I'll start by showing that the formulation of Lagrangian and Hamiltonian Mechanics, thus far, allows us to determine several conservation laws. Consider, for example, the homogeneity of space. Space is homogeneous if the motion (or time-evolution) of a particle (or system thereof) is independent of absolute position. That is, the potential does not vary with absolute position (it can still vary with the vector distance between two particles, as an interaction potential, for example!)

If we make a transformation

`\mathbf{r}\rightarrow\mathbf{r}+\delta\mathbf{r}`

, then the Lagrangian will also transform as `L \rightarrow L+\delta L`

. For a single particle, we can Taylor expand as follows:

L(\mathbf{r}+\delta\mathbf{r},\mathbf{v}) = L(\mathbf{r},\mathbf{v})+\frac{\partial L}{\partial x}\delta x+\frac{\partial L}{\partial y}\delta y+\frac{\partial L}{\partial z}\delta z

Which we can use to write

`\delta L = \frac{\partial L}{\partial \mathbf{r}}\cdot\delta\mathbf{r}`

`\frac{\partial L}{\partial \mathbf{r}}`

is a vector quantity; each component is the derivative of L with respect to the corresponding coordinate of r. For a single particle, then, `\frac{\partial L}{\partial \mathbf{r}}=\nabla L`

.Homogeneity of space requires that

`\delta L = 0`

. Since `\delta \mathbf{r}`

is arbitrary (and therefore not necessarily zero), we have that

\frac{\partial L}{\partial q_i} = 0 ~~~~~~ (\star)

This holds only if L does not depend on absolute position, otherwise there would be a contribution

`\delta L`

from many of the possible choices of `\delta\mathbf{r}`

. Spatial dependence of e.g. V(x) implies spatial variation of L, and momentum would not be conserved.The Euler-Lagrange Equation applies for each coordinate in the vector r. The sum of these Euler-Lagrange Equations (ELEs) means that

`(\star)`

requires that:

\begin{multiline*}

\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} = 0 \\

\Rightarrow p_i = \frac{\partial L}{\partial \dot{q}_i} ~~\mathrm{remains~constant}

\end{multiline*}

We have, therefore, demonstrated the conservation of momentum as a result of requiring translational invariance. That is, any canonical momenta whose conjugate coordinates do not appear explicitly in the Lagrangian are conserved.

Turning once again to time symmetries, let us re-derive the conservation of energy. If the Lagrangian is homogeneous in time, i.e.

`L(q,\dot{q})~~\mathrm{not}~~L(q,\dot{q},t)`

, then:

\frac{dL}{dt} = \sum_i \frac{\partial L}{\partial q_i}\dot{q}_i + \sum_i\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i

As L does not depend explicitly on time, there is no term

`\frac{\partial L}{\partial t}`

on the RHS. Sunstituting `\frac{\partial L}{\partial q_i}`

from the ELE,

\frac{dL}{dt} = \sum_i\dot{q}_i\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_i}+\sum_i\frac{\partial L}{\partial\dot{q}_i}\ddot{q}_i = \sum_i\frac{d}{dt}\left(\dot{q}_i\frac{\partial L}{\partial\dot{q}_i}\right)

\begin{multiline*}

\Rightarrow \frac{d}{dt}\sum_i\left( \dot{q}_i\frac{\partial L}{\partial\dot{q}_i}-L \right) = 0 \\

\Rightarrow H = \sum_i \dot{q}_i\frac{\partial L}{\partial \dot{q}_i} - L~~~\mathrm{remains~constant}

The conservation of energy holds for any motion in a non-time-varying external field V(x)

We turn now to the isotropy of space, and show that angular momentum is conserved due to rotational invariance of the Lagrangian. Consider rotation by an angle

`|\delta\theta|`

(with a direction given by `\delta\theta`

) about a vector. For small rotations, `\mathbf{r}\rightarrow \mathbf{r}+\delta\mathbf{r}`

, with `\delta\mathbf{r} = \delta\theta\times r`

. Each component of the velocity is also transformed by this rotation, `\delta\mathbf{v}=\delta\theta\times\mathbf{v}`

.For a single body, we now impose the requirement that the Lagrangian be unchanged under such a rotation (i.e. we require space to be isotropic).

