Maxwell velocity distribution

By replacing $p = mv$ in Eq.  we obtain at once the distribution of the translational velocities of the particles of a perfect gas:

$$p(v) dv= 4\pi \left (\frac{m}{2\pi k_{\rm B}T}\right )^{3/2} v^2 e^{-mv^2/2k_{\rm B}T} dv,$$ (4.19)

We can also obtain the distribution of each single component of the velocity:

$$p(v_x)dv_x = A_x e^{-m v_x^2/2k_{\rm B}T} dv_x = \left (\frac{m}{2 \pi k_{\rm B}T} \right )^{1/2} e^{-m v_x^2/2k_{\rm B}T} dv_x$$ (4.20)

and similarly for $p(v_y)$ and $p(v_z)$. The normalisation constant $A_x$ is obtained by integrating Eq.  over all possible values of $v_x$:

$$A_x^{-1} = \int_{-\infty}^{\infty} e^{-m v_x^2/2k_{\rm B}T} dv_x = \left (\frac{2 \pi k_{\rm B}T}{m} \right )^{1/2}.$$ (4.21)

We see from  that the most probable value for any component of the velocity is zero. This means that if we consider an element of volume $d v_x d v_y d v_z$ in velocity space, then the most probable velocity will be found when this volume is centred around zero, or in other words it is more probable to find a particle at rest than with any other velocity. However, if we are not interested in the direction of the velocity but only in its modulus, then this is not the case, because a zero velocity can only be realised with all components of the velocity equal to zero but any other finite value $v$ can be obtained by any vector lying on a sphere of radius $v$. The number of these vectors is proportional to the area of the sphere, $4 \pi v^2$, and so the probability of finding a particle with velocity $v$ is given by the Maxwell distribution .

Eqs.  and  can be expressed also in terms of the energy $E = p^2/2m$:

$$p(E)dE = \frac{2}{\sqrt{\pi}}\frac{E^{1/2}}{ (k_{\rm B}T)^{3/2}} e^{-E/k_{\rm B}T} dE,$$ (4.22)

which shows that the degeneracy of the energy $E$ is proportional to $\sqrt{E}$. Fig.  shows the probability density $p(E)$ and its decomposition in the density of states and Boltzmann factor. The increase in the density of states with increasing energy is the reason why $p(E)$ has a maximum at a value of the energy greater than zero. To find this maximum we take the derivative of $p(E)$ and set it to zero. We have:

$$\frac{d p(E)}{dE} = \frac{1}{\sqrt{\pi}}\frac{E^{-1/2}}{ (k_{\rm ... ...}T} -\frac{2}{\sqrt{\pi}}\frac{E^{1/2}}{ (k_{\rm B}T)^{5/2}} e^{-E/k_{\rm B}T},$$ (4.23)

and setting $d p(E) /dE = 0$ gives:

$$E^{-1/2} -\frac{2}{k_{\rm B}T} E^{1/2} = 0; \quad \quad E_{max} = \frac{k_{\rm B}T}{2}.$$ (4.24)

Figure: Energy distribution $p(E)$ (solid black) and its decomposition in the density of states $\sqrt{E}$ (dotted red) and Boltzmann factor (dashed blue).

Let us calculate the average energy:

$$\bar{E} = \int_0^\infty E p(E)dE = \frac{2}{\sqrt{\pi}}\frac{1}{ (k_{\rm B}T)^{3/2}}\int_0^\infty E^{3/2} e^{-E/k_{\rm B}T} dE.$$ (4.25)

We make a change of variable $\epsilon = E/k_{\rm B}T$ and the integral becomes

$$\bar{E} = \frac{2}{\sqrt{\pi}} k_{\rm B}T\int_0^\infty \epsilon ^... ...epsilon = \Gamma(5/2) \frac{2}{\sqrt{\pi}} k_{\rm B}T = \frac{3}{2} k_{\rm B}T.$$ (4.26)

With similar manipulations we can obtain the average velocity and the most probable velocity.

We want to work out now the probability distribution for the energy of a system of $N$ identical, non-interacting particles. We are interested in the distribution of the total energy, rather than the energies of each single particles, because this is a quantity that is more easily accessible experimentally (for example, via its connection to the specific heat). In order to do that it is useful to go back to the distribution of momenta, and rewrite Eq.  in cartesian rather than polar coordinates. To do that we replace the volume of the shell with radii $p$ and $p+dp$, equal to $4\pi p^2 dp$, with the element of volume $d^3{\bf p}$, and we write the probability for the system to have total energy $E = \sum_i p_i^2/2m$ as the product of probabilities:

$$p(E) dE = p({\bf p}_1,\dots,{\bf p}_N)d^3{\bf p}_1\dots d^3{\bf p}_N =$$

$$\frac{1}{(2\pi m k_{\rm B}T)^{3N/2}} e^{-p_1^2/2mk_{\rm B}T} \dots e^{-p_N^2/2mk_{\rm B}T} d^3{\bf p}_1\dots d^3{\bf p}_N =$$

$$e^{-E/k_{\rm B}T} g(E) dE.$$ (4.27)

To work out the density of states $g(E)$ we use dimensional analysis and the condition of normalisation of the probability density. Each factor $d^3{\bf p}$ in  has the dimensions of $E^{3/2}$, and so $g(E)$ must contain a factor $E^{3N/2-1}$. The condition of normalisation requires:

$$\int_0^\infty A E^{3N/2-1} e^{-E/k_{\rm B}T} dE = 1,$$ (4.28)

which gives

$$A^{-1} = (k_{\rm B}T)^{-3N/2} \int_0^\infty e^{-\epsilon} \epsilon^{3N/2-1} d\epsilon = (k_{\rm B}T)^{3N/2} \Gamma(3N/2),$$ (4.29)

and so

$$p(E) dE = \frac{1}{(k_{\rm B}T)^{3N/2} \Gamma(3N/2)} E^{3N/2-1} e^{-E/k_{\rm B}T} dE.$$ (4.30)

We can now calculate the average energy:

$$\bar{E} = \int_0^\infty E p(E) dE = \frac{1}{(k_{\rm B}T)^{3N/2} \Gamma(3N/2)} \int_0^\infty E^{3N/2} e^{-E/k_{\rm B}T} dE =$$

$$\frac{3}{2} N k_{\rm B}T,$$ (4.31)

which is simply $N$ times the average energy of a single particle as obtained in , as expected. This result, and the one in , could also have been obtained in a much more direct way as $\bar{E} = - \partial \ln Z / \partial \beta$, but it was instructive to compute it also from . To obtain the typical fluctuations of the energy we can calculate $(\Delta E)^2 = \bar{E^2} - \bar{E}^2$. We have:

$$\bar{E^2} = \int_0^\infty E^2 p(E) dE =$$

$$\frac{1}{(k_{\rm B}T)^{3N/2} \Gamma(3N/2)} \int_0^\infty E^{3N/2+1} e^{-E/k_{\rm B}T} dE =$$

$$(k_{\rm B}T)^2 \frac{\Gamma(3N/2+2)}{\Gamma(3N/2)} = \left(\frac{3}{2}N + 1\right ) \frac{3}{2} N(k_{\rm B}T)^2 =$$

$$\left(\frac{3}{2}Nk_{\rm B}T \right )^2 + \frac{3}{2} N(k_{\rm B}T)^2,$$ (4.32)

from which we obtain $(\Delta E)^2 = {3\over 2} N (k_{\rm B}T)^2$. The relative fluctuations are therefore $\Delta E / \bar{E} = \sqrt{2/3 N}$, which for large $N$ are completely negligible and the total energy is effectively constant. We found this result already in Eq. .