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Fundamental question on GR: Basis vector meaning

  1. Aug 23, 2011 #1
    This is a very basic question in understanding General Relativity, but the answer still eludes me.

    The simples way I can state it is: "what exactly represent the 4 basis vector at a certain point on the manifold?"

    But let me explain myself.
    Let's take a 4 dimensional Minkowsky space.
    In this context the 4 basis vectors are of simple interpretation: 3 of them represent the "direction" of the "rulers" an observer use to measure thing and one indicates the time "direction" (ok, this is not 100% correct, but you got the juice). Now any Lorentz transformation mantain the metric in the minkowsky form, and this is ok. I can picture the basis vector of an observer who moves with a certain speed with respect to me and by going in his referment frame I can picture my "old" basis vector: this is a long way to say I have changed basis.
    To make this the clearest possible way imagine that a minkowsky space with basis vectors [itex]\vec{e}_t\,,\vec{e}_x\,,\vec{e}_y\,,\vec{e}_z[/itex];and write down the basis vector for an observer who experiment a boost in the x direction.
    His basis vector will be [itex]\vec{e}'_t=a\vec{e}_t+b\vec{e}_x[/itex] and [itex]\vec{e}'_x=c\vec{e}_t+d\vec{e}_x[/itex], and [itex]\vec{e}'_y=\vec{e}_y\,,\vec{e}'_z=\vec{e}_z[/itex] and so i can say at which pace his clock goes ecc.
    The point is that even in this example the basis vector have a precise meaning.

    Now, the question!

    "In a curved spacetime what is the meaning of the basis vector at a certain point?"
    Do they still indicate the direction of the observer's rulers and the pace of his clock?

    Now take any curved manifold.
    Here begins the confusion, becaouse when they say that a certain change of basis will put locally the metric in the minkowsky form to me it's like they say that if i want to see the spacetime locally flat I have to set my rulers in a certain direction and set my clock to a certain pace.
    Or, put in another way, think of a certain basis vector at a point. Now, change basis to make the metric locally flat. Now you have two basis vector at that point, one corresponding to the first observer (suppose that's me) which doesn't see spacetime flat and another observer (let's call him Adam) whic instead does.

    We both are at the same point in spacetime, but Adam moves respect to me with a certain (constant, since we are looking at things in infinitesimal interval) speed, with a certain pace set on his clock.
    Now, is this that allow him to see spacetime locally flat? i mean, the only fact that he moves with a certain speed and/or sets his rulers in a particular direction is enough to make the local flatness manifests itself?

    This seems strange to me. I don't think it is really like that!!
    So I am asking myself: does the basis vector at a certain point mean exactly what I have in mind??
    Or have I misunderstood the concept of basis transformation? So that they do not mean what I've wrote (this is a restatement of the question above!)

    Thanks a lot for the attention fellas!!!
    Last edited: Aug 23, 2011
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  3. Aug 23, 2011 #2


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    I lost you here. You aren't making a change of basis to get to the minkowski metric locally. In general, you can't find a coordinate transformation that puts the metric on a curved space in minkowski form. The metric reducing to the minkowski metric locally is a statement of how an open ball in the neighborhood of a point P will be isomorphic to minkowski space - time. Could you explain what you mean?
  4. Aug 23, 2011 #3
    I lost YOU here :tongue2: !!

    If i'm not mistaking it is a basic concept of GR that a change of basis will reduce the metric to the minkowsky form at first order at a certain point on the manifold, and in that point spacetime IS flat in an infinitesimal neighbour. (It think it's called euqivalence principle, but i could easily be mistaking on the name).

    This is a consequence of what you say, that for each point on a smoot manifold there is a tangent plane.
  5. Aug 23, 2011 #4


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    Yes, you are right; the important part is that the coordinate system must be found at the point P on the n - manifold. That's why I was confused before because you said a change of basis will make the metric locally flat and I couldn't decide if you were somehow talking about a global transformation or one at P. Thanks for setting me straight =D.

