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Transition from Flat Minkowski to GR

  1. Feb 28, 2014 #1
    Hello friends ,
    I have some conceptual problems in understanding the difference between Minkowski spacetime and the spacetime of general relativity. The general spacetime of GR is defined as a smooth manifold which is locally like Minkowski spacetime . What does this statement mean ?

    Does it mean that at a point I can always find a co-ordinate chart such that the associated basis of tangent vectors have an inner product which is (1,1,1,-1)? If that is the case , then I can always use such charts at all points of the smooth manifold and make it locally flat everywhere ? What happens if I had used some different chart ? Will the inner products at the point change to some crazy numbers like (2,2,7,4 ) ?

    Infact , lets take a simple example of a 2-sphere of unit radius.. the natural metric is (1, sin^2 theta) ..how do I mathematically show that it is locally flat ie. have a metric (1,1) ?
    Thanks for any help ...
     
  2. jcsd
  3. Feb 28, 2014 #2

    tiny-tim

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    hello clerk! :smile:
    every space is locally flat (except at a singularity) …

    tangent spaces are flat, and every space at a particular point is locally like its tangent space at that point, ie locally flat :wink:

    (the (1,1) part of your question, i don't understand … isn't it obvious that it's locally (1,1) ?)
    yes! :smile:
    as i said, every space is locally flat everywhere

    as for crazy numbers, you can always change the coordinates locally to make the metric whatever you like, provided you don't change the signs
     
  4. Feb 28, 2014 #3
    Thanks a lot , Tiny Tim !!
    In the 2 -sphere example , I wanted to understand the required co-ordinate transformation that would reduce the metric (1, sin^2 theta) to (1,1) ..if it is locally flat , I should be able to explicitly show that , right ?
    Also please will you see if my following statements are correct .. I want to ensure that I understand the things correctly..
    There exists a co -ordinate chart for every point of spacetime such that the associated basis of tangent vectors have an inner product which is (1,1,1,-1). Had I used a different chart , I would have got a different metric but with the same index. So in the GR texts where we see that the authors make a co-ordinate redefinition to give the metric a different form , he/she is just using a different co-ordinate chart to describe the same geometry..this new co-ordinate chart may not cover the whole spacetime because if it did then we could have made a co-ord transformation to make the metric flat everywhere globally ..this inability to use a single co-ordinate chart to cover the full spacetime is due to curvature of spacetime - which is gravity ..?
     
  5. Mar 1, 2014 #4

    pervect

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    You can transform the coordinates with ##\theta' = k \theta## where k is some constant. But - you might want to read up on non-coordinate basis. (I assume you have a text that you can look it up in). Basis vectors are not necessarily tied to coordinates.
     
  6. Mar 1, 2014 #5

    tiny-tim

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    hi clerk! :smile:
    correct :smile:
    no, there are no flat coordinate charts!

    "locally flat" means that in the limit it is flat, ie the smaller the chart region gets, the nearer it is to flat

    the only coordinate charts that are flat are of spaces which are flat :wink:
     
  7. Mar 1, 2014 #6
    I happend to be reading about the same subject recently trying to verify terminology. I found some nice clarifications here:

    http://en.wikipedia.org/wiki/Tangent_space

    I concluded tangent vectors live in tangent spaces. So a tangent space is a flat coordinate chart in tiny tim's explanation. {I hope!}
     
  8. Mar 1, 2014 #7

    tiny-tim

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    no, as i understand it, a chart is a coordinate map of a region of the space

    eg you can cover the earth's surface with two charts, one of the whole surface minus the north pole, and one of the whole surface minus the south pole (with any sensible projection you choose) :wink:

    (and yes, a tangent vector is an element of a tangent space …

    if the original space can be embedded in a higher-dimensional space, then the tangent space at a point is homomorphic to the tangent plane at that point in that higher-dimensional space)
     
  9. Mar 1, 2014 #8
    ah! had not thought about that....thanks....
     
  10. Mar 1, 2014 #9
    @all - thanks a lot for the stimulating discussions ..
    Just one more clarification - in the popular QFT text Anthony Zee makes the comment that under suitable restrictions , the effect of a gravitational field is equivalent to a co-ordinate transformation - how exactly do I see that ? (Reference chapter : Field theory in curved spacetime , 1st paragraph)
     
  11. Mar 1, 2014 #10

    WannabeNewton

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    1. The term "locally flat coordinates" is a horrible misnomer. Nothing is flat in a set of "locally flat coordinates". A better term would be "locally inertial coordinates". Note that we can always diagonalize the metric tensor at any given point-this is just part of the finite dimensional spectral theorem. The really important property of "locally inertial coordinates" is the vanishing of the Christoffel symbols at the origin of the coordinate system.

    2. Basis vectors in general have nothing to do with coordinate systems. Only the holonomic bases do but we can easily work with non-holonomic bases and in fact do so all the time because it is the latter which allow us to properly describe physics relative to different observers. Non-holonomic bases are what we use for local Lorentz frames.

    3. The effect of a gravitational field is definitely not equivalent to a coordinate transformation. Actually what in the world does that even mean? He (Zee) says this is the principle of general covariance but that's certainly not true. At best he could be talking about the fact that in a locally inertial coordinate system the first order effects of a gravitational field vanish, such as gravitational redshift.

    It's almost sacrilegious to equate the gravitational field, which is a purely geometric object, with coordinate systems.

