Misinterpretation of the Maxwell velocity distribution?

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I suddenly have a problem with the famous Maxwell velocity distribution. The maximum of this bell-shaped curve is commonly interpreted as the most probable velocity for a particle (see for instance https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution: "The most probable speed, vp, is the speed most likely to be possessed by any molecule..")
But according to Boltzmann's distribution the most probable energy for a particle is zero (decreasing exponential: P(E=Ei) = Exp(-Ei/kT)/Σ), so we should expect that the most probable velocity for a randomly picked particle is zero as well? but the probability of velocity zero is zero in the Maxwell distribution!
thank you in advance if you can enlighten me

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The Boltzmann factor is $e^{-E/(k_B T)}$, but to get the Maxwellian distribution, this gets multiplied by the density of states per unit energy interval $n(E)$. If you look over the derivation carefully, you will see how this factor $n(E)$ arises in the computation of the number of states that is first counted in momentum space. The factor $n(E)$ is proportional to $E^{1/2}$, making the distribution function equal to zero at $E=0$.

Indeed and my question is about the physical meaning of this Maxwellian distribution.
Very concretely: the most probable speed of any particle is $\sqrt{2k_BT/m}$ or zero ?

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The most probable speed is $v_{mp}=\sqrt{\frac{2k_BT}{m}}$. The average speed $\bar{v}=\sqrt{\frac{8 k_B T}{\pi \, m}}$, and the r.m.s. velocity is $v_{rms}=\sqrt{\frac{3 k_B T}{m}}$. It can be helpful to memorize each of these. $\\$ And no, the most probable speed is not zero, basically for the same reason that the most probable energy is not zero either. $\\$ Edit: It might interest you that the average velocity is zero. $\\$ It might also interest you, because the density of states is uniform w.r.t. volume on such a graph, if we make a 3-D graph with x-velocity, y-velocity, and z-velocity on the 3 axes, and then plotted (with density proportional to the shading) a probability density function on it=the Boltzmann factor, then you would indeed find the most probable velocity would, in fact, be at $\vec{v}=(v_x,v_y, v_z)=(0,0,0)$.

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The most probable speed is vmp=√2kBTm
this result is indeed the peak of the Maxwell-Boltzmann distribution but is it really the most probable speed of a molecule? I don't think the ordinate of this distribution is the probability of velocity. If that were the case, this distribution would also predict that there are no stopped molecules in the population, which I do not find obvious. Do you think that no molecule is motionless in a box of air in equilibrium?

the most probable energy is not zero
Boltzmann energy probability is $P(E=Ej)= e^{-Ej/(k_B T)}/\sum_i e^{-Ei/(k_B T)}$. It is maximal when $Ej=0$ isn't it?

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I recommend you study this some more. See also https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Rate_Laws/Gas_Phase_Kinetics/Maxwell-Boltzmann_Distributions $\\$ To answer your last question, the sum is over the states $j$. The result is the distribution function. In counting states, they are counted in momentum space. See also https://www.physicsforums.com/threads/boltzmann-vs-maxwell-distribution.918232/#post-5788233 In determining the density of states, see posts 2 and 4 of this "link". In counting the density of states in energy space, $E=p^2/(2m)$, the result is a continous function rather than a discrete one, and this continuous function is proportional to $E^{1/2}$. The result is that the density of states that get counted in the sum over $j$ has value zero at $E=0$.

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vanhees71
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You must be careful to distinguish "speed" and "velocity". Anyway, velocity is not the best concept in kinetic theory, because as all of physics from a fundamental point of view everything happens in phase space, and there momentum rather than the velocity is the "natural" variable.

For an ideal gas the Boltzmann distribution in non-relativistic approximation indeed reads
$$f(\vec{p})=\exp[-(\vec{p}^2/(2 m)-\mu)/T],$$
where $\mu$ is the chemical potential and $T$ the temperature. This distribution gives the number of particles in a phase-space cell $\mathrm{d}^3 r \mathrm{d}^3p$ to be
$$\mathrm{d} N = \mathrm{d}^3 r \mathrm{d}^3p \frac{1}{(2 \pi \hbar)^3} f(\vec{p}).$$
Now if you want distributions for other quantities, which are functions of phase-space positions $(\vec{r},\vec{p})$ you have to consider the appropriate Jacobians for the phase-space volume.

