General relativity
Dr. Gyula Bene
Department for Theoretical Physics, Loránd Eötvös University
Pázmány Péter sétány 1/A, 1117 Budapest
4. week
Riemannian (curvature tensor)
Parallel displacement of a vector along a closed curve. Curvature
tensor. Symmetries of the Riemannian. Bianchi identity. Weil tensor, Ricci
tensor, scalar curvature. Example: calculating the Riemannian on a curved two
dimensional surface. Number of independent tensor components in two, three and
four dimensions.
Curvature tensor
Problem: what local quantity signals that spacetime is curved?
Parallel displacement of a vector: components in a locally Minkowskian
spacetime are unchanged. In curved spacetime:
$$\begin{align}DA^i=0\end{align}$$
During parallel displacement along a geodesics ($Du^i=0$)
the angle made with the tangent of the trajectory ($u^i$)is constant.
Parallel displacement along a closed curve:
|
Fig. 4.1. On a curved surface parallel
displacement along a closed curve does not reproduce the original vector.
$$\begin{align}
\Delta A_k=\oint \Gamma^i_{\phantom{1}kl}A_i dx^l
\end{align}$$
$$\begin{align}
\frac{\partial A_i}{\partial x^l}=\Gamma^n_{\phantom{1}il}A_n
\end{align}$$
Stokes's theorem:
$$\begin{align}
\oint A_i\;dx^i=\int df^{ki}\frac{\partial A^i}{\partial x^k}
=
\frac{1}{2}\int df^{ki}\left(\frac{\partial A^i}{\partial x^k}-\frac{\partial A^k}{\partial x^i}\right)
\end{align}$$
where
$$\begin{align}df^{ki}=dx^{(1) k}dx^{(2) i}-dx^{(1) i}dx^{(2) k}\end{align}$$
stands for the surface of the parallelogram spanned by the vectors $dx^{(1) i}$ and $dx^{(2) i}$.
$$\begin{align}
\Delta A_k=\frac{1}{2}\left[
\frac{\partial \left(\Gamma^i_{\phantom{1}km} A_i\right)}{\partial x^l}-
\frac{\partial \left(\Gamma^i_{\phantom{1}kl} A_i\right)}{\partial x^m}
\right]\Delta f^{lm}
\end{align}$$
$$\begin{align}
\Delta A_k=\frac{1}{2}{\mathcal R}^i_{\phantom{1}klm}A_i \Delta f^{lm}
\end{align}$$
$$\begin{align}
{\mathcal R}^i_{\phantom{1}klm}=
\frac{\partial \Gamma^i_{\phantom{1}km} }{\partial x^l}
-\frac{\partial \Gamma^i_{\phantom{1}kl} }{\partial x^m}
+\Gamma^i_{\phantom{1}nl}\Gamma^n_{\phantom{1}km}-\Gamma^i_{\phantom{1}nm}\Gamma^n_{\phantom{1}kl}
\end{align}$$
Expressing the Riemannian in terms of Christoffel symbols:
$$\begin{align}
{\mathcal R}^i_{\phantom{1}klm}A_i&=
\frac{\partial \left(\Gamma^i_{\phantom{1}km} A_i\right)}{\partial x^l}-
\frac{\partial \left(\Gamma^i_{\phantom{1}kl} A_i\right)}{\partial x^m} \\
&=\left(\frac{\partial \Gamma^i_{\phantom{1}km}}{\partial x^l}-
\frac{\partial \Gamma^i_{\phantom{1}kl} }{\partial x^m}\right) A_i
+\Gamma^i_{\phantom{1}km} \Gamma^n_{\phantom{1}il}A_n- \Gamma^i_{\phantom{1}kl}\Gamma^n_{\phantom{1}im}A_n \\
&=\left(\frac{\partial \Gamma^i_{\phantom{1}km}}{\partial x^l}-
