Introduction to radio astronomy

Question 1

optical and radio astronomy correspond to the two

Answer 1

electromagnetically transparent windows in the Earth’s atmosphere

Question 2

optical wavelengths

Answer 2

0.4 to 0.8 microns

Question 3

radio wavelengths

Answer 3

1mm to 30m

Question 4

photon counting is not an option in radio astronomy so we can almost always think

Answer 4

classically in terms of waves and electric fields rather than photons

Question 5

the transition to quanta occurs in

Answer 5

sub-mm wave radio astronomy

Question 6

in radio astronomy we usually work in terms of

Answer 6

spectral flux density (usually just called flux)

Question 7

flux

Answer 7

symbol S
units Wm^-2Hz^-1

Question 8

1 jansky

Answer 8

10^-26 Wm^-2Hz^-1

Question 9

maximum power out=

Answer 9

S A delta v

Question 10

flux density corresponds loosely to

Answer 10

apparent magnitude in optical astronomy, in the sense that it is a measure of the strength of the signal we see from the source, at Earth

Question 11

extended sources

Answer 11

sources that do not appear as just a point on the sky

Question 12

extended sources have both a

Answer 12

total flux density s(v) and a flux density per steradian B(v)

Question 13

B(v) is a function of

Answer 13

position (theta, thi) on the sky

a map of B(v) across the sky is an image of the source

Question 14

B(v) is also known as

Answer 14

specific intensity
spectral intensity
sky brightness
surface brightness

Question 15

units of B(v)

Answer 15

Wm^-2Hz^-1sr^-1

Question 16

over a small patch of sky, the intensity of a sky pixel of solid angle d omega giving a flux density dS is

Answer 16

B=dS/d omega

Question 17

total flux density is given by

Answer 17

S = integral of B(theta,thi) d omega

Question 18

a source of uniform intensity B and total solid angle omega has a total flux

Answer 18

S=B omega

Question 19

why does the measured surface brightness of an object not depend on its distance from the observer

Answer 19

B= delta S/ delta omega

both an inverse square so B is not a function of r

Question 20

usually in radio astronomy, hv«

Answer 20

kbT for a blackbody source

Question 21

rayleigh jeans law

Answer 21

exp(hv/kbt) = 1 + hv/kbT

B(v) = 1kbTv^2/c^2

Question 22

can think of a radio telescope as

Answer 22

mapping the temperature of the sky

Question 23

most radio sources do not emit via a thermal blackbody mechanism so

Answer 23

can’t assign a temperature to them

Question 24

brightness temperature or effective temperature

Answer 24

a blackbody at Tb would have the same surface brightness as the source at the frequency in question

Question 25

Tb is the real thermodynamic temperature of the source if and only if

Answer 25

the object is a blackbody

T»hv/kb