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About coordinate transformations in general

  1. Sep 17, 2010 #1
    I want to make sure of my understanding of coordinate transformations.

    First of all, is it true that if [tex]\[{x_i}\][/tex] is a coordinate system on a manifold, then

    [tex]\[{q_j} = {q_j}({x_i})\][/tex]

    is a coordinate transform from "x" space to "q" space? If so can "x" be a flat space and "q" a curved space? Thanks.
  2. jcsd
  3. Sep 18, 2010 #2
    q must be congruous & compatible for the transform to be valid & not just local. Whether to call a manifold 'flat' is by one's personal choice.
  4. Sep 18, 2010 #3
    Usually whether the manifold is flat or not - that depends on the curvature tensor. Curvature tensor is a tensor - it transforms in a linear way. If it is zero in one coordinate system, it is zero in any other coordinate system.

    But you can have your own definition of being flat or not-flat. It needs to be given first in order to answer your question. Give a simple example of what is on your mind - and then you will certainly get some help.
  5. Sep 18, 2010 #4


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    The [tex]\[{q_j} \][/tex] can not be arbitrary. The mapping from x's to q's must at least be a homeomorphism but often it is required that it have more structure e.g. it must be a diffeomorphism.

    The geometry of a space does not depend upon coordinates but rather upon measurement of distance and angle. This requires further structure such as a Riemannian metric.

    Often when one parameterizes a region on a manifold, the parameter domain is just a subset of Euclidean space and so is flat while the region on the surface may curve around in a higher dimensional space and thus is curved. But this has nothing to do with the coordinates transformations themselves but rather with the geometry of the regions.
  6. Sep 19, 2010 #5
    I'm trying to understand how the curvature (derived from the metric, and thus the metric) can be inherant in the manifold and also how you can have your own definition of "flat".

    Can the metric be defined apart from a coordinate systems, say, d(p1,p2), where p1 and p2 are points on the manifold somehow specified without coordinates and d(p1,p2) is the distance between p1 and p2? But then again a distance is a number, which can be arbitrarily scaled, right? Or is the curvature scale invariant?
  7. Sep 19, 2010 #6
    The curvature tensor is scale invariant. But not the so called "scalar curvature".

    Moreover, "flatness" depends on the "parallel transport". Two metrics differing by a constant scale factor define the same parallel transport. Parallel transport has a simple geometrical interpretation and can be defined without referring to coordinates. The only thing you need are "differentiable curves" and "tangent vectors". You do not even need a "metric".
  8. Sep 19, 2010 #7


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    A Riemannian metric is independent of the coordinate system. The distance between points is derived from the metric and for a complete manifold is equal to the length of a distance minimizing geodesic.

    The curvature 2 forms are intrinsic because they are derived from the metric and not from the particualr realization of the manifold in another manifold e.g. in Euclidean space.
  9. Sep 23, 2010 #8
    For example, consider the projection of the hemisphere onto a plane. Here we have

    [tex]\[x = r \cdot \sin \theta \cdot \cos \varphi \][/tex]

    [tex]\[y = r \cdot \sin \theta \cdot \sin \varphi \][/tex]

    for fixed radius r, and the inverse,

    [tex]\[\theta = {\cos ^{ - 1}}(\frac{{\sqrt {{r^2} - {x^2} - {y^2}} }}{r})\][/tex]

    [tex]\[\varphi = atan 2(y,x)\][/tex],

    where atan2 is a variant of arctan, only it covers all quadrants.

