# General Mechanics/Alternate Formulations of Newton's Laws
## Reformulating Newton

In the last chapter we saw how to reformulate Newton's laws as a set of second order ordinary differential equations using arbitrary coordinates:
    ${\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {q}}_{i}}}-{\frac {\partial L}{\partial q_{i}}}=0$
It is often far easier to solve a problem by starting with the Lagrangian than by explicitly using Newton's laws. The Lagrangian is also more useful for theoretical analysis.
However, the Lagrangian approach is not the only useful way to reformulate Newton. There is a second, related, approach which is also highly useful for analysis; namely, writing the equations of motion as a set of first order ordinary differential equations in a particularly natural way.
To do this we will use a standard technique, familiar from your calculus studies.

## Derivation

Consider the function
    $H(q,p,t)={\dot {q}}_{i}p_{i}-L(q,{\dot {q}},t)$
where the $q_{i}$ are the generalised coordinates, the $p_{i}$ are a set of new variables whose meaning we will soon see, $H$ is to be a function of the $p$'s and $q$'s alone, and the summation convention is being used.
Then we have
    $dH={\dot {q}}_{i}dp_{i}+p_{i}d{\dot {q}}_{i}-{\frac {\partial L}{\partial {\dot {q}}_{i}}}d{\dot {q}}_{i}-{\frac {\partial L}{\partial q_{i}}}dq_{i}-{\frac {\partial L}{\partial t}}dt\quad (1)$
but, since H is a function of the p‘s and q‘s alone we can also write
    $dH={\frac {\partial H}{\partial q_{i}}}dq_{i}+{\frac {\partial H}{\partial p_{i}}}dp_{i}+{\frac {\partial H}{\partial t}}dt\quad (2)$
For these two equations to both be true the coefficients of the differentials must be equal.
Equation (2) does not contain a term in $d{\dot {q}}_{i}$, so the coefficient of that term in equation (1) must be zero. I.e,
    $p_{i}={\frac {\partial L}{\partial {\dot {q}}_{i}}}$
which gives us a definition of the p$_i$. Using this in Lagrange's equations gives
    ${\dot {p}}_{i}={\frac {\partial L}{\partial q_{i}}}$
and equation (1) simplifies to
    $dH={\dot {q}}_{i}dp_{i}-{\dot {p}}_{i}dq_{i}-{\frac {\partial L}{\partial t}}dt$
A comparison of coefficients with equation (2) now gives the desired set of first order equations for the motion,
    ${\dot {q}}_{i}={\frac {\partial H}{\partial p_{i}}}\quad {\dot {p}}_{i}=-{\frac {\partial H}{\partial q_{i}}}$
These are called Hamilton's equations and H is called the Hamiltonian.

