1 Lecture 26

<ul style="display: flex;"><li style="flex:1">PHYS 445 Lecture 26 - Black-body radiation I </li><li style="flex:1">26 - 1 </li></ul>Lecture 26 - Black-body radiation I What's Important: ··number density energy density Text: Reif In our previous discussion of photons, we established that the mean number of photons with energy is i1n = (26.1) ieß - 1 iHere, the questions that we want to address about photons are: ···what is their number density what is their energy density what is their pressure? Number density n Eq. (26.1) is written in the language of discrete states. The first thing we need to do is replace the sum by an integral over photon momentum states: Si ® òd 3p Of course, it isn't quite this simple because the density-of-states issue that we introduced before: for every spin state there is one phase space state for every h 3, so the proper replacement more like Si ® (1/h 3) òd 3p òd 3r The units now work: the left hand side is a number, and so is the right hand side. But this expression ignores spin - it just deals with the states in phase space. For every photon momentum, there are two ways of arranging its polarization (i.e., the orientation of its electric or magnetic field vectors): where the photon momentum vector is perpendicular to the plane. Thus, we have Si ® (2/h 3) òd 3p òd 3r. (26.2) Assuming that our system is spatially uniform, the position integral can be replaced by the volume òd 3r = V. © 2001 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited. PHYS 445 Lecture 26 - Black-body radiation I The total number of photons is then 26 - 2 <ul style="display: flex;"><li style="flex:1">V</li><li style="flex:1">d3p </li></ul>N =2 <ul style="display: flex;"><li style="flex:1">.</li><li style="flex:1">(26.3) </li></ul>ßh3 e - 1 ò<ul style="display: flex;"><li style="flex:1">This integral is written in terms of energy </li><li style="flex:1">and momentum p; for photons, these </li></ul>quantities are related through Einstein's relation (m = 0) = pc. Dividing N by V gives the integrated number density of photons ng, which can be written in an integral form as 2h3 d 3p n = (26.4) ßpc òe- 1 or as with [number per unit volume between p and p+dp] = ng(p) d 3p 2 d3p n (p)d3p = <ul style="display: flex;"><li style="flex:1">.</li><li style="flex:1">(26.5) </li></ul>h3 eßpc - 1 Let's integrate Eq. (26.4) for an isotropic system, where òanglesd 3p = 4pp 2dp: 8p p2dp n = .ßpc <ul style="display: flex;"><li style="flex:1">h3 </li><li style="flex:1">e</li></ul>- 1 òChanging variables to x = ßpc gives ¥8p x2dx n = (ßhc)3 0 e - 1 xòThe integral has the form of a Reimann Zeta function, and is ¥x2dx =2 (3) =2 · 1.202 xò0 e - 1 Thus, 3æk T hc öøBn =1.202· 16p (26.6) èExample The universe is filled with relic radiation left over from the Big Bang with a temperature of 2.7 K. What is ng of this radiation? ö3 ø1.38 ´ 10-23 · 2.7 æçn =1.202· 16p 6.63 ´ 10-34 · 3.0 ´ 108 è= 4.0´ 108 m- 3 =400cm- 3 © 2001 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited. <ul style="display: flex;"><li style="flex:1">PHYS 445 Lecture 26 - Black-body radiation I </li><li style="flex:1">26 - 3 </li></ul>Energy density u We saw that the number of photons per unit volume between p and p+dp is 2 d3p h3 eßpc - 1 n (p)d3p = .3These photons have an energy of pc, giving an energy density ug(p)d p in this momentum range of 2 d 3p h3 eßpc - 1 u (p)d3p =pc . Taking the integral of this over momentum gives the total energy density ug for an isotropic system, 8p p2dp 8pc p3dp u = pc =.h3 eßpc - 1 h3 e- 1 ßpc <ul style="display: flex;"><li style="flex:1">ò</li><li style="flex:1">ò</li></ul>where òanglesd 3p = 4pp 2dp as before. Once again changing variables to x = ßpc 4¥ 4 ¥ 8pc ækT x3dx ækT x3dx <ul style="display: flex;"><li style="flex:1">öø ò </li><li style="flex:1">ø ò </li></ul>ö<ul style="display: flex;"><li style="flex:1">u = </li><li style="flex:1">=8phc </li></ul>x<ul style="display: flex;"><li style="flex:1">h3 </li><li style="flex:1">c</li></ul>0 ex - 1 hc 0 e - 1 <ul style="display: flex;"><li style="flex:1">è</li><li style="flex:1">è</li></ul>The integral is equal to ¥4x3dx p =.xò0 e - 1 15 Thus, 8p (pkBT)4 u = <ul style="display: flex;"><li style="flex:1">.</li><li style="flex:1">(26.7) </li></ul>15 (h c)3 Example u = Find the energy density of photons at room temperature T = 293 K. 8p (p· 1.38 ´ 10- 23 · 293)4 =5.5 ´ 10- 6 J/m3. 15 (6.63 ´ 10-34 · 3.0 ´ 108)3 Mean energy per photon From Eqs. (26.6) and (26.7) we can determine the mean energy per photon 8p (kBT)4 5·415 (h c)3 u np<ul style="display: flex;"><li style="flex:1">=</li><li style="flex:1">=</li></ul>kBT =2.7kBT (26.8) (kBT )3 1.202· 30 1.202· 16p (hc)3 Notice once again how kBT sets the energy scale. © 2001 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited.

1 Lecture 26

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