Lagrangian

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Lagrangian

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http://en.wikipedia.org/wiki/Lagrangian

The Lagrangian, $L$ , of a dynamical system is a function that summarizes the dynamics of the system. It is named after Joseph Louis Lagrange. The concept of a Lagrangian was originally introduced in a reformulation of classical mechanics known as Lagrangian mechanics. In classical mechanics, the Lagrangian is defined as the kinetic energy, $T$ , of the system minus its potential energy, $V$ . In symbols,

$L = T - V.\quad$

Under conditions that are given in Lagrangian mechanics, if the Lagrangian of a system is known, then the equations of motion of the system may be obtained by a direct substitution of the expression for the Lagrangian into the Euler–Lagrange equation, a particular family of partial differential equations.

The Lagrange formulation

Importance

The Lagrange formulation of mechanics is important not just for its broad applications, but also for its role in advancing deep understanding of physics. Although Lagrange only sought to describe classical mechanics, the action principle that is used to derive the Lagrange equation is now recognized to be applicable to quantum mechanics.

Physical action and quantum-mechanical phase are related via Planck's constant, and the principle of stationary action can be understood in terms of constructive interference of wave functions.

The same principle, and the Lagrange formalism, are tied closely to Noether's theorem, which relates physical conserved quantities to continuous symmetries of a physical system.

Lagrangian mechanics and Noether's theorem together yield a natural formalism for first quantization by including commutators between certain terms of the Lagrangian equations of motion for a physical system.

Advantages over other methods

The formulation is not tied to any one coordinate system -- rather, any convenient variables $\varphi_i(s)$ may be used to describe the system; these variables are called "generalized coordinates" and may be any independent variable of the system (for example, strength of the magnetic field at a particular location; angle of a pulley; position of a particle in space; or degree of excitation of a particular eigenmode in a complex system). This makes it easy to incorporate constraints into a theory by defining coordinates which only describe states of the system which satisfy the constraints.

If the Lagrangian is invariant under a symmetry, then the resulting equations of motion are also invariant under that symmetry. This is very helpful in showing that theories are consistent with either special relativity or general relativity.

Equations derived from a Lagrangian will almost automatically be unambiguous and consistent, unlike equations just thrown together from various sources.

"Cyclic coordinates" and conservation laws

An important property of the Lagrangian is that conservation laws can easily be read-off from it. E.g., if the Lagrangian $\mathcal L$ depends on the time-derivative $\dot q_i$ of a generalized coordinate, but not on $q i$ itself, then the generalized momentum $p_i:=\frac{\partial\mathcal L}{\partial\dot q_i}$ is a conserved quantity. This is a special case of Noether's theorem, see below. Such coordinates are called "cyclic".

To repeat: the conservation of the generalized momentum $p_2:=\frac{\partial\mathcal L}{\partial\dot q_2}$ , say, can be directly seen if the Lagrangian of the system is of the form

$\mathcal L(q_1,q_3,q_4, ...;\dot q_1,\dot q_2,\dot q_3,\dot q_4, ...;t)\,.$

As an aside: If the time, t, does not appear in $\mathcal L$ , then the conservation of the Hamiltonian follows. This is the energy conservation unless the potential energy depends on velocity, as in electrodynamics. More details can be found in any textbook on theoretical mechanics.

Explanation

The equations of motion are obtained by means of an action principle, written as:

$\frac{\delta \mathcal{S}}{\delta \varphi_i} = 0\,.$

where the action, $\mathcal{S}$ , is a functional of the dependent variables $\varphi_i(s)$ with their derivatives and s itself

$\mathcal{S}\left[\varphi_i, \frac{\partial \varphi_i} {\partial s}\right] = \int{ \mathcal{L} \left[\varphi_i [s], \frac{\partial \varphi_i [s]}{\partial s^\alpha}, s^\alpha\right] \, \mathrm{d}^n s }$

and where $s = \{ s^\alpha \} \!$ denotes the set of n independent variables of the system, indexed by $\alpha = 1, 2, 3, \ldots, n .$

The equations of motion obtained from this functional derivative are the Euler–Lagrange equations of this action. For example, in the classical mechanics of particles, the only independent variable is time, t. So the Euler-Lagrange equations are

$\frac{d}{d t}\frac{\partial\mathcal L}{\partial\dot \varphi_i} = \frac{\partial\mathcal L}{\partial\varphi_i} .$

Dynamical systems 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

In the rectangular coordinate system

Suppose we have a three-dimensional space and the Lagrangian

$L(\vec{x}, \dot{\vec{x}}) \ = \ \frac{1}{2} \ m \ \dot{\vec{x}}^2 \ - \ V(\vec{x})$ .

