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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 = ----- = ---------------------------------------------------- = 1.24 10 19 1.60 10 --m
34 8 6

--m

1

eV

Also, to determine frequency and wavelength given energy, use =h hc = ----1.60 10 = -- = ------------------------- 34 h 6.62 10 hc = ----- = 1.24 10 6 19

------ = 2.41 10 eV -----eV
1

14

------ Hz eV
4 1

m = 1.24 10 -----eV

Angstroms

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

Photon (Hz) Typical microwave Typical far infrared (Angstroms) (metres) 0.20 100 10
6

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

= 10 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

4

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.

P R
c

r The flux density received at point P is: F= I cos d
c

=

I cos sin d d = I sin
2

=2 I
0

cos sin d

c

=

R2 I ----r2

That is, we recover the inverse square law. 3.
F

d = size of circular aperture

I

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 4 d = -- d 2 cos 4

Hence the (small) solid angle subtended by the aperture at P is -- d 2 cos 4 d2 = ---------------------- = -- ----- cos3 -2 4L L sec 2

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Next, the flux density incident on the back of the camera at P is d2 = I -- ----- cos -2 4L -- cos4 4 = I ----------------f2

F=

I cos d

I cos

4

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 J = ----- I d 4 1 = -2
c

I sin d

0

1 = -- I 1 cos 2

c

1 R2 = -- I 1 1 ----2 r2

1/2

(b) Flux through the surface of the sphere is: n F= I
2 0 0

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 R2 = 4
2

IR

2

Therefore, L I = -------------4 2R2 (c) The energy density per unit frequency is 4 4 u = ----- J = ----c c 1 R2 -- I 1 1 ----2 r2
1/2 1/2

L R2 = --------------- 1 1 ----2 cR 2 r2 L --------------4 cR 2

Hence, the total (i.e. frequency integrated) energy density is u L --------------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
2

and this is equal to the total luminosity, L . Hence

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u (e) For L = 3. 83 10
26

c

4 r2 = L
11

L u = -------------4 cr 2 m = 2.8 10
13

W and r = 1.50 10 u = 4.5 10
6

Jm

3

eV m

3

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

x2 y2 -------- + -------- = ln 2 2 2 2x 2y

=

8 ln 2

x

8 ln 2

y

(b) The flux per beam is F Now, if I x y Ixy
B

xy =

I x y B x x y y dx dy

varies slowly within the extent of the beam, then we put

I x y and F
B

xy

I x y B x x y y dx dy = I x y

A

and I xy (c) For a Gaussian beam A= x2 y2 exp -------- + -------- dx dy = 2 22 2x y = ----------------- then 8 ln 2
xy x y

FB x y ------------------A

2

x

2

y

=2

xy

(d) Since

x

y

2 A = ----------8 ln 2

= ----------4 ln 2 ----1
x y

-------180

1 ----------3600

2

---------------------------------------- Str square arcsecond

xy

= 2.66 10

11

----- Str 1

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(e) For an unresolved source represented by I x y = A F
B

x

y the flux per beam is

xy =

A

x

y B x x y y dx dy = A B x y

The peak flux per beam is therefore F 00 =AB00 =A The flux density of the source is F= where A x y cos dx dy

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

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

B

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 -------------------------- = 2.18 10 11 2.66 10
2 26 16

Str

IA IB

Wm
17

2

Hz

1

Str Hz

1

10 26 2.9 10 ---------------------------------------- = 1.09 10 11 2.66 10

Wm

2

1

Str

1

7. We are required to prove that · dn n t n n t t ------ ----------------------------------- = ---------------------------------------------------------2 dt 1 t n 1 t n This is fairly straightforward. d d 1 t n ------ n n t n n t ------ 1 t n dn n t dt dt ------ ----------------------------------- = -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------dt 1 t n 1 t n 2 Look at the various parts of this: d ------ n dt n t d = ------ n dt d ------ 1 dt t t n n = t · t =n n · t n · t

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Combining terms:

· dn n t n t t n t 1 t n ------ ----------------------------------- = ------------------------------------------------------------------------------------------------dt 1 t n 1 t n 2 · n n t t = ---------------------------------------------------------2 1 t n

8. 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 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 x E x* , E x E cos 2 sin sin sin 2
* E x E x sin * cos E x E x + * cos E x E x * E x E x + sin * y x y

=

cos sin

sin cos

E E

x y

etc.This just involves a bit of algebra
* Ey Ey * Ey Ey * Ey Ey * Ey Ey

cos cos 2 sin 2 cos

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

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 * * * * E x E y E x E y = E x E y E y E x We can then express the transformations for the Stokes parameters in matrix form: I Q U V 1 0 = 0 cos 2 0 sin 2 0 0 0 sin 2 cos 2 0 0I 0Q 0U 1V

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