# I Average kinetic energy in 1 dimension according to the M-B Distribution

#### JohnnyGui

Summary
Integrating the 1 dimensional MB Distribution in terms of translational kinetic energy up to infinity, does not yield $\frac{1}{2}k_BT$ as it should be.
Summary: Integrating the 1 dimensional MB Distribution in terms of translational kinetic energy up to infinity, does not yield $\frac{1}{2}k_BT$ as it should be.

The format for the 3 dimensional Maxwell-Boltzmann Distribution is $A\cdot e^{-\frac{E}{k_BT}} \cdot g(E)$ in which $A$ can be derived using normalization (integration up to $\infty$ must be 1) and $g(E)$ is the degeneracy according to $g(E)=\frac{V\pi 2^{2.5} \cdot m^{1.5}}{h^3}$

The 3 dimensional average kinetic energy $\bar E$ of a particles system can then be calculated by multiplying this MB distribution with $E$ and integrating it over $\infty$, which yields:
$$\bar E = \int_0^{\infty} \frac{2}{\sqrt \pi} \cdot (\frac{1}{k_BT})^{\frac{3}{2}} \cdot e^{-\frac{E}{k_BT}} \cdot \sqrt{E} \cdot E \cdot dE = \frac{3}{2}k_BT$$
The format for the 1 dimensional MB distribution (e.g. the x-coordinate) is $A\cdot e^{-\frac{E_x}{k_BT}}$ where $A$ is derived by normalizing the integration to 1, which gives $A=\frac{1}{k_BT}$. When calculating the 1 dimensional average energy $E_x$, this MB distribution should also be multiplied by the energy $E_x$ and integrated up to $\infty$ which gives:
$$\bar E_x=\int^{\infty}_0 \frac{1}{k_BT}\cdot e^{-\frac{E_x}{k_BT}}\cdot E_x\cdot dE = k_BT$$
But this should be $\frac{1}{2}k_BT$. The peculiar thing is that when writing $E_x$ in terms of $\frac{1}{2}mv_x^2$ within the format $A \cdot e^{-\frac{E_x}{k_BT}}$, then normalizing the $A$ to that, and multiplying the formula with $\frac{1}{2}mv_x^2$ and integrating up to $\infty$, then one would indeed get $\frac{1}{2}k_BT$
$$\int^{\infty}_0\frac{\sqrt{2m}}{\sqrt{\pi k_BT}}\cdot e^{-\frac{mv_x^2}{2k_BT}}\cdot \frac{1}{2}mv_x^2 \cdot dv=\frac{1}{2}k_BT$$
But it wasn't necessary for the 3 dimensional MB distribution to write the format down in terms of $v$ to get the correct average kinetic energy.

Why does the 1 dimensional MB distribution in terms of $E_x$ give an incorrect average energy and how would one realise that this is the wrong way to do it?

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#### vanhees71

Gold Member
The momentum distribution function in 1D is
$$f(p)=\frac{V}{(2 \pi \hbar)} \exp \left (-\frac{p^2}{2m k_B T} \right).$$
The total number of particles is
$$N=\int_{\mathbb{R}} \mathrm{d} p f(p) = \frac{V}{\hbar} \sqrt{\frac{k_B m T}{2 \pi}}$$
and the total internal energy
$$U=\int_{\mathbb{R}} \mathrm{d} p f(p) \frac{p^2}{2m} = \frac{N}{2} k_B T,$$
as it should be.

If you want to transform the momentum distribution to an energy distribution you have to use the correct transformation laws. Most simple is to use the formal definition,
$$\tilde{f}(E)=\int_{\mathbb{R}} \mathrm{d} p f(p) \delta \left (E-\frac{p^2}{2m} \right)=\int_{\mathbb{R}} \mathrm{d} p f(p) \frac{m}{|p|} f(p) [\delta(p-\sqrt{2m E})+\delta(p+\sqrt{2mE})]=\frac{V}{\pi \hbar} \sqrt{\frac{m}{E}} \exp \left (-\frac{E}{k_{\text{B}} T} \right).$$
Now of course the same results come out for total number and internal energy as before,
$$N=\int_0^{\infty} \mathrm{d} E \tilde{f}(E), \quad U=\int_0^{\infty} \mathrm{d} E f(E) E.$$

#### JohnnyGui

The momentum distribution function in 1D is

f(p)=V(2πℏ)exp(−p22mkBT)​
Thanks. I got 3 questions regarding this formula.

