Intuition The big pictureImagine a box of gas. Every molecule is flying around with some speed — some slow, some fast.
Nobody assigned the speeds ; they emerged from billions of random collisions.
The Maxwell-Boltzmann (MB) distribution answers: if I grab one random molecule, how likely
is its speed to be near v v v
The shape comes from a tug-of-war :
High speed is energetically unlikely — the Boltzmann factor e − m v 2 / 2 k B T e^{-mv^2/2k_BT} e − m v 2 /2 k B T High speed has more "room" in velocity space — the surface area 4 π v 2 4\pi v^2 4 π v 2 v v v
Multiply these two competing effects and you get a curve that rises, peaks, then falls.
Definition Speed distributionf ( v ) d v f(v)\,dv f ( v ) d v v v v v + d v v+dv v + d v ∫ 0 ∞ f ( v ) d v = 1 \int_0^\infty f(v)\,dv = 1 ∫ 0 ∞ f ( v ) d v = 1
We want the explicit form of f ( v ) f(v) f ( v ) T T T
Intuition Why start with components, not speed?The three velocity components v x , v y , v z v_x, v_y, v_z v x , v y , v z independent (no force couples them) and
isotropic (no direction is special). Independence + isotropy almost forces the answer.
HOW. Let the probability of finding v x v_x v x d v x dv_x d v x g ( v x ) d v x g(v_x)\,dv_x g ( v x ) d v x y , z y,z y , z
F ( v x , v y , v z ) = g ( v x ) g ( v y ) g ( v z ) . F(v_x,v_y,v_z) = g(v_x)\,g(v_y)\,g(v_z). F ( v x , v y , v z ) = g ( v x ) g ( v y ) g ( v z ) . By isotropy, F F F magnitude v = v x 2 + v y 2 + v z 2 v=\sqrt{v_x^2+v_y^2+v_z^2} v = v x 2 + v y 2 + v z 2 v x 2 + v y 2 + v z 2 v_x^2+v_y^2+v_z^2 v x 2 + v y 2 + v z 2
g ( v x ) g ( v y ) g ( v z ) = Φ ( v x 2 + v y 2 + v z 2 ) . g(v_x)\,g(v_y)\,g(v_z) = \Phi(v_x^2+v_y^2+v_z^2). g ( v x ) g ( v y ) g ( v z ) = Φ ( v x 2 + v y 2 + v z 2 ) .
Intuition Why this forces an exponentialA product on the left equals a function of a sum on the right.
The only smooth function turning "products into sums of squares" is the exponential:
if ln g ( v x ) + ln g ( v y ) + ln g ( v z ) \ln g(v_x)+\ln g(v_y)+\ln g(v_z) ln g ( v x ) + ln g ( v y ) + ln g ( v z ) v x 2 + v y 2 + v z 2 v_x^2+v_y^2+v_z^2 v x 2 + v y 2 + v z 2 ln g ( v x ) \ln g(v_x) ln g ( v x ) linear in v x 2 v_x^2 v x 2 .