\frac{\partial \Gamma^i_{\phantom{1}kl} }{\partial x^m}
+\Gamma^n_{\phantom{1}km} \Gamma^i_{\phantom{1}nl}- \Gamma^n_{\phantom{1}kl}\Gamma^i_{\phantom{1}nm}
\right) A_i
\end{align}$$
We made use that
$$\begin{align}
\frac{\partial A_i}{\partial x^l}=\Gamma^n_{\phantom{1}il}A_n
\end{align}$$
The derivation is valid for an arbitrary vector $A_i$, hence
$$\begin{align}{\mathcal R}^i_{\phantom{1}klm}=\frac{\partial \Gamma^i_{\phantom{1}km}}{\partial x^l}-
\frac{\partial \Gamma^i_{\phantom{1}kl} }{\partial x^m}
+\Gamma^n_{\phantom{1}km} \Gamma^i_{\phantom{1}nl}- \Gamma^n_{\phantom{1}kl}\Gamma^i_{\phantom{1}nm}\end{align}$$
Properties of the Riemannian
Change of a contravariant vector along an infinitesimally small closed surface
$$\begin{align}
\Delta A^k=-\frac{1}{2}{\mathcal R}^k_{\phantom{1} ilm}A^i \Delta f^{lm}
\end{align}$$
Proof:
$$\begin{align}\Delta \left(A^kB_k\right)=0\end{align}$$
$$\begin{align}
\Delta \left(A^kB_k\right)&=B_k\Delta A^k+A^k\Delta B_k \\
&=B_k\Delta A^k+A^k\frac{1}{2}{\mathcal R}^i_{\phantom{1}klm}B_i \Delta f^{lm} \\
&=B_k\left(\Delta A^k+A^i\frac{1}{2}{\mathcal R}^k_{\phantom{1}ilm} \Delta f^{lm}\right)
\end{align}$$
Since $B_k$ is arbitrary, we have
$$\begin{align}
\Delta A^k=-\frac{1}{2}{\mathcal R}^k_{\phantom{1}ilm}A^i \Delta f^{lm}\end{align}$$
Q.E.D.
Difference of second covariant derivatives taken in different orderings is
proportional to the Riemannian I.
$$\begin{align}
A_{i;k;l}-A_{i;l;k}=A_m {\mathcal R}^m_{\phantom{1}ikl}
\end{align}$$
Proof:
$$\begin{align}
A_{i;k;l}&=A_{i;k,l}-\Gamma^n_{\phantom{1}il}A_{n;k}-\Gamma^n_{\phantom{1}kl}A_{i;n} \\
&=A_{i,k,l}-\frac{\partial }{\partial
x^l}\left(\Gamma^n_{\phantom{1}ik}A_{n}\right)-\Gamma^n_{\phantom{1}il}A_{n,k}+\Gamma^n_{\phantom{1}il}\Gamma^m_{\phantom{1}nk}A_m-\Gamma^n_{\phantom{1}kl}A_{i,n}+\Gamma^n_{\phantom{1}kl}\Gamma^m_{\phantom{1}in}A_m \\
&=A_{i,k,l}-\frac{\partial \Gamma^n_{\phantom{1}ik}}{\partial
x^l}A_{n}-\Gamma^n_{\phantom{1}ik}A_{n,l}
-\Gamma^n_{\phantom{1}il}A_{n,k}-\Gamma^n_{\phantom{1}kl}A_{i,n}
+\Gamma^n_{\phantom{1}il}\Gamma^m_{\phantom{1}nk}A_m+\Gamma^n_{\phantom{1}kl}\Gamma^m_{\phantom{1}in}A_m\\
A_{i;l;k}&=A_{i,l,k}-\frac{\partial \Gamma^n_{\phantom{1}il}}{\partial
x^k}A_{n}-\Gamma^n_{\phantom{1}il}A_{n,k}
-\Gamma^n_{\phantom{1}ik}A_{n,l}-\Gamma^n_{\phantom{1}lk}A_{i,n}
+\Gamma^n_{\phantom{1}ik}\Gamma^m_{\phantom{1}nl}A_m+\Gamma^n_{\phantom{1}lk}\Gamma^m_{\phantom{1}in}A_m \\
A_{i;k;l}-A_{i;l;k}&=\left(\frac{\partial \Gamma^m_{\phantom{1}il}}{\partial
x^k}-\frac{\partial \Gamma^m_{\phantom{1}ik}}{\partial
x^l}+\Gamma^n_{\phantom{1}il}\Gamma^m_{\phantom{1}nk}-\Gamma^n_{\phantom{1}ik}\Gamma^m_{\phantom{1}nl}\right)A_m \\
&={\mathcal R}^m_{\phantom{1}ikl}A_m\end{align}$$
Q.E.D.