1. Isn't there a way to directly derive the 1D energy distribution instead of via the 1D momentum distribution?

2. Why is it wrong to simply substitute the $-\frac{p^2}{2m}$ parameter by $E$ to transform $f(p)$ it into energy distribution $f(E)$?

3. Is $f(p)$ the probability density or the number of particles at a certain $p$? If it is the first, then how can integration as you did in the next section give the total number of particles $N$? Should't the integration simply give $1$?

#### mfb

Mentor
These functions are "per small interval in their parameter" (loosely speaking). dE and dp do not have a linear relation and if you go from one distribution to the other you have to take this into account.

#### vanhees71

Gold Member
The reason for using phase-space variables $f(x,p)$ is that in these variables the a-priori probability measure is particularly simple, namely for each pair of position and momentum variables the number of states is $\mathrm{d} x \mathrm{d} p/(2 \pi \hbar)$. For a derivation, see sect. 1.5 in

In a many-body system $\mathrm{d} N=f(x,p) \mathrm{d} x \mathrm{d} p/(2 \pi \hbar)$ is the number of particles in the phase-space volume (in this case of 1D motion the phase space is in fact a 2D plane, of course) $\mathrm{d} x \mathrm{d}p$.

The reason, why you cannot simply substitute $E=p^2/(2m)$ in the momentum-distribution function is that it doesn't transform as a scalar field under phase-space diffeomorphisms but as a scalar density of rank 1, i.e., $\mathrm{d} N$ is the scalar quantity, and under phase-space diffeomorphisms the phase-space volume element is multiplied by the Jacobian determinant of the transformation. Formally that's most easily achieved using the trick with the $\delta$ distribution shown above.

#### JohnnyGui

In a many-body system dN=f(x,p)dxdp/(2πℏ)dN=f(x,p)dxdp/(2πℏ)\mathrm{d} N=f(x,p) \mathrm{d} x \mathrm{d} p/(2 \pi \hbar) is the number of particles in the phase-space volume (in this case of 1D motion the phase space is in fact a 2D plane, of course) dxdpdxdp\mathrm{d} x \mathrm{d}p.
So $\mathrm{d} x \mathrm{d} p/(2 \pi \hbar)$ is the number of states in an infinitesimally small 2D plane of position and momentum space and $f(x,p)$ gives the number of particles at position $x$ and momentum $p$?

#### vanhees71

Gold Member
$f(x,p) \mathrm{d} x \mathrm{d} p/(2 \pi \hbar)$ gives the number of particles in an infinitesimal phase-space "volume" $\mathrm{d} x \mathrm{d} p$ around the phase-space point $(x,p)$. That's the crucial point of this debate: $f(x,p)$ is mathematically a phase-space density and transforms as such under transformations of the phase-space coordinates.

#### JohnnyGui

$f(x,p) \mathrm{d} x \mathrm{d} p/(2 \pi \hbar)$ gives the number of particles in an infinitesimal phase-space "volume" $\mathrm{d} x \mathrm{d} p$ around the phase-space point $(x,p)$. That's the crucial point of this debate: $f(x,p)$ is mathematically a phase-space density and transforms as such under transformations of the phase-space coordinates.
I'm not really familiar with phase-space formulas (I'm already surprised that they are dependent on the position $x$ of a particle while the MB Distribution doesn't have such a variable) and I

I'm mostly familiar with formulas that give the probabilities of finding a particle, be it in terms of energy, momentum or speed. There is a link here that converts the 3D MB Distribution in terms of velocity into a 1D MB Distribution in terms of $v_x$. When rewriting that 1D formula in terms of 1D energy $E_x$ I'm having an extra factor of $1/2$ compared to the correct formula. Does that mean that when it is in terms of $v_x$ it takes into account all possible directions but when converted in terms of energy then somehow you're merely taking 1 direction into account for which you need to add a factor of 2?

