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Cts approximation, delta function integration, stat mech

  1. Mar 25, 2017 #1
    1. The problem statement, all variables and given/known data

    g(E) question.png

    2. Relevant equations


    3. The attempt at a solution

    So cts approx holds because ##\frac{E}{\bar{h}\omega}>>1##

    So
    ##\sum\limits^{\infty}_{n=0}\delta(E-(n+1/2)\bar{h} \omega) \approx \int\limits^{\infty}_{0} dx \delta(E-(x+1/2)\bar{h}\omega) ##

    Now if I do a substitution ##x'=x\bar{h}\omega## to loose the ##\bar{h}\omega## multiplying the ##x## , ##dx'=\bar{h}\omega dx##

    I get
    ## dx' \frac{1}{\bar{h}\omega}\int\limits^{\infty}_{0}\delta(x'-(E-\frac{1}{2}\bar{h}\omega)) ##

    Now, if I denote the region that ##x'## is integrated over by ##D## I get that this is:
    ##= \frac{1}{\bar{h}\omega} ## if ##x'=E-1/2\bar{h}\omega \in D##
    ##= 0 ## if ##x'=E-1/2\bar{h}\omega \notin D##

    The solution however has:

    gE sol.png

    ##= \frac{1}{\bar{h}\omega} ## if ##E>1/2\bar{h}\omega ##
    ##= 0 ## if ##E<1/2\bar{h}\omega ##

    Excuse me if I'm being stupid but I have no idea how we have converted the requirements of a certain value of ##x'## to lie inside the region of integration or not, which I believe is the definition of the delta function, to inequalities imposed on ##E##?


    Many thanks in advance
     
  2. jcsd
  3. Mar 27, 2017 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Don't you mean for [itex]dx'[/itex] to be inside the integral? If so, you're basically done. For any [itex]x_0[/itex],

    [itex]\int_0^{\infty} \delta(x' - x_0) dx' = 0[/itex] if [itex]x_0 < 0[/itex]
    [itex]\int_0^{\infty} \delta(x' - x_0) dx' = 1[/itex] if [itex]x_0 >0[/itex]

    So we have the particular case [itex]x_0 = E-\frac{1}{2}\bar{h}\omega[/itex]
     
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