Learn About Tetrads: Definition, Meaning & Problems Solved

[m4r35n357]

I've attempted to learn about tetrads on several occasions, but the dry mathematical definitions in eg. MTW, Carrol's notes etc. leave me no wiser on matters such as:
What problems would they allow me to solve?
Is there any useful physical meaning to them?
Basically the stuff I've seen sort of assumes the reader already knows what they are and why they are useful.
Pointers to good introductory texts welcome as usual ;)

 

[bapowell]

They are necessary for including spinors (fermions) in general relativity. If I recall correctly, Weinberg does a decent job of motivating them.

 

[Mentz114]

Tetrads are transformation matrices ( not tensors, even though they are wriiten as mixed rank-2 tensors) that change the basis of vectors. They do linear transformations, so the new vector has components that are linear combinations of the original components. For example, L_a^A is a tetrad, so each index refers a different basis, which is why I've distiguished them as a and A.

<br /> L_a^A = \left[ \begin{array}{cccc}<br /> P &amp; 1 &amp; 0 &amp; 0 \\\<br /> 0 &amp; R &amp; 0 &amp; 0 \\\<br /> S &amp; 0 &amp; 1 &amp; 0 \\\<br /> 0 &amp; 1 &amp; 0 &amp; Q \end{array} \right],\ \ V^a=(v_0,v_1,v_2,v_3)<br />
<br /> L_a^A V^a = (Pv_0+v_1,Rv_1,Sv_0+v_2, v_1+Qv_3) = V^A<br />
Mostly tetrads are used to change basis between global and local coordinate systems in the frame field formalism.

This article is useful,

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

This tetrad
<br /> (F_a^A)^T = \left[ \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\\<br /> -\sqrt{2}\,\sqrt{\frac{m}{r}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; r &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; r\,sin\left( \theta\right) <br /> \end{array} \right]<br />
transforms the Minkowski metric to the Schwarzshild metric in G-P coordinates.

g_ab = F_a^A F_b^B η_AB

[correction - it is the transpose of what I've given, so I added some tex]

 

[m4r35n357]

This article is useful,

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

This tetrad
<br /> F_a^A = \left[ \begin{array}{cccc}<br /> -1 &amp; 0 &amp; 0 &amp; 0\\\<br /> -\sqrt{2}\,\sqrt{\frac{m}{r}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; r &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; r\,sin\left( \theta\right) <br /> \end{array} \right]<br />
transforms the Minkowski metric to the Schwarzshild metric in G-P coordinates.

g_ab = F_a^A F_b^B η_AB

Yes, thanks, I've read the Wikipedia article, it's probably the most understandable so far, but it would seem that I'll just have to keep plugging away at understanding it . . . eg, how the metric "encodes" a tetrad, hmmm.
BTW is the tetrad above really supposed to be asymmetric?

 

[Mentz114]

how the metric "encodes" a tetrad, hmmm.
BTW is the tetrad above really supposed to be asymmetric?

Yes. It is doing the transformation of the differentials dt' = dt - V dr.

I've got Maxima routines to do this stuff.

 

[atyy]

Orthonormal tetrads provide the local orthonormal reference frame of an observer - for example, a free falling observer should have a local Minkowski frame. Writing the metric in terms of tetrads makes it possible to incorporate the "local" indices of fermions, as bapowell pointed out above.

http://www.astro.umd.edu/~miller/teaching/astr680s09/lecture07.pdf
http://particlephd.wordpress.com/2009/02/07/the-spin-connection/ (called "vielbein" in ND, also "vierbein" in 4D)

 

[m4r35n357]

http://www.astro.umd.edu/~miller/teaching/astr680s09/lecture07.pdf
http://particlephd.wordpress.com/2009/02/07/the-spin-connection/ (called "vielbein" in ND, also "vierbein" in 4D)

Thanks, the first reference looks a little more accessible than the second ;)

 

[haushofer]

I would motivate tetrads in the following way.