    EDIT: Now that I reread that paragraph I see that you had been pretty unambiguous as to what you meant by change of basis. My bad completely. Its like the statement: "change of basis to put the metric in flat form" rings some kind of alarm that blocks the ability to read haha.
    Last edited: Aug 23, 2011
  6. Aug 23, 2011 #5


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    Now this I don't get: if they are in the same neighborhood of P then what is stopping both of them from finding coordinate transformations at P such that [itex]g_{\mu \nu } = \eta _{\mu \nu } + O[(x^{\alpha })^{2}][/itex]?
  7. Aug 23, 2011 #6
    Well, all of the vectors you think of in GR lie in the tagent planes to the point they're attached at.
    But this doesn't mean that the spacetime will look flat to all observers (always at first order, i may omit this from now on :smile:); you need the first derivatives of the metric to vanish, and that restricts your choice.

    I guess that when you have find such a coordinate system performing a Lorentz trasformation will keep the metric locally minkowsky; but i'm not sure so i forward the question to someone who knows it better.

    That said, there is nothing stopping nobody: i just want to know if it's only the particular direction of the rulers of Adam and the pace on his clock that allow him to see the spacetime flat (to first order); while I, standing in his exact same point in spacetime but with a different coordinate system and so different basis vectors (BUT always lying in the tangent space at the point), won't.
  8. Aug 24, 2011 #7


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    Let's back up a bit. In flat space, we can chose an arbitrary set of basis vectors [itex]\vec{e}_a\,,\vec{e}_b\,,\vec{e}_c\,,\vec{e}_d[/itex], but to make our life easier, what we generally do is chose a set of orthonormal basis vectors.

    Your notation for [itex]\vec{e}_t\,,\vec{e}_x\,,\vec{e}_y\,,\vec{e}_z[/itex] and your post seems to suggest to me that somehow you are thinking that any set of basis vectors must be orthonormal. This is not the case.

    To make life much easier for ourself, we usually introduce a set of basis vectors [itex]\vec{e}_t\,,\vec{e}_x\,,\vec{e}_y\,,\vec{e}_z[/itex] with the properties

    [itex]\vec{e}_t\ \, \cdot \, \vec{e}_x = 0, \vec{e}_t\ \, \cdot \, \vec{e}_y=0, \vec{e}_t\ \, \cdot \, \vec{e}_z = 0[/itex]
    [itex] \vec{e}_x\ \, \cdot \, \vec{e}_y = 0, \vec{e}_x\ \, \cdot \, \vec{e}_z = 0, \vec{e}_y\ \, \cdot \, \vec{e}_z = 0[/itex], i.e. the basis vectors are orthogonal.

    In our flat space-time, if we have orthogonal basis vectors, the metric tensor will be diagonal. If we don'[t have orthogonal basis vectors, the metric tensor will not be diagonal. But the space-time will still be flat! It's just that we chose non-orthogonal basis vectors.

    The normal part of orthonormal just means that [itex]\vec{e}_t\ \, \cdot \, \vec{e}_t = -1, \vec{e}_x\ \, \cdot \, \vec{e}_x = 1, \vec{e}_y\ \, \cdot \, \vec{e}_y = 1, \vec{e}_z\ \, \cdot \, \vec{e}_z = 1[/itex].

    (Note: You might have different sign conventions, in general though you have the dot products being + or - 1, it looked ugly when I wrote it that way though).

    With orthonormal basis vectors the metric tensor for our flat space-time is not only diagonal, but the absolute value of the coefficients is unity.
  9. Aug 24, 2011 #8
    You're absolutely right pervect, in fact the flatness/curveness of a manifold cannot be detected with the metric alone, is't requested for the Riemann tensor to vanish.

    But you made me think.