Then, the Euler–Lagrange equation is:

$\frac{d~}{dt} \ \left( \, \frac{\partial L}{\partial \dot{x}_i} \, \right) \ - \ \frac{\partial L}{\partial x_i} \ = \ 0$

where $i = 1,2,3$ .

The derivation yields:

$\frac{\partial L}{\partial x_i} \ = \ - \ \frac{\partial V}{\partial x_i}$

$\frac{\partial L}{\partial \dot{x}_i} \ = \ \frac{\partial ~}{\partial \dot{x}_i} \, \left( \, \frac{1}{2} \ m \ \dot{\vec{x}}^2 \, \right) \ = \ \frac{1}{2} \ m \ \frac{\partial ~}{\partial \dot{x}_i} \, \left( \,\dot{x}_i \, \dot{x}_i \, \right) = \ m \, \dot{x}_i$

$\frac{d~}{dt} \ \left( \, \frac{\partial L}{\partial \dot{x}_i} \, \right) \ = \ m \, \ddot{x}_i$

The Euler–Lagrange equations can therefore be written as:

$m\ddot{\vec{x}}+\nabla V=0$

where the time derivative is written conventionally as a dot above the quantity being differentiated, and $\nabla$ is the del operator.

Using this result, it can easily be shown that the Lagrangian approach is equivalent to the Newtonian one.

If the force is written in terms of the potential $\vec{F}=- \nabla V(x)$ ; the resulting equation is $\vec{F}=m\ddot{\vec{x}}$ , which is exactly the same equation as in a Newtonian approach for a constant mass object.

A very similar deduction gives us the expression $\vec{F}=\mathrm{d}\vec{p}/\mathrm{d}t$ , which is Newton's Second Law in its general form.

In the spherical coordinate system

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

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

Then the Euler–Lagrange equations are:

$m\ddot{r}-mr(\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2)+V' =0,$

$\frac{\mathrm{d}}{\mathrm{d}t}(mr^2\dot{\theta}) -mr^2\sin\theta\cos\theta\dot{\varphi}^2=0,$

$\frac{\mathrm{d}}{\mathrm{d}t}(mr^2\sin^2\theta\dot{\varphi})=0.$

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

Despite the use of standard variables such as $x$ , the Lagrangian allows the use of any coordinates, which do not need to be orthogonal. These are "generalized coordinates".

Lagrangian of a test particle

A test particle is a particle whose mass and charge are assumed to be so small that its effect on external system is insignificant. It is often a hypothetical simplified point particle with no properties other than mass and charge. Real particles like electrons and up-quarks are more complex and have additional terms in their Lagrangians.

Classical test particle with Newtonian gravity

Suppose we are given a particle with mass $m \!$ kilograms, and position $\vec{x}$ meters in a Newtonian gravitation field with potential $\zeta \!$ joules per kilogram. The particle's world line is parameterized by time $t\!$ seconds. The particle's kinetic energy is:

$T[t] = {1 \over 2} m \dot{\vec{x}}[t] \cdot \dot{\vec{x}}[t]$

and the particle's gravitational potential energy is:

$V[t] = m \zeta [\vec{x} [t],t] .$

Then its Lagrangian is $L \!$ joules where

$L[t] = T[t] - V[t] = {1 \over 2} m \dot{\vec{x}}[t] \cdot \dot{\vec{x}}[t]- m \zeta [\vec{x} [t],t] .$

Varying $\vec{x}\!$ in the integral (equivalent to the Euler–Lagrange differential equation), we get

$0 = \delta\int{L[t] \, \mathrm{d}t} = \int{\delta L[t] \, \mathrm{d}t}$

$= \int{m \dot{\vec{x}}[t] \cdot \dot{\delta \vec{x}}[t]- m \nabla \zeta [\vec{x} [t],t] \cdot \delta \vec{x}[t] \, \mathrm{d}t}.$

Integrate the first term by parts and discard the total integral. Then divide out the variation to get

$0 = - m \ddot{\vec{x}}[t] - m \nabla \zeta [\vec{x} [t],t]$

and thus

$m \ddot{\vec{x}}[t] = - m \nabla \zeta [\vec{x} [t],t] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$

is the equation of motion — two different expressions for the force.