You know that the equivalence principle allows one to, locally in spacetime, transform the metric g_{\mu\nu} to the flat metric \eta_{AB} (because locally spacetime looks flat). This means that you should always be able to write
<br /> \eta_{AB} = e^{\mu}{}_A e^{\nu}{}_B g_{\mu\nu} \ \ \ \ \ (1)<br />
where e^{\mu}{}_A is just the coordinate transformation bringing you to the local metric,
<br /> e^{\mu}{}_A \equiv \frac{\partial x^{\mu}}{\partial x^A} <br />
This coordinate transformation can be thought of as the transformation which brings you to the tangent space at a point. The relation (1) can of course be inverted to give
<br /> g_{\mu\nu} = e_{\mu}{}^A e_{\nu}{}^B \eta_{AB}<br />
Now the geometry of spacetime is encoded in the tetrad. This tetrad transforms under local Lorentz transformations Lambda,
<br /> e_{\mu}{}^{&#039;A} = \Lambda^{A}{}_ B e_{\mu}{}^B<br />
and general coordinate transformations
<br /> e_{&#039;\mu}{}^{A} = \frac{\partial x^{\nu}}{\partial x^{&#039;\mu}} e_{\nu}{}^{A}<br />
Now, matter is described by fermions. Fermions, being half-integer representations of the Lorentz group, are not tensors. However, what you now can do is to transform to the tangent space, where you can perfectly well define spinorial representations of the Lorentz group! As such, tetrads allow you to describe Fermions.

 

[Muphrid]

Something related to tetrads is the "displacement gauge function" of Gauge Theory Gravity in geometric algebra and calculus. See http://arxiv.org/abs/gr-qc/0405033 by Lasenby, Doran, and Gull. Here's a general rundown:

The general transformation law for tangent and cotangent vectors under coordinate system change is well known: you can derive, for instance, that under a transformation f(x) = x&#039;, a tangent vector like dx/d\tau transforms as

\underline f^{-1} \left(\frac{dx&#039;}{d\tau}\right) = \frac{dx}{d\tau}

where \underline f denotes the Jacobian.

What Lasenby et. al. do is introdude a "displacement gauge function" to make the above equation frame-independent. They posit a linear operator \underline h such that

\underline h&#039;^{-1}(\dot x&#039;) = \underline h^{-1}(\dot x)

(They actually switch one set of primes around, but I consider this very confusing.) It's clear that this operator must have a transformation law of \underline h&#039;^{-1} = \underline h^{-1} \underline f^{-1}.

As with tetrads in general, the idea here is that the operator \underline h^{-1} acts on vectors e_\nu in a flat space and returns vectors g_\nu such that g_\mu \cdot g_\nu = g_{\mu \nu}. All this happens in a fundamentally flat vector space, however. What determines "curvature" the way we usually think of it comes from how the \underline h function can't be reduced to the identity everywhere, no matter what coordinate system transformations we try.

Lasenby et. al. leverage the \underline h function to rewrite GR in a language that is completely freed from the metric. The \underline h function is the more fundamental quantity. For example, consider the stress energy tensor for an ideal fluid:

\underline T(g_\mu) \cdot g_\nu = (p + \rho) (u \cdot g_\mu)(u \cdot g_\nu) + p g_\mu \cdot g_\nu

With just a little algebraic shuffling, one can see that this defines a general operator,

\underline T(a) = (p + \rho) (u \cdot a) u + p a

This enables one to work entirely in the e_\mu frame instead of the g_\mu frame that is usually done.

Using the \underline h function allows one to write guage-invariant Lagrangians for relativistic QM as well--something that is beyond the scope of my experience, but I get the impression it's a major motivating factor for this framework. Again, to me however the really impressive part is that they're able to write the Einstein equations without reference to the metric thanks in large part to what working with the \underline h field gives them the flexibility to do.

 

[m4r35n357]

Thanks folks, should have known there wouldn't be an easy answer, will digest at my pace . . . ;)

 

[Ben Niehoff]

Is there any useful physical meaning to them?

Absolutely! A tetrad is a set of local orthonormal frames. Which is another way of saying it's a set of local measuring rods. Any time you want to know the size of some object as measured by a local observer, or the wavelength of a photon, etc., you contract against a tetrad. Since a tetrad is a set of vectors of unit length, contracting against a tetrad gives you a real, coordinant-invariant measurement.

What problems would they allow me to solve?

Treating the tetrad as a vector-valued 1-form,

e^a = e^a{}_\mu \, dx^\mu,
one can succinctly write the manifold's curvature information as follows, using Cartan's "structure equations":

\begin{align*} T^a &amp;= de^a + \omega^a{}_b \wedge e^b, \\ R^a{}_b &amp;= d\omega^a{}_b + \omega^a{}_c \wedge \omega^c{}_b, \end{align*}
where T^a is the torsion (a vector-valued 2-form), and R^a{}_b is the Riemann tensor (a matrix-valued 2-form). \omega^a{}_b is called the "connection 1-form", and tells you how to take covariant derivatives. After specifying T^a, you solve the first equation for \omega^a{}_b, and then you use the second equation to compute the curvature. This "method of moving frames" is very efficient.

