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What are tetrads for?

  1. Jun 27, 2012 #1
    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 ;)
     
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  3. Jun 27, 2012 #2

    bapowell

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    They are necessary for including spinors (fermions) in general relativity. If I recall correctly, Weinberg does a decent job of motivating them.
     
  4. Jun 27, 2012 #3

    Mentz114

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    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, LaA is a tetrad, so each index refers a different basis, which is why I've distiguished them as a and A.

    [tex]
    L_a^A = \left[ \begin{array}{cccc}
    P & 1 & 0 & 0 \\\
    0 & R & 0 & 0 \\\
    S & 0 & 1 & 0 \\\
    0 & 1 & 0 & Q \end{array} \right],\ \ V^a=(v_0,v_1,v_2,v_3)
    [/tex]
    [tex]
    L_a^A V^a = (Pv_0+v_1,Rv_1,Sv_0+v_2, v_1+Qv_3) = V^A
    [/tex]
    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
    [tex]
    (F_a^A)^T = \left[ \begin{array}{cccc}
    1 & 0 & 0 & 0\\\
    -\sqrt{2}\,\sqrt{\frac{m}{r}} & 1 & 0 & 0\\\
    0 & 0 & r & 0\\\
    0 & 0 & 0 & r\,sin\left( \theta\right)
    \end{array} \right]
    [/tex]
    transforms the Minkowski metric to the Schwarzshild metric in G-P coordinates.

    gab = FaA FbB ηAB

    [correction - it is the transpose of what I've given, so I added some tex]
     
    Last edited: Jun 27, 2012
  5. Jun 27, 2012 #4
    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?
     
  6. Jun 27, 2012 #5

    Mentz114

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

    I've got Maxima routines to do this stuff.
     
  7. Jun 27, 2012 #6

    atyy

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    Last edited: Jun 27, 2012
  8. Jun 27, 2012 #7
  9. Jun 27, 2012 #8

    haushofer

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    I would motivate tetrads in the following way.

    You know that the equivalence principle allows one to, locally in spacetime, transform the metric [itex]g_{\mu\nu}[/itex] to the flat metric [itex]\eta_{AB}[/itex] (because locally spacetime looks flat). This means that you should always be able to write
    [tex]
    \eta_{AB} = e^{\mu}{}_A e^{\nu}{}_B g_{\mu\nu} \ \ \ \ \ (1)
    [/tex]
    where [itex]e^{\mu}{}_A[/itex] is just the coordinate transformation bringing you to the local metric,
    [tex]
    e^{\mu}{}_A \equiv \frac{\partial x^{\mu}}{\partial x^A}
    [/tex]
    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
    [tex]
    g_{\mu\nu} = e_{\mu}{}^A e_{\nu}{}^B \eta_{AB}
    [/tex]
    Now the geometry of spacetime is encoded in the tetrad. This tetrad transforms under local Lorentz transformations Lambda,
    [tex]
    e_{\mu}{}^{'A} = \Lambda^{A}{}_ B e_{\mu}{}^B
    [/tex]
    and general coordinate transformations
    [tex]
    e_{'\mu}{}^{A} = \frac{\partial x^{\nu}}{\partial x^{'\mu}} e_{\nu}{}^{A}
    [/tex]
    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.
     
  10. Jun 27, 2012 #9
    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 [itex]f(x) = x'[/itex], a tangent vector like [itex]dx/d\tau[/itex] transforms as

    [tex]\underline f^{-1} \left(\frac{dx'}{d\tau}\right) = \frac{dx}{d\tau}[/tex]

    where [itex]\underline f[/itex] 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 [itex]\underline h[/itex] such that

    [tex]\underline h'^{-1}(\dot x') = \underline h^{-1}(\dot x)[/tex]

    (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 [itex]\underline h'^{-1} = \underline h^{-1} \underline f^{-1}[/itex].

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

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

    [tex]\underline T(g_\mu) \cdot g_\nu = (p + \rho) (u \cdot g_\mu)(u \cdot g_\nu) + p g_\mu \cdot g_\nu[/tex]

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

    [tex]\underline T(a) = (p + \rho) (u \cdot a) u + p a[/tex]

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

    Using the [itex]\underline h[/itex] 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 [itex]\underline h[/itex] field gives them the flexibility to do.
     
  11. Jun 27, 2012 #10
    Thanks folks, should have known there wouldn't be an easy answer, will digest at my pace . . . ;)
     
  12. Jun 27, 2012 #11

    Ben Niehoff

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    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.

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

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

    [tex]\begin{align*} T^a &= de^a + \omega^a{}_b \wedge e^b, \\ R^a{}_b &= d\omega^a{}_b + \omega^a{}_c \wedge \omega^c{}_b, \end{align*}[/tex]
    where [itex]T^a[/itex] is the torsion (a vector-valued 2-form), and [itex]R^a{}_b[/itex] is the Riemann tensor (a matrix-valued 2-form). [itex]\omega^a{}_b[/itex] is called the "connection 1-form", and tells you how to take covariant derivatives. After specifying [itex]T^a[/itex], you solve the first equation for [itex]\omega^a{}_b[/itex], 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

    [tex]\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*}[/tex]
    In the absence of torsion, the first line is Bianchi's algebraic identity; the second line is Bianchi's differential identity.
     
  13. Jun 27, 2012 #12

    TSny

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    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
     
  14. Jun 27, 2012 #13

    Ben Niehoff

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    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.
     
  15. Jun 28, 2012 #14

    atyy

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    That's very interesting. I looked at Nakahara's book after hearing that Charles Kane had learnt 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.
     
  16. Jun 28, 2012 #15

    haushofer

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    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.
     
  17. Jun 28, 2012 #16
    Is the Poincare algebra related to the "restricted Lorentz group" that is usually considered GR gauge group?
     
  18. Jun 28, 2012 #17
    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:
    Code (Text):
    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)!
     
    Last edited: Jun 28, 2012
  19. Jun 28, 2012 #18
    Nevermind, I didn't notice you were referring to supergravity.
     
  20. Jun 28, 2012 #19

    Mentz114

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

    Or use this function
    Code (Text):
    /********************************************************************/
    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.
     
    Last edited: Jun 28, 2012
  21. Jun 28, 2012 #20
    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 ;)
     
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