%From: Lisa C Jeffrey
%Date: Mon, 25 Nov 1996 08:54:53 -0500
%Subject: Faddeev lecture 4
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\begin{document}
\title{Lecture 4:\\
Singular Lagrangians}
\author{Ludwig Faddeev}
\date{7 November 1996}
\maketitle
%\renorm
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In other lectures, considerable attention has been focused on how to construct
the symplectic structure on the space of classical solutions,
starting from the Lagrangian.
In this lecture we will instead focus on the
fixed time Hamiltonian formalism,
and we will see that the transformation which maps the Lagrangian
into the Hamiltonian is simply a change of variables. The lecture will
chiefly treat the case of singular Lagrangians. Our aim is to present
the first order formalism, in which the Lagrangian depends linearly
on time derivatives of the dynamical variables.
\nc{\qdot}{{\dot{q} }}
In the usual Lagrangian formalism, we begin with the configuration space $M$:
coordinates on $M$ will be denoted $q$. Coordinates on the tangent
bundle $TM $ are then given by $q$ and $\dot{q}$, and the Lagrangian is
a function $\calL(q, \qdot)$.
The action is then defined by
$$ A = \int \calL (q(t), \qdot(t) ) \, dt. $$
Applying the standard variational principle we see that the extrema
of the action are given by the {\em Euler-Lagrange equation}
\beq \label{el} \frac{d}{dt} \frac{\partial \calL}{\partial \qdot} -
\frac{\partial \calL}{\partial q} = 0. \eeq
This is ordinarily a system of
equations of {\em second order}
since the Euler-Lagrange equation involves the term
$\ddot{q} {\partial^2 \calL}/{\partial \qdot \partial \qdot}$.
The Lagrangian is said to be {\em nondegenerate} if the Hessian
${\partial^2 \calL}/{\partial \qdot \partial \qdot}$ is
invertible.
In the Hamiltonian formalism, on the other hand, we work with
$T^* M \times \RR$ (where
$T^* M$ is the cotangent bundle of configuration space, or the
phase space).
On this there is a natural 1-form
\beq \label{oneform} p dq - H(p,q) dt, \eeq
where we
have introduced $(q, p )$ as natural coordinates on $T^* M$.
The corresponding Lagrangian is written as
\beq \label{lagstd} l = p \qdot - H(p,q), \eeq
and the action is
$$ A = \int (p \qdot - H(p,q)) dt. $$
The Euler-Lagrange equation applied to $A$ then gives Hamilton's
equations:
\beq \dot{p} = - \frac{\partial H}{\partial q},
~~~~\dot{q} = \frac{\partial H}{\partial p}. \eeq
Note that
the Lagrangian $l$ is linear in time derivatives, and the Hessian
of $l$ is identically zero. Thus our system is of first order.
In the standard fashion, we have converted a system of second order
differential equations into a system of first order differential
equations, at the expense of increasing the number of variables.
To explicitly see how the Hamiltonian formalism
is derived from the Lagrangian formalism, we introduce
$v = \qdot$ and treat $q$ and $v$ as independent variables.
The Lagrangian then depends on $q$, $v$,
$\qdot$ and $\dot{v}$, and has the following form:
\beq l = \frac{\partial \calL}{\partial v} (\qdot - v) + \calL(q, v). \eeq
Comparing with (\ref{lagstd})
one finds that the Hamiltonian is given by
$$ H = v \frac{\partial \calL}{\partial v} - \calL, $$
and we obtain the Hamiltonian formalism by performing the Legendre
transformation
$$ (q, v) \mapsto (q,p) $$
where we have defined
$$ p = \frac{\partial \calL}{\partial v}. $$
On a general manifold $\Gamma$ with coordinates $\xi$, we would
write the Lagrangian as
\beq l = \sum_a f_a(\xi) \dot{\xi}^a - \phi(\xi). \eeq
The one-form on $\Gamma \times \RR$ corresponding to (\ref{oneform})
would be written as
$$ \sum_a f_a(\xi) d \xi^a - \phi dt = \omega -
\phi dt, $$
and it gives rise to a two-form $ \Omega = d \omega$. (Here $d$ denotes
the de Rham differential on $\Gamma$, rather than on $\Gamma \times \RR$.)
