Understanding the Tetrad Formalism in General Relativity

[Dale]

An example: page 49 of "General Relavity" by Wald, "noncoordinate, orthonormal basis of smooth vector fields $\left( e_\mu \right)^a$ ... $\left\{\left( e_\mu \right)^a \right\}$ is called a tetrad.

Excellent, Wald is certainly authoritative!

 

[Orodruin]

An example: page 49 of "General Relavity" by Wald, "noncoordinate, orthonormal basis of smooth vector fields $\left( e_\mu \right)^a$ ... $\left\{\left( e_\mu \right)^a \right\}$ is called a tetrad.

How I feel lost during x-mas holidays without my set of GR literature ... (left in office)

 

[Higgsono]

But if a Tetrad is smoothly varying, then one could obviously parameterize it using a set of coordinate curves or a "grid" on the Manifold. So that at each point, the tangent vectors to the coordinate curves would be orthogonal to each other. Is this not possible?

 

[Orodruin]

But if a Tetrad is smoothly varying, then one could obviously parameterize it using a set of coordinate curves or a "grid" on the Manifold. So that at each point, the tangent vectors to the coordinate curves would be orthogonal to each other. Is this not possible?

No, the tetrad is not necessarily commuting. It will not be the same thing following the flow of $e_1$ and then $e_2$ as following $e_2$ and then $e_1$.

 

[Higgsono]

No, the tetrad is not necessarily commuting. It will not be the same thing following the flow of $e_1$ and then $e_2$ as following $e_2$ and then $e_1$.

I did not understand this. On the upper half of the sphere, it is certainly possible to find coordinate curves such that the tangent vectors to those curves are everywhere orthogonal.

 

[Orodruin]

I did not understand this. On the upper half of the sphere, it is certainly possible to find coordinate curves such that the tangent vectors to those curves are everywhere orthogonal.

Yes, that does not mean that the corresponding holonomic basis is orthonormal. It certainly is not the case in standard spherical coordinates.

 

[pervect]

But if we consider only the upper half of the 2 sphere, then we can choose a coordinate system, for instance spherical coordinates, so that the metric components are constant everywhere. Now I have constructed a tetrad, but this is no different then just choosing a suitable coordinate system. I don't understand, what am I missing?

If you remove the north and south poles from the a sphere, you can have an orthonormal vector field everywhere but at the excluded points (the north and south pole).

If you define sphereical coordinates $r, \theta, \varphi$ in a usual manner so that the conversion to cartesian coordinates is

$$x = r \sin \theta \cos \varphi \quad y = r \sin \theta \sin \varphi \quad z = r \cos \theta$$

then the associated spherical line element is $dr^2 + r^2 d\theta^2 + r^2\sin^2 \theta \, d\varphi^2$
(This is from memory, it was unexpectedly difficult to look up. But you can use algebra and the chain rule to compute dx^2+dy^2+dz^2 and find the line element yourself if you want to go to all the effort. and if it doesn't seem obvious by inspection).

If you set r=1, you can use $\theta, \varphi$ as coordinates on the unit sphere.

Let $\partial_\theta = \frac{\partial}{\partial \theta}$ and $\partial_\varphi = \frac{\partial}{\partial \varphi}$.

Recall the length^2 of a vector is $g_{\mu\nu} x^\mu x^\nu$, so the length^2 of $\partial_\theta$ is $r^2$, and the length^2 of $\partial_\varphi$ is $r^2 \sin^2 \theta$.

Then with r=1, $\partial_\theta$ is of unit length, $\partial_\phi$ is not. But we can normalize $\partial_\phi$ so that it is of unit length.

The fact that the metric is diagonal indicates that $\partial_\theta$ and $\partial_\varphi$ are orthogonal, i.e. their dot-product is zero.

The metric is singular at $\theta = \pi / 2$, so we need to exclude the poles.

If we interpret the above to the surface of a spherical globe, $\partial_\theta$ points north and is of unit length, while $\partial_\varphi$ points east. The coordinate basis vector $\partial_\varphi$ isn't of unit length we need to multiply it by $1/\sin \theta$# to normalize it to have a unit-length vector.

Going from 2 dimensions to 4 to get a true tetrad isn't that difficult if the 2-d case is understood, but I'll leave it for the reader.

 

[Higgsono]

If you remove the north and south poles from the a sphere, you can have an orthonormal vector field everywhere but at the excluded points (the north and south pole).

If you define sphereical coordinates $r, \theta, \varphi$ in a usual manner so that the conversion to cartesian coordinates is

$$x = r \sin \theta \cos \varphi \quad y = r \sin \theta \sin \varphi \quad z = r \cos \theta$$

then the associated spherical line element is $dr^2 + r^2 d\theta^2 + r^2\sin^2 \theta \, d\varphi^2$
(This is from memory, it was unexpectedly difficult to look up. But you can use algebra and the chain rule to compute dx^2+dy^2+dz^2 and find the line element yourself if you want to go to all the effort. and if it doesn't seem obvious by inspection).

