An antenna
is a device for converting
electromagnetic radiation into
electrical currents or vice-versa, depending on whether it is being
used for receiving or for transmitting. In passive radio
astronomy we use antennas for receiving. However, it is often easier to
calculate the properties of transmitting antennas. Fortunately, most
characteristics of an antenna acting as a transmitter are unchanged if
the antenna is used for receiving, so we often use the analysis of
transmitting antennas to understand the receiving antennas used in
radio astronomy.
The simplest antenna is a short
(meaning much smaller than one
wavelength) dipole
antenna
which is shown above as two colinear conductors
(e.g., wires) driven at the gap between them by a current source
(a transmitter). The properties of an antenna may depend on the
frequency $\nu$ or wavelength $\lambda$, so it is simplest to suppose
the driving current $I$ varies sinusoidally
with
angular frequency $\omega = 2 \pi \nu$:
$$I = I_0 \cos(\omega t)$$
where $I_0$ is the peak current driven into each element of the
dipole. It is
computationally convenient to replace the trigonometric function
$\cos(\omega t)$ with its exponential equivalent, the real part of
$$e^{-i \omega t} = \cos(\omega t) - i \sin(\omega t)$$
so the driving
current can be rewritten as
$$I = I_0 e^{-i \omega t}~.$$
The driving current accelerates charges in the antenna, so we can use
Larmor's formula to calculate the radiation from the antenna by
converting from the language of charges and acclerations to
time-varying currents.
Recall that
current
is defined as the time derivative of charge:
$$I
\equiv {d q \over d t}~.$$
Along a wire, the current is the amount of
charge flowing past any point per unit time. For a wire on the $z$-axis
$$I = {d q \over d t} = {d q \over d z} {d z \over d t} = {d q \over d
z} v~,$$
where $v$ is the instantaneous flow velocity of the charges.
Note that $v \ll c$ normally; it is a
common misconception to believe $v \approx c$ because electrical
signals travel down wires at nearly the speed of light. This is
similar to what happens in a garden hose already filled with an
incompressible fluid—water. When the faucet is turned on, water
flows from the other end of the hose almost immediately, even though
individual water molecules are moving slowly along the hose. As
an example, consider a current of 1 ampere flowing through a copper
wire with cross section $\sigma = 1$ mm$^2$. The number density
of free electrons in the wire is about equal to the number density of
copper
atoms, $n \approx 10^{29}$ m$^{-3}$. In mks units, the charge of
an electron is
$$-e \approx 4.80 \times 10^{-12} {\rm ~statcoul}
\times {1 {\rm ~coul} \over 3 \times 10^9 {\rm ~statcoul}}
\approx 1.60 \times 10^{-19} {\rm ~coul}$$
One Ampere is one Coulomb per second, so the number $N$ of electrons
flowing past any point along the wire in one second is
$$N = {I \over \vert e \vert} \approx
{1 {\rm ~coul~s}^{-1} \over 1.60 \times 10^{-19}{\rm ~coul}}
\approx 6.25 \times 10^{18}{\rm ~s}^{-1}$$
The average electron velocity is only
$$ v \approx {N \over \sigma \rho}
\approx { 6.25 \times 10^{18}{\rm ~s}^{-1} \over 10^{-6}{\rm ~m}^2
\times 10^{29}{\rm ~m}^{-3}} \approx 6 \times 10^{-5} {\rm ~m~s}^{-1}
\ll c$$
Thus the nonrelativistic Larmor equation can be used to calculate the
radiation from a wire.