So ln g ( v x ) = a − b v x 2 \ln g(v_x) = a - b\,v_x^2 ln g ( v x ) = a − b v x 2
g ( v x ) = A e − b v x 2 . g(v_x) = A\,e^{-b v_x^2}. g ( v x ) = A e − b v x 2 . We need b > 0 b>0 b > 0
Normalization ∫ − ∞ ∞ g ( v x ) d v x = 1 \int_{-\infty}^{\infty} g(v_x)\,dv_x = 1 ∫ − ∞ ∞ g ( v x ) d v x = 1 ∫ − ∞ ∞ e − b x 2 d x = π / b \int_{-\infty}^\infty e^{-bx^2}dx = \sqrt{\pi/b} ∫ − ∞ ∞ e − b x 2 d x = π / b
A π b = 1 ⇒ A = b π . A\sqrt{\tfrac{\pi}{b}} = 1 \quad\Rightarrow\quad A = \sqrt{\tfrac{b}{\pi}}. A b π = 1 ⇒ A = π b . Fix b b b We need b b b 1 2 k B T \tfrac12 k_B T 2 1 k B T
⟨ 1 2 m v x 2 ⟩ = 1 2 k B T ⇒ ⟨ v x 2 ⟩ = k B T m . \left\langle \tfrac12 m v_x^2 \right\rangle = \tfrac12 k_B T
\quad\Rightarrow\quad \langle v_x^2\rangle = \frac{k_B T}{m}. ⟨ 2 1 m v x 2 ⟩ = 2 1 k B T ⇒ ⟨ v x 2 ⟩ = m k B T . For the Gaussian g ( v x ) = A e − b v x 2 g(v_x)=A e^{-bv_x^2} g ( v x ) = A e − b v x 2 ⟨ v x 2 ⟩ = 1 2 b \langle v_x^2\rangle = \dfrac{1}{2b} ⟨ v x 2 ⟩ = 2 b 1
⟨ v x 2 ⟩ = 1 / ( 2 b ) \langle v_x^2\rangle = 1/(2b) ⟨ v x 2 ⟩ = 1/ ( 2 b ) ⟨ v x 2 ⟩ = ∫ x 2 e − b x 2 d x ∫ e − b x 2 d x = 1 2 π b − 3 / 2 π b − 1 / 2 = 1 2 b . \langle v_x^2\rangle = \dfrac{\int x^2 e^{-bx^2}dx}{\int e^{-bx^2}dx}
= \dfrac{\tfrac12\sqrt{\pi}\,b^{-3/2}}{\sqrt{\pi}\,b^{-1/2}} = \dfrac{1}{2b}. ⟨ v x 2 ⟩ = ∫ e − b x 2 d x ∫ x 2 e − b x 2 d x = π b − 1/2 2 1 π b − 3/2 = 2 b 1 . Why this step? Variance of a Gaussian is the second moment over the zeroth moment.
Setting 1 2 b = k B T m \dfrac{1}{2b}=\dfrac{k_BT}{m} 2 b 1 = m k B T
b = m 2 k B T , g ( v x ) = ( m 2 π k B T ) 1 / 2 e − m v x 2 / 2 k B T . \boxed{\,b = \frac{m}{2k_B T}\,}, \qquad
g(v_x) = \left(\frac{m}{2\pi k_B T}\right)^{1/2} e^{-\,m v_x^2/2k_B T}. b = 2 k B T m , g ( v x ) = ( 2 π k B T m ) 1/2 e − m v x 2 /2 k B T .
Intuition The geometric factorThe speed distribution counts all velocity vectors of magnitude v v v , which form a
spherical shell of radius v v v d v dv d v 4 π v 2 d v 4\pi v^2\,dv 4 π v 2 d v v v v
The joint density is the product of three identical Gaussians:
F ( v ⃗ ) d 3 v = ( m 2 π k B T ) 3 / 2 e − m ( v x 2 + v y 2 + v z 2 ) / 2 k B T d v x d v y d v z . F(\vec v)\,d^3v = \left(\frac{m}{2\pi k_BT}\right)^{3/2}
e^{-\,m(v_x^2+v_y^2+v_z^2)/2k_BT}\,dv_x\,dv_y\,dv_z. F ( v ) d 3 v = ( 2 π k B T m ) 3/2 e − m ( v x 2 + v y 2 + v z 2 ) /2 k B T d v x d v y d v z . Switch to spherical shells, d 3 v → 4 π v 2 d v d^3v \to 4\pi v^2\,dv d 3 v → 4 π v 2 d v
Let α = m 2 k B T \alpha = \dfrac{m}{2k_BT} α = 2 k B T m f ( v ) ∝ v 2 e − α v 2 f(v)\propto v^2 e^{-\alpha v^2} f ( v ) ∝ v 2 e − α v 2
