Physics 576 Stellar Astrophysics Prof. James Buckley. Lecture 2 Radiation

Transcription

1 Physics 576 Stellar Astrophysics Prof. James Buckley Lecture 2 Radiation

2 Reading/Homework Assignment Read chapter 1, sections 1.1, 1.2, 1.5 Homework will be assigned on Thursday.

3 Radiation

4 Radiation A quick review of some E&M and thermodynamics Maxwell s equations (in MKS units): $%&&'$"&!&$%&&'$!&(&$)&&'$*! " # $)'$* B =0 % E = ρ 0 B = µ 0 o E t + µ 0 J E = B ( ) t ( E)= ( E) 2 E #'((%#!(!(#'((%#)(*(#$((%#& )! " #$%#& t B = µ E t 2 ' 2 E µ0 0 2 E t 2 =0 Wave Equation

5 Electromagnetic Waves Wave equation has a general solutions that can be written as a weighted sum of particular solutions (basis functions) such as plane wave solutions of the form: E = E 0 sin( k r ωt) B = B 0 sin( k r ωt) E = B t k E 0 = ω B 0 k, B, E are mutually perpendicular

6 Phase Velocity 1 E(x 1,t 1 )=1 0.5 t = t 1 E = E 0 sin(kx ωt) E x,t E(x 2,t 2 )=1 E x,t x x = x 1 x = x 2 t = t 2 kx 1 ωt 1 = π/2 kx 2 ωt 2 = π/2 k(x 2 x 1 )=ω(t 2 t 1 ) v phase x 2 x 1 t 2 t 1 v phase = ω k = 1 µ x c = 1 µ0 0 in all frames!

7 Accelerating Charge

8 Dipole Antennas If you drive a transmission like at high frequencies, charge and current can not keep up, producing waves of charge and current down the line As long as the lines are close together, no radiation, but if you spread apart the two wires of the transmission line, power is radiated into space

9 Dipole Radiation My Mathematica solution of Maxwell s equations for radiated power from a dipole radiation source

10 Black Body Radiation Interactions of electric fields of photons with charges in the wall result in acceleration of charges, and re-radiation, thermal equilibrium Radiation emerging from small hole is black-body radiation

11 1D Standing Waves Solutions in cavity are actually standing waves. Above, is a mathematical solution for traveling, reflected and resultant standing wave solutions in a 1-D cavity. Below I show 3-D standing waves in a rectangular cavity n x =1,n y =1,n z =1 n x =3,n y =3,n z =1

12 Density of States L L E x E y E z 2 1 c 2 2 E t 2 =0 L boundary conditions E = cos(k x x) cos(k y y) cos(k z z) cos(ωt) k 2 = k 2 x + k 2 y + k 2 z = ω2 c 2 z da =4πk 2 dk da = 4 k dk " 2 k x = n xπ L,k y = n yπ L,k z = n zπ L k dk y how many states lie within dk or k? dn = 4πk2 dk (π/l) 3 x k!!!!!""# k= π/l L ρ(ν) = dn dν dν = 4πν2 dν c 3

13 Ultraviolet Catastrophe! More and more standing wave modes (per unit frequency) at higher and higher frequencies (i.e., more modes in UV than red) A guess from statistical mechanics is that each unique frequency corresponds to a unique mode of the system and should get an equal share kt of the energy. This results in the Ultraviolet catastrophe where one expects the radiated power per unit frequency or wavelength interval to grow without bound at shorter wavelengths

14 Max Planck A 42 year old German physicist saved the world from the ultraviolet capacity, using nothing but a powerful new constant of nature! Planck s postulate: Any physical entity with one degree of freedom whose ``coordinate is a sinusoidal function of time (i.e., executes simple harmonic oscillations) can possess only a total energy E=n h Applying this to electromagnetic radiation in a cavity, the energy of any standing wave mode is E=n h

15 Discussion Question (a) (b) Starting with the configuration of pool balls at left, which of the final states (a or b) are more likely?

16 Thermodynamics All microstates are equally probable Can define macroscopic states (e.g., one macroscopic state could correspond to 9 balls on the left side and 10 balls on the right) The state of thermodynamic equilibrium is the macroscopic state with the larges number of microstates (w) associated with it. The entropy of the system is can be defined as: S k ln(w) So the state of thermodynamic equilibrium corresponds to the state of maximum entropy.

17 Boltzman Distribution Consider a system of 5 (classical) particles with total energy 4Δε. Assume that each particle can only have an energy that is n times some unit Δε Division label ε=0 ε=δε ε=2δε ε=3δε ε=4δε Number of distinguishable divisions !/3!=20 3 5!/(3!2!)=10 4 5!/(2!2!)=15 5 5!/4!=5 n(e = 0) = = 145 n(e = E) = = 70 n(e =2 E) = = 35 n(e =3 E) =1 20 = 20 n(e =4 E) =1 5=5 n(e =5 E) =0 Probability of one particle taking a large share of the available energy is exponentially suppressed

18 Boltzman factor Say you have a system with N 1 particles with a total energy E Intuitively, you expect that the average energy of a particle will be E/N and that is unlikely for one particle to have much more energy than this. Assume that any exact distribution of energies (a microstate) is equally likely Now, say that one particle has energy E 1 and the other E 2 and the remainder of the system has E E 1 E 2 to share Say that the probability of finding one particle with energy E 1 is P (E 1 ) the probability of finding another particle with energy E 2 is P (E 2 ) (and this doesn t depend on E 1 )so the probability of finding two particles in such states is P(E 1 )P (E 2 ) But this must be the same as the probability of the rest of the system having E (E 1 + E 2 ) call this function of (E 1 + E 2 ),Q(E 1 + E 2 ) if N 1, probability of one particle with (E 1 + E 2 ) is also Q(E 1 + E 2 )

19 Boltzman Factor So we have P (E 1 )P (E 2 ) P (E 1 + E 2 ) The only kind of function that behaves this way is an exponential : P (E) =e αe Evaluating the average energy, and defining Temperature, T E = 0 E e αe de 0 e αe de = 1 α = kt And we obtain the famous Boltzman factor : P (E) = 1 kt e E/kT