From the derivation of Larmor's
formula, recall that
$$E_\bot = {q \dot{v}
\sin\theta \over r c^2}$$
so in the short dipole
$$E_\bot = \int_{z =
-l/2}^{+l/2} {d q \over d z} d z {\dot{v} \sin\theta \over r c^2}$$
For
a sinusoidal driving current,
$$\dot{v} = -i \omega v$$
and
$$E_\bot =
{-i \omega \sin\theta \over r c^2} \int_{-l/2}^{+l/2} {d q \over d z} v
dz$$
$$E_\bot = {-i \omega \sin\theta \over r c^2} \int_{-l/2}^{+l/2} I
d z$$
That is, the radiated electric field strength $E_\bot$ is proportional
to
the integral of the current distribution along the antenna. The current
at the center is just the driving current $I = I_0 e^{-i \omega t}$ and
the current must drop to zero at the ends of the antenna, where the
conductivity goes to zero. For a "short"
antenna, meaning $l \ll \lambda$, the current drop from the center to
the end can be approximated by the linear function satisfying those end
conditions:
$$I(z) = I_0 e^{- i \omega t} \biggl[1 - {\vert z \vert
\over (l / 2)} \biggr]$$
Then
$$\int_{-l/2}^{+l/2} I dz = {I_0 l \over
2}$$
and
$$E_\bot = {-i \omega \sin\theta \over r c^2} {I_0 l \over 2}
e^{-i \omega t}$$
Substituting $\omega = 2 \pi c / \lambda$ gives
$$E_\bot = {-i 2 \pi c \sin\theta I_0 l \over \lambda r c^2 2} e^{-i
\omega t}$$
$$E_\bot = {-i \pi \sin\theta \over c} I_0 {l \over
\lambda} {e^{ -i \omega t} \over r}$$
Since $\vert\vec{E_\bot}\vert =
\vert\vec{H_\bot}\vert$ (cgs), the time-averaged Poynting flux (power
per unit area) is
$$\langle{S}\rangle = {c \over 4 \pi} \langle E_\bot^2 \rangle ,$$
and
$$\langle{S}\rangle = {c \over 4 \pi} \biggl({I_0 l \over \lambda} \,
{\pi \over c} \biggr)^2 {\sin^2\theta \over r^2} \biggl( {1 \over
2}\biggr)$$
because $\langle cos^2 (\omega t) \rangle = 1/2$. Note that
the radiation from a short dipole has the same polarization and the
same doughnut power
pattern (the power pattern is the angular distribution of
radiated power, often normalized to unity at the peak)
$$\bbox[border:3px blue solid,7pt]{P \propto \sin^2\theta}\rlap{\quad
\rm {(3A1)}}$$
as radiation from an accelerated charge
because all of the charges in the dipole are being accelerated along
one short line. From the observer's point of view, the power
received depends only on
the projected (perpendicular
to the line of sight) length $(l \sin\theta)$ of the dipole, another
example of "what
you see is what you get."
The time-averaged total power emitted
is obtained by integrating the
Poynting flux over the surface area of a sphere of any radius $r \gg l$
centered on the antenna:
$$\langle P \rangle =
\int\langle S \rangle dA = {c \over 4 \pi} \biggl({I_0 l \over \lambda}
\, {\pi \over c} \biggr)^2 \biggl({1 \over 2}\biggr) \int_{\phi =
0}^{2\pi} \int_{\theta = 0}^\pi {\sin^2\theta \over r^2} r \sin\theta d
\phi r d\theta$$
$$\langle P \rangle = {c \over 4 \pi} \biggl({I_0 l
\over \lambda} \, {\pi \over c} \biggr)^2 \biggl({1 \over 2}\biggr) 2
\pi \int_{\theta = 0}^\pi \sin^3\theta d\theta$$
Recall that
$\int_0^\pi \sin^3\theta d\theta = 4/3$, so the time-averaged power
radiated by a short dipole is
$$\bbox[border:3px blue solid,7pt]{\langle P \rangle = {\pi^2 \over 3
c} \biggl({I_0 l \over
\lambda}\biggr)^2 }\rlap{\quad \rm {(3A2)}}$$
where $I_0 \cos (\omega t)$ is the driving current, $l$ is the total
length of the dipole, and $\lambda$ is the wavelength.
Radiation Resistance
The power flowing through a circuit is
$P = V \times I$, where $V$ is the voltage (defined as energy per unit
charge) and $I$ is the current (defined as charge flow per unit time),
so $P$ has units of energy per unit time. The physicist George
Simon Ohm observed that the current flowing through most resistive
materials is proportional to the applied voltage, so many objects have
a well-defined resistance defined
by $R = V / I$ (Ohm's
law). For them, $P = V \times I = I^2 R =
V^2 / R$.
