# Local Lorentz Frames

1. May 25, 2013

### WannabeNewton

Hey guys. I'm having doubts about something in a GR text I read (Gravitation - T.Padmanabhan). In general, for Lorentzian 4-manifolds, we call a basis of smooth vector fields $(e_{\mu})^{a}$ (that is, a set of smooth vector fields that define a basis for the tangent space at each event of space-time) a tetrad if $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ i.e. $(e_{0})^{a}(e_{i})_{a} = 0,(e_{0})^{a}(e_0)_{a} = -1, (e_{i})^{a}(e_{j})_{a} = \delta_{ij}$ so the basis of smooth vector fields are also orthonormal. In Padmanabhan's text, when defining Fermi-Walker transport, he says (I tweaked the notation a bit) "Consider an observer who is moving along some worldline (which is not necessarily a geodesic) with the 4-velocity $u^{a}$. Let $(e_{\mu})^{a}$ be a tetrad of basis vectors transported along the trajectory of this observer...The first condition we will impose is that the basis should satisfy the standard relations, $(e_{\mu})^{a}(e_{\nu})_{a} = g_{\mu\nu}$."

This seems to go against the very definition of a tetrad (as an orthonormal basis for the tangent space at each point on the observer's worldline) because the way it is defined in this text, the basis vectors need not be orthonormal. What gives? He later says that Fermi-Walker transport ensures that the tetrad remains orthonormal from the initial erected tetrad (which is of course true in a general context) but he doesn't even define it to be orthonormal to start with so what's up with that?!

My next question involves the use of the term "local Lorentz frame / local inertial frame" in GR. Here is the issue: consider an observer moving along an arbitrary worldline and let $p$ be an event on the observer's worldline. Now by a procedure much like the Gram-Schmidt scheme, we can always find a basis $(e_{\mu})^{a}|_p$ for $T_p M$ such that $g_{\mu\nu}|_p = \eta_{\mu\nu}$ with respect to this basis i.e. we can always find a point-wise tetrad $(e_{\mu})^{a}|_p$ for $T_p M$. Now I've seen some sources (including Padmanabhan) call this point-wise tetrad a local Lorentz frame / locally inertial reference frame for the observer at $p$ which I find incorrect because from what I've learned, the concept of a locally inertial reference frame at $p$ requires that the frame be derived from a coordinate system $\{x^{\mu}\}$, which we call locally inertial coordinates, that the observer must setup in a neighborhood $U$ of $p$ (and when I say the frame is derived from $\{x^{\mu}\}$ I mean $(e_{\mu})^{a}|_p = (\partial_{\mu})^{a}|_p$) such that $\partial_{\alpha}g_{\mu\nu}|_p = 0$ (this further ensures that the metric tensor is approximately the Minkowski metric for any $q\in U$ to first order).

This physically ensures that the locally inertial reference frame at $p$ is that of a freely falling observer (because it is precisely the freely falling observers who are locally inertial) i.e. $u^{a}\nabla_{a}u^{b}|_p = (e_{0})^{a}\nabla_{a}(e_{0})^{b}|_p = (\partial_{0})^{a}\nabla_{a}(\partial_{0})^{b}|_p = \Gamma^{b}_{ac}|_p (\partial_{0})^{c}(\partial_{0})^{a}|_p = 0$
where I have used the fact that in any reference frame of an observer at an event $p$ on his/her worldline, the 4-velocity $u^{a}|_p = (e_0)^a|_p$.

So you can see why I am confused when Padmanabhan, and others, call the tetrad $(e_{\mu})^{a}|_p$ at an event $p$ on an arbitrary observer's worldline (i.e. a point-wise tetrad which is not necessarily derived from a locally inertial coordinate system about $p$) a local Lorentz frame / locally inertial frame since no restriction is placed on the observer being in free fall. What am I missing here? Thanks in advance for any help.

