Intuition The big picture (WHY this exists)In Newtonian mechanics we deal with F ⃗ = m a ⃗ \vec{F} = m\vec{a} F = m a x y z xyz x y z r ⃗ i \vec{r}_i r i generalized coordinates q j q_j q j generalized momentum p j p_j p j generalized force Q j Q_j Q j whatever quantities make Newton's law look the same in the new coordinates .
Definition Generalized coordinatesA set of independent variables q 1 , q 2 , … , q n q_1, q_2, \dots, q_n q 1 , q 2 , … , q n n n n r ⃗ i = r ⃗ i ( q 1 , … , q n , t ) \vec{r}_i = \vec{r}_i(q_1,\dots,q_n,\,t) r i = r i ( q 1 , … , q n , t )
WHY this form? Because the constraints are baked in: a pendulum bob has r ⃗ = ( ℓ sin θ , − ℓ cos θ ) \vec r = (\ell\sin\theta, -\ell\cos\theta) r = ( ℓ sin θ , − ℓ cos θ ) θ \theta θ x 2 + y 2 = ℓ 2 x^2+y^2=\ell^2 x 2 + y 2 = ℓ 2
The generalized velocity is q ˙ j = d q j / d t \dot q_j = dq_j/dt q ˙ j = d q j / d t v ⃗ i = d r ⃗ i d t = ∑ j ∂ r ⃗ i ∂ q j q ˙ j + ∂ r ⃗ i ∂ t . \vec{v}_i = \frac{d\vec r_i}{dt} = \sum_j \frac{\partial \vec r_i}{\partial q_j}\dot q_j + \frac{\partial \vec r_i}{\partial t}. v i = d t d r i = ∑ j ∂ q j ∂ r i q ˙ j + ∂ t ∂ r i .
The cleanest definition comes from virtual work . Imagine displacing each particle by an infinitesimal virtual amount δ r ⃗ i \delta \vec r_i δ r i consistent with constraints, at fixed time . The work done by the applied forces F ⃗ i \vec F_i F i
δ W = ∑ i F ⃗ i ⋅ δ r ⃗ i . \delta W = \sum_i \vec F_i \cdot \delta \vec r_i. δ W = ∑ i F i ⋅ δ r i .
Now express δ r ⃗ i \delta \vec r_i δ r i ∂ / ∂ t \partial/\partial t ∂ / ∂ t δ r ⃗ i = ∑ j ∂ r ⃗ i ∂ q j δ q j . \delta \vec r_i = \sum_j \frac{\partial \vec r_i}{\partial q_j}\,\delta q_j. δ r i = ∑ j ∂ q j ∂ r i δ q j .
Substitute and swap the order of summation:
δ W = ∑ j ( ∑ i F ⃗ i ⋅ ∂ r ⃗ i ∂ q j ⏟ ≡ Q j ) δ q j . \delta W = \sum_j \left( \underbrace{\sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j}}_{\equiv\, Q_j}\right)\delta q_j. δ W = ∑ j ≡ Q j i ∑ F i ⋅ ∂ q j ∂ r i δ q j .
Definition Generalized force Q j = ∑ i F ⃗ i ⋅ ∂ r ⃗ i ∂ q j so that δ W = ∑ j Q j δ q j . \boxed{\,Q_j = \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j}\,}\qquad\text{so that}\qquad \delta W = \sum_j Q_j\,\delta q_j. Q j = i ∑ F i ⋅ ∂ q j ∂ r i so that δ W = ∑ j Q j δ q j .
Q j Q_j Q j Q j Q_j Q j δ q j \delta q_j δ q j its units are (energy / unit of q j q_j q j
If q j q_j q j Q j Q_j Q j
If q j q_j q j angle → Q j Q_j Q j torque (N·m)!
