---
title: "Lagrangian mechanics"
sort_title: "Lagrangian mechanics"
date: 2021-07-01
categories:
- Physics
- Classical mechanics
layout: "concept"
---

**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](/know/concept/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:

$$\begin{aligned}
    \boxed{
        p_n \equiv \pdv{L}{\dot{q}_n} \qquad F_n \equiv \pdv{L}{q_n}
    }
    \qquad \implies \qquad
    \dv{p_n}{t} = F_n
\end{aligned}$$

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:

$$\begin{aligned}
    F_n = \pdv{L}{q_n} = 0
    \qquad \implies \qquad
    \dv{p_n}{t} = 0
\end{aligned}$$

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:

$$\begin{aligned}
    J[u]
    = \int_{t_0}^{t_1} L \dd{t}
    = \int_{t_0}^{t_1} \! \int_{-\infty}^\infty \mathcal{L} \dd{x} \dd{t}
\end{aligned}$$

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