E.g., if you want the distribution for speed you need
$$v^2=\vec{p}^2/m^2,$$
$$\mathrm{d}^3 p = \mathrm{d} p \mathrm{d}^2 \Omega p^2.$$
Now
$$p^2=m^2 v^2 \; \Rightarrow \; \mathrm{d}p p=m^2 \mathrm{d} v v.$$
Then you get
$$\mathrm{d}^3 p =\mathrm{d} v \mathrm{d}^2 \Omega m^3 v^2,$$
which implies
$$\mathrm{d} N = \mathrm{d}^3 r \mathrm{d} v 4 \pi m^3 v^2 \frac{1}{(2 \pi \hbar)^3} \exp[-(m v^2/2-\mu)/T].$$
Note that the proper treatment of the Jacobian always yields the right dimension: $\mathrm{d} N$ must be dimensionless, and that's the case for both the phase-space distribution as well as the speed distribution derived from it (thanks to the Planck action quantum $\hbar$ in the formula!).

I hadn't seen that thread thank you; my question is indeed close to Toby's:
The particle energy E is distributed over a Boltzmann distribution. In a purely translational system like a monoatomic gas, all the energy is kinetic ($E=mv^2/2$), so $v$ is not expected to follow a bell curve but a half gaussian.

By (mis?)interpreting the Maxwell bell curve as the probabilities of values of $v$, the immobile molecules are absent from the system, whereas with the half Gaussian, the most frequent value of $v$ for a molecule is zero: a radically different result, hence my question: do you think that motionless molecules are present or absent in a gas?

vanhees71
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Again you must be careful. The energy distribution is again different. Here you need to transform from $\vec{p}$ to $E$. Since
$$E=p^2/(2m) \; \Rightarrow \; \mathrm{d} E=\mathrm{d} p p/m$$
you get
$$\mathrm{d}^3 p = \mathrm{d} p \mathrm{d}^2 \Omega p^2 =\mathrm{d} E \mathrm{d}^2 \Omega m \sqrt{2m E},$$
and integrating over the solid angle you get
$$\mathrm{d} N =\mathrm{d}^3 r \mathrm{d} E 4 \pi m \sqrt{2mE} \frac{1}{(2 \pi \hbar)^3} \exp[-(E-\mu)/T].$$
I don't understand what you mean by "half Gaussian". The Boltzmann distribution is a distribution for $\vec{p}$ and not $p$. See my previous posting!

Distributions are named distributions not only because of the physical meaning of distributions as some quantity per unit space or unit phase space but also mathematically, because they transform under variable transformations with the Jacobian of the transformation. The Jacobian is thus very important. Otherwise you get the problems you are asking about!

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You can use the speed distribution result to compute the fraction using an integral, that have speeds that are very small, e.g. speed $v<.01 \sqrt{\frac{k_B T}{m}}$. You can compare that to the fraction that have $\sqrt{\frac{2 k_BT}{m}}<v<\sqrt{\frac{2 k_B T}{m}}+.01 \sqrt{\frac{k_B T}{m}}$. You will find the fraction with small speeds to be very small, perhaps surprisingly small. I think Heisenberg's uncertainty principle keeps us from saying that they are ever completely motionless. $\\$ And change the .01 to a .001 and you will see an even more dramatic effect. The fraction will be extremely small for the molecules that have that slow of a speed compared to the number that are at the same size interval, but at the most probable speed.

I don't understand what you mean by "half Gaussian"
$P(v)= \sqrt{\frac{2m}{\pi k_{B}T}} \ \large{e}^{-\frac{mv^{2}}{2 k_{B}T}}$

the fraction with small speeds to be very small, perhaps surprisingly small.
To be very clear: do you mean "the fraction of particles with small speeds is very small ? Thanks for that clear answer, actually in agreement with the widespread interpretation of Maxwell's distribution... but I have trouble convincing myself of it!