Last edited: May 25, 2013
2. May 26, 2013

### ForMyThunder

It's most likely a typo: should be eta and not g. This, $g(e_\mu,e_\nu)=(e_\mu)^a(e_\nu)_a=g_{\mu\nu}$, is the definition of the components of $g$, the $g_{\mu\nu}$.

The book "Gravitation" by Misner, Thorne and Wheeler gives the definition for local Lorentz frame as you have described; with the Christoffel symbols of the coordinate system at the point zero. I don't have a copy of Padmanabhan's text, so I am unable to comment further.

3. May 26, 2013

### WannabeNewton

Yeah I was hoping it was just a typo and not an error in my understanding. Thanks for clearing that up. So the "standard relations" amongst the tetrad alluded to by Padmanabhan's text should be $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ (which is, for example, how Wald defines a tetrad) and not $g_{\mu\nu}$ correct?

As for the local Lorentz frame part, look at the very first page of this: http://web.mit.edu/edbert/GR/gr2.pdf
They say that "Orthonormal bases correspond to locally inertial frames" which is definitely not as strong as the definition of a locally inertial frame in Carrol, MTW etc. and physically it makes no sense to simply claim any observer who erects an orthonormal frame (tetrad) at each event on his/her worldline is locally inertial but according to that document, such a frame is locally inertial which would make the observer a freely falling one thus giving rise to the issues I described in the OP. Thanks again for the response Thunder.

4. May 26, 2013

### ForMyThunder

Yes.

What I get from reading the document, and I may be wrong in this, is that it is saying "given any point $p\in M$ and any orthonormal basis $e_\mu\in T_p M$, there is an inertial coordinate system (that is, such that the Christoffel symbols are 0) such that the coordinate frame coincides with the orthonormal basis at the point $p$". This is done using normal coordinates http://en.wikipedia.org/wiki/Normal_coordinates.

5. May 26, 2013

### WannabeNewton

Yes geodesic normal coordinates are how I learnt about the construction of locally inertial coordinate systems and the associated locally inertial reference frames as well but in the document I linked, in the very first paragraph where I drew the quote from post #3, I did not see anything about the requirement that a local inertial frame at an event be derived from a locally inertial coordinate system about that event, which is what I was criticizing. I'm wondering if it links you to a different document? The section is titled "Orthonormal Bases, Tetrads, and Commutators".

Regardless, are we in agreement that simply specifying a tetrad at a given event on an observer's worldline is not enough to claim that the tetrad corresponds to a locally inertial frame, and that we further require that the tetrad be derived from a locally inertial coordinate system in order to claim that it does indeed correspond to a locally inertial frame at the given event? This way only freely falling observers may erect locally inertial frames, in accordance with the equivalence principle, and not any arbitrary observer.

6. May 26, 2013

### ForMyThunder

I believe the term "correspond" is causing the confusion. To me, this just means that there is some sort of mapping from one set to the other, in this case from the set of orthonormal bases at p to the set of locally inertial reference frames. It doesn't exactly give a definition; in his first set of notes, he says "there exist locally inertial reference frames, corresponding to locally ﬂat coordinates carried by freely
falling observers". He never actually defines them rigorously, just in a way to give the reader an intuitive idea about them (which I hate, being a mathematician at heart).

Does this at all answer your question? I have a very bad habit of going off on tangents.

7. May 26, 2013

### WannabeNewton

I would say so yes, thanks; I absolutely hate the non-rigorous accounts as much as you do and this is why I try to stick to Wald as much as possible and rely heavily on it as my primary source. But to sum it up, how do you stand with regards to my last statement:

8. May 26, 2013

### Bill_K

No, absolutely not! The relationship is correct as given.

The quantities that specify a tetrad eμa may be regarded in several ways. They are an example of hybrid quantities, quantities that have indices of two different types. For fixed a, eμa represents a contravariant vector in the usual sense. Just as the gμν may be used to raise and lower tensor indices, ηab may be used to raise and lower the internal Lorentz indices, Va = ηabVb.