    The point is, here

    [tex]\[\theta = \theta (x,y)\][/tex] and [tex]\[\varphi = \varphi (x,y)\][/tex]

    This would appear to be a coordinate transformation from flat space of [tex]\[(x,y)\][/tex], to a curved space of [tex]\[(\theta ,\varphi )\][/tex]. Am I missing something?
    Last edited: Sep 23, 2010
  10. Sep 24, 2010 #9
    It can be viewed in three ways:

    1) you have a flat space, with flat parallel transport in coordinates (x,y), you have a curved hemisphere with the natural parallel transport of the sphere and in coordinates [tex](\theta,\varphi)[/tex] , you have a map from a flat Riemannian manifold to a non-flat one.
    2) Both [tex](x,y)[/tex] and [tex](\theta,\varphi)[/tex] are just different coordinates on the open unit disc of the flat plane
    3) Both [tex](x,y)[/tex] and [tex](\theta,\varphi)[/tex] are just different coordinates on the curved hemisphere
  11. Sep 28, 2010 #10
    I'm trying to reconcile these statements with others that say that the curvature cannot change with a change of coordinates, which would preclude a coordinate transform from flat to curved space.

    Is it possible that there may be contexts in which these statements apply? For example, I'm wondering about the difference between a "re-parameterization" and a "diffeomorphism". I can certainly imagine that a re-parameterization of a surface is not going to change its curvature. But then General Relativity talks about diffeomorphism invariance, where the laws of physics do not change with a diffeomorphism, the entire point of which is that a change in curvature does not change physics. Is this diffeomorphism a coordinate transformation as described above? Thanks.
  12. Sep 29, 2010 #11
    The entire point is that a change of coordinates does not change the curvature. Changing the curvature changes physics. That's the entire point.

    Curvature is coded in the curvature tensor, not in coordinates.
  13. Oct 10, 2010 #12
    It seems to me that it would be a straightforward procedure to prove that curvature is invariant wrt coordintate changes since we have:

    The scalar curvature is

    [tex]\[R = {g^{\sigma \nu }}{R_{\sigma \nu }}\][/tex]

    where [tex]\[{R_{\sigma \nu }}\][/tex] is the Ricci tensor

    [tex]\[{R_{\sigma \nu }} = {R^\rho }_{\sigma \rho \nu }\][/tex]

    where [tex]\[{R^\rho }_{\sigma \mu \nu }\][/tex] is the Rieman curvature tensor

    [tex]\[{R^\rho }_{\sigma \mu \nu } = {\partial _\mu }{\Gamma ^\rho }_{\nu \sigma } - {\partial _\nu }{\Gamma ^\rho }_{\mu \sigma } + ({\Gamma ^\rho }{_\mu _\lambda }{\Gamma ^\lambda }_{\nu \sigma } - {\Gamma ^\rho }{_\nu _\lambda }{\Gamma ^\lambda }_{\mu \sigma })\][/tex]

    where [tex]\[{\Gamma ^\rho }_{\nu \sigma }\][/tex] is the Christoffel symbol

    [tex]\[{\Gamma ^\rho }_{\nu \sigma } = \frac{1}{2}{g^{\rho \alpha }}(\frac{\partial }{{\partial {q^\sigma }}}{g_{\alpha \nu }} + \frac{\partial }{{\partial {q^\nu }}}{g_{\alpha \sigma }} - \frac{\partial }{{\partial {q^\alpha }}}{g_{\nu \sigma }})\][/tex]

    where g is the metric tensor such that

    [tex]\[{\rm{g = }}{{\rm{J}}^T}{\rm{J}}\][/tex]

    where J is the jacobian

    [tex]\[{\rm{J}} = \left[ {\begin{array}{*{20}{c}}
    {\frac{{\partial {q^1}}}{{\partial {x^1}}}}&{\frac{{\partial {q^1}}}{{\partial {x^2}}}}& \cdots &{\frac{{\partial {q^1}}}{{\partial {x^n}}}}\\
    {\frac{{\partial {q^2}}}{{\partial {x^1}}}}&{\frac{{\partial {q^2}}}{{\partial {x^2}}}}& \cdots &{\frac{{\partial {q^2}}}{{\partial {x^n}}}}\\
    \vdots & \vdots & \ddots & \vdots \\
    {\frac{{\partial {q^m}}}{{\partial {x^1}}}}&{\frac{{\partial {q^m}}}{{\partial {x^2}}}}& \cdots &{\frac{{\partial {q^m}}}{{\partial {x^n}}}}
    \end{array}} \right]\][/tex]

    where [tex]\[{q^\mu } = {q^\mu }({x^i})\][/tex] is the coordinate transformation.

