# Clarification of concepts: Tetrads (Vielbeins), etc

1. Nov 22, 2014

### Breo

Hello folks,

I have doubts about the next concepts in gravitation: Tetrads, dual basis, normal basis, standard basis, local basis, orthogonal basis, FIDO, FREFO, Frames, Diffeomorphisms, Affinne conection, etc.

If I am not wrong, "frame" is a general concept of n-tuplets of vectors defining in a standard basis (standard basis is the basis on the tangent space, right?) $$\partial x^{a}$$ for eveerypoint a tangent space. WE can then find the dual basis $$dx$$ (how? just exchanging partials by exterior derivatives?) and a differential form instead the original vector (how to get it, again?) which both, defines a cotangent vector. The frames in Relativity are always with an orthonormal set of coordinates.

But I just read an exercise which aks for write an adequate dreibein of a metric in terms of differential forms and dual basis of vectors in the tangent space. Tangent space could be a mistake? I think should be cotangent space as is asking for dual.

Local basis is the general term for frame defined basis? Standard basis and normal basis are the same thing?

FIDO is a fixed observer and FREFO is a free falling observer (we should remember here the principle of equivalence) but I do not know how to compute this. I mean, I imagine a "guy" standing in some point of the spacetime observing another guy who is moving (FREFO) like the case of a guy standing at one point in the street whom sees a guy in a car moving away. Is the same thing with metric implications?

Diffeomorphisms are a required condition is need in order to give consistency to the Riemannian geometry so is something we use in the actions to give consistency invariance? I knwo the mathematical definition of diffeomorphism but I do not "draw" in my mind the real consequence in physical terms.

Afinne conection is the gamma symbol, a general case of connection in differential geometry without metrics. Christoffel symbols are the affine connection due to a metric. I

Another thing that pulled me off is when I saw the Einstein - Rosen metric:

$$ds² = \alpha^2 dt² - \beta^4(d\rho^2 - \rho^2 d\theta^2 - sin^2\theta d\phi^2) ;$$ I had use alpha and beta for brieving

And then the metric tensor matrix like this:
$$g_{00}= \alpha^2; \space g_{11} = - \beta^2; \space g_{22} = - \rho^2; \space g_{33} = -sin^2 \theta; \space$$
But why then in the metric ds formula is a common factor of beta^4 ? I do not understand.

I know those are a lot of questions. If you can clarify me at least one of them I would be very grateful :)

Thank you very much in advance.

2. Nov 22, 2014

### Matterwave

You might get better luck if you try to learn well one concept before jumping onto the next one. For example, if you don't have a good grounding of what a tangent space and basis vectors are, or of the dual space (co-tangent space), it will be hard for you to understand well an Affine connection. It will also be very hard for someone to address all your questions without writing just a huge block of text.

Is there a text-book or source that you are following? Perhaps you can start off with one topic and then ask some specific questions on that one topic instead of asking 6 different topics in one thread.

I guess I can give a very cursory overview of some (not all) of the things you brought up. A tangent space (at a point) is made up of all tangent vectors at a point. Just like any vector space, you can use whatever set of n linearly independent vectors to form your basis, where n is the dimension of the tangent space/manifold. There arises a "natural" set of basis vectors called the "coordinate basis" which are built from the local coordinate chart you used to describe the neighborhood of the base point of your tangent space. So if I have a set of coordinates $x^\mu$ which describe the neighborhood, then my set of basis coordinate vectors are (I'm going to use Wald's abstract index notation since I find that to be the clearest set) $(\partial_\mu)^a$ where the directional derivatives are taken along my coordinate lines (lines where all my other coordinates are constant).

Dual to this basis is a set of one-forms whose action on my basis vectors give me a kronecker delta. In other words, a set of dual basis to my coordinate basis are given by $(\omega^\mu)_a$ where $(\omega^\mu)_a(\partial_\nu)^a=\delta^\mu_{~\nu}$. One can check that for my coordinate basis, the dual basis is given by $(\omega^\mu)_a=(dx^\mu)_a$.

I don't have to use a coordinate basis; however, I can use any basis I want. However, if I choose some funky basis, I might just make things way harder on myself. There is one other kind of basis that is often useful, when there is a metric on the manifold, which is an orthonormal basis (if we are in 4-dimensions as we do in general relativity, we often call this a tetrad, sometimes the word vierbeinn is also used). These are the basis for which $g_{ab}(\partial_\mu)^a(\partial_\nu)^b=\delta_{\mu\nu}$. A basis such as this will represent physically the frame of an observer who is using rigid rods and attached clocks to measure distances and time.