And eμa may be used to convert indices from one type to the other. We call this latter process strangulation. Given a vector Vμ, its strangled components in the Lorentz basis are Va = eμaVμ. Conversely, the vector may be unstrangled, Vμ = eμaVa. Consistency is guaranteed by the orthonormality relation eμaeμb = δab, and completeness relation, eμaeνa = δμν .

gμν, ηab and eμa may all be regarded as different components of the same object. In this sense, eμa therefore is just a different way of expressing the metric information, and its differentiation leads to curvature information, the Riemann tensor. In this process in place of the Christoffel symbols we use instead the Ricci rotation coefficients, e[μ,ν]a.

Last edited: May 26, 2013
9. May 26, 2013

### George Jones

Staff Emeritus
Given any event $p$ and any unit timelike element $\bf{u}$ of $T_p$, there always exists a timelike geodesic through $p$ that, at $p$, has tangent $\bf{u}$.

Last edited: May 26, 2013
10. May 26, 2013

### WannabeNewton

Bill I'm confused because Wald and Padmanabhan give the exact same equation just Wald has the Minkowski metric on the right and the other has the general metric tensor on the right. Why is it ok to say $(e_{\mu})^{a}(e_{\nu})_{a} = g_{\mu\nu}$ and still claim the tetrad is that of an orthonormal frame (here $\mu$ labels the basis vectors and $a$ labels the abstract index of a single basis vector).

George, yes that is true but Padmanabhan makes explicit reference to an arbitrary accelerated observer in space-time and talks about local Lorentz frames at events on this observer's worldline, which I cannot agree with (page 183 second paragraph since you have the book) since an arbitrary observer cannot set up locally inertial coordinates at events on his/her worldline and this obstructs him/her from erecting a locally inertial frame / local Lorentz frame.

Thanks for the responses guys.

11. May 26, 2013

### Ben Niehoff

No. A frame is a collection of vectors at a point. A "tetrad" field (I hate that terminology...far too 4d-centric; I call them orthonormal frame fields) is a collection of vector fields that are orthonormal at every point. But there is no requirement that each vector field be integrable (i.e., that it can be written $\partial / \partial x^\mu$ in some coordinates).

Even so, at each point, the vector fields comprise an orthonormal frame, which is the same thing as a local Lorentz frame. The vectors at $p$ can always be written, at $p$ only, as the coordinate vector fields of some geodesic normal coordinates. But this does not hold in a neighborhood of $p$. And that's fine.

Anyway, I see this mistake a lot in this forum, so I urge you to learn it correctly the first time: A frame is a set of basis vectors at a point. A frame field is a collection of vector fields such that they form a basis at every point. Frame fields can be orthonormal or otherwise. A coordinate system is not a frame; a coordinate system is a way to label points by mapping to $R^n$. But a coordinate system does allow you to define a coordinate frame, which is the collection of partial derivatives $\partial / \partial x^\mu$. Other kinds of frames may or may not be coordinate frames, depending.

12. May 26, 2013

### WannabeNewton

I didn't say the coordinate system was the frame though, I said the local Lorentz / locally inertial frame is derived from the locally inertial coordinate system (i.e. the basis vectors constituting the frame at a given event coincide with the coordinate vector fields of the chart at said event). This is, for example, how Carroll defines it on page 74 at the top. And when I said tetrad, I meant the tetrad evaluated at the event in question, so sorry for not clarifying that.

Can I say local Lorentz frames the same as instantaneously co-moving locally inertial frames? I.e. a locally inertial observer at that event co-moving with the arbitrary observer at that event?

EDIT: Thanks for the reply Ben!

Last edited: May 26, 2013
13. May 26, 2013

### George Jones

Staff Emeritus
Yes, but the tetrad for an accelerated observer at p is also the tetrad for a an inertial observer at p This is all that Padmanabhan means, and he is not incorrect. At a different q on the accelerated observer's worldline, there will a different comoving inertial observer.

