Presentation is loading. Please wait.

Presentation is loading. Please wait.

Multipole expansion of the radiation fields

Similar presentations


Presentation on theme: "Multipole expansion of the radiation fields"— Presentation transcript:

1 Multipole expansion of the radiation fields
Consider an electrical system –charges and currents- varying in time. And, for ease of treatment, with both the charge and the current distributions with a sinusoidal (harmonic) time dependence: oscillating charges. Such limitations in our scope do not entail any loss of generality: any arbitrary time dependence – which satisfies the condition of charge conservation, of course - can be expanded into a Fourier series. Then the problem is solved (in principle) for any frequency; and the solutions are convoluted again with the original Fourier spectrum. Where, as usual, the physical values of the measurable quantities are the real parts of the complex variables. In such situation the scalar and vector potentials, as well as the E and B fields, will also have a harmonic dependence on time. We have already seen the general solution of the retarded potentials as a function of their sources, i.e. of ρ and J . In this case of a harmonic dependence on time they take the form: With the usual notation for the wave number k=ω/c Advanced EM - Master in Physics

2 Advanced EM - Master in Physics 2011-2012
The magnetic field is given by: and, in the region where there are neither charges nor currents, the 4th Maxwell law reads: From this equation we obtain: It is not necessary to calculate the scalar potential. And It is not necessary to know the charge density ρ. The problem of solving the previous 3 equations (for A, B and E) given a current distribution J(r’)·e(iωt) is mathematically analogous to the problem of electrostatics, for which a general solution has already been found in the multipole expansion. In these lectures we have not insisted on the most general solution (it can be found in the book by Jackson), but limited ourselves to the case of systems with symmetry for rotations around an axis. The are two differences wrt the electrostatic case: In electrostatics the function which had to be expanded in multipoles was , while in this case (quantities varying in time) it is : 2. In electrostatics we had to expand a scalar. Now, it is a vector. Advanced EM - Master in Physics

3 Advanced EM - Master in Physics 2011-2012
We will now also restrict somewhat the system under study: the system size must be much smaller than the oscillation wavelength. where λ is the wavelength of the emitted radiation and d is the “size” of the charge system. We can then divide the space free of charges (“outside of the charge region”) in three zones: is the proximity zone. is the intermediate zone is the remote (or radiation) region. Before diving into the calculation, a question: what does it mean the request that ? Well, even if this condition is not satisfied, it is always possible (in principle) to subdivide the charge region into smaller volumes where the condition is satisfied. A very common application of this treatment is in the field of antennas, especially of medium and long waves. A natural application is the emission of light from atoms and molecules (as long as it can be treated with classical theory). Visible light has wavelengths around 500 nm = 5·10-5 cm, the typical size of atoms is around 5·10-8 cm. Advanced EM - Master in Physics

4 Linear antenna fed at the center
A last remark: the conditions that both currents and charges oscillate at the same frequency does not imply that they do so with the same phase, which could raise problems with charge conservation. There can be of course a phase factor between the two. And, as a matter of fact, the charge conservation condition in this case is: Oscillations of charge and current are out of phase by An example: Linear antenna fed at the center The current is: Where the + (-) sign is applies to positive (negative) z. And the linear charge density: Advanced EM - Master in Physics

5 Advanced EM - Master in Physics 2011-2012
We return now to looking for the solutions of our “general” case. Solutions of the equation for the vector potential A(r), once the current density are known: In the nearest region, the fields are quasi-static, in the sense that we can approximate the exponential With the constant value of “1”. This is due to the fact that the distance is in that region much smaller than the wavelength. The potentials are the static potentials ( no measurable retardation effect), slowly varying in time. In the “radiation region” the exponential above oscillates rapidly, and determines the shape of the vector potential. Moreover, one can approximate the factor with 1/r and take it out from the integral over dV’. And, it is also possible to replace with If we define We then have the ideal function to approximate with a Taylor’s series And A(r) becomes a series expansion: in which each term Ai corresponds to the power ( n·r’)(i-1). Since kr’<<1 (remember that d/λ<<1!) the series converges rapidly. Advanced EM - Master in Physics

