1. May 19, 2015

### space-time

For the Godel metric (in Cartesian coordinates), I derived the Einstein tensor Gμν as well as its inverse Gμν in a coordinate basis. I tried converting the inverse into an orthonormal basis using a technique for this that was taught to me on another thread long ago. When I tried using said technique, I got a very strange result. One of my basis vectors (e3 to be precise) ended up having an imaginary term. Needless to say, this was wrong, as I looked up what the orthonormal basis inverse Einstein tensor Gμν should be on Wiki (http://en.wikipedia.org/wiki/Gödel_metric) , and it did not match up to what I had derived. I decided that I needed to study to see if there was any other technique for converting to an orthonormal basis. My studies led me to two wiki pages:

http://en.wikipedia.org/wiki/Gödel_metric (this is the one from above)
http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

Now according to the 2nd wiki, to derive a basis vector ea (this should have an arrow over it, but I don't know how to put one there), the formula is:

ea= eajxj

An example of the usage of this formula is shown in the 1st wiki:
e0 = ω√2 * ∂t

Now here is where my problem comes in:

1. First of all, the above example contains within it ∂t. I see where the ω√2 term comes in. However, what exactly are they differentiating with respect to time? The metric tensor for this metric doesn't contain a single t term in it and neither does the term ω√2 . Differentiating anything that I just mentioned with respect to time would result in a basis vector e0 = 0.

2. Their 3rd basis vector e3 = 2ω(e-xz - ∂t). Why does the calculation of this basis vector involve a both a time derivative and a spatial derivative when the other basis vectors only involved derivatives with respect to one coordinate?

Thank you. P.S: In case you would like to see the metric tensor:

g00 = -1/2ω2
g03 and g30 = -ex/2ω2
g11 = 1/2ω2
g22 = 1/2ω2
g33 = -e2x/4ω2

2. May 19, 2015

### fzero

The coframe field can be read off directly from the metric:

$${e^0}_\mu dx^\mu = \frac{1}{\sqrt{2}\omega} ( dt + e^x dz), \ldots.$$

The coefficients ${e^\mu}_a$ are obtained by computing the appropriate matrix inverse, but frame fields are by definition vector fields, so we need to dot the components into a basis of vector fields: $e_a = {e^\mu}_a \partial_\mu$. Alternatively one can use the natural pairing between vector fields and 1-forms to compute the frame fields directly.

Vector fields can act on each other, on functions, on forms, etc. The equation there is an operator equation and doesn't have to be acting on anything in particular. The presence of vector fields there is just a consequence of the definition of frame fields.

3. May 19, 2015

### space-time

Would you mind expanding that example equation of yours all the way out? I think I might understand your advice a little better then. As the equation is written right now in the above post, I don't understand how exactly you derived eνμ. All I understand thus far from what you posted is that e00 = 1/ω√2 and e03 = ex/ ω√2 . I don't know how you got those values either. I have the inverse metric tensor, but I don't get exactly what you did with it in order to derive those coefficients.

4. May 19, 2015

### fzero

Sure. I will use Greek indices $\mu,\nu,\ldots = 0,1,2,3$ for the regular coordinates and latin indices $a,b,\ldots = 0,1,2,3$ for the local flat coordinates. The flat coordinates get raised and lowered with the flat metric $\eta_{ab} = \text{diag}(-1,1,1,1)$.

The idea of the coframes $e^a = {e^a}_\mu dx^\mu$ is to rewrite our metric in the form

$$ds^2 = g_{\mu\nu} dx^\mu dx^\nu = \eta_{ab} e^a e^b.$$

The Goedel metric is almost diagonal, so it's rather easy to write down coframes:

$$\begin{split} ds^2 & = \frac{1}{2\omega} \left[ - (dt + e^x dz)^2 + dx^2 + dy^2 + \frac{1}{2} e^{2x} dz^2 \right] \\ & = - (e^0)^2 + (e^1)^2 +(e^2)^2 +(e^3)^2 . \end{split}$$

hence the claim that

$$e^0 = {e^0}_\mu dx^\mu = \frac{1}{\sqrt{2}\omega} ( dt + e^x dz), \ldots , e^3 =\frac{1}{2\omega} dz .$$

These are 1-forms on our spacetime, but we can also think of them as (co)vectors

$$\mathbf{e}^0 = \left( \frac{1}{\sqrt{2}\omega}, 0,0, \frac{e^x}{\sqrt{2}\omega} \right), \ldots , \mathbf{e}^3 =\left(0,0,0,\frac{e^x}{2\omega}\right).$$

(Technically the coframes are covector fields. If we specify a point on the manifold, then the value of the covector field at that point is a covector. Since this is a physics subforum I won't be so rigorous.)

