57 Radiating Systems, Multipole Fields and Radiation
In this lecture, we'll analyze the electromagnetic fields radiated by a compact time-varying charge and current distribution. The radiated field amplitudes decay with distance as The charges and currents also have "near field" contributions going as essentially the time-dependent version of the fields from static charges and steady currents, but here we'll only consider the far (radiated) fields.
(Note: In fact, the fields are relevant in the radiation zone (far away) for one thing: analyzing radiated angular momentum (which we’re not going to do here). A component in the direction of radiation will couple with a perpendicular component to give a Poynting vector perpendicular to the direction of radiation, but the angular momentum radiated has a factor from .)
The three relevant length scales are the size of the generating distribution, call it a typical wavelength of the radiation, say and the distance at which we're detecting the radiation, say We'll always assume
Since the problem is linear, it is enough to solve it for a single frequency of oscillation. As we shall see, that greatly simplifies the analysis. So (in the next section) we’ll take
But first, a brief review of the basic Green’s function we need (repeating to some extent material from lecture 39, hopefully worth refreshing at this point).
Formal Solution of the Wave Equations: the Green’s Function
Recall that in the Lorenz gauge
the equations for the scalar and vector potentials generated by a charge and current distribution are
To get the full picture of potentials generated by time-dependent charge and current distributions, we need the Green’s function with a delta function source in both space and time:
This is, with
and the integral over is just the delta function in time, that is,
The choice is the retarded Green’s function. It's geometrically equivalent to the expanding spherical pulse going radially outward from a flash of light at point at time
Solving with this Green’s function:
where the “ret” for retarded means measuring at the (earlier) time at which a light signal emitted from would reach at
Radiation at a Single Frequency
In general, these equations are difficult to solve (see lecture 39), because we need to know the detailed development of the current distribution at each point since we're adding the contributions with the time delay, and that varies from point to point. However, if the whole distribution is oscillating at the same frequency, all that's changing in time is the phase, and the phase of the signal from any point is just given by the distance from that point to the observer.
That is, for
the phase delay is just and writing
Furthermore, the vector potential is all we need in the radiation zone: we can drop since the magnetic field is given by and the electric field is with
Taking the origin to be in the compact region of currents and sources (shaded in above figure), for from the figure,
(We've replaces by in the denominator, a good approximation provided but we can't neglect the phase factor except in the case the wavelength much greater than the dimensions of the oscillating current, and we don't want to impose that extra restriction here.)
The natural way to approach the problem systematically is to expand the exponential:
The Leading Term: Electric Dipole
The radiation is in general dominated by the first term in the exponential expansion (although symmetries could make this term zero):
(Remember all this has an understood It's like doing the time-independent Schrodinger equation.)
This is called the "electric dipole term". Obviously, it's only nonzero if at some instant in time there is a net current in some direction, but then half a cycle later, it will be flowing the other way, so the charge is sloshing backwards and forwards. But that's what happens in an oscillating dipoleand it's easy the see they're mathematically equivalent:
where we used an integration by parts of the three separate vector components, followed by using Remember all terms have time dependence so the current and dipole are out of phase by as a simple picture of oscillating charge suggests.
In terms of the dipole, then, the radiation vector potential
The fields follow from with (this is the impedance of the vacuum, about 377 ohms. See the discussion of waveguide impedance in the previous lecture).
The only operates on the factor and we drop the (near field) term, giving
Exercise: draw a sketch! is always parallel to the dipole, so must be azimuthal, and is perpendicular to
The time averaged power per solid angle
(Remember are both going down as )
The total power is (using )
Notice this goes as the fourth power of the frequency!
Landau's Treatment of Dipole Radiation from a Very Compact Charge Distribution
Landau (Classical Theory of Fields, 4th edition, page 189) discusses the case of a radiating dipole charge distribution small in size compared to the wavelength of the radiation, so
and here we take independent of
We can sum over the charges in this system and write (following Landau, we use dots for time differentiation)
Now we need to find Here Landau exploits the fact that in the radiation zone, at some instant in time, only varies in the direction, it’s a function of so we have (Check this!) Furthermore, in fact only depends on in the combination so is equivalent to Visualize the wave moving outwards to see how this can be.
This is a general resultit is not at this point restricted to single frequency motion (but it does depend on the charge distribution size being small compared with the wavelength). The magnetic field only depends on the instantaneous acceleration of the dipole at the appropriate earlier time. As before, from Maxwell’s equations and the differential instantaneous radiation intensity
(This differs by a factor of two from the formula in the last section, that was an average for an oscillating field, this is the momentary intensity.)
