Kinetic energy

Kinetic energy is a form of energy that an object gains due to motion. For a point mass of mass mm, velocity v\mathbf{v} and linear momentum p=mv\mathbf{p}=m\mathbf{v}, it is

K=12m∣v∣2=∣p∣22mK=\frac{1}{2}m\lvert \mathbf{v} \rvert^{2} =\frac{\lvert \mathbf{p} \rvert ^{2}}{2m}

In many cases, such as an extended object like a rigid body, the kinetic energy also likely includes other contributions, such as rotational energy, vibrational energy and so on. For instance, if the point mass it also rotating around an axis, it also gains additional rotational kinetic energy, which is described in the simplest form through the inertia tensor I\mathcal{I} as

K=12ω×IωK=\frac{1}{2}\boldsymbol{\omega}\times \mathcal{I}\boldsymbol{\omega}

In a system of NN point masses, it is simply the sum of all individual kinetic energies:

K=12∑i=1Nmi∣vi∣2K=\frac{1}{2}\sum_{i=1}^{N} m_{i}\lvert \mathbf{v}_{i} \rvert ^{2}

The sum of kinetic and potential energy is called the mechanical energy of the body or system.

In analytical mechanics

Kinetic energy naturally comes up quite often in analytical mechanics. In this context, a few additional details are useful to know:

  • it depends on the state of the body, that is, it is a function of both generalized coordinates, their velocities and (possibly) time: (q1,…,qn,q˙1,…,q˙n,t)≡(q,q˙,t)(q_{1},\ldots,q_{n},\dot{q}_{1},\ldots,\dot{q}_{n},t)\equiv(q,\dot{q},t) where nn is the number of degrees of freedom the body has;
  • it is a dynamical variable of the form T:R2n+1↦R, (q,q˙,t)↦T(q,q˙,t)T:\mathbb{R}^{2n+1}\mapsto \mathbb{R},\ (q,\dot{q},t)\mapsto T(q,\dot{q},t).

For consistency with typical notation, we'll use ri(q,t)\mathbf{r}_{i}(q,t) and vi(q,q˙,t)\mathbf{v}_{i}(q,\dot{q},t) as the position and velocity of the ii-th point mass in a system of NN of them, each with nn degrees of freedom. Velocities are given as the total derivative of r\mathbf{r} in time:

vi(q,q˙,t)=∑j=1n∂ri∂qj(q,t)q˙j+∂ri∂t(q,t)\mathbf{v}_{i}(q,\dot{q},t)=\sum_{j=1}^{n} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} }(q,t) \dot{q}_{j}+\frac{ \partial \mathbf{r}_{i} }{ \partial t } (q,t)

(If the velocity is not explicitly dependent on time, the last term disappears.) Since ∣v∣2=v⋅v\lvert \mathbf{v} \rvert^{2}=\mathbf{v}\cdot \mathbf{v}, the kinetic energy then is

T=12∑i=1Nmi(∑j=1n∂ri∂qjq˙j+∂ri∂t)⋅(∑k=1n∂ri∂qkq˙k+∂ri∂t)=12∑i=1Nmi(∑j,k=1nq˙jq˙k∂ri∂qj⋅∂ri∂qk+2∑l=1nq˙l∂ri∂ql⋅∂ri∂t+∂ri∂t⋅∂ri∂t)=12∑j,k=1n(∑i=1Nmi∂ri∂qj⋅∂ri∂qk)⏟≡ajk(q,t)q˙jq˙k+∑l=1n(∑i=1Nmi∂ri∂ql⋅∂ri∂t)⏟≡bl(q,t)q˙l+12∑i=1Nmi∂ri∂t⋅∂ri∂t⏟≡c(q,t)\begin{align} T&=\frac{1}{2}\sum_{i=1}^{N} m_{i}\left( \sum_{j=1}^{n} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \dot{q}_{j} +\frac{ \partial \mathbf{r}_{i} }{ \partial t } \right)\cdot\left( \sum_{k=1}^{n} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{k} } \dot{q}_{k} +\frac{ \partial \mathbf{r}_{i} }{ \partial t } \right) \\ &=\frac{1}{2}\sum_{i=1}^{N} m_{i}\left( \sum_{j,k=1}^{n} \dot{q}_{j}\dot{q}_{k}\frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} }\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{k} } +2\sum_{l=1}^{n} \dot{q}_{l}\frac{ \partial \mathbf{r}_{i} }{ \partial q_{l} } \cdot \frac{ \partial \mathbf{r}_{i} }{ \partial t } +\frac{ \partial \mathbf{r}_{i} }{ \partial t } \cdot \frac{ \partial \mathbf{r}_{i} }{ \partial t } \right) \\ &=\frac{1}{2}\sum_{j,k=1}^{n} \underbrace{ \left( \sum_{i=1}^{N} m_{i} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{k} } \right) }_{\equiv a_{jk}(q,t) }\dot{q}_{j}\dot{q}_{k}+ \sum_{l=1}^{n} \underbrace{ \left( \sum_{i=1}^{N} m_{i}\frac{ \partial \mathbf{r}_{i} }{ \partial q_{l} }\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial t } \right) }_{ \equiv b_{l}(q,t) }\dot{q}_{l}+ \frac{1}{2}\underbrace{ \sum_{i=1}^{N} m_{i}\frac{ \partial \mathbf{r}_{i} }{ \partial t } \cdot \frac{ \partial \mathbf{r}_{i} }{ \partial t } }_{ \equiv c(q,t) } \end{align}

