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Vectors in index notation

  1. Feb 2, 2013 #1
    given the vector in the first equation below, does that necessarily imply the third equation, as shown?
    [tex]{{u}_{a}}{{e}^{a}}={{x}_{a}}{{e}^{a}}[/tex]
    [tex]{{u}_{a}}{{e}^{l}}g_{l}^{a}={{x}_{a}}{{e}^{l}}g_{l}^{a}[/tex]
    [tex]{{u}_{a}}{{e}^{l}}={{x}_{a}}{{e}^{l}}[/tex]
     
  2. jcsd
  3. Feb 2, 2013 #2
    Hi redstone
    No
    In the first line, you have a single scalar equation, what you have written is that the dot product of u with e is the same as the dot product of x with e.
    That of course does not mean that u = x unless your space is of dimension 1
    In the third line, you have many equations saying that all components of u and x for any given index give the same result when multiplied by any component (but the same for u and x) of e, which lets you conclude that u = x
     
  4. Feb 2, 2013 #3

    vela

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    That's not true. What if ##e^l=0## for all ##l##?

    Regarding the original post, it looks like you're trying to cancel ##g^a_l## from both sides of the equation. You can't do that.
     
  5. Feb 2, 2013 #4
    Hi vela
    well, of course if e is the null vector there isn't much to conclude neither in the first equation nor the third.
    I wasn't trying to tell some general always valid truth, I was just trying to show how the first equation couldn't possibly lead to the third without just saying that, indeed, you can't do the second step :)
     
  6. Feb 3, 2013 #5

    CompuChip

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    The choice of e as opposed to any other letter suggests that [itex]e^l[/itex] is not an arbitrary vector but a basis of the vector space. Although vela has a point that we can't really assume that until redstone gives us a bit more information on what he is actually looking for.
     
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