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
curved 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 curved 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
\[\begin{align}
\Gamma^1_{00}&=-r\omega^2/c^2\\
\Gamma^1_{20}&=\Gamma^1_{02}=-r\omega/c\\
\Gamma^1_{22}&=-r\\
\Gamma^2_{10}&=\Gamma^2_{01}=\omega/(cr)\\
\Gamma^2_{12}&=\Gamma^2_{21}=1/r
\end{align}\]
\[\begin{align}
g_{00} &= 1-r^2\omega^2/c^2 \\
g_{11} &= g_{33}=-1 \\
g_{22} &= -r^2 \\
g_{02} &= g_{20}=-r^2\omega/c\;.\phantom{forgmetrik1}
\end{align}\]
\[\begin{align}
\Gamma_{000}&=\frac{1}{2}\left(g_{00,0}+g_{00,0}-g_{00,0}\right)=0\\
\Gamma_{00\alpha}&=\frac{1}{2}\left(g_{00,\alpha}+g_{0\alpha,0}-g_{0\alpha,0}\right)=\delta_{\alpha
1}\left(-\frac{r\omega^2}{c^2}\right)\\
\Gamma_{\alpha 00}&=\frac{1}{2}\left(g_{\alpha 0,0}+g_{\alpha 0,0}-g_{00,\alpha}\right)=\delta_{\alpha
1}\frac{r\omega^2}{c^2}\\
\Gamma_{0\alpha\beta}&=\frac{1}{2}\left(g_{0\alpha,\beta}+g_{0\beta,\alpha}-g_{\alpha\beta,0}\right)=\left(\delta_{\alpha
2}\delta_{\beta 1}+\delta_{\alpha
1}\delta_{\beta 2}\right)\left(-\frac{r\omega}{c}\right)\\
\Gamma_{\alpha\beta 0}&=\frac{1}{2}\left(g_{\alpha\beta,0}+g_{\alpha
0,\beta}-g_{\beta 0,\alpha}\right)=\left(\delta_{\alpha
2}\delta_{\beta 1}-\delta_{\alpha
1}\delta_{\beta 2}\right)\left(-\frac{r\omega}{c}\right)\\
\Gamma_{\alpha\beta\gamma}&=\frac{1}{2}\left(g_{\alpha\beta,\gamma}+g_{\alpha
\gamma,\beta}-g_{\beta \gamma,\alpha}\right)=\left(\delta_{\alpha
2}\delta_{\beta 2}\delta_{\gamma 1}+\delta_{\alpha
2}\delta_{\beta 1}\delta_{\gamma 2}-\delta_{\alpha
1}\delta_{\beta 2}\delta_{\gamma 2}\right)\left(-r\right)
\end{align}\]
\[\begin{align}
g = det\left(g_{ik}\right)=-r^2
\end{align}\]
\[\begin{align}
g^{00} &= 1 \\
g^{02} &= g^{20}=-\omega/c\;.\phantom{forgmetrik1}\\
g^{22} &= \omega^2/c^2-1/r^2 \\
g^{11} &= g^{33}=-1
\end{align}\]
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