Intuition The big pictureKinetic energy is "born" in Cartesian coordinates as the simple 1 2 m v 2 \frac12 m v^2 2 1 m v 2 generalized coordinates q i q_i q i v 2 v^2 v 2 q ˙ i \dot q_i q ˙ i quadratic-ish polynomial in the velocities — sometimes with extra terms that depend on position and even on time. Understanding where each term comes from is the whole game.
A system of N N N r a \mathbf r_a r a a = 1 , … , N a=1,\dots,N a = 1 , … , N n n n generalized coordinates q 1 , … , q n q_1,\dots,q_n q 1 , … , q n 3 N 3N 3 N r a = r a ( q 1 , … , q n , t ) . \mathbf r_a = \mathbf r_a(q_1,\dots,q_n,\,t). r a = r a ( q 1 , … , q n , t ) . t t t moving/time-dependent (rheonomic).
The total kinetic energy is always
T = ∑ a 1 2 m a r ˙ a ⋅ r ˙ a . T = \sum_a \tfrac12 m_a\, \dot{\mathbf r}_a \cdot \dot{\mathbf r}_a. T = ∑ a 2 1 m a r ˙ a ⋅ r ˙ a . r ˙ a \dot{\mathbf r}_a r ˙ a q ˙ i \dot q_i q ˙ i
Step 1 — Differentiate the position map (chain rule).
Because r a = r a ( q , t ) \mathbf r_a = \mathbf r_a(q,t) r a = r a ( q , t ) r ˙ a = ∑ i = 1 n ∂ r a ∂ q i q ˙ i + ∂ r a ∂ t . \dot{\mathbf r}_a = \sum_{i=1}^{n}\frac{\partial \mathbf r_a}{\partial q_i}\,\dot q_i \;+\; \frac{\partial \mathbf r_a}{\partial t}. r ˙ a = ∑ i = 1 n ∂ q i ∂ r a q ˙ i + ∂ t ∂ r a .
Why this step? The chain rule says each velocity component is "how position changes per coordinate" × \times × ∂ t r a \partial_t \mathbf r_a ∂ t r a
Step 2 — Substitute into T T T
T = ∑ a 1 2 m a ( ∑ i ∂ r a ∂ q i q ˙ i + ∂ r a ∂ t ) ⋅ ( ∑ j ∂ r a ∂ q j q ˙ j + ∂ r a ∂ t ) . T = \sum_a \tfrac12 m_a\left(\sum_i \frac{\partial \mathbf r_a}{\partial q_i}\dot q_i + \frac{\partial \mathbf r_a}{\partial t}\right)\cdot\left(\sum_j \frac{\partial \mathbf r_a}{\partial q_j}\dot q_j + \frac{\partial \mathbf r_a}{\partial t}\right). T = ∑ a 2 1 m a ( ∑ i ∂ q i ∂ r a q ˙ i + ∂ t ∂ r a ) ⋅ ( ∑ j ∂ q j ∂ r a q ˙ j + ∂ t ∂ r a ) .
Why this step? We literally plug the dot product into the same expression and expand. The square of a sum gives three kinds of terms: q ˙ q ˙ \dot q\,\dot q q ˙ q ˙ q ˙ × ( ∂ t ) \dot q\times(\partial_t) q ˙ × ( ∂ t ) ( ∂ t ) × ( ∂ t ) (\partial_t)\times(\partial_t) ( ∂ t ) × ( ∂ t )
Step 3 — Collect by powers of velocity. Expanding the product:
T = T 2 + T 1 + T 0 \boxed{T = T_2 + T_1 + T_0} T = T 2 + T 1 + T 0
T 2 = 1 2 ∑ i , j M i j ( q , t ) q ˙ i q ˙ j , M i j = ∑ a m a ∂ r a ∂ q i ⋅ ∂ r a ∂ q j T_2 = \tfrac12\sum_{i,j} M_{ij}(q,t)\,\dot q_i\dot q_j,\qquad M_{ij} = \sum_a m_a\,\frac{\partial \mathbf r_a}{\partial q_i}\cdot\frac{\partial \mathbf r_a}{\partial q_j} T 2 = 2 1 ∑ i , j M ij ( q , t ) q ˙ i q ˙ j , M ij = ∑ a m a ∂ q i ∂ r a ⋅ ∂ q j ∂ r a
T 1 = ∑ i a i ( q , t ) q ˙ i , a i = ∑ a m a ∂ r a ∂ q i ⋅ ∂ r a ∂ t T_1 = \sum_{i} a_i(q,t)\,\dot q_i,\qquad a_i = \sum_a m_a\,\frac{\partial \mathbf r_a}{\partial q_i}\cdot\frac{\partial \mathbf r_a}{\partial t} T 1 = ∑ i a i ( q , t ) q ˙ i , a i = ∑ a m a ∂ q i ∂ r a ⋅ ∂ t ∂ r a
T 0 = 1 2 ∑ a m a ( ∂ r a ∂ t ) 2 . T_0 = \tfrac12\sum_a m_a\left(\frac{\partial \mathbf r_a}{\partial t}\right)^2. T 0 = 2 1 ∑ a m a ( ∂ t ∂ r a ) 2 .
