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I Requesting clarification about metric tensor

  1. Dec 15, 2017 #1
    I am a little bit confused about the metric tensor and would like some feedback before I proceed with my learning of GR.

    So I understand that metric tensor describes the geometry of the space itself. I also understand that the components of the metric tensor (any tensor for that matter) come into existence only when the coordinate system is selected. However, consider the equations for the components of the metric tensor, viz.,
    $$
    g_{ij} =
    \frac{ \partial{x'^1}}{\partial{x^i}} \frac {\partial{x'^1}}{\partial{x^j}} +
    \frac{ \partial{x'^2}}{\partial{x^i}} \frac {\partial{x'^2}}{\partial{x^j}} +
    \frac{ \partial{x'^3}}{\partial{x^i}} \frac {\partial{x'^3}}{\partial{x^j}}
    $$
    This equation implies that the metric tensor is not just about the space itself, not even about ##a## chosen coordinate system alone, but is a function of ##two## coordinate systems, the primed and the unprimed. In particular, if we choose, say, the spherical polar coordinate system, typically we choose the Cartesian coordinate system as the primed one, and derive the metric tensor. But the above equation ##-## I think ##-## holds for any two coordinate systems and there is nothing stopping us from choosing the cylindrical coordinates instead of Cartesian as the primed system. So the metric tensor is really a function of ##two## coordinate system, in addition to describing the geometry of the space itself?

    I understand the reasoning for choosing the Cartesian coordinate system as the primed one, namely that the differential length in it is evaluated as ##ds^2=dx^2+dy^2+dz^2##, but it appears that the Cartesian coordinates are being given a special status. Can someone clarify to help me understand this better?

    Thanks.
     
  2. jcsd
  3. Dec 15, 2017 #2

    PeterDonis

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    "Come into existence" is not a good way to describe it. The metric tensor exists independently of any coordinate system. Selecting coordinates defines components of the tensor for those particular coordinates.

    Where are you getting these equations from?
     
  4. Dec 15, 2017 #3
    Ok, I'll take your suggestion that "Selecting coordinates defines components of the tensor for those particular coordinates."
    I had meant numerical values or component expressions become more concrete when coordinates are fixed, but your description is better.

    The equation is from Daniel Fleisch's "student's guide to vectors and tensors" (pg 144, eqn 5.15)
     
  5. Dec 15, 2017 #4

    PeterDonis

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    Are you sure it's an equation for the components of the metric tensor? It looks like an equation for a coordinate transformation, which is not the same thing.
     
  6. Dec 15, 2017 #5

    Ibix

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    Fleisch p144:
    https://books.google.co.uk/books?id..._selected_pages&cad=2#v=onepage&q=144&f=false

    He seems to be expressing ##g_{ij}=e_i\circ e_j##, where the es are basis vectors in some coordinate system. So I think Peter is correct to say this is a kind of transformation.

    Unfortunately p143 doesn't seem to be part of the preview, so it's a little difficult to tell...
     
  7. Dec 15, 2017 #6

    Orodruin

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    I am on my phone and cannot open the preview right now, what is his definition of ##\circ##? Is it just the standard inner product? In that case it is a particular expression for the metric tensor on a Euclidean space with ##\vec e_i = \partial_i \vec x##.

    The expression given in the OP seems to me to be the expression for the metric tensor components in an arbitrary set of coordinates ##x## based on its components in a Cartesian coordinate system ##x’##.
     