From Ohm's law for time-varying
currents,
$$\langle P \rangle = \langle I^2 \rangle R$$
If $I = I_0 \cos(\omega t)$,
$$\langle P \rangle = {I_0^2 R \over 2}$$
The radiation
resistance of
an antenna is defined by
$$\bbox[border:3px blue solid,7pt]{R \equiv {2 \langle P \rangle \over
I_0^2}}\rlap{\quad \rm {(3A3)}}$$
For a short dipole, the radiation resistance is
$$R = {2 \pi^2 \over 3 c} \biggl( {l \over \lambda} \biggr)^2$$
Example: A "half-wave" dipole has
length $l = \lambda / 2$. This is the length of a resonant dipole
antenna. Resonant antennas are used in most real applications
because the impedance of a resonant antenna is resistive; nonresonant
antennas have large capacitive or inductive reactances as well. A
half-wave dipole doesn't strictly satisfy our criterion $l \ll \lambda$
for
being "short," so the current distribution along the
dipole is actually closer to sinusoidal than linear and our calculated
radiation resistance will
not be exact. Proceeding nonetheless,
$$R \approx {2 \pi^2 \over 3\times 3\times 10^{10}{\rm ~cm~s}
^{-1}} \biggl({1 \over 2}\biggr)^2 \approx 5.5 \times 10^{-11}{\rm
~s~cm}^{-1}$$
Engineers and real test instruments use the mks "Ohm" (symbol $\Omega$)
as the unit of
resistance. The conversion factor is 1 Ohm = $(10^{-11} /
9)$ s cm$^{-1}$, so
$$R \approx 5.5 \times 10^{-11}{{\rm s} \over {\rm cm}} {9 {\rm ~cm}
\over
10^{-11} {\rm ~s}} ~\Omega \approx 50 ~\Omega$$
[The actual radiation resistance of a half-wave dipole is about 73
$\Omega$.]
The radiation resistance $R_0$ of free
space can be obtained from the relations
$$\vert \vec{S} \vert = {c \over 4 \pi} E^2 {\rm ~\quad and \quad~} P =
{V^2 \over R}~.$$
Since the electric field $E$ is just the voltage per unit length $V /
l$ and the flux is the power per unit area $l^2$,
$$\vert \vec{S} \vert = { c V^2 \over 4 \pi l^2} = {V^2 \over 2 R_0
l^2}$$
$$ R_0 = {4 \pi \over c} = {4 \pi \over 3 \times 10^{10} {\rm
~cm~s}^{-1}}$$
Converting to mks units yields the resistance in Ohms:
$$ R_0 = {4 \pi \over 3 \times 10^{10} {\rm ~cm~s}^{-1} \times
{1/9}\times 10^{-11} {\rm ~sec~cm}^{-1}~\Omega^{-1}} = 120 \pi ~\Omega
\approx 377 ~\Omega~.$$
Since a black hole is a perfect
absorber of radiation, its impedance must also be $120 \pi ~\Omega$ to
match that of free space. A black hole spinning in an external
magnetic field can generate electrical power with a voltage/current
ratio of $120 \pi ~\Omega$, and this process may be important in
powering quasars (see Blandford & Znajek 1977, MNRAS, 179, 433).
The Power Gain of a Transmitting Antenna
The power
gain $G(\theta,\phi)$ of a
transmitting antenna is defined as the power transmitted per unit solid
angle divided by the power transmitted
per unit solid angle from an isotropic antenna driven with equal total
power. Frequently the value of $G$ is expressed logarithmically in
units of decibels (dB):
$$G({\rm dB}) \equiv 10 \times \log_{10}(G)$$
For a lossless antenna, energy
conservation requires that the gain averaged over all directions be
$$\langle G \rangle
\equiv {\int_{\rm sphere} G d \Omega \over \int_{\rm sphere
} d \Omega} = {4 \pi \over 4 \pi}$$ or
$$\bbox[border:3px blue solid,7pt]{\langle G \rangle = 1}\rlap{\quad
\rm {(3A4)}}$$
Consequently
$$ \int_{\rm
sphere} G d \Omega = \int_{\rm sphere} 1 d \Omega = 4 \pi~$$
for any
antenna. Different lossless antennas may radiate with different
directional patterns but
cannot alter the total amount of power radiated. Consequently, the gain
of a lossless
antenna depends only on the angular distribution of radiation from that
antenna. In general, an antenna having peak gain $G_{\rm max}$ must
beam most of its
power into a solid angle $\Delta \Omega$ such that
$$\Delta \Omega \approx {4
\pi \over G
_{\rm max}}~.$$
Thus the higher the gain, the narrow the beam or power
pattern.
