ˆn ) 3 t = q apple ˆn [(ˆn ) a] ˆn ( ˆn ) 3 K := t i t t =[ˆn E( z)] i t t (8.3.83) ˆn ( ˆn ) 3 (8.3.84) E(ż) = q ( )K t E( z) = q ˆn (K a) t B(ż) =ˆn E(ż) (8.3.85) B( z) =ˆn E( z) (8.3.86) 438

23 8.3 The eletromagneti field of a single eletri harge The Coulomb field Let us observe that in the ase of the uniform motion only the Coulomb field remains. This problem has been already solved, however, no expliit expression for retarded fundamental solution of d Alembert equation was used. We have found that E(t, x) =q 3 ( sin # ) 3/ B = E (8.3.87) where =, = ˆn aording to Fig.8.8 Clearly, the existene of the Coulomb field is independent on the fat what Figure 8.8: elation between and is the value of aeleration of the harged partile. The Coulomb part the eletromagneti field E(t, x) and B(t, x) is given terms of position and veloity of the partile at retarded time t. The only differene is that without aeleration the position of the partile at instant t is given by a vetor whereas in the ase of non-zero aeleration the partile is somewhere else. The Coulomb part of eletri field vetors have diretion of the vetor. It beame lear now that this orientation has nothing to do with the atual position ow the partile. In partiular, for a 6= there is no a partile at all at i.e. at the origin of. We shall show that the formulas (8.3.87) are equivalent to expressions (8.3.85). Considering that sin # = b and sin # = b we obtain sin # = sin # sin # = ˆn. (8.3.88) 439

24 8. ELECTOMAGNETIC ADIATION The eletri field reads E(t, x) =q 3 ( sin # = q {z} (ˆn ) ˆn ˆn 3 3 {z } 3 ) 3/ t ( ˆn sin #) 3/ t = q ˆn ( ˆn + ( sin #)) {z } 3/ (ˆn ) = q ( )(ˆn ) [( ˆn ) ] 3/ t t = q ( )(ˆn ) ( ˆn ) 3 t (8.3.89) The expression for a magneti field follows from this result almost immediately if one notes that so we get (ˆn )= ˆn = ˆn = ˆn (ˆn ) (8.3.9) B(t, x) =ˆn E t (8.3.9) It shows that both formulas are perfetly the same. The retarded potentials give a deeper insight into the physis of the problem. We an see that an atual position of the partile has nothing to do with the form of the Coulomb field at t. In order to determine the eletri field only the retarded position has to be taken into aount The radiated power The eletromagneti radiation is a proess in whih the aelerated eletri harges looses the energy. The energy emitted by the eletri harge into the solid angle d in diretion of the vetor ˆn during the interval dt has the value dp (t )dt. The expression dp (t ) is alled the radiation power. Considering an elementary surfae with the element da = ˆn d we have equality of energies dp (t )dt = dp (t)dt. (8.3.9) The retarded time t is related with time t in whih fields E(t) and B(t) are evaluated t = t + (t ) (8.3.93) 44

25 8.3 The eletromagneti field of a single eletri harge what gives dt dt =+ d(t ) dt (8.3.94) Taking derivative wrt. t of the equation (t ) = (t ) (t ) we find 58 where (t )=x z(t ). Then (t ) d(t ) dt = (t )ˆn d(t ) dt d(t ) dt = ˆn(t d ) dt (x z(t )) = ˆn(t ) v(t )= ˆn(t ) (t ) (8.3.95) 58 It follows that dt dt = ˆn (8.3.96) Now we an determine the radiated power P (t ) taking into aount that dp (t)dt = apple dp (t) dt dt dt = dp (t )dt. (8.3.97) Comparing (8.3.9) and (8.3.97) we find the radiated power dp (t ) in terms of the power deteted at (t, x) dp (t )=( ˆn )dp (t) (8.3.98) where the power P (t) is given in terms of the Poynting vetor S(t) dp (t) =S(t) da = S(t) ˆn d (8.3.99) 586 where S = 4 E B = [E (ˆn E)] = 4 4 [ˆn E (ˆn E) ]. (8.3.) Sine the eletri field satisfies E( z) ˆn =then dp (t) = 4 E d = = q (ˆn a)k 4 3 (ˆn = q 4 q 4 ˆn (K a) t K)a t d 4 3 [(ˆn a) K +(ˆn K) a (ˆn a)(ˆn K)(a K)] t d. (8.3.) d 44

