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Transcript

We have now calculated all the intermediate derivatives which we need for calculating the fields E and B. We will now write down the procedure to follow and then the results. We start by making clear HOW we will carry all the derivatives through. ( ) t 'cos t . t ' t ' 1 A 1 A t ' c t c t ' t A A ( A) t 'cos t ( ) t ' t ' Note that some derivatives at constant t’ are not zero: q (ε' v' / c) ()t 'const v'ε' 2 2 | r r ' | (1 ) c ε' r r ' and also [ ] 0 x x | r r ' | Inserting all these various derivatives, gradients and curls in the equations for the fields, we obtain at the end the analytical expressions for E and B, which turn out to be especially simple: Advanced EM - Master in Physics 2011-2012 1 The FIELDS OF Lienard-Wiechert q[a'(a'ε' )ε' ] q ε'(a'β' ) q (ε'β' )(1 '2 ) E(r, t ) 2 3 c | r r ' | (1 β'ε' ) | r r ' |2 (1 β'ε' )3 q(β'ε' )(1 ' ) qa' q[ε'(a'β' )] B(r, t ) ε' | r r ' |2 (1 β'ε' )3 c 2 | r r ' | (1 β'ε' )3 2 Well,… after all, they do not look so simple. We should look at them carefully and in detail. The first remark is that B is just the vector product of B ε'E ε’ and E. This simple move already halves the complication. Next step, we shall look only at E, confident that whatever we find will also apply to B. And, what we find is that E is the sum of two terms, which exhibit significant differences. The first term has a dependence on distance as 1/r: It is the radiation!! And, second remark, which is also not a casual fact: it is proportional to the charge acceleration a’ (“retarded”, of course). The other term has the usual 1/r2 dependence, and it is the static field attached to the charge. Advanced EM - Master in Physics 2011-2012 2 Another expression (Actually, more than two formulations exist…) for the fields exist, and is due to Feynman (FeynmanII 21-1; Lovitch-Rosati pp 399-400). It can be obtained by leaving unchanged the derivatives wrt t ( I mean, not resolving their indirect dependence on the “retarded” parameters), but calculating instead the gradient (derivatives wrt x, y, z). [ ] ε' r ' d ε' 1 d 2ε' E q 2 2 2 2 r' c dt r ' c dt B ε'E ( ) In this formulation, the acceleration of the charge is implicitly stated inside the term 2 1 d ε' c 2 dt 2 ε' Since is a unitary vector, its derivatives can only be orthogonal to its direction. But its direction is the direction towards the observer. Therefore, the electric field can only be orthogonal to the direction of propagation of the wave. This property is also evident from the first formulas given here: the two parts of the radiation term are both orthogonal to the direction of sight. That is where the transversity of the electromagnetic waves comes from. q[a'(a'ε' )ε' ] q ε'(a'β' ) q (ε'β' )(1 '2 ) E(r, t ) 2 3 c | r r' | (1 β'ε' ) | r r' |2 (1 β'ε' )3 Radiation term: acceleration fields Advanced EM - Master in Physics 2011-2012 Velocity fields 3 q[a'(a'ε' )ε' ] q ε'(a'β' ) q (ε'β' )(1 '2 ) E(r, t ) 2 3 c | r r' | (1 β'ε' ) | r r' |2 (1 β'ε' )3 On which basis was decided that one term was the radiation and the other an electrostatic type of field? It was decided on the basis of the dependence from the distance “r”: this is 1/r in one case and 1/r2 in the other. Note moreover that B being equal to the vector product of ε’ and E must be orthogonal to both. Another remark: E and B are orthogonal….. The wave parts of E and B, of course! For the other parts, customarily called the velocity fields, they can take other directions! (Just think of a light ray passing near a magnet….). In the derivation of the formulas we can also see the reason why, having the potentials a 1/r dependence, their derivatives had also a part with 1/r dependence instead of only 1/r2. The reason can be found in the form of the Lienard – Wiechert scalar potential: q r ' (1 β 'ε' ) In the case of an electrostatic potential, taking a derivative of this potential to obtain the fields only gives an 1/r2 dependence. But in this case taking derivatives wrt space gives two terms, one is the derivative of 1/r (which is 1/r2) and the other 1/r times the derivative of 1 / 1 β'ε' This part is dependent on the movement of the charge, its velocities and accelerations. Just imagine p.ex. a rotating charge which, seen from a side, has position, velocity and acceleration with sinusoidal behaviour. Advanced EM - Master in Physics 2011-2012 4 Main forms of the Lienard -Wiechert fields We have already seen