14. May 26, 2013

### WannabeNewton

Ok thanks this clears it up. So we are basically taking a co-moving locally inertial observer at that event and taking his local Lorentz frame and passing it on to the original accelerated observer at that specific event? So this is basically the same thing as instantaneously co-moving inertial references frames for accelerated observers in SR-sort of like instantaneously co-moving locally inertial observers :p?

15. May 26, 2013

### George Jones

Staff Emeritus
Yes.

16. May 26, 2013

### WannabeNewton

So this is what he (Padmanabhan) means when he says the local Lorentz frames are different at each event. Cool, thanks. And when he says "treating Lorentz transformations as..." is he talking about the infinitesimal Lorentz transformation to go from one co-moving locally inertial observer to the very next one at the following event with the boost being the velocity of the very next co-moving observer, at the following event, with respect to the first co-moving observer to whom the Lorentz transformation is being applied?

17. May 26, 2013

### George Jones

Staff Emeritus
By definition, the components of the metric $g$ with respect to any basis $\left\{e_\mu\right\}$ (which may or may not be orthonormal) are $g_{\mu\nu} = g\left( e_\mu ,e\nu \right)$.

18. May 26, 2013

### Bill_K

I think you've got the orthonormality condition and the completeness condition reversed. To get the norm of a vector you must contract on μ, whereas you're contracting on a. Orthonormality says that eμa eμb = δab.

Also, eμa eνa = ημν is not even a covariant statement. There is no such thing as ημν, unless you're restricting the discussion to Minkowski coordinates, which we are not!

19. May 26, 2013

### George Jones

Staff Emeritus
I will have to look at at what Padmanabhan says, but, right now, I am on my way out to brunch with my family, then grocery shopping, then...

20. May 26, 2013

### WannabeNewton

You are certainly correct George but perhaps if I give the reference in Wald, it might direct you better to where my confusions are coming from. On page 49 section 3.4b, titled "Orthonormal Basis (Tetrad) Methods", he says "Thus, we may wish to introduce a 'nonholonomic' i.e. noncoordinate, orthonormal basis of smooth vector fields $(e_{\mu})^{a}$ satisfying $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$..." but you are certainly correct that $g(e_{\mu},e_{\nu}) = g_{\mu\nu}$. How should I differentiate between the two different expressions given?

EDIT: Ah ok, well thanks for all the help! Have fun at brunch, I look forward to hearing from ya.

EDIT 2: Sorry Bill, I literally just saw your post. I gave the reference in Wald where I got that from so maybe you could take a look and see if/where I'm interpreting it incorrectly? I'll read the rest of your post in the meantime, thanks! Carroll gives a similar thing on page 112 expression 3.66 (he calls it the 'canonical form').

Last edited: May 26, 2013
21. May 26, 2013

### Fredrik

Staff Emeritus
For any set $\{e_\mu\}_{\mu=0}^3$ of vector fields, and any coordinate system x defined on a a relevant subset of spacetime, we have
$$g(e_\mu,e_\nu) =g\left((e_\mu)^\rho\partial_\rho,(e_\nu)^\sigma\partial_\sigma\right) =g_{\rho\sigma}(e_\mu)^\rho(e_\nu)^\sigma =\left(g\otimes e_\mu\otimes e_\nu\right)\left(\partial_\rho,\partial_\sigma,dx^\rho,dx^\sigma\right)$$ In abstract index notation, the right-hand side is written as
$$g_{ab}(e_\mu)^a(e_\nu)^b = (e_\mu)^a (e_\nu)_a.$$ So if the $e_\mu$ are the basis vectors associated with a coordinate systems, or if we have simply defined the notation $g_{\mu\nu}$ by $g_{\mu\nu}=g(e_\mu,e_\nu)$, then the condition $(e_\mu)^a (e_\nu)_a=g_{\mu\nu}$ doesn't tell us anything at all about the $e_\mu$. It's just a condition satisfied by any four vector fields.

