What is a Killing field?

  • #1
775
17
Summary:
How do I know whether or not something is a Killing field?
I'm trying to understand Killing fields and I've hit a bump right away. Suppose I'm in a plane and using the Cartesian coordinate system, and I consider the swirly vector field [itex] \vec{v}=(-y,x) [/itex]. It is easy to show that [itex] \nabla _{y} v^x + \nabla_{x} v^y = -1+1=0 [/itex]. But if I convert this to polar coordinates I get [itex] \vec{v}=(0,1) [/itex] and [itex] \nabla _{\theta} r + \nabla_{r}\theta = -r + 1/r \ne 0 [/itex] . So either I'm computing my covariant derivative components wrong, I'm misusing Killing's equation or (and I doubt this) whether or not a vector field is a Killing field depends on one's choice of coordinates.

Any feedback would be appreciated!
 
Last edited:

Answers and Replies

  • #2
775
17
From above:
[tex]
\nabla _{\theta}v^r + \nabla _{r}v^{\theta} = \frac {\partial v^r}{\partial \theta} + \Gamma ^r_{r \theta} v^r + \Gamma ^r_{\theta \theta} v^{\theta} +
\frac {\partial v^\theta}{\partial r} + \Gamma ^{\theta}_{r r} v^r + \Gamma ^\theta_{\theta r} v^{\theta} =
0 + 0 + (-r)(1) + 0 + 0 + (1/r)(1)
[/tex]
 
  • #3
Ibix
Science Advisor
Insights Author
2020 Award
8,320
7,747
Killing's equation is ##\nabla_{(\mu}V_{\nu)}=0##. Notice anything about your index placement?
 
  • #4
775
17
Oh i see, i'm using vectors instead of covectors! Will switch and recalculate in the morning. Thanks Ibex. :)
 
  • #6
775
17
Epilogue: I was wondering, how could I make such a simple mistake? Well, one of my few GR books is David McMahon's Relativity Demystified and sure enough on the first page of Chapter 8, immediately after introducing Killing's equation, he says, "Note that this equation also holds for contravariant components; i.e., [itex] \nabla _{b} X^a + \nabla _{a} X^b = 0 [/itex]." I've read elsewhere that the Demystified books are famous for containing lots of errors.
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,343
7,214
Epilogue: I was wondering, how could I make such a simple mistake? Well, one of my few GR books is David McMahon's Relativity Demystified and sure enough on the first page of Chapter 8, immediately after introducing Killing's equation, he says, "Note that this equation also holds for contravariant components; i.e., [itex] \nabla _{b} X^a + \nabla _{a} X^b = 0 [/itex]." I've read elsewhere that the Demystified books are famous for containing lots of errors.
That equation makes absolutely no sense since there is an index mismatch between the terms.
 
  • #8
Ibix
Science Advisor
Insights Author
2020 Award
8,320
7,747
I've read elsewhere that the Demystified books are famous for containing lots of errors.
Looks like it. As Orodruin says, the indices don't match. You can certainly find some covector field ##\kappa_a## such that ##\nabla_{(a}\kappa_{b)}=0## then calculate ##K^a=g^{ab}\kappa_b##, but ##\nabla_{(a}K^{b)}## is nonsense. I think the bracket notation for (anti-)symmetrisation is meant to remind you of that by prompting you to think "shouldn't those brackets be in line...?"

Note that, as you yourself found, you can get away with it in Cartesian coordinates on Euclidean space because the trivial metric means that ##x_a=g_{ab}X^b=\delta_{ab}X^b=X^a## (even if it makes me a bit shivery typing that last equality). It's still wrong in some sense, but it works anyway in that special case.
 
  • #9
775
17
Thanks! Although I understand the equation now, I'm still unclear on the concept. For example, the wikipedia entry "Killing vector field" says, "Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. 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."

In a plane with Cartesian coordinates, the vector field [itex] \vec{v}=(x,y) [/itex] is a Killing vector field since [itex] \nabla_{x} v_{y} + \nabla_{y} v_{x}=0+0=0 [/itex], yet if I place a circle on this plane with its center at the origin and allow each point to move at the same speed in the direction of the Killing vector at that point, the circle will expand. Isn't this distorting distances? In fact it will expand anywhere in the plane since this field everywhere has non-zero divergence. Perhaps they mean that the object's size can change but not its shape.

I'm also unclear about the connection between Killing vectors and Killing vector fields. If a Killing vector is simply any vector such that as one moves along it, the components of the metric don't change (as with a point at rest in a Schwarzschild gravitational field since the components of the Schwarzschild metric don't include time), then in a Cartesian plane every vector is a Killing vector since the metric [itex] ds^2 = dx^2 + dy^2 [/itex] is the same everywhere. Yet every vector field in a Cartesian plane isn't a Killing vector field, since for example for [itex] \vec{v}=(x,x) [/itex] we have [itex] \nabla_{x} v_{y} + \nabla_{y} v_{x}=0+1 \ne 0 [/itex].
 
