Constant of motion - Aetherwisp

A constant of motion is a quantity that does not change over time and over a trajectory of motion. Formally, a constant of motion is defined as a function $I:\mathbb{R}^{N}\to \mathbb{R}$ , $\mathbf{x}\to I(\mathbf{x})$ , for which, given an autonomous system of ODEs $\dot{\mathbf{x}}=f(\mathbf{x})$ and a starting condition $\mathbf{x}_{0}$ , we have $I(\mathbf{x}(t;\mathbf{x}_{0}))=I(\mathbf{x}_{0})$ for all $t$ and all solutions $\mathbf{x}(t;\mathbf{x}_{0})$ of the system. Its time derivative is zero everywhere: $\frac{d}{dt}I(\mathbf{x}(t;\mathbf{x}_{0}))=0$ .

The locus of points in $\mathbb{R}^{N}$ that satisfies the equation $I(\mathbf{x})=\rho$ , where $\rho$ is a constant, is a hypersurface in $\mathbb{R}^{N}$ , specifically a level set. For example, in $\mathbb{R}^{3}$ , the quantity $I(\mathbf{x})=x^{2}+y^{2}+z^{2}=\rho=R^{2}$ is a sphere. The constant of motion $I$ defines a family of disjoint (i.e. non-intersecting) hypersurfaces dependent on the parameter $\rho$ . The union of all of these hypersurfaces is $\mathbb{R}^{N}$ . Formally, such a family is known in differential geometry as a foliation of $\mathbb{R}^{N}$ .

The benefit of finding constants of motion is that they add constraints to the parameters of a system. Specifically, each constant reduces the degrees of freedom, and therefore the dimensionality of the system, by one. The more constants of motion are known beforehand, the fewer parameters need to be solved to find a trajectory. For instance, a three-dimensional system with two known constants of motion simplifies down to a one dimensional system.

Examples#

> $$v(t)=-\omega x_{0}\sin \omega t+v_{0}\cos \omega t

The total energy of the particle is $I(x,v)=\frac{1}{2}v^{2}+ \frac{1}{2}\omega ^{2}x^{2}$ . To prove that this is, in fact, a constant of motion, we just substitute $x$ and $v$ with the solutions above:
$I(x(t),v(t))&= \frac{1}{2}(-x_{0}\sin \omega t+v_{0}\cos \omega t)^{2} + \frac{1}{2}\omega ^{2}\left( x_{0}\cos \omega t+ \frac{v_{0}}{\omega}\sin \omega t \right)^{2} \\ &=\frac{\omega ^{2}x_{0}^{2}}{2}\sin ^{2}\omega t-\omega x_{0}v_{0}\sin \omega t\cos \omega t+ \frac{v_{0}^{2}}{2}\cos ^{2}\omega t+ \\$

&+ \frac{\omega ^{2}x_{0}^{2}}{2}\cos ^{2}\omega t+\omega x_{0}v_{0}\cos \omega t\sin \omega t + \frac{v_{0}^{2}}{2}\sin ^{2}\omega t \

&=\frac{\omega ^{2}x_{0}^{2}}{2}\underbrace{ (\sin ^{2}\omega t+\cos ^{2}\omega t) }{ 1 }+ \omega x{0}v_{0}\underbrace{ (-\sin \omega t\cos \omega t+\cos \omega t\sin \omega t) }{ 0 }+ \ & + \frac{v{0}^{2}}{2}\underbrace{ (\cos ^{2}\omega t+\sin ^{2}\omega t) }{ 1 } \ &=\frac{\omega ^{2}x{0}^{2}}{2}+ \frac{v_{0}^{2}}{2}=I(x_{0},v_{0}) \end{align}