If you set r=1, you can use $\theta, \varphi$ as coordinates on the unit sphere.

Let $\partial_\theta = \frac{\partial}{\partial \theta}$ and $\partial_\varphi = \frac{\partial}{\partial \varphi}$.

Recall the length^2 of a vector is $g_{\mu\nu} x^\mu x^\nu$, so the length^2 of $\partial_\theta$ is $r^2$, and the length^2 of $\partial_\varphi$ is $r^2 \sin^2 \theta$.

Then with r=1, $\partial_\theta$ is of unit length, $\partial_\phi$ is not. But we can normalize $\partial_\phi$ so that it is of unit length.

The fact that the metric is diagonal indicates that $\partial_\theta$ and $\partial_\varphi$ are orthogonal, i.e. their dot-product is zero.

The metric is singular at $\theta = \pi / 2$, so we need to exclude the poles.

If we interpret the above to the surface of a spherical globe, $\partial_\theta$ points north and is of unit length, while $\partial_\varphi$ points east. The coordinate basis vector $\partial_\varphi$ isn't of unit length we need to multiply it by $1/\sin \theta$# to normalize it to have a unit-length vector.

Going from 2 dimensions to 4 to get a true tetrad isn't that difficult if the 2-d case is understood, but I'll leave it for the reader.

Thanks! But then we have a constant metric, because our basis is orthonormal right? This means that the Christoffel symbols are zero, hence the Riemann tensor is also zero and we have no curvature. But obviously there is curvature, so what is wrong?

 

[Orodruin]

This means that the Christoffel symbols are zero, hence the Riemann tensor is also zero and we have no curvature.

This is only true if your basis is holonomic. As we have already discussed, an orthonormal basis on the sphere is not the coordinate basis of any coordinate system.

 

[Higgsono]

This is only true if your basis is holonomic. As we have already discussed, an orthonormal basis on the sphere is not the coordinate basis of any coordinate system.

aha ok, thanks for clarifying.

 

[strangerep]

[...] But then we have a constant metric, because our basis is orthonormal right? This means that the Christoffel symbols are zero, hence the Riemann tensor is also zero and we have no curvature. But obviously there is curvature, so what is wrong?

You're ignoring the object of anholonomicity $\Omega$, defined via: $$[e_i \,,\, e_j] ~=~ \Omega^k_{~ij} \, e_k ~,$$ where the $e_i$ are the basis elements of the tetrad field.

For a coordinate (a.k.a. holonomic) basis, $\Omega=0$. Otherwise, it appears as a skewsymmetric part of the connection, and hence appears in the curvature tensor (and in the current case, stops the curvature tensor from vanishing).

Here's an example from the spherical polar case. Consider the gradient of a scalar function $f$.$$\nabla f ~=~ \partial_x f \; {\bf \hat x} + \partial_y f \; {\bf \hat y} + \partial_z f \; {\bf \hat z} ~=~ \partial_r f \; {\bf \hat r}+ \frac{1}{r}\,\partial_\theta f \; {\bf \hat\theta}+ \frac{1}{r \sin\theta}\,\partial_\phi f \; {\bf \hat \phi} ~.$$Although, e.g., $$[\partial_x ~,~ \partial_y] ~=~ 0 ~,~~~ \mbox{etc} ~,$$we find $$\left[ \partial_r \,,\, \frac{1}{r}\,\partial_\theta \right] ~=~ -\,\frac{1}{r^2}\, \partial_\theta ~+~ \frac{1}{r}\, \partial_r \partial_\theta ~-~ \frac{1}{r}\,\partial_\theta \partial_r ~=~ -\,\frac{1}{r^2}\, \partial_\theta ~\ne~ 0 ~.$$

 

[Higgsono]

You're ignoring the object of anholonomicity $\Omega$, defined via: $$[e_i \,,\, e_j] ~=~ \Omega^k_{~ij} \, e_k ~,$$ where the $e_i$ are the basis elements of the tetrad field.

For a coordinate (a.k.a. holonomic) basis, $\Omega=0$. Otherwise, it appears as a skewsymmetric part of the connection, and hence appears in the curvature tensor (and in the current case, stops the curvature tensor from vanishing).

Here's an example from the spherical polar case. Consider the gradient of a scalar function $f$.$$\nabla f ~=~ \partial_x f \; {\bf \hat x} + \partial_y f \; {\bf \hat y} + \partial_z f \; {\bf \hat z} ~=~ \partial_r f \; {\bf \hat r}+ \frac{1}{r}\,\partial_\theta f \; {\bf \hat\theta}+ \frac{1}{r \sin\theta}\,\partial_\phi f \; {\bf \hat \phi} ~.$$Although, e.g., $$[\partial_x ~,~ \partial_y] ~=~ 0 ~,~~~ \mbox{etc} ~,$$we find $$\left[ \partial_r \,,\, \frac{1}{r}\,\partial_\theta \right] ~=~ -\,\frac{1}{r^2}\, \partial_\theta ~+~ \frac{1}{r}\, \partial_r \partial_\theta ~-~ \frac{1}{r}\,\partial_\theta \partial_r ~=~ -\,\frac{1}{r^2}\, \partial_\theta ~\ne~ 0 ~.$$

ok thanks, I feel like a noob, where can I read about this stuff? Any book you can recommend?