    Now, i know that i can use whichever basis i want (if I choose it to be orthonormal i will have the minkowsky form); and so, for example, in flat space i can decide to use a polar basis (which is still orthogonal but not orthonormal) or in general any maximal set of linearly independent vector.

    So i am pushed to say that in flat space the observer who uses polar basis, and therefore doesn't see the metric as minkowskian, has nothing less to an observer which instead does.

    Appliyng this concept on a curved manifold, I'd say that the observer A who sees the space as locally flat (so the metric for him is minkowsky at first order) doesen't have anything more than an observer B which instead has any other kind of metric; it's like A has a particular vierbein which (can I say accidentally?) allow him to see space (infinitesimally) around him flat.
    The observer A doesn't have to have particular speed, pace, ecc..; he only has a "lucky" combination of vectors which make the metric flat at first order.

    I instead thought that the basis vector that allowed him to see spacetime (locally) flat had a particular physical meaning.

    Let me make an explicit example (please!).
    Take a 3 dimensional manifold - 1 time dimension and 2 spatial dimesion.
    For semplicity take the metric to be [tex]g_{\mu\nu}=\left(\begin{array}{ccc}-1&0&0\\
    0&\gamma_{ij}&\\ 0&&\end{array} \right)[/tex]so that the timelike basis vector is orthonormal to all spaceike basis vector, which form the purely spatial metric [itex]\gamma_{ij}[/itex] on the spatial hypersurfaces.

    Now take the spatial hypersurfaces to be spherical, with unit radius, and choose polar coordinates [itex](\theta,\phi)[/itex], so that the metric is[tex]g_{\mu\nu}=\left(\begin{array}{ccc}-1&0&0\\
    0&1&0\\ 0&0&sin^2\theta\end{array} \right)[/tex]

    Now spacetime is certainly curved, and the observer i choose to write down the metric with doesen't see flat spacetime locally flat except at the point(s) [itex]\mathcal{P}[/itex] on the manifold given by [itex]\mathcal{P}=(t,\theta=0,\phi)[/itex] (here phi is left free, so actually this works all along a curve, but it's only a particular case:smile:) where the metric becomes minkowsky and all of the first derivatise vanish (ONLY at [itex]\mathcal{P}[/itex]).

    Now, does the kinematical/dynamical status of this observer actually mean something physical or is the fact that i can found an observer who sees spacetime locally (as in this case) flat it a mathematical feature that is only used to simplify the calulations?

    To complete the example take a point [itex]\mathcal{R}=(t,\hat{\theta}, \phi)[/itex] with a certain fixed theta.
    Now, if I change the coordinate system to the new primed set [itex](t',\theta ',\phi ')[/itex]
    related to the old by the transformations
    t&= &t'\\
    \theta &=&\theta'\\
    \phi&= &\frac{\phi'}{sin\hat\theta}

    and i calculate the metric with the basis vectors that these transformation induce i should get

    0&1&\\ 0&0&\frac{sin^2\theta}{sin^2\hat\theta}

    now, at the point [itex]\mathcal{R}[/itex], where [itex]\theta=\hat\theta[/itex] the metric, again, is in the minkowsky form, and its first derivatice vanish as well: the observer who uses these coordinates at [itex]\mathcal{R}[/itex] will see the spacetime as locally flat.

    But then again, are some particular physical proprieties that allow him to see the spacetime as flat (at first order)??
    Or, instead, the transformation I've made don't implies particular physical meanings but it is only a useful mathematical trick that semplifies my calculations??

    I hope I explained myself!!!

    Thanks for the attention!!!!
    Last edited: Aug 24, 2011
  10. Aug 24, 2011 #9


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    (Sorry about the other post)
    So is your overall question basically: is the equivalence principle the physical reason for why [itex]\partial _{\alpha }g_{\mu \nu } = 0[/itex], to 1st order, in the neighborhood of a point P?
  11. Aug 24, 2011 #10
    My question is: Does the observer who sees the spacetime locally flat in his infinitesimal neighbour have some particular PHYSICAL proprieties that allow him to see spacetime flat
    is the fact that I can always find a transformation which makes the metric minkowsky to first order simply a MATHEMATICAL trick that is used only to simplify the calculations?

