Intuition The big idea in one breathIf the Lagrangian does not care about a particular coordinate (you can shift it freely and nothing changes), then something stays constant forever . That "something" is the momentum conjugate to the coordinate. A symmetry of the Lagrangian becomes a conservation law.
Definition Cyclic (ignorable) coordinateA generalized coordinate q k q_k q k cyclic (also called ignorable ) if it does not appear explicitly in the Lagrangian:
∂ L ∂ q k = 0 \frac{\partial L}{\partial q_k} = 0 ∂ q k ∂ L = 0 q ˙ k \dot q_k q ˙ k coordinate itself is absent.
WHY the name "cyclic"? Historically it came from systems where the ignored coordinate was an angle (like ϕ \phi ϕ 0 → 2 π 0\to 2\pi 0 → 2 π
We start from the Euler–Lagrange equation — the master equation of analytical mechanics:
d d t ( ∂ L ∂ q ˙ k ) − ∂ L ∂ q k = 0 \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_k}\right) - \frac{\partial L}{\partial q_k} = 0 d t d ( ∂ q ˙ k ∂ L ) − ∂ q k ∂ L = 0
Why this step? This is just Lagrange's equation for coordinate q k q_k q k every coordinate, cyclic or not.
Now define the generalized (conjugate) momentum :
Why call it momentum? For L = 1 2 m x ˙ 2 − V ( x ) L = \tfrac12 m\dot x^2 - V(x) L = 2 1 m x ˙ 2 − V ( x ) p x = m x ˙ p_x = m\dot x p x = m x ˙ generalizes this idea to any coordinate.
Substitute the definition into Euler–Lagrange:
d d t p k = ∂ L ∂ q k \frac{d}{dt}\,p_k = \frac{\partial L}{\partial q_k} d t d p k = ∂ q k ∂ L
Why this step? ∂ L ∂ q ˙ k \frac{\partial L}{\partial \dot q_k} ∂ q ˙ k ∂ L p k p_k p k p ˙ k \dot p_k p ˙ k
Now impose that q k q_k q k cyclic , i.e. ∂ L ∂ q k = 0 \dfrac{\partial L}{\partial q_k}=0 ∂ q k ∂ L = 0
d p k d t = 0 ⟹ p k = constant \boxed{\frac{dp_k}{dt}=0 \quad\Longrightarrow\quad p_k = \text{constant}} d t d p k = 0 ⟹ p k = constant
must be true (Feynman-style WHY)
The motion is a "balance sheet": p ˙ k \dot p_k p ˙ k ∂ L / ∂ q k \partial L/\partial q_k ∂ L / ∂ q k L L L q k q_k q k no force pushing along q k q_k q k → \to → → \to →
L = 1 2 m ( x ˙ 2 + y ˙ 2 ) L = \tfrac12 m(\dot x^2 + \dot y^2) L = 2 1 m ( x ˙ 2 + y ˙ 2 )
Is x x x ∂ L / ∂ x = 0 \partial L/\partial x = 0 ∂ L / ∂ x = 0 Why? No x x x Conjugate momentum: p x = ∂ L / ∂ x ˙ = m x ˙ p_x = \partial L/\partial \dot x = m\dot x p x = ∂ L / ∂ x ˙ = m x ˙ Why this step? Differentiate L L L x ˙ \dot x x ˙ Conclusion: m x ˙ = const m\dot x = \text{const} m x ˙ = const
Polar coordinates: L = 1 2 m ( r ˙ 2 + r 2 ϕ ˙ 2 ) − V ( r ) L = \tfrac12 m(\dot r^2 + r^2\dot\phi^2) - V(r) L = 2 1 m ( r ˙ 2 + r 2 ϕ ˙ 2 ) − V ( r )
