Boltzmann Distribution

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Homework Statement



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. Homework 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.
 
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Answers and Replies

  • #2
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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?
 
  • #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?
 
  • #4
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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.
 
  • #5
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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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