General relativity
Dr. Gyula Bene
Department for Theoretical Physics, Loránd Eötvös University
Pázmány Péter sétány 1/A, 1117 Budapest
5. week
Action integral of gravity
Short review of classical field theory. Action, Lagrangian density,
Euler-Lagrange equation of motion, conservation theories. Indeterminacy of the
energy-momantum tensor. Action integral of gravity.
Short review of classical field theory
$q(x,y,z,t)$: field component (eg. component of electric field
strength, or component of metric tensor)
$\Lambda\left(q,q_{,i}\right)$
: Lagrangian density (depends on field components and their first derivatives
with respect to spatial coordinates and time). Here
$q_{,i}=\frac{\partial q}{\partial x^i}$
$$\begin{align}
S=\int \Lambda\left(q,q_{,i}\right)d\Omega
\end{align}$$
(
$d\Omega=c\;dV\;dt$
)
$$\begin{align}
\delta S=\int \left[\frac{\partial \Lambda}{\partial q}\delta q
+\frac{\partial \Lambda}{\partial q_{,i}}\delta q_{,i}\right]d\Omega=
\int \left[\frac{\partial \Lambda}{\partial q}\delta q
+\frac{\partial}{\partial x^i}\left(\frac{\partial \Lambda}{\partial
q_{,i}}\delta q\right)
-\delta q\frac{\partial}{\partial x^i}\frac{\partial \Lambda}{\partial
q_{,i}}
\right]d\Omega=0
\end{align}$$
Euler-Lagrange equation of motion:
$$\begin{align}
\frac{\partial}{\partial x^i}\frac{\partial \Lambda}{\partial
q_{,i}}- \frac{\partial \Lambda}{\partial q}=0
\end{align}$$
Energy and momentum conservation:
Since the Langrangian density does not depend explicitely on spatial coordinates and
time, we have
$$\begin{align}
\frac{\partial \Lambda}{\partial x^i}=\frac{\partial \Lambda}{\partial q}\frac{\partial q}{\partial x^i}
+ \frac{\partial \Lambda}{\partial q_{,k}}\frac{\partial q_{,k}}{\partial x^i}
\end{align}$$
Applying the equation of motion:
$$\begin{align}
\frac{\partial \Lambda}{\partial x^i}=\frac{\partial}{\partial x^k}\frac{\partial \Lambda}{\partial
q_{,k}}q_{,i}
+ \frac{\partial \Lambda}{\partial q_{,k}}q_{,k,i}=\frac{\partial}{\partial x^k}
\left(q_{,i}\frac{\partial \Lambda}{\partial
q_{,k}}\right)
\end{align}$$
Zero method:
$$\begin{align}
\frac{\partial}{\partial x^k}\left(q_{,i}\frac{\partial \Lambda}{\partial
q_{,k}}-\delta^k_i \Lambda\right)=0
\end{align}$$
Energy-momentum tensor:
$$\begin{align}
T^k_i=q_{,i}\frac{\partial \Lambda}{\partial
q_{,k}}-\delta^k_i \Lambda
\end{align}$$
Integrating continuity equation
$\frac{\partial T^k_i}{\partial x^k}=0$
from $t_1$ to $t_2$ over time:
$$\begin{align}
\frac{d P^i}{dt}=0
\end{align}$$
where
$$\begin{align}
P^i=\int T^{ik}dS_k
\end{align}$$
stands for the conserved four-momentum.
$T^{ik}$ is indetermined:
$$\begin{align}
T^{ik}+ \frac{\partial}{\partial x^l}\psi^{ikl}
\end{align}$$
also satisfies the continuity equation if
$\psi^{ikl}=-\psi^{ilk}$
.
This indeterminacy does not influence the conserved four-momentum, since
$$\begin{align}
\int \frac{\partial \psi^{ikl}}{\partial x^l}dS_k=
\frac{1}{2} \int \left(dS_k\frac{\partial \psi^{ikl}}{\partial x^l}-dS_l\frac{\partial \psi^{ikl}}{\partial x^k}\right)
=\frac{1}{2} \int \psi^{ikl}df^*_{kl}=0
\end{align}$$
where
$df^*_{kl}=\epsilon_{klmn}df^{mn}$
stands for the dual of the surface element
$df^{mn}=dx^{(1)m}dx^{(2)n}-dx^{(1)n}dx^{(2)m}$
.