    EDIT: I checked out the Weinberg reference in Zee and nowhere does Weinberg say that the effect of a gravitational field is equivalent to a coordinate transformation (whatever that means). All he says is, if a law of physics is generally covariant and holds in the absence of gravitation then it will hold in the presence of gravitation as well. Actually this is not precisely true as stated-it is only true if the law is only zeroth order or first order in ##\nabla_{\mu}##. But the basic idea is if we go to a locally inertial coordinate system then ##\nabla_{\mu} \rightarrow \partial_{\mu}##. If we then write down a law of physics in generally covariant form using ##\partial_{\mu}## in this locally inertial system then it will also hold under ##\nabla_{\mu}## i.e. even when gravitation is present.
     
    Last edited: Mar 1, 2014
  12. Mar 1, 2014 #11
    If more of Zee's statements in his nutshell series are like this one it should send up a red flag about those books.
     
  13. Mar 1, 2014 #12
    WannabeNewton : I agree with you .. I was never able to digest that statement.. but at the same time felt that may be I am missing some deep secret of GR , feel more relaxed now.. :)
     
  14. Mar 2, 2014 #13

    stevendaryl

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    Maybe he's talking about parallel transport? If you take a ball and let it drift in freefall, it might start with a velocity 4-vector in which the spatial components are all zero. But if you wait a second, it will acquire a nonzero spatial component to its velocity 4-vector. That sort of looks like there is a boost operator relating 4-vectors in one region of spacetime to 4-vectors in a neighboring region of spacetime.

    Mathematically, if [itex]X^\mu[/itex] is the displacement vector connecting two neighboring regions of spacetime, and [itex]U^\mu[/itex] is a velocity 4-vector for an object in freefall that drifts from one region to the other, then in the second region, it will have a velocity 4-vector given approximately by:

    [itex]\tilde{U}^\mu = L^\mu_\nu U^\nu[/itex]

    where

    [itex]L^\mu_\nu = \delta^\mu_\nu + \Gamma^\mu_{\nu \lambda} X^\lambda[/itex]

    It looks a little like an infinitesimal coordinate change.
     
  15. Mar 2, 2014 #14

    pervect

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    A tangent space is indeed flat - the usual example is if you have a manifold which is the surface of a sphere, the tangent space is a plane that's tangent to the sphere.

    The tangent space (the plane) isn't the manifold (the sphere). But there's a mapping from the tangent space (the plane) to the manifold (the sphere) called the exponential map that can map vectors from the flat tangent space to displacements on the not-necessarily-flat manifold.
     
  16. Mar 2, 2014 #15

    PeterDonis

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    MTW say something similar in Chapter 7 (basically that the effects of a gravitational field are equivalent to a uniform acceleration of the coordinates), but they say it's the equivalence principle, not the principle of general covariance.

    I think part of the problem is the term "gravitational field", which is ambiguous, as has been discussed in a number of previous threads on PF. As far as I can tell, Zee here is using it to mean "coordinate acceleration due to gravity", which is similar to MTW's usage in the passage I just mentioned. That is not a geometric object, and can, of course, be transformed away by a change of coordinates. (Of course, if that is indeed what Zee meant, he could have just said so directly, instead of using the ambiguous term "gravitational field".)

    The geometric object is *tidal* gravity, which the term "gravitational field" is sometimes used to refer to, but I don't think Zee means it that way here.

    You also have to include ##\sqrt{-g}## in the spacetime integration measure (which Zee notes). The rest of the discussion in the chapter of Zee in question basically says the same thing Weinberg is saying (though he talks more about replacing ##\eta_{\mu \nu}## with ##g_{\mu \nu}##, which also needs to be done to correctly generalize to curved spacetime). I agree that Zee's remark about a "gravitational field" being equivalent to a coordinate transformation isn't helpful; it could have been left out altogether without any loss of understanding, since the rest of his discussion covers the necessary points.
     
  17. Mar 2, 2014 #16

    PeterDonis

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    I've only read QFT in a Nutshell (the book under discussion here), and it has helped me better understand quite a few aspects of QFT, so I don't think a red flag is warranted for that book. I would be interested in any information about the second "nutshell" book that specifically covers GR.
     
  18. Mar 2, 2014 #17

    WannabeNewton

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    But that isn't a coordinate transformation. Parallel transport moves a vector along a curve from one tangent space to another. A coordinate transformation simply changes the components of a vector in the same tangent space by changing the coordinate basis.
     
  19. Mar 2, 2014 #18

    WannabeNewton

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    Right if it's with regards to the equivalence principle then I can agree if, as you say below, we interpret "gravitational field" to mean the coordinate acceleration due to gravity or, as is made apparent in the Newtonian limit, the Christoffel symbols.

    It's very run of the mill so you aren't going to get any insights out of it that you don't already possess. I'm not a fan of the book simply because I much prefer books that stick to coordinate-free (abstract index or index-free) calculations whenever possible and only use coordinates when absolutely necessary, such as Wald. Zee's GR book abuses coordinates at every bend. It's quite reminiscent of Weinberg's book which I also dislike.
     
  20. Mar 2, 2014 #19

    PeterDonis

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    That was kind of the impression I was getting from reading reviews of it. Oh, well.

    I think part of this is the particle physicist viewpoint as contrasted with the relativist's viewpoint. Particle physicists seem to want to think of everything in terms of coordinates.

    Although, as a counterpoint to this, Zee's QFT book does a fair bit of analysis (for example, in the section on non-Abelian gauge theory) in terms of differential forms, with nary a coordinate in sight. So whatever is going on is probably more complicated than just "particle physicists prefer coordinates".
     
  21. Mar 2, 2014 #20

    WannabeNewton

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    It might be that the GR book in question was meant to be accessible to undergrads, in which case coordinates would be much friendlier than the sea of indices in coordinate-free calculations.

    I haven't had much experience with Zee's QFT book but the writing in the book is absolutely phenomenal. It's the only physics book apart from Griffiths' electrodynamics that I can read as if I were reading a novel.
     
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