Mathematical Difference Between Mean Free Path vs RMS Free Path?

[RagincajunLA]

I was wondering if there is a mathematical difference between the RMS free path and the mean free path of molecules in an ideal gas. For example, When I calculate the mean free path, I use use the average velocity and the scattering rate which is a function of the average velocity. I then multiply the two and the average velocities cancel to produce the mean free path.

When I go to calculate the RMS free path, I assume I use the RMS velocity and the RMS scattering rate which is based on the RMS velocity. When I multiply these two, the RMS velocity cancels and I am left with the value as the mean free path. Is this correct? I assumed they would be different. My equations are...

A = \bar{v}\tau
\tau = \frac{1}{\sqrt{2}n \pi d^{2} \bar{v} }

A_{rms} = v_{rms}\tau_{rms}
\tau_{rms} = \frac{1}{\sqrt{2}n \pi d^{2} v_{rms} }​

 

[UltrafastPED]

Try reading this: http://teacher.pas.rochester.edu/phy121/lecturenotes/Chapter18/Chapter18.html

 

[sophiecentaur]

I was wondering if there is a mathematical difference between the RMS free path and the mean free path of molecules in an ideal gas. For example, When I calculate the mean free path, I use use the average velocity and the scattering rate which is a function of the average velocity. I then multiply the two and the average velocities cancel to produce the mean free path.

When I go to calculate the RMS free path, I assume I use the RMS velocity and the RMS scattering rate which is based on the RMS velocity. When I multiply these two, the RMS velocity cancels and I am left with the value as the mean free path. Is this correct? I assumed they would be different. My equations are...

A = \bar{v}\tau
\tau = \frac{1}{\sqrt{2}n \pi d^{2} \bar{v} }

A_{rms} = v_{rms}\tau_{rms}
\tau_{rms} = \frac{1}{\sqrt{2}n \pi d^{2} v_{rms} }​

If the bulk of the gas isn't moving then the average velocity is Zero, surely.

 

[TSC]

I think the root mean square speed is just an approximation to the mean speed. In equilibrium condition, it is a good approximation.
Mean speed is not mean velocity. In equilibrium mean velocity is zero.

 

[sophiecentaur]

It's a lot more than that. The mean square of the velocity is the mean kinetic energy of a set of particles with the same Mass. That corresponds to temperature, which is far more significant than just an 'approximation'.
The RMS Voltage in an AC circuit is the value of the DC equivalent voltage that would supply the same Power to a resistor.
RMS is a very useful quantity to get your head around. (The Standard Deviation of a statistical distribution is an excellent measure of the statistical 'spread' of a set of values.)

 

[TSC]

Arguable, in the determination of mean free path. it is the mean speed that is more accurate rather than the root mean square speed. The mean speed is quite different from the root mean square speed.

 

[sophiecentaur]

Arguable, in the determination of mean free path. it is the mean speed that is more accurate rather than the root mean square speed. The mean speed is quite different from the root mean square speed.

? more accurate?. In what way? Do you mean easier to measure accurately?
If you knew the statistics of the speed distribution (and you would have to, in any case) then you could deduce one from the other, in any case, depending upon which you were able to measure.

But i agree that, to know the time between collisions, it is the mean speed that counts. Does the speed influence the spacing, though? Surely, that is a function of the number of particles per unit volume.

 

[TSC]

Since we agree on that, I have to add, it is the mean relative speed.

 

[joseamck]

I think of it as if a distribution is Gaussian or Poisson... then which one is better to use? and in what sense does accurate now means? just saying

 

[sophiecentaur]

To return to the OP, I wonder what physical significance the RMS free path has. Apart from the fact that it may be 'possible to calculate', it strikes me as not having any particular meaning at all.
There are much quantities that would be of much more interest:
For instance, the frequency of collisions would be related to temperature (RMS speed) and MFP. Now, that has some physical significance. And the RMS speed, as mentioned before, would relate to temperature (always a relevant quantity)..

 

[TSC]

Mean free path = mean speed * mean time.
This sounds so intuitive physically, that is why mean speed is more meaningful.
Note that mean free path is going to affect your viscosity, which has to be decided by experiment.
If you use rms speed, you may not agree with experiment.

 

[sophiecentaur]

Mean free path = mean speed * mean time.
This sounds so intuitive physically, that is why mean speed is more meaningful.
Note that mean free path is going to affect your viscosity, which has to be decided by experiment.
If you use rms speed, you may not agree with experiment.

What I don't understand about this is how you intend to measure 'mean time'. You seem to be confusing a dependent variable with an independent variable. What experiments will give you mean time and mean speed, to let you calculate mean path?
Aamof, I even wonder whether the statistical distributions of time and speed lead to the valid use of your equation.

 

[TSC]

This is done by Clausius in 1858.
Imagine you are a molecule with a smiling face waiting for other molecules to hit. The area of your face is call the scattering cross section. Imagine all these molecules move toward you with a mean speed, and with the number density, and the fact that all these molecules lie within a certain length of tube with a cross sectional area equals to the scattering cross section, you can calculate the mean time = time between molecules hitting your face.
This is how I interpret Clausius's mean free path.