The total power on integrating over solid angle is
Radiation from an Accelerating Particle
From the formula above, we can immediately find the power emitted by a single accelerating particle: imagine the dipole to consist of a fixed charge and a close by, but moving, opposite charge at so Then writing the acceleration of the moving charge we have
This is called the Larmor formula. We’ll discuss the important relativistic generalization later in the course.
A Simple Dipole Example: Short Center-Fed Linear Antenna
Note that Jackson's example here is a short antenna, This means that the charge distribution on each branch can be taken to be uniform, since the charge distribution equilibrates almost at the speed of light. (Not completely obvious, but true.)
For discussion and animations of a dipole antenna both receiving and transmitting, click here. Note that in this lecture, following Jackson, we are in the short dipole approximation.
Taking the antenna along the axis, the origin at its center, the current is the real part of
Charge conservation ensures the linear current decrease,
Hence the dipole moment
The total power radiated from a short center-fed linear dipole antenna is therefore
Recalling that we just found for an oscillating dipole of strength the radiation went as , this result is at first glance surprising. The point is that here there is a fixed input current, not a fixed dipole strength. From the charge conservation equation, at higher frequencies, meaning faster movement of the charge in the wires, less charge is needed to get the same current, so the dipole strength is down by a factor
The power produced in a DC resistance is Analogously, the radiation resistance of an antenna is defined as
The Next Order Term: Magnetic Dipole and Electric Quadrupole Radiation
The next term in the expansion is
(but we'll drop the higher order term in the bracket).
Using the vector identity
we can write the integrand as a sum of parts symmetric and antisymmetric in :
Following Jackson, we'll begin with the second term, the antisymmetric part, which turns out to be
Magnetic Dipole Radiation
You may recall from magnetostatics that the magnetic moment from a current in a loop of wire is
and the natural generalization to a continuum current distribution is
(In terms of particles, this would be )
Comparing with the equation for above, we see the antisymmetric current distribution contributes
This is magnetic dipole radiation. The magnetic field (to leading order) is
Exercise: Verify this.
Comparing Electric Dipole and Magnetic Dipole Fields
Recall now that the magnetic field from an electric dipole was and the electric field from an electric dipole is
So the electric dipole radiation and the magnetic dipole radiation are essentially duals of each other, the electric and magnetic fields being swapped. Of course, the units are different, and there's a sign difference in the corresponding Maxwell equations, so to get from the electric dipole to the magnetic takes
The angular power distribution in magnetic dipole radiation is precisely that from an electric dipole, losing the factor to correct for the different units of electric and magnetic dipoles,
The total power, using and total angle is is
Electric Quadrupole Radiation
Returning to the equation
we’ll now focus on the part symmetric in, the first term.
With 20-20 hindsight, we use the following identity:
Since the current distribution is compact, on integrating over all of space the left-hand side is zero.
Then taking the dot product of the result with the constant unit vector gives
This is the symmetric function we need.
Using charge conservation this symmetric part gives rise to the vector potential (in the radiation zone, dropping terms)
The fields in the radiation zone are
The quadrupole moment tensor is defined as
so the integral giving the magnetic field in the radiation zone can be written
defining the vector by The magnetic field in the radiation zone is
and the electric field is
The power radiated per unit solid angle in a given direction is , that is,
and using (prove them!)
we find the power to be
We should mention that this is sometimes written emphasizing that the only relevant behavior is the rate of change of acceleration at the retarded time.
To picture the radiation, we need some information about the quadrupole. All dipoles look alike they're a single charge plus its opposite slightly displaced, we can always take it along the axis, if we're only looking at one dipole. But a quadrupole is a dipole plus its opposite slightly displaced, and different directions of displacement, relative of the direction of the original dipole, give different charge distributions and different radiation patterns.
Let's look at some examples of for different directions of dipole displacement.
We begin with a dipole pointing along the axis, and displace it along the axis. Represent this by the charge distribution at and and at the origin. This gives the quadrupole tensor
We split it this way because the first (identity) tensor won't contribute to the radiation distribution
we'll just get
Another kind of quadrupole is the laterally displaced opposite dipole. This would give different radiation patterns if rotated about the axis shown, or about an axis through the center of the square perpendicular to the plane.
These are left as an exercise for the reader.