`\delta L = \sum_i\left( \frac{\partial L}{\partial q_i}\cdot\delta r_i + \frac{\partial L}{\partial\dot{q}_i}\cdot\delta v_i \right) = 0`

We can replace

`\frac{\partial L}{\partial v_i}`

by the vector canonical momentum `p_i`

, and `\frac{\partial L}{\partial q_i}`

by `\dot{p}_i`

, leaving:`\begin{multiline*}`

\left(\dot{\mathbf{p}}\cdot\delta\mathbf{r} + \mathbf{p}\cdot\delta\mathbf{v}\right) = 0 \\

\Rightarrow \dot{\mathbf{p}}\cdot\left(\delta\theta\times \mathbf{r}\right) + \mathbf{p}\cdot\left(\delta\theta\times\mathbf{v}\right) = 0

\end{multiline*}

Since

`\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right) = \mathbf{b}\cdot\left(\mathbf{c}\times\mathbf{a}\right)`

,

\begin{multiline*}

\delta\theta\cdot\left(\left[\mathbf{r}\times\dot{\mathbf{p}}\right]+\left[\mathbf{v}\times\mathbf{p}\right]\right)=0\\

\Rightarrow\delta\theta\cdot\frac{d}{dt}\left(\mathbf{r}\times\mathbf{p}\right)=0\end{multiline*}

Since

`\delta\theta`

is arbitrary, this requires that `\mathbf{r}\times\mathbf{p}`

does not change in time, hence angular momentum is a conserved quantity.Hamilton's Equations

Using the ideas presented above, I'm going to take a moment to derive Hamilton's Equations, which will prove useful later on.

Consider changes in the Lagrangian L, according to

`dL = \sum_i\frac{\partial L}{\partial\dot{q}_i}\,d\dot{q}_i + \sum_i\frac{\partial L}{\partial q_i}\,dq_i`

This can be written:

`dL=\sum_ip_i\,d\dot{q}_i+\sum_i\dot{p}_i\,dq_i`

since

`\frac{\partial L}{\partial q_i}=\dot{p}_i`

and `\frac{\partial L}{\partial\dot{q}_i}=p_i`

.Using,

`\sum_i p_i\,d\dot{q}_i = d\left(\sum_i p_i q_i\right) - \sum_i\dot{q}_i\,dp_i`

,

d\left(\sum_i p_i\dot{q}_i - L\right) = -\sum_i\dot{p}_i\,dq_i+\sum_i\dot{q}_i\,dp_i

The argument of the differential on the left is the Hamiltonian, H,

`H(q,p,t)=\sum_i p_i\dot{q}_i - L`

, therefore:`dH = -\sum_i\dot{p}_i\,dq_i + \sum_i\dot{q}_i\,dp_i`

From here, we can obtain Hamilton's Equations:

`\begin{align*}`

\dot{q}_i &= \frac{\partial H}{\partial p_i}\\

\dot{p_i} &= \frac{\partial H}{\partial q_i}

\end{align*}

For m coordinates (and m momenta), Hamilton's Equations form a system of 2m first-order differential equations, compared to the m second-order equations in the Lagrangian treatment.

The total time derivative,

\frac{dH}{dt}=\frac{\partial H}{\partial t}+\sum_i\frac{\partial H}{\partial q_i}\dot{q}_i+\sum_i\frac{\partial H}{\partial p_i}\dot{p}_i

Substituting Hamilton's equations for

`\dot{q}_i, \dot{p}_i`

, the last two terms cancel, so`\frac{dH}{dt} = \frac{\partial H}{\partial t}`

and if H does not depend explicitly on time,

`\frac{dH}{dt}=0`

and energy is conserved!Noether's Theorem

The three conserved quantities above were shown to be related to the invariance of the Lagrangian under some symmetry transformation:

- Translational invariance (homogeneity of space) ==> Conservation of momentum
- Rotational invariance (isotropy of space) ==> Conservation of angular momentum
- Time invariance (homogeneity of time) ==> Conservation of energy

Noether's Theorem states that any differentiable symmetry of the Action (integral of the Lagrangian) of a physical system has a corresponding conservation law.

To every differentiable symmetry generated by local actions, there corresponds a conserved current.

`Symmetry' here, refers to the covariance of the form of a physical law with respect to a Lie group of transformations; the conserved quantity is known as a charge and the flow carrying it as a current (c.f. electrodynamics).

Noether's Theorem, which I will discuss in more detail at a later date, is another critical component used to build gauge theories. The key thing to remember right now is that a symmetry (invariance) of the Lagrangian corresponds to a conserved quantity; we can use this result to look for the underlying symmetry behind quantities we know to be conserved (for example, electric charge).

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