# Noncoordinate basis

1. Dec 5, 2005

### matness

Can anybody give me an explanation about non coordinate bases and its importance?

2. Dec 6, 2005

### HallsofIvy

All that means is that you are not required to have a coordinate-system in order to define a basis for a vector space. I have seen the phrase "non-coordinate" used in Tensors and always assumed it just meant "coordinate free"- that is the equations themselves did not depend on a particular coordinate system.

3. Dec 6, 2005

### robphy

Lectures 4 and 5 at http://ocw.mit.edu/OcwWeb/Physics/8-962Spring2002/LectureNotes/index.htm [Broken] may be enlightening.

Last edited by a moderator: May 2, 2017
4. Dec 6, 2005

### George Jones

Staff Emeritus
The tangent space at each point $p$ of an n-dimensional differentiable manifold is an n-dimensional vector space and thus has an infinite number of bases. If $p$ is contained in a patch covered by the coordinates $\{x^{\mu}\}$, then
$$\{ \frac{\partial}{\partial x^{\mu}}\}$$
is a coordinate basis for the tangent space. However, not all vectors in the
tangent space are tangent to (any) coordinate curves.

In general relativity, spacetime is modeled by a 4-dimensional differentiable manifold. Coordinate bases are very useful for calculations, but non-coordinate orthonormal bases are useful for physical interpretation.

As an example, consider the Schwarzchild metric
$$\mathbf{g}=\left( 1-\frac{2M}{r}\right) \mathbf{dt}\otimes\mathbf{dt}-\left( 1-\frac{2M}{r}\right) ^{-1}\mathbf{dr}\otimes\mathbf{dr}-r^{2}\left( \sin^{2}\theta\mathbf{d\theta}\otimes\mathbf{d \theta}+\mathbf{d\phi}\otimes\mathbf{d\phi}\right).$$
An observer hovering at constant $r$, $\theta$, and $\phi$ has orthonormal basis vectors
\begin{align*} \mathbf{e}_{0} & =\left( 1-\frac{2M}{r}\right) ^{-\frac{1}{2}} \frac{\partial}{\partial t}\\ \mathbf{e}_{1} & =\left( 1-\frac{2M}{r}\right) ^{\frac{1}{2}}\frac {\partial}{\partial r}\\ \mathbf{e}_{2} & =\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}\\ \mathbf{e}_{3} & =\frac{1}{r}\frac{\partial}{\partial\phi}, \end{align*}
while an observer freely falling from rest at infinity has orthonormal basis
vectors
\begin{align*} \mathbf{e}_{0}^{\prime} & =\left( 1-\frac{2M}{r}\right) ^{-1}\frac {\partial}{\partial t}-\left( \frac{2M}{r}\right) ^{\frac{1}{2}}\frac{\partial}{\partial r}\\ \mathbf{e}_{1}^{\prime} & =-\left( \frac{2M}{r}\right) ^{\frac{1}{2} }\left( 1-\frac{2M}{r}\right) ^{-1}\frac{\partial}{\partial t} +\frac{\partial}{\partial r}\\ \mathbf{e}_{2}^{\prime} & =\frac{1}{r\sin\theta}\frac{\partial} {\partial\theta}\\ \mathbf{e}_{3}^{\prime} & =\frac{1}{r}\frac{\partial}{\partial\phi}. \end{align*}
The two orthonormal bases are related by
\begin{align*} \mathbf{e}_{0}^{\prime} & =\left( 1-\frac{2M}{r}\right) ^{-\frac{1}{2} }\mathbf{e}_{0}-\left( \frac{2M}{r}\right) ^{\frac{1}{2}}\left( 1-\frac {2M}{r}\right) ^{-\frac{1}{2}}\mathbf{e}_{1}\\ \mathbf{e}_{1}^{\prime} & =-\left( \frac{2M}{r}\right) ^{\frac{1}{2} }\left( 1-\frac{2M}{r}\right) ^{\frac{-1}{2}}\mathbf{e}_{0}+\left( 1-\frac{2M}{r}\right) ^{-\frac{1}{2}}\mathbf{e}_{1}\\ \mathbf{e}_{2}^{\prime} & =\mathbf{e}_{2}\\ \mathbf{e}_{3}^{\prime} & =\mathbf{e}_{3}. \end{align*}
It turns out that the relative velocity between the 2 observers is given by
$$v=-\left( \frac{2M}{r}\right) ^{\frac{1}{2}},$$
so the relation between the 2 bases is given by the Lorentz transformation
\begin{align*} \mathbf{e}_{0}^{\prime} & =\frac{\mathbf{e}_{0}+v\mathbf{e}_{1}} {\sqrt{1-v^{2}}}\\ \mathbf{e}_{1}^{\prime} & =\frac{v\mathbf{e}_{0}+\mathbf{e}_{1}} {\sqrt{1-v^{2}}}. \end{align*}
Lorentz transformations are useful is general relativity as well as special relativity!

Last edited: Jul 5, 2011
5. Dec 6, 2005

### pervect

Staff Emeritus
Let's take a very simple example. Suppose you have polar coordinates, r and $\theta$.

