Integral of Maxwell-Boltzmann

[Monty Hall]

I'm reading http://www.ndsu.edu/fileadmin/physics.ndsu.edu/Wagner/LBbook.pdf (pg 12, 3.2.2) about the lattice Boltzmann method. I'm speaking of his specfic form of the Maxwell-Boltzman and his claim that the integral of equilibrium distribution is equal to n

$$\int f^0 = n$$

$$f^0(v)=\frac{n}{(2\pi\theta)^{3/2}}e^{-(v-u)^2/2\theta}$$

But when I use gaussian integrals I don't get n but rather $$\frac{n}{2\pi\theta}$$. What am I missing?

 

[zhermes]

From a quick glance, your answer seems correct; if the integral was to be just n, then the leading denominator would have to be

$$\sqrt{2 \pi \theta}$$

 

[Monty Hall]

Thanks for your reply. When I look at the wikipedia version of the maxwell-boltzman, I see they do a triple integral over basically the same equation, there I get N. From how the author wrote the equation, there was no indication that he was speaking of velocity and not speed like bold face or overarrows. Plus it doesn't help that I'm not familiar with statistical mechanics either).

 

[zhermes]

That makes sense; poor form that they didn't make that more clear--either they should have had all 3 serifs, or vectors over their velocities or something.

 

[CellCoree]

Thank you for sharing your thoughts on the integral of Maxwell-Boltzmann and its relationship to the lattice Boltzmann method. It is important to note that the integral of the equilibrium distribution is indeed equal to n, as stated in your quote. However, it is also important to consider the assumptions and conditions that are necessary for this to hold true.

The Maxwell-Boltzmann distribution is a probability distribution that describes the speed of particles in a gas at equilibrium. It assumes that the particles are non-interacting and that the system is in thermal equilibrium. In the context of the lattice Boltzmann method, this distribution is used to model the behavior of fluid particles in a lattice grid.

The integral of the Maxwell-Boltzmann distribution, as shown in your quote, is indeed equal to n. However, this is only true if the integral is taken over the entire velocity space, which is from -∞ to +∞. When using Gaussian integrals, it is important to note that the integral is taken from 0 to +∞. This is because the Maxwell-Boltzmann distribution is only valid for positive velocities.

So, when you use Gaussian integrals to calculate the integral of the Maxwell-Boltzmann distribution, you are only considering a portion of the velocity space and not the entire distribution. This is why you are getting \frac{n}{2\pi\theta} instead of n.

In summary, the integral of the Maxwell-Boltzmann distribution is equal to n, but this is only true if the integral is taken over the entire velocity space. When using Gaussian integrals, it is important to consider the limits of integration and the assumptions of the Maxwell-Boltzmann distribution. I hope this helps clarify any confusion.