  • #10
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,343
7,214
Thanks! Although I understand the equation now, I'm still unclear on the concept. For example, the wikipedia entry "Killing vector field" says, "Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. 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."

In a plane with Cartesian coordinates, the vector field [itex] \vec{v}=(x,y) [/itex] is a Killing vector field since [itex] \nabla_{x} v_{y} + \nabla_{y} v_{x}=0+0=0 [/itex], yet if I place a circle on this plane with its center at the origin and allow each point to move at the same speed in the direction of the Killing vector at that point, the circle will expand. Isn't this distorting distances? In fact it will expand anywhere in the plane since this field everywhere has non-zero divergence. Perhaps they mean that the object's size can change but not its shape.

I'm also unclear about the connection between Killing vectors and Killing vector fields. If a Killing vector is simply any vector such that as one moves along it, the components of the metric don't change (as with a point at rest in a Schwarzschild gravitational field since the components of the Schwarzschild metric don't include time), then in a Cartesian plane every vector is a Killing vector since the metric [itex] ds^2 = dx^2 + dy^2 [/itex] is the same everywhere. Yet every vector field in a Cartesian plane isn't a Killing vector field, since for example for [itex] \vec{v}=(x,x) [/itex] we have [itex] \nabla_{x} v_{y} + \nabla_{y} v_{x}=0+1 \ne 0 [/itex].
No, (x,y) - i.e., the position vector - is not a Killing field. In Cartesian coordinates ##\nabla_ i = \partial_i## and ##v_i = v^i##. Hence,
$$
\nabla_x v_x + \nabla_x v_x = 2
$$
and therefore the Killing equation is not satisfied. The Killing equation must hold regardless of the indices you put in.
 
  • #11
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,343
7,214
As for the second question, there is no such thing as a Killing vector. It is just a shorthand for saying Killing vector field. A vector in itself cannot determine a symmetry.

It is not the case that (x,x) generates a symmetry. What is meant by a symmetry of the metric is that the pushforward of the metric tensor under the transformation returns the metric itself. For a continuous symmetry generated by a vector field ##X##, this is equivalent to ##\mathcal L_X g = 0##, i.e., the Lie derivative of the metric is zero. For certain conditions on the connection (I don't remember by heart the exact conditions, but the Levi-Civita connection does satisfy them), this turns out to be equivalent to the Killing equation.
 
  • #12
775
17
$$
\nabla_x v_x + \nabla_x v_x = 2
$$
I'm confused here. It looks like you took the partial derivative of [itex]v_x[/itex] with respect to x and added it to the partial derivative of [itex]v_x[/itex] with respect to x. Why is [itex]v_y[/itex] not part of this equation?
 
  • #13
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,343
7,214
I'm confused here. It looks like you took the partial derivative of [itex]v_x[/itex] with respect to x and added it to the partial derivative of [itex]v_x[/itex] with respect to x. Why is [itex]v_y[/itex] not part of this equation?
Because I am taking the x-x component of the Killing equation, which does not involve ##v_y##.
 
  • #14
Ibix
Science Advisor
Insights Author
2020 Award
8,320
7,747
I'm also unclear about the connection between Killing vectors and Killing vector fields.
As Orodruin said, Killing vector is shorthand for Killing vector field. He also gave the formal definition in terms of pushforwards, but a way to think of a KVF is that everywhere it points from a point towards a point that is somehow the same. The translation KVFs in Euclidean space thus look like (1,0) because a translation needs to be the same direction and distance everywhere. And you can see why Killing vectors at a point don't make sense - you need the transformation to be specified everywhere in order to be able to say "everywhere is the same under this transformation".
I'm confused here. It looks like you took the partial derivative of [itex]v_x[/itex] with respect to x and added it to the partial derivative of [itex]v_x[/itex] with respect to x. Why is [itex]v_y[/itex] not part of this equation?
Remember that Killing's equation has two free indices. You have ##n^2## equations in ##n## dimensions, of which ##n(n+1)/2## are linearly independent.
 
  • #15
775
17
Remember that Killing's equation has two free indices. You have ##n^2## equations in ##n## dimensions, of which ##n(n+1)/2## are linearly independent.
Ah! My mistake. I confess I was feeling insecure about how I was interpreting those repeating indices. I'm familiar with the Einstein convention but somehow here it confused me. Maybe because so far everywhere I've looked it's been called Killing's equation instead of Killing's equations, the way one usually sees "Einstein's field equations" (plural) even though as a shorthand it's always shown as a single equation.

Alrighty, thank you both. Will now try to understand Lie derivatives to get a deeper grasp of this matter.
 

Related Threads on What is a Killing field?

Replies
8
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
8K
Replies
4
Views
887
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
7
Views
4K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
8
Views
7K
Top