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General Relativity math, Tangent vectors on manifolds

  1. Jun 20, 2011 #1
    I am currently going through some online notes on differential geometry and general relativity. So far I have been following pretty well until I got to 3.4 cont'd letter (e) of http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec3.html" [Broken]document and the definition directly after. My first question is, is yi(t) many different paths on the manifold or is each i a component of one path on the manifold? Also in the definition after the example I think yj are the ambient components of the actual manifold, not the path, and xi are the local components of the manifold, not the path. Is this correct? And if so how do you take the partials if the number of components of the ambient coordinates are usually one more then the number of components of the local coordinates?

    If my questions do not make sense, just ask and I will try to make them more clear, and if this is not the right place for this just direct me where to go. Also if anyone had any better suggestions on places to learn this stuff I am happy to hear it.
     
    Last edited by a moderator: May 5, 2017
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  3. Jun 20, 2011 #2

    pervect

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    The later, it's one path.

    I'm not following this. The yj are just coordinates. A complete set of coordinates specifies a point on the manifold, just like it does on a map. Though a purist would note that a point is different from the coordinates that specify it.

    I suspect you're confused somehow about the first defintion, but I can't quite grasp how you're confused, so maybe I'm wrong and I'm confused. Anyway....

    In loose langauge, the first definition just says that a smooth path is just a mapping from a line segment on R to a set of points on the mainfold (I'd call it a curve on the manifold, but that's sort of what we're trying to define), such that every point on the line corresponds to one point on the manifold, said point specified by its coordinates, yj, and the mapping is "smooth". They haven't gotten into the details of "smoothness" in the notes, so I won't get into it either.
     
    Last edited by a moderator: May 5, 2017
  4. Jun 20, 2011 #3

    atyy

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    Any point on the N-dimensional manifold is specified by N coordinates {x1...xN}.

    Roughly, to specify a curve, which is 1-dimensional, we need N-1 equations to "get rid" of N dimensions. These are the N equations {x1=f1(t) ... xN=fN(t)}, which we think of as N-1 equations by notionally eliminate t to obtain N-1 relations among N coordinates.

    As an explicit example, in 2D with coordinates {x,y} a circle can be written as {x=cost,y=sint} or x2+y2=1.

    Semantic note: many define a curve to be a parametrized path so that {x=cost,y=sint} is a curve, but x2+y2=1 is a path.
     
    Last edited: Jun 20, 2011
  5. Jun 20, 2011 #4
    Thanks for the reply.

    I am pretty sure I understand the basic definition of a path. Where I am running into trouble is I thought that the ambient coordinates consist of s coordinates and the local coordinates consist of n coordinates where s = n +1. Basically the manifold is embedded into a one higher dimension. However, in that example (e) they assign xj based off of if i = j, however I thought that j > i since there are more y coordinates then x coordinates. Therefore how can you compare the two and make a Kronecker delta which I thought was supposed to be a square matrix.
     
  6. Jun 20, 2011 #5

    atyy

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    i,j are just indices which he uses for both local and ambient coordinates, so whether they run from 1...n or 1...s depends on whether local or ambient coordinates are being indexed (I think). So in 3.4e the Kronecker delta indices run from 1...n, the dimension of the manifold, not the ambient space.
     
    Last edited: Jun 20, 2011
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