Understanding the Tetrad Formalism in General Relativity

[Higgsono]

Correct me if I'm wrong. But my understanding is the following.

Introducing a tetrad, means introducing an orthonormal basis of smooth vector fields, satisfying

$(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ at each point. That is, we define a set of 4 vector fields such that they are orthogonal at each point on the manifold.

But then my question is. Why can't we just choose a coordinate system such that our coordinate basis $\Big\{\frac{\partial}{\partial x^{\mu}}\Big\}$ is orthonormal at each point? Why do we need to introduce a vector field on top of our coordinate system to have a basis that is orthonormal at each point?

I suspect that it has to do with the number of charts we want to cover the manifold with. But I'm not sure. For instance on a sphere we can use the coordinate basis because it is everywhere orthonormal, and so we do not need to introduce a tetrad by adding additional vector fields.

 

[stevendaryl]

Correct me if I'm wrong. But my understanding is the following.

Introducing a tetrad, means introducing an orthonormal basis of smooth vector fields, satisfying

$(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ at each point. That is, we define a set of 4 vector fields such that they are orthogonal at each point on the manifold.

But then my question is. Why can't we just choose a coordinate system such that our coordinate basis $\Big\{\frac{\partial}{\partial x^{\mu}}\Big\}$ is orthonormal at each point? Why do we need to introduce a vector field on top of our coordinate system to have a basis that is orthonormal at each point?

I suspect that it has to do with the number of charts we want to cover the manifold with. But I'm not sure. For instance on a sphere we can use the coordinate basis because it is everywhere orthonormal, and so we do not need to introduce a tetrad by adding additional vector fields.

You can choose a coordinate system so that at a single point, the coordinate basis is orthonormal. But if the curvature is nonzero, then away from that point, it will no longer be orthornormal. ("Orthonormal" isn't really the right word, since that implies that each vector has norm +1. You mean that the base vectors have norm $\pm 1$)

 

[stevendaryl]

What I think may be the simplest example showing that it is impossible to do what you want, consider the surface of the Earth. The usual coordinate system is $\theta, \phi$ where $\phi$ is longitude and $\theta$ is latitude. At the equator, the basis vectors for those coordinates are orthonormal (or can be made orthonormal by scaling appropriately). But near the North and South poles, the coordinate system is greatly distorted from being Cartesian, so they don't have orthonormal basis vectors.

 

[Higgsono]

What I think may be the simplest example showing that it is impossible to do what you want, consider the surface of the Earth. The usual coordinate system is $\theta, \phi$ where $\phi$ is longitude and $\theta$ is latitude. At the equator, the basis vectors for those coordinates are orthonormal (or can be made orthonormal by scaling appropriately). But near the North and South poles, the coordinate system is greatly distorted from being Cartesian, so they don't have orthonormal basis vectors.

But how do you choose a vector field so that they remain orthonormal everywhere on the sphere? Do we not have the same problem here? And we can always cover the surface on a sphere with 2 coordinate systems.

 

[Orodruin]

The easiest way to see that such a coordinate system can only exist if the manifold is flat is that the metric components would all be constant. This would mean that all Christoffel symbols (of the Levi-Civita connection) would be zero, since they can be expressed in terms of the inverse metric and the partial derivatives of the metric components - if the metric components are constant these derivatives would all be zero. Since the components of the Riemann curvature tensor are expressed in terms of the connection coefficients, they would all be zero - implying that the manifold (with the Levi-Civita connection) is flat if such coordinates exist.

What I think may be the simplest example showing that it is impossible to do what you want, consider the surface of the Earth. The usual coordinate system is $\theta, \phi$ where $\phi$ is longitude and $\theta$ is latitude. At the equator, the basis vectors for those coordinates are orthonormal (or can be made orthonormal by scaling appropriately). But near the North and South poles, the coordinate system is greatly distorted from being Cartesian, so they don't have orthonormal basis vectors.

Not sure that this is a good example, it is just an example of a coordinate system where this does not happen - not a proof that no such system can exist.

But how do you choose a vector field so that they remain orthonormal everywhere on the sphere? Do we not have the same problem here? And we can always cover the surface on a sphere with 2 coordinate systems.

You cannot find a global orthonormal basis on the sphere.
Edit: See https://en.wikipedia.org/wiki/Hairy_ball_theorem

 

[Dale]

But how do you choose a vector field so that they remain orthonormal everywhere on the sphere?