Example: What is the power gain of a
short dipole? It is sufficient to recall only the angular dependence of
the short-dipole power pattern
$$\langle S \rangle \propto \sin^2\theta~,$$
where $\theta$ is the angle from the dipole axis. Thus
$$G \propto \sin^2\theta = G_0
\sin^2\theta~.$$
The maximum gain $G_0$ is determined by energy
conservation:
$$\int_{\rm sphere}
G d\Omega = \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi G_0
\sin^2\theta
d\theta \sin\theta d\phi = 4 \pi~.$$
$$2 \pi G_0 \int_0^\pi \sin^3\theta d \theta =
4 \pi ~.$$
Recall that $\int_0^\pi \sin^3\theta d \theta = 4/3$ so
$$G_0 = {4 \pi
\over 2 \pi}{3 \over 4} = {3 \over 2}$$
and
$$G(\theta, \phi) = {3
\sin^2\theta \over 2}
~.$$
In dB, the maximum gain $G_0$ of a short dipole is
$$G_0 = 10 \log_{10}(3/2) \approx 1.76{\rm ~dB}.$$
Note that $G(\theta, \phi)$ is
independent of the antenna length in
wavelengths so long as $l \ll \lambda$ because the short-dipole power
pattern is independent of $l$. Varying $l \ll \lambda$
affects only
the radiation resistance.
The Effective Area of a Receiving
Antenna
How can we characterize antennas used
for receiving, as in radio astronomy, rather than for transmitting? The
receiving counterpart of transmitting power gain is the effective
area
or effective
collecting area
of an antenna.
Imagine an ideal antenna that collects
all of the radiation falling on it from a distant point source and
converts it to electrical power—a "rain gauge" for
collecting photons. The total spectral power that it collects
will be the
product of its geometric area $A$ and the incident spectral power per
unit area, or flux density $S$. By analogy, if any real antenna
collects spectral power $P_\nu$, its
effective area $A_{\rm e}$ is defined by
$$\bbox[border:3px blue solid,7pt]{ A_{\rm e} \equiv {P_\nu \over
S_{\rm
(matched)}}}\rlap{\quad \rm {(3A5)}}$$
where $S_{\rm (matched)}$ is the flux density in the "matched"
polarization.
What does matched polarization mean? Any electromagnetic wave can be decomposed into two orthogonal polarized components. For example, the transverse electric field can be resolved horizontal and vertical components, or horizontal and vertical linear polarizations. If the horizontal and vertical electric fields are equal in amplitude and $90^\circ$ out of phase, the radiation is circularly polarized. Any radio wave can also be decomposed into left- and right-circular polarizations. If the wave is essentially random (noise generated by blackbody radiation for example), the two orthogonal components will vary rapidly in intensity but have equal powers when averaged over long times. Such radiation is called unpolarized. Blackbody radiation is unpolarized. Most radio astronomical sources are unpolarized or nearly so.
Any antenna with a single output
collects only one of the two polarizations from an electromagnetic
wave. For example, a linear dipole antenna
collects radiation only from the linear polarization whose electric
field is
parallel to the antenna wires. Electric fields perpendicular to the
antenna wires do not produce currents in the antenna, so the linear
dipole is completely
insensitive to the linear polarization perpendicular to the
wires. An pair of crossed dipoles would be needed to collect
power from both orthogonal polarizations simultaneously.