An affine connection is basically a rule, given at every point, of how to parallel transport vectors to nearby tangent spaces. An affine connection allows you to compare two vectors which are in different tangent spaces (tangent spaces to different points), given a path between the two tangent spaces. No additional information is needed to get an affine connection. When we have a metric on our manifold, we can force affine geodesics (curves for which their tangent vectors are parallel transported along themselves) to match metric geodesics (curves of maximal/minimal proper length) to get a metric-compatible connection, called a metric connection. The Christoffel symbols give the correction terms for a special kind of affine connection, called a Levi-Civita connection, which is a symmetric metric connection (in other words a metric connection with no torsion). Christoffel symbols are coordinate dependent, they are not tensors. They basically tell you how your coordinate basis vectors are twisting and turning as you move along a path.

I don't know anything about the Einstein Rosen metric, so maybe someone else will help you with that.

3. Nov 22, 2014

### Breo

Thank you very much for the reply :).

I am reading mostly Carroll's book. Do you recommend me any other one?

That observer is a FIDO?

I am still confused about the statement of that exercise. Could be a typo?

Last edited: Nov 22, 2014
4. Nov 22, 2014

### Matterwave

Carroll's notes are good, but they are not introductory level. If you are confused about the mathematics, you could consider a more introductory level text like Hartle or Schwartz.

The observer could be carrying any motion, fixed or inertial, and their local rigid frame will be described by a tetrad :) (AFAIK, I'm not an expert in tetrad methods).

I don't know, you'd have to be more specific. You an maybe scan the page of the book that asks you this exercise?

5. Nov 23, 2014

### pervect

Staff Emeritus
My main point - you need to understand vectors and dual vectors BEFORE you try to tackle tensors. So a review of vectors, aka vector spaces, is advised. I will do my best.

If you have an old linear algebra textbook, reread the section on what a vector space is. I will give an informal description which I hope makes sense, elements of a vector space can be multiplied by scalars, and added together. I have omitted some of the other conditions, like associativity for brevity, a textbook will fill in that gap better than my memory.

Now, partial derivative operators can be multiplied by scalars and added together and meet all the other conditions a vector space is supposed to, so they form a vector space. This is the tangent space.

Your linear algebra textbook will also define the dual space, which is a map from vectors to scalars. It will also prove that this dual space is a vector space, and that it has the same dimensional its as the original space. The other important point is that the dual of a dual vector space is the original vector space. A frequently under mentioned point is that vectors and duals transform differently under a change of basis, hence the distinction. Vectors and duals have the same dimension, but they transform differently. And when you apply duality twice, you get the original space back.

The dual space to the tangent space is the co tangent space. Note also that the dual of the co tangent space is the tangent space. Thus a map from a dual vector to a scalar transforms the same way as a vector does.

So if $\partial_x$ is a vector, a map from $\partial_x$ to a scalar is a dual vector. Notationally, $\partial/\partial x$ gives unity when applied to dx, so that is right.

Note that while dx always has a scalar value, it is only a dual vector when is taken as an operator that is applied to a vector and returns a scalar. So the range of dx is always a scalar, but only when it's domain is an operator on vectors is it a dual vector. This makes the notation very slightly ambiguous, you get used to it. In Matterweves notation, aka abstract index notation, (dx) has no index when it is a scalar, we give it a covariant tensor index when it is a map from a vector to a scalar thus $(dx)_i$.

To appreciate this tensor notation you need to know that a tensor of rank m,n is a map from m dual vectors and n vectors to a scalar. So you need to understand both vectors and dual vectors, before you can understand tensors, it is a prerequisite to understand duality to understand tensors.

As far as FIDO goes, I am drawing a blank. Looking at the online notes I don't see the acronym defined, if you could look up what the initials stand for it would be helpful.

6. Nov 23, 2014

### Breo

Sorry for the low quality. My room has low brightness.

7. Nov 23, 2014

### Matterwave

Ah, a driebein (although, before now I have never heard of such a term) is a triad...in other words a tetrad but in 3-D only. There is no typo in that problem statement. The tangent space is the space of vectors. The Co-tangent space is the space of one forms. They are dual to one another. Because the problem is asking you to first find the one-form basis, and then find the "dual" to those (which are vectors), then it is correct in asking you to find the "dual basis of vectors in the tangent space". Vectors live in the tangent space, not the co-tangent space.

8. Nov 23, 2014

### Breo

Thank you very much pervect! :)

I did not understand this paragraph well, do you know any specific book which could be good to learn that? I do not want to disturb you more while asking you for examples or so.

FIDO means fiducial observer, an observer at rest in a given gravitational field. But I do not know what this exactly means. I can not make a picture in my head. By the other hand, I can imagine a FREFO falling.... but to where? because of the gravity of the given place?...

9. Nov 23, 2014

### Breo

So the last sentence about computing the Hodge is just what we must do in order to get the dual basis of vectors in the tangent space? I thought it was the second pard of the problem xD

So it is just to find the vielbein as always, like:

$$e^1 = r sin \theta d\phi$$ which is a 1-form, and then computing the hodge dual?

Is just that?

10. Nov 23, 2014

### Matterwave

I'm a little confused on why you are working on such an advanced problem when you obviously have some very fundamental issues you should understand first.

