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Nature of Geodesic

  1. Jul 1, 2015 #1
    Hello!
    Please help: A world line is given to us. It is known that it is a geodesic. The metric, however, is not known. Since we don't know the metric, it should not be possible to tell whether the geodesic is spacelike/timelike/null (Right?)
    But since the geodesic is known (x,y,z,t), we can find out ∂x/∂t, ∂y/∂t, ∂z/∂t. That is, we can find out the speed of the object which will travel that geodesic. Depending on whether it is greater than, equal to or less than 'c' (light speed), we can say that the geodesic is spacelike, null or timelike. Thus, it seems we can tell this without knowing the metric. What is wrong with this?

    Thank you!
     
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  3. Jul 1, 2015 #2

    Mentz114

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    I don't think being a geodesic or not is relevant to whether the 4-velocity is timelike, spacelike or null because ##u^\mu u_\mu## depends only on ##u## and not on its curvature.

    So, if you are given the contravariant components you need the metric to find the covariant components and ##u^\mu u_\mu##.
     
  4. Jul 1, 2015 #3

    stevendaryl

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    The conclusion that [itex]|\frac{\partial x}{\partial t}| < c[/itex] for a timelike path is only necessarily true in Cartesian coordinates, and you need to know the metric in order to know whether your coordinates are Cartesian. For example, in polar coordinates [itex]r, \theta, \phi[/itex], [itex]\frac{\partial \theta}{\partial t}[/itex], the "speed" in direction [itex]\theta[/itex], can be greater than [itex]c[/itex].[/QUOTE]
     
    Last edited: Jul 1, 2015
  5. Jul 1, 2015 #4

    Mentz114

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    Just to let you know that your post needs an edit ( at the time of writing this).
     
  6. Jul 1, 2015 #5
    I was just thinking of the 3-velocity. Its components are clearly ∂x/∂t, ∂y/∂t, ∂z/∂t. I can find these since the (x,y,z,t) is given to me and I don't need the metric for that. Thus I can find the magnitude of the 3-velocity (without knowing the metric) and compare it with 'c' (the speed of light).
     
  7. Jul 1, 2015 #6

    stevendaryl

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    You can't find the magnitude of a vector without using the metric.

    In 2D spatial coordinates, let me call them [itex]p[/itex] and [itex]q[/itex], what is the magnitude of the "velocity vector" [itex]\vec{v} = (\frac{dp}{dt}, \frac{dq}{dt})[/itex]? If [itex]p[/itex] and [itex]q[/itex] are cartesian coordinates, then

    [itex]|\vec{v}|^2 = \frac{dp}{dt}^2 + \frac{dq}{dt}^2[/itex]

    On the other hand, if they are polar coordinates (with [itex]q[/itex] radial and [itex]p[/itex] angular), then

    [itex]|\vec{v}|^2 = \frac{dq}{dt}^2 + q^2 \frac{dp}{dt}^2[/itex]

    In general,
    [itex]|\vec{v}|^2 = g_{qq} \frac{dq}{dt}^2 + g_{qp} \frac{dq}{dt}\frac{dp}{dt} + g_{pq} \frac{dp}{dt}\frac{dq}{dt} + g_{pp} \frac{dp}{dt}^2[/itex]
     
    Last edited: Jul 1, 2015
  8. Jul 1, 2015 #7

    Mentz114

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    You need all four components to get ##u^\mu u_\mu##. There are no shortcuts to a general covariant result.
     
  9. Jul 1, 2015 #8
    Ok. So, if the components of the 3-velocity are v(x), v(y) and v(z) then is it not true that the magnitude of the "speed" V is given by V^2 = v(x)^2 + v(y)^2 + v(z)^2 ?
    I mean does the validity of the above equation depend on what the metric is?
     
