Maxwell-Boltzmann Distribution of Velocities

Distribution of Velocities: The Maxwell-Boltzmann Distribution

• Introduction to Maxwell-Boltzmann Distribution:

  • The Maxwell-Boltzmann distribution describes the distribution of velocities for particles in a system, specifically, the number of particles moving at certain speeds.

  • This topic is complex, with different functions for various systems, but in classical physics, a single, derivable distribution is used.

• Review of Particle Distribution with Height:

  • Previously, we established that the number of particles at a certain height ($y$) is given by: $N(y) = N_0 e^{-U/kT}$.

  • Here, $N_0$ is the initial number of particles, $U$ is the potential energy, $k$ is Boltzmann's constant, and $T$ is the temperature.

  • The potential energy at that height ($U = mgy$ for gravitational potential energy) governs the particle distribution.

• Initial Hypothesis for Velocity Distribution:

  • Based on the height distribution, it might be intuitive to assume that the number of particles moving at a certain speed would be proportional to the kinetic energy, analogous to potential energy for height.

  • Hypothesized initial form: $N(v) \propto N_0 e^{-KE/kT}$ or $N(v) \propto N_0 e^{-mv^2/(2kT)}$.

  • This makes sense due to conservation of energy: particles reaching a height $y$ (potential energy $mgy$) initially had sufficient kinetic energy ($1/2 mv^2$).

• Critique of the Initial Hypothesis:

  • Case 1: Velocity $v = 0$- If $v = 0$, then $N(0) = N_0 e^0 = N_0$.

    • This implies that the maximum number of particles are perfectly still (have zero velocity).

    • This is physically incorrect for a gas, which is a chaotic system; no particles should be perfectly still (zero particles at zero velocity).

  • Case 2: Velocity $v = \infty$- If $v = \infty$, then $N(\infty) = N_0 e^{-\infty} \to 0$.

    • This result is physically correct: we expect very few, if any, particles to be moving infinitely fast.

  • Conclusion: The simple exponential form is flawed because it predicts a maximum at $v=0$, which contradicts the nature of gases.

• Expected Graphical Representation of Velocity Distribution:

  • A graph of the number of particles ($N(v)$) versus velocity ($v$) should exhibit the following characteristics:- At $v = 0$, $N(v)$ must be $0$ (no particles are perfectly still).

    • The number of particles should rise to a peak (corresponding to a most probable velocity, often near the root-mean-square (RMS) velocity).

    • As $v$ approaches infinity, $N(v)$ should decrease and approach $0$.

• **Defining the Maxwell-Boltzmann Distribution Function, $f(v)$: **

  • The distribution is best described by a function, $f(v)$, representing the fraction of particles moving at velocity $v$.

  • More accurately, it describes the number of particles within a range of velocities, from $v$ to $v + dV$.

  • The number of particles ($dN$) within this range is given by: $dN = f(v) dV$.

  • For example, asking for particles at exactly $300 \text{ m/s}$ is difficult; asking for particles between $280 \text{ m/s}$ and $320 \text{ m/s}$ (i.e., $300 \pm 20 \text{ m/s}) is more practical.

• Normalization of the Distribution Function:

  • If we integrate $f(v)$ over all possible velocities (from $0$ to $\infty$), we should obtain the total number of particles in the system, $N$.

  • Thus, the normalization condition is: $\int_{0}^{\infty} f(v) dV = N$.

• The Correct Maxwell-Boltzmann Distribution Function:

  • The correct form for the Maxwell-Boltzmann distribution function in classical physics is given by:
    

    $f_{MB}(v) = C v^2 e^{-mv^2/(2kT)}$
    

    where $C$ is a constant.

  • Explanation of the $v^2$ term:- When $v = 0$, the $v^2$ term ensures that $f_{MB}(0) = 0$, which correctly reflects that no particles are perfectly stationary.

    • Velocity is a vector ($v<em>x, v</em>y, v<em>z$). The $v^2$ term (which is $v</em>x^2 + v<em>y^2 + v</em>z^2$ for speed) accounts for velocities in all three spatial directions, preventing the need to deal with complex sines and cosines or individual components.