GR textbooks often make tetrads appear very tedious, writing out mountains of indices without explaining any organizing principles. I think the reason is that physicists in general are stubborn in their ways and insist on doing differential geometry the way it was done in the 19th century.

The whole point of the method of moving frames is to hide the extraneous indices by treating things with the language of differential forms. Once you understand how it works, it requires less writing and less memorizing.

It also deals very elegantly with torsion, which the "index gymnastics" method does not. Quick, what are Bianchi's identities in the presence of torsion? Most people can't remember. With Cartan's structure equations, you don't need to remember anything; just take the exterior derivative of each equation to obtain

\begin{gather*} R^a{}_b \wedge e^b = dT^a + \omega^a{}_b \wedge T^b, \\ dR^a{}_b + \omega^a{}_c \wedge R^c{}_b - R^a{}_c \wedge \omega^c{}_b = 0. \end{gather*}
In the absence of torsion, the first line is Bianchi's algebraic identity; the second line is Bianchi's differential identity.

 

[TSny]

Ben Niehoff,

Do you know of a good reference on tetrads for those of us who are just beginning to learn about them but who have some background in GR from studying the standard texts for physicists?
This thread has convinced me that the subject of tetrads is worth learning.

Thanks

 

[Ben Niehoff]

I learned quite a lot about differential geometry from Nakahara's book. Besides tetrads, it also has lots of stuff about topology, fiber bundles, gauge theories, complex geometry, and stuff. If all you want to learn about are tetrads, then the book has about 12 pages you can use, and tons of stuff you don't need. But if you are curious to do some other reading, the rest of it is all very interesting.

It's also pretty abstract. He doesn't have any sections at all about relating the math of GR to experiment. I have to admit, MTW is a decent book for that (but I find its exposition of the math frustratingly tedious). My favorite GR book is Carroll's. But Nakahara is great for just general differential geometry.

 

[atyy]

I learned quite a lot about differential geometry from Nakahara's book. Besides tetrads, it also has lots of stuff about topology, fiber bundles, gauge theories, complex geometry, and stuff. If all you want to learn about are tetrads, then the book has about 12 pages you can use, and tons of stuff you don't need. But if you are curious to do some other reading, the rest of it is all very interesting.

It's also pretty abstract. He doesn't have any sections at all about relating the math of GR to experiment. I have to admit, MTW is a decent book for that (but I find its exposition of the math frustratingly tedious). My favorite GR book is Carroll's. But Nakahara is great for just general differential geometry.

That's very interesting. I looked at Nakahara's book after hearing that Charles Kane had learned all the topology he knew from there. You're the second professional I know who's recommended it. It's very clearly written that even a biologist (like me) can understand it.

 

[haushofer]

I also like Sean Carroll's treatment of tetrads in his GR notes.

By the way, one more reason to know about tetrads: GR can be obtained from a gauging procedure of the Poincaré algebra. The gauge field of the translations can then naturally be interpreted as the Vielbein (whereas the gauge field of the Lorentz transformations turns out to be the spin connection). Minimal Supergravity can also be obtained in this way, which is an alternative to the (tedious) Noether procedure. Even Newton-Cartan theory, formulated in terms of tetrads, can be obtained this way.

 

[TrickyDicky]

By the way, one more reason to know about tetrads: GR can be obtained from a gauging procedure of the Poincaré algebra.

Is the Poincare algebra related to the "restricted Lorentz group" that is usually considered GR gauge group?

