General relativity
Dr. Gyula Bene
Department for Theoretical Physics, Loránd Eötvös University
Pázmány Péter sétány 1/A, 1117 Budapest
2. week
Covariant differentiation
In case of general coordinate transformations the transformation rule of
vectors is different at different points. Therefore, the derivative of a
vector field, being the difference between vectors at two different points,
does not transform as a tensor. Indeed, for a contravariant vector component
we have
$$A'^i=\frac{\partial x'^i}{\partial x^k}A^k$$
and
$$\frac{\partial A'^i}{\partial x'^j}=\frac{\partial x'^i}{\partial
x^k}\frac{\partial x^m}{\partial
x'^j}\frac{\partial A^k}{\partial x^m} + \frac{\partial^2 x'^i}{\partial x^k\partial x^m}\frac{\partial x^m}{\partial x'^j}A^k\;.$$
The extra term is due to the coordinate dependence of the transformation
rule.
In order to obtain a generalization of the derivative which transforms as a
tensor, we have to substract vectors at the same point. Hence, we have to
parallel displace the vector $A^i(x^j)$ to $x^j+dx^j$ and then substract it
from the vector $A^i(x^j+dx^j)$.
Parallel displacement
Thus we are led to the problem of parallel displacement of vectors. In order to
solve it, we extend the method already applied: we define the transformation
from a curvilinear system of coordinates to an inertial (Minkowsky) frame in a small
neighborhood of a point. Then we perform a tiny parallel displacement in the Minkowsky frame
where components remain unchanged. At the end point of the displacement we
switch back to the curvilinear frame.
Suppose the origins of the curvilinear and the Minkowsky frame coincide. Then
we have in a small neighbourhood
$$\begin{align}
x'^i=A^i_jx^j+\frac{1}{2}B^i_{jk}x^jx^k+\mathcal{O}\left(\left(x^j\right)^3\right)\phantom{ph1}
\end{align}$$
Here unprimed coordinates $x^j$ stand for curvilinear coordinates while primed
coordinates $x'^i$ are coordinates of the inertial frame. Clearly,
the constant coefficients $B^i_{jk}$ are symmetric in their lower
indices.
For coordinate differentials we get
$$\begin{align}
dx'^i=A^i_jdx^j+B^i_{jk}x^jdx^k+\mathcal{O}\left(\left(x^j\right)^2\right)\;.\phantom{dtransz1}
\end{align}$$
This implies that the corresponding interval is
$$\begin{align}
ds^2=\left(dx'^0\right)^2-\left(dx'^\alpha\right)^2&=\left(A^0_j A^0_k+A^0_jB^0_{nk}x^n+A^0_kB^0_{nj}x^n\right)
dx^jdx^k\\&-\left(A^\alpha_j
A^\alpha_k+A^\alpha_jB^\alpha_{nk}x^n+A^\alpha_kB^\alpha_{nj}x^n \right)dx^j
dx^k\\&=\left(g_{jk}(0)+\frac{\partial g_{jk}}{\partial x^n}x^n\right)dx^jdx^k\;.
\end{align}$$
The last expression contains the Taylor expansion of the metric to first
order. Then, beyond the previous (zeroth order) relations, we also have
$$\begin{align}
\frac{\partial g_{jk}}{\partial x^n}=A^0_jB^0_{nk}+A^0_kB^0_{nj}-A^\alpha_jB^\alpha_{nk}-A^\alpha_kB^\alpha_{nj}\;.
\end{align}$$
The number of equations equals to the number of unknowns, that is 40.