The integrand disappears on the surface pushed to the infinity.
Angular momentum conservation (case of scalar field):
Four tensor of angular momentum:
$$\begin{align}
J^{ik}=\int\left(x^idP^k-x^kdP^i\right)= \int\left(x^iT^{kl}-x^kT^{il}\right)dS_l
\end{align}$$
Angular momentum conservation is equivalent with vanishing of divergence of
angular momentum density:
$$\begin{align}
0=J^{ik}(t_2)-J^{ik}(t_1)=\oint\left(x^iT^{kl}-x^kT^{il}\right)dS_l=\int\frac{\partial}{\partial x^l}\left(x^iT^{kl}-x^kT^{il}\right)d\Omega
\end{align}$$
$$\begin{align}
\rightarrow \quad
\frac{\partial}{\partial x^l}\left(x^iT^{kl}-x^kT^{il}\right)=0
\end{align}$$
But
$$\begin{align}
\frac{\partial}{\partial x^l}\left(x^iT^{kl}-x^kT^{il}\right) =
x^i\frac{\partial T^{kl}}{\partial x^l}-x^k\frac{\partial T^{il}}{\partial x^l}+\delta_l^iT^{kl}
-\delta_l^kT^{il}=T^{ki}-T^{ik}
\end{align}$$
hence angular momentum conservation implies symmetry of $T^{ki}$.
Prelimiaries: some identities
Derivative of a determinant:
$$\begin{align}
\frac{\partial g}{\partial x^k}=\frac{\partial }{\partial x^k}\epsilon^{i_0i_1i_2i_3} g_{0i_0}g_{1i_1}g_{2i_2}g_{3i_3}=\epsilon^{i_0i_1i_2i_3} \frac{\partial g_{0i_0}}{\partial x^k}g_{1i_1}g_{2i_2}g_{3i_3}+...
\end{align}$$
Coefficient of
$\frac{\partial g_{0i_0}}{\partial x^k}$
is
$$\begin{align}
\epsilon^{i_0i_1i_2i_3} g_{1i_1}g_{2i_2}g_{3i_3}\;,
\end{align}$$
the signed subdeterminant which belongs to row 0 and column $i_0$, -ik
oszlophoz tartozó előjeles aldetermináns, i.e., $g\;g^{0i_0}$. (Similarly for
other terms.)
Thus
$$\begin{align}
\frac{\partial g}{\partial x^k}=g\;g^{ij}\;\frac{\partial g_{ij}}{\partial x^k}\;.
\end{align}$$
Since
$g_{ij}g^{ij}=\delta^j_j=4$
,
$$\begin{align}
g^{ij} \frac{\partial g_{ij}}{\partial x^k} =-g_{ij}\frac{\partial g^{ij}}{\partial x^k}
\end{align}$$
Applying the definition of Christoffel's symbols, namely,
$$\begin{align}
\Gamma^i_{kl}= \frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{ml}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m}\right)
\end{align}$$
we have
$$\begin{align}
\Gamma^i_{ki}= \frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^i}+\frac{\partial g_{mi}}{\partial x^k}-\frac{\partial g_{ki}}{\partial x^m}\right)=\frac{1}{2}g^{im}\frac{\partial g_{mi}}{\partial x^k}=\frac{1}{2g}\frac{\partial g}{\partial x^k}
\end{align}$$
Covariant divergence of a vector:
$$\begin{align}
A^i_{;i}\equiv \frac{D A^i}{Dx^i}= \frac{\partial A^i}{\partial x^i}+\Gamma^i_{ki}A^k=\frac{\partial A^i}{\partial x^i}
+\frac{1}{2g}\frac{\partial g}{\partial x^k}A^k=\frac{1}{ \sqrt{-g}}\frac{\partial \left(\sqrt{-g}\;A^i\right)}{\partial x^i}
\end{align}$$
Action integral of gravity
Stricly speaking, the laws of gravitational field cannot be deduced from other
branches of physics. One may apply just analogies: one expects second order
partial differential equations (i.e., at most first derivatives in the
Lagrangian density), the Lagrangian density is a scalar, field components are
the components of the metric tensor.
Difficulty: no scalar can be compiled from the firs derivatives of the metric
(Christoffel's symbols). The only scalar at hand is the Ricci scalar
(${\mathcal R}$) . This, however, contains second derivatives as well.