 

[m4r35n357]

This tetrad
<br /> (F_a^A)^T = \left[ \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\\<br /> -\sqrt{2}\,\sqrt{\frac{m}{r}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; r &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; r\,sin\left( \theta\right) <br /> \end{array} \right]<br />
transforms the Minkowski metric to the Schwarzshild metric in G-P coordinates.

g_ab = F_a^A F_b^B η_AB

[correction - it is the transpose of what I've given, so I added some tex]

Heh, didn't see your correction at first, so I just got polar Minkowski space. If I transpose one of the tetrads (like below), I get the expected GP metric:

fAa: matrix([1, -sqrt(Rs/r), 0, 0], [0, 1, 0, 0], [0, 0, r, 0], [0, 0, 0, r * sin(%theta)]);
fBb: matrix([1, 0, 0, 0], [-sqrt(Rs/r), 1, 0, 0], [0, 0, r, 0], [0, 0, 0, r * sin(%theta)]);
nuAB: matrix([-1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]);
Gab: fAa . fBb . nuAB;

and if I transpose both, I am back to polar Minkowski space. Did I do right?
[EDIT] no I didn't, the signs of the GP metric are wrong (even after fixing the bad signs)!

 

[TrickyDicky]

Is the Poincare algebra related to the "restricted Lorentz group" that is usually considered GR gauge group?

Nevermind, I didn't notice you were referring to supergravity.

 

[Mentz114]

Heh, didn't see your correction at first, so I just got polar Minkowski space. If I transpose one of the tetrads (like below), I get the expected GP metric:
...
...
and if I transpose both, I am back to polar Minkowski space. Did I do right?
[EDIT] no I didn't, the signs of the GP metric are wrong (even after fixing the bad signs)!

In terms of your matrices, the correct transformation is fAa.nuAB.transpose(fAa).

Or use this function

/********************************************************************/
trans2(L,M) := block(  [N, a,b,m,n],
/*** 
transform rank-2 M with matrix L as in L.M.transpose(L)
****/
N:zeromatrix(4,4),
for a thru 4 do for b thru 4 do 
for m thru 4 do for n thru 4 do 
N[a,b]:N[a,b]+L[a,m]*L[b,n]*M[m,n],
return(N)
)$

by calling trans2(fAa,nuAB);

You got some real quality replies to your question. Frame fields are important because that's how we can relate this complicated theory to what happens to an observer, which is the bottom line.

 

[m4r35n357]

You got some real quality replies to your question. Frame fields are important because that's how we can relate this complicated theory to what happens to an observer, which is the bottom line.

Yes, I strongly suspected this, but had no idea how to do it. Thanks all who contributed, I can tell the answers were good but I'm not able to fully understand them all yet. I do have a slightly better idea now, but still need to read around more to fully get it.
IMO this is a huge hole in GR teaching resources online, it would be lovely if Kevin Brown were to do a piece about it on Mathpages, his writings are more at my level ;)

 

[m4r35n357]

This tetrad
<br /> (F_a^A)^T = \left[ \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\\<br /> -\sqrt{2}\,\sqrt{\frac{m}{r}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; r &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; r\,sin\left( \theta\right) <br /> \end{array} \right]<br />
transforms the Minkowski metric to the Schwarzshild metric in G-P coordinates.

I think I shall remain stuck until I understand how this tetrad is derived. Can anyone enlighten me?

 

[Muphrid]

In deriving the displacement gauge function for a spherically symmetric perfect fluid, you get:

\begin{align*}<br /> \overline h(e^t) &amp;= f_1 e^t \\<br /> \overline h(e^r) &amp;= g_2 e^t + g_1 e^r \\<br /> \overline h(e^\theta) &amp;= r \hat e^\theta \\<br /> \overline h(e^\phi) &amp;= r \sin \theta \; \hat e^\phi<br /> \end{align*}<br />

Where f_1, g_1, g_2 are unknown functions of r,t. Choosing their forms is the gauge choice necessary to choose a specific coordinate system. There are also some additional constraints. Define G such that (f_1 \partial_t + g_2 \partial_r) g_1 = G g_2. (This operator, f_1 \partial_t + g_2 \partial_r, is called \ell_t \equiv e_t \cdot \overline h(\nabla), and it works like a directional derivative operator.) The relevant constraints are then

\begin{align*}<br /> g_1^2 - g_2^2 &amp;= 1 - 2M/r \\<br /> f_1 &amp;= \exp \left(\int^r -G/g_1 \; dr\right)<br /> \end{align*}

Finally, there is one degree of gauge freedom to choose these functions. The choice that corresponds to P-G coordinates is to choose g_1 = 1 everywhere. g_2 = -\sqrt{2M/r} and f_1 = 1 immediately follow, and you get the P-G tetrad.

