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High Energy Astrophysics Solutions to Exercises
1. The relationship between energy and wavelength is hc 6.62 10 3.0 10 ­6 = ----- = ---------------------------------------------------- = 1.24 10 ­ 19 1.60 10 --- - m Also, to determine frequency and wavelength given energy, use 1.60 10 14 = h = -- = -------------------------- ------ = 2.41 10 ------ Hz ­ 34 eV eV h 6.62 10 hc hc ­6 = ----- = ----- = 1.24 10 ------ eV
­1 4 m = 1.24 10 ------ eV ­1 ­ 19 ­ 34 8

--- m

­1

eV

Angstroms

Hence the following table (the primary quantity is given in bold):

Photon Typical microwave Typical far infrared

(Hz)

(Angstroms)

(metres) 0.20 100 10 = 10
­6

(eV) 6.2 10 1.1 10 0.62 2.48 12.4
­6 ­2

­4

Typical near infrared Typical optical Typical UV Soft X-ray Hard X-ray TeV -ray 2.41 10 2.41 10 2.41 10
17 19 26

2 10 5000 1000 12.4 0.124 1.24 10
­8

­6 ­7

5 10 10

­7 ­9

1.24 10 1.24 10 1.24 10

1 keV 100 keV 1TeV

­ 11 ­ 18

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2.

R

c r

P

The flux density received at point P is: F =





I cos d =





I cos sin d d

= 2I

0



c

cos sin d = I sin2

c

R2 = I ----r2 That is, we recover the inverse square law. 3.
F


I


d = size of circular aperture

L

Circular aperture L sec d d cos Projected aperture d cos d

First calculate the solid angle of rays hitting P the back of the camera. Viewed from the point P the aperture appears as an ellipse with major and minor axes d and d cos . The area of the aperture is therefore A -- d cos d = -- d 2 cos 4 4 Hence the (small) solid angle subtended by the aperture at P is 2 -- d cos 4 d2 = ---------------------- = -- ----- cos3 4 L2 L sec 2



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Next, the flux density incident on the back of the camera at P is

F =





4 -- cos d 4 4 = I ----------------I cos d I cos = I -- ----- cos 4 L2 f2
2

where the focal ratio f = L d . 4. (a) The mean intensity at a particular point, a radius r from the centre of the sphere is: 1 1 J = ----- I d = -4 2


0



c

1 1 R2 I sin d = -- I 1 ­ cos c = -- I 1 ­ 1 ­ ----- 2 2 r2

1/2

(b) Flux through the surface of the sphere is: n I


F =

0 0

2





I cos sin d d = I



Hence, the total energy per unit time through the surface of the sphere (i.e. the Luminosity) is: L = F 4 R 2 = 4 2 I R
2

Therefore, L I = -------------42R2 (c) The energy density per unit frequency is 4 4 1 R2 u = ----- J = ----- -- I 1 ­ 1 ­ ----- - c c2 r2 L R2 = --------------- 1 ­ 1 ­ ----- 2 cR 2 r2 L --------------4 cR 2 Hence, the total (i.e. frequency integrated) energy density is L u --------------4 cR 2 (d) This expression can also be derived as follows. The total energy passing through a sphere of radius r is u c 4r and this is equal to the total luminosity, L . Hence
2 1/2 1/2

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L u c 4 r 2 = L u = -------------4 cr 2 (e) For L = 3. 83 10
26

W and r = 1.50 10 u = 4.5 10
­6

11

m = 2.8 10
13

Jm

­3

eV m

­3

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

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(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

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Combining terms:

· d n n t n ­ tt n ­ t1 ­ t n ------ ----------------------------------- = ------------------------------------------------------------------------------------------------dt 1 ­ t n 1 ­ t n 2 · n n ­ t t = --------------------------------------------------------- 1 ­ t n 2

8. (a) In order to determine the transformation of the Stokes parameters we can utilise the * transformations 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 y x y

=

cos sin E ­ sin cos E

x y

We then evaluate with the results: E x E x* = E x E y* = * Ey E x = * Ey E y =

all of the combinations E E , E E
* cos 2 E x E x + ­ sin cos E ­ sin cos E * sin 2 E x E x ­

x

* x

x

* y

etc.This just involves a bit of algebra

* * * sin cos E x E y + sin cos E x E y + sin 2 E y E y * 2 * 2 * x E x + cos E x E y ­ sin E x E y + sin cos E y E * 2 * 2 * x E x ­ sin E x E y + cos E x E y + sin cos E y E * * * sin cos E x E y ­ sin cos E x E y + cos 2 E y E y

* y * 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 transformations for the Stokes parameters in matrix form: I


Q U V We can write this as



1 0 0 = 0 cos 2 sin 2 0 ­ sin 2 cos 2 0 0 0 I' = I

0I 0Q 0U 1V

Q' = cos 2 sin 2 Q U' ­ sin 2 cos 2 U V' = V 8 (b) The frame in which U' = 0 is given by

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U' = ­ Q sin 2 + U cos 2 = 0 U tan 2 = --Q power-law does not continue indefinitely for low frequencies ­ it turns over quite abruptly. This signature is often one that is sought by observers in ascertaining the importance of freefree absorption.

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