## Physical meaning of H and p

To see what H and the p$_i$'s actually are, let's consider a few typical cases.
First, let's look at a free particle, one subject to no forces. This will let us see what the p$_i$'s mean, physically.
If we use Cartesian coordinates for the free particle, we have
    $L=T={\frac {1}{2}}m\left({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}\right)$
The p$_i$'s are the differential of this with respect to the velocities.
    $p_{x}={\frac {\partial L}{\partial {\dot {x}}}}=m{\dot {x}}\quad p_{y}={\frac {\partial L}{\partial {\dot {y}}}}=m{\dot {y}}\quad p_{z}={\frac {\partial L}{\partial {\dot {z}}}}=m{\dot {z}}$
These are the components of the momentum vector.
If we use cylindrical coordinates, we have
    $L=T={\frac {1}{2}}m\left({\dot {r}}^{2}+r^{2}{\dot {\theta }}^{2}+{\dot {z}}^{2}\right)$
and
    $p_{r}={\frac {\partial L}{\partial {\dot {r}}}}=m{\dot {r}}\quad p_{\theta }={\frac {\partial L}{\partial {\dot {\theta }}}}=mr^{2}{\dot {\theta }}\quad p_{z}={\frac {\partial L}{\partial {\dot {z}}}}=m{\dot {z}}$
This time, p$_z$ is the component of momentum is the z direction, p$_r$ is the radial momentum, and p$_θ$ is the angular momentum, which we've previously seen is the equivalent of momentum for rotation.
Since, in these familiar cases, the p$_i$'s are momenta, we generalise and call the p$_i$'s conjugate momenta.
Note that from Hamilton's equations, we see that if H is independent of some coordinate, q, then
    ${\dot {p}}_{q}=-{\frac {\partial H}{\partial q}}=0$
so the momentum conjugate to q is conserved. This connection between conservation laws and coordinate independence lies at the heart of much physics.
Secondly, to see the physical meaning of H, we'll consider a particle moving in a potential V, described using Cartesian coordinates.
This time the Lagrangian is
    $L=T-V={\frac {1}{2}}m\left({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}\right)-V(x,y,z)$
We find that the p$_i$'s are
    $p_{x}={\frac {\partial L}{\partial {\dot {x}}}}=m{\dot {x}}\quad p_{y}={\frac {\partial L}{\partial {\dot {y}}}}=m{\dot {y}}\quad p_{z}={\frac {\partial L}{\partial {\dot {z}}}}=m{\dot {z}}$
so, again, they are the momenta
Now lets calculate H.
    ${\begin{matrix}H&=&{\dot {x}}p_{x}+{\dot {y}}p_{y}+{\dot {y}}p_{y}&-&L&&\\&=&m\left({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}\right)&-&{\frac {1}{2}}m\left({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}\right)&+&V\\&=&{\frac {1}{2}}m\left({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}\right)&&&+&V\\&=&T&&&+&V\end{matrix}}$
so H is the total energy, written as a function of the position and momentum.
This is not always true, but it will be true for all common systems which conserve energy.
In general, if we know how the energy of a systems depends on speeds and positions, we know everything about the system.
Physicists will often start work on a problem by writing down an expression for H, or L, and never bother calculating the actual forces.

## Deeper meaning

If we compare Hamilton's equations,
    ${\dot {q}}_{i}={\frac {\partial H}{\partial p_{i}}}\quad {\dot {p}}_{i}=-{\frac {\partial H}{\partial q_{i}}}$
with the equations of geometrical optics.
    ${\dot {x}}_{i}={\frac {\partial \omega }{\partial k_{i}}}\quad {\dot {k}}_{i}=-{\frac {\partial \omega }{\partial x_{i}}}$
we see that both sets of equations have the same form.
If we made a substitution of the form,
    $H=\alpha \omega \quad p=\alpha k$
for any α, the two sets of equations would become identical.
Now, geometrical optics is an approximation to the full solution of any wave equation, valid when the wavelength is small.
Since classical mechanics is just like geometrical optics, classical mechanics could also be small wavelength approximation to a more accurate wave theory.
This is not a proof, could be is not the same as is, but it does show that classical mechanics is compatible with wave theories, contrary to intuition.
We will see later that this is not a coincidence, classical mechanics really is a small wavelength approximation to a more accurate wave theory, namely quantum mechanics, and that, in that theory, the energy of matter waves is proportional to frequency and the momentum to wave number, just as the substitution above requires.
Hamilton's equations can also be used to construct a mathematical gadget, called a Poisson bracket, which lets us write the equations of classical mechanics in a way which makes the connections with quantum mechanics even clearer, but an explanation of how would be beyond the scope of this book.
Since classical mechanics is just like geometrical optics, there should be an analog to Fermat's principle, that light takes the quickest path, and there is.
Where light takes the quickest path from a to b, matter takes the path of least action, where the action is defined as
    $\int _{a}^{b}L(x,{\dot {x}},t)\,dt$
It can be proved that this integral takes extremal values if and only if the system obeys Lagrange's equations. Saying matter takes the path of least action is equivalent to saying it obeys Lagrange's equations, or Newton's laws.
Furthermore, it turns out that in general relativity, the action is proportional to the time taken, so matter is also really taking the quickest path between two points.