Again, this is all from Lasenby, Doran, and Gull. The full derivation (I've obviously skipped over a lot in terms of how you even get to the spherically symmetric perfect fluid's gague field) is a bit involved, but I think this should point out how one can solve a certain, general case of system and then remove any remaining gauge degrees of freedom to arrive at an answer, which corresponds to choosing a specific coordinate system. That is the basic idea of the process, even if the fine details of what the constraints are and how you get the general case solved are really beyond the scope of a forum thread.

 

[m4r35n357]

Sorry to appear ungrateful for your answer, Muphrid, but I genuinely didn't understand a single word of that :(
I know the GP metric, I just need to get from the metric to the tetrad that I quoted.

 

[Muphrid]

Well, in general if you're trying to work from a metric to a tetrad, it's like you're trying to figure out a matrix such that the matrix times its own transpose is the metric. Perhaps there is a general way to do such a thing, but I don't know it.

What I posted is the result of solving the Einstein equations for a tetrad directly, as opposed to solving for a metric and trying to work down to a tetrad.

 

[m4r35n357]

Well, in general if you're trying to work from a metric to a tetrad, it's like you're trying to figure out a matrix such that the matrix times its own transpose is the metric. Perhaps there is a general way to do such a thing, but I don't know it.

What I posted is the result of solving the Einstein equations for a tetrad directly, as opposed to solving for a metric and trying to work down to a tetrad.

Yes, thanks, it did seem to go back a few steps more than I expected! I'll take the time to try to understand that and your earlier reply in due course, but in the meantime I'm just puzzling over that tiny step . . . I've seen it done for Schwarzschild (on Wikipedia), but that didn't generate an asymmetrical tetrad.

 

[Muphrid]

Yeah, I mean, it would be nice if any general metric could be walked down to a tetrad instead of having to go back to Einstein's equations and reworking them in terms of tetrads instead of the metric. As part of my own interest, I tried to decompose a metric/gauge choice used in numerical relativity to a tetrad, and really I was reduced to looking at the (a)symmetries and guessing until something jumped out at me.

I think tetrads or something like them are the forward-thinking way to approach GR, but decades of results are written in terms of the metric for a particular geometry and resist easy conversion. There's probably been more work done to convert those results to tetrad form than I'm aware of, but still, I can't help but feel it's a relative dearth.

 

[Mentz114]

If you already have the GP metric, you can look for a transformation of the differentials
dt' = a dt + b dr, dr' = c dr. (Because we know that's what is required )

So the tetrad is ( only the t,r part is shown)
<br /> L = \left[ \begin{array}{cc}<br /> a &amp; b \\\<br /> 0 &amp; c<br /> \end{array} \right]<br />
and L_a^A L_b^B η_AB is
<br /> \left[ \begin{array}{cc}<br /> {b}^{2}-{a}^{2} &amp; bc \\\<br /> bc &amp; c^2<br /> \end{array} \right]<br />
setting the above equal to g_ab we now we have 3 equations in 3 unknowns which has solution ( amongst 4 actually)
<br /> a=1,b=-\frac{\sqrt{2}\,\sqrt{m}}{\sqrt{r}},c=1

There are probably better ways that always work, as others have said here and elsewhere.

 

[Ben Niehoff]

I find it easiest not to think in terms of matrices. Write the metric out as a line element,

ds^2 = g_{\mu\nu} dx^\mu dx^\nu,
and play around with algebra until this looks like a sum of 4 squares (one of which gets a minus sign, due to the -+++ signature),

ds^2 = -(e^0)^2 + (e^1)^2 + (e^2)^2 + (e^3)^2.
Now you have your frame fields.

 

[Mentz114]

I find it easiest not to think in terms of matrices. Write the metric out as a line element,

ds^2 = g_{\mu\nu} dx^\mu dx^\nu,
and play around with algebra until this looks like a sum of 4 squares (one of which gets a minus sign, due to the -+++ signature),

ds^2 = -(e^0)^2 + (e^1)^2 + (e^2)^2 + (e^3)^2.
Now you have your frame fields.

Yep. If you put σ₀ = a dt + b dr and σ₁ = c dr, then
<br /> -(\sigma_0)^2 + (\sigma_1)^2 <br />
gives the same algebraic problem.

 

[m4r35n357]

Yep. If you put σ₀ = a dt + b dr and σ₁ = c dr, then
<br /> -(\sigma_0)^2 + (\sigma_1)^2 <br />
gives the same algebraic problem.

OK thanks, I think I have enough clues now to sort this out.