Maxwell-Boltzmann distribution of speeds — most probable, mean, rms: Visual walkthrough

Intuition What we are about to do

The parent note handed you the finished formula for $f (v)$ . Here we grow it from nothing — no formula assumed. By the end you will see every symbol appear and know exactly why it is there. We use ideas from Kinetic Theory of Gases, Average Kinetic Energy and Temperature, and the Boltzmann Distribution as our raw material.

First, three words you must own before line one:

Speed ::: how fast a molecule moves, a single positive number in metres per second. Written $v$ . It is never negative — there is no such thing as $- 300$ m/s of speed.
Fraction ::: a piece of the whole, between $0$ and $1$ . "0.02 of the molecules" means 2 out of every 100.
Velocity ::: speed plus direction. In 3D it is three numbers $(v_{x}, v_{y}, v_{z})$ — how fast to the East, to the North, and Up. Speed is how long that arrow is.

Step 1 — Speed lives on a number line with a wall at zero

WHAT. Draw a horizontal axis for speed $v$ , starting at $0$ on the left and running to the right forever. On it, drop a dot for every molecule in a tiny box of gas.

WHY. Before asking "how many molecules move this fast," we must picture where speeds are allowed to sit. The single most important geometric fact is the wall at $v = 0$ : no dot can go to the left of zero, but nothing stops a dot far to the right. This one-sided fence is the seed of the curve's lopsided shape.

PICTURE. In the figure, the amber wall is the forbidden edge. Notice the dots pile up a little way from the wall and thin out to the right — no left-hand mirror image exists.

Step 2 — How likely is a given energy? The Boltzmann factor

WHAT. Give each molecule an energy tag. A molecule of mass $m$ moving at speed $v$ carries kinetic energy $ε = fixed number \frac{1}{2} mass of one molecule m speed squared v^{2} .$ The Boltzmann Distribution says the relative chance of a molecule owning that much energy is $e^{- ε / k T} = e^{- m v^{2} /2 k T},$ where each symbol does this:

$e^{(\dots)}$ ::: the exponential — a smooth "dimmer switch" that starts at $1$ and dims toward $0$ as its input grows negative.
the minus sign ::: makes bigger energy give a smaller number — high energy is rarer.
$k$ ::: Boltzmann's constant, the conversion between temperature and energy.
$T$ ::: absolute temperature; a bigger $T$ is a bigger "energy budget," so the switch dims more slowly.

WHY this tool and not another? We need a rule that says "each extra bit of energy is harder to afford, and the penalty is milder when it's hot." The exponential is the unique function whose rate of dimming is proportional to its own value — the natural mathematics of "each equal step costs an equal factor." That is precisely the physics of energy sharing (see Average Kinetic Energy and Temperature).

PICTURE. The curve starts tall at $v = 0$ (cheap, low energy) and slides down toward zero for fast molecules. On its own it would say "slowest is most common" — the peak would sit on the wall. That is wrong, and Step 3 fixes it.

Step 3 — How many ways? The velocity-shell surface $4 π v^{2}$

WHAT. Move from speed to velocity space: a 3D room whose axes are $v_{x}, v_{y}, v_{z}$ . A single molecule is one point in this room. Its speed is its distance from the centre: $v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2} (Pythagoras in 3D) .$ So all molecules with the same speed $v$ sit on a sphere of radius $v$ . "Speed between $v$ and $v + d v$ " means a thin spherical shell. The number of ways to land in that shell is its volume: $surface area of the sphere 4 π v^{2} shell thickness d v .$

$4 π v^{2}$ ::: the surface area of a sphere of radius $v$ — it grows as $v^{2}$ .
$d v$ ::: an infinitesimally thin skin.

WHY this tool? The Boltzmann factor tells us how likely one velocity is; it does not count how many different directions give the same speed. A big sphere has vastly more surface than a small one, so there are far more distinct velocity arrows that all have a large speed. That geometric bounty is the factor $v^{2}$ . It grows from zero and pushes the population away from the wall.