The equation of motion becomes
$$ \Omega \dot{\xi} = \frac{\partial \phi}{\partial \xi}. $$
The condition that the Lagrangian is nonsingular is equivalent to the
condition that the
two-form $\Omega$ is invertible: for this to hold, it is
necessary that $\Gamma$ be even dimensional. For important Lagrangians
in physics (for instance the Yang-Mills Lagrangian and the
Lagrangian for electrodynamics) the two-form $\Omega$ is in fact not
invertible. Let us investigate this situation in more detail.
Invoking the Darboux theorem,
on $\Gamma $ we write the coordinates
$\xi$ as $(p, q, z)$ in such a way that the one-form $\omega$ becomes
$$ \omega = p dq + d \theta.$$
The total derivative $d \theta$ may be discarded, and after this the
one-form corresponding to the
Lagrangian becomes
$$ l = p dq - \phi(p,q, z) dt, $$
where
$\frac{\partial \phi}{\partial z} = 0 . $
The variables $z$ may be divided into ``excludable'' variables (which will
ultimately be discarded) and others which enter linearly. After
eliminating the excludable variables, the Lagrangian becomes
\beq \label{elimexcl} l = p dq - \phi(p, q) dt - \sum_\alpha \lambda^\alpha
\varphi_\alpha (p,q). \eeq
Here, the $\lambda^\alpha $ play the role of {\em Lagrange multipliers}
while the $\varphi_\alpha(p,q)$ are constraints. Among the
equations of motion we now obtain the
constraints
$$\varphi_\alpha(p,q) = 0, $$
so we may substitute $p = p(\eta), $ $ q = q(\eta)$ where
$\eta$ are coordinates on the manifold
$$ \Gamma^l = \{ (p,q): \varphi_\alpha(p,q) = 0 \} . $$
We thus obtain a new phase space $\Gamma^l $ on which the
excludable variables and the Lagrange multipliers have been eliminated,
so that the variables $z$ no longer appear and we are left with
a subspace parametrized by the variables $(p,q)$ where the
constraints $\varphi_\alpha(p,q) = 0 $ have been imposed.
The new phase space $\Gamma^l$ may have a nondegenerate symplectic
form, and then our
Hamiltonian reduction is done.
If not, we repeat the above process until Lagrange multipliers
no longer appear,
in which case we have reduced to a phase space with a
nondegenerate symplectic form.
Explicit solution of the constraints could be complicated,
so it is instructive to understand what properties of the
singular Lagrangian (\ref{elimexcl}) guarantee that it will
become nonsingular after the constraints are imposed. There are two
important instances of this.
We can distinguish two different types of constraints. A collection
of constraints $\{ \varphi_a \}$ are called
{\em first class} if the Poisson brackets of all constraints
vanish on the zero locus of the constraints: in other words, if
\beq \{ \varphi_a, \varphi_b \}|_{\varphi = 0 } = 0 . \eeq
We claim that if $\{ \phi, \varphi_a\}|_{\varphi = 0 } = 0 $ for
all $a$, then in the next step of the procedure one obtains a proper
nondegenerate Lagrangian. More precisely, the submanifold
$\Gamma|_{\varphi = 0 } $ may be fibered so that the base is symplectic
and the fibers are isotropic with respect to the Poisson bracket
$\{ \cdot, \cdot \}$. One may summarize the procedure by saying that
``first class constraints kill twice'': when one restricts to the
zero locus of the constraints $\varphi_a$,
one must also set to zero the variables
which are canonically conjugate to the $\varphi_a$, in
other words one must set to zero all Poisson brackets with
the $\varphi_a$. If there are
$N$ constraint equations $\varphi_a, $ $ a = 1, \dots, N$,
then imposing the constraints reduces
the dimension of the phase space by $2N$.