Thus for an unpolarized
source,
$$S_{\rm (matched)} = {S \over 2} ~.$$
Just as energy conservation implies
that all lossless transmitting
antennas have the same average power gain, all lossless receiving
antennas have the same average collecting area. This average
collecting area can be calculated via another thermodynamic thought
experiment.
Imagine an antenna inside a cavity in
full thermodynamic equilibrium at temperature $T$ connected through a
transmission line to a matched resistor
in a second cavity at the same temperature. Suppose further that the
connection
contains a filter that passes only a narrow range of frequencies
between $\nu$ and $\nu +
d \nu$. Since this entire system is in thermodynamic equilibrium, no
net power can
flow between the antenna and the resistor. Otherwise, one cavity would
heat up and the other would cool down, in violation of the second law
of thermodynamics. Thus the
total spectral power from all directions collected by the antenna
$$P_\nu = A_{\rm e} S_{\rm
(matched)} = {S \over 2} = \int_{4\pi} A_{\rm e}(\theta,\phi) {B_\nu
\over 2} d\Omega$$
must equal the Nyquist spectral power $P_\nu = kT$ produced by the
resistor. Using the Rayleigh-Jeans approximation
$$B_\nu = {2 k
T \over \lambda^2}$$
gives
$$P_\nu = {2 k T \over 2 \lambda^2} \int_{4
\pi} A_{\rm e} d \Omega = k T $$
$$\int_{4 \pi} A_{\rm e} d \Omega =
\lambda^2$$
The average collecting area is defined by
$$\langle A_{\rm e} \rangle \equiv {\int_{4
\pi} A_{
\rm e} d \Omega \over \int_{4 \pi} d \Omega }~.$$
The effective collecting area of a
receiving antenna is independent of its environment, so this
result applies for any type of radiation, not just blackbody
radiation. Without using
Maxwell's equations we have obtained the remarkable result true for
all
lossless antennas:
$$\bbox[border:3px blue solid,7pt]{\langle A_{\rm e} \rangle =
{\lambda^2 \over 4 \pi}}\rlap{\quad \rm {(3A6)}}$$
In the case of an isotropic antenna,
the effective collecting area in any direction equals the average
collecting area:
$$A_{\rm e}(\theta, \phi) = \langle A_{\rm e} \rangle = {\lambda^2
\over 4 \pi}$$
Consequently, a nondirectional antenna operating at a very short
wavelength $\lambda$ will have a very small effective
collecting area and poor sensitivity for reception. For this reason,
operators of satellite broadcast services, the new satellite FM radio
service for example,
prefer to transmit at relatively long wavelengths (10 to 20
cm) and
are much less interested in competing with radio astronomers for the
short-wavelength end of the radio spectrum. Likewise, practical radio
telescopes
can be constructed from arrays of dipoles at long wavelengths ($\lambda
> 1$ m), but not at
short wavelengths where the number of dipoles needed to produce useful
collecting
areas is too large.
Reciprocity Theorems
Many antenna properties are the same for both transmitting and receiving. It is often easier to calculate the gain of a transmitting antenna than the collecting area of a receiving antenna, and it is often easier to measure the receiving power pattern than to measure the transmitting power pattern of a large radio telescope. Thus this receiving/transmitting "reciprocity" greatly simplifies antenna calculations and measurements. Reciprocity can be understood via Maxwell's equations or by thermodynamic arguments.
Burke & Smith (1997) state the electromagnetic case for reciprocity clearly: "An antenna can be treated either as a receiving device, gathering the incoming radiation field and conducting electrical signals to the output terminals, or as a transmitting system, launching electromagnetic waves outward. These two cases are equivalent because of time reversibility: the solutions of Maxwell's equations are valid when time is reversed."
The
strong reciprocity theorem:
If a voltage is
applied to the
terminals of an antenna A and the current is measured at the terminals
of another antenna B, then an equal current (in both amplitude and
phase) will appear at the terminals of A if the same
voltage is applied to B.
can be formally derived from Maxwell's
equations [see a partial derivation in Rohlfs & Wilson Sections
5.4] or by
network analysis [see Kraus "Antennas", p. 252].
