The kinetic-energy function

1. Geometry inside the kinetic energy

The term $mr^2\dot\theta^2/2$ in polar coordinates is not an extra interaction. It is ordinary translational kinetic energy expressed in variables whose unit displacements correspond to different physical distances. Understanding this structure explains generalized momentum, the inertial coefficients in Lagrange's equations, and the conditions under which the energy function equals $T+V$.

This chapter uses the chain rule and dot product from Unit 1. Coefficients may depend on coordinates and time, but are held fixed when differentiating with respect to generalized velocities.

2. Expansion into quadratic, linear, and constant parts

For each particle, write

The vector $\mathbf e_{ai}$ describes the positional response to coordinate $q_i$, and $\mathbf w_a$ describes transport caused by the prescribed time dependence of the constraints. Squaring gives

Multiplying by $m_a/2$ and summing yields

where

The subscripts indicate degree in the velocities. $T_0$ does not mean zero kinetic energy: a bead can have $\dot q=0$ relative to a moving support while still moving in the laboratory. If the transformation has no explicit time dependence, all $\mathbf w_a$ vanish and $T=T_2$.

3. Symmetry and positivity of the inertia coefficients

Because the dot product is symmetric, $M_{ij}=M_{ji}$. For any real numbers $z_i$,

For positive masses and regular independent coordinates, a nonzero vector $z$ produces a nonzero displacement of at least one particle. The right-hand side is then strictly positive. Thus the matrix $M$ is positive definite in a regular coordinate patch and invertible there.

Differentiating the quadratic expression uses its symmetry:

Hence $\partial^2T/\partial\dot q_i\partial\dot q_j=M_{ij}$. For a velocity-independent $V$, this is also the velocity Hessian of $L$. Invertibility will permit replacing velocities by momenta in Hamiltonian mechanics. A coordinate singularity, such as polar coordinates at $r=0$, can make this representation singular even though the Cartesian mechanics is regular.

4. Euler's identity and the energy function

A homogeneous function of degree $s$ satisfies $f(\lambda\dot q)=\lambda^s f(\dot q)$. Differentiate both sides with respect to $\lambda$ at $\lambda=1$:

Apply this separately to $T_2,T_1,T_0$:

For $L=T-V(q,t)$, $p_i=\partial T/\partial\dot q_i$, so

For time-independent transformations, $T_1=T_0=0$ and $\mathcal E=T+V$. Whether that quantity is conserved is a second question, decided by $\dot{\mathcal E}=\sum_iQ_i\dot q_i-\partial_tL$. The equality with mechanical energy and the constancy along a motion require different checks.

5. The kinetic-energy derivative identity

One further property explains why differentiating a scalar can recover the projection of acceleration. With constant masses,

Take a total time derivative:

From differentiating the transformation and commuting smooth mixed partial derivatives, $\partial\dot{\mathbf r}_a/\partial q_i=d\mathbf e_{ai}/dt$. Therefore

Subtract the equal velocity terms to obtain

This is a kinetic-energy identity, valid also for smooth explicitly time-dependent transformations. Its use to derive Lagrange's equations from D’Alembert's principle is placed in the optional appendix, respecting the syllabus distinction.

6. Applications

Example 1 · Polar kinetic energy

For $x=r\cos\theta$, $y=r\sin\theta$, one finds $\mathbf e_r=(\cos\theta,\sin\theta)$ and $\mathbf e_\theta=(-r\sin\theta,r\cos\theta)$. Their dot products are $1,0,r^2$. Thus $M_{rr}=m$, $M_{r\theta}=0$, $M_{\theta\theta}=mr^2$, and

Here $p_r=m\dot r$ and $p_\theta=mr^2\dot\theta$. Direct calculation gives $\dot r p_r+\dot\theta p_\theta=m\dot r^2+mr^2\dot\theta^2=2T$. The angular momentum coefficient is a moment of inertia, not a mass. At $r=0$, the angular coordinate ceases to distinguish positions.

