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Astronomy 9603b - 1st Problem List and Assignment

Your solution to problems 1, 4, 5, and 6 has to be handed in on Friday, Feb. 17, 2012.

1. Knowing that the deuterium atom is composed of a single electron and a deuteron (i.e., its nucleus is made of one proton and one neutron), and that these have spins of 1 2 and 1, respectively. a) Will this atom exhibit a hyperfine structure? Justify your answer.

b) If so, can you say between what states will a transition equivalent to the 21-cm line of the hydrogen atom take place?

Solution. a) Since this atom has both electronic and nuclear spins, then there will an interaction between the two that will contribute the atom’s Hamiltonian. This interaction will allow a differentiation between the energy levels of wave functions exhibiting the same amount of orbital angular, which would have been indistinguishable otherwise. b) The 21-cm line is caused by a transition between the n = 1 , F = 1→ 0 states of the hydrogen atom. Because of its different isospin (i.e., nuclear) spin the equivalent deuterium line will take place between different values of F . For this atom, any permissible value of this quantum number must range between I − S ≤ F ≤ I + S , (1.1) or 1 2 ≤ F ≤ 3 2 , where I and S are the quantum numbers for the electronic and nuclear spins, respectively. It, therefore, follows that this transition can only happen between the n = 1 , F = 3 2→ 1 2 atomic states. This is the 92-cm line of deuterium, which was only recently confidently detected (c.f., Rogers et al. 2005, ApJ, L41).

2. Suppose that a gas composed of a single type of particles of mass m in thermal equilibrium exhibits a Maxwellian velocity distribution

n v( )∝ e−mv2

2kBT dv, (2.1) when the velocity is measured relative to a well-defined orientation. In equation (2.1) n v( ) is the number of particles having a velocity confined to a range dv centered on v , kB is the Boltzmann constant, and T is the temperature of the gas.

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a) Assuming that there are a total of N particles in the gas, find the normalization constant missing in equation (2.1).

b) Determine the standard deviation σ 2 for the velocity along this direction.

c) Can you relate the result obtained in b) to a spectral line width potentially observed for a given molecular species and transition in a molecular cloud?

Solution. a) If we integrate over possible values for the velocity we must have that

N = NA

e−−

mv2

2kBT dv−∞

∞

∫ , (2.2) where A is the normalization factor. We surmise that

A = e−mv2

2kBT dv−∞

∞

∫ , (2.3) and

A2 = e−mv2

2kBT dv−∞

∞

∫ ⋅ e−mw2

2kBT dw−∞

∞

∫

= 2π e−mz2

2kBT z dz0

∞

∫

= −2πkBTm

e−mz2

2kBT

0

∞

=2πkBTm

,

(2.4)

where we used z2 ≡ v2 + w2 when changing to polar coordinates. b) Since the mean velocity is zero, the standard deviation for the velocity is determine with the following integral

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σ 2 ≡ v − v( )2

=1A

e−mv2

2kBT v2dv−∞

∞

∫

= −kBTAm

vd e−mv2

2kBT⎛

⎝⎜

⎞

⎠⎟

−∞

∞

∫

= −kBTAm

ve−mv2

2kBT

−∞

∞

− e−mv2

2kBT dv−∞

∞

∫⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

=kBTAm

e−mv2

2kBT dv−∞

∞

∫=kBTm.

