# Why is the most probable energy different from the speed?

• I

## Summary:

The MB Distribution shows that the highest probable energy level is different from the highest probable speed. I have a hard time understanding this physically because energy is tied to speed. Is there a way to show that this is possible physically?

## Main Question or Discussion Point

It appears that the most probable energy level according to the Maxwell-Boltzmann distribution is not equal to the most probable speed squared multiplied by $\frac{1}{2}m$. The most probable speed has a different value.
$$E_{max} = \frac{k_BT}{2}$$
$$v_{max} = \sqrt{\frac{2k_BT}{m}}$$

I am trying to understand this physically rather than mathematically, but I have a hard time comprehending this.

An energy level with the highest probability means it has the most particles out of all other energy levels, therefore that same largest number of particles has the speed corresponding to that energy level.

If I’m imagining energy levels as containers and I draw a specific number of particles in each container, there is no way I can distribute the particles in a way to have the most probable speed in an energy container different from the most probable energy. Energy is tied to speed.

The only way I can think of how this can be possible is if in the continuous approach a certain energy level covers a range of different speed values. If that’s the case then I’d deduce further that the reason is because an infinitesimally small $dE$ covers a larger range of speed values dan an infinitesimally small speed $dv$ does.

Is this reasoning correct?

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PeterDonis
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Where are you getting your formulas for $E_{max}$ and $v_{max}$ from?

vanhees71
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You take averages, and thus [EDIT corrected in view of #5]
$$\langle |\vec{v}| \rangle=\left \langle \sqrt{\vec{v}^2} \right \rangle \neq \sqrt{\langle \vec{v}^2 \rangle}.$$

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Staff Emeritus
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Note that vanhees' comment is not only applicable to this one problem. This happens all the time with distributions.

• vanhees71
Ibix
You take averages, and thus
$$\langle |\vec{v}| \rangle=\left \langle \sqrt{\vec{v}}^2 \right \rangle \neq \sqrt{\langle \vec{v}^2 \rangle}.$$
Minor quibble - shouldn't the middle expression have the square under the square root?

• vanhees71
vanhees71
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of course.

• Ibix
phyzguy
If I’m imagining energy levels as containers and I draw a specific number of particles in each container, there is no way I can distribute the particles in a way to have the most probable speed in an energy container different from the most probable energy. Energy is tied to speed.
Of course there is. Try it. Here's a simple Python program with a triangular distribution. <v> is different from sqrt(<v^2>). The output is:
Mean velocity = 6.3313
Mean energy = 45.1507
Velocity of mean energy = 6.7194

Python:
import numpy as np

left = 0.0
mode = 9.0
right = 10.0
nSamples = 1000000
velocities = []
energies = []
for i in range(nSamples):
v =  np.random.triangular(left, mode, right)
velocities.append(v)
energies.append(v * v)

vbar = np.mean(velocities)
ebar = np.mean(energies)

print("Mean velocity = %.4f"%vbar)
print("Mean energy = %.4f"%ebar)
print("Velocity of mean energy = %.4f"%(np.sqrt(ebar)))

• vanhees71
PeroK
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Major quibble "most probable" is not the same as "average".

• Tazerfish and JohnnyGui
Where are you getting your formulas for EmaxEmax and vmaxvmax from?
They can be found here and here. For the 2nd link, please search for the terms "most probable".

@vanhees71 @Ibix @Vanadium 50 @phyzguy I'm aware that the average speed is not the same as the root-mean-square speed. I don't see how this has a link with the most probable speed. Please see @PeroK 's post #8 above.

The formula for those averages also differ from the most probable speed as shown here.

Major quibble "most probable" is not the same as "average".
This is exactly what I thought.

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vanhees71
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The "most probable speed", i.e., the $v$, where the probability distribution $f(v)$ has a maximum is in general again different from the other two considered "average speeds".

• Tazerfish
PeroK
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They can be found here and here. For the 2nd link, please search for the terms "most probable".

@vanhees71 @Ibix @Vanadium 50 I'm aware that the average speed is not the same as the root-mean-square speed. I don't see how this has a link with the most probable speed.

