Calculating the killing vector fields for axial symmetry

In summary, the conversation discusses the calculation of killing vector fields for axial symmetry in a flat disk galaxy. The participants mention using cylindrical coordinates and the difficulties in finding non-trivial killing vector fields. They also discuss the importance of understanding the metric and using it to determine the killing vector fields. The conversation ends with a mention of particles in flat universes.
  • #1
tut_einstein
31
0
hi,

i need to calculate the killing vector fields for axial symmetry for a project so i can study the galaxy rotation curves. i am assuming the galaxy to be a flat disk, in addition to being axially symmetric. so i figured that the killing vector fields with respect to which the metric manifold on the galaxy will be unchanged are:


firstly, I'm using cylindrical coordinates: (t,r,z,phi) t=time (3+1)

d(phi) (for axial symmetry) where phi is the angle and
d(z) because the galaxy is a flat disk

i don't know if there will be other killing vector fields. the problem is that i don't know of a systematic way to calculate them.

thanks for the help!
 
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  • #2
dz won't be a KVF because a flat disc is not symmetric under z-translation: it would have to be an 'infinite cylinder' sort of object to satisfy this. Depending on what you're doing, dt might be a KVF though (if the setup is stationary).
 
  • #3
Does your project require you to know the non - trivial killing vector fields (if any) for this metric? Do you have the metric in hand btw? You can tell the trivial (usually the important ones) killing fields just by looking at the metric and seeing which coordinates are cyclic but if you really need to find all of them the obvious, yet rather painful (depends on your metric but for 3 + 1 space - time this method is always painful by hand) is to solve killing's equations [itex]\bigtriangledown _{\alpha }\xi _{\beta } + \bigtriangledown _{\beta }\xi _{\alpha } = 0[/itex]. I don't think there is any simper way to find the non - trivial ones and I use the word simple loosely.
 
  • #4
WannabeNewton said:
Do you have the metric in hand btw? You can tell the trivial (usually the important ones) killing fields just by looking at the metric and seeing which coordinates are cyclic

Things are more subtle than this. As an example, consider Minkowski space. In an inertial coordinate system, the four Killing vectors for spacetime translations are obvious, but the six equally important Killing vectors for rotations and boosts are not immediately obvious.

The metric may even be given in a form in which no Killing vectors are obvious. If the metric is given, then
WannabeNewton said:
yet rather painful (depends on your metric but for 3 + 1 space - time this method is always painful by hand) is to solve killing's equations [itex]\bigtriangledown _{\alpha }\xi _{\beta } + \bigtriangledown _{\beta }\xi _{\alpha } = 0[/itex]. I don't think there is any simper way to find the non - trivial ones and I use the word simple loosely.

It sounds, however, like tut_einstein might want to solve the inverse problem: given some symmetries (stationary, axisymmetric?), what des the metric look like? For a treatment of Killing vectors and stationary axisymmetric spacetimes, I recommend section 20.10 from An Introduction to general Relativity and Cosmology by Plebanski and Krasinski,

https://www.amazon.com/dp/052185623X/?tag=pfamazon01-20
 
  • #5
George Jones said:
Things are more subtle than this. As an example, consider Minkowski space. In an inertial coordinate system, the four Killing vectors for spacetime translations are obvious, but the six equally important Killing vectors for rotations and boosts are not immediately obvious.

Point taken. I guess since people automatically associate the isometries of the poincare group with minkowski space, looking at the respective metric wouldn't really be much of a help at all in such a case. I, however, have never seen a metric where not even one killing vector was immediately obvious from the presence of a cyclic coordinate (if not in one coordinate system then in some other). It would be awesome if you could give me an example of one or point me to one, thanks.
 
  • #6
WannabeNewton;3372857I said:
however, have never seen a metric where not even one killing vector was immediately obvious from the presence of a cyclic coordinate (if not in one coordinate system then in some other). It would be awesome if you could give me an example of one or point me to one, thanks.

For any Killing vector field there always exists a cyclic coordinate system, and I didn't write anything that contradicts this. I wrote
George Jones said:
The metric may even be given in a form in which no Killing vectors are obvious.

Consider the Euclidean two-dimensional plane, which has three Killing vector field, two for translations and one for rotations. In elliptical coordinates, the Killing vectors are not obvious. The Euclidean metric expressed with respect to these coordinates is given by third equation of section Scale factors of

http://en.wikipedia.org/wiki/Elliptic_coordinate_system.

That this metric is for the Euclidean plane might not be obvious to someone (like me) who is unfamiliar with elliptical coordinates.
 
  • #7
George Jones said:
For any Killing vector field there always exists a cyclic coordinate system, and I didn't write anything that contradicts this. I wrote
Oh no I wasn't saying you contradicted it, my apologies if it came out that way. I was just asking if there was some example where a metric could be put into any coordinate system without ever making at least one of the killing vector fields obvious.
 
  • #8
Sorry, I didn't mean to sound so harsh.

(PS I haven't forgotten about particles in flat universes.)
 

What is a killing vector field?

A killing vector field is a vector field on a Riemannian manifold that preserves the metric at every point. This means that the metric tensor of the manifold is invariant under the flow of the killing vector field.

What is axial symmetry?

Axial symmetry is a type of symmetry in which an object or system can be rotated about an axis without changing its overall appearance or properties. In mathematics, this means that the object or system has a killing vector field in the direction of the axis of rotation.

Why is it important to calculate killing vector fields for axial symmetry?

Calculating killing vector fields for axial symmetry allows us to understand and analyze the symmetries of a given system or object. This can be useful in various fields such as physics, engineering, and mathematics.

How do you calculate killing vector fields for axial symmetry?

To calculate the killing vector fields for axial symmetry, we use the Killing equation, which relates the killing vector field to the metric tensor of the manifold. We also use the symmetry properties of the object or system in question to simplify the calculations.

What are some applications of calculating killing vector fields for axial symmetry?

Some applications of calculating killing vector fields for axial symmetry include studying the behavior of physical systems with axial symmetry, solving differential equations in physics and engineering, and understanding the symmetries of mathematical objects such as surfaces and manifolds.

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