The condition $(e_\mu)^a (e_\nu)_a=\eta_{\mu\nu}$ on the other hand, is the definition of what it means for $\{e_\mu\}_{\mu=0}^3$ to be an orthonormal set. The condition can also be written as $g(e_\mu,e_\nu)=\eta_{\mu\nu}$, since we always have $(e_\mu)^a (e_\nu)_a=g(e_\mu,e_\nu)$.

Edit: The proof of $g(e_\mu,e_\nu)=(e_\mu)^a (e_\nu)_a$ above would have been clearer if I had done this:
$$g(e_\mu,e_\nu) =g\left((e_\mu)^\rho\partial_\rho,(e_\nu)^\sigma\partial_\sigma\right) =g_{\rho\sigma}(e_\mu)^\rho(e_\nu)^\sigma =(e_\mu)^\rho(e_\nu)_\rho=(e_\mu)^a (e_\nu)_a.$$ The last step follows from the calculation I did in my explanation of abstract index notation below.

Last edited: May 26, 2013
22. May 26, 2013

### dextercioby

It's confusing for me: the Greek indices are world indices and the Latin ones Minkowskian ?

23. May 26, 2013

### Bill_K

OMG Fredrik, this is absolutely false!!

Orthonormality is eμ0 eμ0 = 1, eμ0eμ1 = 0, and so on, ten equations. Do you see, I hope, that each equation is summed on μ?? These are ten vector dot products, each one of them equal to either 0 or 1.

The condition you're confusing this with is eμ0 eν0 + eμ1 eν1 + eμ2 eν2 + eμ3 eν3 = ημν (whatever ημν is supposed to mean in a general coordinate system!) Do you see that this set of equations is summed on the Lorentz index a and is totally different from the other one??

Last edited: May 26, 2013
24. May 26, 2013

### Fredrik

Staff Emeritus
I don't know what that means, but the convention used by Wald is to e.g. write a tensor of type (1,2) as $T^a{}_{bc}$. The latin indices just tell us what type of tensor we're dealing with. The components of this tensor in a coordinate system would be written as $T^\mu{}_{\nu\rho}$.

Also, when you see something like $\omega_a X^a$, it should be interpreted as what would be written as
$$(\omega\otimes X)(\partial_\mu,dx^\mu)=\omega(\partial_\mu)X(dx^\mu) =\omega(\partial_\mu)dx^\mu(X) =\omega(\partial_\mu)X(x^\mu) =\omega_\mu X^\mu$$ in the index-free notation. So we have $\omega_a X^a=\omega_\mu X^\mu$, but it takes a small calculation to see it. (Note that I used the notation X both for a tangent vector and the corresponding "dual dual vector").

Last edited: May 26, 2013
25. May 26, 2013

### WannabeNewton

As per Wald's notation, the Greek indices label which basis vector we are talking about and the Latin ones denote each basis vector using the abstract index notation. So $(e_0)^{a}$ is the time-like basis vector etc. and he says the orthonormal basis of smooth vector fields $(e_{\mu})^{a}$ satisfy $(e_{\mu})^{a}(e_{\nu})_{a} = \eta_{\mu\nu}$ which I interpreted as saying $(e_{0})^{a}(e_{0})_{a} = -1, (e_{i})^{a}(e_{j}) _{a} = \delta_{ij}, (e_{0})^{a}(e_{i})_{a} = 0$ where $i,j=1,2,3$. I thought this was the orthonormality condition but I may be wrong. I mean this is basically just the statement that $g_{ab}(e_{\mu})^{a}(e_{\nu})^{b} = \eta_{\mu\nu}$ which is how I thought tetrads were always defined (so that the associated orthonormal frame at each event can be physically thought of as a clock for the "time axis" and three meter sticks held along mutually orthogonal "spatial axes"). This is also how Carroll defines it in appendix J, expression J.1. which is also what Fredrik wrote (hi Fredrik, how's it goin pal xP)