    I'm thinking that if someone went through all the tedious partial derivatives they might prove that the scalar curvature is invariant, or at least under what conditions of the partials that it remains invariant. Has anyone seen such a proof ever worked out? Or would someone like to work it out? I'm not sure I have that much skill or patience. Thanks.
    Last edited: Oct 10, 2010
  14. Oct 10, 2010 #13
    If you calculate the curvature tensor with your metric, you will get

    \[{R^\rho }_{\sigma \mu \nu } = 0
  15. Oct 15, 2010 #14
    Prof. Kleinert, in his book, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, pages 769-772, says there are conditions on [tex]\[{q^\mu } = {q^\mu }({x^i})\][/tex] which determine whether it is a transformation from flat to flat space or from flat to curved space. He finds that the affine connection can be written:

    [tex]\[{\Gamma _{\lambda \nu }}^\mu = {\partial _i}{q^\mu } \cdot {\partial _\lambda }{\partial _\nu }{x^i} = \frac{{\partial {q^\mu }}}{{\partial {x^i}}} \cdot \frac{\partial }{{\partial {q^\lambda }}}\left( {\frac{{\partial {x^i}}}{{\partial {q^\nu }}}} \right)\][/tex]

    And when this is put into Riemann cuvature tensor,

    [tex]\[{R_{\mu \nu \lambda }}^\kappa = {\partial _\nu }{\Gamma ^\kappa }_{\lambda \mu } - {\partial _\lambda }{\Gamma ^\kappa }_{\nu \mu } + ({\Gamma ^\kappa }{_\nu _\omega }{\Gamma ^\omega }_{\lambda \mu } - {\Gamma ^\kappa }{_\lambda _\omega }{\Gamma ^\omega }_{\nu \mu })\][/tex]

    he gets,

    [tex]\[{R_{\mu \nu \lambda }}^\kappa = {\partial _a}{q^\kappa } \cdot \left( {{\partial _\mu }{\partial _\nu } - {\partial _\nu }{\partial _\mu }} \right){\partial _\lambda }{x^a}\][/tex]

    He then writes, "A transformation for which [tex]\[{x^a}\left( q \right)\][/tex] have commuting derivatives, while the first derivatives [tex]\[{\partial _\mu }{x^a}\left( q \right)\][/tex] do not, carries a flat-space region into a purely curved one."

    Does anyone have experience with the proof of this statement? Thanks.
    Last edited: Oct 15, 2010
  16. Nov 8, 2010 #15
    When dealing with whether we can have a coordinate transformation from flat to curved space, part of the problem may be that we may be used to considering coordinate transformations as transition maps in the overlap of two coordinate charts on the same manifold. So, of course, the curvature is not going to change since that's dealing with different charts on the same manifold. The overlap is still only on one manifold which only has a unique curvature field in that overlap region.