Why this step? Grouping by how many q ˙ \dot q q ˙ T 2 T_2 T 2 quadratic in velocities, T 1 T_1 T 1 linear , T 0 T_0 T 0 none .
Intuition When constraints don't move, time disappearsIf r a \mathbf r_a r a no explicit time dependence (∂ r a / ∂ t = 0 \partial \mathbf r_a/\partial t = 0 ∂ r a / ∂ t = 0 scleronomic constraint), then a i = 0 a_i = 0 a i = 0 T 0 = 0 T_0 = 0 T 0 = 0 T = T 2 = 1 2 ∑ i , j M i j ( q ) q ˙ i q ˙ j T = T_2 = \tfrac12\sum_{i,j}M_{ij}(q)\,\dot q_i\dot q_j T = T 2 = 2 1 ∑ i , j M ij ( q ) q ˙ i q ˙ j pure homogeneous quadratic form in the velocities. This is the case in 95% of textbook problems.
A homogeneous quadratic function obeys Euler's theorem : ∑ i q ˙ i ∂ T / ∂ q ˙ i = 2 T \sum_i \dot q_i \,\partial T/\partial \dot q_i = 2T ∑ i q ˙ i ∂ T / ∂ q ˙ i = 2 T ∑ i p i q ˙ i − L \sum_i p_i\dot q_i - L ∑ i p i q ˙ i − L H = T + V H=T+V H = T + V
Worked example 1. Plane polar coordinates (x = r cos θ , y = r sin θ x=r\cos\theta,\ y=r\sin\theta x = r cos θ , y = r sin θ m m m q = ( r , θ ) q=(r,\theta) q = ( r , θ ) ∂ r ∂ r = ( cos θ , sin θ ) , ∂ r ∂ θ = ( − r sin θ , r cos θ ) . \frac{\partial \mathbf r}{\partial r}=(\cos\theta,\sin\theta),\quad \frac{\partial \mathbf r}{\partial \theta}=(-r\sin\theta,\ r\cos\theta). ∂ r ∂ r = ( cos θ , sin θ ) , ∂ θ ∂ r = ( − r sin θ , r cos θ ) . Why? Differentiate the position map w.r.t. each coordinate.
Now the metric components:
M r r = m ( cos 2 + sin 2 ) = m M_{rr}=m(\cos^2+\sin^2)=m M r r = m ( cos 2 + sin 2 ) = m M θ θ = m r 2 ( sin 2 + cos 2 ) = m r 2 M_{\theta\theta}=m\,r^2(\sin^2+\cos^2)=mr^2 M θ θ = m r 2 ( sin 2 + cos 2 ) = m r 2 M r θ = m ( cos θ ) ( − r sin θ ) + m ( sin θ ) ( r cos θ ) = 0 M_{r\theta}=m(\cos\theta)(-r\sin\theta)+m(\sin\theta)(r\cos\theta)=0 M r θ = m ( cos θ ) ( − r sin θ ) + m ( sin θ ) ( r cos θ ) = 0
Why M r θ = 0 M_{r\theta}=0 M r θ = 0 The radial and angular basis vectors are orthogonal — the cross terms cancel.
T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) \boxed{T=\tfrac12 m(\dot r^2 + r^2\dot\theta^2)} T = 2 1 m ( r ˙ 2 + r 2 θ ˙ 2 )
Worked example 2. Bead on a rotating wire (rheonomic — givesT 1 , T 0 T_1,T_0 T 1 , T 0 ω \omega ω q = r q=r q = r θ = ω t \theta=\omega t θ = ω t x = r cos ω t , y = r sin ω t . x=r\cos\omega t,\quad y=r\sin\omega t. x = r cos ω t , y = r sin ω t . ∂ r ∂ r = ( cos ω t , sin ω t ) , ∂ r ∂ t = ( − r ω sin ω t , r ω cos ω t ) . \frac{\partial \mathbf r}{\partial r}=(\cos\omega t,\sin\omega t),\qquad \frac{\partial \mathbf r}{\partial t}=(-r\omega\sin\omega t,\ r\omega\cos\omega t). ∂ r ∂ r = ( cos ω t , sin ω t ) , ∂ t ∂ r = ( − r ω sin ω t , r ω cos ω t ) . Why ∂ t r ≠ 0 \partial_t\mathbf r\neq0 ∂ t r = 0 The wire's imposed rotation enters explicitly — that's a moving constraint.
