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Definition of local inertial frame

  1. Sep 3, 2011 #1
    I have a question I wanted to clear up. According to the definition of a "local inertial" frame in GR, you must use a coordinate system that locally looks Cartesian, right? I mean if you had a coordinate system with a basis that wasn't orthogonal, then it would not be considered a local inertial frame, right?

    I ask because outside of relativity, I would normally say that whether a frame is inertial or not wouldn't depend on the coordinate system you choose. If your frame is in uniform motion, then it's inertial regardless of which way you point your meter stick.
     
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  3. Sep 3, 2011 #2

    Fredrik

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    I haven't seen this explained in detail anywhere, but I haven't looked really hard. The literature speaks of several different kinds of coordinate systems, "Riemannian normal coordinates", "Fermi normal coordinates", etc., that all seem to correspond to the local (i.e. comoving) inertial coordinate systems of SR. Most of them seem to have the word "normal" in the name, so I think it would be appropriate to define a "normal" coordinate system at a point p as a coordinate system x such that

    a) [itex]x(p)=0[/itex]
    b) [itex]g_{\mu\nu}(p)=\eta_{\mu\nu}[/itex] (where [itex]\eta[/itex] is the Minkowski metric)
    c) [itex]g_{\mu\nu,\,\rho}(p)=0[/itex] (where the comma indicates partial differentiation)

    Then we can start defining classes of "normal" coordinate systems by imposing additional conditions. I don't think there's one specific coordinate system that deserves a name like "the comoving inertial coordinate system" more than any other.

    You might want to look up the term "local Lorentz frame" in MTW. I don't remember exactly what they said, but I think they described a way to define one of these "normal" coordinate systems.
     
  4. Sep 3, 2011 #3
    I'm going to try to give some very explicit and clear cut answers because I have always wondered about parts of your questions as well.

    So we have a common discussion point for frames of reference:

    Wikipedia words it this way,

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

    Notice there is NO reference to a particular type of reference frame. I'd loosely define "proper" here as meaning "local" for this discussion, but it has a more specific definition.

    A local reference frame is explained this way:

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

    Again note it is space (time) to which curvature criteria applies, not the selected coordinate reference frame. Eucledean space is flat multi dimensional space and we can use Cartesian( or other) coordinates to define our measurement parameters...."non-Euclidean" means curved space time.

    To your questions:

    From the above we can conclude, "No". You can use ANY convenient coordinate system...Cartesian or polar or spherical for example.


    "No" again because it's the motion of the coordinate system not its type (Cartesian or Polar or spherical or cylindrical) that makes it inertial or not. You can usually use coordinate transformations to get from one coordinate system to another, say: x = r cosθ and y = r sinθ to go from polar to Cartesian. The Lorentz transformation in relativity is a special one used for homogeneous space



    This is accurate inside relativity too. You last sentence here says it all.


    In general relativity, an inertial reference frame is only an approximation that applies in a region that is small enough for the curvature of space to be negligible. So space is curved over larger distances but we keep our observations to only a small region....analogous to how we measure a mile of highway via GPS,for example, as if it were flat even though there are 'ripples'..hills, valleys, and curvature of tghe earth.

    edit: "Long range" measurements [in curved space] have unique problems....see my next post.
     
    Last edited: Sep 3, 2011
  5. Sep 3, 2011 #4
    I was checking my own notes on frames of reference and came across this, maybe from Pervect, in these forums:

    Definition of an inertial frame:

    Notice, again, it has nothing to do with which coordinate system is chosen.

    and:
    (I don't know who posted this but I liked it enough to include in my own notes.)
     
  6. Sep 3, 2011 #5

    Fredrik

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    These descriptions make it sound like a "frame" is a physical object. A coordinate system is a function from a subset of spacetime into [itex]\mathbb R^4[/itex], and a frame (or a frame field) is an assignment of a basis for the tangent space at p to each point p in some subset of spacetime. They are both mathematical terms.

    I believe that what we're discussing is if there's a natural way to associate a coordinate system or a frame with the world lines that represent the motion of an object.
     
    Last edited: Sep 3, 2011
  7. Sep 3, 2011 #6
    Thanks everyone for your replies. I'm replying to Fredrik because he's given the definition that has me thinking. The definition is

    So let's say I choose a coordinate system with the y-axis rotated by some arbitrary angle [itex]\psi[/itex] in the x-y plane . In other words,

    [itex]\hat{x'}=\hat{x}[/itex]
    [itex]\hat{y'}=\hat{y}\cos\psi + \hat{x}\sin\psi[/itex]
    [itex]\hat{z'}=\hat{z}[/itex] .

    There's nothing stopping me from doing this, right? Then, unless I'm misunderstanding something (which is always quite possible), the metric [itex]g_{\mu\nu}[/itex] is

    [itex]\left(\begin{array}{cccc}
    -1 & 0 & 0 & 0 \\
    0 & 1 & \sin\psi & 0 \\
    0 & \sin\psi & 1 & 0 \\
    0 & 0 & 0 & 1
    \end{array}\right)
    [/itex] .

    Here, even if we're only looking at things locally, [itex]\psi[/itex] is an arbitrary angle, so it doesn't have to be small. So it would seem to me that this metric is not [itex]\eta_{\mu\nu}[/itex], and (b) above fails. So it's not an inertial frame.

    I'd definitely appreciate anyone who could clear this up for me.
     
  8. Sep 3, 2011 #7

    DrGreg

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    I think the problem is that there is no universally agreed definition of "frame". Some people use it to mean "coordinate system" (see post #5). In the context of special relativity, some people use it specifically to mean "inertial coordinate system". Others use it to mean "frame field" (see post #5), which, in GR, is not the same thing.