    Thanks for the patience guys!!!!

    Last edited: Aug 24, 2011
  12. Aug 24, 2011 #11


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    What I'm saying is that your ability to find a coordinate transformation such that the metric reduces to the minkowski metric in the neighborhood of that point is, in terms of the physics, a consequence of the equivalence principle. Within that open ball, one cannot differentiate between the gravitational field and acceleration so one cannot identify the tidal forces that are indicative of curvature.
  13. Aug 24, 2011 #12
    You don't seem to be correctly understanding my question!!!
    I think pervect has..

    Ps: the manifold i'm considering is DEFINITELY not an open ball!!

    thanks a lot anyway for your generosity and disponibility, you're great!
  14. Aug 24, 2011 #13


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    I don't think you are understanding me. Pick a point P on the 4 - manifold. The open ball in the neighborhood of P is isomorphic to [itex]\mathbb{R}^{4}[/itex]. From this you can say that [itex]\partial _{\alpha }g_{\mu \nu } = 0[/itex] in this neighborhood. The physical analogue of this is that the observer at P cannot tell the difference between the gravitational field and acceleration. Without any way of detecting tidal forces, he says that [itex]\frac{D^{2}\xi ^{\alpha }}{D\tau ^{2}} = 0[/itex] and from this he concludes that [itex]R^{\alpha }_{\beta \mu \nu } = 0[/itex]. This is all, of course, to first order and is all a consequence of the equivalence principle.
  15. Aug 24, 2011 #14


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    I'm still having a hard time figuring out Tedd's question.

    The problem is that "locally flat" seems to mean something specific to Tedd, but I'm not quite sure what it is, it seems to be different from the way I understand it (which is based on MTW's presentation of the term).

    Specifically, going to the example:

    What's special about these particular points that makes spacetime "locally flat" there?

    The only thing I can see is that the metric is digaonal with the diagonal coefficients being (-1, 1, 1 ) there. Is that what you mean by "locally flat", Tedd?
  16. Aug 25, 2011 #15
    First of all I apologize for my poor explaination!

    To wannabenewton:
    ah, i see now what open ball you were referring to!!
    But still that's not my question. Ok, a locally inertial observer will see geodesics as straight lines but that's not my question!
    Maybe i will be more clear in the answet to pervect

    To pervect:
    with locally flat at a point P I mean that if [itex]g_{\mu\nu}[/itex] is the metric of the observer with some certain basis vector at P you have [tex]g_{\mu\nu}(P)=\eta_{\mu\nu}\,\,\,,\partial_\alpha g_{\mu\nu}(P)=0\,\,\,,\partial_\beta \partial_\alpha g_{\mu\nu}(P)\neq 0[/tex]
    where [itex]\eta_{\mu\nu}[/itex] is the minkowsky metric.

    Now in my example in those particular point, with the basis chosen, the metric is locally flat, with the definition of local flatness given above.

    if you take instead a generic point on the sphere [itex]P=(t,\hat\theta,\phi)[/itex] to "make" the metric locally flat in that point you have to change basis vectors with the transformation i wrote in the example (which is merely a renormalization).

    So the points themselves doesn't have anything in particular, I only choose the basis vector at those points that makes the space looks flat at first order (always with the definition above)

    PS: looking back at my example i see now a mistake: when i say that

    I'm actually meaning at the point [itex]\mathcal{P}=(t,\theta=\frac{\pi}{2},\phi)[/itex] !

    Distraction misktake...
  17. Aug 25, 2011 #16


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    I think your definition of local flatness is what's confusing you. Consider just the first part of your definition.


    By this definition, a perfectly flat plane may or may not meet your requirements for "locally flat" depending on your choice of basis vectors. Which is very awkward at the least. Where did this definition come from?