  8. Dec 15, 2017 #7
    I'll post some more details later, please hold on.
     
  9. Dec 15, 2017 #8
    y.jpg
    Attached is the picture of pg 144-45.

    What Fleisch says is that in the transformation eqn.
    $$
    \begin{bmatrix}
    dx'^1\\
    dx'^2\\
    dx'^3
    \end{bmatrix} =
    \begin{bmatrix}
    \frac{ \partial{x'^1}}{\partial{x^1}} \frac{\partial{x'^1}}{\partial{x^2}} \frac{\partial{x'^1}}{\partial{x^3}}\\
    \frac{ \partial{x'^2}}{\partial{x^1}} \frac{\partial{x'^2}}{\partial{x^2}} \frac{\partial{x'^2}}{\partial{x^3}}\\
    \frac{ \partial{x'^3}}{\partial{x^1}} \frac{\partial{x'^3}}{\partial{x^2}} \frac{\partial{x'^3}}{\partial{x^3}}
    \end{bmatrix}
    \begin{bmatrix}
    dx^1\\
    dx^2\\
    dx^3
    \end{bmatrix}
    $$
    each column of the transformation matrix is the component of the original, unprimed basis vector, expressed in the new, primed coordinate system, which you can see is true if you refer to eqn 1.20 of his book. Therefore, ##g_{ij}=e_i \circ e_j## actually becomes the eqn that I quoted in the OP:
    $$
    g_{ij} =
    \frac{ \partial{x'^1}}{\partial{x^i}} \frac {\partial{x'^1}}{\partial{x^j}} +
    \frac{ \partial{x'^2}}{\partial{x^i}} \frac {\partial{x'^2}}{\partial{x^j}} +
    \frac{ \partial{x'^3}}{\partial{x^i}} \frac {\partial{x'^3}}{\partial{x^j}}
    $$
    I was able to correctly derive the metric tensor of 2D and 3D polar and spherical coordinate systems based on this understanding. See my notes on pg 145 in the picture.

    But irrespective of this eqn, given the spherical polar coordinate system, for example, the components of the metric tensor assumes a Cartesian system for evaluation. That's why I am thinking that, in addition to the geometry of the space itself, there are ##two## coordinate systems involved in determining the metric tensor.

    Secondly, consider the Christoffel symbol (quoted below, from another book):
    $$
    \Gamma^\lambda_{\ \mu\nu}\equiv \frac{\partial{x^\lambda}}{\partial{X^\alpha}} \frac{\partial^2{X^\alpha}}{\partial{x^\mu} \partial{x^\nu}}
    $$
    ##X^\mu## and ##x^\mu## refer to basis vectors of two different reference frames. In this affine connection, definitely there are two ref frames/coordinate systems involved. We also know that ##\Gamma## can be completely expressed in terms of components of metric tensor, ##g##, as ##\Gamma^\lambda_{\ \mu\nu}=
    \frac{1}{2}g^{\lambda\rho}[
    \frac{ \partial{g_{\rho\nu}}}{\partial{x^\mu}}+
    \frac{ \partial{g_{\rho\mu}}}{\partial{x^\nu}}-
    \frac{ \partial{g_{\mu\nu}}}{\partial{x^\rho}}]## (The derivatives are wrt the unprimed basis vectors). So I suspected that the metric tensor components have to have ##two## coordinate systems involved. Hence the original question, "So the metric tensor is really a function of ##two## coordinate system, in addition to describing the geometry of the space itself?"

    I hope I am not way off in my understanding...
     
  10. Dec 15, 2017 #9

    PeterDonis

    Staff: Mentor

    Yes.

    No. The equation ##g_{ij} = e_i \circ e_j## is for the components of the metric tensor in a single coordinate system. What you have written down is an expression that involves two coordinate systems. They're not the same.

    Sorry, that's not in accordance with the PF rules. If you want us to critique your derivation (which might be a good idea since it can't possibly be doing what you think it's doing given the above), you need to use the PF LaTeX feature to post it directly here.

    No, they don't. You can write the metric tensor in any coordinates you like. Where are you getting this from?

    This is not correct.

    I think we need to hold off on discussing anything else until you have a correct understanding of the metric tensor.
     
  11. Dec 15, 2017 #10

    PeterDonis

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    Expanding on this a bit: I realize that the equation in the OP appears in the book, but I am frankly confused about what the authors can possibly mean. It seems like they are just inviting confusion and misunderstanding. Possibly someone who is more familiar with this textbook than I am can clarify their approach.
     
  12. Dec 15, 2017 #11

    Orodruin

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    This is not correct. You can use the Euclidean space as a grounds for developing an intuition for the metric and other concepts, but in the more general case of an arbitrary manifold, the implications and discussions will generally go in the other direction. A metric tensor is just a (0,2) tensor on the manifold that is (a) symmetric, (b) positive definite, (c) if ##g(X,X) = 0## then ##X = 0##. The metric defines distances and angles on the manifold, which includes defining concepts such as inner products between tangent vectors. By definition the components are given by ##g_{ij} = g(\partial_i,\partial_j)##. One way of uniquely defining the metric tensor is to give its components in a particular coordinate system. There is no need to introduce a second one. In the example you have come across, the metric tensor is defined in the original Cartesian coordinate system and your equation is just the transformation of those components to spherical coordinates. This does not mean you could not have defined it in spherical coordinates to start with.