(2.5)

where we integrated by parts and used the result from equation (2.4). This result could have been easily guessed from the well-known definition of a Maxwellian distribution as a function of the standard deviation. c) When a molecular spectral line emanating from a region of a molecular cloud in thermal equilibrium is observed, the line profile will have a Maxwellian distribution and an intensity proportional to the number of molecules present within the column of gas probed by the telescope beam (assuming the line is optically thin). The width of the line profile as quantified by its standard deviation will thus be given by the square root of equation (2.5)

3. Assuming that the motion of the gas in some region of the interstellar medium is due to thermal and turbulent processes, show that width of an observed spectral line σ l is given by σ l

2 = σ T2 +σ th

2 , (3.1) where σ T and σ th are, respectively, the turbulent and thermal velocity dispersions along the line-of-sight. To derive equation (3.1) you must further assume that the two types of processes (i.e., turbulent and thermal) are statistically independent, as well as the individual motion of the particles composing the gas. Solution. If v is the random variable associated with the velocity of a given particle along the line-of-sight, then

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v = vT + vth , (3.2) where vT and vth are, respectively, the random variables associated with the turbulent and thermal components of the particle’s motion. We can then calculate

σ l2 ≡ vl − vl( )2

= vl2 − 2 vl vl + vl

2

= vl2 − vl

2 ,

(3.3)

where stands for the average. We now need the following quantities

vl

2 = vT + vth2

= vT2 + 2 vT vth + vth

2 , (3.4)

since we assume statistical independence for the turbulent and thermal processes, and

vl2 = vT + vth( )2

= vT2 + 2 vT vth + vth

2 . (3.5)

Inserting equations (3.4) and (3.5) into equation (3.3) we find

σ l2 = vT

2 + 2 vT2 vth + vth

2( ) − vT 2 + 2 vT vth + vth 2( )= vT

2 − vT2( ) + vth2 − vth 2( )

= σ T2 +σ th

2 .

(3.6)

We, therefore, see that the square of the line width is a sum of the squares of the different velocity dispersions.

4. The angular resolution of a telescope at a given wavelength can be defined by calculating the diffraction pattern from a plane wave impinging on an aperture of the same dimension as the telescope. a) Assume that a plane wave is normally incident on a circular aperture of radius R and

show that the corresponding diffraction pattern (for the intensity) at a distance much greater than R is given by

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I θ( ) = I 0( ) 2J1 kRsin θ( )⎡⎣ ⎤⎦kRsin θ( )

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

2

, (4.1)

where θ is the angle between the axis of symmetry of the aperture and the vector linking the centre of the aperture to the observation point, and k = 2π λ with λ the wavelength of the plane wave.

b) Show that the angular distance between the intensity I 0( ) at the centre (where it is maximum) of the diffraction pattern and the first minimum is

Δθ 1.22 λ

D, (4.2)

with D = 2R the diameter of the aperture. When applied to a telescope this is the so-called diffraction limit.

c) Expand I θ( ) I 0( ) from equation (4.1) up to the second order and compare it to the corresponding expansion for the equation of a Gaussian beam

I θ( )I 0( ) e

−θ2

2σ 2 (4.3)

to show that the full-width-half-magnitude (FWHM) ΔΩ of the Gaussian beam is

ΔΩ ≈ 1.06 λD. (4.4)

[Note: You may need the following equations to solve this problem

Jn x( ) =12π

e nϕ− x sin ϕ( )⎡⎣ ⎤⎦dϕ−π

π

∫dxdx

⎛⎝⎜

⎞⎠⎟

m

xα Jα x( )⎡⎣ ⎤⎦ = xα −mJα −m x( )

Jn x( ) =−1( )m

m! Γ m + n +1( )m=0∞

∑ x2⎛⎝⎜

⎞⎠⎟2m+n

.

(4.5)

] Solution. a) The amplitude and phase changes of a plane wave after it has diffracted from a pinhole and traveled a distance d is given by e− jkd d . If we treat each point of the circular

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aperture as a pinhole, then for an observer located a distance r R away from the aperture in a direction θ as defined above we find that d r + R + ρ sin ϕ( )⎡⎣ ⎤⎦sin θ( ), (4.6) with ρ and ϕ the polar coordinates mapping the area delimited by the aperture. Integrating over the aperture we have for the amplitude of the diffraction pattern

