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Radiation Lecture 39: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay So far we have discussed propagation of electromagnetic waves both in free space and in waveguides without worrying about how they are produced. In the following two lectures we will discuss the production of electromagnetic waves. Once produced they carry their energy and momentum and propagate in free space, may be to be received by an antenna. An antenna is a physical device which transmits or receives electromagnetic waves. What causes a source to emit radiation? We have seen that charges at rest creates electrostatic field. A steady current, on the other hand is a source of magnetic field. Neither of these lead to radiation. However, it is known that accelerating charges emit radiation. To discuss this in some detail, we fall back on our workhorse, viz., the Maxwellโs equations. Let us recall the potential formulation of the electromagnetic field that we had discussed earlier. Using the expression for Faradayโs law, ๐ป × ๐ธโ = โ โ ๐๐ต ๐๐ก We rewrite this in terms of vector potential ๐ด, ๐ป × (๐ธโ + ๐๐ด )=0 ๐๐ก This enablesus to express the quantity within the bracket as gradient of a scalar function V, ๐ธโ = โ๐ป๐ โ ๐๐ด ๐๐ก Using Lorentz gauge condition, ๐ปโ ๐ด+ 1 ๐๐ =0 ๐ 2 ๐๐ก the Maxwellโs equations can be expressed as a pair of homogeneous wave equations, ๐ป2๐ โ 1 ๐2๐ ๐ =โ 2 2 ๐ ๐๐ก ๐ 1 (1) ๐ป2๐ด โ 1 ๐2๐ด = โ๐0 ๐ฝ ๐ 2 ๐๐ก 2 (2) Since the structure of both equations is same, we can solve either of them and write the solution to the remaining equation by similarity. We will solve the equation for the vector potential. The method of solution depends on defining a โGreenโs functionโ. Let us define a time Fourier transform through the equation, ๐ด(๐ฅ , ๐ก) = 1 โ โซ ๐ด(๐ฅ , ๐) ๐ ๐๐๐ก ๐๐ 2๐ โโ the inverse of which is โ ๐ด(๐ฅ , ๐) = โซ ๐ด(๐ฅ , ๐ก) ๐ โ๐๐๐ก ๐๐ก โโ Substituting in Eqn. (2) ๐ป2๐ด โ 1 ๐2๐ด 1 โ ๐2 2 = โซ + (๐ป ) ๐ด(๐ฅ , ๐) ๐ ๐๐๐ก ๐๐ ๐ 2 ๐๐ก 2 2๐ โโ ๐2 = โ๐0 ๐ฝ(๐ฅ , ๐ก) ๐0 โ =โ โซ ๐ฝ(๐ฅ , ๐) ๐ ๐๐๐ก ๐๐ 2๐ โโ Taking inverse Fourier transform for both sides, we get ๐2 (๐ป 2 + 2 ) ๐ด(๐ฅ , ๐) = โ๐0 ๐ฝ(๐ฅ , ๐) ๐ ๐ ๐ Defining ๐ = , we get, (๐ป 2 + ๐ 2 )๐ด(๐ฅ , ๐) = โ๐0 ๐ฝ(๐ฅ , ๐) To obtain the solution of this equation, we define a subsidiary equation for a function known as the Greenโs function โโโโฒ ) = โ4๐๐ฟ 3 (๐ฅ โ ๐ฅ โโโโฒ ) (๐ป 2 + ๐ 2 )๐บ(๐ฅ โ ๐ฅ Thus the Greenโs function satisfies a very similar equation in which the inhomogeneous term of the original equation has been replaced by a delta function. In terms of the Greenโs function the