Action, the Hamiltonian, and the Lagrangian

For most people mechanics is synonymous with Newton's Laws. Almost everyone learns $F=ma$ in school. However, there are actually other formulations of mechanics. These alternate approaches are heavily utilized in modern physics. And central to these formulation's is the concept of action. Action $S$ can seem like a nebulous quantity. However, each part of the differential of action is actually an just a familiar physical quantity.

Let's look at the total differential of Action
$$dS=\underset{i}{\sum}\frac{\partial S}{\partial q_{i}}dq_{i}+\frac{\partial S}{\partial t}dt$$
Here $q_{i}$ are the spatial coordinates usally denoted $i=\{1,2,3\}$ for $x$,$y$,$z$. The magnitude of the last component $\frac{\partial S}{\partial t}$ is equal to the energy, or in this case the more formal Hamiltonian. The Hamiltonian is
$$H=T+V $$
where $T$ is the kinetic energy and $V$ is the potential. So $H$ is just the total energy, at least for all closed systems.  This relation is called the Hamiltonian-Jacobi Equation
$$\frac{\partial S}{\partial t}=-H$$
Now the other components of $dS$ represent the momentum of the system, that is
$$\frac{\partial S}{\partial q_{i}}=p_{i}$$
Where $p_{i}$ is the momentum in the $q_{i}$ direction. Now let's define the Lagrangian $L$
$$L=T-V$$
And note $\underset{i}{\sum}p_{i}\dot{q_{i}}=2T$. $\dot{q_{i}}$ denotes the time derivative of the position $q_{i}$, i.e. the velocity in the $q_{i}$ direction. Ok, let's divide the action differential $dS$ by $dt$
$$\frac{dS}{dt}=\underset{i}{\sum}\frac{\partial S}{\partial q_{i}}\frac{dq_{i}}{dt}+\frac{\partial S}{\partial t}=\underset{i}{\sum}p_{i}\dot{q_{i}}-H=L$$
So the time derivative of Action is actually the Lagrangian! The last portion of the calculation can be used to transform between the Lagrangian and the Hamiltonian, it's called the Legendre Transform
$$\underset{i}{\sum}p_{i}\dot{q_{i}}-H=L$$
And we can integrate the equation, showing
$$S=\int Ldt+C$$
Action is often defined as strictly $S=\int Ldt$, but we won't be so picky. If we integrate the Hamiltonian Jacobi equation we can also see
$$S=-\int Hdt+W$$
Where $W$ is not explicitly dependent on time. We can even investigate the nature of $W$
 $$W=S+\int Hdt=\int Ldt+C+\int Hdt=C+\int(L+H)dt=C+\underset{i}{\sum}\int p_{i}dq_{i}$$
 When $C=0$, this is sometimes called the abbreviated action.

The Euler-Lagrange Equation

Ok, so now we know all that, but what does that have to do with calculations? Well to do calculations in Lagrangian mechanics we'll need the Euler-Lagrange equation, it allows us to get from the Lagrangian to equations of motion, the sort of $F=ma$ equations familiar in Newtonian mechanics. We can derive the Euler-Lagrange equation using the principle of least action, technically stationary action. This principle says that the path taken by a body is the one that minimizes the action, and more technically the path along which the action is stationary.  To do this we are going to find the path $q(t)$ that minimizes the action between two times, $t_1$ and $t_2$.