DrClaude
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Yes, the fraction of motionless particles is vanishingly small. If you have trouble visualising it, think of the following.

Start with a box (you can keep it 2D for simplicity) in which you some balls, in the absence of gravity and friction. The balls can move around, collide with each other and the walls. Now, start with all balls motionless, except for one to which you impart a great force. Can you see how, after a relatively short while, you basically will not have a single ball standing still?

vanhees71
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$P(v)= \sqrt{\frac{2m}{\pi k_{B}T}} \ \large{e}^{-\frac{mv^{2}}{2 k_{B}T}}$
This is a Gaussian (normalized) distribution, i.e., it gives the probability to find a particle with one component of the velocity $v$. Here $v \in \mathbb{R}$. It's not the correct probability distribution for the speed, which is $|\vec{v}| \geq 0$.

To be very clear: do you mean "the fraction of particles with small speeds is very small ? Thanks for that clear answer, actually in agreement with the widespread interpretation of Maxwell's distribution... but I have trouble convincing myself of it!
I mean what I wrote: You have to carefully take into account the Jacobian when changing the variable of which you want to describe the distribution (no matter whether it's a probability distribution or a particle-number distribution).

Yes, the fraction of motionless particles is vanishingly small. If you have trouble visualising it, think of the following.

Start with a box (you can keep it 2D for simplicity) in which you some balls, in the absence of gravity and friction. The balls can move around, collide with each other and the walls. Now, start with all balls motionless, except for one to which you impart a great force. Can you see how, after a relatively short while, you basically will not have a single ball standing still?
I see but if all the balls move, none is energyless, which is not predicted by Boltzmann's energy distribution over the particles?
Consider another way to see the balls in the box: all are moving and colliding, but following a collision, a fast ball can be brutally slowered and even stop ?

it gives the probability to find a particle with one component of the velocity $v$.
It is the velocity norm (or the component along the trajectory) such that $Ec=mv^2/2$

DrClaude
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I see but if all the balls move, none is energyless, which is not predicted by Boltzmann's energy distribution over the particles?
You get a similar distribution for energy, since there is a simple relation between speed and kinetic energy. I think you are forgetting to factor in the degeneracy factor (or density of states). You get a similar situation when consider the distribution of rotational states of molecules in a gas, see https://commons.wikimedia.org/wiki/File:Distribution_of_rotational_states_of_nitrogen_gas_at_298_K.png

Consider another way to see the balls in the box: all are moving and colliding, but following a collision, a fast ball can be brutally slowered and even stop ?
The probability of a collision where one of the balls would stop is extremely small.

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I see but if all the balls move, none is energyless, which is not predicted by Boltzmann's energy distribution over the particles?
Consider another way to see the balls in the box: all are moving and colliding, but following a collision, a fast ball can be brutally slowered and even stop ?
You need to remember that there is motion in 3-dimensions for these particles. If the motion of the particles were all in one dimension, there would be a tremendously higher probability that the particle would come to a stop after a collision. If it was just one-dimensional motion, the speed distribution function would be the half-Gaussian that you previously mentioned. $\\$ You could observe the motion in just one dimension and have an artificial speed function for that one-dimensional motion, and I think that might be exactly what you are trying to do. $\\$ In computing the speed distribution function, all 3 dimensions are part of that distribution. In the process, it picks up a $v^2$ factor on the Boltzmann factor in the conversion of counting states: $dp_x \, dp_y \, dp_z=4 \pi \, p^2 \, dp$ with a $(4 \pi ) \, m^3$ factored out to give $v^2 \, dv$.

