a naive question about killing vectors

[xiaoke]

The definition of killing vector is that the Lie algebra of any metric
in the correspoding direction is zero. Generally if the metric has a
cyclic coordinate x^{/mu},then there is a corresponding killing vector
in the tangent direction.

[ Mod. note: Presumably, "Lie algebra" in the above should be "Lie
derivative". -ik ]

Is there any example that a space has killing vectos but do not have
cyclic coordinate in the matrics?

[Jonathan Thornburg]

xiaoke <xiaokeishere@gmail.com> wrote:
> Is there any example that a space has killing vectos but do not have
> cyclic coordinate in the matrics?

Yes, any translation symmetry. For example, Minkowski spacetime
has the 4 (orthonormal) Killing vectors (in Cartesian coordinates)
$\partial / \partial_t$
$\partial / \partial_x$
$\partial / \partial_y$
$\partial / \partial_z$
but none of these coordinates are cyclic.

ciao,

--
-- Jonathan Thornburg <jthorn@aei.mpg.de>
Max-Planck-Institut fuer Gravitationsphysik (Albert-Einstein-Institut),
Golm, Germany, "Old Europe" http://www.aei.mpg.de/~jthorn/home.html
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

[Robert Low]

Jonathan Thornburg wrote:
> xiaoke <xiaokeishere@gmail.com> wrote:
>>Is there any example that a space has killing vectos but do not have
>>cyclic coordinate in the matrics?
> Yes, any translation symmetry. For example, Minkowski spacetime
> has the 4 (orthonormal) Killing vectors (in Cartesian coordinates)
> $\partial / \partial_t$
> $\partial / \partial_x$
> $\partial / \partial_y$
> $\partial / \partial_z$
> but none of these coordinates are cyclic.

I *think* he means cyclic in the sense that the term is used
in classical mechanics: i.e. the coordinate does not occur
in the metric coefficients.

In which case, the answer is no. Given
a metric with a Killing vector, you can always take
a coordinate system in which the Killing vector is
\partial/\partial x for some coordinate x, and the
metric coefficient are then independent of x.
(At least, you can do this in some neighbourhood
of each point.)

[carlip-nospam@physics.ucdavis.edu]

Robert Low <mtx014@coventry.ac.uk> wrote:
> Jonathan Thornburg wrote:
>> xiaoke <xiaokeishere@gmail.com> wrote:
>>>Is there any example that a space has killing vectos but do not have
>>>cyclic coordinate in the matrics?
>> Yes, any translation symmetry. For example, Minkowski spacetime
>> has the 4 (orthonormal) Killing vectors (in Cartesian coordinates)
>> $\partial / \partial_t$
>> $\partial / \partial_x$
>> $\partial / \partial_y$
>> $\partial / \partial_z$
>> but none of these coordinates are cyclic.

> I *think* he means cyclic in the sense that the term is used
> in classical mechanics: i.e. the coordinate does not occur
> in the metric coefficients.

> In which case, the answer is no. Given
> a metric with a Killing vector, you can always take
> a coordinate system in which the Killing vector is
> \partial/\partial x for some coordinate x, and the
> metric coefficient are then independent of x.
> (At least, you can do this in some neighbourhood
> of each point.)

That's true, for any single Killing vector. It might
also be worth noting, though, that if you have two
noncommuting Killing vectors, you can't simultaneously
find coordinates for which one is \partial/\partial x
and the other is \partial/\partial y.

So, while any space that has a Killing vector has a
corresponding cyclic coordinate, a space can have (for
example) ten Killing vectors but fewer than ten cyclic
coordinates.

Steve Carlip