#### JohnnyGui

I'm not really familiar with phase-space formulas (I'm already surprised that they are dependent on the position $x$ of a particle while the MB Distribution doesn't have such a variable) and I

I'm mostly familiar with formulas that give the probabilities of finding a particle, be it in terms of energy, momentum or speed. There is a link here that converts the 3D MB Distribution in terms of velocity into a 1D MB Distribution in terms of $v_x$. When rewriting that 1D formula in terms of 1D energy $E_x$ I'm having an extra factor of $1/2$ compared to the correct formula. Does that mean that when it is in terms of $v_x$ it takes into account all possible directions but when converted in terms of energy then somehow you're merely taking 1 direction into account for which you need to add a factor of 2?
Correction to my question regarding the extra factor of $\frac{1}{2}$.
Is it because, in terms of $v_x$, the formula calculates the probability of that velocity only in 1 direction, while when converting it into $E_x$, you should also take the opposite possible direction into consideration and therefore the probability in terms of $E_x$ should be multiplied by a factor of $2$ yielding:
$$f(E_x) = \sqrt{\frac{1}{\pi E_x k_B T}} \cdot e^{-\frac{E_x}{k_BT}} \cdot dE_x$$

#### vanhees71

Gold Member
I'm not really familiar with phase-space formulas (I'm already surprised that they are dependent on the position $x$ of a particle while the MB Distribution doesn't have such a variable) and I

I'm mostly familiar with formulas that give the probabilities of finding a particle, be it in terms of energy, momentum or speed. There is a link here that converts the 3D MB Distribution in terms of velocity into a 1D MB Distribution in terms of $v_x$. When rewriting that 1D formula in terms of 1D energy $E_x$ I'm having an extra factor of $1/2$ compared to the correct formula. Does that mean that when it is in terms of $v_x$ it takes into account all possible directions but when converted in terms of energy then somehow you're merely taking 1 direction into account for which you need to add a factor of 2?
Of course, the MB distribution for global thermal equilibrium of an ideal gas, neglecting external fields like gravity on Earth, doesn't depent on position since the equilibrium state is a homogeneous and isotropic spatial distribution of particles, but it's still a phase-space distribution and transforms as such under changes of the phase-space coordinates.

I don't know, where you factor 1/2 comes from. Have you checked my calculation? Maybe you have forgotten that for each energy $E=p^2/(2m)$ there are two solutions for $p$, i.e., $p=\pm \sqrt{2mE}$? Otherwise changing from $p$ to $v=p/m$ just leads to the corresponding factor due to $\mathrm{d} p=m \mathrm{d} v$, i.e.,
$$\tilde{f}(x,v) \mathrm{d} x \mathrm{d} v=f(x,p) \mathrm{d} x \mathrm{d} p=f(x,p) m \mathrm{d} x \mathrm{d} v,$$
from which
$$\tilde{f}(x,v)=f(x,m v)m.$$

#### JohnnyGui

but it's still a phase-space distribution and transforms as such under changes of the phase-space coordinates.
I think I need to delve deeper into phase-space coordinates.

I don't know, where you factor 1/2 comes from. Have you checked my calculation? Maybe you have forgotten that for each energy E=p2/(2m)E=p2/(2m)E=p^2/(2m) there are two solutions for ppp, i.e., p=±√2mEp=±2mEp=\pm \sqrt{2mE}? Otherwise changing from ppp to v=p/mv=p/mv=p/m just leads to the corresponding factor due to dp=mdvdp=mdv\mathrm{d} p=m \mathrm{d} v, i.e.,
It was indeed the factor of $2$ that I had to multiply the formula with because of the 2 different directions for a particular velocity.

