# What's the covariant derivative of a normalized, timelike Killing vector?

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.

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

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.