In order to determine coefficients $B^i_{jk}$ let us introduce the quantity
$$\begin{align}
b_{jnk}=A^0_jB^0_{nk}-A^\alpha_jB^\alpha_{nk}\;,\phantom{ph2}
\end{align}$$
which is symmetric in indices $n$, $k$. We may write then
$$\begin{align}
\frac{\partial g_{jk}}{\partial x^n}&=b_{jnk}+b_{knj}\\
\frac{\partial g_{kn}}{\partial x^j}&=b_{kjn}+b_{njk}\\
\frac{\partial g_{nj}}{\partial x^k}&=b_{nkj}+b_{jkn}
\end{align}$$
Here the second and third lines follow from the first one via a cyclic
permutation of indices. Substracting the third equation from the sum of the
first and the second ones, we have
$$\begin{align}
b_{kjn}=\frac{1}{2}\left(\frac{\partial g_{jk}}{\partial x^n}+\frac{\partial g_{kn}}{\partial x^j}-\frac{\partial g_{nj}}{\partial x^k}\right)\;.\phantom{ph3}
\end{align}$$
Multiplying this with the inverse matrix of $A^i_j$ we get the coefficients
seeked for.
Parallel displacement of $dx^i$ is performed in three steps:
- Transformation at $-x^i$ to the inertial frame
$$\begin{align}
dx'^i=A^i_jdx^j-B^i_{jk}x^jdx^k+\mathcal{O}\left(\left(x^j\right)^2\right)\;.\phantom{dtransz1}
\end{align}$$
- Parallel displacement in the inertial frame to the origin ( $dx'^i$ unchanged)
- Inverse transform at the origin back to the curvilinear coordinate system
$$\begin{align}
dx'^i=A^i_jd\tilde{x}^j
\end{align}$$
$d\tilde{x}^j$ is the result of parallel displacement in curvilinear coordinates.
Since
$$\begin{align}
A^0_jA^0_k-A^\alpha_jA^\alpha_k=g_{jk}\;,\phantom{ph4}
\end{align}$$
we get
$$\begin{align}
A^0_kdx'^0-A^\alpha_kdx'^\alpha=g_{kj}d\tilde{x}^j\;.
\end{align}$$
Multiplying with $g^{jk}$ we have
$$\begin{align}
d\tilde{x}^j=g^{jk}\left(A^0_kdx'^0-A^\alpha_kdx'^\alpha\right)\;.
\end{align}$$
Inserting the expression of $dx'^i$:
$$\begin{align}
d\tilde{x}^j&=g^{jk}\left(A^0_kA^0_ldx^l-A^\alpha_kA^\alpha_ldx^l-A^0_kB^0_{nl}x^ndx^l+A^\alpha_kB^\alpha_{nl}x^ndx^l\right)\\
&=g^{jk}g_{kl}dx^l-g^{jk}b_{knl}x^ndx^l\\
&=dx^j-\frac{1}{2}g^{jk}\left(\frac{\partial g_{kn}}{\partial x^l}+\frac{\partial g_{kl}}{\partial x^n}-\frac{\partial g_{nl}}{\partial x^k}\right)x^ndx^l
\end{align}$$
Accordingly, the change of components of an arbitrary contravariant vector
$C^j$ due to a parallel displacement $\delta x^n$:
$$\begin{align}
\delta C^j=-\frac{1}{2}g^{jk}\left(\frac{\partial g_{kn}}{\partial
x^l}+\frac{\partial g_{kl}}{\partial x^n}-\frac{\partial g_{nl}}{\partial
x^k}\right)C^l\delta x^n\;.
\end{align}$$
Christoffel's symbols:
$$\begin{align}
\Gamma^j_{nl}=\frac{1}{2}g^{jk}\left(\frac{\partial g_{kn}}{\partial
x^l}+\frac{\partial g_{kl}}{\partial x^n}-\frac{\partial g_{nl}}{\partial
x^k}\right)\;, \phantom{ph5}
\end{align}$$
hence
$$\begin{align}
\delta C^j=-\Gamma^j_{nl}C^l\delta x^n\;.\phantom{ph5a}
\end{align}$$
Christoffel's symbols are symmetric in their lower indices. They are not
tensor components, since they vanish in Minkowsky frame.