I'm reading this article now regarding converting a 1D to a 3D MB Distribution. It is deriving a formula to calculate an infinitesimally small volume $dV$ on the surface of a sphere with radius $v$ in velocity space. It states that $dV = dv_xdv_zdv_z$. I have drawn this small volume piece in velocity space myself and deduced the following:

So I have deduced that $dV = \sin(\theta) \cdot v \cdot d\phi \cdot v \cdot d\theta \cdot dv$. This is also what the link states. However, I really can't see how this is the same as $dV = dv_xdv_zdv_z$. According to my drawing, I have deduced that
$$dv_z = \sin(\theta)\cdot v\cdot d\phi\cdot \sin(\phi)$$
$$dv_x = \sin(\theta)\cdot v\cdot d\phi\cdot \cos(\phi)$$
$$dv_y = \sin(\theta) \cdot v \cdot d\theta$$
But I can't equate them to make $dV = \sin(\theta) \cdot v \cdot d\phi \cdot v \cdot d\theta \cdot dv$.

#### vanhees71

Gold Member
First of all, it doesn't make sense to derive the 3D distribution from the 1D distribution, because the 3D distribution provides more information than the 1D distribution. The logic is as follows:

Suppose you have the distribution for $\vec{v}$. Let's call it $P(\vec{v})=P(v_x,v_y,v_z)$. Then the probability distribution for just one component is given by
$$P_1(v_x)=\int_{\mathbb{R}^2} \mathrm{d} v_y \mathrm{d} v_z P(v_x,v_y,v_z).$$
The meaning is: The number $\mathrm{d} N$ of particles contained in a small volume element $\mathrm{d} V_v=\mathrm{d} v_x \mathrm{d} v_y \mathrm{d} v_z$ in velocity space is given by
$$\mathrm{d} N = P(\vec{v}) \mathrm{d} V_v.$$
Now you can use any coordinates you want. The volume element $\mathrm{d} V_v$ in terms of Cartesian velocity components read $\mathrm{d} v_x \mathrm{d} v_y \mathrm{d} v_z$.

This is easy to understand: The volume element written in some coordinates is determined by the coordinate lines, i.e., the lines, where you keep two of the three coordinates fixed and let the 3rd one run. In the case of Cartesian coordinates this leads to a grid in 3D space which defines little rectangular boxes, which of course have the volume $\mathrm{d} v_x \mathrm{d} v_y \mathrm{d} v_z$.

Now you want the volume element in terms of spherical coordinates. To get it, you first write down the vector's Cartesian coordinates in terms of the these coordints:
$$\vec{v}=\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \begin{pmatrix} v \sin \vartheta \cos \varphi \\ v \sin \vartheta \sin \varphi \\ v \cos \vartheta \end{pmatrix}.$$
Now consider the coordinate lines. They again lead to a specific grid of the volume, making up the volume as infinitesimal little boxes (not necessarily rectangular, but for spherical coordinates in fact they are, because spherical coordinates are orthogonal curvilinear coordinates) spanned by the tangent vectors on the three coordinate lines in the specific point. These vectors are given by
$$\mathrm{d} \vec{b}_v=\mathrm{d} v \partial_v \vec{v}=\mathrm{d} v \begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix} \\ \mathrm{d} \vec{b}_{\vartheta} = \mathrm{d} \vartheta \partial_{\vartheta} \vec{v} = \mathrm{d} \vartheta v \begin{pmatrix} \cos \vartheta \cos \varphi \\ \cos \vartheta \sin \varphi \\ -\sin \vartheta \end{pmatrix}\\ \mathrm{d} \vec{b}_{\varphi} = \mathrm{d} \varphi v \begin{pmatrix} -\sin \vartheta \sin \varphi \\ \sin \vartheta \cos \varphi \\ 0 \end{pmatrix}.$$
Now the volume of this little box, spanned by these three trangent vectors is
$$\mathrm{d} V_v = (\vec{b}_v \times \vec{b}_{\vartheta}) \cdot \vec{b}_{\varphi}=\mathrm{d} v \mathrm{d} \vartheta \mathrm{d} \varphi v^2 \sin \vartheta.$$
This is of course not the same as $\mathrm{d} v_x \mathrm{d} v_y \mathrm{d}_z$, because this is another little box than the one provided by the coordinate lines of the spherical coordinates, but it's a valid volume element in defining some 3D grid of the velocity space.

