Derivative Maxwell boltzmann distribution

  • Thread starter giraffe
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  • #1
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Homework Statement


i need to show that the peak of the maxwell boltzmann distribution is equal to 1/2 kt.

Homework Equations


maxwell boltzmann distribution according to modern physics 3rd edition by kenneth kramer.

ill try to do my best with this

[itex] N(E)= \frac{2N}{√∏} \frac{1}{(kT)^\frac{3}{2}} E^\frac{1}{2} e^\frac{-E}{kT}[/itex]

N is the total number of molecules while N(E) is the distribution function (with units energy to the -1) defined so that N(E) dE is the number of molecules dN in the energy interval dE at E. dn=N(E)dE

The Attempt at a Solution



so i need to take the derivative and set that equal to 0 and hope i get 1/2kt. im having trouble with the derivative itself. im taking the derivative with respect to E so everything else is considered a constant. so to try to make this easier i took all that junk in front of the E and said it is just some constant a. that allowed me to go through and do the product rule. after that, ive been trying to simplify it but am getting nowhere. need some advice on how to do this properly as i believe im not.

thanks a bunch
 

Answers and Replies

  • #2
ZetaOfThree
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Hi giraffe! Do you think you could show some more work so that we can see where exactly you are getting stuck?
 
  • #3
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i can try.

as i mentioned [itex] a = \frac{2N}{√∏} \frac{1}{(kT)^\frac{3}{2}} [/itex] which is just a constant since we are doing this with respect to E

so now you have [itex] (aE^ \frac{1}{2})(e^\frac{-E}{kT}) [/itex]

follow the product rule: derivative of first * second + first *derivative of second

[itex] \frac{aE^ \frac{-1}{2} e^ \frac{-E}{kT}}{2} - \frac{aE^ \frac{1}{2} e^ \frac{-E}{kT}}{kT} [/itex]

after that ive tried to combine the fractions and cancel some stuff out but nothing works.
 
  • #4
ZetaOfThree
Gold Member
110
23
Good job so far. Now remember that ##\frac{aE^ \frac{-1}{2} e^ \frac{-E}{kT}}{2} - \frac{aE^ \frac{1}{2} e^ \frac{-E}{kT}}{kT}=0## since we are talking about maximizing (the derivative must equal zero). Do you think you could do some cancellations to simplify?
 
  • #5
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ahhh...silly me...thanks...if i add the 2nd term over then the numerators will cancel out leaving 1/2 on left and 1/kT on right. multiply by kT and you have the 1/2kT=0 which is what i want.

thanks again for helping me with that simple mistake.
 
  • #6
ZetaOfThree
Gold Member
110
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ahhh...silly me...thanks...if i add the 2nd term over then the numerators will cancel out leaving 1/2 on left and 1/kT on right. multiply by kT and you have the 1/2kT=0 which is what i want.

thanks again for helping me with that simple mistake.

Whoa! The E's don't cancel since there's ##E^{-1/2}## on the left and ##E^{1/2}## on the right.
 
  • #7
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Correct. Rearranging you'll get 1/2kT=E^1/4 which is the final derivative. Set E=0 to find max.
 
  • #8
ZetaOfThree
Gold Member
110
23
Not correct. When you multiply both sides by ##E^{1/2}## you add the exponents of ##E##, not multiply them (which is what I think you did). If you then solve for ##E##, you find the value of ##E## that maximizes ##N(E)##. You don't have to take another derivative.
 

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