formal solution of the equation for the vector potential can be written as a convolution, ๐ด(๐ฅ , ๐) = ๐0 โโโโฒ , ๐)๐3 ๐ฅโฒ โซ ๐บ(๐ฅ โ โโโ ๐ฅ โฒ )๐ฝ(๐ฅ 4๐ To see how it comes about, 2 ๐0 โโโโฒ , ๐)๐3 ๐ฅ โฒ โซ (๐ป 2 + ๐ 2 )๐บ(๐ฅ โ โโโ ๐ฅ โฒ )๐ฝ(๐ฅ 4๐ ๐0 โโโโฒ , ๐)๐ 3 ๐ฅ โฒ = โ๐0 ๐ฝ(๐ฅ , ๐) = โซ โ 4๐๐ฟ 3 (๐ฅ โ โโโ ๐ฅ โฒ )๐ฝ(๐ฅ 4๐ (๐ป 2 + ๐ 2 )๐ด(๐ฅ, ๐) = Remember that in the above, the primed coordinate โโโ ๐ฅโฒ is for the source and the unprimed quantity refers to the point where the field is being evaluated. The solution must be spherically symmetric about the source point, i.e. depend on ๐ = |๐ฅ โ โโโ ๐ฅโฒ|. With this substitution, the equation for the Greenโs function becomes, (๐ป 2 + ๐ 2 )๐บ(๐ ) = โ4๐๐ฟ 3 (๐ ) First consider ๐ โ 0, for which the equation becomes (๐ป 2 + ๐ 2 )๐บ(๐ ) = 0 Expressing the Laplacian in spherical coordinates, as there is no dependence on angles, 1 ๐2 (๐ ๐บ) + ๐ 2 ๐บ = 0 ๐ ๐๐ 2 the solution for which is seen to be ๐ ๐บ~๐ ±๐๐๐ so that the Greenโs function, excepting at the origin has the solution as a linear combination of an outgoing wave and an incoming wave. ๐บ=๐ด ๐ ๐๐๐ ๐ โ๐๐๐ +๐ต ๐ ๐ What about the solution at the origin? Let us take a small sphere of radius ๐ 0 around the origin. Since the argument of the delta function is included within the sphere, we have, โซ (๐ป 2 + ๐ 2 )๐บ(๐ )๐3 ๐ = โ4๐ โซ ๐ฟ 3 (๐ )๐3 ๐ = โ4๐ ๐ 0 ๐ 0 Since ๐บ(๐ ) is singular only at the origin, we can calculate the integral over it within the sphere, by replacing it by the expression obtained for it except at the origin, ๐ 0 ๐ด ๐ต 1 2 ๐ 02 โซ ๐ 2 ๐บ(๐ )๐3 ๐ = โซ ๐ 2 ( + ) ๐3 ๐ = 4๐๐ 2 (๐ด + ๐ต) โซ ๐ ๐๐ = 4๐(๐ด + ๐ต)๐ 2 โ0 ๐ ๐ 2 0 ๐ as ๐ 0 โ 0. Thus, 3 โซ ๐ป 2 ๐บ(๐ )๐3 ๐ = โ4๐ ๐ 0 As ๐ 0 โ 0, ๐บ = ๐ด+๐ต , ๐ so that ๐ป 2 ๐บ(๐ ) = (๐ด + ๐ต) ๐ป 2 1 = โ4๐(๐ด + ๐ต)๐ฟ 3 (๐ ) ๐ Substituting this into the preceding integral, we get ๐ด+๐ต =1 Thus we have, ๐ โโโโโฒ | ๐๐|๐ฅโ๐ฅ ๐ฅโฒ| |๐ฅ โ โโโ ๐บ(๐ฅ โ โโโ ๐ฅโฒ) = ๐ outgoing wave โโโโโฒ | โ๐๐|๐ฅโ๐ฅ ๐ฅโฒ| { |๐ฅ โ โโโ incoming wave We will take the outgoing solution and proceed further. ๐ด(๐ฅ , ๐) = ๐0 โโโโฒ , ๐)๐3 ๐ฅ โฒ โซ ๐บ(๐ฅ โ โโโ ๐ฅ โฒ )๐ฝ(๐ฅ 4๐ โโโโโฒ | ๐๐|๐ฅ โ๐ฅ โ ๐0 ๐ โโโโฒ , ๐กโฒ)๐ ๐๐๐ก โฒ ๐๐กโฒ๐3 ๐ฅ โฒ = โซ โซ ๐ฝ(๐ฅ โฒ 4๐ |๐ฅ โ โโโ ๐ฅ | โโ Let us take the inverse Fourier transform of the above to get ๐ด(๐ฅ , ๐ก). It may, however, be remembered that ๐ = ๐ ๐ has ๐ dependence, ๐ด(๐ฅ , ๐ก) = โ โ ๐๐ 1 ๐0 ๐3 ๐ฅ โฒ โโโโโฒ โโโโฒ , ๐ก โฒ ) โซ ๐ ๐๐๐ก โฒ ๐ ๐ |๐ฅโ๐ฅ | ๐ โ๐๐๐ก ๐๐ โซ โซ ๐๐ก โฒ ๐ฝ(๐ฅ 2๐ 4๐ |๐ฅ โ โโโ ๐ฅ โฒ | โโ โโ = โ 1 ๐0 ๐3 ๐ฅ โฒ ๐ โโโโฒ , ๐ก โฒ )2๐๐ฟ (๐ก โฒ โ ๐ก + |๐ฅ โ โโโ โซ โซ ๐๐ก โฒ ๐ฝ(๐ฅ ๐ฅ โฒ |) โฒ โโโ 2๐ 4๐ |๐ฅ โ ๐ฅ | โโ ๐ = ๐0 ๐3 ๐ฅ โฒ ๐ โโโโฒ , ๐ก โ |๐ฅ โ โโโ โซ ๐ฝ (๐ฅ ๐ฅ โฒ |) โฒ 4๐ |๐ฅ โ โโโ ๐ ๐ฅ| Parallely, one can write, for the scalar potential 1 ๐3 ๐ฅ โฒ ๐ โโโโฒ , ๐ก โ |๐ฅ โ โโโ ๐(๐ฅ , ๐ก) = โซ ๐ (๐ฅ ๐ฅ โฒ |) 4๐๐0 |๐ฅ โ โโโ ๐ ๐ฅโฒ| 4 Note that the potential at time t at the position where it is being calculated is given in terms of current ๐ โโโโฒ |. The potential is influenced by the wave which came in the source at an โearlier timeโ ๐ก โ |๐ฅ โ ๐ฅ ๐ from the source at a time which is earlier by the time taken by the electromagnetic wave to travel this distance. This is known as the โretardedโ potential. The other solution is the incoming wave solution which would give to an โadvancedโ solution. Example : Localized oscillation of charge or current Let us consider a localized source of charge or current, given by โโโโฒ , ๐ก โฒ ) = ๐ฝ(๐ฅ โโโโฒ )๐ โ๐๐๐ก ๐ฝ(๐ฅ โฒ โโโโฒ , ๐ก โฒ ) = ๐(๐ฅ โโโโฒ )๐ โ๐๐๐ก โฒ ๐(๐ฅ ๐ด(๐ฅ , ๐ก) = = ๐0 ๐3 ๐ฅ โฒ ๐ โโโโฒ , ๐ก โ |๐ฅ โ โโโ โซ ๐ฝ (๐ฅ ๐ฅ โฒ |) โฒ 4๐ |๐ฅ โ โโโ ๐ ๐ฅ| ๐0 ๐3 ๐ฅ โฒ ๐ โโโโฒ ) exp (โ๐๐ (๐ก โ |๐ฅ โ โโโ โซ ๐ฝ(๐ฅ ๐ฅ โฒ |)) 4๐ |๐ฅ โ โโโ ๐ ๐ฅโฒ| โโโโโฒ | ๐๐|๐ฅ โ๐ฅ ๐0 โ๐๐๐ก ๐ โโโโฒ )๐3 ๐ฅโฒ = ๐ โซ ๐ฝ(๐ฅ โฒ โโโ 4๐ |๐ฅ โ ๐ฅ | Thus we have, โโโโโฒ | ๐๐|๐ฅ โ๐ฅ ๐0 ๐ โโโโฒ )๐3 ๐ฅโฒ ๐ด(๐ฅ ) = โซ ๐ฝ(๐ฅ 4๐ |๐ฅ โ โโโ ๐ฅโฒ| and, similarly, โโโโโฒ | ๐๐|๐ฅโ๐ฅ 1 ๐ โโโโฒ )๐3 ๐ฅโฒ ๐(๐ฅ ) = โซ ๐(๐ฅ 4๐๐0 |๐ฅ โ โโโ ๐ฅโฒ| ๐ where ๐ = ๐ . There are three regions of interest, assuming that the source is confined to a small region of space with typical dimension d. The distance at which the field is being measured is R= |๐ฅ โ โโโ ๐ฅ โฒ | and the wavelength of the radiation is ๐. 1. Near field for which ๐ โช ๐ โช ๐ 2. Intermediate zone for which ๐ โช ๐ โ ๐ 3. Radiation zone or Far field with ๐ โช ๐ โช ๐ 5 Our primary interest will be in the radiation zone, which we will discuss in detail in the next lecture. We will have very little to say about the intermediate zone and its analysis is very complicated. Consider the near field for which ๐ โช ๐ โช ๐. In this region ๐๐ = replace ๐ โโโโโฒ | ๐๐|๐ฅโ๐ฅ 2๐๐ ๐ โช 1 as a result of which we can โ 1. The expression for the vector potential becomes ๐ด(๐ฅ ) = ๐0 1 โโโโฒ )๐3 ๐ฅโฒ โซ ๐ฝ(๐ฅ 4๐ |๐ฅ โ โโโ ๐ฅโฒ| which is the familiar expression for the vector potential in magnetostatics. The only difference from the magnetostatic case is the time variation of ๐ด(๐ฅ , ๐ก) which is simply to multiply the expressions obtained in the static case by ๐ โ๐๐๐ก . Thus we can say that the situation is โquasi-staticโ and we can borrow all our results from magnetostatics to discuss this region. Radiation Lecture 39: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Tutorial Assignment 1. Consider a thought experiment in which a charge q appears abruptly at the origin at time ๐ก = ๐ก1 and disappears equally abruptly at time ๐ก = ๐ก2 . Find the retarded potentials for the problem. Determine the corresponding fields. 