From 805718880c936d778c99fe0d5cfdb238342a83c7 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 1 Jul 2021 22:21:26 +0200
Subject: Expand knowledge base

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+---
+title: "Lagrangian mechanics"
+firstLetter: "L"
+publishDate: 2021-07-01
+categories:
+- Physics
+
+date: 2021-07-01T18:44:43+02:00
+draft: false
+markup: pandoc
+---
+
+# 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](/know/concept/calculus-of-variations/).
+
+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.
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