Maxwells velocity distribution

In summary, the author is asking why a radial distribution is better than a volume distribution for particles with average speeds. The author also asks if it is possible to use a pdf or cdf method to get the same information.
  • #1
rabbed
243
3
On this page: http://galileo.phys.virginia.edu/classes/252/kinetic_theory.html
From the text beginning with "Let us now figure out the distribution of particles as a function of speed.",
whats the reasoning behind multiplying a radial distribution with a volume?
Is it the same thing as integrating the PDF radially, but only for one instantaneous value instead of carrying out the full integration?
Since we have a radial distribution and a formula relating spherical volume to the radius, shouldn't it be possible to use the PDF or CDF method to derive the volume distribution or is that not what we want?
 
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  • #2
To give you a simple example: to find the mass inside an arbitrary volume V, where the density ##\rho## is not uniform, you divide the volume into lots of infinitesimal small boxes with volume dV (called a volume element). The volume element at position (x,y,z), within V, has volume dV=dxdydz and mass dm=##\rho##(x,y,z)dxdydz
The total mass is found by adding up the masses of all the small boxes in the volume ... which is what "integrate" means: ##m= \int_V dm##

How would you go about using a pdf or cdf method do get to the same information?

But note: the author does not want to do a volume integral. What the author wants is the number of particles with speed between v and v+dv.
What pdf or cdf method did you have in mind for that?
 
  • #3
But there is a coordinate transformation going on also?
We know that the fraction of particles at point (x,y,z) is N * X_PDF(x)*dx * Y_PDF(y)*dy * Z_PDF(z)*dz
The fraction of particles at radius r is N * R_PDF(r)*dr
To go between these steps, do we need to use the volume element for spherical coordinates, r^2*sin(a)?
 
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  • #4
The number (not fraction) of particles in the box (in velocity coordinates) is given by: ##dv_xdv_ydv_z## is $$dn = NA^3e^{-Bv^2}dv_xdv_ydv_z : v^2=v_x^2+v_y^2+v_z^2$$

The fraction of particles in that box is dn/N.

What the author wants is the number dN between v and v+dv. Thus dN = Nf(v)dv so f(v)dv is the fraction of particles with speeds between v and v+dv.
There is no need to change to spherical coordinates here - the author exploits the symmetry.

It's like if you have a sphere in x-y-z coords, you know the volume between ##r=\sqrt{x^2+y^2+z^2}## and ##r+dr## is going to be ##dV=4\pi r^2\; dr## without having to change to spherical coordinates and integrating over the angles. You can do that if you like: you get the same result.

The author is just deriving the formula in terms of speed rather than velocity. call that a change in coordinates if you like.
 
  • #5
Hm, could you say the following:
N * R_PDF(r)*dr is NOT the number of particles inside either one or all volume elements of the spherical shell at radius r.
What is needed is the number of particles inside all volume elements of the spherical shell at radius r.
Instead, N * R_PDF(r)*dr is the number of particles inside the cubic volume element at a point (x,y,z) where r = sqrt(x^2+y^2+z^2).
That is why we need to multiply N * R_PDF(r)*dr with the number of cubic volume elements contained in the spherical
shell at radius r - 4*pi*r^2 number of elements?
 
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  • #6
Not really following your notation.
"R_PDF(r)"?

But that sounds kinda right. You can do that if the particle number density is spherically symmetric.
Take care - the author isn't talking about a literal volume.
 
  • #7
Thanks, I think I understand.
Sorry, with R_PDF(r) I mean the radial/speed distribution.

Can you also say that the maxwell distribution gives a constant gas pressure no matter if you randomize the average speed radially or from independent cartesian coordinates, which is the reason to have such a radial distribution and uniform direction instead of distributing average velocity as uniform points on a sphere or as uniform direction and uniform volume or any other distribution?

It kind of blurs out the dimension boundaries and difference in cartesian and spherical spaces in a way.

There are so many layers of thought to this distribution!
 
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1. What is Maxwell's velocity distribution?

Maxwell's velocity distribution is a probability distribution that describes the distribution of molecular velocities in a gas at a given temperature. It was first developed by James Clerk Maxwell in the 19th century and is based on the kinetic theory of gases.

2. How is Maxwell's velocity distribution calculated?

Maxwell's velocity distribution is calculated using the Maxwell-Boltzmann distribution, which takes into account the mass of the molecules, temperature of the gas, and the universal gas constant. The formula is: f(v) = (m/2πkT)^3/2 * 4πv^2 * e^(-mv^2/2kT), where m is the mass of the molecule, k is the Boltzmann constant, T is the temperature, and v is the velocity.

3. What is the significance of Maxwell's velocity distribution?

Maxwell's velocity distribution is significant because it provides a quantitative understanding of the behavior of gas molecules. It helps explain why gases at different temperatures have different kinetic energies and how this affects the overall properties of the gas, such as pressure and temperature.

4. How does Maxwell's velocity distribution relate to the ideal gas law?

Maxwell's velocity distribution is used to derive the ideal gas law, which describes the relationship between the pressure, volume, and temperature of an ideal gas. The distribution shows that at a given temperature, the average kinetic energy of gas molecules is directly proportional to the temperature, which is a key component of the ideal gas law.

5. Can Maxwell's velocity distribution be applied to all gases?

No, Maxwell's velocity distribution is most accurate for gases that behave as ideal gases, meaning they have negligible molecular interactions and occupy a large volume. Real gases may deviate from this distribution due to factors such as intermolecular forces and molecular size. However, the distribution can still provide a good approximation for many real gases under certain conditions.

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