Thus the same distribution $P(\vec{v})$ can be written as a distribution for $v$, $\vartheta$, and $\varphi$. Of course you have to refer to the corresponding grid given by the coordinate lines, i.e., you have
$$\mathrm{d} N=\tilde{P}(v,\vartheta,\varphi) \mathrm{d} v \mathrm{d} \vartheta \mathrm{d} \varphi =P[\vec{v}(v,\vartheta,\varphi)] \mathrm{d} v \mathrm{d} \vartheta \mathrm{d} \varphi v^2 \sin \vartheta,$$
i.e., you get
$$\tilde{P}(v,\vartheta,\varphi) =P[\vec{v}(v,\vartheta,\varphi)] v^2 \sin \vartheta.$$
Of course this generalizes to more than three dimensions: The above used "parallelipedal product" is nothing else than the Jacobi determinant,
$$\mathrm{d}^3 v = \mathrm{d} v \mathrm{d} \vartheta \mathrm{d} \varphi \mathrm{det} \frac{\partial (v_x,v_y,v_z)}{\partial(v,\vartheta,\varphi)},$$
where in our case $q_1=v$, $q_2=\vartheta$, and $q_3=\varphi$.
The general transformation rule for an arbitrary multi-variable density function thus gets
$$\tilde{f}(q')=f[q(q')] \mathrm{det} \frac{\partial (q_1,q_2,\ldots,q_n)}{\partial (q_1',q_2,',\ldots, q_n')}.$$

#### JohnnyGui

Explanation
Thank you so much for the detailed explanation. I got it now regarding $dV$.

First of all, it doesn't make sense to derive the 3D distribution from the 1D distribution, because the 3D distribution provides more information than the 1D distribution.
But it is possible according to the link right?
Sorry again, but I had one last question about deducing a 1D distribution from the 3D distribution. I tried doing this but I'm not sure if my reasoning is correct, could you please find the fault in my logic in the following?

The 3D probability distribution density in terms of energy $f(E)$ of finding a particle with an energy between $E$ and $E + dE$ is as stated:
$$f(E) = 2 \cdot \sqrt{\frac{E}{\pi}} \cdot \bigg(\frac{1}{k_BT}\bigg)^{\frac{3}{2}} \cdot e^{-\frac{E}{k_BT}}$$
This formula takes into account the number of quantum states, which is basically the number of ways each 1 dimensional energy variable can change while keeping the sum $E_x + E_y + E_z = E$ constant. Thus, if you divide $f(E)$ by the number of quantum states $g(E)$, you'd get the probability density of finding a particle with energy $E$ but with specific values combination of $E_x$, $E_y$ and $E_z$. Let's call the probability density of a particle with specific values of the dimensional energies $f(|E|)$:
$$f(|E|) = \frac{f(E)}{g(E)}$$
The number of quantum states is known to be $g(E) = \frac{V \cdot 2^{2.5}\pi \cdot m^{1.5}}{h^3} \cdot \sqrt{E}$. Writing the whole formula for $f(|E|)$ out and simplifying eventually gives:
$$f(|E|) = \frac{h^3}{V} \cdot \bigg(\frac{1}{2m\pi \cdot k_BT}\bigg)^{\frac{3}{2}} \cdot e^{-\frac{E}{k_BT}}$$
This formula is the product of the probability density of each of the 3 dimensional specific valued energies, so $f(|E|) = f(|E_x|) \cdot f(|E_y|) \cdot f(|E_z|)$.
Since formula of each 1 dimensional probability density must have the same format, I can see from the formula $f(|E|)$ that each dimensional probability density distribution should look like:
$$f(|E_n|) = \frac{h}{L} \cdot \sqrt{\frac{1}{2m\pi \cdot k_B T}} \cdot e^{-\frac{E_n}{k_B T}}$$
We know that $f(|E|)$ is the probability density for a particle with specific values of $|E_x|$, $|E_y|$ and $|E_z|$. In order to deduce the probability density for a particle with a specific value of only one of the 3 dimensional energies, for example $|E_x|$, we'd have to keep the $f(|E_x)$ fixed while running an integration for $f(E_y)$ and $f(E_z)$ Thus:
$$f(|E_x|,E_y, E_z) = f(|E_x|) \cdot \int_0^{\infty} f(E_y) \cdot dE_y \cdot \int_0^{\infty} f(E_z) \cdot dE_z$$
I would expect this formula to give the 1D MB Distribution for a specific energy dimension. Writing the formula all out gives me:
$$f(|E_x|,E_y, E_z) = \frac{h^3}{V} \cdot \frac{1}{2m\pi} \cdot \sqrt{k_BT} \cdot e^{-\frac{E_x}{k_BT}}$$
However, the 1D MB Distribution in terms of energy should be:
$$f(|E_x|, E_y, E_z)= \sqrt{\frac{1}{\pi E_x \cdot k_BT}} \cdot e^{-\frac{E_x}{k_BT}}$$
In which exact step in my reasoning did I go wrong?

