Clarification of concepts: Tetrads (Vielbeins), etc

[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 brievingAnd 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.

 

[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.

 

[Breo]

Thank you very much for the reply :).

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

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.

That observer is a FIDO?

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.

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

 

[Matterwave]

Thank you very much for the reply :).

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

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.

That observer is a FIDO?

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

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?

 

[pervect]

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.

 

[Breo]

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

 

[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.

 

[Breo]

Thank you very much pervect! :)

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(dx)_i.

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.

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.

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?...

 

[Breo]

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.

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?

 

[Matterwave]

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?

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...

 

[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?

 

[Matterwave]

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?

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.

 

[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. I am very grateful for all your help today :)

 

[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 $(r d\theta)_a(\partial_\theta)^a$?

 

[Breo]

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 $(r d\theta)_a(\partial_\theta)^a$?

1 and r?

 

[Matterwave]

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

 

[Breo]

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

 

[Matterwave]

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

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

 

[Breo]

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

 

[Matterwave]

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

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 $(e^\mu)_a(e_\nu)^a=\delta^\mu_{~\nu}$.

 

[Breo]

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

 

[Matterwave]

No, it is just some arbitrary vector.

 

[pervect]

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.

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

Perhaps this is a good time to [make a] note that most references are not sufficiently picky to
distinguish between “dx”, the informal notion of an infinitesimal displacement, and
“dx”, the rigorous notion of a basis one-form given by the gradient of a coordinate function.

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

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?...

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".

There's a way to talk about this without introducing coordinates, but I'm not sure it'd be helpful to talk about at this point. If you really want to know, ask.

 

[Breo]

No, it is just some arbitrary vector.

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 $de^3$...

 

[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?

 

[Breo]

Yes, but I thought it was wrong as you didnt say me it was ok xD I forgot to wrote it in that notation! I will keep that in mind!

 

[Breo]

Thank you Pervect and Matterwave again! I think I am disentangling my missunderstandings.

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".

There's a way to talk about this without introducing coordinates, but I'm not sure it'd be helpful to talk about at this point. If you really want to know, ask.

Mmm so we can say a FIDO is at rest in a stationary metric while the FREFO is "taking away" by the metric, then, by a proper time? I heard something about FREFO are defined on vielbeins, is this true?

I love gravitation so I really have curiosity about what are you referring to in your last statement :)

 

[Breo]

I would appreciate every taught about these [FREFO and FIDO]. Also, where can I find a clear and/or deeply approach to the subject?Thank you in advance.

 

[PAllen]

Free falling obsersers seems obvious. The way I've seen fiducial observers used is if you have coordinates such that the time coordinate corresponds to a particular class of observers (rather not not being observationally defined at all), then the observers 'defining' the time coordinate are 'fiducial'. In the Schwarzschild solution, these would be static observers at spatial infinity [for the external, vacuum solution, to be more precise].

 

[Breo]

Can you give me an example of when we can "use" frefos and when fidos?

For instance, if I have two particles interacting gravitationally how a FREFO and how FIDO observiers would measure their acceleration magnitude and direction?

 

[PAllen]

Can you give me an example of when we can "use" frefos and when fidos?

For instance, if I have two particles interacting gravitationally how a FREFO and how FIDO observiers would measure their acceleration magnitude and direction?

I don't know what you are getting at. Any coordinates or any observer can be used for any problem. The only difference is mathematical complexity. Is that what you are getting at?

 

[Breo]

Yes, any difference. Until now, I only have seen definitions so I want to see an example in practice to get a the mechanical picture in my mind.

So, how a FIDO, and then how a FREFO, would measure the particle problem? I want to see the differences

 

[Breo]

Is FIDO seated on a stationary metric and FREFO on tetrads?

 

[PAllen]

Is FIDO seated on a stationary metric and FREFO on tetrads?

Yes to the second. Concpetually, I don't see FIDO as limited to a stationary metric. I've seen the term fiducial observer used in cosmological solutions, which are certainly not stationary. FYI, I've never seen your acronyms used.

 

[Breo]

Free falling observer measures proper time while Fiducial observer measures natural time?

 

[Matterwave]

Free falling observer measures proper time while Fiducial observer measures natural time?

