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High Energy Astrophysics Solutions to Exercises
10.We can put I
a+5 ----------2

I = ---- 3

+3

I = ---- 3

a+5 ----------2

and when the factor is raised to the 2 a + 5 power, this just gives a term of the dimensions of frequency. It can easily be shown that c I ----- ----me 3 eB is dimensionless. Since ----- has the dimensions of frequency (it is the cyclotron frequency), me then the dimensions are OK. 11.The total energy density and pressure are given by: B2 = 1 + c E e + -------2 0 p
tot

e B2 = 1 + c p ---- + -------3 2 0

Hence the expression for the minimum pressure follows from the expression for the minimum energy density by the substitution 1 + cp 1 + c E -----------------3 The corresponding expression for the minimum pressure magnetic field is: B Hence, B p-min 1 + cp ------------- = ---------------------B e-min 3 1 + cE For c p = c E = 0 and a = 2 B B
2 ----------a+5 p-min

me a + 1 1 + cp c - = ----- ----------- ------------------ C 2 1 a ----3 e 2 me

I - f a ---------- 12 L

2 ----------a+5

1 2/7 = -- = 0.73 . p-min e-min 3 The subsidiary condition on the particle pressure follows from

1

p tot 1 + c p e B p p B = -----------------+ ----- = + ----B 3 B 0 B 0 p p B = ----B 0

Also, analogously to the minimum energy case: pp --------- mc 2 so that a+1 4 1 p p a + 1 1 1B ---+ ----------- -- = 0 ---- ----- + ----------- = 0 p p = -----------2B a+1 pp B 2 B p p 0 The total minimum pressure: p Hence,
2 p tot,min B p-min 1 + cp --------------- = -------------- = ---------------------2 3 1 + cE tot,min B e-min 4 ----------a+5 tot,min 2 B p-min 4 a+5 = ----------- + 1 -------------- = ----------a+1 2 0 a+1 2 B p-min ------------- 2 0 a+1 ----------2 0 1 + cp e 2 1 I = ------------------ C 2 1 a ------- ---------- f a 1 2 - 0 c L 3

B2 -------- 2 0

For example, for c p = c E = 0 and a = 2 , p tot,min 1 --------------- = -- 3 tot,min
4/7

= 0.53

12. (i) The minimum energy density solution can be calculated using a simple modification of the spreadsheet on the course web-site. The only difference is that in the template, the surface brightness in physical units is calculated from the flux per beam and the beam dimensions, whereas one needs to use the plotted values of mJy arcsec 2 to estimate I in physical units. That is, I I I 10 29 19 ---------------------------------------------- = ------------------------------ ----------------------------------- = 4.25 10 ------------------------------ 2 Hz 1 Sr 1 2 2 mJy arcsec mJy arcsec 2 W m 1 -------- ----------- 180 3600 I used the plotted values in figures 14 and 16 of the paper to estimate the FWHM, , and the surface brightness in mJy arcsec 2 to calculate the minimum energy magnetic field, the corresponding particle energy density and the total minimum energy density, although I only asked for the minimum pressure. I have used c E = c P = 0 appropriate for a electron-positron plasma. However, the numbers don't vary all that much if you put c E = c p = 1 , for example.

2

( arcsec )

( arcsec )

I

B B

e-min p-min

nT nT

p 10 p p 10

12 12

J m 3 N m 2

tot 10 p tot 10

12 12

J m 3 N m 2

( mJy arcsec 2 ) 1.09 1.20 0.35 0.69 0.88 0.41

25 W 50 W 100 W 25 E 50 E 100 E

7.4 11.8 15.6 7.4 13.2 16.9

1.9 1.4 1.7 1.3 1.1 0.82 1.7 1.2 1.5 1.1 1.1 0.84

1.8 0.96 1.4 0.63 0.50 0.27 1.4 0.75 0.91 0.50 0.65 0.35

3.2 1.7 2.6 0.78 1.1 0.61 2.5 1.3 2.5 1.3 1.2 0.64

(ii) The pressure of the thermal plasma is nkT where n is the total (ion plus electron) number density and T is the temperature. Hence the minimum pressure required to confine the jet is n
min

p tot,min p tot,min 10 12 T = --------------- = ----------------------------------------- ----------------------------- ------- 23 12 N m 2 10 7 kT 1.38 10 10 7 10 p tot,min T 3 = 7.2 10 ----------------------------- ------- 12 N m 2 10 7 10
1

