What are Killing vectors?

[snoopies622]

TL;DR

What is the difference between two adjacent vectors in a Killing field and two vectors separated by a parallel transport?

I don't have an intuitive feel for Killing vectors.

Wikipedia says, " . . . more simply, the flow generates a symmetry, in the sense that moving each point of an object the same distance in the direction of the Killing vector will not distort distances on the object."

That just sounds like parallel lines to me. Is there a difference?

 

[Dale]

There are also Killing vectors that represent rigid rotations. And Born rigid accelerations too.

 

[snoopies622]

I found this quote on a different PF thread:

For parallel transport, you move the vector in a direction that is always parallel to the curve. For Lie transport, you move the vector so that its direction is always perpendicular to the curve.

Thoughts? It sounds like this restricts the direction a vector can be Lie transported in a way that would not be a case for parallel transport (which, it seems to me, could be any direction).

 

[PeterDonis]

I found this quote on a different PF thread

Please give a link to the thread. (If you use the PF quote feature to quote posts, the link will automatically be provided for you.)

 

[PeterDonis]

I don't have an intuitive feel for Killing vectors.

Wikipedia says, " . . . more simply, the flow generates a symmetry, in the sense that moving each point of an object the same distance in the direction of the Killing vector will not distort distances on the object."

That just sounds like parallel lines to me. Is there a difference?

Consider this example: vectors in a Euclidean plane.

Parallel transport in this case is simple: the direction the vector points stays the same no matter where you transport it in the plane. So if the vector starts out pointing "due North" (in the positive $y$ direction as Cartesian coordinates are normally chosen), it will always point due North.

The Euclidean plane also has 3 Killing vector fields (or more precisely a 3 parameter group of them, but we won't need to get into complexities like that here). Two of them are trivial for this discussion: straight line translation in the $x$ direction or the $y$ direction. Lie transport of a vector along these Killing vector fields works the same as parallel transport for this simple case: the direction the vector points stays the same. However, this is a very rare special case.

The third Killing vector field, however, works differently. It represents rigid rotations about some chosen origin point. Lie transport along this Killing vector field does not work like parallel transport: it changes the direction of vectors. Basically, think of moving a vector around a circle, and keeping the angle of the vector with the tangent to the circle constant. The vector will do a complete 360 degree rotation as it goes once around the circle. This is much more typical of Lie transport as compared with parallel transport.

 

[snoopies622]

Please give a link to the thread. (If you use the PF quote feature to quote posts, the link will automatically be provided for you.)

It comes up at the top of this search, but strangely not finding the quote in the cited PF thread itself.

https://www.google.com/search?q=lie+parallel+perpendicular+always+curve+transport&sca_esv=ca37d0d638fdd5cb&sca_upv=1&sxsrf=ADLYWIKr3AyGy5V8P0pfZW7WIj3im8XGuA:1715652125032&ei=HcZCZqq3AZav5NoPksKW0Ao&oq=lie+parallel+perpendicular+always+curve+transport&gs_lp=EhNtb2JpbGUtZ3dzLXdpei1zZXJwIjFsaWUgcGFyYWxsZWwgcGVycGVuZGljdWxhciBhbHdheXMgY3VydmUgdHJhbnNwb3J0MgUQIRigATIFECEYoAEyBRAhGKABSLo1ULcQWKwxcAF4AZABAJgBmgKgAZwNqgEFMC43LjO4AQPIAQD4AQGYAgugAtsOwgIKEAAYsAMY1gQYR8ICBRAhGJ8FmAMAiAYBkAYIkgcFMS43LjOgB_Aw&sclient=mobile-gws-wiz-serp

 

[snoopies622]

The third Killing vector field, however, works differently. It represents rigid rotations about some chosen origin point. Lie transport along this Killing vector field does not work like parallel transport: it changes the direction of vectors. Basically, think of moving a vector around a circle, and keeping the angle of the vector with the tangent to the circle constant. The vector will do a complete 360 degree rotation as it goes once around the circle. This is much more typical of Lie transport as compared with parallel transport.

Thanks Peter, i think i'm getting a feel for it now. Like the way a set of particles in a stone move when the stone is either translated or rotated.

 

[PeterDonis]

It comes up at the top of this search, but strangely not finding the quote in the cited PF thread itself.

Which is why you should always quote other PF posts when you want to refer to them--because that ensures that you are referring to something that is actually posted somewhere on PF, as opposed to something a search engine confabulated.

 

[Orodruin]

Which is why you should always quote other PF posts when you want to refer to them--because that ensures that you are referring to something that is actually posted somewhere on PF, as opposed to something a search engine confabulated.

It is not quotable for mere mortals, presumably because the thread is locked. Providing a direct link to the thread would therefore be the best possible option.

https://www.physicsforums.com/threads/lie-transport-vs-parallel-transport.17064/

 

[snoopies622]

Follow up: It looks like — just going by my geometric intuition here — that Killing fields must have zero divergence. Can one please give an example of a vector field with zero divergence that is not a Killing field?

 

[PeterDonis]

It looks like — just going by my geometric intuition here — that Killing fields must have zero divergence.

Zero covariant divergence, yes.

Can one please give an example of a vector field with zero divergence that is not a Killing field?

The 4-velocity field describing static observers in Schwarzschild spacetime. In Schwarzschild coordinates this is

$$
U = \left( \frac{1}{\sqrt{1 - \frac{2M}{r}}}, 0, 0, 0 \right)
$$

 

[snoopies622]

hmm i thought the path of a static observer in Schwarzschild spacetime was a Killing vector, since the metric doesn't change as one simply advances in time.

https://physics.stackexchange.com/questions/214434/killing-vectors-in-schwarzschild-metric?rq=1

 

[PeterDonis]

i thought the path of a static observer in Schwarzschild spacetime was a Killing vector

The worldlines of static observers are the same as the integral curves of the timelike Killing vector field, but the actual vector field describing the 4-velocities of those observers is not a Killing vector field.

 

[Orodruin]

Zero covariant divergence, yes.

To put this into context, the Killing condition $\mathcal L_K g = 0$ may be rewritten
$$
\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0
$$
It is therefore clear that multiplication by the inverse metric $g^{\mu\nu}$ immediately results in
$$
2\nabla_\mu K^\mu = 0
$$

The reverse implication is clearly not true. It should be evident that the contraction with the metric cannot be inverted generally and a counter example was already provided.

but the actual vector field describing the 4-velocities of those observers is not a Killing vector field.

… as evident from the Killing field being $\partial_t$ and $g(\partial_t,\partial_t) = g_{tt} \neq 1$.

 

[PeterDonis]

It is therefore clear that multiplication by the metric $g_{\mu\nu}$ immediately results in
$$
2\nabla_\mu K^\mu = 0
$$

I assume you mean contraction with the inverse metric $g^{\mu \nu}$?

 

[Orodruin]

I assume you mean contraction with the inverse metric $g^{\mu \nu}$?

Yes, indeed. Fixing.