Special relativistic test particle with electromagnetism

In special relativity, the form of the term which gives rise to the derivative of the momentum must be changed; it is no longer the kinetic energy. It becomes:

$- m c^2 \frac{d \tau[t]}{d t} = - m c^2 \sqrt {1 - \frac{v^2 [t]}{c^2}}$

$= -m c^2 + {1 \over 2} m v^2 [t] + {1 \over 8} m \frac{v^4 [t]}{c^2} + \dots$

(In special relativity, the energy of a free test particle is $m c^2 \frac{dt}{d \tau [t]} = \frac{m c^2}{\sqrt {1 - \frac{v^2 [t]}{c^2}}} = +m c^2 + {1 \over 2} m v^2 [t] + {3 \over 8} m \frac{v^4 [t]}{c^2} + \dots$ )

where $c \!$ meters per second is the speed of light in vacuum, $\tau \!$ seconds is the proper time (i.e. time measured by a clock moving with the particle) and $v^2 [t] = \dot{\vec{x}}[t] \cdot \dot{\vec{x}}[t].$ The second term in the series is just the classical kinetic energy. Suppose the particle has electrical charge $q\!$ coulombs and is in an electromagnetic field with scalar potential $\phi \!$ volts (a volt is a joule per coulomb) and vector potential $\vec{A}$ volt seconds per meter. The Lagrangian of a special relativistic test particle in an electromagnetic field is:

$L[t] = - m c^2 \sqrt {1 - \frac{v^2 [t]}{c^2}} - q \phi [\vec{x}[t],t] + q \dot{\vec{x}}[t] \cdot \vec{A} [\vec{x}[t],t]$

Varying this with respect to $\vec{x}$ , we get

$0 = - \frac{d}{d t}\left(\frac{m \dot{\vec{x}}[t]} {\sqrt {1 - \frac{v^2 [t]}{c^2}}}\right) - q \nabla\phi [\vec{x}[t],t] - q \partial_t{\vec{A}} [\vec{x}[t],t]- q \dot{\vec{x}}[t] \cdot \nabla\vec{A} [\vec{x}[t],t]+ q \nabla{\vec{A}} [\vec{x}[t],t] \cdot \dot{\vec{x}}[t]$

which is

$\frac{d}{d t}\left(\frac{m \dot{\vec{x}}[t]} {\sqrt {1 - \frac{v^2 [t]}{c^2}}}\right) = q \vec{E}[\vec{x}[t],t]+ q \dot{\vec{x}}[t] \times \vec{B} [\vec{x}[t],t]$

which is the equation for the Lorentz force where

$\vec{E}[\vec{x},t] = - \nabla\phi [\vec{x},t] - \partial_t{\vec{A}} [\vec{x},t]$

$\vec{B}[\vec{x},t] = \nabla \times \vec{A} [\vec{x},t]$

General relativistic test particle

In general relativity, the first term generalizes (includes) both the classical kinetic energy and interaction with the Newtonian gravitational potential. It becomes:

$- m c^2 \frac{d \tau[t]}{d t}$

$= - m c \sqrt {- g_{\alpha\beta}[x[t]] \frac{d x^{\alpha}[t]}{d t} \frac{d x^{\beta}[t]}{d t}} .$

The Lagrangian of a general relativistic test particle in an electromagnetic field is:

$L[t] = - m c \sqrt {- g_{\alpha\beta}[x[t]] \frac{d x^{\alpha}[t]}{d t} \frac{d x^{\beta}[t]}{d t}} + q \frac{d x^{\gamma}[t]}{d t} A_{\gamma}[x[t]] .$

