Exercises — Kinetic energy in generalized coordinates

The rule: $T_1$ and $T_0$ appear only when $\partial \mathbf{r}/\partial t\ne 0$, i.e. when the position map literally contains $t$. (a) No $t$ in $x=r\cos \theta ,y=r\sin \theta$ → only $T_2$. (b) Map is $x=r\cos \omega t$: contains $t$ → $T_2$ and $T_0$; $T_1=0$ here because the one moving coordinate's basis vector is perpendicular to the drift (shown in L2.2). (c) The pivot position contains the imposed clock term $A\cos (\Omega t)$ (driving frequency $\Omega$), so $t$ appears explicitly → in general all three, $T_2,T_1,T_0$.

From the parent worked example: $M_{rr}=m$, $M_{\theta \theta }=m{r}^2$, $M_{r\theta }=M_{\theta r}=0$. $M=\left(\begin{array}{cc}m& 0\\ 0& m{r}^2\end{array}\right).$ It is symmetric ($M_{r\theta }=M_{\theta r}$) and it depends on position through $r^2$: the "effective mass" for angular motion grows as you move outward — a lever-arm effect.

Differentiate the map with respect to each coordinate (that is the whole method — $M_{ij}$ is dot products of these): $\frac{\partial \mathbf{r}}{\partial r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ),$ $\frac{\partial \mathbf{r}}{\partial \theta }=(r\cos \theta \cos \varphi ,r\cos \theta \sin \varphi ,-r\sin \theta ),$ $\frac{\partial \mathbf{r}}{\partial \varphi }=(-r\sin \theta \sin \varphi ,r\sin \theta \cos \varphi ,0).$ Now take dot products, using ${\cos }^2+{\sin }^2=1$ repeatedly:

• $M_{rr}=m({\sin }^2\theta {\cos }^2\varphi +{\sin }^2\theta {\sin }^2\varphi +{\cos }^2\theta )=m({\sin }^2\theta +{\cos }^2\theta )=m.$
• $M_{\theta \theta }=m{r}^2({\cos }^2\theta {\cos }^2\varphi +{\cos }^2\theta {\sin }^2\varphi +{\sin }^2\theta )=m{r}^2.$
• $M_{\varphi \varphi }=m{r}^2{\sin }^2\theta ({\sin }^2\varphi +{\cos }^2\varphi )=m{r}^2{\sin }^2\theta .$
• Off-diagonals: e.g. $M_{r\theta }=mr(\sin \theta \cos \theta {\cos }^2\varphi +\sin \theta \cos \theta {\sin }^2\varphi -\sin \theta \cos \theta )=0$. Similarly $M_{r\varphi }=M_{\theta \varphi }=0$ (the spherical basis is orthogonal). $\boxed{T=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2+r^2{\sin }^2\theta {\dot{\varphi }}^2\right)}$

Since the coordinate is $r$, we differentiate the map with respect to $r$ (that plays the role of $q$ throughout): $\frac{\partial \mathbf{r}}{\partial r}=(\cos \omega t,\sin \omega t),\frac{\partial \mathbf{r}}{\partial t}=(-r\omega \sin \omega t,r\omega \cos \omega t).$ $T_2$: $M_{rr}=m({\cos }^2\omega t+{\sin }^2\omega t)=m$, so $T_2=\frac{1}{2}m{\dot{r}}^2$. $T_1$: $a_r=m[\cos \omega t(-r\omega \sin \omega t)+\sin \omega t(r\omega \cos \omega t)]=0$ — the two products are equal and opposite. This is $T_1=0$ because $\partial \mathbf{r}/\partial r$ points along the wire while $\partial \mathbf{r}/\partial t$ points across it (perpendicular); their dot product vanishes. $T_0=\frac{1}{2}m(r\omega {)}^2$. $\boxed{T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\omega }^2}$

The figure below draws this: the straight wire (white) at some instant, the bead (yellow dot), the yellow arrow $\partial \mathbf{r}/\partial r$ pointing outward along the wire (your free direction), and the blue arrow $\partial \mathbf{r}/\partial t$ pointing sideways (the spin dragging the bead around). The pink right-angle mark shows they meet at $90^{\circ }$ — that perpendicularity is exactly why their dot product, and hence $T_1$, is zero.

