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High Energy Astrophysics Solutions to Exercises

5. (a) The beam is defined by x2 y2 B x y = exp ­ -------- + -------- 2 2 2 2 x y The FWHM contour is defined by: 1 x2 y2 B x y = -- -------- + -------- = ln 2 2 2 2 2 x 2 y The axes of this ellipse are therefore: x y = 8 ln 2 x 8 ln 2 y (b) The flux per beam is
B F x y =



I x y B x ­ x y ­ y dx dy

Now, if I x y varies slowly within the extent of the beam, then we put I x y I x y and
B F x y I x y B x ­ x y ­ y dx dy = I x y A

and
B F x y I x y ------------------A

(c) For a Gaussian beam A=



x2 y2 exp ­ -------- + -------- dx dy = 2 2 2 2 x y

2 x 2 y = 2 x

y

x y (d) Since x y = ----------------- then 8 ln 2 x y 2 12 A = ----------- x y = ----------- -------- ----------- ---------------------------------------- Str 4 ln 2 180 3600 square arcsecond 8 ln 2 = 2.66 10
­ 11

x y ----- ----- Str 1 1

1


(e) For an unresolved source represented by I x y = A x y the flux per beam is
B F x y =



A x y B x ­ x y ­ y dx dy = A B x y

The peak flux per beam is therefore F 0 0 = A B 0 0 = A The flux density of the source is F =




A x y cos dx dy

where is the angle between the ray and the point on the sky. Because of the delta-function, F = A
B

Hence, for an unresolved source, the flux per beam F source.

is equal to the flux density of the

6. From the contour image of Cygnus A, we have: · Point A = Contour 7 => 0.04 14.5 Jy/beam = 0.58 Jy/beam · Point B = Contour 2 => 0.002 14.5 Jy/beam = 2.9 10
­2

Jy/beam
­ 11

The beam is circular with a FWHM of 1 arcsecond. Hence A = 2.66 10 Hence the respective surface brightnesses are given by: 0.58 10 ­ I A -------------------------- = 2.18 10 ­ 11 2.66 10
­2 ­ 26 16

Str

Wm
17

­2

Hz

­1

Str Hz

­1

2.9 10 10 ­26 ­ I B ---------------------------------------- = 1.09 10 ­ 11 2.66 10 7. We are required to prove that

Wm

­2

­1

Str

­1

· d n n t n n ­ t t ------ ----------------------------------- = ---------------------------------------------------------dt 1 ­ t n 1 ­ t n 2 This is fairly straightforward. d d 1 ­ t n ------ n n t ­ n n t ------ 1 ­ t n d n n t dt dt ------ ----------------------------------- = -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2 dt 1 ­ t n 1 ­ t n Look at the various parts of this: d ------ n n t = dt d ------ 1 dt d · · ------ n t n ­ t = n t n ­ t dt · ­ t n = ­ t n

2


Combining terms

· d n n t n ­ t t n ­ t 1 ­ t n ------ ----------------------------------- = ------------------------------------------------------------------------------------------------dt 1 ­ t n 1 ­ t n2 · n n ­ t t = --------------------------------------------------------- 1 ­ t n 2

8. In order to determine the transformation of the Stokes parameters we can utilise the trans* formations of the electric field in a wave to determine the quantities E E . For a rotation by in the plane of the wave: E E i.e.
E x = cos E x + sin E y E y = ­ sin E x + cos E x * x x * y y x y

=

cos sin E ­ sin cos E

x y

We then evaluate all of the combinations E E , E E etc.This just involves a bit of algebra with the results: * * * * E x E x* = cos 2 E x E x + sin cos E x E y + sin cos E y E x + sin 2 E y E y * * * * E x E y* = ­ sin cos E x E x + cos 2 E x E y ­ sin 2 E y E x + sin cos E y E y * * * * * E y E x = ­ sin cos E x E x ­ sin 2 E x E y + cos 2 E y E x + sin cos E y E y * * * * * E y E y = sin 2 E x E x ­ sin cos E x E y ­ sin cos E y E x + cos 2 E y E y One then has to form the combinations that give the Stokes parameters: * * * E x E x* + E y E y = E x E x + E y E y * * * * * E x E x* ­ E y E y = cos 2 E x E x ­ E y E y + sin 2 E x E y + E y E x * * * * * * E x E y + E x E y = ­ sin 2 E x E x ­ E y E y + cos 2 E x E y + E y E x * * * * Ex Ey ­ Ex Ey = Ex Ey ­ Ey Ex We can then express the transofrmations for the Stokes parameters in matrix form: I


Q U V



1 0 0 = 0 cos 2 sin 2 0 ­ sin 2 cos 2 0 0 0

0I 0Q 0U 1V

3