If the four space-time coordinates $x^{\alpha}\!$ are given in arbitrary units (i.e. unit-less), then $g_{\alpha\beta}\!$ meters squared is the rank 2 symmetric metric tensor which is also the gravitational potential. Also, $A_{\gamma}\!$ volt seconds is the electromagnetic 4-vector potential. Notice that a factor of c has been absorbed into the square root because it is the equivalent of

$c\, \sqrt {1 - \frac{v^2 [t]}{c^2}} =\sqrt {- ( - c^2 + v^2 [t])} .$

Note that this notion has been directly generalized from special relativity

Lagrangians and Lagrangian densities in field theory

The time integral of the Lagrangian is called the action denoted by $S$ .
In field theory, a distinction is occasionally made between the Lagrangian $L$ , of which the action is the time integral:

$\mathcal{S} = \int{L \, \mathrm{d}t}$

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

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

The Lagrangian is then the spatial integral of the Lagrangian density. However, $\mathcal{L}$ 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 $\vec x$ is incorporated into the index $i$ or the parameters $s$ in $\varphi_i(s)$ . Quantum field theories in particle physics, such as quantum electrodynamics, are usually described in terms of $\mathcal{L}$ , and the terms in this form of the Lagrangian translate quickly to the rules used in evaluating Feynman diagrams.

Selected fields

To go with the section on test particles above, here are the equations for the fields in which they move. The equations below pertain to the fields in which the test particles described above move and allow the calculation of those fields. The equations below will not give you the equations of motion of a test particle in the field but will instead give you the potential (field) induced by quantities such as mass or charge density at any point $[\vec{x},t]$ . For example, in the case of Newtonian gravity, the Lagrangian density integrated over space-time gives you an equation which, if solved, would yield $\zeta [\vec{x},t]$ . This $\zeta [\vec{x},t]$ , when substituted back in equation (1), the Lagrangian equation for the test particle in a Newtonian gravitational field, provides the information needed to calculate the acceleration of the particle.

Newtonian gravity

The Lagrangian (density) is $\mathcal{L}$ joules per cubic meter. The interaction term $m \zeta \!$ is replaced by a term involving a continuous mass density $\mu \!$ kilograms per cubic meter. This is necessary because using a point source for a field would result in mathematical difficulties. The resulting Lagrangian for the classical gravitational field is:

$\mathcal{L}[\vec{x},t] = - \mu [\vec{x},t] \zeta [\vec{x},t] - {1 \over 8 \pi G} (\nabla \zeta [\vec{x},t])^2$

where $G \!$ meters cubed per kilogram second squared is the gravitational constant. Variation of the integral with respect to $\zeta \!$ gives:

$0 = - \mu [\vec{x},t] \delta\zeta [\vec{x},t] - {2 \over 8 \pi G} (\nabla \zeta [\vec{x},t]) \cdot (\nabla \delta\zeta [\vec{x},t]) .$

Integrate by parts and discard the total integral. Then divide out by $\delta\zeta \!$ to get:

$0 = - \mu [\vec{x},t] + {1 \over 4 \pi G} \nabla \cdot \nabla \zeta [\vec{x},t]$

and thus

$4 \pi G \mu [\vec{x},t] = \nabla^2 \zeta [\vec{x},t]$

which yields Gauss's law for gravity.

Electromagnetism in special relativity

The interaction terms $- q \phi [\vec{x}[t],t] + q \dot{\vec{x}}[t] \cdot \vec{A} [\vec{x}[t],t]$ are replaced by terms involving a continuous charge density $\rho \!$ coulombs per cubic meter and current density $\vec{j} \!$ amperes per square meter. The resulting Lagrangian for the electromagnetic field is:

$\mathcal{L}[\vec{x},t] = - \rho [\vec{x},t] \phi [\vec{x},t] + \vec{j} [\vec{x},t] \cdot \vec{A} [\vec{x},t] + {\epsilon_0 \over 2} {E}^2 [\vec{x},t] - {1 \over {2 \mu_0}} {B}^2 [\vec{x},t] .$

Varying this with respect to $\phi \!$ , we get

$0 = - \rho [\vec{x},t] + \epsilon_0 \nabla \cdot \vec{E} [\vec{x},t]$

which yields Gauss' law.