$\frac{\partial \mathbf{r}}{\partial q}=(1,kq).$ $M_{qq}=m(1\cdot 1+kq\cdot kq)=m(1+k^2{q}^2).$ No $t$ in the map, so $T=T_2$: $\boxed{T=\frac{1}{2}m(1+k^2{q}^2){\dot{q}}^2}$ Notice the effective mass $m(1+k^2{q}^2)$ grows away from the bottom: on a steeper part of the parabola the same $\dot{q}$ carries more real speed.

$T_1$ measures the overlap $\partial \mathbf{r}/\partial q\cdot \partial \mathbf{r}/\partial t$: "does the direction you can freely move share any component with the direction the constraint is dragging you?" For a wire that only rotates, your free direction (radially, along the wire) is exactly perpendicular to the drag direction (tangential, from the spin). Perpendicular ⇒ dot product $=0$ ⇒ $T_1=0$. This is exactly the perpendicular pair of arrows drawn in the L2.2 figure (yellow along the wire, blue across it).

To make $T_1\ne 0$, let the constraint drag you partly along your free direction. Example: a wire being translated along its own length, $x=q+vt$ (coordinate $q$, the wire slides in the $x$-direction at speed $v$). Then $\partial \mathbf{r}/\partial q=(1,0)$ and $\partial \mathbf{r}/\partial t=(v,0)$ are parallel, giving $a_q=mv\ne 0$ and $T_1=mv\dot{q}$ (this is parent Example 3).

With $T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\theta }}^2$: $\frac{\partial T}{\partial \dot{r}}=m\dot{r},\frac{\partial T}{\partial \dot{\theta }}=m{r}^2\dot{\theta }.$ $\dot{r}(m\dot{r})+\dot{\theta }(m{r}^2\dot{\theta })=m{\dot{r}}^2+m{r}^2{\dot{\theta }}^2=2\left(\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\theta }}^2\right)=2T.✓$ Why it matters: this factor of $2$ is exactly why, for scleronomic systems, the energy function ${\sum }_i{p}_i{\dot{q}}_i-L$ equals $2T_2-(T_2-V)=T+V$, giving $H=T+V$. See Energy function and Hamiltonian and Generalized momentum and conserved quantities.

First build $T$. With $\dot{x}=\dot{q}+v$ and $\dot{y}=\dot{w}$, $T=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)=\frac{1}{2}m[(\dot{q}+v{)}^2+{\dot{w}}^2]=\frac{1}{2}m({\dot{q}}^2+{\dot{w}}^2)+mv\dot{q}+\frac{1}{2}m{v}^2.$ So $T_2=\frac{1}{2}m({\dot{q}}^2+{\dot{w}}^2)$, $T_1=mv\dot{q}$, $T_0=\frac{1}{2}m{v}^2$, coordinates $(q,w)$. $\frac{\partial T}{\partial \dot{q}}=m\dot{q}+mv,\frac{\partial T}{\partial \dot{w}}=m\dot{w}.$ $\dot{q}(m\dot{q}+mv)+\dot{w}(m\dot{w})=m{\dot{q}}^2+mv\dot{q}+m{\dot{w}}^2.$ Compare with $2T=m{\dot{q}}^2+m{\dot{w}}^2+2mv\dot{q}+m{v}^2$: $\left({\sum }_i{\dot{q}}_i\frac{\partial T}{\partial {\dot{q}}_i}\right)-2T=-mv\dot{q}-m{v}^2=-(T_1+2T_0).$ Why: Euler's theorem gives degree $\times$ term: $T_2\to 2T_2$, $T_1\to 1\cdot T_1$, $T_0\to 0$. So $\sum {\dot{q}}_i{\partial }_{{\dot{q}}_i}T=2T_2+T_1$, which differs from $2T=2T_2+2T_1+2T_0$ by exactly $-(T_1+2T_0)$. The leftover is produced by the time-dependent ($T_1,T_0$) pieces — precisely why $H\ne T+V$ for rheonomic systems.

