Categories: Classical mechanics, Physics.

# Lagrangian mechanics

**Lagrangian mechanics** is a formulation of classical mechanics,
which is equivalent to Newton’s laws,
but offers some advantages.
Its mathematical backbone is the
calculus of variations,
and hence it is built on the **principle of least action**,
which states that the path taken by a system
will be a minimum of the **action** (i.e. energy cost) of that path.

For a moving object with position $x(t)$ and velocity $\dot{x}(t)$, we define the Lagrangian $L$ as the difference between its kinetic and potential energies:

$\begin{aligned} \boxed{ L(x, \dot{x}, t) \equiv T - V = \frac{1}{2} m \dot{x}^2 - V(x) } \end{aligned}$From variational calculus we then get the Euler-Lagrange equation, which in this case turns out to just be Newton’s second law:

$\begin{aligned} \dv{}{t}\Big( \pdv{L}{\dot{x}} \Big) = \pdv{L}{x} \qquad \implies \qquad m \ddot{x} = - \pdv{V}{x} = F \end{aligned}$But compared to Newtonian mechanics, Lagrangian mechanics scales better for large systems. For example, to describe the dynamics of $N$ objects $x_1(t), ..., x_N(t)$, we only need a single $L$ from which the equations of motion can easily be derived. Getting these equations directly from Newton’s laws could get messy.

At no point have we assumed Cartesian coordinates: the Euler-Lagrange equations keep their form for any independent coordinates $q_1(t), ..., q_N(t)$:

$\begin{aligned} \dv{}{t}\Big( \pdv{L}{\dot{q}_n} \Big) = \pdv{L}{q_n} \end{aligned}$We define the **canonical momentum conjugate** $p_n(t)$
and the **generalized force conjugate** $F_n(t)$ as follows,
such that we can always get Newton’s second law:

But this is actually a bit misleading, since $p_n$ need not be a momentum, nor $F_n$ a force, although often they are. For example, $p_n$ could be angular momentum, and $F_n$ torque.

Another advantage of Lagrangian mechanics is that
the conserved quantities can be extracted from $L$ using Noether’s theorem.
In the simplest case, if $L$ does not depend on $q_n$
(then known as a **cyclic coordinate**),
then we know that the “momentum” $p_n$ is a conserved quantity:

Now, as the number of particles $N$ increases to infinity, variational calculus will give infinitely many coupled equations, which is obviously impractical.

Such a system can be regarded as continuous, so the $N$ functions $q_n$ can be replaced by a single density function $u(x,t)$. This approach can also be used for continuous fields, in which case the complex conjugate $u^*$ is often included. The Lagrangian $L$ then becomes:

$\begin{aligned} L(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t) = \int_{-\infty}^\infty \mathcal{L}(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t) \dd{x} \end{aligned}$Where $\mathcal{L}$ is known as the **Lagrangian density**.
By inserting this into the functional $J$
used for the derivation of the Euler-Lagrange equations, we get:

This is simply 2D variational problem, so the Euler-Lagrange equations will be two PDEs:

$\begin{aligned} 0 &= \pdv{\mathcal{L}}{u} - \pdv{}{x}\Big( \pdv{\mathcal{L}}{u_x} \Big) - \pdv{}{t}\Big( \pdv{\mathcal{L}}{u_t} \Big) \\ 0 &= \pdv{\mathcal{L}}{u^*} - \pdv{}{x}\Big( \pdv{\mathcal{L}}{u_x^*} \Big) - \pdv{}{t}\Big( \pdv{\mathcal{L}}{u_t^*} \Big) \end{aligned}$If $\mathcal{L}$ is real, then these two Euler-Lagrange equations will in fact be identical.

Finally, note that for abstract fields, the Lagrangian density $\mathcal{L}$ rarely has a physical interpretation, and is not unique. Instead, it must be reverse-engineered from a relevant equation.

## References

- R. Shankar,
*Principles of quantum mechanics*, 2nd edition, Springer.