Difference of second covariant derivatives taken in different orderings is
proportional to the Riemannian II.
$$\begin{align}
A^i_{;k;l}-A^i_{;l;k}=-A^m {\mathcal R}^i_{\phantom{1}mkl}
\end{align}$$
Proof:
$$\begin{align}A^i_{;k;l}&=A^i_{;k,l}+\Gamma^i_{\phantom{1}nl}A^n_{;k}-\Gamma^n_{\phantom{1}kl}A^i_{;n} \\
&=A^i_{,k,l}+\frac{\partial }{\partial
x^l}\left(\Gamma^i_{\phantom{1}nk}A^{n}\right)+\Gamma^i_{\phantom{1}nl}A^n_{,k}+\Gamma^i_{\phantom{1}nl}\Gamma^n_{\phantom{1}mk}A^m-\Gamma^n_{\phantom{1}kl}A^i_{,n}-\Gamma^n_{\phantom{1}kl}\Gamma^i_{\phantom{1}mn}A^m \\
&=A^i_{,k,l}+\frac{\partial \Gamma^i_{\phantom{1}nk}}{\partial
x^l}A^{n}+\Gamma^i_{\phantom{1}nk}A^n_{,l}+\Gamma^i_{\phantom{1}nl}A^n_{,k}+\Gamma^i_{\phantom{1}nl}\Gamma^n_{\phantom{1}mk}A^m-\Gamma^n_{\phantom{1}kl}A^i_{,n}-\Gamma^n_{\phantom{1}kl}\Gamma^i_{\phantom{1}mn}A^m\\
A^i_{;l;k}&=A^i_{,l,k}+\frac{\partial \Gamma^i_{\phantom{1}nl}}{\partial
x^k}A^{n}+\Gamma^i_{\phantom{1}nl}A^n_{,k}+\Gamma^i_{\phantom{1}nk}A^n_{,l}+\Gamma^i_{\phantom{1}nk}\Gamma^n_{\phantom{1}ml}A^m-\Gamma^n_{\phantom{1}lk}A^i_{,n}-\Gamma^n_{\phantom{1}lk}\Gamma^i_{\phantom{1}mn}A^m\end{align}$$
$$\begin{align}
A^i_{;k;l}-A^i_{;l;k}&=-\left(\frac{\partial \Gamma^i_{\phantom{1}ml}}{\partial
x^k}-\frac{\partial \Gamma^i_{\phantom{1}mk}}{\partial
x^l}+\Gamma^i_{\phantom{1}nk}\Gamma^n_{\phantom{1}ml}
-\Gamma^i_{\phantom{1}nl}\Gamma^n_{\phantom{1}mk}
\right)A^m \\
&=-{\mathcal R}^i_{\phantom{1}mkl}A^m\end{align}$$
Q.E.D.
