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Home , Cross section (physics), Mean free path

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Order of Magnitude Astrophysics - a.k.a. Astronomy 111

Photon Opacities in Matter

If the for the relevant process that scatters or absorbs radiation given by σ and the number of scatters is n, then the of a is given by l = (nσ)−1. In the case of radiation, it is conventional to define a quantity κ (called opacity) such that

α ≡ nσ ≡ ρκ (1)

where ρ is the mass density of the scatterers. The of a system of size R is defined to be R τ ≡ αR = . (2) l The opacity for the is provided mainly by three different processes: (1) by free electrons; (2) the free-free absorption of photons, and (3) the bound-free transitions induced by matter by the photons that are passing through it.

Electron Scattering

The simplest case is the one in which the charged is accelerated by an electromagnetic wave that is incident upon it. Consider a charge q placed on an electromagnetic wave of amplitude E. The wave will induce an acceleration a ≈ (qE/m), causing the charge to radiate. The power radiated will be 2q2a2 2q4 P = = E2. (3) 3c3 3m2c3 – 2 –

Because the incident power in the electromagnetic wave per unit area is S = cE2/(4π), the

scattering cross section (for electrons with m = me) is !2 P 8π q2 σ ≡ ≈ . × −25 2, T = 2 6 7 10 cm (4) S 3 mec which is the Thomson scattering cross section. This cross section governs the basic scattering phenomena between charged and radiation. The corresponding mean free path for

−1 photons through a plasma is lT = (neσT ) , and the Thomson scattering opacity, defined to be

κ ≡ (neσT /ρ), is ! ! ne σT 2 −1 κT = = 0.4 cm g (5) np mp for ionized hydrogen with ne = np.

Random Walks and Diffusion

Diffusion: Macroscopic Theory

Diffusion is the random migration of or small particles arising from the motion of thermal energy. A particle at a temperature T has, on average, a kinetic energy associated with

1 movement along each axis of 2 kBT. Einstein showed in 1905 that this is true regardless of the size 1 2 of the particle. A particle of mass m and velocity vx along the x-axis has a kinetic energy 2 mvx. 1 2 1 This quantity fluctuates, but on the average 2 < mvx >= 2 kBT, where<> denotes an average over time or over an ensemble of similar particles. From this relationship we compute the mean-square

2 velocity < vx >= kBT/m, which we can use to estimate the velocity of, for example, a of the protein lysozyme. Lysozyme has a molecular weight of 1.4 × 104g. This is the mass of one mole, or 6 × 1023 molecules. The mass of one molecule is then m = 2.3 × 10−20g. The value of

2 1/2 o o 3 < vx > at 300 K (27 C) is 1.3 × 10 cm/s. This is a sizable speed. If there were no obstructions, the molecule would cross the classroom in about 1 second. Since the protein is not in but immersed in an aqueous medium, it does not go very far before it bumps into molecules of – 3 –

water. As a result, it is forced to wonder around: to execute a random walk. If a number of such molecules were confined initially they would wonder about in all directions and spread out. This is simple diffusion.

1d Random Walk

In order to characterize diffusive spreading, it is convenient to reduce the problem to its barest essentials and consider motion along one dimension. Let’s assume that the particles start at t = 0 at a position x = 0 and execute a random walk. Each particle steps to the right or to the left a distance l. The probability of going to the right at each step is 1/2 and the probability of going to the left is 1/2. That is, successive steps are statistically independent. We also assume that each particle moves independently of all others.

These rules have two consequences: (i) the particles go nowhere on the average and (ii) their root-mean-square displacement is proportional not to the time (or number of ) but to the square-root of the time. It is possible to establish these propositions by using an iterative

procedure. Consider an ensemble if N particles. Let be xi(N) the position of the ith particle after Nth steps:

xi(N) = xi(N − 1) ± l. (6)

The sign + will apply to roughly half of the particles and the − sign to the other half. The mean displacement of the particles after the Nth step is then

