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Frame fields vs. Coordinate bases

  1. May 2, 2007 #1
    What is the difference, if any, between frame fields and coordinate bases?
  2. jcsd
  3. May 2, 2007 #2

    George Jones

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    They are, in general, different.

    A frame field is a set of four vector fields such that at each event in a region of spacetime, the vector fields evaluated at that event form an orthonormal basis for the tangent space. Frame fields are particularly useful for calculating physical effects.

    A coordinate basis is derived from a coordinate system over a region of spacetime. To find a vector that is a member of a coordinate basis, hold fixed three coordinates well letting the remaining coordinate vary. This produces a curve in spacetime, and the tangent vector to this curve is a member of a coordinate basis (think partial derivative). One curve can be produced for each coordinate. Cooridinate bases are useful for many calculations, including those that exploit symmetry.

    If a coordinate basis is a frame field over a region of spacetime, then that region of spacetime is flat (i.e., looks like a region of Minkowski spacetime).

    Frame fields can be derived from coordinate bases by doing something analogous to Gram-Schmidt. A simple example is Schwarzschild.
    Last edited: May 2, 2007
  4. May 2, 2007 #3


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    An example might help.

    In polar coordinates, one often defines unit vectors in the r and theta directions as [tex]\hat{r}[/tex] and [tex]\hat{\theta}[/tex].

    This is a frame field. It's a non-coordinate basis.

    The detailed mathematical treatment involves identifying vectors with partial derivatives or derivative operators - I don't know if you've seen this sort of treatment or not.

    With this sort of treatment, a coordinate basis is just the set of partial derivatives with respect to any coordinate system.

    It also turns out (if I recall correctly) that the derivative operators representing a coordinate basis commute - those that represent non-coordinate bases do not commute. That is how you could for instance show that the hatted vectors I alluded to either couldn't possibly be a coordinate basis no matter what coordinates you picked.
  5. May 2, 2007 #4

    Chris Hillman

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    Coordinate bases versus frame fields

    I think pervect had the right idea: study some simple but nontrivial examples. But I'll offer some general discussion:

    Start with a smooth manifold M, i.e. we have some way to define derivatives of functions on M and we have tangent spaces at each point p. A "coordinate" on some neighborhood U of p is nothing but a monotonic nonconstant smooth function on U. If the tangent spaces have dimension d, a "coordinate chart on U" is nothing but a collection of d smooth functions whose gradients are pairwise non-parallel on U. Then the level surfaces form "transversely intersecting" nets. (Think of deforming a picture with three sets of mutually orthogonal planes in E^3). A "vector field on U", [itex]\vec{X}[/itex], is nothing but a first order linear homogeneous partial differential operator. At each point p, we can consider [itex]\vec{X}[/itex] and the coordinate vector fields [itex]\partial_{x^1}, \, \partial_{x^2}, \dots[/itex] to live in the tangent space at p, which is a dimensional vector space. Thus, by linear algebra we can express [itex]\vec{X}_p[/itex] as a linear combination of the coordinate basis vectors. The coefficients are the "components" with respect the coordinate basis.

    It is important to realize that the most useful "coordinates" will have geometric interpretations which are coordinate-free, that is, we can define these nice coordinates in a noncircular fashion. This kind of geometric reasoning is neccessary when we are trying to construct a sufficiently general chart for something like stationary axisymmetric spacetimes.

    Forget spacetime for a moment and just think about noncartesian charts on ordinary euclidean space. Now, coordinate bases have many convenient properties and conventional index gymnastics computations uses this kind of basis. The great thing about tensor equations is that they are true in any coordinate chart. But the bad thing about tensor components is that they are "unphysical". For example, if we have an electric field vector and some observers, these observers can certainly define directions at their location and measure both magnitude and direction of the field. Now, in index gymnastics we can compute the field vector. Bu the coordinate vectors don't have unit length, so to compare with observation we need to need to rescale them and then rotate the result to obtain the components measured by our observer using his arbitrary unit directions "in space".

    For example, consider a cylindrical chart, in which the line element becomes
    [tex]ds^2 = dz^2 + dr^2 + r^2 \, d\phi^2, \;
    -\infty < z < \infty, \; 0 < r < \infty, \; -\pi < \phi < \pi
    Here, the coordinate basis vectors are [itex]\partial_t, \, \partial_r, \partial_\phi[/itex]. The last is not a unit vector, however. So to construct "local frame vectors" at a given point, we need to rescale it:
    \vec{e}_1 = \partial_z , \;
    \vec{e}_2 = \partial_r , \;
    \vec{e}_3 = \frac{1}{r} \, \partial_\phi
    If you draw a picture, you'll see these particular frame vectors are "aligned" with the chart. We can apply a smooth section in SO(3)-bundle to rotate this frame into the Cartesian frame, which is both a coordinate basis and a frame field (only possible in flat space!):
    \vec{f}_1 = \partial_x , \;
    \vec{f}_2 = \partial_y , \;
    \vec{f}_3 = \partial_z

    In spacetime, we can use a smooth section in SO(1,3)-bundle to rotate/boost one frame into another. Now this is very useful because this is just what we need to compare the physical experience of observers in different states of motion whose world lines pass near some event p. In other posts at PF I have given some very detailed examples in which I compared the physical experience of some important families of observers in the Schwarschild vacuum: static observers who use their rocket engine to hover in place, Lemaitre obsevers who fall in freely and radially "from rest at infinity", Novikov observers who fall in freely and radially from rest at r=r_0, slowfall observers who maintain just the right outward thrust which would hold them up against gravity if they lived in a Newtonian universe, and who therefore slowlyl fall in radially because "Einsteinian gravity is stronger" (because gtr is a nonlinear field theory in which gravitational field energy gravitates), and so on.

    But the fact that frame fields, not coordinate bases, give components of a multicomponent object which an observer might actually measure, so that we can check theory against experiment, is not the only reason for prefering them. It turns out that most elementary computations are just plain easier if you use Cartan's "moving frame" formalism. See MTW, the book by Flanders cited at http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html#TensorCalc[/url] and the book by Frankel cited at [url]http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html#physback [Broken] for many examples. Using this formalism, you can compute covariant derivatives and so on directly, but if you don't know it, you can always convert from any coordinate frame to any frame field by applying the appropriate linear transformation at each event.

    There is never much point in listing components of a vector or tensor with respect to a coordinate basis since these have no convenenient geometrical or physical interpretation! So trying to interpret them will only lead to confusion. Rather, to understand the local geometry and physics you have to choose some family of observers, in fact some frame field, compute the components in this frame, and interpret the results in terms of the physical experience of these observers, in some thought experiment. It doesn't matter how you obtain the frame components, although as I said Cartan's methods are almost always easiest, as long as you obtain them before trying to interpret a vector or tensor.
    Last edited by a moderator: Apr 22, 2017 at 5:20 PM
  6. May 3, 2007 #5
    Thank you for all your responses.
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