Is ϕ \phi ϕ ∂ L / ∂ ϕ = 0 \partial L/\partial\phi = 0 ∂ L / ∂ ϕ = 0 r r r V V V Conjugate momentum: p ϕ = ∂ L / ∂ ϕ ˙ = m r 2 ϕ ˙ p_\phi = \partial L/\partial\dot\phi = m r^2 \dot\phi p ϕ = ∂ L / ∂ ϕ ˙ = m r 2 ϕ ˙ Why? Only the r 2 ϕ ˙ 2 r^2\dot\phi^2 r 2 ϕ ˙ 2 ϕ ˙ \dot\phi ϕ ˙ Conclusion: p ϕ = m r 2 ϕ ˙ = const p_\phi = mr^2\dot\phi = \text{const} p ϕ = m r 2 ϕ ˙ = const angular momentum L z L_z L z
L = 1 2 m ( x ˙ 2 + y ˙ 2 ) − m g y L = \tfrac12 m(\dot x^2 + \dot y^2) - mgy L = 2 1 m ( x ˙ 2 + y ˙ 2 ) − m g y
∂ L / ∂ x = 0 \partial L/\partial x = 0 ∂ L / ∂ x = 0 x x x p x = m x ˙ = const p_x = m\dot x = \text{const} p x = m x ˙ = const Why? gravity acts only vertically.∂ L / ∂ y = − m g ≠ 0 \partial L/\partial y = -mg \ne 0 ∂ L / ∂ y = − m g = 0 y y y not cyclic → p y = m y ˙ p_y = m\dot y p y = m y ˙ p ˙ y = − m g \dot p_y = -mg p ˙ y = − m g Why this step? The vertical force − m g -mg − m g ∂ L / ∂ y \partial L/\partial y ∂ L / ∂ y
L = 1 2 m ( θ ˙ 2 R 2 + R 2 sin 2 θ ϕ ˙ 2 ) L = \tfrac12 m\big(\dot\theta^2 R^2 + R^2\sin^2\theta\,\dot\phi^2\big) L = 2 1 m ( θ ˙ 2 R 2 + R 2 sin 2 θ ϕ ˙ 2 )
ϕ \phi ϕ p ϕ = m R 2 sin 2 θ ϕ ˙ = const p_\phi = mR^2\sin^2\theta\,\dot\phi = \text{const} p ϕ = m R 2 sin 2 θ ϕ ˙ = const Why? Symmetry about the polar axis.
Common mistake Steel-manning the common errorsMistake A: "If q ˙ k \dot q_k q ˙ k Why it feels right: both involve "missing." The fix: cyclic means q k q_k q k q ˙ k \dot q_k q ˙ k ϕ ˙ \dot\phi ϕ ˙ L L L ϕ \phi ϕ
Mistake B: "Cyclic coordinate ⇒ that coordinate is constant."
Why it feels right: "something is conserved." The fix: it's the conjugate momentum p k p_k p k not q k q_k q k ϕ \phi ϕ m r 2 ϕ ˙ mr^2\dot\phi m r 2 ϕ ˙
Mistake C: "Energy is the conjugate momentum of time, so time being absent gives momentum conservation."
Fix: time absence (∂ L / ∂ t = 0 \partial L/\partial t = 0 ∂ L / ∂ t = 0 Hamiltonian/energy , a different theorem (Beltrami / energy function h h h
Recall Feynman: explain to a 12-year-oldImagine you're on a perfectly flat, endless ice rink. Sliding left or right, the rink looks exactly the same — it doesn't "care" where you are sideways. Because nothing changes when you move sideways, nothing slows you down sideways: your sideways speed stays the same forever. That "doesn't-care" is a cyclic coordinate , and the unchanging sideways oomph is the conserved momentum . Whenever the world looks the same after you move or turn, something stays constant.
"No coordinate in L L L
Or: C yclic → C onserved (C onjugate momentum). Three C's in a row.