6 Advanced EM - Master in Physics 2011-2012
Let us now take A1 (corresponding to the constant value 1 in the expansion) We need to compute this integral. It can be written as a parts integral. This passage is a parts integral, see next page This integral is an old friend: it is the momentum of ELECTRIC DIPOLE of the charge density distribution. Note how it has become a constant, since it does not depend any longer on r or on t. The fields are obtained via the curls of the potential: the unitary vector n, defined in the previous page, appears, it is minus the direction of observation from point P. This term has a 1/r2 dependence on r: it is the velocity field, not interesting to us here. This is the field of radiation This is only the field of radiation part of the electric field Advanced EM - Master in Physics

7 Advanced EM - Master in Physics 2011-2012
Calculation of That requires the integration of a vector, i.e. three scalar integrals. Let us do the calculation for Jx only. In the last passage we have assigned to the surface integral the value “zero”. This is because the surface encompasses all the volume of charge and current and, of course, outside that volume the current distribution is null. We can extend to the three components of the vector potential what we have shown only for Jx , and then obtain: Advanced EM - Master in Physics

8 Advanced EM - Master in Physics 2011-2012
From the formulas of the previous page it is evident that the fields: Are orthogonal to each other Are orthogonal to the direction of propagation Have a 1/r dependence on distance |E|= |B| We write again the formula for the electric field: We do recognize here something very similar to the radiation electric field generated by a charge at rest that we had treated time ago, in the derivation of the Larmor formula: -q +q An OSCILLATING DIPOLE can be seen in its simplest mode as a single charge oscillating along an axis (z-axis here) . Another charge –q would reset the total charge of the system, and turn it into a better looking dipole. It has, of course, no influence on the radiation: the dipole field is generated by the oscillating charge, not by the fixed one. Advanced EM - Master in Physics

9 Advanced EM - Master in Physics 2011-2012
From the comparison of the two formulas for the dipole radiation of an electric field shown in the previous page we recognize that the radiation emitted by an oscillating dipole is the same as that generated by a charge oscillating with an acceleration equal to: where the primed quantities indicate that they are retarded. And, by looking at the figure of the previous page, we easily deduce that: We will remark explicitly that in the exact formulas for the fields also terms in r-2, r-3 …... are present. They belong of course to the fields of velocity. We shall not discuss them here any further. Dipole radiation Let us take a look at the formulas which give the electric dipole radiation fields. They determine the directions and sizes of the fields. Advanced EM - Master in Physics

10 Advanced EM - Master in Physics 2011-2012
What about the direction of the fields? Let us resort to the image of the dipole oscillating up and down in the Earth center: up and down means on the line connecting the two poles. And let us examine how the fields are directed along any spherical surface centered on the dipole. Of course they lay on the sphere surface, since they are perpendicular to the line of sight, which is also the sphere radius. The system in which these directions are along the coordinate axes is that of the spherical coordinates [r,θ,φ] which, however, is not easy to visualize – until we realize that it is the first system we have learnt – in primary school! Well then, the radiation of ELECTRIC PIPOLE has the electric field aligned with the meridians, and the magnetic field is aligned with the parallels. The radiation of MAGNETIC DIPOLE (we shall talk about it later) instead has the B field along the meridians, the electric field along the parallels. This picture shows the electric field as a function of θ. Note that its amplitude – and the amplitude of B as well, of course – is proportional to sinθ. The irradiated power – of the utmost importance in a world full of electromagnetic signals- is proportional to E2. Given that the electric dipole term is the first in a series expansion, it is obviously the term which will allow the largest power emitted per power spent in the emitting antenna. Advanced EM - Master in Physics

11 Advanced EM - Master in Physics 2011-2012
The emitted power can be easily computed with the Poynting’s vector or, simpler, with the Larmor formula, using the equivalence formula between charge acceleration and oscillating dipole moment given two pages ago. Where W is the power irradiated averaged over a full period τ=2π/ω. If we want to calculate directly the dipole momentum of the antenna shown in page 4: Where the “i” factor in the denominator indicates that the oscillation of the dipole is out of phase by 90º wrt the current in the antenna. The emitted power is –as a function of the current peak and of the physical dimensions of the radiating system (and for k·d<<1 !!!): We have so far only dealt with the first term in the series expansion of the electric field. The next term, A2, will be the sum of two contributions, the magnetic dipole and the electric quadrupole Advanced EM - Master in Physics


Download ppt "Multipole expansion of the radiation fields"

Similar presentations


Ads by Google