Phys 344 Ch 7 Lecture 6 April 6 th, HW29 51,52 HW30 58, 63 HW31 66, 74. w ε w. π ε ( ) ( )

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1 Phys Ch 7 Lecture 6 April 6 th,009 1 Mon. /6 Wed. /8 Fri. /10 Mon. /1 Review 7. Black Body Radiation he Rest 7. Debye Solids 7.6 Bose-Einstein Condensate HW9 1, HW0 8, 6 HW1 66, 7 HW6,7,8 HW9,0,1 Equipment: Blowtorch & metal wire (to heat up.) Blackbody Radiation -lab. o ariacs and lightbulbs Handheld spectrometers Optical pyrometer he Plank Spectrum = = = dnw/ n p/ w n p / wnw / np / w( ) d w w 8 π 1 8 π = g( ) n p/ w( ) d = d = d or β = d β β We can do this integral to see how much energy we ve got per unit volume otal Energy x ( ) ( ) = d dx = = Γ ζ = β x ( ) e 1 ( ) e 1 ( ) ( ) 0 ch ch 0 ch 1 ch o Here, we recognized the integral from Appendix B. Sometimes one talks not about the total energy density of the volume, but the energy density at a particular photon energy: u ( ) β ch e ( ) 1 hen again, it s interesting to see where all the energy is; which states contribute the most energy to the system. o here s a balance: As we go higher in energy, there are more photons states with the same energy and each photon has more energy; then again, the average occupancy of higher energy states decreases with higher energy. β 1 β e 1 e β max

2 Phys Ch 7 Lecture 6 April 6 th,009 Demo: Blackbody spectrum. Lab (If I can get it to work) Wien s Law Looking at the spectrum, there s clearly a peak energy / frequency / wavelength. We can find it by maximizing the function. = 0 β e β β β β ( ) β β 1 e β = 0 = 0 he two outer factors easily give 0 at = 0, where the plot is indeed minimized, the inner factor though is not analytic. β = 0 β 1 e a value for β can be found by first plotting the left β 1 ( 1 e ) = β side and the right side and seeing where they intersect, then running the numbers until the two sides are close enough to equal. his hc happens around β =.8 =.8k =. 8k. λ It is this relationship that underlies optical pyrometery. Without having to touch thermometer to object, you can tell how hot a radiator is if you identify the most popular wavelength. his works on the sun, a lightbulb filament, even you and I (if you filter out the visible spectrum light). Demo: Optical Pyrometer See how matching colors (peak wavelengths) allows us to determine temperature. Prep for Pr. 1: One part of the problem asks what would be the filament temperature with the maximum visible efficiency, i.e., with as much of its radiated light in the visible range as possible. Graphically, what would the distribution look like? he peak is in the middle of the visible range. S S 1 Pr. 1: Einstein coefficients of Spontaneous and Induced emission & adsorption. Since they addressed this in quantum, hopefully looking at it again in this class will help them to make connections. Imagine a gas of atoms. Consider any two internal states, s 1 and s, of an atom. Let s be the higher energy state so that (s ) (s 1 ) =. his is then the energy of a photon admitted or adsorbed during a transition between the two states, ph = hf = (s ) (s 1 ). o make things simple, let s say that state is the only

3 Phys Ch 7 Lecture 6 April 6 th,009 one that feed into state 1 and vice versa. Let s consider the rate at which population 1 dn changes: 1. dt hree mechanisms o Spontaneous Emission: Atoms in state could spontaneously drop down into state 1, and in the process emit a photon. he rate at which this happens in the gas should depend on the population of atoms in state, N. = AN dt spon. emission o Adsorption: It could change by adsorbing photons of frequency f and transitioning to state. his rate should be proportional to the population in state 1, N 1, and the population of photons of frequency f, which in turn is proportional to the energy density at that frequency, u(f) = BN 1u( f ) dt adsorb o Stimulated Emission: It could be that when a photon of the transition frequency comes by, f, it stimulates an atom in state to drop down to state 1. he rate with which this happens in the gas should be proportional to the population of atoms in state, N, and the photons of the right frequency, which is proportional to u(f). = B Nu( f ) dt stimemmis. Adding these all together, o = AN BN 1u( f ) + B Nu( f ) dt Einstein s relations between the coefficients o In equilibrium (so these processes are allowed to run enough to balance the populations), the ratio of the two atomic populations should simply be N ( 1) β hfβ the ratio of their probabilities: = e = e. And there should be N1 no net change in either population: hf β = 0 = AN BN 1u( f ) + B N u( f ) so A = ( Be B ) u( f ) dt hfβ hf o Putting in our u(f), we have A = ( Be B ), the easiest way ( ) hfβ c e 1 to make this temperature independent is if B=B, then we have: hfβ hf A = B( e 1) = B hf. ( ) hfβ ( ) c c Significance o Applying perturbation theory to Quantum mechanics (pag of Liboff s Intro to Q.M.) tells us what B=B is, so then Einstein s reasoning allows one to determine A, and so the rate of spontaneous emission. he perturbation is the electric field oscillating at f. his gives rise r r to an oscillating perturbation on the Hamiltonian: H = d E(t) (dipole times oscillating electric field).