The frame fields are also associated to vectors and since they are dual to the coframes, we have the relationship

$$\mathbf{e}^a \cdot \mathbf{e}_b = \delta^a_b,$$

which should be compared with the relationship ${e^a}_\mu {e_b}^\mu =\delta^a_b.$ Which relationship you use is a matter of taste, but here we can see easily that

$$\begin{split} & \left( \frac{1}{\sqrt{2}\omega}, 0,0, \frac{e^x}{\sqrt{2}\omega} \right) \cdot ( -1 , 0 , 0 , e^{-x} ) = 0, \\ & \left(0,0,0,\frac{e^x}{2\omega}\right) \cdot ( -1 , 0 , 0 , e^{-x} ) = \frac{1}{2\omega}, \end{split}$$

so we have a frame vector

$$\mathbf{e}_3 = 2\omega( -1 , 0 , 0 , e^{-x} ),$$

that corresponds to a frame vector field

$$e_3 = 2\omega \left( -\partial_t + e^{-x} \partial_z \right).$$

We could use this method to work out all of the frame vectors $e_a$ and it would be equivalent to performing the matrix inverse on the ${e^a}_\mu$.

5. May 21, 2015

### space-time

Alright, I'm starting to understand this method more, but there are a few potential problems that are confusing me:

1. Shouldn't e0= 1/squrt(2ω) * (dt + exdz)? You seem to only have the 2 in the square root and not the ω, which doesn't make sense to me because if - (e0)2 = -1/2ω * (dt + exdz)2 then the square root of all of this (without the negative) is 1/squrt(2ω) * (dt + exdz).

2. From what I see, (e3)2 = 1/4ω * e2xdz2. Therefore e3 should equal 1/(2 * squrt(ω)) * exdz
However, you have 1/2ω * dz . Why is this?

6. May 21, 2015

### fzero

I made a typo in writing down the metric above. The prefactor in the version from the wiki is $1/(2\omega^2)$. The coframes match that choice.

7. May 22, 2015

### space-time

I see. Well that fixes the problem with e0 and most of the problem with e3, but shouldn't e3 still have an ex in the numerator along with the dz?

8. May 22, 2015

### stevendaryl

Staff Emeritus
Did anyone answer your question? This is one of the strangest conventions of modern differential geometry when you first see it, but it makes sense when you think about it long and hard. In a coordinate basis, the basis vector in the $x$ direction, $e_x$, is identified with the operator $\frac{\partial}{\partial x} = \partial_x$, which acts on scalar fields.

In the old-fashioned 3D vector calculus that people learn prior to relativity, there is a "gradient" operator $\vec{\nabla}$ that can be combined with a vector $\vec{A}$ to produce a "directional derivative": $\vec{A} \cdot \vec{\nabla}$. It works on any scalar function $F(x,y,z)$ as follows:

$(\vec{A} \cdot \vec{\nabla}) F = A^x \partial_x F + A^y \partial_y F + A^z \partial_z F$

The modern convention is to say that a vector (or a vector field, more precisely) simply IS a directional derivative. Vector $\vec{A}$ is simply defined to be an operator such that

$\vec{A} F = A^x \partial_x F + A^y \partial_y F + A^z \partial_z F$

If you write $\vec{A}$ out in components as $A^x e_x + A^y e_y + A^z e_z$, where $e_x, e_y$ and $e_z$ are basis vectors, this has the implication that

$e_x = \partial_x$, $e_y = \partial_y$, $e_z = \partial_z$

Now, getting back to GR, if you have a local coordinate system $x^\mu$, then you can write any vector field $A$ as linear combination of basis vectors:

$A = \sum_\mu A^\mu e_\mu = \sum_\mu A^\mu \partial_\mu$

In particular, the elements in the tetrad $e_a$ are vectors, too, so they can be written in terms of a coordinate basis as:

$e_a = \sum_\mu e_a^\mu e_\mu = \sum_\mu e_a^\mu \partial_\mu$

The $\partial_\mu$ isn't acting on anything; this is just an equation relating operators.

