Why most probable energy and speed do not correspond?

[meowts]

TL;DR Summary

(mv(most probable)^2)/2 != E(most probable)

Greetings. I am studying M.C.T. and Thermodynamics. A question has arisen concerning the most probable kinetic energy. As we know, the most probable kinetic energy is equal to E(m.p.) = kT/2, and the average is equal to E(mean.) = 3kT/2. Moreover, if we take velocities that seem to correspond to these energies, then everything holds for E(mean) and E(mean) = mv(mean square)^2/2, but this equality does not hold for the most probable energy. The physics teacher asks for a physical justification, i.e. the following explanation: "In the transition from F(v) to F(E), F(v) is not multiplied by a constant, but by dv/dE, so the maximum shifts" does not suit him. Please help explain why everything works with the average energy, but with the most probable energy there is such a shift. Thank you.

 

[sophiecentaur]

"Most Probable" is the Mode of a distribution (the peak of the curve). For a skewed distribution, the Mode and Mean do not necessarily coincide.

 

[meowts]

"Most Probable" is the Mode of a distribution (the peak of the curve). For a skewed distribution, the Mode and Mean do not necessarily coincide.

why?

 

[sophiecentaur]

This link could convince you; it's a matter of definition. The three terms which are used to describe a form of 'average' and they each are most suitable for specific cases of data distribution.

If you don't like that link then try Googling "Mean, Mode and Median".

 

[mjc123]

The question, as I understand it, was not about the difference between mean and mode, but why the mean speed corresponds to the mean energy, but the most probable speed doesn't correspond to the most probable energy.

First, it's wrong: the mean speed does not correspond to the mean energy. Mean energy corresponds to the mean value of v², which is not equal to the square of the mean of v. Likewise the most probable energy corresponds to the most probable value of v², which is not the square of the most probable value of v.

Your teacher is wrong; the solution is mathematical, not physical, because the problem is mathematical, not physical. It's about how we define the probability. For a continuous distribution, the probability of getting exactly any particular value is zero. We have to define an interval, and say that the probability of the energy being between E and E+dE is F(E)dE. For constant intervals dE, this is a maximum when F(E) is a maximum.

Because E is directly proportional to v², a constant interval dE corresponds to a constant interval d(v²), and F(v²) is simply equal to F(E) multiplied by a constant, and has a maximum at the value of v² corresponding to the value of E for which F(E) is a maximum.

But E is not directly proportional to v, so a constant interval dE does not correspond to a constant interval dv. For a given interval, the probability F(E)dE = F(v)dv, and so
F(v) = F(E)(dE/dv)
F(v) has a different functional form from F(E), and has a maximum at a value of v not corresponding to the maximum of F(E). The probability F(v)dv is a maximum for this value of v for constant intervals dv, and this is what we mean by "most probable v". This difference in the definitions of most probable E and v is the source of the apparent discrepancy.

 

[sophiecentaur]

First, it's wrong: the mean speed does not correspond to the mean energy.

Agreed - the mean velocity is zero, in any case and using the absolute value of v is hardly ever advisable. The distribution curves for v and v² are different in all systems.

TL;DR Summary: (mv(most probable)^2)/2 != E(most probable)

Please help explain why everything works with the average energy, but with the most probable energy there is such a shift.

The way I read it, the question is why the mean and most probable energies are not the same. There's a bit of a red herring in the post where plain v is discussed but that's not a suitable way to describe what's happening, in any case (is it?). The answer to the above quote is just that you can't 'expect' mean and mode to be the same; if they were then why use the two values to discuss any distribution? (Except for a symmetrical, single humped distribution, of course but that's a special)

 

[jbriggs444]

Agreed - the mean velocity is zero, in any case and using the absolute value of v is hardly ever advisable. The distribution curves for v and v2 are different in all systems.

Advisable or not, if we are to compare mean speed to mean energy, we are comparing the average of the absolute value to the average of the square. Though I do agree and was taught to avoid absolute value.

As you understand perfectly well, the distributions for $v$, $|v|$ and $v^2$ will all be different.

If we are going to compare apples to apples, then we would not compare the mean of $|v|$ to the mean of $v^2$. The one is a speed and the other is roughly an energy. We would want to take the square root before comparing. *voila* -- a root mean square average.

A root mean square mean is different from the regular mean: RMS = $\sqrt{\frac{\sum v^2}{n}}$
A harmonic mean is different from a regular mean. HM = $\frac{n}{\sum \frac{1}{v}}$
A geometric mean is different from a regular mean. GM = $\sqrt[n]{\prod v}$ = $e^{(\frac{\sum \ln v}{n})}$

One can see a whole family of means lurking here: The "$f$"-mean = $f^{-1}(\frac{\sum f(x)}{n})$

 

[sophiecentaur]

we are comparing the average of the absolute value to the average of the square.

But the OP specifically compares mean with most likely (mode). Whether you use Energy or Speed graphs, the two 'averages' will differ. Imo the OP is at a basic level and my basic statement about statistics is the appropriate one.
Does your list of alternative forms of average help to answer the OP's "why" question?

 

[jbriggs444]

Does your list of alternative forms of average help to answer the OP's "why" question?

It does not answer the why question.

It does suggest that if there are many flavors of average, they must sometimes yield different values. Else, what is the point?