T 2 T_2 T 2 M r r = m M_{rr}=m M r r = m T 2 = 1 2 m r ˙ 2 T_2=\tfrac12 m\dot r^2 T 2 = 2 1 m r ˙ 2 T 1 T_1 T 1 a r = m ∂ r ∂ r ⋅ ∂ r ∂ t = m [ cos ω t ( − r ω sin ω t ) + sin ω t ( r ω cos ω t ) ] = 0 a_r=m\,\frac{\partial\mathbf r}{\partial r}\cdot\frac{\partial\mathbf r}{\partial t}=m[\cos\omega t(-r\omega\sin\omega t)+\sin\omega t(r\omega\cos\omega t)]=0 a r = m ∂ r ∂ r ⋅ ∂ t ∂ r = m [ cos ω t ( − r ω sin ω t ) + sin ω t ( r ω cos ω t )] = 0 T 0 T_0 T 0 1 2 m ( r ω ) 2 \tfrac12 m(r\omega)^2 2 1 m ( r ω ) 2 T = 1 2 m r ˙ 2 + 1 2 m r 2 ω 2 \boxed{T=\tfrac12 m\dot r^2 + \tfrac12 m r^2\omega^2} T = 2 1 m r ˙ 2 + 2 1 m r 2 ω 2 T 1 = 0 T_1=0 T 1 = 0 T 0 ≠ 0 T_0\neq0 T 0 = 0 T 0 T_0 T 0 potential and produces the centrifugal effect.
T 1 T_1 T 1 x = q + v t x = q + vt x = q + v t v v v x x x y = y= y = w w w ∂ r / ∂ q = ( 1 , 0 ) \partial\mathbf r/\partial q=(1,0) ∂ r / ∂ q = ( 1 , 0 ) ∂ r / ∂ t = ( v , 0 ) \partial\mathbf r/\partial t=(v,0) ∂ r / ∂ t = ( v , 0 ) a q = m ( 1 ) ( v ) = m v a_q = m(1)(v)=mv a q = m ( 1 ) ( v ) = m v T 1 = m v q ˙ T_1=mv\,\dot q T 1 = m v q ˙ T 0 = 1 2 m v 2 T_0=\tfrac12 mv^2 T 0 = 2 1 m v 2 T = 1 2 m ( q ˙ 2 + w ˙ 2 ) + m v q ˙ + 1 2 m v 2 = 1 2 m [ ( q ˙ + v ) 2 + w ˙ 2 ] . T=\tfrac12 m(\dot q^2+\dot w^2)+mv\dot q+\tfrac12 mv^2 = \tfrac12 m[(\dot q+v)^2+\dot w^2]. T = 2 1 m ( q ˙ 2 + w ˙ 2 ) + m v q ˙ + 2 1 m v 2 = 2 1 m [( q ˙ + v ) 2 + w ˙ 2 ] . Why? That's just 1 2 m x ˙ 2 \tfrac12 m\dot x^2 2 1 m x ˙ 2 T 0 + T 1 + T 2 T_0+T_1+T_2 T 0 + T 1 + T 2
Common mistake "T is always quadratic in the velocities."Why it feels right: Every intro problem (T = 1 2 m r ˙ 2 + 1 2 m r 2 θ ˙ 2 T=\frac12 m\dot r^2+\frac12 mr^2\dot\theta^2 T = 2 1 m r ˙ 2 + 2 1 m r 2 θ ˙ 2 The fix: That's only true for scleronomic (time-independent) constraints. The moment a constraint moves on its own schedule (∂ t r ≠ 0 \partial_t\mathbf r\neq0 ∂ t r = 0 T 1 T_1 T 1 T 0 T_0 T 0 r a \mathbf r_a r a t t t
M i j M_{ij} M ij Why it feels right: In Cartesian coordinates the mass is just the number m m m The fix: M i j = M i j ( q ) M_{ij}=M_{ij}(q) M ij = M ij ( q ) M θ θ = m r 2 M_{\theta\theta}=mr^2 M θ θ = m r 2 r r r T T T q q q
Common mistake Forgetting the cross term inq ˙ i q ˙ j \dot q_i\dot q_j q ˙ i q ˙ j i ≠ j i\neq j i = j twice in ∑ i , j \sum_{i,j} ∑ i , j 1 2 \frac12 2 1
Recall Feynman: explain it to a 12-year-oldImagine your toy car can only move along bent tracks. To say how fast it's really moving, you'd normally use plain left-right-up-down speed. But it's easier to say "how fast is it going along the track ". Math lets us trade one for the other. If the track itself is being yanked around (someone is spinning the whole table), then even a car sitting still on the track is actually moving in the room — so we add extra speed terms. That extra stuff is the T 1 T_1 T 1 T 0 T_0 T 0 1 2 ( stuff ) × ( speed ) 2 \frac12 (\text{stuff})\times(\text{speed})^2 2 1 ( stuff ) × ( speed ) 2
2-1-0: Quadratic, Linear, Lonely "
T 2 T_2 T 2 Q uadratic in q ˙ \dot q q ˙ T 1 T_1 T 1 L inear (only if constraints move), T 0 T_0 T 0 L onely / no velocity (only if constraints move). Both linear & lonely vanish when time isn't in the map .