G θ( ) = e− jk r+ R+ρ sin ϕ( )⎡⎣ ⎤⎦ sin θ( ){ }

dρdρdϕ

−π

π

∫0R

∫

e− jk r+R sin θ( )⎡⎣ ⎤⎦

re− jkρ sin ϕ( )sin θ( )ρdρdϕ

−π

π

∫0R

∫

= 2π e− jk r+R sin θ( )⎡⎣ ⎤⎦

rJ0 kρ sin θ( )⎡⎣ ⎤⎦ρdρ0

R

∫ ,

(4.7)

where we have used the first of equations (4.5) in the last step. Moreover, the second of equations (4.5) provides us with

ddx

xJ1 x( )⎡⎣ ⎤⎦ = xJ0 x( ), (4.8) such that with x = kρ sin θ( ) we have

J0 kρ sin θ( )⎡⎣ ⎤⎦ρdρ0R

∫ =1

k sin θ( )⎡⎣ ⎤⎦2 J0 x( )xdx0

kR sin θ( )∫

=RJ1 kRsin θ( )⎡⎣ ⎤⎦

k sin θ( ) . (4.9)

Inserting this result in equation (4.7) we get

G θ( ) = 2πR2 e

− jk r+R sin θ( )⎡⎣ ⎤⎦

rJ1 kRsin θ( )⎡⎣ ⎤⎦kRsin θ( )

= 2G 0( ) e− jkR sin θ( )

rJ1 kRsin θ( )⎡⎣ ⎤⎦kRsin θ( ) ,

(4.10)

where we used the last of equations (4.5) for θ → 0 to determine G 0( ) . The intensity I θ( ) ≡ G θ( ) 2 is therefore

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I θ( ) = I 0( ) 2J1 kRsin θ( )⎡⎣ ⎤⎦kRsin θ( )

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

2

. (4.11)

b) Since the first zeros on either sides of J1 x( ) are located at x = ±3.8317 , then the diffraction limit is

Δθ 3.8317kR

1.22 λD. (4.12)

c) From the last of equations (4.5) we find that

I θ( )I 0( ) ≈ 1−

kRθ2

⎛⎝⎜

⎞⎠⎟2

, (4.13)

while

e−

θ2

2σ 2 ≈ 1− θ2σ

⎛⎝⎜

⎞⎠⎟2

. (4.14)

Equating these last two relations we find that

σ ≈ 2kR

=2λ

πD. (4.15)

However, since the FWHM ΔΩ = 8 ln 2( )σ for a Gaussian beam we find that

ΔΩ ≈ 1.06 λD. (4.16)

We see from equations (4.12) and (4.16) that Δθ and ΔΩ are practically equivalent for specifying the resolution of a telescope (within this rather crude approximation, that is).

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5. Consider a globally neutral but weakly ionized plasma. Assuming that every charged particle has the elementary charge e , we write the equations of motion for the neutral, positive ion, and electron components of the gas as

mnnndvndt

= nµiν i v i − vn( ) + nµeνe ve − vn( ) + nnfn

mindv idt

= ne E + v ic× B⎛

⎝⎜⎞⎠⎟− nµiν i v i − vn( )

mendvedt

= −ne E + vec× B⎛

⎝⎜⎞⎠⎟− nµeνe ve − vn( ),

(5.1)

respectively. In equations (5.1) nn and n are the volume density for the neutrals and ions (and electrons), µi and µe are the reduced mass for ion/neutral and electron/neutral collisions, ν i and νe are the (constant) ion and electron collision rates with the neutrals, and fn accounts for other forces acting on the neutral fluid (e.g., pressure gradients and self-gravity). Because of the assumed low ionization (i.e., n nn ) and the dominating nature of electromagnetic interactions we have neglected in the last two equations momentum transfers due to collisions between electrons and ions, as well as any other forces acting on the ion and electron fluids.

