Blackbody radiation (Text 2.2)

Transcription

1 Blabody radiation (Text.) How Raleigh and Jeans model the problem:. Next step is to alulate how many possible independent standing waves are there per unit frequeny (ν) per unit volume (of avity). It is easier to alulate the number of possible independent standing waves (states) per unit first, sine is related to ν. z -spae d Note that the states are paed together lose together uniformly in the -spae, beause L is large. y With n1, olume per state in -spae L x L y L z LxLyL z ( volume of avity) Therefore, the total number of possible standing waves within the spherial shell 1 olume of spherial shell 8 olume per state x How many possible states are there within this thin spherial shell between radii and +d? Eah state represent a possible standing wave defined by wave vetor ( x, y, z ) in the avity. There are two possible polarizations for a transverse standing wave: possible polarizations 1/8 of a omplete sphere E E

2 Blabody radiation (Text.) How Raleigh and Jeans model the problem:. (Con t) The total number of possible standing waves within the spherial shell 1 olume of spherial shell 8 olume per state 1 4 d 8 d But is related to ν: νλ ν/ /λ ν/ Therefore, the total number of standing waves between ν and ν+ is: d ν Definition: Density of states G(E) is the number of states per unit energy E per unit volume of sample, Number of states between energy E and E+dE In present ase, G(E)dE Density of standing waves G(ν) is the number of possible standing waves per unit frequeny ν per unit volume of avity, Number of possible standing waves between frequenies ν and ν + G(ν)d ν Equal G( ν ) G( ν ) or G( ν ) (Eq.(.1) of text)

3 Blabody radiation (Text.) How Rayleigh and Jeans model the problem:. Density of wave funtion allows us to alulate the total of any physial quantity (say, f(ν ) that is a funtion of frequeny: Total f for the whole system f( ν )G( ν ) An example is energy E(ν), but how does E depends on ν? Rayleigh assumed the lassial law of equipartition energy. He said, one dimensional waves always have degrees of freedom, one for potential energy (x) and the other for ineti energy (v). In ase of eletromagneti wave, these two degree of freedom are derived from eletri field and magneti field. 0 Classially, eah degree of freedom has an energy of B T/. So the energy for eah standing wave is B T: E(ν) B T B is the Boltzman onstant: B J/K or me/k This is THE fatal assumption! Boltzman

4 Blabody radiation (Not in text) If you understand the previous slide, you should be able to answer the following sample GRE problem: A three-dimensional harmoni osillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of the osillator is (A) 1/T (B) T (C) /T (D) T (E) 6T

5 Blabody radiation (Text.) How Raleigh and Jeans model the problem: 4. Final touh. If we define the spetral density as the energy radiated per unit frequeny per unit volume. u( ν ) u( ν ) G( ν )E( ν ) u( ν ) G( ν )E( ν ) B T Rayleigh-Jeans Formula Ha! Ha! I got it! Note how this follows Wien's Law : u( ν ) u( λ) dλ u( λ) u( ν ) dλ ν - λ dλ λ BT 8 BT u( λ) λ λ λ 8 BλT 5 λ f ( λt) with 5 λ I am satisfied! f ( λt) 8 λt B Rayleigh Wien

6 Blabody radiation (Text.) But don t be happy so soon! Raleigh-Jeans formula blow up as ν!

7 Blabody radiation (Text.) Those density of state stuff is just simple ounting and arithmeti, and I an t see anything wrong in it. To mae the theoretial urve fit the experimental urve, the energy of the standing waves has to depend on ν. In other wards, the equipartition law has to be disarded. After trying many times, I find E( ν ) (h is a onstant) e B T 1 will produe a perfet math between theoretial urve and experimental data. Why don t you all the onstant h under my name? Max Plan h Js

8 Blabody radiation (Text.) Following Plan s suggestion : u( ν ) E( ν ) u( ν ) T e 1 8 h u( ν ) B B Although the final result is orret, but the way we present the derivation is ind of misleading and atually wrong (after we now the right physis). The proper way to do it is: u( ν ) 1 u( ν ) BT e 1 8 h ν u( ν ) (Eq.(.4 in text)) BT e 1 In other words, we made two mistaes in the original derivation : 1. We over simplify n( ν )E( ν ) into a single E( ν ) and ignore the statistial term n( ν ). n( ν )G( ν )E( ν ). E( ν ) should be instead of our more ompliated form. (The ompliated stuff omes from n( ν ).) e ν T 1 (Eq.(.4 in text)) This equation math experimental data perfetly! Note that this equation also follow Wien s Law! I am satisfied also! Wien One mistae will mae things wrong, but two mistaes may mae things orret!