26 8. ELECTOMAGNETIC ADIATION Substituting K we get dp (t) = q 4 3 ( ˆn ) 6 [(ˆn a) (ˆn ) +( ˆn ) a (ˆn a)( ˆn )(ˆn a a)] t d = q 4 3 ( ˆn ) 6 6 4(ˆn a) ( + ˆn ) +( ˆn ) a {z } ( )+ ( ˆn ) (((((((((( (ˆn a) ( ˆn ) + (ˆn a)( ˆn )(ˆn ) i t d. (8.3.) Plugging this result to the formula (8.3.98) we obtain the expression for angular distribution of radiated power where dp (t )= apple W (t a a)( a), ˆn) := +(ˆn ( ˆn ) 3 ( ˆn ) 4 ( q 4 3 W (t, ˆn)d (8.3.3) ) (ˆn a) ( ˆn ) 5 t. (8.3.4) Veloity and aeleration parallel In present ase the vetors of aeleration and veloity of the harged partile are olinear what an be ast in the form a =. It leads to the following expression ˆn (K a) = ˆn (ˆn a) (ˆn a)ˆn a = ( ˆn ) 3 ( ˆn ) whih gives dp (t )= q a (ˆn a) 4 3 ( ˆn ) 5 t d. (8.3.5) We introdue the spherial oordinates taking versor ẑ as being parallel to the vetor ẑ := /. Similarly a = aẑ. The versor ˆn form an angle # with. then ˆn a = a os #. The power radiated into an elementary solid angle d =sin#d#d'reads i.e. dp (t )= q a sin # 4 3 ( os #) 5 t d. (8.3.6) Note, that this formula does not depend on the mutual orientation between aeleration and veloity (the sign of a does not appear). The funtion W (t, ˆn) is shown in Fig.8.9 where we 44

27 8.3 The eletromagneti field of a single eletri harge (a) (b) () Figure 8.9: The angular distribution of the radiated power: (a) = =., (b) =.5, () onsider three different values of the veloity. In order to obtain a total value of the radiated power one has to integrate the last expression over the angles # and ' what gives q a sin3 # P (t ) = ( ) d#. (8.3.7) 4 3 ( os #)5 t 589 Defining new oordinate u := os # we obtain q a P (t ) = 3 du u ). u)5 ( ( The last integral an be alulated with the help of new variable y := ( du ( u ) = u)5 = + y(4 dy ( y) y5 3y) 3( 3 y = ) + 3 = u dy + 4. )3 3 ( ( y5 y)

28 8. ELECTOMAGNETIC ADIATION 58 The radiated power reads P (t )= q a 3 3 ( ) 3 6. (8.3.8) ret 58 Notie, that for small veloities this formula simplifies to the following one P (t )= q a 3 3. (8.3.9) Veloity and aeleration perpendiular Another interesting situation ours when aeleration and veloity are mutually perpendiu- lar vetors i.e. (8.3.4) dp (t )= a =. The radiated power an be obtained diretly from (8.3.3) and apple q a 4 3 ( ˆn ) 3 ( ) (ˆn a) ( ˆn ) 5 t d. (8.3.) We introdue the instantaneous Cartesian frame suh that ẑ := / and ˆx := a/a. Taking into aount that ˆn = os #, ˆn a = a sin # os ' we get the angular distribution of radiated power dp (t )= apple q a 4 3 ( os #) 3 ( ) a sin # os ' ( os #) 5 d. (8.3.) ret This distribution learly depends on the angle '. It vanishes for ' = n and os # = irradiated power reads apple P (t )= q a sin # 4 3 d# d' ( os #) 3 ( where d' os ' =. We hange a variable to u := os # apple P (t )= q a 4 3 The integral read du ( u) 3 ( ) a sin 3 # os ' ( os #) 5 ret. A total (8.3.) ) du u ( u) 5. (8.3.3) ret du ( u) 3 = ( ) du u ( u) 5 = 4 3 ( )