Symmetries of the Riemannian
$$\begin{align}
{\mathcal R}_{iklm}=g_{in}{\mathcal R}^n_{\phantom{1}klm}
&=
\frac{1}{2}\left(\frac{\partial^2 g_{im}}{\partial x^k \partial x^l}
+\frac{\partial^2 g_{kl}}{\partial x^i \partial x^m}
-\frac{\partial^2 g_{il}}{\partial x^k \partial x^m}
-\frac{\partial^2 g_{km}}{\partial x^i \partial x^l}
\right) \\
&+g_{np}\left(\Gamma^n_{\phantom{1}kl}\Gamma^p_{\phantom{1}im}-\Gamma^n_{\phantom{1}km}\Gamma^p_{\phantom{1}il}\right)\phantom{riemann_kov}
\end{align}$$
Proof:
$$\begin{align}{\mathcal R}_{iklm}&=g_{in}{\mathcal R}^n_{\phantom{1}klm}=
g_{in}\left(\frac{\partial \Gamma^n_{\phantom{1}km} }{\partial x^l}
-\frac{\partial \Gamma^n_{\phantom{1}kl} }{\partial x^m}
+\Gamma^n_{\phantom{1}pl}\Gamma^p_{\phantom{1}km}-\Gamma^n_{\phantom{1}pm}\Gamma^p_{\phantom{1}kl}\right) \\
&=\frac{\partial \Gamma_{ikm} }{\partial x^l}
-\frac{\partial \Gamma_{ikl} }{\partial x^m}-\Gamma^n_{\phantom{1}km}g_{in,l}+
\Gamma^n_{\phantom{1}kl}g_{in,m}+g_{in}\left(\Gamma^n_{\phantom{1}pl}\Gamma^p_{\phantom{1}km}-\Gamma^n_{\phantom{1}pm}\Gamma^p_{\phantom{1}kl}\right)\end{align}$$
$$\begin{align}
\Gamma_{ikm}&=\frac{1}{2}\left(g_{ik,m}+g_{im,k}-g_{km,i}\right)\\
\Gamma_{ikl}&=\frac{1}{2}\left(g_{ik,l}+g_{il,k}-g_{kl,i}\right)\\
g_{in,l}&=\Gamma^p_{\phantom{1}il}g_{pn}+\Gamma^p_{\phantom{1}nl}g_{ip}\\
g_{in,m}&=\Gamma^p_{\phantom{1}im}g_{pn}+\Gamma^p_{\phantom{1}nm}g_{ip}\end{align}$$
$$\begin{align}
{\mathcal R}_{iklm}&=\frac{1}{2}\left(g_{ik,ml}+g_{im,kl}-g_{km,il}-g_{ik,lm}-g_{il,km}+g_{kl,im}\right) \\
&-\Gamma^n_{\phantom{1}km}\Gamma^p_{\phantom{1}il}g_{pn}-\Gamma^n_{\phantom{1}km}\Gamma^p_{\phantom{1}nl}g_{ip}+\Gamma^n_{\phantom{1}kl}\Gamma^p_{\phantom{1}im}g_{pn}+\Gamma^n_{\phantom{1}kl}\Gamma^p_{\phantom{1}nm}g_{ip} \\
&+g_{in}\left(\Gamma^n_{\phantom{1}pl}\Gamma^p_{\phantom{1}km}-\Gamma^n_{\phantom{1}pm}\Gamma^p_{\phantom{1}kl}\right)\end{align}$$
I.e.
$$\begin{align}
{\mathcal R}_{iklm}&=\frac{1}{2}\left(\frac{\partial^2 g_{im}}{\partial x^k\partial x^l}+
\frac{\partial^2 g_{kl}}{\partial x^i\partial x^m}-\frac{\partial^2 g_{km}}{\partial x^i\partial x^l}-\frac{\partial^2 g_{il}}{\partial x^k\partial x^m}\right) \\
&+g_{pn}\left(\Gamma^n_{\phantom{1}kl}\Gamma^p_{\phantom{1}im}
-\Gamma^n_{\phantom{1}km}\Gamma^p_{\phantom{1}il}\right)\end{align}$$
Q.E.D.
Antisymmetry with respect to the exchange of the first and second, or the
third and fourth indices
$$\begin{align}
{\mathcal R}_{iklm}=-{\mathcal R}_{kilm}=-{\mathcal R}_{ikml}
\end{align}$$
Symmetry with respect to the exchange of the first and second paires of indices
$$\begin{align}
{\mathcal R}_{iklm}={\mathcal R}_{lmik}
\end{align}$$
Cyclic sum of any three indices vanishes
$$\begin{align}
{\mathcal R}_{iklm}+{\mathcal R}_{imkl}+{\mathcal R}_{ilmk}=0
\end{align}$$
Proof:
$$\begin{align}{\mathcal R}_{iklm}+{\mathcal R}_{imkl}+{\mathcal R}_{ilmk}&=
