Energy Distribution: quick revision, 19 bites

The Core Idea: Most Probable Energy

Not all energies are equally likely.

The **energy distribution function** $f(E)$ describes the probability that a particle in a system (like a gas) has energy $E$. For a classical ideal gas at temperature $T$, the **Maxwell-Boltzmann distribution** gives $f(E) \propto \sqrt{E} \, e^{-E/k_B T}$, where $k_B$ is the Boltzmann constant. This function peaks at…

Takeaway. Energy distribution is not uniform; it peaks at a low energy but has a high-energy tail.

What is an Energy Distribution Function?

A probability map for energies.

An **energy distribution function** $f(E)$ is a mathematical function that gives the probability density of finding a particle with energy $E$. It is defined such that $f(E) \, dE$ is the fraction of particles with energy between $E$ and $E + dE$. The function is normalized: $\int_0^\infty f(E) \, dE = 1$.

Takeaway. It tells you how likely each energy value is for a particle in the system.

The Boltzmann Factor

The exponential that rules thermal physics.

The **Boltzmann factor** $e^{-E/(k_B T)}$ is the core of the distribution. It tells us that states with higher energy are exponentially less likely to be occupied. This factor appears in all classical statistical mechanics and explains why particles tend to occupy lower energy states at finite temperatures.

Takeaway. Higher energy states are exponentially suppressed by the Boltzmann factor.

The Full Maxwell-Boltzmann Energy Distribution

The exact formula, normalized.

This is the **normalized** version, meaning the integral over all energies from 0 to infinity equals 1. The factor $\frac{2}{\sqrt{\pi}} \left(\frac{1}{k_B T}\right)^{3/2}$ ensures proper normalization.

Takeaway. The distribution combines a $\sqrt{E}$ density of states with the exponential Boltzmann factor.

Why the √E Factor?

More states at higher energy.

The $\sqrt{E}$ factor arises from the **density of states** $g(E)$ in three dimensions. In a 3D gas, the number of states with energy between $E$ and $E+dE$ grows as $\sqrt{E}$. So even though the Boltzmann factor decreases with energy, the number of available states increases, creating a peak.

Takeaway. The √E factor reflects the growing number of states with energy, balancing the exponential drop.

Most Probable vs. Average Energy

They are not the same!

The **most probable energy** is the peak of $f(E)$, found by setting $df/dE = 0$, giving $E_{mp} = \frac{1}{2} k_B T$. The **average energy** is $\langle E \rangle = \frac{3}{2} k_B T$, which is three times larger. This difference is due to the long high-energy tail.

Takeaway. Average energy is always higher than the most probable energy for this distribution.

At Room Temperature

Numbers you can feel.

At $T = 300$ K, $k_B T \approx 0.026$ eV. The most probable energy is $E_{mp} = 0.013$ eV, and the average energy is $\langle E \rangle = 0.039$ eV. These are tiny compared to typical chemical bond energies (~1 eV), which is why most molecules are not dissociated at room temperature.

Takeaway. Thermal energies at room temperature are small, around tens of meV.

Chemical Reaction Rates

Why reactions speed up with heat.

In chemistry, the **Arrhenius equation** $k = A e^{-E_a/(k_B T)}$ is a direct consequence of the Boltzmann factor. Only molecules with energy above the activation energy $E_a$ can react. The fraction of such molecules is given by the tail of the energy distribution, which increases exponentially with temperature.

Takeaway. Reaction rates depend on the high-energy tail of the distribution, which grows with temperature.

Confusing Speed and Energy Distributions

They look different!

The **speed distribution** $f(v)$ is proportional to $v^2 e^{-mv^2/(2k_B T)}$, while the **energy distribution** $f(E)$ is proportional to $\sqrt{E} \, e^{-E/(k_B T)}$. They are related by $E = \frac{1}{2} m v^2$, but the shapes differ. Don't use the speed formula when you need the energy one.

Takeaway. Always check whether you need the speed or energy distribution.

Normalization Check

Always verify the integral.

When using an energy distribution, always check that $\int_0^\infty f(E) \, dE = 1$. This ensures you have the correct normalization constant. For the Maxwell-Boltzmann distribution, the constant is $\frac{2}{\sqrt{\pi}} \left(\frac{1}{k_B T}\right)^{3/2}$.

Takeaway. A quick normalization check can catch many errors.

Remember the Peak

Half kT, three halves kT.

Remember: **most probable** energy = $\frac{1}{2} k_B T$, **average** energy = $\frac{3}{2} k_B T$. The '1/2' and '3/2' are easy to mix up. Think: the peak is at half, the average is at three-halves.

Takeaway. Peak at ½ kT, average at 3/2 kT.

Historical Origin

Maxwell and Boltzmann's legacy.

The distribution was first derived by **James Clerk Maxwell** in 1860 for speeds, and later generalized by **Ludwig Boltzmann** in the 1870s. It was a cornerstone in establishing the field of statistical mechanics and explaining thermodynamics from microscopic principles.

Takeaway. This distribution is a foundational result in physics.

Validity Limits

Not for quantum gases.

The Maxwell-Boltzmann distribution applies only to **classical** ideal gases where quantum effects are negligible. For **fermions** (e.g., electrons) or **bosons** (e.g., photons), you must use Fermi-Dirac or Bose-Einstein statistics instead. At low temperatures or high densities, quantum effects become important.

Takeaway. Use Maxwell-Boltzmann only for classical, non-interacting particles.

Deriving the Most Probable Energy

Set derivative to zero.

Takeaway. The most probable energy is found by maximizing ln f, which is often easier.

Energy vs. Speed Distribution

Two sides of the same coin.

Takeaway. Energy and speed distributions are related but have different shapes and peaks.

Atmospheric Escape

Why light gases escape Earth.

The **high-energy tail** of the distribution explains why lighter gases like hydrogen and helium can escape Earth's gravity. A small fraction of molecules have speeds exceeding the escape velocity, and over geological time, this leads to significant loss of these gases from the atmosphere.

Takeaway. The tail of the energy distribution drives atmospheric escape.

Using the Distribution in Calculations

A step-by-step approach.

Takeaway. Averages are computed via integrals over the distribution.

Temperature Dependence

Hotter means broader and higher.

As temperature $T$ increases, the distribution **broadens** and the peak shifts to higher energies. The most probable energy scales linearly with $T$, and the fraction of particles with high energy increases dramatically because of the exponential factor. This is why chemical reactions speed up with temperature.

Takeaway. Temperature controls both the peak and the width of the energy distribution.

Common Integral You'll Need

Memorize this for exams.

This integral is used to compute averages like $\langle E \rangle$ and $\langle E^2 \rangle$. For example, $\langle E \rangle = \frac{3}{2} k_B T$ comes from this with $n = 3/2$.

Takeaway. Know the gamma integral to quickly compute averages.