What is a time-like killing vector?

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SUMMARY

A time-like Killing vector is defined by the Killing equation, which states that the covariant derivative of the vector field satisfies v_{i;j} + v_{j;i} = 0. This indicates the existence of a coordinate system where the metric tensor remains invariant under transformations, specifically isometries. A vector is classified as time-like if it meets the condition g_{ij} v^{i} v^{j} > 0, assuming a metric signature of (+,-,-,-). An example of a system exhibiting a time-like Killing vector is a stationary black hole.

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kurious
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What is a time-like killing vector?
 
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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 existence 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 (+,-,-,-).
 
Last edited:
kurious said:
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.

Atheist said:
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
 
Last edited:
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.
 
kurious said:
What is a time-like killing vector?

Killing vectors are generated by isometries. Isometries are transformations which leave lengths unchanged. For a more technical definition, see.

http://mathworld.wolfram.com/Isometry.html

A time-like Killing vector means, roughly speaking, that the distances in the system are unchanged as time increases (i.e by a time translation). Since the distances are defined by the mteric tensor, g_ab, this means that the components of the metric tensor are unchanged by time.

A stationary black hole is an example of a system with a time-like Killing vector.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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