\frac{1}{2}\left(\frac{\partial^2 g_{im}}{\partial x^k\partial x^l}+
\frac{\partial^2 g_{kl}}{\partial x^i\partial x^m}-\frac{\partial^2
g_{km}}{\partial x^i\partial x^l}-\frac{\partial^2 g_{il}}{\partial x^k\partial
x^m}\right) \\
&+\frac{1}{2}\left(\frac{\partial^2 g_{il}}{\partial x^m\partial x^k}+
\frac{\partial^2 g_{mk}}{\partial x^i\partial x^l}-\frac{\partial^2
g_{ml}}{\partial x^i\partial x^k}-\frac{\partial^2 g_{ik}}{\partial x^m\partial
x^l}\right) \\
&+\frac{1}{2}\left(\frac{\partial^2 g_{ik}}{\partial x^l\partial x^m}+
\frac{\partial^2 g_{lm}}{\partial x^i\partial x^k}-\frac{\partial^2
g_{lk}}{\partial x^i\partial x^m}-\frac{\partial^2 g_{im}}{\partial x^l\partial
x^k}\right) \\
&+g_{pn}\left(\Gamma^n_{\phantom{1}kl}\Gamma^p_{\phantom{1}im}
-\Gamma^n_{\phantom{1}km}\Gamma^p_{\phantom{1}il}\right) \\
&+g_{pn}\left(\Gamma^n_{\phantom{1}mk}\Gamma^p_{\phantom{1}il}
-\Gamma^n_{\phantom{1}ml}\Gamma^p_{\phantom{1}ik}\right) \\
&+g_{pn}\left(\Gamma^n_{\phantom{1}lm}\Gamma^p_{\phantom{1}ik}
-\Gamma^n_{\phantom{1}lk}\Gamma^p_{\phantom{1}im}\right)=0\end{align}$$
Q.E.D.
Ricci tensor
$$\begin{align}
{\mathcal R}_{ik}=g^{lm}{\mathcal R}_{limk}={\mathcal R}^m_{\phantom{1}imk}
\end{align}$$
$$\begin{align}
{\mathcal R}_{ik}={\mathcal R}_{ki}
\end{align}$$
Invariant curvature (Ricci scalar)
$$\begin{align}
{\mathcal R}=g^{ik}{\mathcal R}_{ik}
\end{align}$$
Two dimensional case
$$\begin{align}
{\mathcal R}=\frac{2{\mathcal R}_{1212}}{\gamma}
\end{align}$$
$$\begin{align}
\frac{{\mathcal R}}{2}=K=\frac{1}{\rho_1\rho_2}
\end{align}$$
where $\rho_1$ and $\rho_2$ stand for the principal curvature radii.
Proof:
Choose a Cartesian coordinate system at a point $P$ of a two dimensional surface,
embedded into a three dimensional Eucledian frame so that
- Point $P$ is the origin
- Axis $z$ is normal to the surface
- Planes $xz$ and $yz$ are the planes of principal curvatures.
The equation of the surface in a small neighborhood of the origin:
$$\begin{align}
z=\frac{x^2}{2\rho_1}+\frac{y^2}{2\rho_2}\;.
\end{align}$$
Surface metric is derived from the distance of two nearby surface points:
$$\begin{align}
ds^2&=dx^2+dy^2+dz^2=dx^2+dy^2+\left(\frac{xdx}{\rho_1}+\frac{ydy}{\rho_2}\right)^2 \\
&=\left(1+\frac{x^2}{\rho_1^2}\right)dx^2+\left(1+\frac{y^2}{\rho_2^2}\right)dy^2+2\frac{xy}{\rho_1\rho_2}dxdy\;,
\end{align}$$
thus
$$\begin{align}
g_{11}&=1+\frac{x^2}{\rho_1^2}\;,\\
g_{12}&=g_{21}=\frac{xy}{\rho_1\rho_2}\;,\\
g_{22}&=1+\frac{y^2}{\rho_2^2}
\;.
\end{align}$$
First derivatives vanish at the origin, and the metric is the unit matrix
there. Then we have
$$\begin{align}
{\mathcal R}_{1212}(0)=\frac{\partial^2 g_{12}}{\partial x \partial y}-\frac{1}{2}\frac{\partial^2 g_{11}}{\partial y^2}-\frac{1}{2}\frac{\partial^2 g_{22}}{\partial x^2}=\frac{1}{\rho_1\rho_2}\;.
\end{align}$$
Q.E.D.
