rybicki & lightman problem 6.4 the following spectrum is observed from a point source at unknown...

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Rybicki & Lightman Problem 6.4 llowing spectrum is observed from a point source at unknown distanc me the source is spherical, emitting synchrotron radiation R = radius of sphere B = magnetic field the space between us and the source is filled with hydrogen which s and absorbs by bound-free transitions, but is unimportant at uencies where the synchroton is optically thin.

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Rybicki & Lightman Problem 6.4

The following spectrum is observed from a point source at unknown distance d

Assume the source is spherical, emitting synchrotron radiation R = radius of sphere B = magnetic fieldAnd the space between us and the source is filled with hydrogen which emits and absorbs by bound-free transitions, but is unimportant at Frequencies where the synchroton is optically thin.

Let the synchrotron source function be

Sν = A(erg cm-2s−1Hz−1)B

B0

⎝ ⎜

⎠ ⎟

−1/ 2ν

ν 0

⎝ ⎜

⎠ ⎟

5 / 2

The absorption coefficient for synchrotron

ανs =C(cm−1)

B

B0

⎝ ⎜

⎠ ⎟

(p+2)/ 2ν

ν 0

⎝ ⎜

⎠ ⎟

−p+4

2

The bound-free absorption coefficient:

ανbf = D(cm−1)

ν

ν 0

⎝ ⎜

⎠ ⎟

−3

Where A, Bo, νo, C and D are constants. p = power-law index of electron energies

a. Find R and B in terms of A, Bo, νo, C, D and the solid angle subtended by the source,

Ω =πR2

d2

⎝ ⎜

⎠ ⎟

When the source is optically thin

Fν ∝ jν = αν Sν

∝ ν −( p−1)/ 2

Here, the source is optically thin in the part ofthe spectrum has

Fν ∝ ν−1/ 2

so

p −1

2=

1

2

p = 2

At ν2 the source becomes optically thick.

So

αν 2

s ds ≈∫ αν 2

s R ~ 1 And

F0 = Sν 2Ω = π

R2

d2 Sν 2

αν 2

s R ~ 1 implies

CB

B0

⎝ ⎜

⎠ ⎟

2ν 2

ν 0

⎝ ⎜

⎠ ⎟

−3

R =1 (Equation 1)

F0 = πR2

d2 Sν 2implies

AB

B0

⎝ ⎜

⎠ ⎟

−1/ 2ν 2

ν 0

⎝ ⎜

⎠ ⎟

5 / 2

Ω = F0

B

B0

⎝ ⎜

⎠ ⎟= A

ν 2

ν 0

⎝ ⎜

⎠ ⎟

5 / 2

Ω F0−1

⎣ ⎢ ⎢

⎦ ⎥ ⎥

2

Substitute into (1), solve for R:

R =C−1 ν 2

ν 0

⎝ ⎜

⎠ ⎟

−7

AΩ F0−1

( )−4

(b) Now find the solid angle and distance to the source

The ν2 part is the optically thick thermal emitter: the Rayleigh-Jeans tail

So ν1 = the frequency at which the hydrogen becomes optically thick

so

αν 1

bf ds ≈∫ αν 1

bf d ≈1 so

d = D−1 ν1

ν 0

⎝ ⎜

⎠ ⎟

3

Ω =πR2

d2

⎝ ⎜

⎠ ⎟

R =C−1 ν 2

ν 0

⎝ ⎜

⎠ ⎟

−7

AΩ F0−1

( )−4

and

So we have

Ω =π A−8 C−2 D2 ν 1

ν 0

⎝ ⎜

⎠ ⎟

−6ν 2

ν 0

⎝ ⎜

⎠ ⎟

−14

F08