# Lagrangian

A Lagrangian $\displaystyle \mathcal{L}[\varphi_i]$ of a dynamical system, named after Joseph Louis Lagrange, is a functional of the dynamical variables $\displaystyle \ \varphi_i(s)$ which concisely describes the equations of motion of the system. The equations of motion are obtained by means of an action principle, written as

$\displaystyle \frac{\delta \mathcal{S}}{\delta \varphi_i} = 0$

where the action $\displaystyle \mathcal{S}[\varphi_i] = \int{\mathcal{L}[\varphi_i(s)]{}\,d^ns},$

$\displaystyle {}{}{}{}\ s_\alpha$ denoting the set of parameters of the system.

The equations of motion obtained by means of the functional derivative are identical to the usual Euler-Lagrange equations. Dynamical system whose equations of motion are obtainable by means of an action principle on a suitably chosen Lagrangian are known as Lagrangian dynamical systems. Examples of Lagrangian dynamical systems range from the (classical version of the) Standard Model, to Newton's equations, to purely mathematical problems such as geodesic equations and Plateau's problem.

## An example from classical mechanics

The concept of a Lagrangian was originally introduced in a reformulation of classical mechanics known as Lagrangian mechanics. In this context, the Lagrangian is usually taken to be the kinetic energy of a mechanical system minus its potential energy.

Suppose we have a three dimensional space and the Lagrangian

$\displaystyle \begin{matrix}\frac{1}{2}\end{matrix} m\dot{\vec{x}}^2-V(\vec{x}).$

Then, the Euler-Lagrange equation is $\displaystyle m\ddot\vec{x}+\nabla V=0$ where the time derivative is written conventionally as a dot above the quantity being differentiated, and $\displaystyle \nabla$ is the del operator.

Using this result we can easily show that the Lagrangian approach is equivalent to the Newtonian one. We write the force in terms of the potential $\displaystyle \vec{F}=- \nabla V(x)$ ; then the resulting equation is $\displaystyle \vec{F}=m\ddot{\vec{x}}$ , which is exactly the same equation as in a Newtonian approach for a constant mass object. A very similar deduction gives us the expression $\displaystyle \vec{F}=d\vec{p}/dt$ , which is Newton's Second Law in its general form.

Suppose we have a three-dimensional space in spherical coordinates, r, θ, φ with the Lagrangian

$\displaystyle \frac{m}{2}(\dot{r}^2+r^2\dot{\theta}^2 +r^2\sin^2\theta\dot{\varphi}^2)-V(r).$

Then the Euler-Lagrange equations are:

$\displaystyle m\ddot{r}-mr(\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2)+V' =0,$
$\displaystyle \frac{d}{dt}(mr^2\dot{\theta}) -mr^2\sin\theta\cos\theta\dot{\varphi}^2=0,$
$\displaystyle \frac{d}{dt}(mr^2\sin^2\theta\dot{\varphi})=0.$

Here the set of parameters $\displaystyle \ s_i$ is just the time $\displaystyle \ t$ , and the dynamical variables $\displaystyle \ \phi_i(s)$ are the trajectories $\displaystyle \vec x(t)$ of the particle.

## Lagrangians and Lagrangian densities in field theory

In field theory, occasionally a distinction is made between the Lagrangian $\displaystyle L$ , of which the action is the time integral

$\displaystyle S = \int{L \, dt}$

and the Lagrangian density $\displaystyle \mathcal{L}$ , which one integrates over all space-time to get the action:

$\displaystyle S [\varphi_i] = \int{\mathcal{L} [\varphi_i (x)]\, d^4x}$

The Lagrangian is then the spatial integral of the Lagrangian density. However, $\displaystyle \mathcal{L}$ is also frequently simply called the Lagrangian, especially in modern use; it is far more useful in relativistic theories since it is a locally defined, Lorentz scalar field. Both definitions of the Lagrangian can be seen as special cases of the general form, depending on whether the spatial variable $\displaystyle \vec x$ is incorporated into the index $\displaystyle i$ or the parameters $\displaystyle s$ in $\displaystyle \varphi_i(s)$ . Quantum field theories in particle physics, such as quantum electrodynamics, are usually described in terms of $\displaystyle \mathcal{L}$ , and the terms in this form of the Lagrangian translate quickly to the rules used in evaluating Feynman diagrams.