1 XN < x (N) >= x (N). (7) i N i i=1 – 4 –

On expressing xi(N) in terms of xi(N − 1), we find

1 XN < x (N) >= [x (N − 1) ± l] (8) i N i i=1 1 XN = x (N − 1) =< x (N − 1) > . (9) N i i i=1 The second term in brackets averages to zero, because its sign is positive for roughly half of the particles and negative for the other half. This equation tells us that the mean position of the particles does not change from step to step. Since particles all start at the origin, where the mean position is zero, the mean position remains zero. That is, the spreading of the particles is symmetrical about the origin. How much do the particles spread? A convenient way of measuring

2 1/2 2 the spread is the root-mean-square displacement < x (N) > . To find < x (N) >, we write xi(N) in terms of xi(N − 1):

xi(N) = xi(N − 1) ± l (10)

2 2 2 xi (N) = xi (N − 1) ± 2lxi(N − 1) + l . (11)

Then we compute the mean,

1 XN < x2(N) >= x2(N), (12) N i i=1 which is 1 XN < x2(N) >= [x2(N − 1) ± 2lx (N − 1) + l2] (13) N i i i=1 =< x2(N − 1) > +l2. (14)

As before, the second term in the brackets averages to zero. Since xi(0) = 0 for all particles i, < x2(0) >= 0. Thus < x2(1) >= l2, < x2(2) >= 2l2, ...., and < x2(N) >= Nl2. We conclude that the mean-square displacement increases with N. The particles execute N steps in a time t = NT, where T is the time between displacements, T = l/vx. This implies that

< x2(t) >= (t/T)l2 = (l2/T)t, (15) – 5 –

where we write x(t) rather than x(N). For convenience we define a diffusion coefficient, D = l2/(2T), in units of cm2/s. This gives < x2 >= 2Dt.

Scattering

A particular useful way of looking at scattering, which leads to important order-of-magnitude estimates, is by means of random walks. It is possible to view the processes of absorption, emission, and propagation in probabilistic terms for a single photon rather than the average behavior of large number of photons. For example, the exponential decay of a beam of photons has the interpretation that the probability of a photon traveling an optical depth τ before absorption is e−τ. Similarly, when radiation is scattered isotropically we can say that a single photon has equal probabilities of scattering into equal solid angles. In this way we can speak of a typical or sample path of a photon, and the measured intensities can be interpreted as statistical averages over photons moving in such paths.

Now consider a photon emitted in an infinite, homogenous scattering region. It travels a displacement r1 before being scattered, then travels in a new direction over a displacement r2 before being scattered, and so on. The net displacement of the photon after N free paths is

R = r1 + r2 + r3 + ··· + rN (16)

We would like to find a rough estimate of the distance | R | traveled by a typical photon. Simple averaging of equation [16] over all sample paths will not work, because the average displacement, being a vector, must be zero. Therefore, we first square equation [16] and then average. This yields the mean square displacement traveled by the photon:

2 2 2 2 2 l∗ ≡ hR i = hr1i + hr2i + ··· + hrNi + +2hr1 · r2i + 2hr1 · r3i + ···+ ···. (17)

Each term involvement the square of a displacement averages to the mean square of the free path of a photon, which is denoted l2. To within a factor of order unity, l is simply the mean free path of – 6 –

a photon. The cross terms in equation [17] involve averaging the cosine of the angle between the directions before and after scattering, and this vanishes for isotropic scattering. (It also vanishes for any scattering with front-back symmetry, as in Thomson or .)Therefore

2 2 l∗ = Nl , (18) √ l∗ = Nl. (19)

The quantity l∗ is the root mean square net displacement of the photon, and it increases as the square root of the number of scatterings. This result can be used to estimate the mean number of scattering in a finite medium. Suppose a photon is generated somewhere within the medium; then the photon will scatter until it escapes completely. For regions of large optical depth the number of scatterings required to do this is roughly determined by setting l∗ ∼ L, the typical size of the medium. From equation [19] we find

L2 N ≈ . (20) l2

Since l is of the order of the mean free path, L/l is approximately the optical thickness of the medium τ. Therefore, we have

N ≈ τ2, (τ  1). (21)

For regions of small optical thickness the mean number of scatterings is small, of order 1−e−τ ≈ τ; that is

N ≈ τ, (τ  1). (22)

For most order-of-magnitude estimates it is sufficient to use N ≈ τ2 + τ or N ≈ max(τ, τ2) for any optical thickness.