# Boltzmann distribution vs. Distribution of energy

Hi folks!

i'm a biologist trying to understand some basics of statistical mechanics. unfortunately, i cannot get over the following problem(s).

A)

in the boltzmann distribution the fraction of particles with energy Ei is given by:

$$\frac{Ni}{N} = \frac{exp(-\beta Ei)}{\sum exp(-\beta Ej)} \:\:\: (1)$$

The most likely state should therefore be Ei = 0 with probability 1/Z.

However, when one derives the distribution of energies via the Maxwell-boltzmann speed distribution one obtains:

$$f_E\,dE = 2\sqrt{\frac{E}{\pi(kT)^3}}~\exp\left[\frac{-E}{kT}\right]\,dE$$ $$\:\:\: (2)$$
.

f(E) goes to zero for E = 0. How is this possible if particles with Ei = 0 are the most frequent species?

B)

In reaction kinetics the reaction constant for one direction is given by:

$$k = A exp(-\beta Ea) \:\:\:(3)$$

where A is the Arrhenius constant and Ea is the hight of the energy barrier for the reaction.

The term $$exp(-\beta Ea)$$ is supposed to correspond to the fraction of particles that are fast enough to get over the energy barrier.

Coming back to equation (2), shouldn't this fraction correspond to $$\int f(E)dE$$ from Ea to infinity? I don't see how one could get this from integrating (2) ?
Respectively, shouldn't taking the sum in equation (1) over all particles with Ei > Ea give this value as well?

What am i missing here?

Tim

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no thermodynamics experts here? :)

anyone?

A)

in the boltzmann distribution the fraction of particles with energy Ei is given by:

$$\frac{Ni}{N} = \frac{exp(-\beta Ei)}{\sum exp(-\beta Ej)} \:\:\: (1)$$

The most likely state should therefore be Ei = 0 with probability 1/Z.
Yes, that's correct. The most likely state will be with the one with Ei = 0. It might help to look at the equation like this:

$$\frac{N_i}{N} = \frac{g_i exp(-\beta E_i)}{\sum g_j exp(-\beta Ej)} \:\:\: (1)$$

where the gj's are the degeneracy of the states (that is, the number of states with the same energy).

However, when one derives the distribution of energies via the Maxwell-boltzmann speed distribution one obtains:

$$f_E\,dE = 2\sqrt{\frac{E}{\pi(kT)^3}}~\exp\left[\frac{-E}{kT}\right]\,dE$$ $$\:\:\: (2)$$
.

f(E) goes to zero for E = 0. How is this possible if particles with Ei = 0 are the most frequent species?
When considering the energy distribution, you have to count up all the states with a particular energy E (or that have energy between E and E + dE if you want to be technical). Since the energy of a single ideal gas particle is E = 1/2 m (vx^2 + vy^2 + vz^2) the energy is determined by magnitude of the velocity. But since velocity is a vector, there will be a multiple states with different velocity vectors, but the same magnitude of velocity and energy. So even though the lowest energy state will have the highest probability, there's only one such state, and a state with slightly higher energy will have a slightly lower probability, but there will be a bunch of states with that energy.

To get there, you have to start by looking at the probability by enumerating each state, and then figure out which states have the same energy. Since the velocities of particles indicate what state they're in, start with the velocity distribution. $$f_v(v_x,v_y,v_z) dv_xdv_ydv_z$$ is the number of particles with velocity between $$(v_x,v_y,v_z)$$ and $$(v_x+dv_x,v_y+dv_y,v_z+dv_z)$$

$$f_v(v_x,v_y,v_z) dv_xdv_ydv_z = \frac{A}{Z} exp\left(- \frac12 \beta m(v_x^2 + v_y^2 + v_z^2) \right)dv_xdv_ydv_z$$

Then you can switch to a speed distribution using $$dv_xdv_ydv_z = 4\pi v^2 dv$$ to get

$$f_v(v) dv = \frac{B}{Z} v^2 exp\left(- \frac12 \beta mv^2 \right)dv$$

(the factor of 4pi is absorbed into the normalization constant B). Now even though the most probable state is the one with zero energy, because as velocity increases there are more states with the same energy the velocity distribution will have a peak at some v > 0. If you are comfortable thinking about things in velocity space, then the number of states with speed between v and v + dv are counted by the volume of a thin spherical shell with radius v: 4 pi v^2 dv

Then to switch to an energy distribution, use $$dv = \frac{dv}{dE}dE = \frac{1}{mv}dE = \frac{1}{\sqrt{2E}}dE$$

$$f_E dE = f_v dv = f_v \frac{1}{\sqrt{2E}}dE = \frac{C}{Z} \sqrt{E} exp\left(- \frac12 \beta E \right)dE$$

(again, constant factors like sqrt(2) have been absorbed into the normalization constant C).

So in the end, while the state with E_i = 0 is most likely to be occupied, when you multiply the number of states at a given energy by the probability of occupying a state with that energy, the most frequency species won't be the ones with E_i = 0, it will be at something higher.

B)
In reaction kinetics the reaction constant for one direction is given by:

$$k = A exp(-\beta Ea) \:\:\:(3)$$

where A is the Arrhenius constant and Ea is the hight of the energy barrier for the reaction.

The term $$exp(-\beta Ea)$$ is supposed to correspond to the fraction of particles that are fast enough to get over the energy barrier.

Coming back to equation (2), shouldn't this fraction correspond to $$\int f(E)dE$$ from Ea to infinity? I don't see how one could get this from integrating (2) ?
Respectively, shouldn't taking the sum in equation (1) over all particles with Ei > Ea give this value as well?

What am i missing here?

Tim
I'm not so sure on this one, because it's been a few years since I've taken pchem or had anything to do with reaction rates. I think you might want to not consider particles with much higher energy than E_a because if their energy is too high, the reaction "won't stick." But I can't really do more then speculate on the answer to your question.

hello kanato,

wow thanks a lot for this crisp explanation! Can't believe Atkins hasn't a paragraph like this :)

thanks again!