9. May 22, 2015

### fzero

Sorry, that's another typo. The factor you want is there in $\mathbf{e}^3$ but I left it out in $e^3$.

10. May 24, 2015

### space-time

-
Ok, so I get what you are saying now. Now, I am just wondering a couple of other things:

1. According to the wiki:
e0= ω√2 * ∂t
e1= ω√2 * ∂x
e2= ω√2 * ∂y
e3= 2ω(e-xz - ∂t)

Now going by the formula: ea= eaμμ , you can calculate that the coefficients eaμ are as follows:

e00 = ω√2
e11=ω√2
e22=ω√2
e33=2ω/ex
e30= - 2ω

Now it is that last one (the e30= - 2ω) that bothers me a bit.

Now if I look at the inverse metric gμν,which has the following elements:
g00= 2ω2
g11= 2ω2
g22= 2ω2
g33= 4ω2/e2x
g03=g30= - 4ω2/ex

(Every other element is 0)

I can easily see that most of the coefficients eaμ are just the square root of their corresponding inverse metric tensor element (example: e00 = ω√2 which is the square root of 2ω2 which is g00).

However, e30= - 2ω which is not the square root of - 4ω2/ex.

This fact, combined with the fact that I only knew for certain how to derive the coefficients eaμ because I already actually had the ea values and the formula ea= eaμμ given to me on the wiki, leads me to ask:

Exactly how would one derive the eaμ coefficients assuming that you did not already know the ea values? I know that fzero said in an earlier post that you use the inverse metric, but you use it how? Is it wrong to say that you simply take the square root of an inverse metric element to get the corresponding coefficient (seeing as how - 2ω is not the square root of - 4ω2/ex)? Why is e3= 2ω(e-xz - ∂t) which has derivatives with respect to multiple axes instead of just one axis like with all of the other ea vectors?

Finally, once I have derived all the ea values, how do I use them to convert Gμν to an orthonormal basis?

11. May 24, 2015

### Mentz114

A tensor is brought into the local basis by the transformation

$T^{ab}={e^a}_\mu {e^b}_\nu T^{\mu\nu}$

Given ${e^a}_\mu$ and looking at it as a transformation matrix, the inverse is ${e_a}^\mu=(({{e^a}_\mu})^{-1})^T$

This relation must be satisfied by the tetrads formed from the basis vectors.

$\eta^{ab}={e^a}_\mu {e^b}_\nu g^{\mu\nu}\rightarrow {e_a}^\mu{e_b}^\nu \eta^{ab}= g^{\mu\nu}$

fzero was writing much the same thing as I was.

Last edited: May 24, 2015
12. May 24, 2015

### fzero

The ${e_a}^\mu$ satisfy $\eta^{ab}{e_a}^\mu{e_b}^\nu = g^{\mu\nu}$, so if the inverse metric were diagonal, the ${e_a}^\mu$ would just be square roots. When there are non-diagonal terms, it is more complicated.

I didn't say to compute the inverse metric, I said that since ${e_a}^\mu{e^b}_\mu = \delta^b_a$, we can say that, as matrices $[{e_a}^\mu] = ([{e^b}_\mu ]^T)^{-1}$ (I think we need the transpose there to be consistent with the placement of indices).

Given the frame vector fields $e_a = {e_a}^\mu\partial_\mu$ we can use the so-called musical isomorphism to write corresponding one-forms ${e^\flat}_a = {e_a}^\mu dx_\mu$. Then

$$g^{\mu\nu} d x_\mu dx_\nu = \eta^{ab} {e^\flat}_a {e^\flat}_b.$$