    Let's consider a transformation of a disk described with (x,y) coordinates with a euclidean metric to a hemisphere with coordinates [tex]\[(\theta ,\phi )\] [/tex]. I'm wondering if such a coordinate transformation can be achieved by use of a larger space, say 3 dimensional Euclidean space, E3. First, transform from the disc to E3 where there is overlap, then transform from E3 to [tex]\[(\theta ,\phi )\] [/tex] coordinates of the hemisphere where again there is overlap. Do the E3 coordinates cancel out?
  17. Nov 8, 2010 #16
    Is Kleinert giving some examples? Just one would do.
  18. Nov 8, 2010 #17
    Actually, it's a 1500 page book, but I have not found such examples. You would think this is a fundamental question that should be studied up front, but if he does address these issues, then his proof is distributed somewhere amoung those 1500 pages. Obviously, the primary example of transformation from flat to curved space is going from a disc to a hemisphere. And the obvious question is does the metric still transform like a tensor in that situation. But I've not seen anywhere on the Web or any book I have where this question is addressed. If it were not possible to do this type of transformation, you'd think they would show it.
  19. Nov 8, 2010 #18
    This is what makes me suspicious. You can easily find examples with functions that are nondifferentiable a certain number of times here and there, but to have it over an open set? I really would like to see one example - otherwise his whole theory may well have an empty content. And any reasonable map from a disk to a hemisphere will be differentaible as many times as you want except, perhaps along one or several radii - if you really want it.
  20. Nov 14, 2010 #19
    Now I'm not so sure a coordinate transformation can go from a flat space to a curved space. For it seems that coordinates are something inscribed on a space, and a coordinate transformation would simply be a change of coordinates on that same space. Perhaps it would be better to a those kind of transformations, a space transformation, not a coordinate transformation. Or it may be that a transformation of the type: [tex]\[{q^\mu } = {q^\mu }({x^i})\][/tex] is always considered to be a "coordinate transformation" since the [tex]\[{q^\mu }\][/tex] are always considered coordinates. Anyone have any experience with this distinction?
  21. Nov 14, 2010 #20
    As long as it is not said explicitly when you just see [tex]q^i(x^\mu)[/tex] it can be interpreted both ways. Most authors are very clear when they write their formulas, some other prefer to be fuzzy - perhaps that is the way they are thinking.
  22. Nov 16, 2010 #21
    Of course, coordinate transformations by themselves are of little use. It's how scalars, vectors, and tensors transform under a change of coordinates that is interesting. It's easy to understand that a scalar function does not change under coordinate changes, since changing the coordinates that describe a point does not change where the point is in relation to the rest of the manifold. And it does not change the value of the function, since it is a function of the point and not the coordinated used to describe the point. There is no change of space, so there is no change for any function on that space.

    However, how does one transform functions when there is a change of the underlying space, say, from a flat space to a curved space? I've read where any manifold can be viewed as a submanifold of a larger dimensional space, [tex]\[{R^d}\][/tex]. So for example, you can have both a 2D disc and a hemisphere being submanifolds of a larger 3D flat space. How does a function defined on a disc transform to a function defined on a hemisphere? It would seem that the 2D function defined on a disc would have to be defined on 3D in order to have values on the hemishpere, right? But even then you don't know where the disc is in relation to the hemisphere inside the 3D space. And you don't know how the 2D function defined on the disc is assigned values in 3D. So how to proceed?

    How do you know you have a unique curved space representation of a flat space function? I'm considering that it might be that the curved space integral of the curved space function must equal the flat space integral of the flat space function. For that seems to be taking into account every minuscule detail of the whole function in both spaces. Does this sound right? Would there be an caveats or qualifications to such integrals? For example, maybe it must be required that the integration over any interval must equal its curved space counterpart, for you can have different functions with the same value of integration over some particular interval. Can the equivalence between these integrals be proven to result in a unique curved space function? Thanks.
    Last edited: Nov 16, 2010
  23. Nov 17, 2010 #22
    Now that I think about it, you could have a flat space disc measured in light years mapped to a hemisphere measured in inches through some space transformation. Then the flat space integration will not equal the curved space integration, because the metric is not the same. So the question remains. How do you know you have a unique curved space representation of a flat space function?

    I suppose f(x)=f(x(q))=f(q), independent of whether q and x are different coordinates on the same space or actually transform from one type of space to another. But dq does not necessarily equal dx, since the metric may be different in q-space than in x-space. So differentials and integrals are not the same. Anyone have any ideas on how to proceed? Does the Jacobian still apply when transforming across different curvature space as it does in curvilinear coordinate transformation? Thanks.
  24. Nov 17, 2010 #23
    Function is a function. If it has a value f(p) on the disk and this point of the disk maps to a point q on a hemisphere, then the function on the hemisphere at q i s equal to its value on the disk at p. Unless this is not a scalar function. It can be a scalar density of some weight. But I think that Kleinert is thinking about diffeomorphisms (that is active deformations) of the manifold that are not twice differentiable. The bad thing is that he did not give even one example. With an example we would see the weak spots in this idea.
  25. Nov 17, 2010 #24
    Yes, that much seems obvious now. I guess I was overcomplicating things because I'm trying to make sure before proceeding. So what is the next thing to transform, derivatives, and then integrals? I would think the differentials, [tex]\[d{x^i}\][/tex] and [tex]\[d{q^\mu }\][/tex], would transforms as:

