Examining the basic principles of radio astronomy means first recognizing we are talking about a window of observation that involves the ionosphere as opposed to the atmosphere. In the first case we have a terrestrial layer transparent to radio waves, in the second case we have a layer transparent to light waves - for optical observation. The basic electromagnetic (EM) spectrum showing the relation of visible light to radio waves is shown below:
Note here that radio waves are at much longer wavelengths. The earliest radio wave observations were made at kilohertz (kHz) frequencies and then gradually were able to also detect sources at megahertz and beyond. All radio astronomy - at whatever frequency- begins with knowing the basic radiation characteristics. To this end we wish to know:
i) The direction it comes from,
ii) The emergent flux,
iii) The polarization
In respect of the flux, consider a small element of surface ds: Then to find the amount of energy which flows per unit solid angle we use the graphic below:
From this we can obtain:
dE v = I cos q ds dw dt dv
Where:
I = f(u,f,q, x, y, z, t)
is defined as the specific intensity or brightness. The units we are making use of here are:
[ergs / cm 2 steradian sec cps]
Integrating the specific intensity over the solid angle:
FLUX/ F v = ò I v (q, f) cos q dw
And over the total solid angle:
F v = Total flux = ò 4p I (q, f) dwf
F v Units [ergs / cm 2 sec cps]
Note: The infinitesimal sold angle dw is equivalent to: sin q dq df
Now we examine the specific situation for a radio telescope antenna. In the most basic kind of radio telescope system we will have:
Which shows a parabolic dish with the receiver at the prime focus. Note that a parabolic dish has the property whereby radio waves reflected off it come to a focus at the same time. This is critical given if the waves arrived at different times the telescope would be essentially useless. The signal obtained would be incoherent which the reader might be able to deduce by the end of this post.
Note that the antenna is essentially some sort of conducting wire or metal rod which is the 'heart' of the telescope. Its basic function is to capture incoming electromagnetic waves and convert them into an electric current in the wire. For maximum efficiency the dimension of the antenna wire or probe should be one fourth the size of the radio wave it is intended to intercept. Thus for the 21 cm line of hydrogen the probe should be 5 cm in length.
There is still the matter of the parameters of the source in relation to the antenna and the solid angle formed. In the sketch below I show a radio source of area A, at a distance D from an antenna of dimension (C) and we wish to find the radio flux reaching the antenna from the source.
Comparison of solid angles, source & antenna
We have in the first instance:
dE v = I v cos q ds dw dE v =
Then the flux reaching C from the radio source is:
F v = dE v / C dv dt = I v A W / C
Now, if b is the solid angle: A/ D 2
And W is the solid angle C/ D 2
Then: A/ C = b / W
The flux then received becomes:
F ' v = I v b
We must use: F ' v = P v / C D u
Since the radio telescope measures Power: P v = dE v / dt
The next important feature when considering the basic principles of radio astronomy is the polarization of the radio source. This is given that the information is considerable and every non-thermal process emits some random polarization which can be described by two components, and electric (E) vector and a magnetic (H) vector:
1) E(y, t) = E 0 sin 2 p u (t – y/c )
2) H(x, t) = H 0 sin 2 p u (t –x/c )
Electromagnetic waves with the above vectors are often represented as in the diagram below:
Where S denotes the Poynting vector: S = E X H. The EM- waves are polarized when their E- field components are preferentially oriented in a particular direction. Basically, the up and down motion of the EM radio waves is converted to the up and down motion of electrons in the antenna wire which gives rise to the current detected at the telescope.
Linearly or horizontally polarized: I.e. the E- vector is confined to one (horizontal) plane
Vertically polarized: I.e. the E- vector is confined to one (vertical) plane
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Circular: The E-vector rotates through 360 degrees
Elliptic: Any polarization not circular or plane, but note in most texts circular is regarded as a limiting case of elliptic polarization.
The maximum power is obtained under the condition:
dP/ dRB = 0 (N.B. In general, R ℓ ≈ 0, and R ℓ << R r
dP/ dR B = d/ dR B [ R B V2 / (R r + R B) 2
= V2 { [(R r + R B) 2 - R B 2(R r + R B)] /(R r + R B)4
dP/ dR B = R r 2 - R B 2 /(R r + R B) 4
Then the available power is:
PA = V2 /4R B = V2 /4R r
The diffuse power is: P r = I R r 2
Let the total area of an antenna be A T then the total power:
P T = A T F
Where F is the energy flux of the incoming wave .
The Gain of an antenna:
For an isotropic lossless antenna the flux emitted is:
F = (1/4p ) 1/ d 2
(Where P is the applied Power )
For an anisotropic antenna:
F = g (1/4p ) 1/ d 2
Where the gain g is subject to the constraint:
ò 4p g dw = 4p
If:
r = ò g dw /4p
For all secondary lobes then the beam efficiency of the antenna is: 1 - r
The relation between the gain of the antenna and its effective aperture (A') such that:
g( (q , φ ) = 4p A' (q , φ)/ l2
Now, every radio telescope receives waves that come from fairly restricted areas of the sky and this is defined by what's called the beam. Every antenna - and by extension radio telescope- has a main lobe called the "main beam" and also side lobes off to the side. The side lobes have different causes but most result from diffraction (Consult the astronomy archives for past answers pertaining to diffraction in optical telescopes). The pattern of a radio telescope beam and typical side lobes are shown below:
Note the side lobes are weak but can still represent a significant fraction of the power a radio telescope receives. Generally 80-90 percent of the signal entering a receiver from a typical dish comes through the main beam, the other 10-20 percent through the side lobes. In the case of the initial radio telescope diagram, try to imagine the wave fronts coming in at an oblique angle instead of directly., e.g.
Here, the arrival of the left side of the wave is later than the right, because it happens to travel an extra distance. If one finds this distance d is such that:
d sin q = l/ 2
Then we have the left -radio wave fully cancelling the right and no signal gets through. Obviously one wants to avoid this condition. Ideally then, we need:
d sin q = 0.
Again, every radio telescope has maximum reception in location and this falls off according to the beam pattern. A convenient way to specify this pattern is by using the beam width and one half the peak (power) height.
From this one can deduce the optimum beam width in terms of the resolution. Thus, if two objects are closer than the beam width then they will blur into one, and the astronomer will be unable to distinguish the sources. This conveys the importance of a narrow beam width, which just means the signal falls off rapidly as one shifts from one side of the source to the other.The narrower the beam width the better the resolution of the radio telescope. In general:
Beam width = Resolution = Wavelength / diameter = l/ D
Hence, if the wavelength is 21 cm (say for the hydrogen line) and the diameter D of the telescope is 10 m (1000 cm) we have: 21cm / 1000 cm = 0.021 arcseconds
See Also:
Brane Space: All Experts Redux: Basics of Radio Telescopes
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