Talk:Killing vector field

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Inconsistent notation[edit]

Please fix the notation for consistency. You start with the Killing vector field "X", then switch to "K", then back to "X", then back to "K".

Positive Ricci curvature[edit]

If the Ricci curvature is positive, then a Killing field must have a zero.

May I ask how come? I had some idea that the Hopf fibration defines a nonvanishing vector field on S3, and that this field acts by isometries, but I might be wrong... (too tired to do the calculations tonight :/ ) Someone here who can help me, please? (A google on "killing field" + Hopf gave me e.g. this paper...see section III) \Mikez 03:06, 21 Dec 2004 (UTC)

Hmm, that seems right to me. Writing the metric on S3 as

one sees that the vector field

which generates the U(1) action in the Hopf fibration, is a Killing field which has norm 1 everywhere (and hence is nonvanishing). -- Fropuff 07:11, 2005 Jan 9 (UTC)

I have now removed that statement. \Mikez 18:58, 7 Feb 2005 (UTC)

Awesome![edit]

Killing Field??? Maths is awesome!!! 130.195.2.100 03:50, 8 August 2006 (UTC)[reply]

We kill vectors there. actually, at the Killing horizon. 67.198.37.16 (talk) 05:08, 8 November 2020 (UTC)[reply]

What?[edit]

Is there anyway this article could be edited so somebody without a degree in math can know what the flying flip it is talking about. I would really like to know, but the article is filled with so much math specific jargon it is impossible without first reading 90 other articles. 88.154.199.46 08:34, 8 June 2007 (UTC)[reply]

It would be a pretty difficult to understand what a Killing field is without knowing what a Riemannian metric is at least. Most people who know that would have or be studying for a degree in mathematics or physics. There is no jargon in the article that I can see but it does rely on specialist terminology. Fortunately this is wikipedia so you can at least click on the links.Billlion 20:17, 10 June 2007 (UTC)[reply]
Wondering who wrote that and was surprised to find it was me! Another answer is to understand what an isometry is eg in Euclidean space a cylinder and a sphere then a Killing vector field is an infinitesimal isometry. That is a way to give some kind lay explanation. Billlion (talk) 14:33, 27 April 2023 (UTC)[reply]

Interesting name[edit]

This article does have an interesting name. 88.154.243.24 08:03, 12 July 2007 (UTC)[reply]

results only for Riemannian spaces?[edit]

The lead makes it sound like the article is going to discuss both Riemannian and semi-Riemannian space, but there are some results in it that I don't think are true for a non-Riemannian signature. "Negative Ricci curvature implies there are no nontrivial (nonzero) Killing fields." I think this is false for semi-Riemannian spaces. E.g., I think general relativity has cosmological solutions with negative Ricci curvature, but these solutions do have Killing vectors such as spatial rotations.--75.83.69.196 (talk) 05:30, 31 January 2011 (UTC)[reply]

Yeah, some of those claims feel smelly. 67.198.37.16 (talk) 05:00, 8 November 2020 (UTC)[reply]

Relation to the fundamental vector field[edit]

Is there any difference between the generalized Killing vector field (see [[1]] and the fundamental vector field? 89.135.20.50 (talk) 09:04, 19 November 2011 (UTC)[reply]

Yes, theres a difference. They're more or less unrelated. Killing fields are isometries, fundamental fields are just the action of a Lie group mapped out onto a manifold. They are not really related in any kind of naive, direct sense. I guess there is some weird deeper relationships for symmetric spacess but that's getting off the mark. 67.198.37.16 (talk) 05:04, 8 November 2020 (UTC)[reply]

Illustration[edit]

A Killing vector field (red) with integral curves (blue) on a sphere.

A Killing field is a function of the metric. What is the metric that defines the Killing field in the picture? I assume there may be many different metrics defined on a sphere, and they will all have different Killing fields. Bartosz (talk) 16:51, 9 November 2014 (UTC)[reply]