The
strong reciprocity theorem implies that the transmitter voltages
$V_{\rm
A}$ and $V_{\rm B}$ are related to the receiver currents $I_{\rm A}$
and $I_{\rm B}$ by
$${V_{\rm A} \over I_{\rm B}} =
{V_{\rm B} \over I_{\rm A}}$$
for any pair of antennas $A$ and $B$.
For most radio astronomical applications, we are not concerned with the detailed phase relationships of voltages and currents, and we can use a weak reciprocity theorem that relates the angular dependences of the transmitting power pattern and the receiving collecting area of any antenna:
The power pattern of an antenna is the same for transmitting and receiving.
That is:
$$\bbox[border:3px blue solid,7pt]{G(\theta, \phi) \propto
A_{\rm e}(\theta, \phi)}\rlap{\quad \rm {(3A7)}}$$
The weak reciprocity theorem can be proven by another simple thermodynamic thought experiment:
An antenna is connected to a matched
load inside a cavity initially in equilibrium at temperature $T$. The
antenna simultaneously receives power
from the cavity walls and transmits power generated by the resistor.
The total
power transmitted in all directions must equal the total power received
from all
directions since no net power can be transferred between the antenna
and the
resistor; otherwise the resistor would not remain at temperature $T$.
Moreover, in any direction, the power received and
transmitted by the
antenna must be the same, else the cavity wall in directions where the
transmitted
power was greater than the received power would rise in temperature and
the
cavity wall in directions of lower transmitted/received power ratio
would cool, leading to a violation of the second law of thermodynamics.
The constant of proportionality
relating $G$ and $A_{\rm e}$ can be
derived from our earlier results for an isotropic antenna
$$\langle A_{\rm e} \rangle = {\lambda^2 \over 4 \pi} {\rm
\qquad and \qquad} \langle G \rangle = 1$$
Thus energy conservation and the weak
reciprocity
theorem imply
$$\bbox[border:3px blue solid,7pt]{A_{\rm e}(\theta, \phi) = {\lambda^2
G(\theta, \phi) \over 4 \pi}}\rlap{\quad \rm {(3A8)}}$$
for any antenna. This extremely useful equation lets us compute the
receiving power
pattern from the transmitting power
pattern and vice versa.
Example: We can use our calculation
of the transmitting power pattern of a short dipole to calculate its
effective collecting area when used as a
receiving antenna:
$$A_{\rm e}(\theta,\phi) = {\lambda^2
G(\theta,\phi) \over 4 \pi} = {\lambda^2 \over 4 \pi} {3 \sin^2\theta
\over 2}$$
$$A_{\rm e} = {3 \lambda^2
\sin^2 \theta \over 8 \pi}$$
Example: What is the power per unit bandwidth $P_\nu$ collected by a short dipole at $\nu = 10$ GHz broadside to ($\theta = 90^\circ$) the Sun, a thermal source whose flux density is $S \approx 1.2 \times 10^6$ Jy?
The broadside ($\sin\theta = 1$)
collecting area of the short dipole is
$$A_{\rm e} = {3 \lambda^2 \over 8 \pi}$$
$$\lambda = {c \over \nu} = {3
\times 10^8 {\rm ~m~s}^{-1} \over 10^{10} {\rm ~Hz}} = 0.03{\rm ~m}$$
so
$$A_{\rm e}
= {3 \times (0.03{\rm ~m})^2 \over 8 \pi} = 1.07 \times 10^{-4}{\rm
~m}^2
\approx 1{\rm ~cm}^2~.$$
Clearly, dipoles or other nearly isotropic
antennas have very small collecting areas at short wavelengths.
For an unpolarized source,
$$P_\nu = A_{\rm e} S_{\rm (matched)}
= {A_{\rm e} S \over 2}$$ $$P_\nu = 1.07 \times 10^{-4} {\rm ~m}^2
\times {1.2 \times 10^6 {\rm ~Jy} \over 2}
\times {10^{-26} {\rm ~W~m}^{-2}{\rm ~Hz}^{-1} \over 1{\rm ~Jy}}$$
$$P_\nu = 6.4
\times 10^{-25} {\rm ~W~Hz}^{-1}$$
Arrays of dipoles do make sense at
long wavelengths. For example, the Long Wavelength Array (LWA)
for $\nu = 20$ to 80 MHz ($\lambda \approx 4$ to 15 m) will consist of
53 stations, each a 100 m $\times$ 100 m array of crossed
dipoles. The effective collecting area of the LWA will be up to
$A_{\rm e} =
4 \times 10^6$ m$^2$ at $\lambda = 15$ m.