  10. Jul 1, 2015 #9

    stevendaryl

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    Think about the simple case of an object orbiting the Earth above the equator at a constant radius [itex]r[/itex] and a constant angular velocity [itex]\frac{d\phi}{dt}[/itex]. What is the speed of this object? It is [itex]v = r \frac{d\phi}{dt}[/itex]. It isn't simply [itex]\sqrt{(\frac{dr}{dt})^2 + (\frac{d\phi}{dt})^2}[/itex]. (That wouldn't even make sense, in terms of units)
     
  11. Jul 1, 2015 #10

    stevendaryl

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    Just to elaborate a little: The fact that we call the coordinates [itex]x[/itex] and [itex]y[/itex], as opposed to [itex]r[/itex] and [itex]\phi[/itex], doesn't mean anything. You can pick whatever names for your coordinates you like, and that doesn't change the mathematics. What makes the difference between Cartesian coordinates and polar coordinates is the metric.
     
  12. Jul 1, 2015 #11

    robphy

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    You know absolutely nothing about the metric? Not even its signature?
    If it is a Lorentz-signature metric, you might not need to know the whole metric to answer your question.

    Do you know one timelike vector at a point along the geodesic? (I guess you must if you are assuming that you can compute a 3-velocity.)
    By the way, to find the magnitude of a 3-velocity, you need at least a spatial metric (as stevendaryl suggests).

    So, the question is: what do you know about this metric?
     
  13. Jul 1, 2015 #12

    Nugatory

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    The metric is in that equation too - you just aren't seeing it because you've written that equation in Cartesian coordinates and in those coordinates the metric components are equal to one: ##v^2=g_{xx}v_x^2+g_{yy}v_y^2+g_{zz}v_z^2=v_x^2+v_y^2+v_z^2##. If you used some other coordinate system (polar coordinates, for example) some of the metric components would have values other than one, and they'd have to appear explicitly in the calculation.
    (You'll also notice that this equation is careless about the distinction between upper and lower indices - that also only works for the special case of Cartesian coordinates).
     
    Last edited: Jul 1, 2015
  14. Jul 1, 2015 #13
    Ok. So I think I understand it now. To compute even the magnitude of the 3-velocity along the geodesic, I will need the metric.
    Now just one more query : In cartesian coordinates in flat space-time, the minimum spatial between two neighbouring points is given by (dx)^2 +(dy)^2 +(dz)^2. But now if we are in curved space-time, the minimum spatial distance between two points is not given by above equation. So if I take two points in our 3D space (in presence of earth's gravity), then the distance measured by placing a straight ruler between these two points is not the minimum possible distance. I am finding this hard to imagine. I mean how can you get a distance smaller than that?
     
  15. Jul 1, 2015 #14

    Nugatory

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    It's easier to see this if you start with a two-dimensional example. On a flat sheet of paper the shortest distance between two points is given by ##\Delta{s}^2=\Delta{x}^2+\Delta{y}^2##, just as you say. But that's not true on the curved surface of the earth. In fact, you can't even lay a rectangular x-y grid down on that curved surface; the closest you can come is lines of latitude and longitude.
     
    Last edited: Jul 1, 2015
  16. Jul 1, 2015 #15

    pervect

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    Yes
    I'm not sure how you draw this conclusion. The first point is that to define distance, you need to eliminate the time from space-time. There are a couple of ways to do this, the easiest way (which turns out to not always be possible) is to define hypersurfaces of constant time, then, with time no longer a factor, you can use your space-time metric to define the distances between points on this spatial hypersurface (this is usually called the induced metric). I won't go into the "hard way", but I'll mention that if you have a rotating frame, the "easy way" I described above won't work.

    The next point is that on this spatial hypersurface (which is a volume element that we've extracted by eliminating time from space-time), the geodesic paths will be in GR (though not in some other theories, but I'll not go into details) the geodesics are the paths of shortest distance.

    It may be helpful to consider the surface of the Earth as an example. The lines of shortest distances between two points are great circles, which are geodesics. Note that lines of constant lattitude on the Earth are not geodesics, they are not great circles and they aren't the shortest distance between two points.
     
  17. Jul 2, 2015 #16
    Thanks Nugatory and pervect!!
     
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