Latitude and longitude are perpendicular, so all you have to do in this case is to divide the coordinate basis vectors by their norm. Not all coordinate systems are orthogonal, so that cannot always be done.

You still miss the poles of course

 

[Orodruin]

Latitude and longitude are perpendicular, so all you have to do in this case is to divide the coordinate basis vectors by their norm. Not all coordinate systems are orthogonal, so that cannot always be done.

This does not provide you with a global orthonormal basis. Those fields would be singular at the poles.

 

[Higgsono]

This does not provide you with a global orthonormal basis. Those fields would be singular at the poles.

Ok, so how do I construct a tetrad on the 2-sphere then?

 

[Orodruin]

Ok, so how do I construct a tetrad on the 2-sphere then?

There is no global tetrad. You can only construct local tetrads. For most purposes, this is enough.

 

[Higgsono]

The easiest way to see that such a coordinate system can only exist if the manifold is flat is that the metric components would all be constant. This would mean that all Christoffel symbols (of the Levi-Civita connection) would be zero, since they can be expressed in terms of the inverse metric and the partial derivatives of the metric components - if the metric components are constant these derivatives would all be zero. Since the components of the Riemann curvature tensor are expressed in terms of the connection coefficients, they would all be zero - implying that the manifold (with the Levi-Civita connection) is flat if such coordinates exist.

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?

 

[Orodruin]

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?

No you cannot. Spherical coordinates do not give you constant metric components.

 

[Higgsono]

No you cannot. Spherical coordinates do not give you constant metriccomponents.

ok, so the difference between using a tetrad or coordinate basis is that a coordinate basis uses the tangent vetors (partial derivatives) from R^n, while a tetrad uses tangent vectors (partial derivatives) defined on the manifold?

 

[Orodruin]

ok, so the difference between using a tetrad or coordinate basis is that a coordinate basis uses the tangent vetors (partial derivatives) from R^n, while a tetrad uses tangent vectors (partial derivatives) defined on the manifold?

No. Both the tetrad and the holonomic basis are tangent vector fields on the manifold. Each can be expressed in terms of the other. The difference is that one is orthonormal and the other holonomic.

 

[Higgsono]

No. Both the tetrad and the holonomic basis are tangent vector fields on the manifold. Each can be expressed in terms of the other. The difference is that one is orthonormal and the other holonomic.

ok, so basically I don't understand the concept of tetrad at all. Don't we need to specify coordinates in order to define a tetrad on the manifold? edit: or maybe I understand it better now. I guess I was confused with the difference between coordinate system and vector field.

 

[Higgsono]

So my understanding is this. Introducing a tetrad on M is basically the same as introducing a coordinate system on M such that the tangent vectors to the coordinate curves are orthogonal.

 

[Orodruin]

So my understanding is this. Introducing a tetrad on M is basically the same as introducing a coordinate system on M such that the tangent vectors to the coordinate curves are orthogonal.

No. Again, this is not correct. Your tetrad does not need to be based on any coordinate system whatsoever. It is just a complete set of vector fields that are orthonormal.

 

[Higgsono]

No. Again, this is not correct. Your tetrad does not need to be based on any coordinate system whatsoever. It is just a complete set of vector fields that are orthonormal.

eeh ok, but how do I define a vector field without coordinates? If the metric should be Minskowski, does that not imply that they are defined using coordinates. How could you otherwise take their inner product?

 

[Dale]

Don't we need to specify coordinates in order to define a tetrad on the manifold?

No. Sometimes you can make a very meaningful tetrad in situations where a coordinate system is problematic. For example it is easy to construct a tetrad for a family of observers on a rotating disk, but because of simultaneity constructing a coordinate system is problematic.

The way I think of it is that the tetrad represents the physics. The coordinates are just labels.

 

[Orodruin]

eeh ok, but how do I define a vector field without coordinates? If the metric should be Minskowski, does that not imply that they are defined using coordinates. How could you otherwise take their inner product?

A vector field is independent of any coordinates you define. It will be the same vector field regardless of what coordinates you have introduced.

The metric is just a (0,2) tensor field, it is a (0,2) tensor at each point. The requirement on the tetrad is that letting the metric take two different tetrad vectors should always give you zero and if you let it take the same it should give you +(-)1.

You do not need coordinates to talk about components of a tensor. What you need is a basis for the tangent space. You can get such a basis from a coordinate system by considering the corresponding holonomic basis and the corresponding components are what is referred to as the components in those coordinates. But the basis you choose need not be holonomic.