• Derivation of the Constant $C$ using Gaussian Integrals:

  • To find $C$, we use the normalization condition:
    

    $N = \int_{0}^{\infty} C v^2 e^{-mv^2/(2kT)} dV$

  • Gaussian Integrals:- A fundamental Gaussian integral identity is: $\int_{0}^{\infty} e^{-ax^2} dx = \frac{1}{2}\sqrt{\frac{\pi}{a}}$.

    • To obtain the form we need (with $x^2 e^{-ax^2}$), we differentiate the above integral with respect to $a$:- Left side: $\frac{d}{da} \int<em>{0}^{\infty} e^{-ax^2} dx = \int</em>{0}^{\infty} -x^2 e^{-ax^2} dx$

      • Right side: $\frac{d}{da} \left(\frac{1}{2}\sqrt{\frac{\pi}{a}}\right) = \frac{1}{2}\sqrt{\pi} \frac{d}{da} (a^{-1/2}) = \frac{1}{2}\sqrt{\pi} (-\frac{1}{2}) a^{-3/2} = -\frac{1}{4}\sqrt{\frac{\pi}{a^3}}$.

      • Equating and removing the negative signs: $\int_{0}^{\infty} x^2 e^{-ax^2} dx = \frac{1}{4}\sqrt{\frac{\pi}{a^3}}$.

  • Substitution and Solving for $C$:- Let $a = \frac{m}{2kT}$.

    • Perform a u-substitution: Let $u = \sqrt{\frac{m}{2kT}} v$.- Then, $v = \sqrt{\frac{2kT}{m}} u$ and $v^2 = \frac{2kT}{m} u^2$.

      • Also, $du = \sqrt{\frac{m}{2kT}} dV$, so $dV = \sqrt{\frac{2kT}{m}} du$.

    • Substitute these into the integral for $N$:
      

      $N = C \int<em>{0}^{\infty} \left(\frac{2kT}{m} u^2\right) e^{-u^2} \left(\sqrt{\frac{2kT}{m}} du\right)$ $N = C \left(\frac{2kT}{m}\right)^{3/2} \int</em>{0}^{\infty} u^2 e^{-u^2} du$

    • Apply the Gaussian integral result (with $a=1$ for $u^2$):
      

      $N = C \left(\frac{2kT}{m}\right)^{3/2} \left(\frac{1}{4}\sqrt{\frac{\pi}{1^3}}\right)$
      

      $N = C \left(\frac{2kT}{m}\right)^{3/2} \frac{\sqrt{\pi}}{4}$

    • Solve for $C$:
      

      $C = 4N \left(\frac{m}{2kT}\right)^{3/2} \frac{1}{\sqrt{\pi}} = 4N \left(\frac{m}{2\pi kT}\right)^{3/2}$

• Final Maxwell-Boltzmann Distribution Function:

  • Substituting $C$ back into the function gives the full Maxwell-Boltzmann distribution:
    

    $f_{MB}(v) = 4\pi N \left(\frac{m}{2\pi kT}\right)^{3/2} v^2 e^{-mv^2/(2kT)}$

  • This equation describes how particles distribute themselves with velocities in a classical gas, representing a "base model" without quantum mechanics.

  • The integral of this function from $0$ to $\infty$ correctly yields the total number of particles, $N$.

• Normalized (Probability) Form of the Distribution:

  • The function can be divided by the total number of particles ($N$) to obtain a probability density function, often denoted by lowercase $f(v)$.

  • $f(v) = \frac{f_{MB}(v)}{N} = 4\pi \left(\frac{m}{2\pi kT}\right)^{3/2} v^2 e^{-mv^2/(2kT)}$

  • The integral of this normalized function over all velocities equals $1$:
    

    $\int_{0}^{\infty} f(v) dV = 1$

  • This $1$ represents $100\%$ probability. It signifies that there is a $100\%$ certainty of finding any particle within the velocity range of $0$ to $\infty$ (since we are considering all possible non-negative speeds).

  • This concept is central to statistical mechanics, where the distribution directly relates to the probability of finding particles at certain speeds or within certain speed ranges.

  • The ideal gas law can even be derived from these statistical first principles.

• Conclusion:

  • The Maxwell-Boltzmann distribution is a complex but fundamental concept in thermodynamics, heavily relying on mathematical tools like Gaussian integrals.

  • While mathematically intensive, it provides profound insights into the behavior of gases.

  • Future topics, such as the First Law of Thermodynamics (mechanical work, energy, and heat), will involve mathematically simpler integrals.