Define a cyclic (ignorable) coordinate. A coordinate
q k q_k q k that does not appear explicitly in
L L L :
∂ L / ∂ q k = 0 \partial L/\partial q_k = 0 ∂ L / ∂ q k = 0 (its velocity
q ˙ k \dot q_k q ˙ k may still appear).
What is conserved when q k q_k q k Its conjugate momentum
p k = ∂ L / ∂ q ˙ k p_k = \partial L/\partial \dot q_k p k = ∂ L / ∂ q ˙ k .
Define generalized (conjugate) momentum. p k ≡ ∂ L / ∂ q ˙ k p_k \equiv \partial L / \partial \dot q_k p k ≡ ∂ L / ∂ q ˙ k .
Derive the conservation law from Euler–Lagrange. d d t ∂ L ∂ q ˙ k − ∂ L ∂ q k = 0 \frac{d}{dt}\frac{\partial L}{\partial \dot q_k} - \frac{\partial L}{\partial q_k}=0 d t d ∂ q ˙ k ∂ L − ∂ q k ∂ L = 0 ; with
∂ L / ∂ q k = 0 \partial L/\partial q_k = 0 ∂ L / ∂ q k = 0 ,
p ˙ k = 0 ⇒ p k \dot p_k = 0 \Rightarrow p_k p ˙ k = 0 ⇒ p k const.
In central-force motion, which coordinate is cyclic and what is conserved? ϕ \phi ϕ is cyclic;
p ϕ = m r 2 ϕ ˙ p_\phi = m r^2 \dot\phi p ϕ = m r 2 ϕ ˙ = angular momentum
L z L_z L z is conserved.
Why is ϕ \phi ϕ ϕ ˙ \dot\phi ϕ ˙ L L L Cyclic refers to absence of the coordinate itself, not its velocity.
Common mistake: does cyclic mean q k q_k q k No — the conjugate momentum
p k p_k p k is constant, not
q k q_k q k .
Which symmetry does a cyclic linear coordinate x x x Translational symmetry ⇒ linear momentum
m x ˙ m\dot x m x ˙ conserved.
What conserves energy instead of momentum? L L L not depending explicitly on time (
∂ L / ∂ t = 0 \partial L/\partial t=0 ∂ L / ∂ t = 0 ) ⇒ Hamiltonian conserved (different theorem).
Euler–Lagrange Equations — the master equation we differentiated.Generalized Momentum — the quantity that gets conserved.Noether's Theorem — the deep generalization: every continuous symmetry ↔ conservation law.Central Force Motion — cyclic ϕ \phi ϕ Hamiltonian Mechanics — cyclic coordinates simplify H H H Conservation of Energy — the time-translation analogue.
partial L over partial qk = 0
dpk/dt = partial L over partial qk
Intuition Hinglish mein samjhoDekho, idea bahut simple hai. Lagrangian L L L q q q q ˙ \dot q q ˙ q k q_k q k L L L q ˙ k \dot q_k q ˙ k q k q_k q k cyclic coordinate kehte hain. Matlab ∂ L / ∂ q k = 0 \partial L/\partial q_k = 0 ∂ L / ∂ q k = 0
Ab Euler–Lagrange equation likho: d d t ∂ L ∂ q ˙ k = ∂ L ∂ q k \frac{d}{dt}\frac{\partial L}{\partial \dot q_k} = \frac{\partial L}{\partial q_k} d t d ∂ q ˙ k ∂ L = ∂ q k ∂ L conjugate momentum p k p_k p k p ˙ k = 0 \dot p_k = 0 p ˙ k = 0 p k p_k p k constant reh gaya. Bas — symmetry se conservation law nikal aaya!
Real examples: free particle me x x x p x = m x ˙ p_x = m\dot x p x = m x ˙ ϕ \phi ϕ p ϕ = m r 2 ϕ ˙ p_\phi = mr^2\dot\phi p ϕ = m r 2 ϕ ˙ angular momentum L z L_z L z momentum p k p_k p k q k q_k q k ϕ \phi ϕ