The second type of constraints $\varphi_a$ are those for which
$\det \{ \varphi_a, \varphi_b \} \ne 0 $: a system of constraints
for which this condition holds is called a system of {\em second
class constraints}. If a Lagrangian
gives rise to a system of second class constraints, the next
step of reduction leads to a nondegenerate Lagrangian. The number of
$\phi_a$ must be $2N$ and the dimension of the phase space is then
reduced by $N$ through imposition of the constraints.
Constrained quantization was originally considered by
Dirac \cite{Dirac}.
\noindent{\bf Example 1: the free scalar field}
Given a field $\varphi(t, \vx) $, and the free field
Lagrangian \beq \label{lagr1}
\calL = \frac{1}{2}(\partial_\mu \varphi)^2, \eeq
canonical quantization normally
specifies that one should introduce the canonically conjugate
momentum $\pi(t, \vx) = \partial_0 \varphi (t, \vx)$, where
$x_0 = t $ and $(x_1, x_2, x_3) = \vx$. The points $\vx$ in space
should be thought of as labelling the field. Notice that the momentum
$\pi$ is not Lorentz invariant.
We may alternatively replace the
Lagrangian (\ref{lagr1}) (which is second order in
time derivatives) by a first order Lagrangian: we do this by
introducing a vector field $\varphi_\mu$ (in other words a
collection of four fields $(\varphi_0, \varphi_1, \varphi_2, \varphi_3)$
which transform according to the standard action of the Lorentz group on
$\RR^4$) and write
\beq \label{lagr2}
l = (\partial_\mu \varphi ) \varphi^\mu - \frac{1}{2} (\varphi^\mu)^2. \eeq
Of course under the substitution
$\varphi_\mu = \partial \varphi/\partial x^\mu $ the Lagrangian in
(\ref{lagr2}) reduces to that in (\ref{lagr1}): however if
we regard $\varphi_\mu$ as independent of $\varphi$, (\ref{lagr2}) is
first order in time derivatives.
(We have introduced the Einstein summation convention, summing over
pairs of repeated indices $\mu$, where
$\mu = \{ 0, k \} $ and $k = 1, 2, 3$ correspond to space coordinates
while $\mu = 0 $ corresponds to the time coordinate.)
We may rewrite (\ref{lagr2}) as
\beq l = (\partial_0 \varphi ) \varphi^0 - (\partial_k \varphi ) \varphi^k
- \frac{1}{2} \varphi_0^2 +\frac{1}{2} \varphi_k^2. \eeq
Under the substitution
$\pi = \varphi_0 $, and solving for $\varphi_k, $
$\varphi_k = \partial_k \varphi$, we obtain
\beq \calL
= (\partial^0 \varphi) \pi - \frac{1}{2} (\pi^2 + (\partial_k \varphi)^2 ),
\eeq
which indeed shows that $\pi$ and $\varphi$ are
canonically conjugate and that the Hamiltonian $H$ has the usual
form
$$ H = \half (\pi^2 + (\partial_k \varphi)^2. ) $$
The variables $\varphi_k$ can be eliminated because they enter
(\ref{lagr2}) quadratically and without time derivatives.
\noindent{\bf Example 2: electromagnetism}
In this example it is important that spacetime has dimension 4.
