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Chapter 10

Chapter 10 The Derivation ofthe Planck FormulaTopicsThe Planck formula for black-body radiation. Revision of waves in a box. Radiationin thermal equilibrium. The equipartition theorem and the ultraviolet photoelectric effect. The wave-particle duality. Quantisation of radiation andthe derivation of the Planck spectrum. The Stefan-Boltzmann IntroductionIn the first lecture, we stated that the energy den-sity of radiation per unit frequency intervalu( )for black-body radiation is described by thePlanckformula(Figure ),u( ) d =8 h 3c31(eh /kT 1)d ( )wherePlanck s constant,h= 10 34J this lecture, we demonstrate why quantum con-cepts are necessary to account for this formula. Theprogramme to derive this formula is as follows. First, we consider the properties of waves ina box and work out an expression for the ra-diation spectrum in thermal equilibrium Spectrum of black body radiationf(x) dx x3ex 1dxin terms of the dimensionless 1 2 3 4 5 6 7 8 9 (x).

A0 sinkr0, where we have shifted the origin to the point ct along the r-axis such that r = r0 + ct. Thus, the expression for the wave after time t is A(r;t) = A0 sinkr0 = A0 sin(kr ¡kct): But, if we observe the wave at a flxed value of r, we ob-serve the amplitude to …

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Transcription of Chapter 10

1 Chapter 10 The Derivation ofthe Planck FormulaTopicsThe Planck formula for black-body radiation. Revision of waves in a box. Radiationin thermal equilibrium. The equipartition theorem and the ultraviolet photoelectric effect. The wave-particle duality. Quantisation of radiation andthe derivation of the Planck spectrum. The Stefan-Boltzmann IntroductionIn the first lecture, we stated that the energy den-sity of radiation per unit frequency intervalu( )for black-body radiation is described by thePlanckformula(Figure ),u( ) d =8 h 3c31(eh /kT 1)d ( )wherePlanck s constant,h= 10 34J this lecture, we demonstrate why quantum con-cepts are necessary to account for this formula. Theprogramme to derive this formula is as follows. First, we consider the properties of waves ina box and work out an expression for the ra-diation spectrum in thermal equilibrium Spectrum of black body radiationf(x) dx x3ex 1dxin terms of the dimensionless 1 2 3 4 5 6 7 8 9 (x).

2 X=h kT1 The Derivation of the Planck Formula2 Application of the law of equipartition of en-ergy leads to theultraviolet catastrophe, whichshows that something is seriously wrong withthe classical argument. Then, we introduce Einstein s deduction thatlight has to be quantised in order to accountfor the observed features of thephotoelectriceffect. Finally, we work out the mean energy permode of the radiation in the box assuming theradiation is quantised. This leads to Planck sradiation calculation indicates clearly the necessity ofintroducing the concepts of quantisation and quantainto Waves in a Box - RevisionLet us revise the expression for an electromagnetic (orlight) wave travelling at the speed of light in some ar-bitary direction, say, in the direction of the vectorr. Ifthe wave has wavelength , at some instant the ampli-tude of the wave in ther-direction isA(r) =A0sin2 r (see Figure ).

3 In terms of thewave vectork, we canFigure The properties of sine andcosine (r) =A0sin(k r) =A0sinkr,where|k|= 2 / . Wave vectors will prove to be veryimportant quantities in what wave travels at the speed of lightcin ther-directionand so, after timet, the whole wave pattern is shifted adistancectin the positiver-direction and the pattern isA0sinkr , where we have shifted the origin to the pointctalong ther-axis such thatr=r +ct. Thus, theexpression for the wave after timetisA(r, t) =A0sinkr =A0sin(kr kct).But, if we observe the wave at a fixed value ofr, we ob-serve the amplitude to oscillate at frequency . There-fore, the time dependence of the wave amplitude is sin(2 t/T)whereT= 1is the period of oscillation of the , the time dependence of the wave at any pointThe Derivation of the Planck Formula3is sin t, where = 2 is theangular frequencyof thewave.

4 Therefore, the expression for the wave isA(r, t) =A0sin(kr t),and the speed of the wave isc= Electromagnetic Modes in a Box More RevisionConsider a cubical box of sideLand imagine wavesbouncing back and forth inside it. The box has fixed,rigid, perfectly conducting walls. Therefore, the electricfield of the electromagnetic wavemust be zero at thewalls of the boxand so we can only fit waves into thebox which are multiples of half a wavelength. The firstfew examples are shown in Figure Waves which can be fitted into abox with perfectly conducting thex-direction, the wavelengths of the waves whichcan be fitted into the box are those for whichl x2=Lwhereltakes any positive integral value, 1, 2, 3,..Similarly, for theyandzdirections,m y2=Landn z2=L,wheremandnare also positive expression for the waves which fit in the box in thex-direction isA(x) =A0sinkxxNow,kx= 2 / xwherekxis the component of thewave-vector of the mode of oscillation in , the values ofkxwhich fit into the box are thosefor which x= 2L/land sokx=2 l2L= lL,whereltakes the valuesl= 1,2,3.

