Learn About Tetrads: Definition, Meaning & Problems Solved

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m4r35n357 said:
OK thanks, I think I have enough clues now to sort this out.
Spoke too soon, OK so I've done the algebra which gives me a set of four (orthonormal?) vectors, yes? So how is the tetrad matrix constructed from these vectors, or is it constructed in some other way? Is it just four column vectors side by side?
I now realize part of my confusion is that I am struggling with alien (to me) terminology, frames/coframes are not concepts that I've dealt with up until now . . .

[EDIT] perhaps I mean four row vectors stacked up . . . ?
[EDIT 2] I have now tried both and obtained the correct metric as in the following Maxima code:
Code: 
kill(all);
fAa: matrix([1, %beta, 0, 0], [0, 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 . nuAB . transpose(fAa);
and
Code: 
kill(all);
fAa: matrix([1, 0, 0, 0], [%beta, 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: transpose(fAa) . nuAB . fAa;
So row and column arrangements both work, depending on which "side" is transposed in the Gab equation. Is there a convention for this, or are both equally (in)correct?
 
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Let T be the matrix transpose operator, then T.T = I where I is the identity. If you multiply (T.A).B.A through with T you get T.(T.A).(T.B).T.A = A.B.(T.A) if B is symmetric.

So they are the same. Tensor operations are not always equivalent to matrix operations, so it's best to use matrix operations with care.
 
Mentz114 said:
Let T be the matrix transpose operator, then T.T = I where I is the identity. If you multiply (T.A).B.A through with T you get T.(T.A).(T.B).T.A = A.B.(T.A) if B is symmetric.

So they are the same. Tensor operations are not always equivalent to matrix operations, so it's best to use matrix operations with care.
Cheers, this is starting to make some kind of sense; I've got the Doran metric in this form now, and am picking away at it to try to understand how the basis vectors work (they are surprisingly simple in this metric for some reason). Boyer-Lindquist is surprisingly difficult to make sense of, but I can live with that for now!
I'd previously been expanding the Doran metric from the River Model paper by hand, but this tetrad approach appears to give the same results, from a simpler starting point. Here it is if you are interested, I think it's right!:
Code: 
("Doran");
kill(all)$
/*
R: sqrt(r^2+a^2)$
%rho: sqrt(r^2+a^2*cos(theta)^2)$
%beta: sqrt(r*Rs/R^2)$
*/
fAa: matrix([1, 0, 0, 0], [%beta * R / %rho, %rho / R, 0, - %beta * R / %rho * a * sin(%theta)^2], [0, 0, %rho, 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: transpose(fAa) . nuAB . fAa;
with R, beta and rho defined within comments.
 
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m4r35n357 said:
Cheers, this is starting to make some kind of sense; I've got the Doran metric in this form now, and am picking away at it to try to understand how the basis vectors work (they are surprisingly simple in this metric for some reason). Boyer-Lindquist is surprisingly difficult to make sense of, but I can live with that for now!
I'd previously been expanding the Doran metric from the River Model paper by hand, but this tetrad approach appears to give the same results, from a simpler starting point. Here it is if you are interested, I think it's right!:
..
..
with R, beta and rho defined within comments.

If you get the Doran metric, it must be right. I would put a 'trigsimp' into save effort.
Code: 
Gab: trigsimp(transpose(fAa) . nuAB . fAa);
and a line to check it
Code: 
Gab - Doran;
 
Mentz114 said:
If you get the Doran metric, it must be right. I would put a 'trigsimp' into save effort.
Yeah, this is only test stuff ATM, and not too messy. I usually try trigsimp and factor on most things these days; sometimes they improve output and other times they make a real mess ;)
Anyway now I can take a look back on the thread and consolidate. Thanks to all who contributed.
 