    [tex]\[d{x^i} = \frac{{\partial {x^i}}}{{\partial {q^\mu }}} \cdot d{q^\mu }\][/tex] and [tex]\[d{q^\mu } = \frac{{\partial {q^\mu }}}{{\partial {x^i}}} \cdot d{x^i}\][/tex]

    Or is it more complicated than that because the metric also changes?
    Last edited: Nov 17, 2010
  26. Nov 20, 2010 #25
    As I think about it some more, as soon as [tex]\[{q_j} = {q_j}({x_i})\][/tex] is given, there is really no choice in how the differentials transform. By simply following the rules for differentiating functions, they must be given as shown above. And it doesn't matter if [tex]\[{q_j} = {q_j}({x_i})\][/tex] transforms between coordinate patches on the same space/manifold or it transforms between entirely different spaces that have different curvatures. Even if lightyears in x-space are transformed to inches in q-space, the [tex]\[\frac{{\partial {q^\mu }}}{{\partial {x^i}}}\][/tex] transform the distance measured in lightyears to the distance measured in inches. It takes into account the difference in the way distance is measured because the units of [tex]\[\frac{{\partial {q^\mu }}}{{\partial {x^i}}}\][/tex] is inches/lightyear, when you multiply it by dx in lightyears you get inches. So it takes into account the different metrics in each space.

    So the next question is how integrals transform. I've recently reviewed the use of the jacobian, at least for transforming 2D surface integrals from one space to another. The jacobian was introduced in the context of curvilinear coordinate transformations, which is just a reparameterization of the same space, where the overlap between each of the coordinate patches covers the entirity of the same space. But in the development, there actually was no qualification between which kinds of spaces to which it applies. The differential area element was defined in terms of the differential length along each of its coordinates. Each end of these differential lengths are points which were transformed to the new space using [tex]\[{q_j} = {q_j}({x_i})\][/tex], which defined differentials in the new space. These new differential line segments were not necessarity orthogonal. So the area of the new parallelogram was calculated using the cross product, where the vectors of the cross product were formed from the new differential line segments. This cross product reproduced the 2D jacobian which was used as a correction factor when transfroming the integral. It seems to me that the transformation of area elements (and integrals in general) only relied on a point for point transformation of differential line segments using [tex]\[{q_j} = {q_j}({x_i})\][/tex], which could just as easily tranform across different spaces as across different coordinate patches.

    As the area element on a curved surface becomes smaller, it becomes better approximated by an area element in the tangent space to that surface. And it seems that's all the jacobian matrix does, transform vectors in one tangent space to vectors in a different tangent space. The example above seems to prove that the jacobian can be used to transform one tangent space to another even between entirely different spaces. Does this seem right? One thing that bothers me is that using the cross product to calulate the area of the new parallelogram assumes that the new parallelogram exists in a flat orthogonal space. Does this assumption mean that the transformation was across different coordinate patches of the same background flat space? Or does it only mean that all spaces are locally flat so that every tangent space is also flat? Another thing that bothers me is that I'm not quite convinced that the transformed integral equals the original integral when the transform is across a different space. It seems the usual development uses the jacobian to transform curvilinear coordinates on the same space and is used precisely because the transformed integral provides an easier way to calculate the original integral to which it is equivalent. But when transforming across different spaces, can one integral in a space measured in lightyears equal a transformed integral in a space measured in inches? I'm still not sure of that.

    And IIRC, all that's necessary to transform any tensor is a linear transformation of the tangent space. Does all this prove that the regualar rules for transforming tensors works between entirely different spaces as well as between coordinate patches on the same space?
    Last edited: Nov 20, 2010
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