In the second step, we just computed the Scalar product. In the third step, we opened up the brackets and inverted the order of the sums in each of the three steps, then defined some convenience functions to show how those terms are fully determined by their indexes. Of note is that ajka_{jk} is quadratic (second order) with respect to q˙\dot{q}, blb_{l} is linear (first order) and cc is constant (zeroth order). We can identify these three separate scaling behaviors with three energy terms:

T=T2+T1+T0T=T_{2}+T_{1}+T_{0}

where the subscript denotes the order of the term. These are

T2=12∑j,k=1najk(q,t)q˙jq˙kT1=∑l=1nbl(q,t)q˙lT0=12c(q,t)\begin{align} T_{2}&=\frac{1}{2}\sum_{j,k=1}^{n}a_{jk}(q,t)\dot{q}_{j}\dot{q}_{k} \\ T_{1}&=\sum_{l=1}^{n} b_{l}(q,t)\dot{q}_{l} \\ T_{0}&=\frac{1}{2}c(q,t) \end{align}

Of the three, T2T_{2} is the most interesting, because if r\mathbf{r} is not explicitly time-dependent, r=r(q)\mathbf{r}=\mathbf{r}(q), then of course ∂ri/∂t=0\partial \mathbf{r}_{i}/\partial t=0 for all ii. But this implies that bl(q,t)=0b_{l}(q,t)=0 and c(q,t)=0c(q,t)=0 everywhere, which deletes the T1T_{1} and T0T_{0} terms. In such a case, the kinetic energy is uniquely determined by T2T_{2}:

T=T2(if r=r(q))T=T_{2}\qquad(\text{if }\mathbf{r}=\mathbf{r}(q))

This occurs, for instance, if motion is subject to fixed constraints, which are a common and useful situation in mechanics.

> (we use $\zeta$ instead of $z$ in cylindrical coordinates just for clarity). We'll want to find velocity in terms of cylindrical ones, so we use > $$\mathbf{v}(q,\dot{q})=\sum_{i=1}^{3} \frac{ \partial \mathbf{r} }{ \partial q_{i} } \dot{q}_{i}

where r=(x,y,z)\mathbf{r}=(x,y,z) and qi=rcos⁡θ,rsin⁡θ,ζq_{i}=r\cos \theta,r\sin \theta,\zeta. Written out in full:

> On the $x$ axis > $$v_{x}=\frac{ \partial x }{ \partial r } \dot{r}+\frac{ \partial x }{ \partial \theta } \dot{\theta}+\frac{ \partial x }{ \partial \zeta } \dot{\zeta}=\cos \theta\ \dot{r}-r\sin \theta \ \dot{\theta}

On the yy axis

> On the $z$ axis > $$v_{z}=\frac{ \partial z }{ \partial r } \dot{r}+\frac{ \partial z }{ \partial \theta } \dot{\theta}+\frac{ \partial z }{ \partial \zeta } \dot{\zeta}=\dot{\zeta}

So our velocity expressed in cylindrical coordinates is

> Our kinetic energy must therefore be > $$\begin{align} > T&=\frac{1}{2}m\lvert \mathbf{v} \rvert ^{2}=\frac{1}{2}m(v^{2}_{x}+v^{2}_{y}+v^{2}_{z}) \\ > &=\frac{m}{2}[\cos ^{2}\theta \dot{r}^{2}+r^{2}\sin ^{2}\theta \dot{\theta}^{2}-2r \dot{r}\dot{\theta}\cos \theta \sin \theta \\ > &\qquad +\sin ^{2}\theta \dot{r}^{2}+r^{2}\cos ^{2}\theta \dot{\theta}^{2}+2r \dot{r}\dot{\theta}\cos\theta \sin \theta+\dot{\zeta}^{2}] > \end{align}

Recalling that sin⁡2θ+cos⁡2θ=1\sin ^{2}\theta+\cos ^{2}\theta=1, the first two columns simplify greatly whereas the third column cancels out. We are left with

#### The kinetic matrix Let's now focus our attention to just the term $T_{2}$ and more specifically $a_{jk}(q,t)$. This quantity is known as the **kinetic matrix**, because it is, in fact, a matrix. If we write $T_{2}$ as

T_{2}=\sum_{j=1}^{n} \dot{q}{j}\sum{k=1}^{n}a_{jk}\dot{q}{k} =\dot{\mathbf{q}}{j}\cdot \mathrm{a},\dot{\mathbf{q}}_{k}

we can see that $\mathrm{a}$ has the shape of a matrix and $a_{jk}$ are its components. It is defined as

a_{jk}=\sum_{i=1}^{N} m_{i} \frac{ \partial \mathbf{r}{i} }{ \partial q{j} } \cdot \frac{ \partial \mathbf{r}{i} }{ \partial q{k} }

It has a few interesting properties: - it is [[Symmetric matrix|symmetric]], $a_{jk}=a_{kj}$. This follows from being defined by a real scalar product. - it is [[Matrix sign definitions|positive definite]], that is, $\mathbf{v}\cdot \mathrm{a}\,\mathbf{v}>0$ for all $\mathbf{v}\neq 0\in \mathbb{R}^{n}$. - it is [[Invertible matrix|invertible]], since it is positive definite.