Note that
$$\begin{align}
B^i_{jk}=A^i_n\Gamma^n_{jk}\;.\phantom{ph6}
\end{align}$$
The change of covariant vector components may be derived from the observation
that the scalar product of a contravariant and a covariant vector is unchanged
during parallel displacement:
$$\begin{align}
0=\delta \left(D_jC^j\right)=D_j\delta C^j+C^j\delta D_j=-D_j\Gamma^j_{nl}C^l\delta x^n+C^l\delta D_l\;.\phantom{ph7}
\end{align}$$
Since this is true for an arbitrary vector
$C^l$, we have
$$\begin{align}
\delta D_l=\Gamma^j_{nl}D_j\delta x^n\;.\phantom{ph5b}
\end{align}$$
The change of tensor components during parallel displacement must follow the
same rule as that of a corresponding product of vector components. Hence, in
the expression of $\delta T^{...j...}_{...k...}$ there is a term
$-\Gamma^j_{nl}\delta x^nT^{...l...}_{...k..}$ for
every contravariant index $j$, and there is a term $\Gamma^l_{nk}\delta
x^nT^{...j...}_{...l..}$ for every covariant index
$k$.
Covariant derivatives
Covariant differential: we substract from vector components at $x^i+dx^i$ vector components parallel
displaced from $x^i$ to $x^i+dx^i$:
$$\begin{align}
DA^j=A^j(x^i+dx^i)-\left(A^j(x^i)-\Gamma^j_{ik}A^kdx^i\right)\approx
\frac{\partial A^j}{\partial x^i}dx^i+\Gamma^j_{ik}A^kdx^i\;,\phantom{ph8}
\end{align}$$
Covariant derivative:
$$\begin{align}
\frac{DA^j}{dx^i}=
\frac{\partial A^j}{\partial x^i}+\Gamma^j_{ik}A^k\;.\phantom{ph9}
\end{align}$$
$DA^j$ is a contravariant vector and $DA^j/dx^i$ is a second rank
tensor of mixed type.
Similarly, for a covariant vector we have
$$\begin{align}
DA_j=A_j(x^i+dx^i)-\left(A_j(x^i)+\Gamma^k_{ij}A_kdx^i\right)\approx
\frac{\partial A_j}{\partial x^i}dx^i-\Gamma^k_{ij}A_kdx^i\;,\phantom{ph8a}
\end{align}$$
and
$$\begin{align}
\frac{DA_j}{dx^i}=
\frac{\partial A_j}{\partial x^i}-\Gamma^k_{ij}A_k\;.\phantom{ph9a}
\end{align}$$
The covariant derivatives of tensors may be derived on the basis of their
change during parallel
displacement. E.g.,
$$\begin{align}
\frac{DT^i_j}{dx^k}=
\frac{\partial T^i_j}{\partial x^k}+\Gamma^i_{lk}T^l_j-\Gamma^l_{jk}T^i_l\;.\phantom{ph10}
\end{align}$$
Some notations:
$$\begin{align}
T^i_{j,k}&\equiv \frac{\partial T^i_j}{\partial x^k}\;,\phantom{ph11a}\\
T^i_{j;k}&\equiv \frac{DT^i_j}{dx^k}\;.\phantom{ph11b}
\end{align}$$
The covariant derivative of the metric tensor is zero. Indeed,
$$\begin{align}
DA_i=g_{ik}DA^k=D\left(g_{ik}A^k\right)=A^kDg_{ik}+g_{ik}DA^k\;,
\end{align}$$
or may be proven by direct calculation:
$$\begin{align}
Dg_{ik}&=g_{ik,n}dx^n-\Gamma^l_{in}g_{lk}dx^n-\Gamma^l_{kn}g_{il}dx^n\\
&=g_{ik,n}dx^n-\frac{1}{2}\left(g_{ik,n}+g_{kn,i}-g_{in,k}\right)dx^n
-\frac{1}{2}\left(g_{ik,n}+g_{in,k}-g_{kn,i}\right)dx^n\equiv 0\;.
\end{align}$$
Precession of a free spherically symmetric top
As an application let us consider the problem of the precession of a free,
spherically symmetric top in stationary gravitational field. By stationary we
mean that the components of the metric are independent of time in the chosen
curvilinear frame. This is not the synonim for static, since in the latter
case the mixed index components $g_{0\alpha}$ vanish, too.