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Well, I got the Schwarzschild, Gullstrand-Painleve and Doran metrics "deconstructed" very easily. For some reason the Boyer-Lindquist metric seems unusually resistant to such analysis. I first tried mutating the Doran metric without success, the tried to factorize the metric into four squares by hand (again unsuccessfully).
Then I stumbled across this paper: http://casa.colorado.edu/~ajsh/phys5770_08/grtetrad.pdf . In section 5.22 it gives the factorization and the inverse vierbein just in precisely the form I was looking for! Unfortunately it does not seem to give the correct metric when applied to the Lorentz frame.
After much experimentation I went back to the ctensor documentation and found there is a way to do the calculation without relying explicitly on matrix multiplications, which I show below:
Code: 
kill(all)$
if get('ctensor,'version)=false then load(ctensor)$
cframe_flag:true$
verbose:true$
("Covariant frame metric");
lfg: matrix([-1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]);
("Inverse frame matrix to metric");
("Minkowski");
fri: matrix([1, 0, 0, 0], [0, 1, 0, 0], [0, 0, r, 0], [0, 0, 0, r * sin(%theta)]);
cmetric()$
lg;
("Schwarzschild");
fri: matrix([sqrt(1 - %beta^2), 0, 0, 0], [0, 1 / sqrt(1 - %beta^2), 0, 0], [0, 0, r, 0], [0, 0, 0, r * sin(%theta)]);
cmetric()$
lg;
("Gullstrand-Painleve");
fri: matrix([1, 0, 0, 0], [%beta, 1, 0, 0], [0, 0, r, 0], [0, 0, 0, r * sin(%theta)]);
cmetric()$
lg;
("Boyer-Lindquist");
fri: matrix([R*sqrt(1-%beta^2)/%rho,0,0,-R*sqrt(1-%beta^2)/%rho*a*sin(%theta)^2],[0,%rho/(R*sqrt(1-%beta^2)),0,0],[0,0,%rho,0],[-a*sin(%theta)/%rho,0,0,R^2*sin(%theta)/%rho]);
cmetric()$
lg;
("Doran");
fri: matrix([1, 0, 0, 0], [%beta * R / %rho, %rho / R, 0, - %beta * R / %rho * a * sin(%theta)^2], [0, 0, %rho, 0], [0, 0, 0, R * sin(%theta)]);
cmetric()$
lg;
Unfortunately, the BL metric still looks wrong to me (in the g_{tt}{}^{} and g_{\phi\phi}{}^{} components). I tend to trust Hamilton's work ;) so I'm a bit puzzled about this. Can anyone spot my error(s)?
 
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m4r35n357 said:
Code: 
("Boyer-Lindquist");
fri: matrix([R*sqrt(1-%beta^2)/%rho,0,0,-R*sqrt(1-%beta^2)/%rho*a*sin(%theta)^2],[0,%rho/(R*sqrt(1-%beta^2)),0,0],[0,0,%rho,0],[-a*sin(%theta)/%rho,0,0,R^2*sin(%theta)/%rho]);
cmetric()$
lg;
Unfortunately, the BL metric still looks wrong to me (in the g_{tt}{}^{} and g_{\phi\phi}{}^{} components). I tend to trust Hamilton's work ;) so I'm a bit puzzled about this. Can anyone spot my error(s)?

Your BL tetrad is correct. You probably need one more trigsimp. The acceleration of the observer in the static frame has one component in the r direction given by
[tex] \frac{m}{r\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}-\frac{{a}^{2}}{{r}^{2}\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}[/tex]
You can see the gravitational and rotational accelerations. If a=0 the result is the same as for the Schwarzschild stationary frame. This is exactly what is expected.

Also I subtracted the result of cmetric() from the Kerr-BL metric from Visser and the result is zero ( after a trigsimp).
 
Ah, thanks for the confirmation, now I see how it all comes together, unfortunately Maxima doesn't have the information to fix it, so I'll need to recast variables a bit. My problems are caused by trying to make BL look more like Doran than it otherwise would, but then so does Hamilton, so I don't feel bad!
 
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Mentz114 said:
The acceleration of the observer in the static frame has one component in the r direction given by
[tex] \frac{m}{r\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}-\frac{{a}^{2}}{{r}^{2}\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}[/tex]
You can see the gravitational and rotational accelerations. If a=0 the result is the same as for the Schwarzschild stationary frame. This is exactly what is expected.
OK, you've got me! Thought I'd take a little while to try to understand this, but after some head-scratching I'm not sure where it comes from or what it's telling me. Would you mind elaborating as I can't work out what acceleration you are referring to or how you are deriving it?
 
m4r35n357 said:
Would you mind elaborating as I can't work out what acceleration you are referring to or how you are deriving it?
The proper acceleration of a 4-velocity uμ is given by the covariant derivative of uμ projected onto uμ, i.e.
[tex] \frac{d^2x^\mu}{d\tau^2}=\nabla_\nu u_\mu u^\nu[/tex]
If the vector is a geodesic then the acceleration is zero, otherwise the frame is experiencing a force.

That is rather informal description. The subject is covered in textbooks and the oft-quoted Wiki article is a decent introduction ( which you've probably read ).
http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

I'll PM you with the gory details of calculating this with Maxima.

[edit]
The acceleration I gave in my previous post is only valid in the equatorial plane [itex]\theta=\pi/2[/itex]. Also, if [itex]mr=a^2[/itex] it's a geodesic, and is the path of a circular orbit where the centripetal force cancels the gravitational field. Isn't that a nice way to find a geodesic !
 
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Ah, that explains it; I haven't done any actual covariant derivative calculations before, just read the theory and gone straight on to the geodesic equations ;)
Yes I've seen that wikipedia article but I find it very confusing, with inconsistent notation and lots of missing steps . . . pity none of my books go into tetrads in any depth.
All I know about tetrads is what I've learned in this forum and what I've tried myself. So having found out what tetrads are, I am only now in a position to attempt to use them, I look forward to the gory details ;)