the expressions by Schwartz and Feynman for the Lienard-Wiechert radiation electric field. They are repeated here, together with the form used by the book of Jackson, which is basically a simplification of Schwartz’s. Schwartz: q[a'(a'ε' )ε' ] q ε'(a'β' ) q (ε'β' )(1 '2 ) E(r, t ) 2 3 c | r r' | (1 β'ε' ) | r r' |2 (1 β'ε' )3 Jackson: E q ε'β' q '{( ε'β' ) β'} r '2 '2 (1 β'ε' )3 cr ' (1 β'ε' )3 Feynman [ ] ε' r ' d ε' 1 d 2ε' E q 2 2 2 2 r' c dt r ' c dt ( ) Advanced EM - Master in Physics 2011-2012 5 Velocity fields Velocity fields are the fields generated by a charge in motion without acceleration, i.e. a charge in uniform motion. I choose my reference system with the x axis coincident with the trajectory of the charge, and put the origin where the charge is at time 0. Then, x = vt is the equation of motion of my charge. I want to calculate the potentials and the fields generated –in point P and at time t- by the motion of that charge. Let’s call R the retarded point (i.e. the point where the charge was when it generated the fields which are seen in P at time t), and A the actual point, i.e. the point where the particle is at the time the potential and fields in P are calculated. I have the usual 2 equations for the retarded time (speed of light of the spheric light shell and position of the charge at the retarded time). r' c r '2 ( x vt' ) 2 y 2 z 2 c 2 (t t ' ) 2 t' t Solving this last equation in t’, we obtain… (1 2 )t ' t x 1 c c ( x vt) 2 (1 2 ) ( y 2 z 2 ) r ' c(t t ' ) Advanced EM - Master in Physics 2011-2012 6 We need to calculate the potential q r ' (1 β'ε' ) q r 'β'r' For that, we start to compute ' v2 r 'β'r' c(t t ' ) ' ( x vt' ) c[t x (1 2 )t ' ] c c And then, we would like to replace t’ with its value as a function of t. But we remember that we had just done that the previous page: (1 2 )t ' t x 1 c c ( x vt) 2 (1 2 ) ( y 2 z 2 ) So, we replace this in that, and arrive to: r'β'r' 1 2 ( x vt 1 2 )2 y 2 z2 Which we can immediately use for the expression of the potentials as functions of only the coordinates of the measuring point P, i.e. x, y, z and t; and NOT of the retarded time or position. ( x, y , z , t ) A β 1 2 ( q x vt 1 2 )2 y Advanced EM - Master in Physics 2011-2012 2 z2 7 In the derivation of this formula we have used only the EofM, but it is obvious that the equations we end up with do smell of Lorentz transformation, even if we have not studied it in detail: The ( x vt 1 2 ) expression really invites us to study the problem with the Lorentz transformation. In conclusion, we have calculated the Lienard-Wiechert potentials of a charge in uniform linear motion. If the motion is directed along the “x” axis, then A=(βΦ,0,0), and it is easy to compute the gradient, curl and time derivative of the potentials in order to find the fields, because in the final formulae the potentials are given as a function of x,y,z,and t, not as functions of x’, t’ and all that. The formulas for the electric field components are: Ex 1 Ax q ( x vt) 2 2 2 2 3 x c t { ( x vt) y z } q ( x vt) 2 ( x vt) 2 y 2 z 2 2 ( x vt) 2 y 2 z 2 Ey qy 2 2 2 2 3 y { ( x vt) y z } q y 2 ( x vt) 2 y 2 z 2 2 ( x vt) 2 y 2 z 2 Ez qz z { 2 ( x vt) 2 y 2 z 2 }3 Advanced EM - Master in Physics 2011-2012 8 Now, first thing to notice is that in the formulas the coordinates of P enter as (x-vt), y and z. And, these are the coordinates of P in a reference frame obtained by the previous one by displacement of the origin along the x axis to bring the origin to the point “A”. Second thing, the direction of the electric fields always point away from “A”: in fact E x x vt ; Ey y Ey Ez y z Therefore, if we make a drawing of the electric field, it seems to stem out of the “actual” position of the charge. But in reality, as we have seen, the fields were “generated” when the charge was in the position “R”, well before “A”. So the fields seem to be flowing out of the point “A”, and they move with the charge, being always centered on it. And here comes the third observation: the electric field – compared to the electrostatic one – has the component along the direction of motion a factor γ2 smaller than in the electrostatic case, while in the other two directions the fields are a factor of γ larger than in the electrostatic case. The field status for a (fast) particle is shown in the figures. Advanced EM - Master in Physics 2011-2012 9 Advanced EM - Master in Physics 2011-2012 10 Advanced EM - Master in Physics 2011-2012 11