3
$\begingroup$

This may be a dumb or vague question:

Is there any criterion that a metric tensor needs to fulfill such that coordinates it is expressed in can be called global. Or alternatively what is the definition of global coordinates? For instance why is one coordinate system for $AdS_n$ called global while others are not?

In particular I am not asking if global coordinates exist but rather how one can check if a set of given coordinates is global. By inspection of the metric tensor or what is the same thing the line element.

$\endgroup$
  • $\begingroup$ This might be helpful < math.stackexchange.com/questions/1110580/… > $\endgroup$ – N. Steinle Sep 16 '18 at 15:41
  • $\begingroup$ Actually, this might answer your question < mathoverflow.net/questions/308925/… > $\endgroup$ – N. Steinle Sep 16 '18 at 15:43
  • $\begingroup$ @N.Steinle you don't need the less-than/greater-than symbols around links. Also, you can also use [text](link) notation for hyperlinks as well. $\endgroup$ – Kyle Kanos Sep 16 '18 at 16:41
  • $\begingroup$ "Matrix tensor" is sort of redundant, since any tensor is a linear operator, and any matrix that is of interest in physics probably transforms as a tensor (or something closely related, like a tensor density). Do you really mean metric tensor? If you don't mean "metric tensor," then I don't understand what the material about a "matrix tensor" has to do with the rest of the question -- what matrix do you have in mind, and why would it be relevant? $\endgroup$ – Ben Crowell Sep 16 '18 at 17:20
  • $\begingroup$ That matrix part was a strange typo. Of course I meant metric I fixed this. Also the links N. Steinle sent are as far as I can see just more related to existence. I was asking however how one could check for a particular choice of coordinates if it is global. $\endgroup$ – Michael Sep 16 '18 at 18:08
2
$\begingroup$

The bare minimum requirement for a coordinate chart to be usable in general relativity is essentially that the metric, expressed as a matrix, is always finite and invertible. That is, both $g_{ij}$ and $g^{ij}$ must exist. This requires both that the metric be nondegenerate (which is a coordinate-independent criterion, basically the metric has to have the right signature) and that there are no coordinate singularities when the metric is expressed componentwise in terms of these particular coordinates.

Depending on what we want to do, we will usually want to impose stricter regularity requirements than this. For example, we probably want the metric, expressed in these coordinates, to be such that we can take the derivatives that are required in order to calculate the Riemann tensor -- otherwise we wouldn't be able to state the Einstein field equations.

A valid set of global coordinates is simply a valid coordinate chart that covers all points in the spacetime. There is no requirement that we work in a global chart or that such a chart exist.

$\endgroup$
  • $\begingroup$ So is a sufficient criterion that the determinant is non-zero at every point(invertibility you mentioned) and that there are no singularities on the range of values variables are defined on? $\endgroup$ – Michael Sep 16 '18 at 21:46
  • $\begingroup$ @Michael: That would be necessary, but not sufficient for most purposes. The second paragraph of my answer describes stricter criteria that would be necessary and sufficient for many purposes in GR. There is no universally agreed upon criterion for what is the right regularity condition, because it depends to some extent on what you're trying to do. E.g., if you look at a mathematically rigorous book like Hawking and Ellis, they describe a whole family of different regularity conditions, which can be dialed up or down as needed. $\endgroup$ – Ben Crowell Sep 17 '18 at 21:15
  • $\begingroup$ Ok sounds good I accepted your answer as the best for this question $\endgroup$ – Michael Sep 18 '18 at 3:12
0
$\begingroup$

Basically, this is due to the topology of spacetime. It is also due to how we represent coordinates.

Take a sphere - for example the earth. If we were to try to represent with the kind of charts we see in books - that is a square - we see we can never do this with just one page - we always need more than one. This is because a square is essentially flat but a sphere is not.

Were we to instead try to represent it with a globe we see straightaway that only one is neccessary.

Now, the mathematical apparatus of differential geometry relies on an atlas of patches and these patches are modelled only on the analogue of squares here - Euclidean spaces. It doesn't allow any other kind.

Thus - if you are modelling spacetime as a sphere - and using standard differential geometry to represent it - your atlas will not be global - in the sense of requiring just one patch - it will always need more than one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.