PICTURE. Two shells are drawn: a tiny inner one (few dots) and a fat outer one (many dots). Same skin thickness, but the outer shell's roomier surface holds far more molecules.

Step 4 — Multiply the two effects: the tug-of-war

WHAT. A shell is populated only if molecules are both allowed there (the $v^{2}$ counting) and willing to pay the energy (the Boltzmann dimming). Multiply them and attach a constant $A$ to be pinned down later: $f (v) = A rising: more ways with speed 4 π v^{2} falling: costlier to be fast e^{- m v^{2} /2 k T} .$

WHY. These two opposing trends — one climbing from zero, one sinking toward zero — must compete. Their product is small near $v = 0$ (nothing wants to be there because there's no room), small at huge $v$ (too expensive), and therefore peaks somewhere in between. That peak is the whole reason the curve has a hump.

PICTURE. The rising $v^{2}$ (cyan), the falling $e^{- m v^{2} /2 k T}$ (amber, dashed), and their product $f (v)$ (white). Watch the white curve inherit its rise from cyan and its fall from amber, cresting where they trade dominance.

Step 5 — Pin down $A$ : the area must equal one

WHAT. Every molecule has some speed, so the total fraction is $1$ : $\int_{0}^{\infty} f (v) d v = 1.$

$\int_{0}^{\infty}$ ::: add up over every allowed speed, from the wall to infinity.
$= 1$ ::: "100 % of molecules are accounted for."

Using the standard integral $\int_{0}^{\infty} v^{2} e^{- α v^{2}} d v = \frac{1}{4} π / α^{3}$ with $α = \frac{m}{2 k T}$ , solving for $A$ gives $A = (\frac{m}{2 π k T})^{3/2} .$

WHY. $A$ is a pure bookkeeping scale. Without it, doubling the number of molecules or changing units would rescale the curve; fixing the area to 1 turns the curve into an honest probability and locks $A$ forever.

PICTURE. Same-shaped curve, but the shaded area under it is stamped " $= 1$ ". Temperature will later stretch it wide or squeeze it tall — but this shaded area never changes.

Step 6 — Find the peak: most probable speed $v_{m p}$

WHAT. The peak is the speed where the curve stops rising and starts falling — its slope is momentarily flat. "Slope" is the derivative $\frac{df}{d v}$ ; flat means it equals zero. Set $α = \frac{m}{2 k T}$ so $f \propto v^{2} e^{- α v^{2}}$ : $\frac{d}{d v} (v^{2} e^{- α v^{2}}) = (from v^{2} 2 v - from the exponential 2 α v^{3}) e^{- α v^{2}} = 0.$ The exponential is never zero, so the bracket vanishes: $2 v (1 - α v^{2}) = 0 \Rightarrow v^{2} = \frac{1}{α} = \frac{2 k T}{m} .$

WHY the derivative? It is the one tool that answers "where does a smooth curve turn around?" A hump's summit is exactly where its slope switches from $+$ to $-$ , i.e. where the derivative crosses zero.

PICTURE. A red dashed vertical line drops from the crest to the axis, meeting it at $v_{m p}$ . The tangent line drawn at the crest is perfectly horizontal.

Step 7 — The other two speeds are averages, not peaks

WHAT. The peak is one molecule's favourite speed. The mean $\overset{v}{ˉ}$ and the rms $v_{r m s}$ ask about the whole population: $\overset{v}{ˉ} = \int_{0}^{\infty} v f (v) d v, \overline{v^{2}} = \int_{0}^{\infty} v^{2} f (v) d v, v_{r m s} = \overline{v^{2}} .$

$\overset{v}{ˉ}$ ::: multiply each speed by its fraction and add — the ordinary average.
$\overline{v^{2}}$ ::: average of the square of speed; $v_{r m s}$ is its square root.