Velocities: ${\dot{x}}_1={\ell }_1\cos {\theta }_1{\dot{\theta }}_1,{\dot{y}}_1={\ell }_1\sin {\theta }_1{\dot{\theta }}_1\Rightarrow {\dot{x}}_1^2+{\dot{y}}_1^2={\ell }_1^2{\dot{\theta }}_1^2.$ ${\dot{x}}_2={\ell }_1\cos {\theta }_1{\dot{\theta }}_1+{\ell }_2\cos {\theta }_2{\dot{\theta }}_2,{\dot{y}}_2={\ell }_1\sin {\theta }_1{\dot{\theta }}_1+{\ell }_2\sin {\theta }_2{\dot{\theta }}_2.$ Square and add, using $\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2=\cos ({\theta }_1-{\theta }_2)$: ${\dot{x}}_2^2+{\dot{y}}_2^2={\ell }_1^2{\dot{\theta }}_1^2+{\ell }_2^2{\dot{\theta }}_2^2+2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2.$ Assemble $T=\frac{1}{2}{m}_1({\dot{x}}_1^2+{\dot{y}}_1^2)+\frac{1}{2}{m}_2({\dot{x}}_2^2+{\dot{y}}_2^2)$: $\boxed{T=\frac{1}{2}(m_1+m_2){\ell }_1^2{\dot{\theta }}_1^2+\frac{1}{2}{m}_2{\ell }_2^2{\dot{\theta }}_2^2+m_2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2.}$ Reading off the metric (recall the $\frac{1}{2}{\sum }_{i,j}$ counts the pair $\{1,2\}$ twice, so the coefficient of ${\dot{\theta }}_1{\dot{\theta }}_2$ equals $M_{12}$, not $2M_{12}$): $M_{11}=(m_1+m_2){\ell }_1^2,M_{22}=m_2{\ell }_2^2,M_{12}=M_{21}=m_2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2).$ This is scleronomic (no $t$) so $T=T_2$. The coupling lives entirely in the position-dependent $M_{12}$.

The figure shows the geometry: the fixed pivot at top, the first rod (white, length $L_1$) to bob $m_1$ at angle ${\theta }_1$ from the vertical dashed line, and the second rod (blue, length $L_2$) hanging from $m_1$ to bob $m_2$ at angle ${\theta }_2$ from its own vertical. The angle difference ${\theta }_1-{\theta }_2$ between the two rods is what appears inside the cosine of $M_{12}$ — when the rods align, that cosine is $1$ and the coupling is strongest.

At ${\theta }_1={\theta }_2$: $\cos ({\theta }_1-{\theta }_2)=1$. With equal masses/lengths: $M=\left(\begin{array}{cc}2m{\ell }^2& m{\ell }^2\\ m{\ell }^2& m{\ell }^2\end{array}\right).$ Positive-definite test (Sylvester): top-left entry $2m{\ell }^2>0$; determinant $\det M=(2m{\ell }^2)(m{\ell }^2)-(m{\ell }^2{)}^2=2m^2{\ell }^4-m^2{\ell }^4=m^2{\ell }^4>0.$ Both leading minors positive ⇒ positive-definite ✓. This guarantees $T_2>0$ for any nonzero velocity, as physics demands (energy of motion can't be negative). See Configuration space and the metric tensor.