Make the further assumption that the inertial terms (i.e., the acceleration) in the second and third of equations (5.1) are negligible. That is, assume that the Lorentz force and the friction forces arising from collisions with the neutral fluid dictate the ion and electron motions. Then show that

mnnndvndt

= −∇B2

8π⎛⎝⎜

⎞⎠⎟+14π

B ⋅∇( )B + nnfn , (5.2)

where j = ne v i − ve( ) is the current density. Solution. With the assumption that the inertial terms in the second and third of equations (5.1) are negligible, then combining these simplified ion and electron fluid equations yields

nµiν i v i − vn( ) + nµeνe ve − vn( ) = 1c j × B, (5.3) where j = ne v i − ve( ) is the current density. Inserting equation (5.3) in the first of equations (5.1) we transform the equation of motion for the neutral fluid to

mnnndvndt

=1cj × B + nnfn . (5.4)

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This last expression renders apparent the coupling of the neutral to the magnetic field through collisions with the charged particles. The strength of the coupling is dependent on the current density and will increase with charge separation.

Assuming that it is appropriate to approximate the Maxwell/Ampère law to

∇ × B = 4πcj, (5.5)

we transform the Lorentz force term of equation (5.3) to

4πcj × B[ ]i = ∇ × B( ) × B⎡⎣ ⎤⎦i

= εijkε jmn ∂mBn( )Bk= δkmδ jn − δknδ jm( ) ∂mBn( )Bk= ∂kBj( )Bk − ∂ jBk( )Bk= ∂kBj( )Bk − 12 ∂ j BkBk( )= B ∇ ⋅B( ) − 1

2∇B2⎡

⎣⎢⎤⎦⎥i.

(5.6)

Inserting this result into equation (5.4) we find

mnnndvndt

= −∇B2

8π⎛⎝⎜

⎞⎠⎟+14π

B ⋅∇( )B + nnfn . (5.7)

6. a) Use the HR diagram of Figure 6.1, as well as any other material (i.e., figures or tables) covered in class to evaluate the main-sequence turn-on ( ton ) and turn-off ( toff ) times for NGC 4755. Explain why ton > toff , and give the approximate time duration of star formation activity for this cluster.

b) Would you expect the above method for estimating the time duration of star formation to work equally well when only lower mass stars populate a cluster? For example, would you expect it to work very well for the Hyades cluster located in Taurus (also shown in Figure 6.1)?

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Figure 6.1 – H-R diagrams for four open clusters associated with molecular clouds. The ZAMSs (broken curves) and the best-fit isochrones are also shown. See Figure 4.20 of Stahler and Palla for a higher resolution figure (if needed).

Solution. a) For this problem we use Figure 1.18 and Table 1.1 of Stahler and Palla (or alternatively Figure 1.11 and Table 1.1 of the lecture notes). According to Figure 6.1 low-mass stars appear on the main sequence in NGC 4755 at a temperature of Teff 10

3.9 K 7,900 K . Referring to Figure 1.18 this corresponds to a star of approximately 1.5M , which requires about 10

8 yr of gravitational contraction to reach the main sequence; we thus have ton ≈ 10

8 yr . We also see from Figure 6.1 that

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more massive stars start leaving the main sequence at Teff 104.3 K 20,000 K , which

from Table 1.1 corresponds to a B2/B3 star ( 8 −10M ) with a main sequence lifetime (or turn-off time) toff ≈ 3×10

7 yr . Since low-mass protostars take a lot of time to contract under gravity and appear on the ZAMS, their high-mass counterparts, who are practically born on the ZAMS in this case (see Figure 1.18), will generally go through their main sequence lifetime and leave the ZAMS during that time; we then expect that ton > toff . For this object ton − toff ≈ 7 ×10

7 yr , which is the duration over which stars have been forming since toff and ton represent the latest and earliest epochs of star formation, respectively. b) It should be apparent that the ton > toff condition will only be met for clusters where there are stars of high enough masses that their main sequence turn-off time toff tms is at least shorter than the contraction time needed for the lowest mass stars to reach the ZAMS. Referring to Figure 1.18 we see that ton < 10

8 yr, which is the approximate main sequence lifetime for a 4M star (B8 spectral type; see Table 1.1). Therefore, we would not expect this technique to work too well for clusters that do not harbor stars at least as massive as this value. This is certainly the case for the Hyades cluster. According to Figure 6.1 main sequence turn-off happens in this cluster at Teff 10

3.9 K 7,900 K , which according to Table 1.1 corresponds to a 2M star (of A5 spectral type). Correspondingly, we find tms ≈ 1.1×10

9 yr for such a star, which is greater than the maximum turn-on time.