We could address the "why" question with examples.

Data: 1, 1, 3, 5, 100

Mean: ( 1 + 1 + 3 + 5 + 100 ) / 5 = 110 / 5 = 22
Median: 3
Mode: 1
Root mean square: $\sqrt{( 1 + 1 + 9 + 25 + 10000 ) / 5} \approx \sqrt{2007.2} \approx 44.80$
Harmonic mean: $\frac{5}{1 + 1 + 1/3 + 1/25 + 1/10000} \approx \frac{5}{2.37} \approx 2.11$
Geometric mean: $\sqrt[5]{1 \times 1 \times 3 \times 5 \times 100} \approx 4.32$

[Almost any data set would do. You want an odd number of values so that the median is easy. You want more than three values because otherwise the median will match the mode. You want something asymmetric, otherwise the median will match the mean. You want the mode to be off center, otherwise the mode will match the median. It is nice if the mean comes to a round number. I was too lazy to pick values to make more round numbers than that].

 

[sophiecentaur]

It does suggest that if there are many flavors of average, they must sometimes yield different values. Else, what is the point?

It would be good if the OP were to take that on board. It would make this discussion worth while.

 

[Vanadium 50]

First, as mentioned earlier, mean is not most likely.

However, there is another issue. The probability of any givenm speed (and speed is different from velocity as well) is zero. We can only speak of a range of energies or speeds. If you bin the data in bins of equal ranges of energies, they are not binned in equal bin ranges of speeds and vice versa. e.g. bins of 0-1,1-2, and 2-3 in speeds corresponds to 0-1, 1-4 and 4-9 in energies.

If the OP were to have posted a plot - with units - this would have been apparent. A good exercise for them would be to find and examine such plots.

 

[vanhees71]

The point is that distributions are, well, distributions, i.e., they transform accordingly when changing the independent variables you are interested in. Obviously you are considering an ideal gas in the non-degenerate (classical) limit. Then the fundamental distribution is the Maxwell-Boltzmann distribution for momentum,
$$f(\vec{p})=\frac{1}{(2 \pi \hbar)^3} \exp[-\vec{p}^2/(2m k T)].$$
That means that in a phase-space element $\mathrm{d}^3 x \mathrm{d}^3 p$ there are (on average)
$$\mathrm{d} N =\mathrm{d}^3 x \mathrm{d}^3 p f(\vec{p})$$
particles.

If you now want the distribution for finding particles in some range of (kinetic) energy, you have to transform both the phase-space-distribution function and the phase-space element to the new variable. For the function it's easy, because $E=\vec{p}^2/(2m)$. So you simply can write
$$f=\frac{1}{(2 \pi \hbar \hbar)^3} \exp[-E/(k T)].$$
For the phase-space volume element you need to express it in spherical coordinates
$$\mathrm{d}^3 p = \mathrm{d} p \mathrm{d} \vartheta \mathrm{d} \varphi p^2 \sin \vartheta.$$
Integrating over the angles you get a facto $4 \pi$ and the distribution for $p=|\vec{p}|$ is obtained from
$$\mathrm{d}N = \mathrm{d}^3 x \mathrm{d} p 4 \pi p^2 \frac{1}{(2 \pi \hbar)^3} exp[-p^2/(2mkT)].$$
So the distribution function in terms of $p$ is
$$\frac{\mathrm{d} N}{\mathrm{d}^3 x \mathrm{d} p} = \frac{4 \pi p^2}{(2 \pi \hbar)^3} \exp[-p^2/(2mkT)].$$
Now $p^2=\vec{p}^2=2 m E$ and thus
$$2m \mathrm{d} E=2 p \mathrm{d} p \; \Rightarrow \; \mathrm{d} p=\frac{m}{p} \mathrm{d} E=\frac{m}{\sqrt{2mE}} \mathrm{d} E$$
and thus
$$\mathrm{d}^3 p =\mathrm{d} E \frac{m}{p} \mathrm{d} \vartheta \mathrm{d} \varphi 2 m E \sin \vartheta= \mathrm{d} E \mathrm{d} \vartheta \mathrm{d} \varphi m \sqrt{2mE} \sin \vartheta.$$
Integrating over the angles gives again a factor $4 \pi$ and thus
$$\mathrm{d} N =\mathrm{d}^3 x \mathrm{d} E 4 \pi m \sqrt{2mE} \frac{1}{(2 \pi \hbar)^3} \exp[-E/(k T)].$$
So the energy-distribution function is
$$\frac{\mathrm{d}N}{\mathrm{d}^3 x \mathrm{d} E}=\frac{4 \pi m \sqrt{2mE}}{(2 \pi \hbar)^3} \exp[-E/(k T)].$$
That's why the most probable momentum $p$, $p_{\text{max}}$ and the most probable energy $E$, $E_{\text{max}}$ do not fulfill $E_{\text{max}}=p_{\text{max}}/(2m)$ (WRONG!), because of the different weight in the distribution functions due to the transformation of both the distribution function and the phase-space distribution element $\mathrm{d}^3 x \mathrm{d}^3p$ between the different independent quantities $p$ and $E$.

 

[sophiecentaur]

The point is that distributions are, well, distributions,

Absolutely. You can forget about the source data and where it came from; the descriptors are what they are and in a good system of statistics, the descriptors and their derivations are distinct.