> It is dependent only on the starting conditions. > > For an even simpler form, let's now set $\omega=1$. The energy therefore becomes $I(x,v)=\frac{1}{2}(x^{2}+v^{2})$. If we set an arbitrary energy quantity $E$, then the system will maintain this total energy throughout its entire motion. The formula for $I$ is well-known: it is a [[circle]]. In fact, each value $E$ describes a circle in $(x,v)$-space (i.e. [[phase space]]). Lower energies make for smaller circles and viceversa. > > ![[Plot 1D Harmonic oscillator constant energy surfaces.svg|70%]] > > As you can see, we reduced a two-dimensional problem (solving for all phase space) into a one-dimensional one (solving only for the constant energy [[Curve|curve]]). > [!example]- 1D conservative system > In a [[conservative system]] (here given with one degree of freedom), the total force $F(x)$ can be given as the derivative of a [[potential energy]] $V(x)$: > $$\frac{F(x)}{m}=f(x)=- \frac{V'(x)}{m}

where $m$ is the mass. $f(x)$ is an acceleration, $f(x)=a=\ddot{x}$ . The total energy therefore is

> where the first term is [[kinetic energy]]. In such a system, this is always a constant of motion. In fact, using the [[Chain rule]] we get > $$\frac{d}{dt}E(x(t),v(t))=\frac{ \partial E }{ \partial x } (x(t),v(t))\dot{x}(t)+\frac{ \partial E }{ \partial v } (x(t),v(t))\dot{v}(t)=\ldots

This system is described by the two ODEs $\dot{x}=v$ and $\dot{v}=f(x)$ . We can substitute these two above

> If $v$ does not depend on $x$, we have > $$\frac{ \partial E(x,v) }{ \partial x } =V'(x),\qquad \frac{ \partial E(x,v) }{ \partial v } =mv

and so

> which proves our point. ### In Lagrangian mechanics Within the context of a Lagrangian system of $n$ [[degrees of freedom]], a constant of motion is a function $I:\mathbb{R}^{2n+1}\to \mathbb{R},(q,\dot{q},t)\to I(q,\dot{q},t)$ for which $\frac{d}{dt}I(q(t),\dot{q}(t),t)=0$ that solves the [[Lagrange equation]]. If we know a constant of motion $I(q,\dot{q},t)$ and, then we can write the following equation:

I(q(t),\dot{q}(t),t)=I_{0}

where $I_{0}$ is some constant. This is a first order [[Ordinary differential equation|ODE]] in $q(t)$ (technically a family of ODEs). If we solve this equation, we can find the motion $q(t)$ just from the constant of motion. In systems with multiple coordinates, the constant of motion only finds the motion of one coordinate. For instance, in a system with coordinates $(r,\theta)$ and zero [[angular momentum]], the angle $\theta$ likely does not appear in the [[Lagrangian]]. In such a case, it is a [[Lagrangian|cyclical coordinate]] and its [[Conjugate momenta|conjugate momentum]] $P_{\theta}=mr^{2}\dot{\theta}$ is a constant of motion, leading to a first-order ODE for $\dot{\theta}$ which can be solved for $\theta(t)$. $r(t)$ would still need be solved manually as usual. > [!example]- Energy > Let's consider the following [[dynamical variable]] for a [[Lagrangian]] $L$: > $$E(q,\dot{q},t)=\sum_{i=1}^{n} \dot{q}_{i}\frac{ \partial L }{ \partial \dot{q}_{i} }(q,\dot{q},t)-L(q,\dot{q},t)

The derivative of this is
$\frac{d}{dt} E(q,\dot{q},t)&=\sum_{i=1}^{n} \left( \ddot{q}_{i}\frac{ \partial L }{ \partial \dot{q}_{i} }+\dot{q}_{i} \frac{d}{dt} \frac{ \partial L }{ \partial \dot{q}_{i} } \right)-\sum_{i=1}^{n} \left( \frac{ \partial L }{ \partial q_{i} } \dot{q}_{i} + \frac{ \partial L }{ \partial \dot{q}_{i} }\ddot{q}_{i} \right)-\frac{ \partial L }{ \partial t } \\ &=\sum_{i=1}^{n} \dot{q}_{i}\left( \frac{d}{dt} \frac{ \partial L }{ \partial \dot{q}_{i} } -\frac{ \partial L }{ \partial q_{i} } \right) - \frac{ \partial L }{ \partial t } \\ (\text{if }q(t)\text{ solves Lag. eq.})&=-\frac{ \partial L }{ \partial t } \end{align}$

