# Noncoordinate basis

#### matness

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

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#### HallsofIvy

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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.

#### robphy

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

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#### George Jones

Staff Emeritus
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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!

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#### pervect

Staff Emeritus
matness said:
Can anybody give me an explanation about non coordinate bases and its importance?
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 $$\frac{\partial}{\partial \theta}$$. 

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 .

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.

#### robphy

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Gold Member
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#### Ratzinger

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.
Weren't one-forms covariant tensors?

#### dextercioby

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

Daniel.

#### pervect

Staff Emeritus
Ratzinger said:
Weren't one-forms covariant tensors?
Probably <checking>, yes, it appears I got that backwards :-(

#### 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?  #### robphy

Homework Helper
Gold Member
bchui said:
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?  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?"

#### Ratzinger

In General Relativity, the manifold is 4-dimensional, so why don;t we simply take and use much simplier symbols in the differential geometry?
note that manifold is only required to look locally like R^n, it may look globally different from R^n

#### 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

#### robphy

Homework Helper
Gold Member
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.

#### 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:

#### George Jones

Staff Emeritus
Gold Member
bchui said:
Hold on, should Schwarzschild be $$S^3\times\Re$$ instead, for $$(r,\theta,\varphi)$$ is 3-dimensional spherical plus $$t\in\Re$$?
$\left\{ r, \theta, \phi \right\}$ 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

#### 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?

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#### George Jones

Staff Emeritus
Gold Member
bchui said:
Schwarzschild is a metric on the manifold
$$M=S^2\times {\bf R}^+\times {\bf R}$$, for we have $$r>0$$ and $$t\in {\bf R}$$
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.

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 $${\bf R}^4$$?
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 $\mathbb{R}^4$.

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

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