2. A long straight wire lying along the z axis carries a current given by 0 (๐ก โค 0) ๐ผ(๐ก) = { ๐ผ0 ๐ก (๐ก > 0) Determine the retarded potentials and the electric and the magnetic fields. Solutions to Tutorial Assignments ๐ฅ 1. Since the charge is static only retarded scalar potential is generated. For time ๐ก < ๐ก1 + ๐ the information that a charge has appeared at the origin has not reached the position x. Thus the potential is zero for such time. Likewise, though the charge has disappeared at time ๐ก2 , this ๐ฅ information would not reach the position x till a later time ๐ก2 + ๐ . Thus the potential is 6 1 ๐(๐ฅ ) = 4๐๐ ๐ 4๐๐0 ๐ 2 ๐ 0๐ฅ ๐ฅ for ๐ก1 < ๐ก โ ๐ < ๐ก2 and zero at other times. The corresponding field is ๐ธโ = ๐ฬ in the same time interval and is zero at other times. 2. The wire being charge neutral, the scalar potential is zero. The vector potential, because of cylindrical symmetry depends only on the distance s of the point of observation P from the wire, and is given by ๐0 ๐ผ(๐ก โฒ ) ๐ด(๐ , ๐ก) = ๐งฬ โซ ๐๐ง 4๐ ๐ r z O P s Consider an element dz at a distance z form the origin (see figure). Information about the ๐ current element would reach the point P at a time ๐ = โ๐ง 2 +๐ 2 ๐ later than the time the current was actually established. Thus the current, as seen by the point P is given by ๐ผ(๐ก โฒ ) = { ๐ ๐ where ๐ก โฒ = ๐ก โ = ๐ก โ โ๐ง 2 +๐ 2 . ๐ 0 (๐ก โฒ โค 0) ๐ผ0 ๐ก โฒ (๐ก โฒ > 0) Now let us look at the range of integration. For an element dz located at a distance z from the origin to influence the field at the point P, its distance from P should be less than ct. Thus, โ๐ง 2 + ๐ 2 < ๐๐ก . Hence the potential is given by โ๐ ๐0 ๐ด(๐ , ๐ก) = 2 ๐ผ0 ๐งฬ โซ 4๐ 0 2 ๐ก 2 โ๐ 2 โ๐ง 2 +๐ 2 ๐ ๐๐ง โ๐ง 2 + ๐ 2 ๐กโ (the factor of 2 in front is because only the upper half of the wire is considered in the integration). The integral is straightforward and gives, ๐ด(๐ , ๐ก) = ๐0 ๐๐ก + โ๐ 2 ๐ก 2 โ ๐ 2 โ๐ 2 ๐ก 2 โ ๐ 2 ๐ผ0 ๐งฬ [๐ก ln ( ] )โ 2๐ ๐ ๐ 7 The electric field is given by ๐ธโ = โ ๐๐ด ๐0 ๐๐ก + โ๐ 2 ๐ก 2 โ ๐ 2 = โ ๐ผ0 ๐งฬ ln ( ) ๐๐ก 2๐ ๐ and the magnetic field is given by โ =โ×๐ด= ๐ต ๐๐ด๐ง ๐0 ๐ฬ = ๐ผ โ๐ 2 ๐ก 2 โ ๐ 2 ฬ ๐ ๐๐ 2๐๐ ๐ 0 Radiation Lecture 39: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Self Assessment Questions 1. An long straight wire lies along the z direction. At time ๐ก = 0, a constant current ๐ผ0 is switched on. Calculate the vector potential and the electric and the magnetic fields. 