To derive the Euler-Lagrange Equation we'll regard the integral $S=\int Ldt+C$ as a function of the path of an object $q(t)$. And we'll note that our result will not depend on $C$ so we set $C=0$. We then say
$$S[q(t)]=\int_{t_{1}}^{t_{2}}L(q(t),\dot{q}(t),t)dt$$
Where the brackets denote a functional: a function of a function. To minimize the functional we'll set the functional derivative of the action equal to zero, that is $\frac{\delta S}{\delta q(t)}=0$. To do this we'll add a function $n(t)$ to $q(t)$ which is the same as $q(t)$ at the end points, that is $q(t_1)=n(t_1)$ and $q(t_2)=n(t_2)$. Then we'll multiply a real number $\epsilon$ by $n(t)$ and take the limit as $\epsilon$ goes to zero. This is called the functional derivative in the direction $\eta(t)$  with respect to $q(t)$.
$$\frac{\delta S}{\delta q(t)}=\underset{\epsilon\rightarrow0}{lim}\frac{S[q(t)+\epsilon\eta(t)]-S[q(t)]}{\epsilon}$$
To carry out this limit, we'll Taylor expand $S[q(t)+\epsilon\eta(t)]$,
 $$S[q(t)+\epsilon\eta(t)]=S[q(t)]+\frac{d}{d\epsilon}S[q(t)+\epsilon\eta(t)]\epsilon+\frac{d^{2}}{d\epsilon^{2}}S[q(t)+\epsilon\eta(t)]\epsilon^{2}+\dots$$
The first term cancels with $-S[q(t)]$, and all the other terms, where epsilon doesn't cancel with the denominator go to zero so we just need the first order term, the derivative relative to epsilon
\[\begin{align*}\underset{\epsilon\rightarrow0}{lim}\frac{S[q(t)+\epsilon\eta(t)]-S[q(t)]}{\epsilon} &=\underset{\epsilon\rightarrow0}{lim}\frac{\frac{d}{d\epsilon}S[q(t)+\epsilon\eta(t)]\epsilon+\frac{d^{2}}{d\epsilon^{2}}S[q(t)+\epsilon\eta(t)]\epsilon^{2}+\dots}{\epsilon} \\ &= [\frac{d}{d\epsilon}(S[q(t)+\epsilon\eta(t)])]_{\epsilon=0}+\underset{\epsilon\rightarrow0}{lim}\frac{d^{2}}{d\epsilon^{2}}S[q(t)+\epsilon\eta(t)]\epsilon+\dots \\ &=[\frac{d}{d\epsilon}(S[q(t)+\epsilon\eta(t)])]_{\epsilon=0}+0 \end{align*}\]
Hence,
\[\begin{align*}\frac{\delta S}{\delta q(t)} &= \underset{\epsilon\rightarrow0}{lim}\frac{S[q(t)+\epsilon\eta(t)]-S[q(t)]}{\epsilon} \\ &= [\frac{d}{d\epsilon}(S[q(t)+\epsilon\eta(t)])]_{\epsilon=0} \\ &=\int_{t_{1}}^{t_{2}}\frac{dq}{d\epsilon}\frac{\partial L}{\partial q}+\frac{d\dot{q}}{d\epsilon}\frac{\partial L}{\partial \dot{q}}+\frac{dt}{d\epsilon}\frac{d\partial}{\partial t}dt \\ &=\int_{t_{1}}^{t_{2}}\eta\frac{\partial L}{\partial q}+\dot{\eta}\frac{\partial L}{\partial \dot{q}}dt\end{align*}\]


Using integration by parts $$[\eta\frac{\partial L}{\partial \dot{q}}]_{t_1}^{t_2}=\int_{t_{1}}^{t_{2}}\dot{\eta}\frac{\partial L}{\partial \dot{q}}dt+\int_{t_{1}}^{t_{2}}\eta\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}dt$$
 So,

$$\int_{t_{1}}^{t_{2}}\eta\frac{\partial L}{\partial q}+\dot{\eta}\frac{\partial L}{\partial \dot{q}}dt=\eta\int_{t_{1}}^{t_{2}}\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}dt+[\eta\frac{\partial L}{\partial \dot{q}}]_{t_1}^{t_2}$$


$[\eta\frac{\partial L}{\partial \dot{q}}]_{t_1}^{t_2}$ is zero since $\eta$  is zero at the endpoints. So $$\eta\int_{t_{1}}^{t_{2}}\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}dt=0$$
Thus,
$$\frac{\partial L}{\partial q}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$$
Which is the Euler Lagrange equation.

Generalized Coordinates

One of the most useful aspects of encoding mechanics in this way is it allows us to use generalized coordinates. Generalized coordinates are non-inertial parameters of a system.

Let's take the triple pendulum below.