But the text is asking you to first compute a "natural" basis of 3 vectors which are ortho-normal, the dual one-form basis for this, and then the Hodge dual of the one-form basis forms...which will be a set of 3 2-forms in 3-D space. I think that is what the question is asking...

11. Nov 23, 2014

### Breo

This is my first year with Gravitation subject and this is an exercise I must to for Monday.

Until today I only calculated dreibeins with the next form: e¹=dt ; e²= Rdθ and e³=rsinθdϕ for spherical coordinates so I am not sure about this exercise. What I used to do is defining the dreibein in natural basis, right? but how can I obtain the dual one-form basis for this?

12. Nov 23, 2014

### Matterwave

The dual basis will be the the one which satisfies the following $(e^\mu)_a (e_\nu)^a = \delta^\mu_{~\nu}$ take into consideration the fact that $(dx^\mu)_a(\partial_\nu)^a=\delta^\mu_{~\nu}$ for a coordinate basis.

Hint: Since you have no terms like $d\theta+d\phi$ in your basis one-forms, it should not be too hard to figure out a dual vector basis.

13. Nov 23, 2014

### Breo

So I must obtain:

$$(e^{\nu})_a = \delta^{\nu}_{\mu} (w^{\mu})_a$$ ?

By the way, how you write the latex formulas in the same line without falling downwards?

EDIT: I replied before you edited... I do not know what to do... I hate to say this but, can you show me a direct example computing one of the dreibeins of the exercise? Do not do it if it disturbs you, please. Don't worry. Im very grateful for all your help today :)

Last edited: Nov 23, 2014
14. Nov 23, 2014

### Matterwave

Use # instead of \$ wraps.

You should be able to do this from inspection. For example, what is $(dr)_a (\partial_r)^a$? Or what is $(r d\theta)_a(\partial_\theta)^a$?

15. Nov 23, 2014

### Breo

1 and r?

16. Nov 23, 2014

### Matterwave

Yes, so how would you modify them to give 1 and 1?

17. Nov 23, 2014

### Breo

Times the inverse. However why we want to get 1?

18. Nov 23, 2014

### Matterwave

Look at the equationin post #12, what happens when $\mu=\nu$?

19. Nov 23, 2014

### Breo

You obtain ones but I do not get what to do with that. I only see trivial restults

20. Nov 23, 2014

### Matterwave

What do you mean? The requirement $(e^\mu)_a(e_\nu)^a=\delta^\mu_{~\nu}$ is not a trivial one, but one that defines what it means that a one form basis is THE dual to a vector basis. Say I have a vector with components $A^\mu$. I can write this vector down like this: $A^a=A^\mu(e_\mu)^a$ (write out the summation if it is confusing to you). Now, if we want to figure out one of its components $A^\nu$, what we can do is act on it with the basis one-forms: $(e^\nu)_a A^a = (e^\nu)_a A^\mu(e_\mu)^a = A^\mu (e^\nu)_a(e_\mu)^a=A^\mu \delta^\nu_{~\mu} = A^\nu$. We could not do this if our basis one forms and basis vectors did not satisfy the condition $(e^\mu)_a(e_\nu)^a=\delta^\mu_{~\nu}$.

21. Nov 23, 2014

### Breo

Is $A^a$ the metric tensor in the exercise?

22. Nov 23, 2014

### Matterwave

No, it is just some arbitrary vector.

23. Nov 23, 2014

### pervect

Staff Emeritus
Try Sean Caroll's online lecture notes, http://xxx.lanl.gov/pdf/gr-qc/9712019v1.pdf , at or around the following section:

The value of either form of "dx" is a scalar. But the first dx is just a scalar and nothing more, the second is something more. It's an operator that has a scalar value, and not just a scalar. You seem to be more comfortable thinking in terms of 'gradient', which is how Sean writes it above, while I prefer to say "one-form".

If you want/need more on vector spaces and their duals, you can try the you-tube video

The concept of "at rest" in a gravitational field only really makes sense when you have a static or stationary metric. The quick-and-dirty way of identifying such a metric is it has some coordinate representation where none of the metric coefficients is a function of time. Then in that coordinate system, "at rest" means "constant coordinates".

24. Nov 23, 2014

### Breo

I think I got it:

$$e^1 = 1 \\ e^2 = R \\ e^3 = Rsin\theta d\phi$$

These are the dreibein in terms of differential forms with the dual basis of the vectors in the tangent space, right?

I had a little doubt about calc $de^1$, $de^2$ and $de^3$...

25. Nov 23, 2014

### Matterwave

You don't calculate $de^1$...$(e^1)_a$ is already a one-form basis, you don't need to try to calculate it's exterior derivative. But I thought you already calculated the $(e^i)_a$ in your previous post? You even wrote down $(e^1)_a = (dr)_a,~(e^2)_a=r(d\theta)_a,~(e^3)_a=r\sin\theta (d\phi)_a$ in your notation?