Doing the integrals (parent note has the Gaussian results): $\overset{v}{ˉ} = \frac{8 k T}{π m}, v_{r m s} = \frac{3 k T}{m} .$

WHY three? The curve is lopsided (Step 1's wall). The peak ignores the tail. The mean feels the long right tail and shifts right of the peak. The rms squares speeds first, so a molecule twice as fast counts four times — the tail dominates even harder, dragging $v_{r m s}$ the furthest right.

PICTURE. All three markers on one curve: $v_{m p}$ (peak, amber), $\overset{v}{ˉ}$ (cyan), $v_{r m s}$ (white), in strict left-to-right order. The right tail is shaded to show what pulls the averages rightward.

Step 8 — Edge and limiting cases (nothing left to surprise you)

WHAT / WHY / PICTURE, four scenarios you might otherwise hit blind:

At exactly $v = 0$ : $f (0) = 0$ . Why: the $v^{2}$ factor kills it — zero surface area on a zero-radius sphere means no ways to stand still. The curve touches the wall, it does not start high there.
As $v \to \infty$ : $f \to 0$ . Why: the exponential beats every power of $v$ ; being enormously fast is impossibly expensive. The tail never quite reaches zero but hugs the axis.
$T \to 0$ (freezing): $v_{m p}, \overset{v}{ˉ}, v_{r m s} \to 0$ together and the curve collapses into a tall spike at the wall — all molecules nearly still. Why: no energy budget to spend on speed.
Heavy vs light gas (same $T$ ): all speeds $\propto 1/ M$ , so a heavy gas gives a tall, narrow, left-shifted curve; a light gas a short, wide, right-shifted one. Why: equal average KE $\frac{3}{2} k T$ means heavy molecules must move slower. This is why H₂ leaks from the atmosphere and N₂ stays.

The figure overlays a cold/heavy curve (spiky, near the wall) and a hot/light curve (flat, far right) — same area $1$ , opposite silhouettes.

The one-picture summary

Everything on one canvas: the rising $v^{2}$ counting (cyan), the falling Boltzmann willingness (amber), their product $f (v)$ (white) with area locked to $1$ , and the three characteristic speeds marked in their fixed order. If you can redraw this from memory, you own the derivation.

Recall Feynman retelling — say it in plain words

Picture a line of speeds with a wall at zero and open road to the right. Two forces fight over where molecules pile up. The first force says "there are more ways to be fast" — because in the 3D room of velocities, a bigger speed means a bigger sphere with more surface, and surface grows like $v^{2}$ ; this pushes the crowd away from the wall. The second force says "but being fast is expensive" — energy costs multiply, and the exponential dims the chance the faster you go; this pulls the crowd back toward the wall. Multiply the two and you get a hump: nearly nobody at rest, nearly nobody blazingly fast, a crest in between. Scale the whole thing so the shaded area is exactly one molecule-full — that fixes the constant. To find the crest, ask where the slope goes flat (the derivative equals zero) and out pops $v_{m p} = 2 k T / m$ . But the crest isn't the average: because the right tail has no left twin, the plain average $\overset{v}{ˉ}$ sits a bit to the right, and the energy-weighted $v_{r m s}$ — which squares speeds and so double-counts the fast tail — sits furthest right of all. Heat it up and the hump slides right and flattens; make the gas heavier and it slides left and sharpens; but the shaded area is always one, because every molecule always has some speed.

Recall Check yourself

Why is $f (0) = 0$ and not maximum? ::: The $4 π v^{2}$ counting factor is zero at $v = 0$ — a zero-radius sphere has no surface, so there are no ways to have zero speed. Which single tool finds the peak, and what condition does it impose? ::: The derivative $df / d v$ ; the peak is where it equals zero (flat slope). Why does $v_{r m s}$ end up largest of the three? ::: Squaring speeds before averaging over-weights the fast right tail (double-speed counts fourfold), pulling it furthest right. What stays constant as you change $T$ or $M$ ? ::: The area under $f (v)$ , always $1$ — all molecules are always counted.