 

[stevendaryl]

But how do you choose a vector field so that they remain orthonormal everywhere on the sphere? Do we not have the same problem here?

Yes, we have the same problem. That was my point: You can't come up with a coordinate system on the sphere where the coordinate basis is always orthonormal.

And we can always cover the surface on a sphere with 2 coordinate systems.

Not an orthonormal one.

 

[stevendaryl]

Not sure that this is a good example, it is just an example of a coordinate system where this does not happen - not a proof that no such system can exist.

Well, it wasn't offered as a proof, it was offered as an example.

 

[stevendaryl]

This does not provide you with a global orthonormal basis. Those fields would be singular at the poles.

Well, are tetrads supposed to be only used within a simply connected region? If so, Dale's construction works fine in a small region of the Earth.

As for globally, doesn't the theorem that you can't comb a sphere imply that there is no way to do it?

[edit] I see that @Orodruin already made this point.

 

[Dale]

Are tetrads required to be smoothly varying from point to point? If not then you can certainly have a tetrad covering the sphere. Just form the usual latitude and longitude tetrad and then choose an arbitrary set of vectors at the poles.

 

[Orodruin]

Are tetrads required to be smoothly varying from point to point?

Typically, yes.

 

[Dale]

Typically, yes.

I couldn’t find an online reference that specified that as a requirement.

 

[vanhees71]

Well, I guess as anything concerning differentiable manifolds you can define smooth tetrad fields within a single map, but an atlas usually will necessarily consist of more than one map to cover the entire manifold. E.g. for the sphere in 3D Euclidean space any atlas contains at least two maps.

 

[Dale]

Well, I guess as anything concerning differentiable manifolds you can define smooth tetrad fields within a single map, but an atlas usually will necessarily consist of more than one map to cover the entire manifold. E.g. for the sphere in 3D Euclidean space any atlas contains at least two maps.

For coordinate maps yes, you are absolutely right. But there is no need for a tetrad to be tied to a coordinate chart at all, so the coverage limitations on charts may not apply to tetrads.

For example, on the sphere you can form the usual latitude and longitude based tetrads everywhere except the poles, and on the poles you can simply pick an arbitrary direction as your $x_1$ direction and then the $x_2$ vector is determined. Then you have a tetrad field which covers the entire sphere whose only problem is smoothness at the poles. I just have not found an authoritative source which stipulates smoothness as a requirement for a tetrad, although it is the usual case.

 

[Orodruin]

I don't see how a non-smooth tetrad would be of practical use. To me, a very important property is to write smooth vector fields as linear combinations $Y = Y^a e_a$, where the $e_a$ are the tetrad fields and the $Y^a$ are smooth functions (components in terms of the tetrad). The nice properties of writing a field like that are quite diminished if $e_a$ are no longer smooth themselves.

 

[vanhees71]

For coordinate maps yes, you are absolutely right. But there is no need for a tetrad to be tied to a coordinate chart at all, so the coverage limitations on charts may not apply to tetrads.

For example, on the sphere you can form the usual latitude and longitude based tetrads everywhere except the poles, and on the poles you can simply pick an arbitrary direction as your $x_1$ direction and then the $x_2$ vector is determined. Then you have a tetrad field which covers the entire sphere whose only problem is smoothness at the poles. I just have not found an authoritative source which stipulates smoothness as a requirement for a tetrad, although it is the usual case.

My point was exactly this. The topology of the manifold is locally defined to be that of $\mathbb{R}^d$, and it's the maps which defines this local topology and thus what continuous means for functions on the manifold. A differential manifold as a whole is defined by atlasses consisting of compatible maps, and the compatibility between laps on overlap regions makes the manifold a whole. Thus you also need to define tetrade fields piecewise on maps with some consistency condition in the overlap regions. I guess at each point in the overlap region the tetrad field defined on one piece must be given by a $\mathrm{SO}(m,n)$ (for relativity it's of course $(m,n)=(1,3)$ or $(m,n)=(3,1)$ depending on your preferred east- or west-coast convention for the fundamental form). However, I'm not aware of any textbook which defines tetrades (or more generally $(m+n)$-beins) globally.

 

[George Jones]

I just have not found an authoritative source which stipulates smoothness as a requirement for a tetrad, although it is the usual case.

I don't see how a non-smooth tetrad would be of practical use.

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."

 

[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.