Difference between "g" and "g^ij"

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In summary, the conversation discusses the difference between the metric ##g## and the inverse metric ##g^{ij}##, with the latter being used to denote the elements on the inverse matrix while the former is typically used to represent the determinant of the matrix. The professor is asking for both the covariant matrix and its inverse, as well as its determinant, which is important for dealing with tensor densities and maintaining invariance under transformations.
  • #1
pabugeater
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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,
 
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  • #2
##g \equiv \text{det}(g_{ij})##.
 
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  • #3
##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:
$$g=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right) dx^i\otimes dx^j = g_{ij} dx^i\otimes dx^j.$$
 
  • #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.
 
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  • #5
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.
 
  • #6
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.
 
  • #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.
 
  • #8
pabugeater said:
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.

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.
 
  • #9
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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What is the difference between "g" and "g^ij"?

"g" and "g^ij" are both mathematical symbols used in the field of differential geometry. "g" refers to the metric tensor, which is a mathematical object that describes the local geometry of a curved space. It is used to calculate distances, angles, and other geometric properties in the space. "g^ij" is the inverse of the metric tensor, and it is used to raise and lower indices in equations. In other words, "g" describes the geometry of a space, while "g^ij" is used as a mathematical tool to manipulate equations in that space.

How are "g" and "g^ij" related?

"g^ij" is the inverse of "g", meaning that multiplying "g" and "g^ij" together will result in the identity matrix. In other words, "g" and "g^ij" are inversely proportional to each other.

Can "g" and "g^ij" be used interchangeably?

No, "g" and "g^ij" serve different purposes and cannot be used interchangeably. While "g" is used to describe the geometry of a space, "g^ij" is used as a mathematical tool to manipulate equations in that space. Using one in place of the other would result in incorrect calculations and interpretations.

What do the indices in "g^ij" represent?

The indices in "g^ij" represent the rows and columns of the inverse metric tensor. The number of indices corresponds to the dimensionality of the space being described.

How is "g^ij" calculated?

The inverse metric tensor, "g^ij", can be calculated by taking the inverse of the metric tensor, "g", using matrix algebra. This involves calculating the determinant of "g" and using it to find the inverse matrix.

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