Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating the killing vector fields for axial symmetry

  1. Jun 24, 2011 #1
    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!
     
  2. jcsd
  3. Jun 24, 2011 #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).
     
  4. Jun 24, 2011 #3

    WannabeNewton

    User Avatar
    Science Advisor

    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.
     
  5. Jun 24, 2011 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
    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/Introduction...623X/ref=sr_1_1?ie=UTF8&qid=1308922683&sr=8-1
     
  6. Jun 24, 2011 #5

    WannabeNewton

    User Avatar
    Science Advisor

    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.
     
  7. Jun 24, 2011 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For any Killing vector field there always exists a cyclic coordinate system, and I didn't write anything that contradicts this. I wrote
    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.
     
  8. Jun 24, 2011 #7

    WannabeNewton

    User Avatar
    Science Advisor

    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.
     
  9. Jun 24, 2011 #8

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sorry, I didn't mean to sound so harsh.

    (PS I haven't forgotten about particles in flat universes.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculating the killing vector fields for axial symmetry
  1. Killing fields (Replies: 6)

  2. Killing vector fields (Replies: 3)

Loading...