Solution: since ${\mathcal R}$ is linear in the second derivatives, we show that
$$\begin{align}
\int {\mathcal R}\sqrt{-g}d\Omega= \int G\sqrt{-g}d\Omega+ \int \frac{\partial \left(\sqrt{-g}\;w^i\right)}{\partial x^i}d\Omega
\end{align}$$
where $G$ contains only first derivatives. Note the $w^i$ does not transform
as a vector.
Find second derivatives in ${\mathcal R}\sqrt{-g}$:
$$\begin{align}
{\mathcal R}\sqrt{-g}=
\sqrt{-g}\;g^{ki}\;R^m_{kmi}=\sqrt{-g}\;g^{ki}\;\left(
{\underline{ \frac{\partial \Gamma^m_{ki} }{\partial x^m}
-\frac{\partial \Gamma^m_{km} }{\partial x^i}}}
+\Gamma^m_{nm}\Gamma^n_{ki}-\Gamma^m_{ni}\Gamma^n_{km}
\right)
\end{align}$$
Second derivatives appear only in the derivatives of Christoffel1s symbols:
$$\begin{align}
\sqrt{-g}\left(g^{km}\;\frac{\partial \Gamma^i_{km} }{\partial x^i}
-g^{ki}\;\frac{\partial \Gamma^m_{km} }{\partial x^i}\right)
=\frac{\partial }{\partial x^i}\left(\sqrt{-g}\left[g^{km}\;\Gamma^i_{km}-g^{ki}\;\Gamma^m_{km}\right]\right)
\end{align}$$
$$\begin{align}
-\left(\Gamma^i_{km}\frac{\partial \left(\sqrt{-g}\;g^{km}\right)}{\partial x^i}-\Gamma^m_{km}\frac{\partial \left(\sqrt{-g}\;g^{ki}
\right)}{\partial x^i}\right)
\end{align}$$
Thus
$$\begin{align}
w^i=g^{km}\;\Gamma^i_{km}-g^{ki}\;\Gamma^m_{km}
\end{align}$$
and
$$\begin{align}
G=g^{ki}\;\left(\Gamma^m_{nm}\Gamma^n_{ki}-\Gamma^m_{ni}\Gamma^n_{km}
\right)
+\frac{\Gamma^m_{km}}{\sqrt{-g}}\frac{\partial \left(\sqrt{-g}\;g^{ki}
\right)}{\partial x^i}-\frac{\Gamma^i_{km}}{\sqrt{-g}}\frac{\partial \left(\sqrt{-g}\;g^{km}\right)}{\partial x^i}
\end{align}$$
Expressing derivatives of the metric tensor in terms of Christoffel's symbols ($g^{ik}_{;l}=0$):
$$\begin{align}
\frac{\partial g^{im}}{\partial x^l}=-\Gamma^i_{kl}\;g^{km}-\Gamma^m_{kl}\;g^{ik}
\end{align}$$
Aplying this, we get:
$$\begin{align}
G=g^{ki}\;\left(\Gamma^m_{nm}\Gamma^n_{ki}-\Gamma^m_{ni}\Gamma^n_{km}
\right)
+\Gamma^m_{km}\;\Gamma^n_{in}\;g^{ki}-\Gamma^m_{km}\;\Gamma^i_{ni}\;g^{nk}-\Gamma^m_{km}\;\Gamma^k_{ni}\;g^{in}\end{align}$$$$\begin{align}
{\underline{-\Gamma^i_{km}\;\Gamma^n_{in}\;g^{km}}}+\Gamma^i_{km}\;\Gamma^k_{ni}\;g^{nm}{\underline{+\Gamma^i_{km}\;\Gamma^m_{ni}\;g^{kn}}}
\end{align}$$
Hence
$$\begin{align}
G=g^{ki}\;\left(\Gamma^m_{ni}\Gamma^n_{km}-\Gamma^m_{nm}\Gamma^n_{ki}
\right)
\end{align}$$
Lagrangian density:
$$\begin{align}
-\frac{c^3}{16\pi k}{\mathcal R}\sqrt{-g}
\end{align}$$
Minus sign ensures that the action integral be positive definite.
$k=6.67\times 10^{-11} \frac{{\rm m^3}}{{\rm kg s^2}}$
stands for the gravity constant.
bene@arpad.elte.hu