# What is a time-like killing vector?

by kurious
Tags: killing, timelike, vector
 P: 653 What is a time-like killing vector?
 P: 38 Unless given further explanation I´d say it´s exactly what the name sais: Killing vector: A vector that fulfies the Killing-equation $$v_{i;j} + v_{j;i} = 0$$. The existence of a Killing-vector implies the existance of a coordinate system where the metric tensor is independent of one of the coordiantes. time-like: A vector v is timelike if $$g_{ij} v^{i} v^{j} >0$$. EDIT: As pmb_phy correctly claims I should mention that above inequality assumes the signature of the metric to be (+,-,-,-).
P: 2,954
 Quote by kurious What is a time-like killing vector?
A few preliminaries - A coordinate transformation which leaves the components of the metric tensor invariant is called an isometry. This means that when the coordinates are change from the primed coordinates, x', to the unprimed coordinates x, the metric tensor remains unchanged, i.e. is the same function of the coordinates. This means

$$g'_{\alpha\beta}(x') = g_{\alpha\beta}(x')$$

For the components of the metric tensor invariant under the isometry we must have

$$g_{\mu\nu} (x) = \frac{\partial x'^{\alpha}}{\partial x^{\beta}}\frac{\partial x'^{\mu}}{\partial x^{\nu}}g(x'(x))$$

Consider the infinitesimal coordinate transformation

$$x' = x^{\alpha} + \epsilon \xi^{\alpha}$$

where $$\xi^{\alpha}(x)$$ is a vector field and $$\epsilon$$ -> 0. For this coordinate transformation to yield an isometry the $$\xi^{\alpha}$$ must satisfy the following equation

$$\xi_{\mu;\nu} + \xi_{\nu;\mu} = 0$$

As Atheist mentioned, this equation is called Killing's equation and the solutions Killing vectors.

 Quote by Atheist time-like: A vector v is timelike if $$g_{ij} v^{i} v^{j} >0$$.
That depends on the signature of the metric tensor.

Pete

 Emeritus PF Gold P: 8,147 What is a time-like killing vector? Perfectly true, Pete, but the definition is still good with the appropriate sign in. For newbies, the semicolon in Atheist's definition denotes covariant derivative, so the equation he gives, called Killing's equation, is a differential equation.
Emeritus