Bianchi's identity
$$\begin{align}
{\mathcal R}^n_{\phantom{1}ikl;m}+{\mathcal R}^n_{\phantom{1}imk;l}+{\mathcal R}^n_{\phantom{1}ilm;k}=0\phantom{bian1}
\end{align}$$
Proof:
$$\begin{align}
{\mathcal R}^n_{\phantom{1}ikl;m}+{\mathcal R}^n_{\phantom{1}imk;l}+{\mathcal R}^n_{\phantom{1}ilm;k}=\frac{1}{2}{\mathcal R}^n_{\phantom{1}ikl;m}E^{klmp}\sqrt{-g}\end{align}$$
where $$\begin{align}E^{klmp}=\frac{1}{\sqrt{-g}}e^{klmp}\end{align}$$
The cyclic sum multiplied by $1/\sqrt{-g}$ is a tensor. Switching to a locally
geodetic (Minkowskian) frame we have
$$\begin{align}
{\mathcal R}^n_{\phantom{1}ikl;m}=\frac{\partial {\mathcal R}^n_{\phantom{1}ikl}}{\partial x^m}
=\frac{\partial^2 \Gamma^n_{\phantom{1}il}}{\partial x^m \partial x^k}
-\frac{\partial^2 \Gamma^n_{\phantom{1}ik}}{\partial x^m \partial x^l}\end{align}$$
This implies
$$\begin{align}
{\mathcal R}^n_{\phantom{1}ikl;m}+{\mathcal R}^n_{\phantom{1}imk;l}+{\mathcal R}^n_{\phantom{1}ilm;k}
&=\frac{\partial^2 \Gamma^n_{\phantom{1}il}}{\partial x^m \partial x^k}
-\frac{\partial^2 \Gamma^n_{\phantom{1}ik}}{\partial x^m \partial x^l} \\
&+\frac{\partial^2 \Gamma^n_{\phantom{1}ik}}{\partial x^l \partial x^m}
-\frac{\partial^2 \Gamma^n_{\phantom{1}im}}{\partial x^l \partial x^k} \\
&+\frac{\partial^2 \Gamma^n_{\phantom{1}im}}{\partial x^k \partial x^l}
-\frac{\partial^2 \Gamma^n_{\phantom{1}il}}{\partial x^k \partial
x^m}=0\end{align}$$
Q.E.D.
Bianchi's identity implies by contracting indices $i$ with $k$ and $n$ with $l$:
$$\begin{align}
0=g^{ik}\left({\mathcal R}^l_{\phantom{1}ikl;m}+{\mathcal R}^l_{\phantom{1}imk;l}+{\mathcal R}^l_{\phantom{1}ilm;k}\right)=-{\mathcal R}_{,m}+2{\mathcal R}^l_{\phantom{1}m;l}
\end{align}$$
or
$$\begin{align}{\mathcal R}^l_{\phantom{1}m;l}=\frac{1}{2}\frac{\partial {\mathcal R}}{\partial x^m}\phantom{bian2}\end{align}$$
Number of components of the Riemannian
Two dimensional case:
a single independent component, eg. ${\mathcal R}_{1212}$.
Three dimensional case:
${\mathcal R}_{\alpha\beta\gamma\delta}$'s
first ($\alpha\beta$) and second
($\gamma\delta$) pairs of indices take on three values each, hence the number
of independent components is the same as that of a symmetric $3\times 3$
matrix, that is 6. (Cyclic sum vanishes automatically, without implying any further restrictions.)
Four dimensional case:
${\mathcal R}_{iklm}$'s first ($ik$) and second
($lm$) pairs of indices take on six values each. The number of independent
components of a $6\times 6$ symmetric matrix is 21. Cyclic sum vanishes
automatically, except when all the four indices are different. In that latter
case we have a single further algebraic restriction among components, reducing
the number of independent components to 20.
Weil's tensor:
$$\begin{align}C_{iklm}={\mathcal R}_{iklm}-\frac{1}{2}{\mathcal R}_{il}g_{km}+\frac{1}{2}{\mathcal R}_{im}g_{kl}+\frac{1}{2}{\mathcal R}_{kl}g_{im}-\frac{1}{2}{\mathcal R}_{km}g_{il}+\frac{1}{6}{\mathcal R}\left(g_{il}g_{km}-g_{im}g_{kl}\right)\end{align}$$
It possesses all the algebraic symmetries of the Riemannian, but contracting
any two indices we get zero (irreducible tensor).
bene@arpad.elte.hu