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Difference between "g" and "g^ij"

  1. Jul 26, 2014 #1
    I have a metric given as a line element differential ds^2. I have written down the 3x3 matrix with the coefficients. I am then asked to determine "g" (which I think i just did by enumerating the matrix) and "g^ij", which I normally would assume means the same thing, but apparently not.

    So what exactly is the difference?

    Many thanks,
     
  2. jcsd
  3. Jul 26, 2014 #2

    WannabeNewton

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    ##g \equiv \text{det}(g_{ij})##.
     
  4. Jul 26, 2014 #3

    Fredrik

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    ##g^{ij}## typically denotes what's on row i, column j of the inverse of the matrix with ##g_{ij}## on row i, column j. (Note that this ensures that ##g^{ij}g_{jk}=\delta^i_k##).

    I would guess that g isn't the determinant here, but I suppose it could be. Has that definition been used in other problems, or in the text? If not, then I think you're just dealing with this:
    $$g=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right) dx^i\otimes dx^j = g_{ij} dx^i\otimes dx^j.$$
     
  5. Jul 26, 2014 #4
    Thank you both for your replies, but I too do not think the determinant is what the prof is looking for. For some context, I am not looking for actual answers to the problems, once I figure out what the prof wants I intend to do all the math myself! I'm freshly back in the classroom after graduating with a BS in biology over 40 years ago, taking GR for the first time, so the learning curve has been steep. This is actually homework from the spring semester that I'm slowing slogging through.

    The only context I have is a problem a bit later when he's asking for both the contravariant and covariant g's (along with connection coefficients and more, see image below, I'm working on problem 5):

    http://www.lehigh.edu/~sol0/HW2.png [Broken]

    So g has off-diagonal elements, and I suppose you are correct in that gij is the inverse matrix.
     
    Last edited by a moderator: May 6, 2017
  6. Jul 26, 2014 #5

    Matterwave

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    Looking at that paper...it looks to me like g is the determinant...

    The professor gives you the covariant ##g_{ij}## and is asking you to find its inverse ##g^{ij}## and its determinant ##g##. That's how I would see the problem anyways.
     
  7. Jul 26, 2014 #6

    Fredrik

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    I agree. It would be weird to ask for both g and ##g_{ij}## if g is just the metric. So it probably is the determinant.
     
  8. Jul 26, 2014 #7
    Thank you all, I'll run with the metric, its inverse and determinant, although it's unclear to me (at this time) why the determinant is important.
     
  9. Jul 26, 2014 #8

    WannabeNewton

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    The metric determinant shows up naturally when dealing with tensor densities, particularly with regards to integration.

    But you won't see this for a while so I wouldn't worry too much about the actual use of ##g## for now.
     
  10. Jul 26, 2014 #9

    ChrisVer

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    For example you have the volume element:
    [itex]dV= d^{4}x = dx^{0} dx^{1} dx^{2} dx^{3} [/itex]
    Suppose you don't work in a flat spacetime, but in a spacetime with a general metric with elements [itex]g_{\mu \nu}[/itex].
    Then you can make a transformation [itex]x^{\mu} \rightarrow \bar{x}^{\mu}= \bar{x}^{\mu}(x^{\nu})[/itex], and see that the volume element is no longer invariant. In order to be invariant you have to allow:
    [itex]dV= \sqrt{|g|} d^{4}x [/itex]
    which is the invariant volume element which you can use for integrations. [itex]|g|[/itex] is the absolute value of the determinant of your metric ([itex]g_{\mu \nu}[/itex]).
     
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