1

So I know the Boltzmann distribution is: $$ P\propto \exp \left(-E / k_BT \right) $$

where $E$ is energy, $k_B$ is the Boltzmann constant and $T$ is the temperature. However, when we replace $E$ for the kinetic energy $1/2 mv^2$ and we get $$ P \propto \exp \left(-mv^2 / 2k_BT \right) \, . $$

This becomes the probability for a particular velocity $\mathbf{v}$, and not for all particles with a set kinetic energy.

Why is this?

DanielSank
  • 24,439
Toby Peterken
  • 1,923
  • 14
  • 30
  • 1
    I am slightly confused by your question. You use $E = \frac{1}{2}m|\mathbf{v}|^{2}$, and therefore your probability distribution is in terms of $|\mathbf{v}|$ and is in fact equivalent in every way to your initial Boltamann distribution, as all particles with velocity $|\mathbf{v}|$ have the same kinetic energy. Perhaps I have misunderstood what you are asking? :) – Thomas Russell Jun 20 '17 at 16:17
  • 1
    It becomes the probability for a particular velocity $v$ and a particular mass $m$ – John Rennie Jun 20 '17 at 16:19

2 Answers2

1

this becomes the probability for a particular velocity v, and not for all particles with a set kinetic energy.

Why is this?

No, I think you are misunderstanding the initial formulation of the Maxwell-Boltzmann distribution. Your first version with $P \propto e^{-E/kT}$ is a probability as a function of variable kinetic energy, $E$, for each particle and a single, scalar temperature, $T$, for the entire distribution.

The second version with $P \propto e^{-m \ v^{2}/2kT}$ is a probability as a function of variable velocity, v, for each particle and a single, scalar temperature, $T$, for the entire distribution.

Thus, in both cases one has $P$ as a function of some variable that is specific to each particle (i.e., $E$ or v) and a scalar (i.e., $T$) that is related to the full width at half max (FWHM) of the distribution.

Side Notes:

  • Be careful not to confuse the velocity distribution with the speed distribution, as they have slightly different coefficients.
  • Technically the conversion from kinetic energy should be done with momentum, not velocities but in the nonrelativistic limit this is a trivial issue.
DanielSank
  • 24,439
honeste_vivere
  • 15,592
  • But my book says speed is distributed by $v^2e^{-\frac{mv^2}{2k_bT}}$, this is a completely different function – Toby Peterken Jun 20 '17 at 18:12
  • Yes, that is what I point out in my Side Notes section. – honeste_vivere Jun 20 '17 at 18:20
  • ok, well why can you split this energy up into different components? – Toby Peterken Jun 20 '17 at 18:23
  • Kind of, but not really (i.e., be careful). This is why I said one should deal with momenta, as energy is a scalar. The completely general Maxwell-Boltzmann distribution is a geometric series for each degree of freedom and can have a different "temperature" for each degree of freedom (really a different thermal spread, not formal thermo temperature). – honeste_vivere Jun 20 '17 at 18:38
  • @TobyPeterken - You might find the following interesting: https://physics.stackexchange.com/q/216819/59023 – honeste_vivere Jun 21 '17 at 16:26
0

Because the formula depends on $v^2$, not $\vec{v}$, it's actually a probability for the particle to have a given speed. However, there's only one speed for each kinetic energy, so it's really the same as the probability for the particle to have the corresponding kinetic energy.

The probability for a particle to have a given velocity is vastly smaller, because there are so many velocities with a given speed (the particle can be moving in a vast number of possible directions).

Rich006
  • 689
  • This is assuming a one-component gas, where all particles have the same mass. If there are two or more different kinds of particles, then there is no longer a bijection between speed and KE. – probably_someone Jun 20 '17 at 16:51
  • The speed distribution is not the same as the velocity distribution for realistic gas (e.g., 3 degrees of freedom). – honeste_vivere Jun 20 '17 at 17:11