Then the tangent space in the coordinate basis given by the vectors $$\frac{\partial}{\partial r}$$ and $$\frac{\partial}{\partial \theta}$$. [1]

The vectors in this tangent space happen to be orthogonal, but not orthonormal. A line element in polar coordinates is $ds^2 = dr^2 + r^2 d \theta^2$, you can see that the metric tensor is not an identity.

Suppose you express the tangent space in terms of orthonormal vectors $\hat{r}$ and $\hat{\theta}$ rather than in terms of the coordinate basis vectors of [1].

The result is very useful for physics. It is always possible to define an orthonormal basis at a point, even if you are unfortunate and have non-orthogonal vectors in the coordinate basis.

Such an orthonormal basis can be defined in general by specifying an array of n one-forms (aka contravariant tensors) that map the vectors of the tangent space to n scalars. These n scalars are the n orthonormal coordinates.

[Definition: a vector is a member of the tangent space, the dual of a vector is a map of a vector to a scalar, this has various names such as "one-form", contravariant vector, etc.]

One usually sees this described as "an orthonormal basis of one-forms".

Note that it is not strictly necessary that a non-coordinate basis be orthonormal, this is just a typical application of the concept.

6. Dec 6, 2005

### robphy

Last edited by a moderator: May 2, 2017
7. Dec 8, 2005

### Ratzinger

Weren't one-forms covariant tensors?

8. Dec 9, 2005

### dextercioby

Ya, they were and it's Schwarzschild...

Daniel.

9. Dec 9, 2005

### pervect

Staff Emeritus
Probably <checking>, yes, it appears I got that backwards :-(

10. Jan 1, 2006

### bchui

In General Relativity, the manifold $$M$$ is 4-dimensional, so why don;t we simply take $$M=\Re^4$$ and use much simplier symbols in the differential geometry?

11. Jan 2, 2006

### robphy

The theory itself allows the possibility for 4-manifolds other than R4. Indeed, there are interesting solutions that don't have R4 as the base manifold. An interesting related question is "How can we determine the topology of spacetime?"

12. Jan 2, 2006

### Ratzinger

note that manifold is only required to look locally like R^n, it may look globally different from R^n

13. Jan 31, 2006

### bchui

Mobius Strip Universe?

So, are there any existing models with M not equals to Re^4, say, for example any theories for the manifold M being a Mobius Strip? :surprised

14. Feb 1, 2006

### robphy

Schwarzschild is R2 x S2.
A Mobius strip is only two-dimensional...but if you cross it with some other 2-d space [to obtain a 4-manifold], it will fail to be either time- or space-orientable...which might be regarded as unphysical.

15. Feb 1, 2006

### bchui

Hold on, should Schwarzschild be $$S^3\times\Re$$ instead, for
$$(r,\theta,\varphi)$$ is 3-dimensional spherical plus $$t\in\Re$$?

Could we possible think of $$M$$ being a higher-dimensional Mobius strip or Klein bottle analog? That should be something to do with "worm-holes" in space and we could possible to do "space travel"?

Last edited: Feb 1, 2006
16. Feb 1, 2006

### George Jones

Staff Emeritus
$\left\{ r, \theta, \phi \right\}$ is not a coordinate system for $S^3$. $S^3$ is the 3-dimensional surface of a 4-dimensional ball; $S^3$ is not a 3-dimensional ball.

$\left\{ \theta, \phi \right\}$ is a coordinate system for (most of) $S^2$, the 2-dimensional surface of a 3-dimensional ball. Actually, since $S^2$ is compact, it cannot be covered in its entirety by a single chart.

Regards,
George

17. Feb 2, 2006

### bchui

:!!) You are right, so Schwarzschild is a metric on the manifold
$$M=S^2\times {\bf R}^+\times {\bf R}$$, for we have $$r>0$$ and $$t\in {\bf R}$$

So, $$M$$ is actually the domain of the parameter values, not the "actual space" we want to describe? For, the "actual space" we want to describe is $${\bf R}^4$$?

Does that applies to Robertson-Walker metric and many others?

Last edited: Feb 2, 2006
18. Feb 2, 2006

### George Jones

Staff Emeritus
One has to be a little careful here, but basically yes.

Also, note that $\mathbb{R}$ and $\mathbb{R}^+$ are diffeomophic via $x \mapsto e^x$, so the underlying manifold for Schwarzschild also can be considered to be $M = S^{2} \times \mathbb{R}^{2}$, justs as robphy said.

No, an n-dimensional manifold is that looks locally like $\mathbb{R}^n$, but not necessarily globally. The "actual space" for Schwarzschild is $M$, not $\mathbb{R}^4$.

The open Friedman-Robertson-Walker universes have underlying manifold $\mathbb{R}^4$, while the closed Friedman-Robertson-Walker universes have underlying manifold $\mathbb{R} \times S^{3}$.

Open and closed refer to 3-dimensional spatial hypersurfaces, and closed means compact, i.e., $S^3$ is compact.

Regards,
George

Last edited: Feb 2, 2006
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