Why does r ˙ a \dot{\mathbf r}_a r ˙ a ∂ r a / ∂ t \partial\mathbf r_a/\partial t ∂ r a / ∂ t Because under rheonomic (moving) constraints the position map
r a ( q , t ) \mathbf r_a(q,t) r a ( q , t ) depends explicitly on time, so the chain rule adds an explicit clock term.
Write the general decomposition of T T T T = T 2 + T 1 + T 0 T=T_2+T_1+T_0 T = T 2 + T 1 + T 0 — quadratic, linear, and velocity-free in the
q ˙ i \dot q_i q ˙ i .
Definition of the mass matrix M i j M_{ij} M ij M i j = ∑ a m a ∂ q i r a ⋅ ∂ q j r a M_{ij}=\sum_a m_a\,\partial_{q_i}\mathbf r_a\cdot\partial_{q_j}\mathbf r_a M ij = ∑ a m a ∂ q i r a ⋅ ∂ q j r a ; symmetric and positive-definite.
When is T T T T = T 2 T=T_2 T = T 2 When constraints are scleronomic (
∂ r a / ∂ t = 0 \partial\mathbf r_a/\partial t=0 ∂ r a / ∂ t = 0 ), making
a i = 0 a_i=0 a i = 0 and
T 0 = 0 T_0=0 T 0 = 0 .
What is T T T 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 ) .
By Euler's theorem, what is ∑ i q ˙ i ∂ T 2 / ∂ q ˙ i \sum_i \dot q_i\,\partial T_2/\partial \dot q_i ∑ i q ˙ i ∂ T 2 / ∂ q ˙ i 2 T 2 2T_2 2 T 2 (homogeneous of degree 2 in velocities).
Physical meaning of T 0 T_0 T 0 1 2 m r 2 ω 2 \tfrac12 mr^2\omega^2 2 1 m r 2 ω 2 — acts like an effective potential giving the centrifugal effect.
Is M i j M_{ij} M ij No, it depends on the generalized coordinates
q q q (e.g.
M θ θ = m r 2 M_{\theta\theta}=mr^2 M θ θ = m r 2 ).
Lagrangian mechanics — L = T − V L=T-V L = T − V T T T Generalized coordinates and constraints — scleronomic vs rheonomic.Euler-Lagrange equations — where ∂ T / ∂ q ˙ \partial T/\partial \dot q ∂ T / ∂ q ˙ ∂ T / ∂ q \partial T/\partial q ∂ T / ∂ q Generalized momentum and conserved quantities — p i = ∂ T / ∂ q ˙ i = ∑ j M i j q ˙ j p_i=\partial T/\partial\dot q_i = \sum_j M_{ij}\dot q_j p i = ∂ T / ∂ q ˙ i = ∑ j M ij q ˙ j Configuration space and the metric tensor — M i j M_{ij} M ij Energy function and Hamiltonian — H = T 2 − T 0 + V H=T_2-T_0+V H = T 2 − T 0 + V H = T + V H=T+V H = T + V
symmetric positive-definite metric
gives partial r partial t
gives partial r partial t
Cartesian KE half m v squared
Position map r_a of q and t
Rheonomic time-dependent constraint
Intuition Hinglish mein samjhoDekho, kinetic energy ki asli definition toh simple hai: T = 1 2 m v 2 T=\frac12 m v^2 T = 2 1 m v 2 q i q_i q i θ \theta θ r r r r ˙ \dot{\mathbf r} r ˙ q ˙ i \dot q_i q ˙ i r ˙ = ∑ i ( ∂ r / ∂ q i ) q ˙ i + ∂ r / ∂ t \dot{\mathbf r}=\sum_i (\partial\mathbf r/\partial q_i)\dot q_i + \partial\mathbf r/\partial t r ˙ = ∑ i ( ∂ r / ∂ q i ) q ˙ i + ∂ r / ∂ t
Jab is poori cheez ka square karke T T T T 2 T_2 T 2 T 1 T_1 T 1 T 0 T_0 T 0 T 2 T_2 T 2 T = 1 2 ∑ M i j q ˙ i q ˙ j T=\frac12\sum M_{ij}\dot q_i\dot q_j T = 2 1 ∑ M ij q ˙ i q ˙ j
Yeh M i j M_{ij} M ij M θ θ = m r 2 M_{\theta\theta}=mr^2 M θ θ = m r 2 r r r 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 ) T T T T 1 T_1 T 1 T 0 T_0 T 0