5 Similar resultsare found in theyandzdirections:ky= mL,kz= us now plot a three-dimensional diagram with axeskx,kyandkzshowing the allowed values ofkx,kyandkz. These form a regular cubical array of points, eachof them defined by the three integers,l, m, n( ). This is exactly the same as the velocity, or mo-mentum, space which we introduced for Derivation of the Planck Formula4 The waves can oscillate in three dimensions but the com-ponents of theirk-vectors,kx,kyandkz, must be suchthat they are associated with one of the points of thelattice ink-space. A wave oscillating in three dimen-sions with any of the allowed values ofl, m, nsatisfiesthe boundary conditions and so every point in the lat-tice represents a possible mode of oscillation of the waveswithin the box, consistent with the boundary Illustrating the values oflandmwhich result in components of wavevectorswhich can fit in the , in three-dimensions, the modes of oscillation canbe writtenA(x, y, z) =A0sin(kxx) sin(kyy) sin(kzz) ( )To find the relation betweenkx, ky, kzand the angularfrequency of the mode, we insert this trial solutioninto the three-dimensional wave equation 2A x2+ 2A y2+ 2A z2=1c2 2A t2.

6 ( )The time dependence of the wave is also sinusoidal,A=A0sin tand so we can find thedispersion relationforthe waves, that is, the relation between andkx, ky, kz,by substituting the trial solution into ( );|k|2= (k2x+k2y+k2z) = the three-dimensional wave-vector. Now,k2=k2x+k2y+k2z= 2L2(l2+m2+n2),and so 2c2= 2L2(l2+m2+n2) = 2p2L2,( )wherep2=l2+m2+ Counting the ModesWe now return to statistical physics. We need thenumber of modes of oscillation in the frequency in-terval to + d . This is now straightforward,since we need only count up the number of lat-tice points in the interval ofk-spacektok+ dkcorresponding to to + d . We carried out asimilar calculation in converting from dvxdvydvzto 4 Derivation of the Planck Formula5In Figure , the number density of lattice pointsis one per unit volume of (l, m, n) space.

7 We areonly interested in positive values ofl,mandnandso we need only consider one-eighth of the sphere ofradiusp. The volume of a spherical shell of radiuspand thickness dpis 4 p2dpand so the number ofmodes in the octant isdN(p) =N(p) dp=(18)4 p/Land dk= dp/L, we finddN(p) =L32 ,L3=Vis the volume of the box andk=2 /c. Therefore, we can rewrite the expressiondN(p) =V2 2k2dk=V2 28 3 2c3d =4 2Vc3d Finally, for electromagnetic waves, we are alwaysNote on the Polarisation of Electro-magnetic WavesIllustrating the electric and magneticfields of an electromagnetic wave. TheEandBfields of the wave are perpendicularto each other and to the direction of prop-agation of the wave. There is an indepen-dent mode of propagation in whichEandBare rotated through 90 with respect tothe direction of propagationC.

8 Any po-larisation of the wave can be formed bythe sum of these two independent modesof two independent modes, orpolarisations,per state and so we have to multiply the result bytwo. Because of the nature of light waves, there aretwo independent states associated with each latticepoint (l, m, n). The final result is that the numberof modes of oscillation in the frequency interval to + d isdN=8 2Vc3d Thus,per unit volume, the number of states isdN=8 2c3d ( )This is areally important The Average Energy per Mode andthe Ultraviolet CatastropheWe now introduce the idea that the waves are inthermodynamic equilibrium at some temperatureT. We showed that, in thermal equilibrium, weaward12kTof energy to each degree of is because, if we wait long enough, there areThe Derivation of the Planck Formula6processes which enable energy to be exchanged be-tween the apparently independent modes of oscil-lation.

9 Thus, if we wait long enough, each mode ofoscillation will attain the same average energyE,when the system is in thermodynamic , the energy density of radiation per unitfrequency interval per unit volume isdu=u( ) d =8 2c3Ed ,u( ) =8 , we established in Section that the averageenergy of a harmonic oscillator in thermal equilib-rium isE=kTand so the spectrum of black-bodyradiation is expected to beu( ) =8 2c3E=8 2kTc3.( )The Rayleigh-Jeans Lawu( ) =8 result isvery bad news the energy densityof radiationdiverges at high frequencies. Einsteinexpressed this result forcibly 0u( ) d = 08 2kTc3d .The Ultraviolet CatastropheAccording to classical physics, the energydensity of black-body radiation diverges athigh frequencies 08 2kTc3d .This is the famous result known as theultravioletcatastrophe the total energy in black-body ra-diation diverges.

10 This expression for the spectralenergy distribution of radiation is known as theRayleigh-Jeans Lawand, although it diverges athigh frequencies, it is in excellent agreement withthe measured spectrum at low frequencies and hightemperatures. This result was derived by LordRayleigh in 1900. This was one of the key prob-lems of classical physics what has gone wrong? The Photoelectric Effect RevisionThe solution was contained in Einstein s revolutionarypaper of 1905 in which he proposed that, under certaincircumstances, light should be regarded as consisting ofThe Derivation of the Planck Formula7a flux of particles, what we now callphotons, each withenergyh , wherehis Planck s constant. He discoveredthis result from a remarkable analysis of the thermody-namics of radiation in the Wien region of the black-bodyspectrum.


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