The single formula Q j δ q j = Q_j\,\delta q_j = Q j δ q j =
Definition Generalized (canonical) momentum p j = ∂ L ∂ q ˙ j \boxed{\,p_j = \frac{\partial L}{\partial \dot q_j}\,} p j = ∂ q ˙ j ∂ L L = T − V L = T - V L = T − V T T T V V V
WHY define it this way? Watch what the Euler–Lagrange equation becomes:
d d t ∂ L ∂ q ˙ j − ∂ L ∂ q j = 0 ⟹ p ˙ j = ∂ L ∂ q j . \frac{d}{dt}\frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j}=0 \;\;\Longrightarrow\;\; \dot p_j = \frac{\partial L}{\partial q_j}. d t d ∂ q ˙ j ∂ L − ∂ q j ∂ L = 0 ⟹ p ˙ j = ∂ q j ∂ L . Newton's second law in disguise : rate of change of (generalized) momentum = (generalized) force. Defining p j p_j p j ∂ L / ∂ q ˙ j \partial L/\partial \dot q_j ∂ L / ∂ q ˙ j
Worked example Recover ordinary momentumFor a free particle L = 1 2 m x ˙ 2 L = \tfrac12 m\dot x^2 L = 2 1 m x ˙ 2 p x = ∂ L / ∂ x ˙ = m x ˙ p_x = \partial L/\partial \dot x = m\dot x p x = ∂ L / ∂ x ˙ = m x ˙
Worked example Angular momentum appears for freePolar coordinates: T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) T = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) T = 2 1 m ( r ˙ 2 + r 2 θ ˙ 2 ) p θ = ∂ L ∂ θ ˙ = m r 2 θ ˙ = L ang . p_\theta = \frac{\partial L}{\partial \dot\theta} = m r^2 \dot\theta = L_{\text{ang}}. p θ = ∂ θ ˙ ∂ L = m r 2 θ ˙ = L ang . Why this step? The coordinate is an angle, so its conjugate momentum is angular momentum — exactly matching the force/torque pattern from §2.
Intuition Conservation falls out (the 80/20 payoff)If L L L not depend on a coordinate q j q_j q j cyclic/ignorable ), then ∂ L / ∂ q j = 0 \partial L/\partial q_j = 0 ∂ L / ∂ q j = 0 p ˙ j = 0 \dot p_j = 0 p ˙ j = 0 p j p_j p j symmetry . No θ \theta θ L L L
System: mass m m m ℓ \ell ℓ g g g θ \theta θ
Step 1 — position. r ⃗ = ( ℓ sin θ , − ℓ cos θ ) \vec r = (\ell\sin\theta,\,-\ell\cos\theta) r = ( ℓ sin θ , − ℓ cos θ ) Why: one DOF, θ \theta θ ∣ r ⃗ ∣ = ℓ |\vec r|=\ell ∣ r ∣ = ℓ
Step 2 — kinetic energy. v ⃗ = ℓ θ ˙ ( cos θ , sin θ ) \vec v = \ell\dot\theta(\cos\theta,\sin\theta) v = ℓ θ ˙ ( cos θ , sin θ ) ∣ v ⃗ ∣ = ℓ θ ˙ |\vec v|=\ell\dot\theta ∣ v ∣ = ℓ θ ˙ T = 1 2 m ℓ 2 θ ˙ 2 T=\tfrac12 m\ell^2\dot\theta^2 T = 2 1 m ℓ 2 θ ˙ 2
Step 3 — potential. V = m g y = − m g ℓ cos θ V = mgy = -mg\ell\cos\theta V = m g y = − m g ℓ cos θ
Step 4 — Lagrangian. L = 1 2 m ℓ 2 θ ˙ 2 + m g ℓ cos θ L = \tfrac12 m\ell^2\dot\theta^2 + mg\ell\cos\theta L = 2 1 m ℓ 2 θ ˙ 2 + m g ℓ cos θ
Step 5 — generalized momentum.
p θ = ∂ L ∂ θ ˙ = m ℓ 2 θ ˙ . p_\theta = \frac{\partial L}{\partial \dot\theta}= m\ell^2\dot\theta. p θ = ∂ θ ˙ ∂ L = m ℓ 2 θ ˙ . Why this step? p θ p_\theta p θ
Step 6 — generalized force. Conservative, so Q θ = − ∂ V / ∂ θ = − m g ℓ sin θ Q_\theta = -\partial V/\partial\theta = -mg\ell\sin\theta Q θ = − ∂ V / ∂ θ = − m g ℓ sin θ Why this step? θ \theta θ Q θ Q_\theta Q θ torque . And − m g ℓ sin θ -mg\ell\sin\theta − m g ℓ sin θ
Step 7 — equation of motion. p ˙ θ = Q θ ⇒ m ℓ 2 θ ¨ = − m g ℓ sin θ ⇒ θ ¨ = − g ℓ sin θ . \dot p_\theta = Q_\theta \Rightarrow m\ell^2\ddot\theta = -mg\ell\sin\theta \Rightarrow \ddot\theta = -\tfrac{g}{\ell}\sin\theta. p ˙ θ = Q θ ⇒ m ℓ 2 θ ¨ = − m g ℓ sin θ ⇒ θ ¨ = − ℓ g sin θ .