Example 2 · A pendulum with a moving pivot

Let the pivot move horizontally along $X(t)$. With $x=X(t)+a\sin\theta$ and $y=-a\cos\theta$,

The three terms are $T_2,T_1,T_0$. With $V=mga(1-\cos\theta)$,

This differs from laboratory $T+V$ by the linear transport term and twice the constant term. The momentum is not simply the bob's angular momentum relative to the pivot. Its extra term records the prescribed support motion in the laboratory kinetic energy.

Example 3 · A bead on a rotating radial wire

A straight horizontal wire through the origin rotates at prescribed angular speed $\Omega$. Choose signed distance $q$ along it, so $\mathbf r=(q\cos\Omega t,q\sin\Omega t)$. Then $\mathbf e_q\cdot\mathbf w=0$, $|\mathbf w|^2=\Omega^2q^2$, and

Lagrange's equation gives $m\ddot q-m\Omega^2q=0$. The energy function is $\mathcal E=m\dot q^2/2-m\Omega^2q^2/2$ and is conserved because this reduced $L$ is autonomous. Laboratory kinetic energy is not conserved: the motor sustaining the rotation supplies work through the wire reaction. Thus autonomy of a reduced Lagrangian need not imply conservation of laboratory mechanical energy.

Example 4 · Mixed kinetic terms

Consider $T=\tfrac12m(\dot q_1^2+2\alpha\dot q_1\dot q_2+\dot q_2^2)$, where both coordinates have dimensions of length and $|\alpha|<1$. The coefficient matrix is $m\begin{pmatrix}1&\alpha\\\alpha&1\end{pmatrix}$, with determinant $m^2(1-\alpha^2)>0$.

The momenta are $p_1=m(\dot q_1+\alpha\dot q_2)$ and $p_2=m(\alpha\dot q_1+\dot q_2)$. Multiplying each by its velocity and adding gives $2T$, including both cross terms. Solving the two equations gives

When $\alpha=1$, only $\dot q_1+\dot q_2$ affects $T$, so the assumed two-velocity description is singular. The restriction is mathematical information about independence, not an arbitrary algebraic inconvenience.

7. Conditions behind the familiar formulas

Before writing $\sum_i p_i\dot q_i=2T$, check that the entire kinetic energy is quadratic homogeneous in the generalized velocities. Moving constraints can add lower-degree terms. Before replacing $\mathcal E$ by $T+V$, check the same condition and the absence of velocity dependence in $V$. These checks will be central to constructing and interpreting the Hamiltonian.

Exercises

Hints and solutions are collected separately.

1. 10.1 Why can $T$ be nonzero when every $\dot q_i=0$?

2. 10.2 Derive the coefficients $M_{ij},b_i,c$ in the kinetic-energy expansion.

3. 10.3 Prove that the regular inertia matrix is positive definite.

4. 10.4 For $T=A\dot q^2+B\dot q+C$, calculate $\dot q\,\partial T/\partial\dot q$ and compare it with $2T$.

5. 10.5 Prove $\mathcal E=T_2-T_0+V$ for velocity-independent $V$.

6. 10.6 For a bead on $\mathbf r=(q,Ut)$ with constant $U$, find $T,p,H$ when $V=0$.

7. 10.7 What fails in polar coordinates at $r=0$?

8. 10.8 Calculate $p_1,p_2$ for $T=\tfrac12m(2\dot q_1^2+2\dot q_1\dot q_2+3\dot q_2^2)$.

9. 10.9 Derive the kinetic-energy identity relating its derivatives to acceleration projection.

10. 10.10 For a pivot moving as $X=Ut$, find the term linear in pendulum angular velocity.

11. 10.11 Does a symmetric inertia matrix have equal diagonal elements?

12. 10.12 For $L=\tfrac12m\dot q^2+B(q)\dot q-V(q)$, show that $B(q)$ cancels from the equation of motion.