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Lord Jestocost
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To be very clear: do you mean "the fraction of particles with small speeds is very small ? Thanks for that clear answer, actually in agreement with the widespread interpretation of Maxwell's distribution... but I have trouble convincing myself of it!
In making the step from the Boltzmann “energy distribution” to the Maxwell “speed distribution”, the Boltzmann energy distribution function must be multiplied by an velocity dependent factor to account for the density of velocity states available to the particles. Have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/maxspe.html#c4

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In making the step from the Boltzmann “energy distribution” to the Maxwell “speed distribution”, the Boltzmann energy distribution function must be multiplied by an velocity dependent factor to account for the density of velocity states available to the particles. Have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/maxspe.html#c4
Yes. And that $v^2$ factor results for the reason I mentioned in post 17.

thank you very much for all your kind answers; I will think about them deeply... but slowly
Regards

mjc123
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Consider the statement in #4:
It might also interest you, because the density of states is uniform w.r.t. volume on such a graph, if we make a 3-D graph with x-velocity, y-velocity, and z-velocity on the 3 axes, and then plotted (with density proportional to the shading) a probability density function on it=the Boltzmann factor, then you would indeed find the most probable velocity would, in fact, be at $\vec{v}=(v_x,v_y, v_z)=(0,0,0)$ .
The most probable velocity - that is, the most probable individual combination of vx, vy and vz values - is (0,0,0). Any other particular combination is less likely, but for a given speed v > 0 there are many combinations that give the same v, e.g. for v = √3 you could have (1,1,1) or (1,-1,-1) or (√2,1,0) or (0,√3,0) or many more. The probability of v in the interval v → v + dv is proportional to 4πv2dv f(v). That is why v = 0 is so improbable.
Likewise in energy terms, the density of states increases with energy, so with appropriate coordinate transformation, the distribution function is proportional to E1/2e-E/kT, and it does not have a maximum at E = 0.

DrClaude
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In making the step from the Boltzmann “energy distribution” to the Maxwell “speed distribution”, the Boltzmann energy distribution function must be multiplied by an velocity dependent factor to account for the density of velocity states available to the particles.
I'm confused by what you are saying here. When I hear "distribution," I figure that this means a probability distribution function. In that case, even the energy distribution will contain a factor $\sqrt{E}$ coming from the density of states (as @mjc123 just mentioned while I was typing this). I think you are meaning the Boltzmann factor in stead of "Boltzmann energy distribution."

The most probable velocity - that is, the most probable individual combination of vx, vy and vz values - is (0,0,0). Any other particular combination is less likely, but for a given speed v > 0 there are many combinations that give the same v, e.g. for v = √3 you could have (1,1,1) or (1,-1,-1) or (√2,1,0) or (0,√3,0) or many more. The probability of v in the interval v → v + dv is proportional to 4πv2dv f(v).
understood. Now imagine a graph whose ordinate would be "fractions of particles" (not density of velocity states) and the abscissa would be "velocity norms" (in which the different 3D combinations you enumerate collapse). According to you, this graph would be bell-shaped or monotonically decreasing?

That is why v = 0 is so improbable
v=0 is indeed reduced to a single combination of 3D components, but to me not improbable because the velocity component combinations are equiprobable only for given norms, owing to the principle of symmetry, while the norms are directly related to the energy distribution

@Kairos For 3-D combinations, what gets compared is the volume of a sphere at the origin of radius $dv$, resulting in $V_1=\frac{4}{3} \pi (dv)^3$,compared to the volume of a spherical shell of surface area $4 \pi \, v^2$ and thickness $dv$, resulting in $V_2=4 \pi \, v^2 \, dv$. The spherical shell has tremendously more volume. The spherical shell also has a non-unity Boltzmann factor, but the overwhelming difference is the volumes of the two. $\\$ In the previous post, you ask, is the curve bell-shaped? The curve is $f(v)=A v^2 e^{-m v^2/(2 k_B T)}$, ($v>0$). It looks a little like a bell, but it is non-Gaussian.
Hi, I rethought this question after your advices. If we momentarily drop the calculations with the shell volumes and return to physics: Boltzmann says that in an ideal monoatomic gas, there is a certain fraction of particles without energy, then after conversion into velocities, all the molecules start moving! (see #12) This calculation would therefore have the magical effect of giving energy to energyless molecules. For me, this paradox is unacceptable and is explained by a wrong interpretation of the Maxwell distribution ordinate: it does not represent the fractions of particles at the different velocities, but velocity densities (a more abstract notion!) where the runs at the same speed from right to left and from left to right are distinguished (as explained in #21). I can't give up the idea that the velocity occupied by the largest number of molecules could be 0 and not $\sqrt{2k_BT/m}$.