Worked example Bead on a rotating wireA bead slides on a frictionless straight wire forced to rotate at constant ω \omega ω θ = ω t \theta=\omega t θ = ω t r r r T = 1 2 m ( r ˙ 2 + r 2 ω 2 ) T=\tfrac12 m(\dot r^2 + r^2\omega^2) T = 2 1 m ( r ˙ 2 + r 2 ω 2 ) Forecast: Is p r = m r ˙ p_r=m\dot r p r = m r ˙ Verify: p r = ∂ L / ∂ r ˙ = m r ˙ p_r=\partial L/\partial\dot r = m\dot r p r = ∂ L / ∂ r ˙ = m r ˙ m r ¨ = m r ω 2 m\ddot r = m r\omega^2 m r ¨ = m r ω 2 q q q T T T not from V V V ∂ T / ∂ r \partial T/\partial r ∂ T / ∂ r Q j Q_j Q j ∂ L / ∂ q j \partial L/\partial q_j ∂ L / ∂ q j − ∂ V / ∂ q j -\partial V/\partial q_j − ∂ V / ∂ q j
Common mistake "Generalized momentum is alwaysm q ˙ m\dot q m q ˙ Why it feels right: p = m v p=mv p = m v Reality: p θ = m ℓ 2 θ ˙ p_\theta = m\ell^2\dot\theta p θ = m ℓ 2 θ ˙ p j = ∂ L / ∂ q ˙ j p_j=\partial L/\partial\dot q_j p j = ∂ L / ∂ q ˙ j
Common mistake "Generalized force = real force component."Why it feels right: for Cartesian q = x q=x q = x Reality: for an angular coordinate it's a torque; in general Q j = ∑ i F ⃗ i ⋅ ∂ r ⃗ i / ∂ q j Q_j=\sum_i \vec F_i\cdot \partial\vec r_i/\partial q_j Q j = ∑ i F i ⋅ ∂ r i / ∂ q j weighted projection with the strange-units weight ∂ r ⃗ i / ∂ q j \partial\vec r_i/\partial q_j ∂ r i / ∂ q j
Common mistake "If a coordinate is cyclic, the velocityq ˙ \dot q q ˙ Why it feels right: sounds like conservation. Reality: it's the momentum p j p_j p j p θ = m r 2 θ ˙ p_\theta=mr^2\dot\theta p θ = m r 2 θ ˙ θ ˙ \dot\theta θ ˙ changes as r r r
Common mistake Forgetting the explicit-time term inδ r ⃗ i \delta\vec r_i δ r i virtual displacements time is frozen, so we drop ∂ r ⃗ i / ∂ t \partial\vec r_i/\partial t ∂ r i / ∂ t δ \delta δ d d d Q j Q_j Q j
Recall Feynman: explain to a 12-year-oldImagine pushing a merry-go-round. To describe it you don't track every plank's x , y x,y x , y twist (torque) instead of plain push, and "how much it's spinning" (angular momentum) instead of plain speed. Lagrangian mechanics has one master recipe: write down energy, and the force-like and momentum-like numbers for whatever variable you chose pop right out — push for a slider, twist for a spinner, automatically.
"P is partial-velocity, Q is project-onto."
p j = ∂ L / ∂ q ˙ j p_j=\partial L/\partial\dot q_j p j = ∂ L / ∂ q ˙ j dot ) and Q j = F ⃗ ⋅ ∂ r ⃗ / ∂ q j Q_j=\vec F\cdot\partial\vec r/\partial q_j Q j = F ⋅ ∂ r / ∂ q j project force onto how position responds to q j q_j q j "Cyclic coordinate ⇒ conserved momentum."
Define generalized momentum p j p_j p j p j = ∂ L ∂ q ˙ j p_j=\dfrac{\partial L}{\partial \dot q_j} p j = ∂ q ˙ j ∂ L , the derivative of the Lagrangian w.r.t. generalized velocity.
Define generalized force Q j Q_j Q j Q j = ∑ i F ⃗ i ⋅ ∂ r ⃗ i ∂ q j Q_j=\sum_i \vec F_i\cdot\dfrac{\partial \vec r_i}{\partial q_j} Q j = ∑ i F i ⋅ ∂ q j ∂ r i , defined so
δ W = ∑ j Q j δ q j \delta W=\sum_j Q_j\,\delta q_j δ W = ∑ j Q j δ q j .
For conservative forces, Q j = ? Q_j=? Q j = ? Q j = − ∂ V ∂ q j Q_j=-\dfrac{\partial V}{\partial q_j} Q j = − ∂ q j ∂ V .
Why does an angular coordinate give a torque, not a force? Because
Q j δ q j Q_j\delta q_j Q j δ q j must equal work; if
q j q_j q j is an angle then
Q j Q_j Q j has units energy/radian = N·m = torque.