> Clearly then, if $\frac{ \partial L }{ \partial t }=0$, then $E$ is a constant of motion. This occurs when $L$ does not explicitly depend on time, that is, $L\equiv L(q,\dot{q})$, and in these cases, the quantity $E$ can always be used as a constant of motion. This property is known as the **time-shift invariance** of $L$. In plain terms, $L$ does not change in time: the dynamics of the system *now* are the same as they were in the past and will always be the same in the future. #### Examples > [!example]- Mechanical system with purely positional forces > For a mechanical system with purely positional forces, if the Lagrangian is explicitly time independent, $L\equiv L(q,\dot{q})$, then we can write > $$L(q,\dot{q})=\frac{1}{2}\sum_{n,k} a_{nk}(q)\dot{q}_{n}\dot{q}_{k}-V(q)=T_{2}-V

$a$ is a Homogeneous function of degree $2$ in $\dot{q}$ and $V$ is a homogeneous function of degree $0$ in $\dot{q}$ . The quantity $E$ is

> We can see that $E$ is the *total* mechanical energy of the system, given by the sum of [[kinetic energy]] and [[potential energy]]. > [!example]- 2D harmonic oscillator > We now imagine a two-dimensional [[harmonic oscillator]] of Lagrangian >$$L=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})- \frac{m\omega ^{2}}{2}(x^{2}+y^{2})

where $\omega ^{2}=k/m$ . Think of it a spring being pushed and pulled away and towards the origin, passing through it every half cycle. It is not rotating, it's just a harmonic oscillator moving diagonally on the plane. This makes it just a harmonic oscillator on a specific axis in 2D. We can find the motion pretty easily from each term of $L$ :

> $$\frac{d}{dt} \frac{ \partial L }{ \partial y } -\frac{ \partial L }{ \partial y } =0\quad\to \quad \ddot{y}=-\omega ^{2}y

These are simple second order ODEs that solve to

> For an illustrating example, we should also solve this in [[polar coordinates]]. If we use the [[Coordinate transformation]] $r=\cos \varphi,\ y=r\sin \varphi$, the Lagrangian becomes > $$L=\frac{m}{2}(\dot{r}^{2}+r^{2}\dot{\varphi}^{2})- \frac{m\omega ^{2}}{2}r^{2}

This Lagrangian does not explicitly depend on $\varphi$ ; this is your hint that there's a constant of motion somewhere in here. The terms of $E$ are

> $$0=\frac{d}{dt} \frac{ \partial L }{ \partial \dot{\varphi} }- \cancel{ \frac{ \partial L }{ \partial \varphi } } \quad\to \quad \frac{ \partial L }{ \partial \dot{\varphi} } =mr^{2}\dot{\varphi}

The latter equation is a constant of motion. If we set $mr^{2}\dot{\varphi}=\ell$ , we can invert to get $\dot{\varphi}=\ell/mr^{2}$ . If we plug this into $\ddot{r}$ we get

> This is an ODE in $r(t)$ of the form $\ddot{r}=f(r)$ and can therefore be solved generally. We can also plug $\dot{\varphi}$ in the polar Lagrangian to get > $$L_\text{eff}(r,\dot{r})=\frac{m\dot{r}^{2}}{2}- \frac{m\omega ^{2}r^{2}}{2}- \frac{\ell^{2}}{2mr^{2}}

This isn't the original Lagrangian and does not even depend on the same variables, but it nonetheless contains the same information to the original. This effective Lagrangian uses the fact that the oscillator is not rotating, so the angular dependency in polar coordinates collapses to a constant of motion and we can get all the information about the system by just solving a single ODE, $\ddot{r}$ , instead of two, $\ddot{x}$ and $\ddot{y}$ .