## Electromagnetic Lagrangian

Generally, in Lagrangian mechanics, the Lagrangian is equal to

$\displaystyle L = T - V$

where T is kinetic energy and V is potential energy. Given an electrically charged particle with mass m and charge q, with velocity v in an electromagnetic field with scalar potential φ and vector potential A, the particle's kinetic energy is

$\displaystyle T = {1 \over 2} m \mathbf{v} \cdot \mathbf{v}$

and the particle's potential energy is

$\displaystyle V = q\phi - {q \over c} \mathbf{v} \cdot \mathbf{A}$

where c is the speed of light. Then the electromagnetic Lagrangian is

$\displaystyle L = {1 \over 2} m \mathbf{v} \cdot \mathbf{v} - q\phi + {q \over c} \mathbf{v} \cdot \mathbf{A} .$

## Lagrangians in Quantum Field Theory

### Quantum Electrodynamic Lagrangian

The Lagrangian density for QED is

$\displaystyle \mathcal{L} = \bar \psi (i \not \!\, D - m) \psi - {1 \over 4} F_{\mu \nu} F^{\mu \nu}$

where ψ is a spinor, $\displaystyle \bar \psi := \psi^\dagger \gamma^0$ ($\displaystyle \gamma^0$ being a gamma matrix), $\displaystyle F^{\mu\nu}$ is the electromagnetic tensor, D is the gauge covariant derivative, and $\displaystyle \not \!\, D$ is Feynman notation for $\displaystyle \gamma^\sigma D_\sigma$ .

### Dirac Lagrangian

The Lagrangian density for a Dirac field is

$\displaystyle \mathcal{L} = \bar \psi (i \not \! \; \partial - m) \psi = 0$ .

### Quantum Chromodynamic Lagrangian

The Lagrangian density for quantum chromodynamics is [1] [2] [3]

$\displaystyle \mathcal{L} = -{1\over 4} F^\alpha {}_{\mu\nu} F_\alpha {}^{\mu\nu} - \sum_n \bar \psi_n (\not\!\, D_\mu + m_n) \psi_n$

where $\displaystyle D_\mu$ is the QCD gauge covariant derivative, and $\displaystyle F^\alpha {}_{\mu\nu}$ is the gluon field strength tensor.

## Mathematical formalism

Suppose we have an n-dimensional manifold, M and a target manifold T. Let $\displaystyle \mathcal{C}$ be the configuration space of smooth functions from M to T.

Before we go on, let's give some examples:

• In classical mechanics, in the Hamiltonian formalism, M is the one dimensional manifold $\displaystyle \mathbb{R}$ , representing time and the target space is the cotangent bundle of space of generalized positions.
• In field theory, M is the spacetime manifold and the target space is the set of values the fields can take at any given point. For example, if there are m real-valued scalar fields, φ1,...,φm, then the target manifold is $\displaystyle \mathbb{R}^m$ . If the field is a real vector field, then the target manifold is isomorphic to $\displaystyle \mathbb{R}^n$ . There is actually a much more elegant way using tangent bundles over M, but we will just stick to this version.

Now suppose there is a functional, $\displaystyle S:\mathcal{C}\rightarrow \mathbb{R}$ , called the action. Note that it is a mapping to $\displaystyle \mathbb{R}$ , not $\displaystyle \mathbb{C}$ ; this has to do with physical reasons.

In order for the action to be local, we need additional restrictions on the action. If $\displaystyle \varphi\in\mathcal{C}$ , we assume S[φ] is the integral over M of a function of φ, its derivatives and the position called the Lagrangian, $\displaystyle \mathcal{L}(\varphi,\partial\varphi,\partial\partial\varphi, ...,x)$ . In other words,

$\displaystyle \forall\varphi\in\mathcal{C}\, S[\varphi]\equiv\int_M d^nx \mathcal{L}(\varphi(x),\partial\varphi(x),\partial\partial\varphi(x), ...,x).$

Most of the time, we will also assume in addition that the Lagrangian depends on only the field value and its first derivative but not the higher derivatives; this is only a matter of convenience, though, and is not true in general! We will make this assumption for the rest of this article.

Given boundary conditions, basically a specification of the value of φ at the boundary if M is compact or some limit on φ as x approaches $\displaystyle \infty$ (this will help in doing integration by parts), the subspace of $\displaystyle \mathcal{C}$ consisting of functions, φ such that all functional derivatives of S at φ are zero and φ satisfies the given boundary conditions is the subspace of on shell solutions.

The solution is given by the Euler-Lagrange equations (thanks to the boundary conditions),

$\displaystyle \frac{\delta}{\delta\varphi}S=-\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}\right)+ \frac{\partial\mathcal{L}}{\partial\varphi}=0.$

Incidentally, the left hand side is the functional derivative of the action with respect to φ.