We start with a vector field $A_\mu(x) $ (the photon field),
which should be thought of as
a 1-form or a connection:
$$ A = \sum_\mu A_\mu dx^\mu. $$
The associated curvature is
$$F = dA,$$ or as a two-form $F_{\mu \nu} dx^\mu \wedge dx^\nu, $ where
\beq \label{curv}
F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. \eeq
The Lagrangian is
\beq \label{lagel1} \calL = \int |F|^2, \eeq
which is second order in time derivatives. A trick due to
J. Schwinger enables us to replace it with a Lagrangian which
is first order in time derivatives: this is to regard
$A$ and $F$ as independent variables. By analogy with our procedure
in Example 1, we write
\beq \label{lagel2}
\calL = (\partial_\mu A_\nu - \partial_\nu A_\mu) F^{\mu \nu}
- \frac{1}{2} F_{\mu \nu}^2. \eeq
Of course (\ref{lagel2}) reduces to (\ref{lagel1}) by restoring
the functional dependence (\ref{curv}) of $F$ on $A$. However
the Lagrangian in (\ref{lagel2}) is linear in time derivatives of the
fields, and also introduces some constraints.
We rewrite (\ref{lagel2}) as
\beq \label{lagel3}
\calL = (\partial_0 A_k) F^{0k} + A_0 (\partial_k F^{0k}) -
F^{ik} (\partial_i A_k - \partial_k A_i) - \frac{1}{2} (F^{0k})^2
+ \frac{1}{2} (F^{ik})^2. \eeq
Here, $F^{0k} = E^k$ are the components of
the {\em electric field } $\vec{E}$, while
$A_k$ are the components of
the {\em vector potential}.
Up to the Hodge star operator $*$
which converts two-forms on $\RR^3$ into one-forms,
the quantity $F^{ik}$ is the
{\em magnetic field} $\vec{H}$: we substitute
\beq \label{magdef} \vec{H} = \nabla \times \vec{A}, \eeq
or
$H_i = \epsilon_{ijk} \nabla_j A_k $
in terms of the totally antisymmetric
tensor $\epsilon$ in three indices which
represents the Hodge star operator in $\RR^3$.
Thus the Lagrangian becomes
\beq \label{lagel4}
\calL = (\partial_0 A_k) E^k - \frac{1}{2} (\vec{E}^2 - \vec{H}^2) +
A_0 \partial_k E^k, \eeq
where we already excluded the independent variables $\vec{H}$, so that
$\vec{H}$ in (\ref{lagel4}) is given by (\ref{magdef}).
On the other hand, $A_k$ and $E^k$ are canonically conjugate variables
(appearing in a term which is first order in time derivatives), and
$A_0$ is a Lagrange multiplier multiplying the
constraint
\beq G = \nabla \cdot E = \partial_k E^k . \eeq
The constraint $G = 0 $ is called {\em Gauss's law}: it tells us
that the divergence of the electric field is zero. This is a first
class constraint, $ \{ G(\vx), G(\vy) \} = 0 $, and we have also
\beq \{ G, \vec{E} \} = \{ G, \vec{H} \} = 0 . \eeq
For any function $\lambda$ on $\RR^3$
we have
$$ \{ A_k, \int G(\vx) \lambda (\vx) d \vx \} = \partial_k \lambda(\vx); $$
equivalently $G$ is a generator of $U(1)$ gauge transformations:
the gauge group $\calG$ is the group of maps from $\RR^3$ to $U(1)$, and
its Lie algebra is the group of maps from $\RR^3$ to $\RR$. An element
$\lambda \in {\rm Lie} (\calG) $
sends $A_k $ to $A_k + \partial_k
\lambda$. The imposition of Gauss's law may be viewed
as symplectic reduction with respect to the action of the $U(1)$ gauge
group $\calG$.
Since the constraint $G$ is of first class, it reduces the degrees
of freedom from three functions $A_k(\vx)$ to two. Thus
(in physical language) light has two polarizations.
The above treats the situation of the electromagnetic field in the
absence of sources. If on the other hand the electric field
interacts with a charged field $J^\mu$, Gauss's law becomes
\beq \label{gaussmod} G = \nabla \cdot E + J_0(\vx) \eeq
where $J_0 $ is the {\em charge density}. If we define
$$ Q = \int_{\RR^3} J_0 dx $$
if we express $Q$ as a flux of the electromagnetic field
via a surface in $\RR^3$ (for instance a sphere $S^2(R)$ of
radius $R$)
we obtain
$$ Q = \lim_{R \to \infty} \int_{S^2(R)} \epsilon_{ikj }
E^k dS_{ij}. $$
Thus the equation
(\ref{gaussmod}) leads to {\em Coulomb's law},
according to which the electromagnetic force from a point
charge at distance $r$ is proportional to
$1/r^2$.