In the absence of torque, the rotation axis of the top is unchanged in a
locally comoving inertial frame. This is equivalent with a parallel
displacement in the time direction. To this aim we need a four vector. The two
(nearby) ends of the top are considered as simultaneous events and their
difference is calculated. Simultaneity in curvilinear coordinates is
equivalent with a time coordinate difference
$$\begin{align}
dx^0=-\frac{g_{0\alpha}}{g_{00}}dx^\alpha\;.\phantom{szink1}
\end{align}$$
Let us denote the spatial part of the four vector by
$n^\alpha$, then the time component is $n^0=-(g_{0\alpha}/g_{00})n^\alpha$.
The change of the axis direction during an elapsed time $\delta x^0$ is
$$\begin{align}
\delta n^\alpha=-\Gamma^\alpha_{k0}n^k\delta x^0\;,
\end{align}$$
which is equivalent with $n^\alpha_{;0}=0$. Note that $n^0_{;0}=0$ is also
satisfied. (Special case of a general rule: a tensorial equation given in
Minkowski spacetime may be generalized to curved spacetime by replacing
derivatives with covariant derivatives.)
We get
$$\begin{align}
\frac{d n^\alpha}{dx^0}=\left(-\Gamma^\alpha_{\beta 0}+\Gamma^\alpha_{0 0}\frac{g_{0\beta}}{g_{00}}\right)n^\beta\phantom{porg1}
\end{align}$$
In case of a rotating frame we have
$$\begin{align}
\Gamma^1_{00}&=-r\omega^2/c^2\\
\Gamma^1_{20}&=\Gamma^1_{02}=-r\omega/c\\
\Gamma^1_{22}&=-r\\
\Gamma^2_{10}&=\Gamma^2_{01}=\omega/(cr)\\
\Gamma^2_{12}&=\Gamma^2_{21}=1/r
\end{align}$$
Then
$$\begin{align}
\dot{n}^1&=\frac{r\omega}{1-r^2\omega^2/c^2}n^2\\
\dot{n}^2&=-\frac{\omega}{r}n^1\\
\dot{n}^3&=0
\end{align}$$
Here dot above quantities means derivation with respect to time $t$.
Introducing the proper quantities
$$\begin{align}
n_r&=n^1\\
n_\varphi&=\frac{r}{\sqrt{1-r^2\omega^2/c^2}}n^2\\
n_z&=n^3\;,
\end{align}$$
we have
$$\begin{align}
\dot{n}_r&=\frac{\omega}{\sqrt{1-r^2\omega^2/c^2}}n_\varphi\phantom{porg2a}\\
\dot{n}_\varphi&=-\frac{\omega}{\sqrt{1-r^2\omega^2/c^2}}n_r\phantom{porg2b}\\
\dot{n}_z&=0\;.\phantom{porg2c}
\end{align}$$
This means that the axis of the top is precessing backwards with angular
velocity $\omega/\sqrt{1-r^2\omega^2/c^2}$ in the frame co-rotating with the
disk. Since it is measured in the time $t$ of the inertial frame in which the
center of the disk is at rest, we can tell that observed from this same
inertial frame, during a whole revolution time $2\pi/\omega$ the axis of the
top (projected to the disk's plane) precesses by an angle
$2\pi(1/ \sqrt{1-r^2\omega^2/c^2}-1)$ opposite to the disk's rotation, i.e.,
the angular velocity of the precession is
$-\omega(1/ \sqrt{1-r^2\omega^2/c^2}-1)$.
(Note that in the limit $r\omega\ll c$ this is approximately
$-r^2\omega^3/(2c^2)$.) This is the Thomas precession, a well known result in
special relativity.
If $g_{0\alpha}=0$ (static field), there is no effect. The stationary
gravitational field of a rotating massive body does have nonzero $g_{0\alpha}$
metric tensor components, so the precession appears (frame dragging, cf.
Gravity Probe B experiment).
bene@arpad.elte.hu