    If you'd asked me a year ago, I'd have given an answer in agreement with Naty1 (which isn't any of the things in the first paragraph): essentially a chosen decomposition of 4D spacetime into 3D space + 1D time, but without specifying a coordinate system for space. Now that I know about "frame fields" I'm aware that the term is somewhat ambiguous.

    Fredrik's definition of "locally inertial" in post #2 is a definition of "locally inertial normal coordinates". Whether or not you think that defines a "locally inertial frame" depends on what you think a frame is.
     
  9. Sep 3, 2011 #8
    I'm going by a definition in Schutz (pg. 156), which basically uses the definition of post #2, and says they will refer to these coordinate systems as "local Lorentz frames" or "local inertial frames." So if I understand you, Schutz has simply defined "local inertial frame" to include an orthonormal basis, but others might call it a "local inertial orthonormal frame" or whatever. The choice of phrase "local inertial frame" seems a bit misleading to me, though, but maybe that's just my own hang-up.

    Thanks

     
  10. Sep 3, 2011 #9

    Fredrik

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    That matrix of metric components looks less complicated than I would have guessed. I think you answered your own question when you said that the word "inertial" is a bit misleading. Maybe that's why these things go by other names as well.
     
  11. Sep 3, 2011 #10

    DrGreg

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    For what it's worth, I make it[tex]
    \left[\begin{array}{cccc}
    -1 & 0 & 0 & 0 \\
    0 & \sec^2 \psi & -\tan \psi \, \sec \psi & 0 \\
    0 & -\tan \psi \, \sec \psi & \sec^2 \psi & 0 \\
    0 & 0 & 0 & 1
    \end{array}\right][/tex](But that doesn't alter the argument)
     
    Last edited: Sep 3, 2011
  12. Sep 3, 2011 #11
    DrGreg, maybe I need to take a step back. I was calculating the metric as

    [itex]g_{\mu\nu} = \vec{e}_\mu\cdot\vec{e}_{\nu}[/itex],

    so I was getting

    [itex]g_{x'x'}=\hat{x} \cdot \hat{x}=1[/itex]
    [itex]g_{y'y'}=(\hat{y}\cos\psi +\hat{x}\sin\psi)\cdot (\hat{y}\cos\psi +\hat{x}\sin\psi) = 1[/itex],
    etc.

    Am I doing something wrong?


     
  13. Sep 3, 2011 #12

    DrGreg

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    Sorry, I misunderstood your notation. I was thinking [itex]\hat{y}[/itex] was a coordinate, not a basis vector. (The hat notation isn't one I ever use myself; I would denote the vector as [itex]\mathbf{j}[/itex] or [itex]\partial_y[/itex] or [itex]\frac{\partial}{\partial y}[/itex], or maybe even [itex]\mathbf{e}_y[/itex].) So I think you were right after all.
     
  14. Sep 3, 2011 #13
    Ah, fair enough. No problem at all. Thanks for your help. I think things seem clear to me now.

     
    Last edited: Sep 3, 2011
  15. Sep 4, 2011 #14

    atyy

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    Uniform motion relative to what?
     
  16. Sep 4, 2011 #15

    Fredrik

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    This is how I do it: The basis vectors associated with the new coordinate system are given by [tex]\begin{pmatrix}\vec e'_0\\ \vec e'_1\\ \vec e'_2\\ \vec e'_3\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & \sin\psi & \cos\psi & 0\\ 0 & 0 & 0 & -1\end{pmatrix}\begin{pmatrix}\vec e_0\\ \vec e_1\\ \vec e_2\\ \vec e_3\end{pmatrix}.[/tex] I want to express this in the form [tex]\vec e'_\mu=M^\rho_\mu\vec e_\rho,[/tex] where the [itex]M^\rho_\mu[/itex] are the components of a matrix, but if we want the upper index to be the row index, we have to define M as the transpose of the matrix above. So we define [tex]M=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & \sin\psi & 0\\ 0 & 0 & \cos\psi & 0\\ 0 & 0 & 0 & -1\end{pmatrix}.[/tex] One way to understand this is to note that [itex]\vec e'_\mu=M^\rho_\mu\vec e_\rho[/itex] is actually the [itex]\mu[/itex] component of the equation [tex]\begin{pmatrix}\vec e'_0 & \vec e'_1 & \vec e'_2 & \vec e'_3\end{pmatrix}=\begin{pmatrix}\vec e_0 & \vec e_1 & \vec e_2 & \vec e_3\end{pmatrix} \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & \sin\psi & 0\\ 0 & 0 & \cos\psi & 0\\ 0 & 0 & 0 & -1\end{pmatrix},[/tex] which is the transpose of the one we started with.

    The components of the metric in the new coordinate system are given by [tex]g'_{\mu\nu}=g(\vec e'_\mu,\vec e'_\nu)=g(M^\rho_\mu \vec e_\rho,M^\sigma_\nu\vec e_\sigma)=M^\rho_\mu M^\sigma_\nu g(\vec e_\rho,\vec e_\sigma)=M^\rho_\mu M^\sigma_\nu \eta_{\rho\sigma}=(M^T\eta M)_{\mu\nu}[/tex] So the matrix of components of the metric in the new coordinate system is [itex]M^T\eta M[/itex]...which to my surprise is actually equal to the result you got. I thought your method looked to simple, but I see now that it makes sense (when the components of the metric in the coordinate system we're transforming from are [itex]\eta_{\mu\nu}[/itex]).
     
    Last edited: Sep 4, 2011
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