2. A conducting wire is bent into a loop of two semicircles, one of radius ๐ 1 and the other ๐ 2 , (< ๐ 1 ) connected by straight segments. The loop carries a current ๐ผ(๐ก) = ๐ผ0 ๐ก, which is directed anticlockwise on the shorter semicircle and clockwise on the other one.. Calculate the vector potential at the centre of the loop. Calculate the electric field at the centre. Can you calculate the magnetic field from the expression of the vector potential? ๐ 1 ๐ 2 8 Solutions to Self Assessment Questions 1. The problem is very similar to Problem 2 of the Tutorial. The vector potential is given by ๐ด(๐ , ๐ก) = ๐0 ๐๐ก + โ๐ 2 ๐ก 2 โ ๐ 2 ๐ผ0 ๐งฬ ln ( ) 2๐ ๐ 2. y ๐ 1 ๐ 2 O x Since the current is switched on at t=0, the (retarded) vector potential at the centre O is given by ๐ ๐0 ๐ผ๐ (๐ก โ ๐ ) ๐0 ๐ผ๐ 1 ๐0 ๐ผ๐ ๐ด= โซ ๐๐ = ๐กโซ ๐๐โ โ โซ ๐๐โ 4๐ ๐ 4๐ ๐ 4๐๐ Since โฎ ๐๐ = 0, the second integral vanishes. We are left with ๐ด = ๐0 ๐ผ๐ 1 ๐กโซ ๐ ๐๐โ . 4๐ Note that the integral is a line integral over the loop, which is directed differently at different points of the loop. Let us take the loop to be in the x-y plane with directions as shown in the figure. Recall that the direction of ๐๐โ is along the direction of the current. The contribution from the two straight sections give โ ๐0 ๐ผ๐ ๐ ๐ก ln ๐ 2 ๐ฬ, 4๐ 1 the negative sign is due to the fact that the current in both the sections are directed along the negative x direction. We now have to perform the integration over the two semicircles. Consider the integration over the bigger semicircle. We can resolve ๐๐ = ๐ 1 ๐๐(cos ๐๐ฬ + sin ๐๐ฬ) . The denominator of the integral also has ๐ 1 so that the integral becomes independent of ๐ 1 (In a similar way, the integral of the smaller semicircle will be independent of ๐ 2 ). The y-component of the integral would vanish by symmetrically placed elements on either side of the y-axis, leaving us with ๐0 ๐ผ0 ๐ก ๐ฬ 4๐ as the value of the integral. The integral over the smaller semicircle is done in a similar way but its direction would be opposite because of the direction of the current. Thus the 9 contribution from the two semicircles would cancel and the vector potential at the centre would be given by โ ๐0 ๐ผ๐ ๐ ๐ก ln ๐ 2 ๐ฬ 4๐ 1 . The electric field is trivial and is given by ๐ธโ = โ ๐๐ด ๐0 ๐ 1 = ๐ผ0 ln ๐ฬ ๐๐ก 4๐ ๐ 2 Since the vector potential has been calculated only at a single point, it is not possible to calculate its curl and hence the magnetic field cannot be determined by the above calculation. 10