Most probable speed v p v_p v p — maximize f f f d d v ( v 2 e − α v 2 ) = 0 \dfrac{d}{dv}\!\left(v^2e^{-\alpha v^2}\right)=0 d v d ( v 2 e − α v 2 ) = 0
2 v e − α v 2 − 2 α v 3 e − α v 2 = 0 ⇒ 1 = α v 2 ⇒ v p = 2 k B T m . 2v e^{-\alpha v^2} - 2\alpha v^3 e^{-\alpha v^2}=0 \Rightarrow 1=\alpha v^2
\Rightarrow \boxed{v_p=\sqrt{\tfrac{2k_BT}{m}}}. 2 v e − α v 2 − 2 α v 3 e − α v 2 = 0 ⇒ 1 = α v 2 ⇒ v p = m 2 k B T . Why? The peak is where the rising v 2 v^2 v 2
Mean speed ⟨ v ⟩ \langle v\rangle ⟨ v ⟩ — ∫ 0 ∞ v f ( v ) d v \int_0^\infty v f(v)\,dv ∫ 0 ∞ v f ( v ) d v
⟨ v ⟩ = 8 k B T π m . \langle v\rangle = \sqrt{\frac{8k_BT}{\pi m}}. ⟨ v ⟩ = π m 8 k B T . RMS speed v r m s v_{rms} v r m s — from ⟨ v 2 ⟩ = ∫ 0 ∞ v 2 f ( v ) d v = 3 k B T m \langle v^2\rangle = \int_0^\infty v^2 f(v)\,dv = \dfrac{3k_BT}{m} ⟨ v 2 ⟩ = ∫ 0 ∞ v 2 f ( v ) d v = m 3 k B T 1 2 m ⟨ v 2 ⟩ = 3 2 k B T \tfrac12 m\langle v^2\rangle=\tfrac32 k_BT 2 1 m ⟨ v 2 ⟩ = 2 3 k B T
v r m s = 3 k B T m . v_{rms}=\sqrt{\frac{3k_BT}{m}}. v r m s = m 3 k B T .
Worked example Example 1 — Most probable speed of N₂ at 300 Km = 28 × 1.66 × 10 − 27 = 4.65 × 10 − 26 m = 28\times1.66\times10^{-27}=4.65\times10^{-26} m = 28 × 1.66 × 1 0 − 27 = 4.65 × 1 0 − 26 k B T = 1.38 × 10 − 23 × 300 k_BT=1.38\times10^{-23}\times300 k B T = 1.38 × 1 0 − 23 × 300 v p = 2 ( 1.38 × 10 − 23 ) ( 300 ) 4.65 × 10 − 26 ≈ 422 m/s . v_p=\sqrt{\frac{2(1.38\times10^{-23})(300)}{4.65\times10^{-26}}}\approx 422\ \text{m/s}. v p = 4.65 × 1 0 − 26 2 ( 1.38 × 1 0 − 23 ) ( 300 ) ≈ 422 m/s . Why this step? v p v_p v p
Worked example Example 2 — Fraction of molecules withv x > 0 v_x>0 v x > 0 g ( v x ) g(v_x) g ( v x ) even Gaussian centered at 0, so exactly
half have v x > 0 v_x>0 v x > 0 Why? Symmetry: no direction preferred ⇒ ⟨ v x ⟩ = 0 \langle v_x\rangle=0 ⟨ v x ⟩ = 0
Worked example Example 3 — Ratiov r m s / v p v_{rms}/v_p v r m s / v p v r m s v p = 3 k B T / m 2 k B T / m = 3 / 2 ≈ 1.225 \dfrac{v_{rms}}{v_p}=\dfrac{\sqrt{3k_BT/m}}{\sqrt{2k_BT/m}}=\sqrt{3/2}\approx1.225 v p v r m s = 2 k B T / m 3 k B T / m = 3/2 ≈ 1.225 Why this step? Temperature and mass cancel — the ratio is universal for any ideal gas.
Recall Before reading on: what happens tof ( v ) f(v) f ( v ) T T T Forecast: peak shifts right and curve flattens/broadens.
Verify: v p ∝ T v_p\propto\sqrt T v p ∝ T and more spread-out speeds. ✓
Common mistake "The probability of speedv v v e − m v 2 / 2 k B T e^{-mv^2/2k_BT} e − m v 2 /2 k B T Why it feels right: The Boltzmann factor genuinely gives the probability of a single
velocity state . So shouldn't speed follow it directly?