We'll assign three coordinate $\alpha$,$\beta$,$\gamma$ to the three pivots. So, if we're going to write the Lagrangian we'll need the potential energy. In this case that's the integral of torque. And torque itself is the force times lever arm. Force on these rods, with linearly distributed mass $\rho$, is just $r\rho g sin \theta$ integrated over the length of the rod.
\[\begin{align*}V &= \int\tau d\theta \\ &= \int\int_0^l r\times Fdrd\theta \\ &= \int\int_0^l rF\sin(\theta) drd\theta \\ &= \int\int_0^l r(\rho g)\sin(\theta) drd\theta \\ &= \int\frac{1}{2}l(l\rho)g\sin(\theta) d\theta \\ &= \frac{1}{2}lmg(-\cos(\theta)+C)\end{align*}\]
We can disregard the c in the potential. So that takes care of the first pendulum, but what about the second? Well the torque is still just due to gravity so we can just add the angles alpha and beta to get the potential there that is, $\frac{1}{2}lmg\cos(\alpha+\beta)$, Finally we do the same adding a gamma for the final arm and we can write the full potential for the first joint
$$V_{\alpha}=-\frac{1}{2}lmg\cos(\alpha)-\frac{1}{2}lmg\cos(\alpha+\beta)-\frac{1}{2}lmg\cos(\alpha+\beta+\gamma)$$
Then $V_{\beta}=-\frac{1}{2}lmg\cos(\alpha+\beta)-\frac{1}{2}lmg\cos(\alpha+\beta+\gamma)$ and $V_{\gamma}=-\frac{1}{2}lmg\cos(\alpha+\beta+\gamma)$. Hence,
$$V_{system}=-\frac{1}{2}lmg\cos(\alpha)-\frac{2}{2}lmg\cos(\alpha+\beta)-\frac{3}{2}lmg\cos(\alpha+\beta+\gamma)$$
So that takes care of the potential, what about the kinetic energy? Well we know that's just $\frac{1}{2} I \dot{\theta}$. So, we just need to calculate the moment of inertia $I$ which is
$$I=\int_{0}^{l}r^{2}\rho dr=\rho l^{3}/3=ml^{2}/2$$
Finally we'll write the Lagrangian
$$L=T-V=\frac{1}{2}I\dot{\alpha}^{2}+\frac{1}{2}I\dot{\beta}^{2}+\frac{1}{2}I\dot{\gamma}^{2}+\frac{1}{2}lmg\cos(\alpha)+\frac{2}{2}lmg\cos(\alpha+\beta)+\frac{3}{2}lmg\cos(\alpha+\beta+\gamma)$$
 Now we can use the Euler-Lagrange equations to calculate the equations of motion
$$\frac{\partial L}{\partial \alpha}=-\frac{1}{2}lmg\sin(\alpha)-\frac{2}{2}lmg\sin(\alpha+\beta)-\frac{3}{2}lmg\sin(\alpha+\beta+\gamma)$$
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{\alpha}}=\frac{d}{dt}I\dot{\alpha}=I\ddot{\alpha}$$
Hence our first equation of motion is,
$$I\ddot{\alpha}=-\frac{1}{2}lmg\sin(\alpha)-\frac{2}{2}lmg\sin(\alpha+\beta)-\frac{3}{2}lmg\sin(\alpha+\beta+\gamma)$$
This agrees with the usual $\tau=I\ddot{\theta}$ from Newtonian Mechanics. We'll get two more equations of motion in the same way
$$I\ddot{\beta}=-\frac{2}{2}lmg\sin(\alpha+\beta)-\frac{3}{2}lmg\sin(\alpha+\beta+\gamma)$$
$$I\ddot{\gamma}=-\frac{3}{2}lmg\sin(\alpha+\beta+\gamma)$$

Hamilton's Equations

Now we'll take a shot at Hamiltonian Mechanics. First we need Hamilton's Equations, they'll get us equations of motion in Hamiltonian Mechanics. Let's take the partial derivative with respect to momentum of the Hamiltonian, that is $\frac{\partial H}{\partial p}$
 $$\frac{\partial H}{\partial p}=\frac{\partial T}{\partial p}+\frac{\partial V}{\partial p}=\frac{\partial}{\partial p}(\frac{p^{2}}{2m})+0=\frac{2(m\dot{q})}{2m}=\dot{q}$$
This is the first of Hamilton's equations. Next let's calculate the partial spacial derivatives $\frac{\partial H}{\partial q}$,
$$\frac{\partial H}{\partial q}=\frac{\partial T}{\partial q}+\frac{\partial V}{\partial q}=0+\nabla V=-F=-\frac{dp}{dt}$$
This is the second of Hamilton's equations, hence Hamilton's Equations are
$$\frac{\partial H}{\partial q}=-\dot{p}$$
 $$\frac{\partial H}{\partial p}=\dot{q}$$
Let's make sure this checks out. For a simple harmonic oscillator. The Hamiltonian would be
$$H=\frac{p^{2}}{2m}+\frac{kx^{2}}{2}$$
And the equation of motion would be
$$\frac{\partial H}{\partial x}=kx=-\dot{p}=-\frac{d}{dt}(m\dot{x})=-m\ddot{x}$$
 Hence the classic spring force is recovered
$$m\ddot{x}=-kx$$

Final Thoughts

There's a lot more in the execution of these theories, introducing Lagrange multipliers to handle constraints in Lagrangian mechanics for instance, but this is the basic framework required to understand these alternative approaches to mechanics.

Interestingly, in Hamiltonian mechanics we didn't have to apply the principle of least action. Its requirement is unique to the Lagrangian approach. However, one can derive from the principle of least action the Newtonian and Hamiltonian approaches. This is why it's often considered the more fundamental physical law, used even in modern quantum field theory.