Dear Colleagues,
The classical form of speed distribution of particles
Ac²exp( -bc²)dc is valid for particles in 1D box or not. ?
Please discuss.
Thanks
N Das
I think, this elementary velocity distribution formula, considering ideal gas particles, is valid for 1D box. But, there could be important amendments in case of real gases, or in presence of different potentials or interactions.
Dear Souvik Bhattacharjee,
I am thinking of free particles in a 1D box.
Say , n number of particles ( identical and distinguishable) particles inside a 1 D box.
Then , dn(c) = A c²exp( -bc²)dc , this distribution is valid or not ?
I can't find the logic of the factor "c²".
Thanks
N Das
Hi Prof. Nityananda Das
The weighing factor c2 in a Maxwell distribution is a weighting speed factor in velocity space, it makes it possible to look for different shapes in the distribution according to their mass, for elements with more molecular mass the speed is smaller than for the light ones, let us say comparing H2 and O2.
The shape of the curve for O2 is more pronounced than for H2, the cause is a bigger mass and therefore smaller speeds for O2 molecules
The c2 factor stands in the integrated equation for f(v) but the dN/N 1D equation does not have it. Please check equations 1.7.5 and 1.7.6, also plot 1.7.3 in the following reference:
The Maxwell Distribution Laws. (2016, April 3). Retrieved May 5, 2021, from https://chem.libretexts.org/@go/page/47223
Best Regards.
Dear Dr. Nityananda Das
The reason of the factor "c2" is already explained well by Dr. Pedro L. Contreras E.
I just want to mention that, the square term is important to maintain the symmetric nature (isotropy) of the distribution function. So, the particles seem similarly contributing to the distribution, whether they move to the left or to the right.
Dear P Contreras,
Thanks for you nice answer.
c² factor comes due to volume in velocity space.
That's why I think that for 2D box , the formula should contain c , and for 1D , only the exp term.
Thanks a lot
N Das
Dear Dr. Nityananda Das
Have you noticed, that if the formula only contain 'c', it will be an odd function. So, integrating it from (-L/2) to (+L/2), L=Length of the box, will give a zero.
Secondly, if only the exponential decay term is there, you will not get any most probable velocity, because there will be no maxima in the plot. It will be a monotonically decreasing function of c.
Thank you, Profs. Souvik Bhattacharjee and Nityananda Das
I taught Statistical Physics for years, and I thank you for the question and answers because is it is interesting to look at other dimensions in Maxwell speed distribution f(v).
Prof. Nityananda Das, the dimensions "speed space rather than vector velocity space" comes from volume V space as you state in the previous post, and it comes exactly from an N-dimensional ball volume **, please see the attached CC Wikipedia commons page.
We ought to remember additionally, that the velocity f(v) distribution is slightly different from the speed f(v) distribution (F. Reif, Berkeley Physics course Vol V, 1967)
It is rather general from a math point of view the N Dim case, I also found a discussion in the N case in math.StackExchange forum ** and I can tell you that is really useful in calculating the number of allowed states Omega in Statistical Physics.
* https://math.stackexchange.com/questions/3709870/n-dimensional-maxwell-boltzmann-distribution
** https://en.wikipedia.org/wiki/Volume_of_an_n-ball
Best Regards.
Identical elastically colliding particles in a 1d box just exchange their velocities, so equilibration may be unreachable.
Interesting, Prof. Miccola Bondarenco
I always taught students in classical mechanics that total P, L and E for a classical system of N particles, follow conservation theorems (3-D) *, but I don't know, how do they get in thermodynamical equilibrium if they are totally isolated and confined to a 1D string.
* L. Landau & E. Lifshitz Mechanics, Butterworth-Heinemann 1982.
Best Regards.
Dear P Contreras,
Particles confined in 1D box and thermally insulated from surrounding, can be thought of a system of a microcanonical ensemble .
Am I wrong in thinking in this manner ?
Thanks
N Das
In one dimension the the speed varies from minus infinity to plus infinity . So average speed is zero .
In 1D gas dn(u) = A exp( -bu²)du
considering u direction only ,
So average velocity will be integration of
A exp( -bu²)udu
with the limit
- ♾️ to + ♾️ instead of 0 to ♾️
and as it is odd function value will be zero. So, average velocity in one dimension is zero.
Please check
Some of the answers above mix up speed and velocity. Speeds are never negative, so they will not average to zero. That is true regardless of dimensionality.
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