2
$\begingroup$

I'm reading The large scale structure of spacetime and in page 72 the author says:

A static metric admits a timelike killing vector $K$. We define the timelike unit vector $V$ as $V=K/f$, where $f^2=-g(K,K)$.

Then, $\nabla_{b}V^a = - V^c V_b \nabla_c V^a$

But the math doesn't check out, and I'm not sure how he got that result.

$\endgroup$
  • $\begingroup$ That identity after the Then, cannot be right, because it has one free index on the LHS and two free indices on the RHS. $\endgroup$ – Jerry Schirmer Aug 19 at 18:54
  • $\begingroup$ You appeared to have lost a time derivative: $\nabla_{b}V^{a}=-\dot V^{a}V_{b}$. Shouldn't you be trying to show $\dot V^{a}=\nabla_{b }V^{a} V^{b}$? $\endgroup$ – Cinaed Simson Aug 20 at 1:09
  • $\begingroup$ The text says: Then $V^a_{;b}=-\dot{V}^aV_b$, where $\dot{V}^a=V^a_{;b}V^b$. $\endgroup$ – MBN Aug 20 at 8:32
  • $\begingroup$ Yes, I know the text says V-dot, but immediately after it gives a formula for V-dot. I've put them together, since that's what I want to know $\endgroup$ – Diana99 Aug 20 at 10:22
1
$\begingroup$

On p. 72 of the referenced Hawking and Ellis book, a timelike Killing vector $\pmb{K}$ of magnitude $K^aK_a = -f^2$, and a unit vector $\pmb{V} = f^{-1}\pmb{K}$ along $\pmb{K}$, with $V^aV_a = -1$, are introduced. Using component notation throughout, we make note of the fact that since $\pmb{K}$ is a Killing vector, its components satisfy the relation \begin{align} K_{b{;c}} + K_{c{;b}} = 0 \quad \Rightarrow \quad K_{c{;b}} = -K_{b{;c}}\,. \tag{Killing} \end{align} Also, taking the covariant derivative of the relation $V^aV_a = -1$ yields \begin{align} V_{a\,;c}V^a + V_aV^a_{\,\,\,;c} = 0\,. \tag{unitV} \end{align} However, $V_aV^a_{\,\,\,;c} = g_{ab}V^b(g^{ad}V_d)_{\,;c} = g^{ad}g_{ab}V^bV_{d\,;c} = \delta^d_bV^bV_{d\,;c} = V^bV_{b\,;c} \equiv V^aV_{a\,;c}$, noting that the covariant derivative of the metric vanishes. Substituting this result in equation (unitV) thus yields $V_{a\,;c}V^a + V_aV^a_{\,\,\,;c} = V_{a\,;c}V^a + V^aV_{a\,;c} = 2V^aV_{a\,;c} = 0$, from which it follows that \begin{align} V_aV^a_{\,\,\,;c} = V_{a\,;c}V^a = 0\,. \tag{covunitV} \end{align} With these preliminaries out of the way, we find for the covariant derivative of $V^a$: \begin{align} V^a_{\,\,\,;b} = (f^{-1}K^a)_{;b} &= (f^{-1}g^{ac}K_c)_{;b}\,,\\ &= -f^{-2}f_{;b}g^{ac}K_c + f^{-1}g^{ac}K_{c{;b}}\,,\\ &= -f^{-1}f_{;b}V^{a} + f^{-1}g^{ac}K_{c{;b}}\,,\\ &= -f^{-1}f_{;b}V^a - f^{-1}g^{ac}K_{b{;c}}\,,\\ &= -f^{-1}f_{;b}V^a - f^{-1}g^{ac}(fV_b)_{;c}\,,\\ &= -f^{-1}f_{;b}V^a - g^{ac}\left[f^{-1}f_{;c}V_b + V_{b\,;c}\right]\,,\\ &= -f^{-1}f_{;b}g^{ac}V_c -g^{ac}f^{-1}f_{;c}V_b + g^{ac}V_{b\,;c}\,, \end{align} where, in the fourth line, we used equation (Killing). Contracting with $V^b$, we obtain \begin{align} V^a_{\,\,\,;b}V^b &= -f^{-1}f_{;b}g^{ac}V_cV^b - g^{ac}f^{-1}f_{;c}V_bV^b + g^{ac}V_{b;c}V^b\,. \end{align} The last term vanishes, according to equation (covunitV), and we replace $V_bV^b = -1$ in the second term, reducing the last equation to \begin{align} V^a_{\,\,\,;b}V^b &= -f^{-1}f_{;b}V^bV^a + f^{-1}f_{;c}g^{ac}\,.\tag{contract1} \end{align} Contracting this result with $V_a$ yields \begin{align} V_aV^a_{\,\,\,;b}V^b &= - f^{-1}f_{;b}V^bV^aV_a + g^{ac}f^{-1}f_{;c}V_a\,,\\ &= f^{-1}f_{;b}V^b + f^{-1}f_{;c}V^c \,,\\ &= 2f^{-1}f_{;b}V^b \,. \end{align} But the left-hand side of this equation vanishes, since $V_aV^a_{\,\,\,;b} = 0$, leaving \begin{align} f^{-1}f_{;b}V^b = 0 \,. \end{align} Using this result in equation (contract1), we see that the first term on the right-hand side vanishes to obtain \begin{align} V^a_{\,\,\,;b}V^b = f^{-1}f_{;c}g^{ac} \equiv \dot{V}^a\,, \end{align} where we note the definition of $\dot{V}^a$ by Hawking and Ellis. Contracting this result with $V_b$ yields their expression for the covariant derivative: \begin{align} V^a_{\,\,\,;b} = -\dot{V}^aV_b\,. \end{align} The last two equations are those asked to be explained in the question. These rather tedious details may be helpful in understanding their origins. Hopefully, I haven't misplaced an index somewhere.

$\endgroup$
  • $\begingroup$ Thank you so much! This answer was exactly what I needed, and it has all the details I was missing $\endgroup$ – Diana99 Aug 27 at 15:05
  • $\begingroup$ You are quite welcome. I'm happy to hear that it was helpful. $\endgroup$ – splitcomplexes Aug 27 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.