7. Let us assume that the specific intensity Iν emanating from some astronomical source can be adequately represented by a blackbody equation

Bν T( ) =2hν 3 c2

ehν kT −1. (7.1)

It follows that the flux can be written as Fν T( ) = Bν T( )ΔΩ, (7.2) with ΔΩ the solid angle of the telescope beam used to measure Fν T( ) . We further assume that the telescope beam is Gaussian, and we have the following definition ΔΩ = 2πσ 2, (7.3) where σ is the “standard deviation” equivalent of the beam (i.e., it is not its FWHM). Using calculations made in Problem 4 it is easy to show that

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ΔΩ = 4π

λD

⎛⎝⎜

⎞⎠⎟2

, (7.4)

with λ and D the wavelength at which the observations are made and the telescope diameter aperture, respectively. a) Since at radio through submillimetre wavelengths astronomer are equally likely to use either a temperature (in degrees K) or the Jansky (1 Jy = 1×10−23 erg cm−2 Hz−1 ) as units to quantify measured astronomical flux (either from a spectra line or continuum emission), use the Rayleigh-Jeans (i.e., long wavelength) approximation for the blackbody equation as well as equations (7.2) and (7.4) to show that

FνTA

∗ 8k

η ν( )πD2 . (7.5)

In equation (7.5) η ν( ) is the beam efficiency at frequency ν (see Chapter 3 of the Lecture Notes) and k the Boltzmann constant. If Fν T( ) is expressed in Jansky and the equivalent antenna temperature TA

∗ in K, then verify that for a 10.4-m diameter telescope

FνTA

∗ 32.5η ν( )

JyK

⎡⎣⎢

⎤⎦⎥. (7.6)

b) Can you comment on the adequacy of using the Rayleigh-Jeans approximation to assign a temperature to spectral line measured at, say, 850 GHz. Solution. a) The Rayleigh-Jeans approximation for the blackbody equation assumes that hν kT ; equation (7.1) then simplifies to

Bν T( ) 2kTν 2

c2

2kTλ 2

. (7.7)

Inserting this relation and equation (7.4) in equation (7.2) we have

Fν T( ) 2kTλ 2

ΔΩ

8kTπD2

. (7.8)

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Since the antenna temperature at a given frequency is defined as TA∗ =η ν( )T we then

find that

FνTA

∗ 8k

η ν( )πD2 . (7.9)

Using k = 1.38 ×10−16 erg K−1 and D = 1040 cm we find that

FνTA

∗ 32.5η ν( )

JyK

⎡⎣⎢

⎤⎦⎥. (7.10)

b) According to equation (7.10) a measured flux of 1000 Jy will yield a temperature T 30.8 K using the Rayleigh-Jeans approximation. But for that temperature at 850 GHz we have

hνkT

= 1.33, (7.11)

and the Rayleigh-Jeans approximation is therefore not appropriate. Instead, inverting equation (7.1) yields

T = hν k

ln 1+ 2hν3

c2⎛⎝⎜

⎞⎠⎟

ΔΩFν

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= hν k

ln 1+ 8hνπD2Fν

⎡

⎣⎢

⎤

⎦⎥

, (7.12)

which gives T 48.5 K for a flux of 1000 Jy at 850 GHz. This value is more representative of the blackbody-equivalent temperature of the astronomical source responsible for the measured flux.