 

[pervect]

Thanks! But then we have a constant metric, because our basis is orthonormal right? This means that the Christoffel symbols are zero, hence the Riemann tensor is also zero and we have no curvature. But obviously there is curvature, so what is wrong?

Chistoffel symbols are usually derived in a coordinate basis, aka a holonomic basis. It's also possible to use tetrad methods, where one uses connection one-forms rather than Christoffel symbols. See for instance Wald, "General Relativity" pg 49, sec 3.4b (if you have that text). But I am not familiar with the details of this process of using connection 1-forms, even though I can spot the section that does in my textbook. Additionally, I can't guarantee that the way described in that particular textbook is the only way of dealing with non-holonomic basis'

At a guess, though, you're probably applying methods derived using the assumption that one has a holonomic (coordinate) basis to a non-holonomic basis to compute the curvature tensor, and that's where you're going wrong.

 

[strangerep]

ok thanks, I feel like a noob, where can I read about this stuff? Any book you can recommend?

Wald has already been suggested.

Also, it wouldn't hurt to do thorough revision of operator in spherical coordinates (pen in hand). I.e., derive from first principles the various operator formulas listed in this Wikipedia article. If you get stuck, there's plenty of youtube videos that walk you through it slowly. (The reason I mention this is that is important to have a well-understood concrete example at hand.)

 

[romsofia]

Theory of gravitational interactions (https://www.springer.com/us/book/9788847026919) chapter 12 covers this. Wald is good, but I actually like how this book covered it better. He has a good appendix on differential forms as well, and the book has solutions for all the problems.

 

[Higgsono]

Theory of gravitational interactions (https://www.springer.com/us/book/9788847026919) chapter 12 covers this. Wald is good, but I actually like how this book covered it better. He has a good appendix on differential forms as well, and the book has solutions for all the problems.

thanks, that book looks great!

 

[vanhees71]

I

The metric is singular at $\theta = \pi / 2$, so we need to exclude the poles.

It's singular at $\theta=0$ and $\theta=\pi$, and that's why we need to exclude the poles defined by the polar axis of the polar coordinate system (in the 3D embedding Euclidean space that's the 3-axis of the choosen Cartesian basis you started with).

 

[vanhees71]

Wald has already been suggested.

Also, it wouldn't hurt to do thorough revision of operator in spherical coordinates (pen in hand). I.e., derive from first principles the various operator formulas listed in this Wikipedia article. If you get stuck, there's plenty of youtube videos that walk you through it slowly. (The reason I mention this is that is important to have a well-understood concrete example at hand.)

It's also worth thinking about why the table about "Coordinate Conversions" is partially wrong (or at least sloppy)!

 

[pervect]

Some assorted comments:

Yes, it's 0 and pi that are singular as vanhees points out for the case I wrote about, since sin(theta)=0 for theta=0 and theta=pi.

Dealing directly with nonholonomic bases is something that can be done, but most textbooks I've seen take the approach of solving the problem in a coordinate basis, then converting to a non-holonomic basis. Of course I haven't seen every textbook.. But it's not uncommon for a textbook to present only the methods for computing curvature that work in a coordinate basis, and in those cases one must convert the answser in the holonomic basis to a nonholonomic basis.

This two-step process may seem painful, but then the third point arises. The third point is that he calculations of computing curvature are so involved that it's usually automated with computer algebra anyway. There's a free program, Maxima, that can do the calculations in a holonomic basis, and several other programs (that aren't free) that can do the calculations in nonholonomic bases directly.

There are some PF articles about using Maxima, it's not terribly straightforwards to use due to the program not following standard GR textbook conventions on such things as the order of components in the Riemann tensor.

 

[Ibix]

There are some PF articles about using Maxima, it's not terribly straightforwards to use due to the program not following standard GR textbook conventions on such things as the order of components in the Riemann tensor.

See https://www.physicsforums.com/insights/solving-einsteins-field-equations-in-maxima/ which has links to a sequence of posts by Chris Hillman and includes some work in tetrads.

Maxima numbers indices from 1, not 0.
$R^a{}_{bcd}$=riem[b,c,d,a]
$\Gamma^a_{bc}$=mcs[b,c,a]

 

[strangerep]

It's also worth thinking about why the table about "Coordinate Conversions" is partially wrong (or at least sloppy)!

Oh, did I miss something? What do you think is wrong?

 

[vanhees71]

What I meant is the sloppy formula for the polar angle in polar coordinates (and the other analogous cases too). First of all one must recall the domains of the coordinates. For polar coordinates a convenient choice is $r \in (0,\infty)$ and $\varphi \in (-\pi,pi]$.

Then the correct formula for the angle in terms of the Cartesian coordinates is
$$\varphi=\arctan_2(y,x)=\text{sign} \; y \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right).$$