    This goes back to what I was saying earlier, but maybe a diagram will help

    Looking at the attached figure, if you chose one particular set of coordinates (x,y) you'll have orthonormal basis vectors [itex]{e}_x[/itex] and [itex]{e}_y[/itex], that is the familiar situation on the left.


    You can write formally (I'm not sure if you've seen the notation
    [tex]e_x = \frac{\partial}{\partial x}\,\,e_y = \frac{\partial}{\partial y}[/tex]
    given your coordinates x,y, this is called a "coordinate basis" when your basis vectors are the partial derivatives of your coordinates.

    I'm not sure if you've been introduced to the notion of basis vectors as partial derivatives, some textbooks take that course of action others don't.

    Pictorially, you can see how [itex]e_x[/itex] represents a notion of "an increment in the x direction" and [itex]e_y[/itex] represents "an increment in the y direction", I hope.

    I can explain in more depth why we the basis vectors are [itex]\frac{\partial}{\partial x}[/itex] rather than , say, dx, if you need to know, but I don't want to digress from the main point too much, so I'll hold off unless you ask.

    Now, how about the diagram on the right? Suppose we make the coordinate substitution:

    X = x, Y = x+y

    If you have Penrose's "road to reality", you can check out his diagram on pg 190, if not, it's no big deal, it was the inspiration for this.

    Then this is a new coordinate system, and the vectors [itex]e_{X}\,\,e_{Y}[/itex] are the coordinate basis for these new coordinates.

    One vector represents holding the X coordinate fixed and varying Y, the other vector represents holding the Y coordinate fixed and varying X.

    Because the goal of using tensors is to allow arbitrary coordinate systems, it's perfectly valid to have non-unit and/or non-orthogonal basis vectors.

    I may have overexplained a bit, actually. Basis vectors do not HAVE to be the coordinate basis vectors, i.e. the partial derivatives, that's just a common choice. You can choose __any__ pair of vectors for your basis vectors on a plane.

    Whether or not your metric is diagonal or not has EVERYTHING to do with what choice of basis vectors you made, and NOTHING to do with flatness.

    Attached Files:

  18. Aug 25, 2011 #17
    Thanks for your attention pervect!!

    Now, i know the basis of differentia geometry, i'm familiar with the tensor formalism, coordinate basis ecc... but..have you never seen my definition of an observer who sees spacetime locally flat???

    I do not mean the spacetime IS flat in that point!!

    I mean that with a particular basis vector set the metric becomes minkowsky to first order, the connection vanish (only in that point) and to this observer (namely to this basis vector set) geodesics that pass by that point are straight lines (always in an infinitesimal leighbour)

    Now, some precisation:
    We are not arguing that if I choose a particular basis the spacetime becomes flat.
    Flatness-non flatness is an intrinsic propriety of the manifold, and only the riemann tensor (and derivaitves) measure that, and the riemann tensor, being a tensor :tongue: doesn't depend on the basis you chose, so the global topology of the manifold is untouchable.

    Yes, to that observer space won't seem locally flat.
    But with my definition you can find some basis vector which instead do make the spacetime look locally flat(with the meaning I give in my definition); BUT being in this case the spacetime flat everywhere we have actually built a global inertial frame (which is instead impossible on curved spacetime).

    If, as I probably have, did not make myself clear, please point where does the confusion start!!!
    (I'm pretty sure my definiton is correct)
  19. Aug 26, 2011 #18


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    Not really - which doesn't mean that someone doesn't do it somewhere. So if you've got a source you want to quote or mention that uses the words this way, feel free to point it out.

    However, everything you're saying seems to me to be a statement about the particular coordinate choice you are using.

    We could reword what you're saying to "locally flat coordinates" and that would make general sense to me,though the adjective "flat" seems to be not-quite-precise. But the way I see this described usually (MTW, for instance) is "Fermi Normal coordinates". Which have all the properties you describe as long as the observer isn't accelerating. I suppose you could add that to be really specific. So you might say " The fermi normal coordinates of a geodesic observer".