The fix: Speed lumps together all velocity vectors of that magnitude. There are
4 π v 2 4\pi v^2 4 π v 2 v v v v 2 v^2 v 2 f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 e 0 = 1 e^0=1 e 0 = 1
v p = ⟨ v ⟩ = v r m s v_p=\langle v\rangle=v_{rms} v p = ⟨ v ⟩ = v r m s Why it feels right: They're all "typical speeds," same order of magnitude.
The fix: The distribution is asymmetric (long high-speed tail). The tail pulls
averages above the peak: v p < ⟨ v ⟩ < v r m s v_p<\langle v\rangle<v_{rms} v p < ⟨ v ⟩ < v r m s
m m m Why it feels right: Tables list molar mass M M M The fix: Either use per-molecule mass m m m k B k_B k B or molar mass M M M R R R v p = 2 R T / M v_p=\sqrt{2RT/M} v p = 2 R T / M k B k_B k B
Recall Feynman: explain to a 12-year-oldPicture a crowd of bumper cars crashing randomly. After a while, a few cars are zooming, a few
are crawling, and most are at a medium speed. If you sort them by speed, you get a hill-shaped
graph: almost none are frozen still (you need some speed), almost none are super-fast (that
takes huge energy), and the bump in the middle is the "usual" speed. Heat the floor so cars push
harder — the whole hill slides toward faster speeds and spreads out. That hill is the
Maxwell-Boltzmann distribution.
Mnemonic Speeds in size order:"Please Make Room" → P eak (most P robable, 2 \sqrt2 2 M ean (8 / π \sqrt{8/\pi} 8/ π R MS (3 \sqrt3 3 "area × energy"
= v 2 ⋅ e − energy / k B T v^2\cdot e^{-\text{energy}/k_BT} v 2 ⋅ e − energy / k B T
What two competing factors shape f ( v ) f(v) f ( v ) Geometric factor
4 π v 2 4\pi v^2 4 π v 2 (more velocity-space room at high
v v v ) vs Boltzmann factor
e − m v 2 / 2 k B T e^{-mv^2/2k_BT} e − m v 2 /2 k B T (energy cost).
Why is f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 The
v 2 v^2 v 2 shell-volume factor vanishes at
v = 0 v=0 v = 0 — no velocity vectors have zero magnitude in a shell of radius 0.
Form of single-component distribution g ( v x ) g(v_x) g ( v x ) ( m 2 π k B T ) 1 / 2 e − m v x 2 / 2 k B T \left(\frac{m}{2\pi k_BT}\right)^{1/2}e^{-mv_x^2/2k_BT} ( 2 π k B T m ) 1/2 e − m v x 2 /2 k B T , a Gaussian centered at 0.
Full MB speed distribution? f ( v ) = 4 π ( m 2 π k B T ) 3 / 2 v 2 e − m v 2 / 2 k B T f(v)=4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2 e^{-mv^2/2k_BT} f ( v ) = 4 π ( 2 π k B T m ) 3/2 v 2 e − m v 2 /2 k B T .
Most probable speed v p v_p v p 2 k B T / m \sqrt{2k_BT/m} 2 k B T / m , from
d d v ( v 2 e − α v 2 ) = 0 \frac{d}{dv}(v^2e^{-\alpha v^2})=0 d v d ( v 2 e − α v 2 ) = 0 .
Mean speed ⟨ v ⟩ \langle v\rangle ⟨ v ⟩ 8 k B T / π m \sqrt{8k_BT/\pi m} 8 k B T / π m .
RMS speed v r m s v_{rms} v r m s 3 k B T / m \sqrt{3k_BT/m} 3 k B T / m .
Order of the three speeds? v p < ⟨ v ⟩ < v r m s v_p<\langle v\rangle<v_{rms} v p < ⟨ v ⟩ < v r m s (ratios
2 : 8 / π : 3 \sqrt2:\sqrt{8/\pi}:\sqrt3 2 : 8/ π : 3 ).