What is natural time? All observers measure their own proper time.

 

[PAllen]

Free falling observer measures proper time while Fiducial observer measures natural time?

Per my post #29, fiducial observers define coordinate time for some chosen coordinates. There is nothing more 'natural' about this time than the proper time for any other observers. If you have coordinates based on a class of fiducial observers, coordinate time is proper time for those particular observers. Free fall observers measure their own proper time as well. All observers measure proper time.

I'm having trouble understanding what question you really want answered. Another observation, that may or may not be helpful, is that in GR there is no such thing as global free fall coordinates. Even if you can extend a free fall observers tetrad to cover a substantial section of spacetime via Fermi-Normal coordinates, only the origin observer is a free fall observer (in general). Fiducial observers, on the other hand, are used to build global coordinates that have some desired properties. Thus, you pick a family of observers that capture some symmetry of the manifold (static observers at infinity for SC geometry; comoving observers for a cosmological solution), and build coordinates such that coordinate time is the proper time measured by the fiducial observers. At least, that is the only sense I've seen the term used. If you have particular reference with a different sense of the term, it would be helpful to refer to that source to frame further questions.

[Edit: Let me add a clarification to the above: In some cases you can build global coordinates based on a family free fall observers (effectively using these as your fiducial observers). However, except for flat Minkowski space, you have to give up on orthogonality of your coordinates to achieve this - the metric will have off diagonal terms. ]

 

[Breo]

Coordinate time, that it was what I meant.

Thank you very much for the answers. Yours helped me a lot to understand what FIDO and FREFO are.

 

[Breo]

After thinking a while about proper time. Can you give me an example with equations or not about how a frefo measures his proper time?

 

[PAllen]

After thinking a while about proper time. Can you give me an example with equations or not about how a frefo measures his proper time?

Integral of the line element along the world line of any observer is proper time. Can you clarify your question, as this fact is like the 1+1=2 of relativity, so I am having a hard time believing this is what you are really asking.

 

[Matterwave]

After thinking a while about proper time. Can you give me an example with equations or not about how a frefo measures his proper time?

There's no equations necessary. Any observer measures his own proper time. That is the definition of proper time.

EDIT: Looks like PAllen is quicker to the draw than I.

 

[Breo]

Yes, sure for a fiducial we take the time from integrating. But for a free falling observer... what formula gives the proper time.

Imagine you have $e^0 = Adt -Bdx^i \\ e^i= Cdx^i$

how to measure the proper time? in terms of the tetrad maybe?

 

[PAllen]

Yes, sure for a fiducial we take the time from integrating. But for a free falling observer... what formula gives the proper time.

Imagine you have $e^0 = Adt -Bdx^i \\ e^i= Cdx^i$

how to measure the proper time? in terms of the tetrad maybe?

I'm glad you clarified, because that gives a more interesting question.

First, note that a coordinate chart can represent all observer world lines that pass through the chart's coverage, and they all give the same proper time for any observer (pretty much by construction of how the metric transforms between charts). So, what I said before is trivially true: represent a free fall world line in any coordinate chart and you compute their proper time by integration; same as any observer. Further, computing which are the free fall observers in any coordinates is (conceptually - can be messy in practice) straightforward: solve the geodesic equation.

However, it seems you are asking how it is done in tetrad formalism. I admit I've never used tetrad formalism for this purpose, and am not very familiar with it, in general (I use coordinates or pure geometric formalisms; never studied tetrads much). I've only looked at tetrad formalism for local interaction, not for any integrated quantity like proper time. Hopefully someone else can explain how this might be done.

 

[Matterwave]

Yes, sure for a fiducial we take the time from integrating. But for a free falling observer... what formula gives the proper time.

Imagine you have $e^0 = Adt -Bdx^i \\ e^i= Cdx^i$

how to measure the proper time? in terms of the tetrad maybe?

For ANY observer, as PAllen pointed out, you get the proper time by integrating along its world line. This is not just for "fiducial observers". As I stated in my last post, the free-fall observer, or any observer for that matter, does not need to use ANY equations to figure out his own proper time, he just needs to look at his watch.

I think you have to be more specific with your question.