    You probably mean that ##X^\mu## and ##x^\mu## are the coordinates in different coordinate systems, the coordinates are not vectors.

    Furthermore, again this is just a part of the general transformation rule for the Christoffel symbols to a general coordinate system from a Cartesian coordinate system on a Euclidean space (where the Christoffel symbols are zero!).

    I am not familiar with this particular textbook. However, I do take a similar approach to acquainting the reader with the metric and Christoffel symbols when I first discuss tensors in general coordinates in a Euclidean space. You can introduce the metric as a tensor with components ##g_{ij} = \vec E_i\cdot \vec E_j## and the Christoffel symbols as ##\partial_i \vec E_j = \Gamma_{ij}^k \vec E_k##, where ##\vec E_i = \partial_i \vec x## in the Euclidean case. Again, this is just to provide the student with some understanding of what the concepts are and how they are used and to get some familiarity with them. Later, when discussing general manifolds, I turn everything around and show that the natural way of doing things is just the other way around, i.e., the metric defines the inner product and not the other way around.
     
  13. Dec 15, 2017 #12

    Orodruin

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    To be clear, what you are doing here is just transforming the metric tensor, using the knowledge that in Cartesian coordinates ##g_{ij} = \delta_{ij}##. It has nothing to do with the definition of the metric tensor itself.
     
  14. Dec 15, 2017 #13
    It could be that the equation holds for orthogonal systems only.
     
  15. Dec 15, 2017 #14

    Orodruin

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    Sorry, but now you are just guessing. The transformation equation is the general transformation of the metric tensor from a Cartesian coordinate system to a general coordinate system. It is just
    $$
    \newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
    g'_{ij} = g_{ab} \dd{x^a}{x'^i} \dd{x^b}{x'^j}
    $$
    with ##g_{ab} = \delta_{ab}## in the Cartesian coordinate system and the sums over ##a## and ##b## explicitly written out.
     
  16. Dec 15, 2017 #15
    Thanks to Orodruin and PeterDonis. I think my mistake was to take a specific example and generalize it. I'll have to be more careful, but thanks!
     
  17. Dec 15, 2017 #16
    I have to clarify that Fleisch's book (a student's guide to vectors and tensors) is one of the best introductory books on tensors; best explanations on contravarient and covarient components. It helped me very well.
     
  18. Dec 15, 2017 #17

    FactChecker

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    A tensor exists as a physical or geometric object without specifying a coordinate system or the associated components. In order to qualify as a tensor, its expression in two coordinate systems must follow the covariant/contravariant transformation rules. Mathematically, a tensor can be considered an equivalence class where the components in any two coordinate systems transform according to the tensor covariant/contravariant transformation rules.

    I think it's similar for a metric tensor. The equation in the OP must be satisfied for any two coordinate systems in order to call it a metric tensor. That does not mean that that equation is necessary to define the tensor or to make it exist.
     
  19. Dec 15, 2017 #18

    Orodruin

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    I never liked the definition of tensors saying ”you have these components that transform like this”. I find it much more elegant to define them in a coordinate free manner (linear maps from n copies of the tangent space to m copies of the tangent space). It becomes much clearer that tensors are single coordinate independent objects and once selecting coordinate systems, the transformation rules follow immediately.

    The equation is just the coordinate transformation from Cartesian coordinates to an arbitrary system so it is only necessary that it holds for transformations from Cartesian coordinates (not from any system).
     
  20. Dec 17, 2017 #19
    Can you please point me to some literature that does this? I would like to see, at least in principle, how the components of the metric tensor for spherical coordinates (for example) can be derived straight away without going through the Cartesian coordinate system.
     
  21. Dec 17, 2017 #20

    stevendaryl

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    Derived starting from what? If all you know is that you have three coordinates, [itex]r, \theta, \phi[/itex], you of course can't derive the metric, because there are infinitely many possible metrics expressible in terms of those coordinates.
     
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