Yes, and that one is clearly not a Killing field of the standard metric of the sphere, so I think it shoud be removed, for it can cause some confusion. — Preceding unsigned comment added by ChicoCaramello (talkcontribs) 07:09, 2 April 2015 (UTC)[reply]
I was wondering the same thing. The text below the (full-size) picture reads as follows:
"A Killing vector field K = sin(phi)*e_theta + cot(theta)*cos(phi)*e_phi and its integral curves of the S2 (minus the poles) with induced metric. Created with POVRAY."
It is signed with "Mrfister - Own work" — but the Mrfister is in red so I don't know if he can be reached to ask him what he was thinking.
Since the "induced metric" is never explained and, for me at least, not easily guessable: I agree that the illustration should be removed — immediately. Ideally it will be replaced soon by a more understandable one.Daqu (talk) 17:13, 25 September 2015 (UTC)[reply]
I've removed the image. Can you come up with any ideas for what would make an appropriate illustration? (I'm thinking a sphere could work with the natural metric, and with the two killing vector fields in different colours. But maybe something completely different would be better?) Cesiumfrog (talk) 22:36, 25 September 2015 (UTC)[reply]
Thank you, Cesiumfrog !
Sure, here's a possibility: Just draw a vector field on the unit sphere tangent to rotational motion, of constant angular velocity ω, about some fixed axis of the sphere. I would "not" put two Killing vector fields on the same picture of a sphere; in fact, they are all so similar to one another that one example should suffice.
E.g., for rotation about the vertical axis, at any point (x,y,z) of the sphere (i.e., such that x2 + y2 + z2 = 1), at angular velocity ω, the vector V(x,y,z) at the point (x,y,z) would be
(*) V(x,y,z) = ω sqrt(x2 + y2) (-y, x, 0).
For a more typical example, I would pick an axis that is not vertical. Ideally, an axis a whose projection P(a) to the web page makes the same angle with the vertical direction on the page as the angle that its direction in space makes with the vertical direction in space.
For example, suppose the projection to the page is just the orthogonal projection to the y- and z-axes of space. Then I would use the rotation in space about the axis that is 15° from the z-axis and 75° from the positive y-axis (which after projection points to the right on the page). This could be obtained by applying the rotation by -15° = -π/12, about the positive x-axis in space, to the vector field defined above at (*).
More specifically, this would be the vector field W defined at the point (x,y,z) in space via
1) rotate (x,y,z) by +15° about the positive x axis, to get
(x',y',z') = (x, cos(π/12)y -sin(π/12)z, sin(π/12)y + cos(π/12)z).
Then
2) evaluate the vector field (*) V at this point (x',y',z') to get
(x", y", z") = V(x',y',z') = ω sqrt(x'2 + y'2) (-y', x', 0).
Finally,
3) apply the rotation about the positive x-axis of angle -π/12 to get the vector field of the
W(x,y,z) = (x", cos(-π/12)y" -sin(-π/12)z", sin(-π/12)y" + cos(-π/12)z")
i.e.,
W(x,y,z) = (x", cos(π/12)y" sin(π/12)z", -sin(π/12)y" + cos(π/12)z").
Of course, this requires choosing an angular velocity ω.Daqu (talk) 17:24, 27 September 2015 (UTC)[reply]
Fair enough, I guess a single killing field on a sphere does seem pedagogically uncluttered (and I was already thinking moderately off-axis myself). It might be good to have a second example also, with something other than rotation of a volume of revolution. I suppose only global rotations/translations work for surfaces embedded in 3D (e.g. no poloidal winding of a torus), so I think a helical vector field on the curved surface of a stout cylinder would be good. (A third possible example would be to emphasise that it is not a global concept and hence the field need not be extendible to cover all the manifold, for example it might only cover one side of a cube or one flat surface of an annular-washer-with-rectangular-crosssection, and on that subregion it could again be a combination of translation with rotation but with the origin or axis of rotation displaced away from the centroid of the surface and volume.) Cesiumfrog (talk) 23:33, 27 September 2015 (UTC)[reply]
Nice idea, to provide another illustration in 3-space. Incidentally, to show the Killing vector field that generates a one-parameter group of screw-translations — as you suggest — it is not necessary to use a cylinder. You could alternatively take a peanut-shaped Oval of Cassini curve, given by say |z-1| |z+1| = 1.02 for z in the complex plane — and then creating the surface obtained by translating this curve up and down in the z direction while simultaneously rotating it about the z-axis by an angle proportional to z (say z).Daqu (talk) 15:49, 28 September 2015 (UTC)[reply]

To answer Bartosz original question, the metric is the plain-old-every-day bog-standard sphere metric . That illustration is based on equations for the Killing fields that appear explicitly in Schwarzschild coordinates#Killing vector fields, and in various other locations over the net. (I've yet to check my paper textbooks) I plugged them into and turned the crank and and verified that they are zero. However, I really hate that picture, there is something very misleading about it. From what I can tell, the choice of as one of the fields is fine, in that its normalized to unit length. But if you want to draw 3D pictures, you have to stick a in front of it. ... and that changes the angles on everything. That is, you have to use to draw the picture ... so, yeah, that picture is just ... wrong... coordinate craziness ... I patched up the article mostly, but its still garbagey with some coordinate craziness in it. ... anyway, it's late at night, so that's all for now. 67.198.37.16 (talk) 07:19, 8 November 2020 (UTC)[reply]

I fixed it up so that it is whole, complete and "makes sense". It turned out to be a small thesis, instead of the 3 short sentences I had hoped to write. Bummer about that. 67.198.37.16 (talk) 20:37, 8 November 2020 (UTC)[reply]

Killing vector fields on the 2-sphere[edit]

The paragraph about killing vector fields on 2-fields mentions the two "obvious" fields

This is not what I find on this physics forum thread If someone has the book mentioned in another thread

(General Relativity Demystified, McMahon) could double check this paragraph for errors in the (incomplete) calculation?  — Preceding unsigned comment added by 88.215.103.183 (talk) 12:55, 22 December 2020 (UTC)[reply] 

Souldn't the Lie bracket of coordinate vectors vanish?[edit]

The "Killing fields on a 2-sphere" subsection in the article sounds informative but this line is difficult for an uninitiated reader (like myself), confusing or perhaps imprecise:

However, moving off of the equator, the two vector fields and are no longer orthonormal, and so, in general one has for a point in general position.

I agree the two vector fields are not orthonormal in general position, but that does not imply the Lie bracket gives non-zero values, and I would think it should vanish where the two vector fields in question are defined. How could it be possible that the Lie bracket of coordinate vectors should be non-vanishing? 240F:7C:CFB3:1:CF39:4DD5:6DD7:4D5C (talk) 03:49, 19 March 2022 (UTC)[reply]

You're right, the bracket vanishes since the basis is coordinate induced. The article is incorrect. The incorrect statement is that is a Killing field. It can be verified that the component of the Killing equation is non-zero. In fact, there's an intuitive view of why this doesn't conserve lengths: if we flow all points in the direction, they move from the north pole to the south pole. This means the points spread out from the north pole, stretching and distorting the metric there.Zephyr the west wind (talk) 21:27, 6 May 2022 (UTC)[reply]