State the "cancellation of dots" identity. ∂ v ⃗ i ∂ q ˙ j = ∂ r ⃗ i ∂ q j \dfrac{\partial \vec v_i}{\partial \dot q_j}=\dfrac{\partial \vec r_i}{\partial q_j} ∂ q ˙ j ∂ v i = ∂ q j ∂ r i , since
v ⃗ i \vec v_i v i is linear in
q ˙ j \dot q_j q ˙ j .
What is a cyclic (ignorable) coordinate and its consequence? One that doesn't appear in
L L L ; then
p ˙ j = ∂ L / ∂ q j = 0 \dot p_j=\partial L/\partial q_j=0 p ˙ j = ∂ L / ∂ q j = 0 , so
p j p_j p j is conserved.
Generalized momentum of plane-polar θ \theta θ T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) T=\frac12 m(\dot r^2+r^2\dot\theta^2) T = 2 1 m ( r ˙ 2 + r 2 θ ˙ 2 ) p θ = m r 2 θ ˙ p_\theta=mr^2\dot\theta p θ = m r 2 θ ˙ = angular momentum.
Why isn't p j p_j p j m q ˙ j m\dot q_j m q ˙ j Because
p j = ∂ L / ∂ q ˙ j p_j=\partial L/\partial\dot q_j p j = ∂ L / ∂ q ˙ j ; units and mass factors depend on the coordinate (e.g.
m ℓ 2 θ ˙ m\ell^2\dot\theta m ℓ 2 θ ˙ for a pendulum).
In virtual work, why drop ∂ r ⃗ i / ∂ t \partial\vec r_i/\partial t ∂ r i / ∂ t Virtual displacements are taken at frozen time, so the explicit time term vanishes.
How does Euler–Lagrange relate p j p_j p j Q j Q_j Q j p ˙ j = ∂ L / ∂ q j \dot p_j=\partial L/\partial q_j p ˙ j = ∂ L / ∂ q j , i.e. rate of change of generalized momentum equals the generalized "force" term — Newton's law restated.
Lagrangian Mechanics — L = T − V L=T-V L = T − V p j p_j p j Q j Q_j Q j Euler-Lagrange Equation — the equation p ˙ j = ∂ L / ∂ q j \dot p_j=\partial L/\partial q_j p ˙ j = ∂ L / ∂ q j Generalized coordinates and constraints — where q j q_j q j r ⃗ i ( q , t ) \vec r_i(q,t) r i ( q , t ) D'Alembert's Principle — virtual work foundation that defines Q j Q_j Q j Noether's Theorem — cyclic coordinate ⇒ conserved p j p_j p j Hamiltonian Mechanics — p j p_j p j Angular momentum — the p θ p_\theta p θ
Generalized coordinates qj
Generalized velocity qj dot
Cancellation of dots identity
Virtual displacement delta ri
Intuition Hinglish mein samjhoDekho, Lagrangian mechanics ka pura idea yeh hai ki ham awkward x , y , z x,y,z x , y , z q j q_j q j θ \theta θ generalized momentum p j = ∂ L / ∂ q ˙ j p_j=\partial L/\partial \dot q_j p j = ∂ L / ∂ q ˙ j generalized force Q j = ∑ i F ⃗ i ⋅ ∂ r ⃗ i / ∂ q j Q_j=\sum_i \vec F_i\cdot \partial \vec r_i/\partial q_j Q j = ∑ i F i ⋅ ∂ r i / ∂ q j
Sabse mast baat: Q j Q_j Q j δ W = ∑ j Q j δ q j \delta W=\sum_j Q_j\,\delta q_j δ W = ∑ j Q j δ q j q j q_j q j Q j Q_j Q j q j q_j q j Q j Q_j Q j torque ban jaata hai! Ek hi formula, dono kaam. Same tarah, pendulum ka p θ = m ℓ 2 θ ˙ p_\theta=m\ell^2\dot\theta p θ = m ℓ 2 θ ˙ m v m v m v
Sabse powerful idea (80/20 wala): agar Lagrangian L L L q j q_j q j dikhta hi nahi (cyclic coordinate), toh p ˙ j = ∂ L / ∂ q j = 0 \dot p_j=\partial L/\partial q_j=0 p ˙ j = ∂ L / ∂ q j = 0 p j p_j p j conserved . Yahi figure-skater wala physics hai — haath andar khinche, r r r p θ p_\theta p θ θ ˙ \dot\theta θ ˙ symmetry → conservation .
Tip: Hamesha steps follow karo — position r ⃗ ( q ) \vec r(q) r ( q ) T T T V V V L = T − V L=T-V L = T − V p j p_j p j Q j Q_j Q j p ˙ j = Q j \dot p_j=Q_j p ˙ j = Q j