Coordinate is $q=\alpha$; the spin $\omega t$ is imposed (rheonomic). Compute the two vectors. $\frac{\partial \mathbf{r}}{\partial \alpha }=(R\cos \alpha \cos \omega t,R\cos \alpha \sin \omega t,R\sin \alpha ),$ $\frac{\partial \mathbf{r}}{\partial t}=(-R\sin \alpha \omega \sin \omega t,R\sin \alpha \omega \cos \omega t,0).$ $T_2$: $M_{\alpha \alpha }=m(R^2{\cos }^2\alpha {\cos }^2\omega t+R^2{\cos }^2\alpha {\sin }^2\omega t+R^2{\sin }^2\alpha )=m{R}^2({\cos }^2\alpha +{\sin }^2\alpha )=m{R}^2.$ So $T_2=\frac{1}{2}m{R}^2{\dot{\alpha }}^2.$ $T_1$: $a_{\alpha }=m{\partial }_{\alpha }\mathbf{r}\cdot {\partial }_t\mathbf{r}=m[R\cos \alpha \cos \omega t(-R\sin \alpha \omega \sin \omega t)+R\cos \alpha \sin \omega t(R\sin \alpha \omega \cos \omega t)+0]$. The first two terms are $\mp m{R}^2\omega \cos \alpha \sin \alpha \cos \omega t\sin \omega t$ — equal magnitude, opposite sign — so $a_{\alpha }=0\Rightarrow T_1=0.$ (Again: free direction along the hoop is perpendicular to the spin-drag direction.) $T_0$: $\frac{1}{2}m[(R\sin \alpha \omega \sin \omega t{)}^2+(R\sin \alpha \omega \cos \omega t{)}^2]=\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha .$ $\boxed{T=\frac{1}{2}m{R}^2{\dot{\alpha }}^2+\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha .}$ Physically, $T_0=\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha$ depends on position only, so in the Lagrangian it behaves like a negative potential — it drives the centrifugal push that flings the bead outward (larger $\sin \alpha$), the classic mechanism that can lift the bead off the bottom above a critical $\omega$.

The figure shows the vertical circular hoop, the spin axis (the vertical diameter) drawn as a dashed line, the bead at angle $\alpha$ from the bottom, and the horizontal radius $R\sin \alpha$ that the bead traces out as the hoop turns — it is that horizontal radius, squared, that produces the ${\sin }^2\alpha$ inside $T_0$.

Effective potential $V_{\text{eff}}(\alpha )=V-T_0=-mgR\cos \alpha -\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha .$ Stationary points: $V_{\text{eff}}^{\prime }(\alpha )=mgR\sin \alpha -m{R}^2{\omega }^2\sin \alpha \cos \alpha =mR\sin \alpha (g-R{\omega }^2\cos \alpha )$. At $\alpha =0$, $\sin \alpha =0$ ⇒ it is always an equilibrium. Its stability: expand the second derivative at $\alpha =0$, $V_{\text{eff}}^{\prime \prime }(0)=mgR\cos 0-m{R}^2{\omega }^2({\cos }^{2}0-{\sin }^{2}0)=mgR-m{R}^2{\omega }^2=mR(g-R{\omega }^2).$ Stable (minimum) requires $V_{\text{eff}}^{\prime \prime }(0)>0\Rightarrow g>R{\omega }^2$. It loses stability when $\boxed{{\omega }_c=\sqrt{\frac{g}{R}}.}$ Above ${\omega }_c$ the bottom becomes a maximum and a new stable angle ${\alpha }_{\star }$ with $\cos {\alpha }_{\star }=g/(R{\omega }^2)$ appears — the bead rides up the spinning hoop. This whole analysis is powered by correctly separating $T_0$; see Euler-Lagrange equations for the equation of motion it feeds.

Which coordinate quantity decides whether $T_1,T_0$ exist? ::: Whether the position map $\mathbf{r}(q,t)$ contains an explicit, imposed $t$ (rheonomic) — if not, $T=T_2$ only. What is $M_{12}$ for the double pendulum? ::: $m_2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2)$. Critical spin for the rotating hoop bead? ::: ${\omega }_c=\sqrt{g/R}$.