4 Phys Ch 7 Lecture 6 April 6 th,009 In the presence of this, the wave function of the atom is not an unperturbed energy eigenstate, but it can be built of a sum of them. If the frequency of the field s oscillation equals the frequency difference between the un-perturbed state the atom was in and another un-perturbed state, then there s a probability that the atom will end up in the new state when the perturbation is over. Prep for Pr. : Given a temperature, you ll find the fraction of energy in the visible spectrum, i.e., you ll evaluate the integral for specific limits. o do this, note that x x e in this range, so you can evaluate a simpler integral. 7.. Specific Heat and Entropy of a Photon Gas π C = = k 1 1 Recall : ds v, N = d, and d = d = Cd, N o So Prep for Pr. : C S( ) = N = π π d = k d = k n = ns w / ns ( ) = s s g( ) ns ( ). d = Prep for Pr. 8: You ll consider neutrinos, which are virtually massless fermions. 7.. he Cosmic Background Radiation Go over experimental and heoretical approaches to this problem. In the early stages of the universe, matter was densely enough packed and was ionized so that it interacted strongly with light. he light was then necessarily in thermal equilibrium with the matter. As the universe expanded and cooled, charged particles were able to combine, forming neutral particles, meanwhile the density of mass decreased. At that point, around 000 K, the light became fairly free to travel. he light from that time is still traveling, creating a background radiation. However, as the universe continues to expand, the light s wavelengths gets stretched more and more red shifting it. his shifts the distribution of photons as if it were cooler and cooler. 7.. Photons Escaping through a Hole At what rate is energy radiated out of a hole? his has application in power transmission and thermometry, since (not surprisingly) this rate is related to temperature. cdt Rdθ R θ A

5 Phys Ch 7 Lecture 6 April 6 th,009 Consider the photons in a differential chunk of volume d = ( cdt)( Rdθ )( Rsin θdφ ) hey account for an energy d = d where is the energy density we ve found, =. 1 he photons in this differential volume are uniformly headed every which way. So the fraction that are headed for the opening is equal to the solid angle the Acosθ opening subtends over the full sphere of possible directions, π. / π. R So, the amount of energy headed for the hole from this small volume is Acosθ A d headed. out = dfheaded. out = cdt Rdθ R sin θdφ = cdt cosθ sin θdθd πr π ( )( )( ) ( ) φ Integrating over the ½ spherical shell of volume at distance R gives all the energy in that shell and headed for the opening. headed. out = π π / π / A A = π π ( cdt) cosθ sin θdθdφ = ( cdt) π sin θd(sin θ ) ( cdt) φ= 0θ = 0 Finally, all this energy, moving at speed c, will go out the hole in time dt, i.e., the rate at which the energy radiates is headed. out A A π P = = c = c = A dt 1 1c h Stefan s Law o P = Aσ was experimentally determined in However, with out Planck s reasoning and introduction of h, this could not be derived, thus σ could not be recognized as the jumble of fundamental constants it is. 7.. Radiation from Other Objects he author argues that a black body of equal size and temperature must radiate equal energy. his is based on the principle when two bodies are in thermal contact, heat will flow from the hotter to the cooler. his in turn follows from the second law of thermodynamics as such a process maximizes entropy. he argument says that if two bodies were of the same area and at the same temperature, then there must be no net heat flow. his means that a cavity and a simple blackbody must radiate the same amount of energy. Furthermore, if we imagine selectively filtering the light, so only a certain band of light can be exchanged, the same reasoning says that the energy in that band of light must be the same for the cavity and the blackbody. So, the whole spectrum must be the same. Emissivity o Say we hold up our cavity and a reflective object, then, some of the light that the cavity shines on the object bounces right off again, not affecting θ = 0 A

6 Phys Ch 7 Lecture 6 April 6 th,009 6 the temperature. he object must be responsible for thermally reradiating only that fraction of the light which is adsorbed, e. hen the power radiated is P = σea. o Another way of thinking of it is that the emissivity quantifies how well the object s thermal energy is translated into light. Prep for Pr.. You ll consider the energy you radiate. In comparing it with the energy you eat, recall that dietary calories are really kilocalories he Sun and the Earth Our favorite blackbody radiator. Here on Earth, 170 W/m of energy are radiated down on us from the sun. Since we re m away, and assuming spherically uniform radiation, we have P P 11 6 I = = P = I πr = (170W / m )π ( m) =.9 10 W A πr Now, the area of the sun is about m. So, assuming e=1, we have everything we need to calculate the sun s temperature from P = σea which yields 800 K. Similarly, we have everything we need to calculate the peak wavelength, around 880 nm. o Note: his is where the most energy is invested. he most popular wavelength is another matter all together. he Earth s Adsorption and Radiation o In a crude model, the Earth has an emissivity of about 0.7. o racking the energy exchange means considering all three players Earth, Sun, and Atmosphere. he atmosphere and Earth radiatively, conductively, and convectively exchange energy. So, the math to track the exchanges gets complicated.