The formula for those averages also differ from the most probable speed as shown here.

Please see @PeroK 's post #8 above.

This is exactly what I thought.
Do all particles have the same mass?

• JohnnyGui
The "most probable speed", i.e., the $v$, where the probability distribution $f(v)$ has a maximum is in general again different from the other two considered "average speeds".
That's what I'm saying. My question in that regard is how it can be physically explained that the most probable energy does not correspond to the most probable speed that belongs to that energy.

@PeroK Yes

vanhees71
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A distribution is a distribution, i.e., it transforms under changes of the independent variable. Let $f(v)$ be the velocity distribution. For the Maxwell distribution, it's $f(v) = N v^2 \exp[-m v^2/(2 kT)]$.

Then you have $E=m v^2/2m$. From this you get
$$\tilde{f}(E) \mathrm{d} E =f(v) \mathrm{d} v.$$
Now $v=\sqrt{2mE}$ and thus $\mathrm{d} v=\mathrm{d} E \sqrt{m/(2E)}$, leading to
$$\tilde{f}(E) \mathrm{d} E = f(v) \sqrt{m/(2 E)} \mathrm{d} E \; \Rightarrow \; \tilde{f}(E)=\sqrt{\frac{m}{2 E}} f(\sqrt{2 m E}).$$

A distribution is a distribution, i.e., it transforms under changes of the independent variable. Let $f(v)$ be the velocity distribution. For the Maxwell distribution, it's $f(v) = N v^2 \exp[-m v^2/(2 kT)]$.

Then you have $E=m v^2/2m$. From this you get
$$\tilde{f}(E) \mathrm{d} E =f(v) \mathrm{d} v.$$
Now $v=\sqrt{2mE}$ and thus $\mathrm{d} v=\mathrm{d} E \sqrt{m/(2E)}$, leading to
$$\tilde{f}(E) \mathrm{d} E = f(v) \sqrt{m/(2 E)} \mathrm{d} E \; \Rightarrow \; \tilde{f}(E)=\sqrt{\frac{m}{2 E}} f(\sqrt{2 m E}).$$
I understand that the formulas are different mathematically, but my question is how this can be imagined physically, for example using my scenario to distribute the particles in such a way to fulfill that criteria.

If that's not possible then how about the following reasoning?
The only way I can think of how this can be possible is if in the continuous approach a certain energy level covers a range of different speed values. If that’s the case then I’d deduce further that the reason is because an infinitesimally small dEdEdE covers a larger range of speed values dan an infinitesimally small speed dvdvdv does.

PeroK
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The only way I can think of how this can be possible is if in the continuous approach a certain energy level covers a range of different speed values. If that’s the case then I’d deduce further that the reason is because an infinitesimally small $dE$ covers a larger range of speed values dan an infinitesimally small speed $dv$ does.

Is this reasoning correct?
Here's a slightly artificial example:

Consider a large finite set of particles, clustered round $v = 0$ and $v = v_0$, with more around $v_0$. If you square these numbers, the cluster round $0$ stays close to zero, but the cluster around $v_0$ spreads out more. And the density around $v^2 = 0$ may be greater than the density around $v^2 = v_0^2$.

To use a bit of simple maths:

$v = \Delta v \approx 0 \ \Rightarrow \ v^2 = (\Delta v)^2 \approx 0$

But:

$v = v_0 + \Delta v \ \Rightarrow \ v^2 = v_0^2 + 2v_0 \Delta v$

Which is spread much further away from $v_0^2$.

• vanhees71
Here's a slightly artificial example:

Consider a large finite set of particles, clustered round $v = 0$ and $v = v_0$, with more around $v_0$. If you square these numbers, the cluster round $0$ stays close to zero, but the cluster around $v_0$ spreads out more. And the density around $v^2 = 0$ may be greater than the density around $v^2 = v_0^2$.