29 8.3 The eletromagneti field of a single eletri harge Figure 8. (a) (b) Figure 8.: The radiated power angular distribution: (a) =., (b) = Finally we obtain the result P (t )= q a 3 3 ( ) 4. (8.3.4) ret Comparing this result with the retilinear motion we an see that looses in irular motion beahaves as 4 whereas in the retilinear motion they are proportional to 6. The aeleration is 445

30 8. ELECTOMAGNETIC ADIATION (a) (b) Figure 8.: The radiated power angular distribution (setion ' =): (a) =.3. =., (b) equal to a = v = where is a radius of the orbit. For ultrarelativisti ase we have P (t )= q 3 ( ) and then we onlude that looses of energy aused by radiation an be redued if we use ael- erator with big radius adiation of systems with many harges etarded potentials In this setion we shall analyze the problem of radiation generated by huge number of eletri harges. The motion of suh harges auses a four-urrent J µ (x). The four-urrent an be desribed by a smooth (non delta-like) funtion of oordinates when the number of harges is 446

31 8.4 adiation of systems with many harges suffiiently big and the harges are lose enough. The retarded potential is given in the form A µ (x) = 4 d 4 x D (x x )J µ (x ) = d 3 x dx (x x ) (x x x x ) x x J µ (x, x ) = d 3 x J µ (x x x, x ) x x, (8.4.5) where ( x x )=. The retarded time t is given by expression t = t The omponents of the four potential read A µ (t, x) = x x. (8.4.6) d 3 x J µ (t x x, x ) x x. (8.4.7) These formula an be onsidered as a starting point to obtain expressions haraterizing the radiation generated by antennas. In ontrary to the ase of a single partile we annot derive exat formulas. For this reason one an to find approximated solutions, valid far from the soure of radiation Potentials in the radiation zone A radiation zone is defined as a region far from the radiating soure. Stritly speaking the distane from the soure must be muh larger than the harateristi size of the soure i.e. x x. The approximated solution an be obtained by means of expansion in the radiation zone. Expanding the expression x x we get x x = x x x + x / = x x x x + x x = x x x x +... Then r := x and ˆn := x/ x we find that / x x = r ˆn x +... (8.4.8) 584 It follows that the retarded time beame t r = t {z } t e + ˆn x +... (8.4.9) 447

32 8. ELECTOMAGNETIC ADIATION Figure 8.3: The soure of an eletromagneti radiation (antenna) The emission time t e is defined as a retarded time of the origin of the referene frame. One an adopt this expression as the most rough approximation of the retarded time. The term ˆn x represent orretions to the retarded time assoiated with the finite size of the radiated system. The partial derivative of the retarded time i t r i t + ˆn x +... i r i n j x j +... = n i + ij n i n j x j +...= n i + r r (xi n i (ˆn x )) +... = n i r [ˆn (ˆn x )] i and then rt = ˆn + r ˆn (ˆn x ). (8.4.) The leading behaviour of A µ in the radiation zone is given by the series of expansion of (8.4.7) A µ (t, x) = = r = r d 3 x J µ (t (r ˆn x +...), x ) r ˆn x +... d 3 x ˆn x ( )J µ ˆn x (t e + d 3 x + r ˆn x r +... J µ (t e, x )+ +...,x ) ˆn t J µ (t e, x ) +... (8.4.) 448