\noindent{\bf Example 4: Gravitation}
In this example we introduce a metric $g^{\mu \nu}$ with the associated
{\em affine connection} $\Gamma_{ij}^k$ and the scalar curvature
(we omit indices for clarity)
$$ R = \partial \Gamma + \Gamma^2. $$
To write a Lagrangian that is first order in time derivatives,
we take $g$ and $\Gamma$ as independent variables: more precisely we write
$$ h ^{\mu \nu} = \sqrt{g} g^{\mu \nu},$$ and write the Lagrangian as
\beq \label{laggrav1} \calL = h (\partial \Gamma + \Gamma^2). \eeq
The connection $\Gamma$ contains first order time derivatives of $h$, so
$\partial \Gamma$ contains second order time derivatives of $h$.
This may be altered by integrating by parts, so that we obtain
\beq \label{laggrav2}
\calL = - \Gamma \partial h + h \Gamma^2. \eeq
This Lagrangian is not manifestly covariant, however it is this
form which is to be
used in the so called {\em asymptotically flat} case.
The definition of asymptotic flatness can be given on different levels of
sophistication; we shall use the most na\"{\i}ve but practical
way of introducing the admissible coordinates. In such
coordinates we have in the vicinity of space infinity
$$ g = g_0 + O (\frac{1}{r} ), $$
$$ \partial g = O (\frac{1}{r^2} )$$
$$\Gamma = O (\frac{1}{r^2}) $$
where $g_0$ is the Minkowski metric.
%Dottie: please insert the text here. -- Lisa Jeffrey
An admissible change of variables is of the form
\beq \label{coords}
X_\mu\to \Lambda_{\mu v}X_r+a_\mu+
b_\mu(x),
\eeq
where $\Lambda_{\mu v}$ is a Lorentz rotation, $a_\mu$
defines a translation and
$$
b_\mu(X)= O\left(\frac{1}{r} \right),\quad
\partial b=O\left(\frac{1}{r^2}\right).
$$
In these coordinates the density (\ref{laggrav2}) is of
order $O\left(\frac{1}{r^4}\right)$ and the action
$$
A=\int\calL\,dx
$$
is invariant with respect to the change of coordinates
(\ref{coords}).
Inspection of the density (\ref{laggrav2}) cubic in the
variables $(h,\Gamma)$, shows that out of
$ 10 + 40 = 50$ variables, the $\Gamma_{00}^\mu$ enter
linearly and 26 other components $\Gamma$ enter
quadratically without time derivatives.
So the Lagrangian (\ref{laggrav2}) takes the form $(\ref{elimexcl})$,
however the
constraints do not commute with the corresponding $\phi$.
So we must solve them explicitly.
Fortunately this is possible and altogether solving the
constraints and eliminating the excludable variables
allows one to express 30 components of $\Gamma$ via the metric
$h^{\mu \nu}$ and 6 components $\Gamma_{ik}^0$.
Let
\begin{eqnarray*}
q^{ik} &=& h^{00}h^{ik}-h^{in}h^{nk}\\
\Pi_{ik} &=& \frac{\Gamma_{ik}^0}{h^{00}}
\end{eqnarray*}
be a (density of weight $+2$) of the first
quadratic form and (a density of weight $-1$) of the
second quadratic form of the surface ${x}_0=0$ embedded
into space-time.
