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Simple GR Einstein Notation Question

  1. Oct 6, 2012 #1
    In reviewing some basic GR (just to keep my old brain sharp), i was looking at the Einstein notation cinvention and was a bit confused. I see how you do the dot product of say:

    ei.ej = δij

    (i.e. 1 or 0)

    But then the book I'm reading talks about ei.ej or ei.ej. Isn't that just the square of ei or ei? I thought the choice of index notation was irrelevant as it just goes over 1, 2, 3?

    Or am I missing something basic? (first time in my life!)

    If it is, why not just call it square?

    Thanks!

    Sterling
     
  2. jcsd
  3. Oct 6, 2012 #2
    Traditional vector algebra doesn't define a squaring operation on vectors.

    The point you should be understanding here is that, because both the indices are down (or up), a factor of the metric (or its inverse) must appear in the results of evaluating these expressions.
     
  4. Oct 6, 2012 #3

    vanhees71

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    I guess [itex]e_j[/itex] are some basis vectors (in the tangent space at each point of the space-time manifold) and [itex]e^j[/itex] are the corresponding dual basis vectors.

    Then the dot product is defined as the (pseudo-)metric on the space-time manifold and you have
    [tex]e^j \cdot e^k=g^{jk}[/tex]
    and
    [tex]e_j \cdot e^k =\delta_j^k[/tex]
    and
    [tex]e_j \cdot e_k = g_{jk}.[/tex]
    Note that these basis vectors are not necessarily holonomous coordinates wrt. to some generalized coordinates.

    A special very useful case for the use of such a system of basis vectors is to choose these orthogonal, i.e., such that [itex]g^{jk}=\eta^{jk}[/itex] with [itex](\eta^{jk})=\mathrm{diag}(1,-1,-1,1)[/itex]. Then the [itex]e^{j}[/itex] are a tetrad (or vierbein) field. For details have a look at the corresponding Wikipedia article

    http://en.wikipedia.org/wiki/Tetrad_(general_relativity)
     
  5. Oct 6, 2012 #4
    I was seeing ei or ei as a matrix, which my Linear Algebra book does allow for squaring. But regardless, I'm incorrect. I looked ahead in the book, and the examples shown obviously are not the square of the matrix. Or dot product by itself. I just don't see how they get their answer. Obviously I am missing something basic about the summation convention. Damn, first time in my life!

    I understand how you get each ei component: (∂r/∂u, ∂r/∂v, ∂r/∂w).

    And of course each ei component: ∇u, ∇v, ∇w

    Ahhhh! Wait a minute! I figured it out. I was treating ei as a three by three matrix multiplied by itself rather than three separate vectors, each of which can be dot producted together to create a three by three matrix! That's giving me the right answer according to the example in the book!

    Thanks, Muphrid! Your insight was key.

    Sterling
     
  6. Oct 6, 2012 #5
    Yes, I did see ei.ej defined as gij and vice versa. I'll get to the General Relativity subjects shortly enough, and it is fascinating. I just want to understand the basics of how to do the mathematics for right now. So far it's pretty simple basic linear algebra and dot products of equations, nothing more than a new way of looking at the basics of these that one learns in a second year of college mathematics. I breezed through multi d calc and linear algebra, but struggled with complex variables and my course on partial diff eq. Hopefully the math won't get too bad. The preface says I have all the pre requisites to conquer this subject! But now that I am in my fifties, I just hope I still have the brain power to conquer it after thirty years of not really using it all. I did go through my old multi-d, diff eq, and linear algebra books before starting this, but not as in depth as I did in college. We'll see. I's still interesting to do such mathematics after so many years!

    Sterling
     
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