Unformatted text preview:

Mention: programming, probably second homeworkInitial question: Suppose that my boy Lord Voldemort waved his wand and stoppedall fusion in the Sun. How long would it take before we noticed the Sun’s light dimming?Radiation forcesBefore heading to scattering, one comment about radiation forces. The book has adiscussion on it (end of §1.4), but I think it’s unnecessarily complicated (and it’s not general,either). The easy way is to remember that force is change in momentum per time, F = dp/dt.The magnitude of the momentum of a photon is p = E/c, where E = ~ω is the photon’senergy. Therefore, to figure out the net force on a particle that is absorbing or scatteringradiation, just (1) compute the incident rate of momentum on the particle, (2) compute therate of momentum after scattering, and (3) take the difference.Scattering and Random WalksWe’re now going to take the first steps towards understanding how radiation interactswith matter. In this lecture we are going to spend substantial time on an idealization:coherent, frequency-independent, isotropic scattering, along with random walks. There aretwo reasons for this. First, it allows us to define a number of important quantities and exploreconcepts in a simple setting, which will provide a basis for more complicated processes.Second, the concepts of random walks and diffusion are easy to understand but are alsofar more general than they appear, so an understanding of them allows order of magnitudeestimates in a rather broad array of problems.So, let’s set up the physical situation: a photon travels along merrily, and at some pointit hits a particle and bounces off. In general the nature of the scattering could depend onlots of things, but for now we’ll assume the scattering is coherent (the photon energy doesnot change), frequency-independent (the rate or cross section of scatterings does not dependon the photon’s frequency), and isotropic (the probability of the photon scattering into aparticular direction is independent of the direction). That’s just to simplify things. Now weneed to define several quantities: cross section, opacity, mean free path, and optical depth.Cross sections.—Ask class: in everyday life, how would you define cross section? Often,simply its geometrical area. If you throw projectiles at me, the ones that hit will hit in myarea in profile, maybe 0.6 m2face-on, and half that edge-on. As long as the projectile is smallcompared to me, its nature doesn’t matter much; a water balloon or a rotten tomato willhave the same effective cross section of interaction. Ask class: can they think of anotherway to define it? Suppose there were to be a steady pelting by rotten tomatoes, but thateveryone had such bad aim that any area had the same probability of being hit as any otherarea. We can then define the tomato flux, Ftom, as the number per area per second. My crosssection is then Rtom/Ftom, where Rtomis the rate at which they hit me in number per second.In the microscopic world, it is this latter definition that is more general and therefore moreuseful.The cross section is not always independent of the nature of the projectiles. Ask class:if the “projectiles” are photons, can they think of an everyday example of this? A pane ofglass is nearly transparent to visible light, but is opaque to infrared, hence the greenhouseeffect. The effective cross section is defined as rate divided by flux, again, making theeffective cross section much less for visible light than for infrared. Similarly, for some reactionX + α → Y + β, we define a cross section σαβ(v) as the number of such reactions per secondper target divided by the incident flux of the projectiles. Then the reaction rate per volumeis σαβ(v)vnαnXcm−3s−1. If the target and projectile are the same, have to divide by twoto avoid double-counting. Ask class: can they think about how to get the average rate ifthere is a distribution of particle velocities? Have to integrate σαβv over that distribution.Opacity.—We can think of the cross section as the “effective area” of a single particle toa particular interaction (scattering in our case). We can also define the opacity, which is thetotal effective area per mass: κ is in cm2g−1. So, for a gram of material, the “effective area”to the given interaction is κ cm2. You might, therefore, be tempted to calculate opacity bysimply taking the total number of particles per gram and multiplying by the cross sectionper particle. Ask class: why is this wrong? Be careful! The particles that count are theones that are involved in the particular interaction, and may be a small fraction of the totalnumber. Example: the cross section to scattering of low-energy photons off of free electronsis the Thomson cross section σT= 6.65 × 10−25cm2. However, if material is neutral (andthere are therefore no free electrons), the opacity to Thomson scattering is zero! I’ll beobnoxiously reiterative about this, since the distinction between opacity and cross section isimportant conceptually.Mean free path.—This is the “average” distance traveled by a photon between interac-tions. For example, if the number density of scatterers is n cm−3and the cross section isσ cm2, then the mean free path (mfp) is l = 1/(nσ). Similarly, if the total mass density isρ g cm−3and the opacity to the scattering process is κ cm2g−1then l = 1/(ρκ). Note that ifyou forget these expressions you can get them from dimensional arguments. If the scatterersare distributed randomly (e.g., not in a lattice) then the probability that a photon will travela distance d or greater before scattering is exp(−d/l), from the Poisson distribution. Youcan verify that this distribution gives a mean distance of l. Ask class: can they give mean example of how a non-random distribution can mess this up? Imagine a perfect cubiclattice. In some directions, one can travel forever without hitting anything. We’ll almostalways deal with gasses or fluids, though, where the assumption of random distribution isgood.Optical depth.—If we imagine a path in a medium, parameterized by the variable s, thenthe optical depth τ isτ =Zdsl(s), (1)where l(s) is the mean free path along each point. In words, if you travel straight in somedirection (that is, you aren’t deflected by scatterings), the optical depth is the number ofmean free paths you have traveled. For example, if the distance is one mean free path, theoptical depth


View Full Document

UMD ASTR 601 - Lecture Notes

Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?