After solving for $\Gamma$, mentioned above, the
Lagrangian (\ref{laggrav2}) reduces to the form
$$
\Pi_{ik}\partial_0 q^{ik}+\frac{1}{h^{00}}C_0(\Pi,q)+
\frac{h^{0k}}{h^{00}}C_k(\Pi,q)+\partial_i\partial_k
q^{ik}
$$
where $C_0$ and $C_k$ are some functionals of $\Pi$,
$q$ and derivatives.
The term $\partial_i\partial_k q^{ik}$ cannot be
discarded in spite of the fact that it is a pure divergence as its
flux through infinity does not vanish.
Indeed this flux
$$
\calM=\lim_{R \to \infty}\int\limits_{S_R}
\epsilon_{k\ell m}\partial_{ik}q^{ik}dS^{\ell n}
$$
defines the full energy $\calM$ of the gravitational
field and a very nontrivial theorem states that
$\calM$ is positive unless the metric is not flat.
Four constraints $C_0$, $C_k$ are of the first class;
let $\eta_k(x)$, $\eta_0(x)$ be the vector field and the
function on the space slice.
We put
\begin{eqnarray*}
C_0(\eta_0) &=& \int C_0(\Pi(x),q(x))\eta_0(x)d^3 x\\
C(\eta) &=& \int C_k(\Pi(x),q(x))\eta^k(x)d^3x
\end{eqnarray*}
to get the brackets
$$
\{C(\eta''),C(\eta')\}=
C([\eta^n,\eta'])
$$
where $[\quad,\quad]$ is the commutator of vector
fields.
This means that $C(\eta)$ are generators of
diffeomorphisms of our space slice.
Furthermore
$$
\{C(\eta),C_0(\eta_0)\}=
C_0(\eta\eta_0)
$$
where $\eta\eta_n$ is the result of the action of the vector
field $\eta$ on the function $\eta_0$.
Finally
$$
\{C_0(\eta''_0),C_0(\eta'_0)\}=
C(\hat{\eta}),
$$
where the vector field $\hat{\eta}$ is given by
$$
\hat{\eta}^n=q^{in}(\partial_i
\eta''_0\eta'_0-\eta''_0\partial_i \,
\eta'_0)
$$
What about the Hamiltonian?
Taking into account the asymptotic condition, it is
more natural to consider
$\left(\frac{1}{h^{0n}}+1\right)$ rather than
$1/h^{nm}$ to be a lagrangian multiplier; thus, the
Hamiltonian
$$
H=C_0 + \partial_i \partial_n q^{in}
$$
differs from the constraint $C_{0}$ only by a term which is a
total divergence.
The explicit expression for $C_0$ shows that $H$ is
quadratic in $\Pi$ and first derivatives of $q$, like all
other Hamiltonians in previous examples.
The $4$ constraints kill $4$ degrees of freedom from the 6
in $q^{ik}$, so gravitons (like photons above) have two
polarizations.
The reduction above is associated with the names
of Dirac and Arnowitt-Deser-Misner [ADM].
The explicit forms of all expressions above in the
coordinates $q^{ik}$, $\Pi_{ik}$ can be found in my
survey \cite{F:grav}.
%% end of inserted text
\noindent{\bf Example 4: the Dirac equation}
In this lecture, lack of time prevents us from
treating the case of the Dirac equation,
in which the field is a spin $1/2$ fermion, and the Lagrangian is
first order in time derivatives from the beginning and is nondegenerate.
\begin{thebibliography}{99}
\bibitem{Dirac} P.A.M. Dirac, {\em Proc. Roy. Soc.} {\bf A 246},
326, 1958; P.A.M. Dirac, {\em Lectures on Quantum Mechanics}, New York
(Yeshiva University), 1964.
\bibitem{F:grav} L.D. Faddeev, Sov. Phys. Uspekhi {\bf 25} (1982) %poss 92?
130.
\bibitem{FS} L.D. Faddeev, A.A. Slavnov, {\em Gauge Fields: An Introduction
to Quantum Theory}, Addison-Wesley (Frontiers in Physics vol. 83), (second
edition), 1991.
\end{thebibliography}
\end{document}