#### vanhees71

Gold Member
This is an unnecessary complicated way to think about this. Just start from the more simple momentum-distribution and integrate out $p_2$ and $p_3$ to get the distribution for $p_1$.

#### JohnnyGui

This is an unnecessary complicated way to think about this. Just start from the more simple momentum-distribution and integrate out $p_2$ and $p_3$ to get the distribution for $p_1$.
My question is more, is it possible to deduce it using my mentioned method? If yes or not, at which step did I reason the wrong way?

#### vanhees71

Gold Member
I don't understand the concept of the reasoning to begin with. Why are you thinking you can infer from a coarser description (i.e., looking only at one momentum component) to the finer description (i.e., looking at all three momentum components)? This is of course in some sense possible due to Galilei/Poincare symmetry, which makes the equilibrium state homogeneous and isotropic (in the center-mass/center-momentum frame). With this additional symmetry taken into account, you can argue in this way, but it's more cumbersome than just to use the complete phase-space description to start with.

Wow guys

#### JohnnyGui

I don't understand the concept of the reasoning to begin with. Why are you thinking you can infer from a coarser description (i.e., looking only at one momentum component) to the finer description (i.e., looking at all three momentum components)? This is of course in some sense possible due to Galilei/Poincare symmetry, which makes the equilibrium state homogeneous and isotropic (in the center-mass/center-momentum frame). With this additional symmetry taken into account, you can argue in this way, but it's more cumbersome than just to use the complete phase-space description to start with.
It is indeed assumed that it is isotropic. but still my reasoning would not lead to the correct formula. Hence me finding this peculiar.

#### JohnnyGui

I don't understand the concept of the reasoning to begin with. Why are you thinking you can infer from a coarser description (i.e., looking only at one momentum component) to the finer description (i.e., looking at all three momentum components)? This is of course in some sense possible due to Galilei/Poincare symmetry, which makes the equilibrium state homogeneous and isotropic (in the center-mass/center-momentum frame). With this additional symmetry taken into account, you can argue in this way, but it's more cumbersome than just to use the complete phase-space description to start with.
There is something that I noticed in this link in the box "Velocity Distribution in One Dimension". It says that the 1D MB Probability distribution has the format of $f(v_z) = A\cdot e^{-\frac{mv_z^2}{2k_BT}}$ and that $A$ can be deduced by reasoning that integration of $f(v_z)$ must be $1$. But how can integration of only one dimensional cover all possible probabilities? Shouldn't the other dimensions ($y$ and $x$) also contribute to the probability? Or is it such that, when integrating $f(v_z)$ up to infinity, you're also taking every possible combination values of $v_z$, $v_y$ and $v_x$ into account, and thus all the probabilities?

"Average kinetic energy in 1 dimension according to the M-B Distribution"

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