Why must g ( v x ) g(v_x) g ( v x ) − v x 2 -v_x^2 − v x 2 Independence factorizes the joint density into a product; isotropy makes it depend only on
v x 2 + v y 2 + v z 2 v_x^2+v_y^2+v_z^2 v x 2 + v y 2 + v z 2 ; only the exponential turns products into sums.
What fixes the constant b = m / 2 k B T b=m/2k_BT b = m /2 k B T Equipartition:
⟨ 1 2 m v x 2 ⟩ = 1 2 k B T \langle\tfrac12 mv_x^2\rangle=\tfrac12 k_BT ⟨ 2 1 m v x 2 ⟩ = 2 1 k B T ⇒
⟨ v x 2 ⟩ = 1 / ( 2 b ) = k B T / m \langle v_x^2\rangle=1/(2b)=k_BT/m ⟨ v x 2 ⟩ = 1/ ( 2 b ) = k B T / m .
How does the curve change with higher T T T Peak shifts right (
v p ∝ T v_p\propto\sqrt T v p ∝ T ), curve broadens and lowers (area conserved).
Boltzmann factor and partition function — the e − E / k B T e^{-E/k_BT} e − E / k B T Equipartition theorem — supplies ⟨ v x 2 ⟩ = k B T / m \langle v_x^2\rangle = k_BT/m ⟨ v x 2 ⟩ = k B T / m b b b Kinetic theory of gases — v r m s v_{rms} v r m s P = 1 3 n m ⟨ v 2 ⟩ P=\tfrac13 n m\langle v^2\rangle P = 3 1 nm ⟨ v 2 ⟩ Gaussian integrals — normalization and momentsEffusion and Graham's law — flux ∝ ⟨ v ⟩ ∝ 1 / m \propto \langle v\rangle \propto 1/\sqrt m ∝ ⟨ v ⟩ ∝ 1/ m Maxwell-Boltzmann vs Fermi-Dirac vs Bose-Einstein — classical limit of quantum statistics
depends only on v squared
product equals sum of squares
multiply by 4 pi v squared shell
Boltzmann factor vs shell area
Gaussian g = A exp -b vx squared
Intuition Hinglish mein samjhoSocho ek box mein gas hai, billions of molecules randomly collide kar rahe hain. Koi
controller nahi hai jo speeds decide kare — phir bhi ek fixed pattern emerge ho jaata hai.
Yahi pattern Maxwell-Boltzmann distribution hai: agar tum randomly ek molecule pakdo, uski
speed v v v
Iska shape do cheezon ki ladai se banta hai. Ek taraf Boltzmann factor e − m v 2 / 2 k B T e^{-mv^2/2k_BT} e − m v 2 /2 k B T geometry factor 4 π v 2 4\pi v^2 4 π v 2 f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 v 2 v^2 v 2
Derivation ka core idea: kyunki v x , v y , v z v_x, v_y, v_z v x , v y , v z g ( v x ) = A e − b v x 2 g(v_x)=A e^{-bv_x^2} g ( v x ) = A e − b v x 2 b b b ⟨ 1 2 m v x 2 ⟩ = 1 2 k B T \langle \tfrac12 m v_x^2\rangle = \tfrac12 k_B T ⟨ 2 1 m v x 2 ⟩ = 2 1 k B T b = m / 2 k B T b=m/2k_BT b = m /2 k B T 4 π v 2 d v 4\pi v^2 dv 4 π v 2 d v
Teen speeds yaad rakho aur derive karna seekho: v p = 2 k B T / m v_p=\sqrt{2k_BT/m} v p = 2 k B T / m ⟨ v ⟩ = 8 k B T / π m \langle v\rangle=\sqrt{8k_BT/\pi m} ⟨ v ⟩ = 8 k B T / π m v r m s = 3 k B T / m v_{rms}=\sqrt{3k_BT/m} v r m s = 3 k B T / m v p < ⟨ v ⟩ < v r m s v_p<\langle v\rangle<v_{rms} v p < ⟨ v ⟩ < v r m s