33 8.4 adiation of systems with many harges Sine the limits of integration do not depend on time then d 3 x@ t f = d dt that A µ (t, x) = r d 3 x J µ (t e, x )+ d r dt d 3 xf. It follows d 3 x (ˆn x )J µ (t e, x )+... (8.4.) We shall express this formula in terms of the eletri and the magneti dipole moments p(t) := d 3 x x J (t, x ) (8.4.3) m(t) := d 3 x x J(t, x ) (8.4.4) 5847 The salar potential takes the form A (t, x) = Q r + rˆn ṗ(t e)+... (8.4.5) where Q = onst. The first integral assoiated with the vetor potential A is equal to the time derivative of the eletri dipole moment. This an be seen as follows: let us onsider the losed surfae S whih ontains all the soures and has not ommon points or intersetions with the region V oupied by the harges and the urrents. There is no ontribution from the soures on suh a surfae, so H S (xi J j )da j =. The lhs of this integral an be transformed as follows I S (x i J j )da j = = = 5848 whih gives V V d 3 j(x i J j )= d 3 j(x i J j ) V d 3 x [ ij J j + x i (@ jj j )] = d 3 x J i d 3 x x t V V d 3 x J i d d 3 x x i = d 3 x J i ṗ i, dt V V V (8.4.6) d 3 x J(t e, x )=ṗ(t e ). (8.4.7) The seond integral an be transformed as follows n i d 3 x x i J j = n i d 3 x[x i J j x j J i ]+n i d 3 x[x i J j + x j J i ] {z } aquadrupolemoment = n i d 3 x[x i J j x j J i ] + = [ˆn m] j + (8.4.8) {z } ijk m k 449

34 8. ELECTOMAGNETIC ADIATION Finally, the vetor potential is given in terms of time derivatives of eletri and magneti dipole moment A(t, x) = r [ṗ(t e) ˆn ṁ(t e )] + (8.4.9) The eletri and the magneti field We drop therms that drop faster than /r i.e. The eletri field takes the form apple Q E = r r + rˆn ṗ(t e) E = ta B = r A (8.4.3) r r r r rni r = r (rt e) ˆn p(t e ) {z } ˆn + r [ p(t e) ˆn m(t e )] r [ p(t e) ˆn m(t e )] + = [(ˆn p)ˆn r {z p +ˆn m]+ (8.4.3) } ˆn (ˆn p) 585 then For the magneti field we get E = r ˆn [ˆn p + m] t e (8.4.3) 5853 then apple B = r r [ṗ(t e) ˆn ṁ(t e )] = r (rt e) [ p(t e ) ˆn m(t e )] + {z } ˆn +... B = r ˆn [ˆn m p] t e (8.4.33) 5854 It follows diretly from (8.4.3) that B = ˆn E. 45

35 8.4 adiation of systems with many harges The angular distribution of radiation The angular distribution of radiation is given by expression dp (t) =(ˆn S)r d = 4 E r d = 4 B r d = 4 3 [ˆn [ˆn p + m]] d = [[ˆn p + m] [ˆn (ˆn p) +ˆn m] ]d 4 3 {z } = 4 3 [(ˆn p) +(ˆn p) m + m (ˆn m) ]d (8.4.34) {z } {z } ˆn ( p m) (ˆn m) 5856 dp (t) = 4 3 [(ˆn p) +(ˆn p) +ˆn ( p m)] te (8.4.35) The total power P (t) reads P (t) = apple 4 3 (ˆn p) d + (ˆn p) d + ˆn ( p m)d (8.4.36) It is onvenient to hoose the spherial oordinates in suh a way that in the first integral the z axis is is in diretion of p, in the seond one with m and in the third one with diretion of p m. Then (ˆn p) d = p d# sin 3 # = 8 3 p (8.4.37) (ˆn m) d = m d# sin 3 # = 8 3 m (8.4.38) ˆn ( p m)d = p m d# sin # os # = (8.4.39) The total power of radiation in the radiation zone has the value P (t) = 3 p + m. (8.4.4) Let us notie that m so ontribution from magneti dipole radiation is muh less important than ontribution from eletri dipole radiation. In the simplest ase of the single eletri dipole we have p = qz, so p = q z = qa. It leads to a total power of a single elementary dipole radiation This expression is known as a Larmor formula. P (t) = q 3 3 a. (8.4.4) 45