Discovering Co-frame Basis Vectors for General Einstein Metric

[jfy4]

OK. Having done this for boosted frames, I guessed that the coframe would be of the form

$$\vec{e_0}= a\ dt\ -\ e03\ dz ,\ \ \ \vec{e_1}= a\ dx,\ \ \ \vec{e_2}= a\ dy,\ \ \ \vec{e_3}= e33\ dz$$

where e03 and e33 are unknown. Working out the sum of the tensor products shown in my previous post gives

$$\left[ \begin{array}{cccc} -{a}^{2} & 0 & 0 & -a\ e03 \\ 0 & {a}^{2} & 0 & 0 \\ 0 & 0 & {a}^{2} & 0 \\ -a\ e03 & 0 & 0 & {e33}^{2}-{e03}^{2} \end{array} \right]$$

and it's trivial to solve for e03 and e33 so this matrix is the required metric.

I've also done this with a frame basis where the target is the inverse metric.

Mentz was kind enough to give me this explanation for co-frame basis for a specific metric that I specified. However, I was wondering if there is a general procedure for finding the co-frame basis vectors for a general Einstein metric. I see the procedure involves the tensor product of the basis vectors with (maybe all?) unknown entries. I am not sure how to pick which parts of these basis-vectors would be unknowns, and which would be knowns. If someone could give me a nice run down that would be great.

Thanks,

 

[Ben Niehoff]

This is actually very easy. First, one point: "the" co-frame basis is not unique; it is only unique up to local SO(1,3) transformations. As for how to find a co-frame basis, my usual method is this:

1. Start with the line element.

2. Think of the line element as a quadratic form in n variables (where n is the dimension of the space). The variables in this case are the differentials dx, dy, dz, etc.

3. Now just use basic algebra: complete the square. The goal is to write your quadratic form as a sum of n squares (for metrics of indefinite signature, some squares will have a minus sign in front).

For the Godel metric, this is quite easy. The line element is

$$ds^2 = a^2 \big( -(dt + e^x \; dz)^2 + dx^2 + dy^2 + \frac12 e^{2x} \; dz^2 \big)$$

You'll notice this is already written as a sum of four squares (with one minus sign, as the signature is -+++). Therefore, you can just read off the co-frame:

$$\mathbf{e}^0 = a (dt + e^x \; dz)$$
$$\mathbf{e}^1 = a \; dx$$
$$\mathbf{e}^2 = a \; dy$$
$$\mathbf{e}^3 = \frac{a}{\sqrt{2}} e^x \; dz$$

Now you have your basis 1-forms. To get the corresponding basis vectors (the frame fields), just use the condition

$$\mathbf{e}^a (\vec{e}_b) = \delta^a_b$$

In this case, this is pretty easy. You should find

$$\vec{e}_0 = \frac1a \; \partial_t$$
$$\vec{e}_1 = \frac1a \; \partial_x$$
$$\vec{e}_2 = \frac1a \; \partial_y$$
$$\vec{e}_3 = \frac{\sqrt{2}}{a} (e^{-x} \; \partial_z - \partial_t)$$

Note: Most of the time, you don't actually need to find the frame; you can just use the co-frame. You can use the co-frame to compute the connection form, curvature form, and Ricci tensor, and simply express them in an orthonormal basis.

Edited to add:

More generically, if your line element is not quite so easy, you can think of this as a matrix problem. The metric can be thought of as a symmetric matrix. The goal is to find the "square root" of this matrix; i.e., a matrix A such that

$$A^T A = G$$

It is easy to show that this is always possible. Since G is symmetric, you can always diagonalize it by an orthogonal matrix. Then the diagonal matrix can be split into two diagonal matrices, each one having the square roots of the eigenvalues as its entries.

 

[jfy4]

Let me give this a try, it does seem easy after you showed me... If I take the canonical coordinates for spherically-symmetric perfect fluid solutions

$$ds^2=r^2d\vartheta^2+r^2\sin^2\vartheta d\varphi^2+e^{2\lambda(r)}dr^2-e^{2\nu(r)}dt^2$$

the following co-frame basis 1-forms are

$$\mathbf{e}^0=e^{\nu(r)}dt$$

$$\mathbf{e}^1=e^{\lambda(r)}dr$$

$$\mathbf{e}^2=rd\vartheta$$

$$\mathbf{e}^3=r\sin\vartheta d\varphi.$$

Now that part I think I have a hold of. But I don't follow how you got the field-frames. Can you do the sum explicitly for me between the basis vectors? It seems from what you wrote that it is just one over the co-frame.

 

[Mentz114]

The goal is to write your quadratic form as a sum of n squares.

That's the bit I've been missing, thanks.

I get the frame fields from the coframe by writing the coframe as a matrix and doing some matrix operations, which is the equivalent of your method but easier to script.

 

[Ben Niehoff]

I get the frame fields from the coframe by writing the coframe as a matrix and doing some matrix operations, which is the equivalent of your method but easier to script.

Yes, the frame fields are just the inverse matrix to the co-frame fields. For simple metrics I find it easier to just make some guesses and modify them as needed.