Varying instead with respect to $\vec{A}$ , we get

$0 = \vec{j} [\vec{x},t] + \epsilon_0 \partial_t \vec{E} [\vec{x},t] - {1 \over \mu_0} \nabla\times \vec{B} [\vec{x},t]$

which yields Ampère's law.

Electromagnetism in general relativity

For the Lagrangian of gravity in general relativity, see Einstein-Hilbert action. The Lagrangian of the electromagnetic field is:

$\mathcal{L}[x] = + J^{\gamma}[x] A_{\gamma}[x]- {1 \over 4\mu_0} F_{\mu \nu}[x] F_{\alpha \beta}[x] g^{\mu\alpha}[x] g^{\nu\beta}[x] \sqrt{\frac{-1}{c^2} \mathrm{det} [g[x]]}.$

If the four space-time coordinates $x^{\alpha}\!$ are given in arbitrary units, then: $\mathcal{L}$ joule seconds is the Lagrangian, a scalar density; $J^{\gamma}\!$ coulombs is the current, a vector density; and $F_{\mu \nu}\!$ volt seconds is the electromagnetic tensor, a covariant antisymmetric tensor of rank two. Notice that the determinant under the square root sign is applied to the matrix of components of the covariant metric tensor $g_{\alpha\beta}\!$ , and $g^{\alpha\beta}\!$ is its inverse. Notice that the units of the Lagrangian changed because we are integrating over $x^0, x^1, x^2, x^3\!$ which are unit-less rather than over $t, x, y, z \!$ which have units of seconds meters cubed. The electromagnetic field tensor is formed by anti-symmetrizing the partial derivative of the electromagnetic vector potential; so it is not an independent variable. The square root is needed to convert that term into a scalar density instead of just a scalar, and also to compensate for the change in the units of the variables of integration. The factor of $\frac{-1}{c^2}$ inside the square root is needed to normalize it so that the square root will reduce to one in special relativity (since the determinant is $- c^2 \!$ in special relativity).

Lagrangians in quantum field theory

Dirac Lagrangian

The Lagrangian density for a Dirac field is:

$\mathcal{L} = \bar \psi (i \hbar c \not\!D - mc^2) \psi$

where $\psi\!$ is a Dirac spinor, $\bar \psi = \psi^\dagger \gamma^0$ is its Dirac adjoint, $D\!$ is the gauge covariant derivative, and $\not\!D$ is Feynman notation for $\gamma^\sigma D_\sigma\!$ .

Quantum electrodynamic Lagrangian

The Lagrangian density for QED is:

$\mathcal{L}_{\mathrm{QED}} = \bar \psi (i \hbar c\not\!D - mc^2) \psi - {1 \over 4\mu_0} F_{\mu \nu} F^{\mu \nu}$

where $F^{\mu \nu}\!$ is the electromagnetic tensor.

Quantum chromodynamic Lagrangian

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

$\mathcal{L}_{\mathrm{QCD}} = \sum_n \bar \psi_n (i \hbar c\not\!D - m_n c^2) \psi_n - {1\over 4} G^\alpha {}_{\mu\nu} G_\alpha {}^{\mu\nu}$

where $D\!$ is the QCD gauge covariant derivative, $n = 1...6$ counts the quark types, and $G^\alpha {}_{\mu\nu}\!$ is the gluon field strength tensor.

Mathematical formalism

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

Examples

In classical mechanics, in the Hamiltonian formalism, $M$ is the one-dimensional manifold $\mathbb{R}$ , 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, $φ 1,...,φ m$ , then the target manifold is $\mathbb{R}^m$ . If the field is a real vector field, then the target manifold is isomorphic to $\mathbb{R}^n$ . There is actually a much more elegant way using tangent bundles over $M$ , but we will just stick to this version.

Mathematical development

Consider a functional, $\mathcal{S}:\mathcal{C}\rightarrow \mathbb{R}$ , called the action. Physical reasons determine that it is a mapping to $\mathbb{R}$ , not $\mathbb{C}$ .

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

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

It is assumed below, in addition, that the Lagrangian depends on only the field value and its first derivative but not the higher derivatives.

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 $\infty$ (this will help in doing integration by parts), the subspace of $\mathcal{C}$ 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),

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

The left hand side is the functional derivative of the action with respect to $φ$ .