To use a bit of simple maths:

$v = \Delta v \approx 0 \ \Rightarrow \ v^2 = (\Delta v)^2 \approx 0$

But:

$v = v_0 + \Delta v \ \Rightarrow \ v^2 = v_0^2 + 2v_0 \Delta v$

Which is spread much further away from $v_0^2$.
Ah, so in the context of my reasoning. A $dE$ starting from low energy values (thus low speeds) would cover more particles of the same speed and a $dE$ starting from higher energy (higher speeds) covers only a portion of the particles with the same speed because they are more spread out when squared?

• vanhees71 and PeroK
@PeroK I noticed an inner clash afterwards, because I also know that $dE = mv \cdot dv$. Doesn't this mean that $dE$ get wider at higher speeds and thus covers the increasing spread of the speed anyway?

PeroK
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@PeroK I noticed an inner clash afterwards, because I also know that $dE = mv \cdot dv$. Doesn't this mean that $dE$ get wider at higher speeds and thus covers the increasing spread of the speed anyway?
Isn't that just another way of saying what's already been said?

Isn't that just another way of saying what's already been said?
I don't think so because I said in post #16 that $dE$ covers only a portion of higher speeds because of the spread, but if $dE$ gets wider at higher speeds, it would compensate for that spread and thus cover all of them anyway. I.e. the "density" within $dE$ stays the same.

PeterDonis
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I said in post #16 that $dE$ covers only a portion of higher speeds because of the spread, but if $dE$ gets wider at higher speeds, it would compensate for that spread and thus cover all of them anyway. I.e. the "density" within $dE$ stays the same.
The fact that these look different to you in vague ordinary language does not mean you have actually proposed two different possible answers to your question. The mathematical fact that $dE = m v dv$ is the answer to your question. If you think that means "gets wider at higher speeds", and you think that's something different from "covers only a portion of higher speeds", then what do you think "covers only a portion of higher speeds" means mathematically that makes it different?

The fact that these look different to you in vague ordinary language does not mean you have actually proposed two different possible answers to your question.
How an answer "looks" is also influenced by how one understands it. I'm not solely basing the difference on the "looks" of my answers, but on my own understandings when writing them.
As I see them, there is a difference since one answer says "covers a portion" and the other says "covers all"

The mathematical fact that dE=mvdvdE=mvdvdE = m v dv is the answer to your question. If you think that means "gets wider at higher speeds", and you think that's something different from "covers only a portion of higher speeds", then what do you think "covers only a portion of higher speeds" means mathematically that makes it different?
"covers only a portion" in the sense that the value $dE$ stays as a constant and does not get wider with increasing speed.

PeterDonis
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one answer says "covers a portion" and the other says "covers all"
The only one who used the words "covers a portion" was you. What mathematical fact are you basing that on?

"covers only a portion" in the sense that the value $dE$ stays as a constant
So what mathematical fact justifies this description? Note that nothing @PeroK said in any of his posts implies this.

The only one who used the words "covers a portion" was you. What mathematical fact are you basing that on?
The fact that he, as well as another user, liked my post #16 containing that very description of $dE$ as a statement I asked about would make me assume that my understanding about $dE$ was correct. Hence me asking about it after.

So what mathematical fact justifies this description? Note that nothing @PeroK said in any of his posts implies this.
Perhaps I based my description on his mathematical fact because I understood it in another way than what he meant?

When reading him talking about the density of speed squared values changes around different $v$'s, I concluded that the reason the probability gets influenced by that is because a $dE$ would eventually not cover all the speed squared values when they're spread more out, thus ignoring a number of particles that had speeds very close to a certain (high) $v$.

I discarded this thought when I noticed that $dE$ gets wider with increasing speeds, thus thinking not ignoring a number of particles that are close to a certain (high) $v$ and and taking them all into account for the probability. Which brought me back to the initial OP question.

PeterDonis
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When reading him talking about the density of speed squared values changes around different $v$'s, I concluded
...the wrong thing. What he meant was simply that $dE = m v dv$.

This is a good example of why it's better to use math, not ordinary language, when talking about physics.

• vanhees71
PeterDonis
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he, as well as another user, liked my post #16
Because they thought you were just trying to say $dE = m v dv$ in different words. If they had known you were describing something different and inconsistent with $dE = m v dv$, they would not have liked your post.

Again, an example of why it's better to just use math instead of vague ordinary language.