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Boltzmann Distribution

  1. Jun 18, 2014 #1
    1. The problem statement, all variables and given/known data

    I have to find the boltzmann ditribution of a 1 dimensional ideal gas.
    The answer is given as:

    [itex]\frac{dn}{n}=\sqrt{\frac{m}{2piKT}}e^{(\frac{-mc^2}{2KT})}[/itex]

    For the second part I have to find the mean kinetic energy.

    2. Relevant equations / Attempt

    For part 1:
    I know how to work out the Boltzmann distribution for a 3D and 2D gas. However, for a 1D gas, I can't figure out what the constant has to be. I know the form to solve it is:

    [itex]\int_0^\infty C e^{(\frac{-mc^2}{2KT})} dv = 1 [/itex] (1)

    Where [itex]C[/itex] is a constant.
    However, when I do this and solve for [itex]C[/itex], I get a factor of 2 in front of my equation. Is there something wrong in my logic here? Am I meant to use a factor infront of my [itex]C[/itex]

    For the second part, I know that I need to get a [itex]v^2[/itex] infront of the exponential, but I cannot figure out how to do this for the 1D case and even for the 3D case.

    Any help would be much appreciated and please tell me if I need to clarify anything.
     
    Last edited: Jun 18, 2014
  2. jcsd
  3. Jun 18, 2014 #2
    New progress:

    Is it right to say that equation (1) should actually be equal to a half since the probability of the particle being on the positive side of the line is actually a hal and not 1?
     
  4. Jun 18, 2014 #3

    Orodruin

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    1. What do you have against negative velocities? Half the phase space consists of negative velocities...

    2. How are thermal averages defined? What is the expression for kinetic energy?
     
  5. Jun 18, 2014 #4
    1. It's the speed distribution, so I am looking for the absolute value of the velocity. This crashes my idea of it being equal to a half I realize.

    2. [itex] Average KE = \frac{1}{2}m\bar{v^2}[/itex]
    I now realize that the equation for [itex]\bar{v^2}[/itex] would be:

    [itex]\bar{v^2}=\int_0^\infty v^2\frac{dn}{n} dv [/itex]

    Then equate to 1 and solve.
     
  6. Jun 18, 2014 #5
    Looking at my OP, I can see that what I said can be a bit confusing, so here is the question.
     

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