Antenna Temperature
A convenient practical unit for the
power per unit frequency output of a receiving antenna is the antenna
temperature $T_{\rm A}$.
Antenna
temperature has nothing to do with the physical temperature of the
antenna as measured by a thermometer; it is only the temperature of a
matched resistor whose
thermally generated power per unit frequency equals that produced by
the antenna. It is
widely used because:
$$\bbox[border:3px blue
solid,7pt]{T_{\rm A} \equiv {P_\nu \over k}
}\rlap{\quad \rm {(3A9)}}$$
Example: What is the increase in the antenna temperature of our short dipole produced by thermal emission from the Sun at $\nu = 10$ GHz?
$$P_\nu = 6.4 \times 10^{-25}
{\rm ~W~Hz}^{-1}$$
so
$$T_{\rm A} = {6.4 \times 10^{-25} {\rm ~W~Hz}^{-1} \over 1.38 \times
10^{-23} {\rm ~J~K}^{-1}}
\approx 0.046 {\rm ~K}$$
Beam
Solid Angle
The beam
solid angle $\Omega_{\rm A}$
of a lossless antenna is defined as
$$\bbox[border:3px blue solid,7pt]{\Omega_{\rm A} \equiv \int_{4 \pi}
P_{\rm n} (\theta, \phi) d
\Omega}\rlap{\quad \rm {(3A10)}}$$
where $P_{\rm n} (\theta, \phi) $ is the power pattern normalized to
unity maximum:
$$P_{\rm n} = {G(\theta, \phi) \over G_{\rm max}}~.$$
$$\Omega_{\rm A} = {1 \over G_{\rm max}} \int_{4 \pi} G d \Omega
= {4 \pi \over G_{\rm max}}$$
The beam solid angle is a useful
parameter for estimating the antenna temperature produced by a compact
source covering solid angle $\Omega_{\rm s}$ and having uniform
brightness temperature $T_{\rm B}$. "Compact" means that the
source is much smaller than the beam so that the variation of $P_{\rm
n}$ is small across the source. The power per unit bandwidth
received from the source by the antenna pointing at it is
$$P_\nu = \int_{4 \pi} A_{\rm e}(\theta, \phi) {B_\nu \over 2} d
\Omega$$
$$P_\nu \approx {\lambda^2 G_{\rm (max)} k T_{\rm B} \Omega_{\rm s}
\over 4 \pi \lambda^2} = k T_{\rm B} {\Omega_{\rm s} \over \Omega_{\rm
A}}~.$$
Thus the antenna temperature $T_{\rm A} = P_\nu / k$ is
$$T_{\rm A} \approx T_{\rm B} {\Omega_{\rm s} \over \Omega_{\rm A}}~.$$
In words, the antenna temperature equals the source brightness
temperature multiplied by the fraction of the beam solid angle filled
by the source.
Main
Beam Solid Angle
The main beam of
an antenna is defined
as the region containing the principal response out to the first zero;
responses outside this region are called sidelobes
or, very far from the
main beam, stray
radiation.
The main beam solid angle
$
\Omega_{\rm MB}$ is defined as
$$ \bbox[border:3px blue solid,7pt]{\Omega_{\rm MB} \equiv \int_{\rm
MB} P_{\rm n} (\theta, \phi) d
\Omega}\rlap{\quad \rm {(3A11)}}$$
The fraction of the total beam solid angle inside the main beam is
called the main
beam efficiency
or, loosely, the beam efficiency.
$$\bbox[border:3px blue solid,7pt]{\eta_{\rm B} \equiv
{\Omega_{\rm MB} \over \Omega_{\rm
A}}}\rlap{\quad \rm {(3A12)}}$$