I have read that the existence of an arbitrary timelike killing field implies that we can find some coordinates such that metric is independent of time in that coordinates. For me independence of metric from time coordinate implies that in that coordinate system all of vectors of the Killing field are simultaneously in $\frac{\partial}{ \partial t }$ form. But in GR book of Sean Carrol it is written that it is generally impossible to find such coordinates that all Killing vectors are in simple derivative form. Can you help me resolve this?

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    $\begingroup$ What do you mean by "all of vectors of the Killing field are simultaneously in $\frac{}{\partial t}$ form"? $\endgroup$
    – NDewolf
    Jan 22, 2020 at 15:25
  • $\begingroup$ Killing vectors form a field. So there is a Killing vector for each point in spacetime. I mean that each of that vectors is purely in time direction. It is similar to Killing field that is associated with translations in time direction in flat Minkowski metric. $\endgroup$
    – Timur9717
    Jan 22, 2020 at 15:30

1 Answer 1


It is in general impossible to find coordinates where multiple Killing fields are simple partial derivatives.

I mean, if you have Killing fields $K_1,...,K_r$ with some relation $$ [K_i,K_j]=C^k_{ij}K_k, $$ then unless $C^k_{ij}=0$, you cannot take all the $K_i$ to be partial derivatives, since those always commute.

But. For the result OP is asking about, only one Killing field is present. It is known that nonvanishing vector fields can always be "straightened", eg. if $X$ is a nonvanishing vector field on the manifold $M$, there exists charts such that $$ X=\frac{\partial}{\partial x^1}. $$

Also, then $$ \mathscr L_X=\frac{\partial}{\partial x^1} $$when expressed in this chart, where $\mathscr L$ is the Lie derivative

Putting the two together, if $K$ is a timelike Killing field that is nonvanishing (if it does vanish somewhere, restrict the domain), there is a coordinate system such that $K=\partial/\partial t$, but since $K$ is Killing, we have $$ \mathscr L_K g=0 \Leftrightarrow\frac{\partial}{\partial t}g_{\mu\nu}=0, $$ which implies that in the straightening coordinate system the metric coefficients $g_{\mu\nu}$ are independent of $t$.

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    $\begingroup$ Perhaps it is also valuable if you give $\mathcal{L}_X$ a name for the OP and also define $\mathfrak{X}$, for the physicists :), just in case... $\endgroup$
    – ohneVal
    Jan 22, 2020 at 15:36

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