    That's more or less what I was trying to point out.

    What was your question, then? We started out talking about basis vectors, and you seemed to have some questions about their significance.

    I think we agree now that in general you can choose your basis vectors any way you want, that you have complete freedom in this regard, but if you want a diagonal metric at some point, you need to choose orthonormal basis vectors at that point.

    I don't think it was mentioned this before, but following on, if you want the connection coefficients (the first derivative of the metric) to vanish as well, you need to parallel transport your initial choice of basis vectors. So this condition, combined with the first one, specifies your basis vectors everywhere in the neighborhood.

    Furthermore, the worldline of the observer must be a geodesic, and there's only one geodesic through a point, so the initial choice of basis vectors specifies the observer's worldline.

    So picking out some point and saying the metric is diagonal there and that the connection coefficeints also vanish is sufficient to describe a unique coordinate system (the coordinate system whose coordinate basis vectors are the ones we've specified), Fermi Normal coordinates.

    We haven't addressed the size limitations on this coordinate choice - it may not necessarily cover all of space-time, it will only cover a region of space-time where geodesics through the original point P don't cross.
  20. Aug 28, 2011 #19
    Ok, i think I can restate my question now!!!

    I looked for the source of my definition, it comes from my professor(!); and you're right, i could not find it as i stated in the books i'm currently using (shutz carroll ecc).

    I can see that all of the books, as you did, refer to a set of orthonormal basis for which the metric in a certain point on the manifold meets the condition i wrote(metric in that point=minkowsky metric, first derivative of the metric in that point=connection in that point=0; ecc..); so actually is of a set of orthonormal basis we're talking about, or more precisely locally inertial frames. Excuse me for my poor definition!!

    But actually my question stays the same.
    It's true in fact that an observer with an orthonormal basis set in a certain point will see the manifold around him as flat( isn't it?), and that's why it's called locally inertial.
    So my question (that kinda answers by itself) were: an observer who sees spacetime around him as (at first order) flat, and thus he has an orhtonormal basis set (or, restating, an oberver in a locally inertial frame), has some phisical difference from the other observers who are not using an orthonormal frame??
    Namely, the fact that he points his rulers in a certain direction (as to make them orthonormal) has some physical meaning?
    Or this observer is exacly like everyone else at that point, except that the metric to him is more simple and thus the calculation are easier, and so choosing this observer to calculate things is like a mathematical trick?

    Now, some additional comments on your kind answer:

    absolutely yes.

    This is sure, but only if you impose a torsion free connection (this is only a precisation), since with such a connection lenght and scalar product are conserverd if you parallel trnasport vectors (in our case basis vector) along the motion on a curve.
    If such a frame is called fermi-normal, well, i learned a new thing today!!

    I do not agree with that, you need also to secify the initial 4velocity of the observer, since finding the geodesic through a point is like solving a initial value problem, and you need first derivative (tangent vector) too.
    EDITED: ops, i see now that you also asked for the basis to be specified---no problem then, the timelike vector is the four velocity of the observer.

    I'm aware of this, but it's not important in this discussion.

    Really, thanks for your time and attention!!!

    PS:Another small question, which i hope won't push the main discussion apart.
    why i need him to follow a geodesics (to be a freely falling observer)?
    Couldn't I do the same for an observer who is accelerating, and thus follows a generic timelike curve?
    If i find a local inertial frame and then i parallel trasnport it along the curve, should't I obtain the same result since the scalar product of the vectors don't vary (always with a metric compatible connection)
  21. Aug 28, 2011 #20


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    [nitpick]There are an infinite number of geodesics through a point, but a point and a direction from that point (e.g. the timelike basis vector) do specify a unique geodesic.[/nitpick] Sorry to nitpick.
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