1

m

3

p tot,min T 3 = 7.2 10 ----------------------------- ------- 12 N m 2 10 7 10

1

cm

3

Given the entries in the above table the minimum no. densities at various points along the jet are:

(arcsec) 25 W 50 W 100 W 25 E 50 E

p 10

tot,min

n
2

min 3 2 3 3 3 3

12

Nm

cm

1.7 0.78 0.61 1.3 1.3

1.2 10 5.6 10 4.4 10 9.4 10 9.4 10

3

(arcsec) 100 E

p 10

tot,min

n
2

min 3 3

12

Nm

cm

0.64

4.6 10

(iii) The answers that one obtains here will differ depending upon the exact assumptions. Here are the parameters that I used:

West lobe Contour representing "average" plateau of surface brightness (i.e. ignore local hot spots or warm spots) Diameter, D (gives distance through lobe as well as volume) Redshift Beam 1
2 p

East lobe 7th contour => 77.8 mJy

11th contour => 272 mJy

10.4

11.7

0.01174 60 60 10 10 0 5.7 10
14 5

c E c

Minimum total energy density Minimum energy

Jm J

3

2.7 10

14

Jm J

3

3.0 10

51

2.0 10
51

51

(iv) The total minimum energy is therefore approximately 5 10 J about an order of magnitude less than Cygnus A. The accreted mass that can be responsible for this energy is given by E where we take 0.1 Thus, M
acc,min min

= M

acc,min

c

2

5 10 5 = ---------------- = 2.8 10 solar masses 2 0.1 c

51

This is the minimum mass of the black hole that is implicated in the radio emission from IC 4296 and is less than that implicated in Cygnus A, because of the lower minimum energy. This is probably indicative of a lower accretion rate, rather than a lower black hole mass, i.e. the mass accreted is a significant underestimate of the mass of the black hole.

4

13. (i) For a uniform (in solid angle) distribution of pitch angles, 1 sin2 = ----4

0

2

0

2 sin2 sin d d = -3

(ii) Hence the mean energy loss rate (averaged over man scatterings) is given by: d B 2 sin2 2 ----- = c T ------------------------- 2 = -dt 0 3 (iii) The energy of an electron is given by: d 2 ----- = -dt 3 This integrates to 11 2 -- ---- = - 0 3
2 T -------- B - t - ---- m e c 0 2 T -------- B - - ----- me c 0 2 2 T -------- B - - ---- m e c 0 2

where 0 is the Lorentz factor at t = 0 . Write this equation as: 0 = ---------------------------------------------------2 T B 2 1 + -- -------- ----- 0 t 3 m e c 0 The time at which the Lorentz factor is reduced to half of its initial value is: t Using 1 1 c 2 = ---------- 0 c = ------0 0 0 c gives t (iv) The critical frequency is 3 eB 3 eB c = -- ----- 2 sin c = ----- ----- 2 sin --2 me 4 me To evaluate the mean critical frequency we calculate 1 sin = -2
syn syn

3 = -2

0 me c --------------- B 2 T

1 0

3 me = -- -------------- B 2 2 0 c T

1 0

0

sin sin d = -4

5

Hence, the critical frequency corresponding to 0 is: 3 eB 0 = ----- ----- -16 m e (v) From the above: and t
syn 1 0 2 0

3 eB = -------- ----- - 16 m e

1/2

1 / 2 0

3 me 3 eB = -- -------------- B 2 ----- ----- - - 16 m e 2 0 c T

1/2

1 / 2 0

1/2 3 3 / 2 em e = --------- -------------------- B 8 0 c T

3 / 2 1 / 2 0

(vi) Evaluating the leading constant gives t leading to the following table: B 1 nT 10 nT 1T
0 syn

= 1.40 10 B

6 3 / 2 1 / 2 0

se c s

t

syn 7

1 GHz 6 10 10
14

4.4 10 yrs 1800 yrs 51 days

Hz

17

Hz

6