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SR, GR and Twin Paradox

  1. Apr 19, 2007 #1
    It is possible to draw world-lines of the two twins in the non-accelerating one (A) frame of ref. The aging of the other (B), with respect to the first, is evaluated integrating its proper time differential (and comparing it with A's proper time).

    My question is (certainly already asked): is there a difference, about our ability to draw those diagrams and make that computation, if B accelerates because of a star's gravitational field and not because of his rocket's engines?

    Could we avoid knowing GR and using SR only, in this case? If the answer is yes, in which other cases (where B is accelerating) would be possible to use SR only?
  2. jcsd
  3. Apr 19, 2007 #2


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    While one does have to use GR when one is in the gravitational field of a massive body, not very much of GR is needed.

    Basically, one has to replace the line element used in SR

    [tex]d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2[/tex]

    by the more general line element for GR

    [tex]d\tau^2 = \sum_{i,j} g_{ij} dx^i dx^j[/tex]

    Specifically, for a single massive body, one would use the Schwarzschild metric, and one would typically write out the sum as

    [tex]d\tau^2 = (1 - r/r_s) dt^2 - \frac{dr^2}{1-r/r_s} - r^2 d\theta^2 - r^2 \sin^2 \theta d\phi^2[/tex]

    which equates [itex]dx^0[/itex]=dt, [itex]dx^1[/itex]=dr, [itex]dx^2[/itex]=[itex]d\theta[/itex] [itex]dx^3[/itex] = [itex]d\phi[/itex]

    and replaces the general metric with the Schwarzschild metric, so that for instance [itex]g_{00}[/itex] = (1-r/[itex]r_s[/itex]) and [itex] g_{11} [/itex]= [tex]\frac{1}{1-r/r_s}[/tex]

    [itex]r_s[/itex] is the Schwarzschild radius

    One solves for [itex]d\tau[/itex] by taking [itex]\sqrt{d\tau^2}[/itex] which is given by the metric, and then one integrates [itex]d\tau[/itex] in the usual manner you describe, just as one does for SR, except that one uses the general metric to compute [itex]d\tau[/itex] rather than the flat Minkowski metric.
    Last edited: Apr 19, 2007
  4. Apr 19, 2007 #3
    So, if twin B (the one who accelerates) doesn't look out of the window, how can he establish that he accelerates because of a grav. field and not because of his rockets and so, that he have to use another metric instead of the Minkowsky one?
  5. Apr 19, 2007 #4


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    An observer in a flat Minkowski space-time does not HAVE to use a Minkowski metric, it's just the easiest and simplest choice. An accelerating observer in a Minkowski space-time might chose to use Rindler coordinates, for instance.

    Take a look at http://www.eftaylor.com/pub/chapter2.pdf

    This has some material from one of Taylor's book I'll quote a brief excerpt:

    So a metric is defined by setting up a number of events in space-time, and assigning them coordinates.

    If the events are close enough to each other, the distance (in GR, the Lorentz interval) will be determined by some quadratic form written in terms of the coordinates. The coefficients of this quadratic form are the metric.

    It might or might not be obvious, but one can recover the notion of distances and times given the Lorentz intervals between all pairs of points. I don't want to digress on how this is done right at the moment, but if you want more info, ask.

    So you can have many metrics that describe the same physical situation, by chosing different coordinates, i.e. assigning different numbers for the coordinates of some physically defined set of events.

    If you know the metric coefficients in some particular neighborhood, there is a way to determine if the underlying space-time is flat, but it's a bit technical. One basically computes the curvature tensor - if the space-time is flat, there will be no curvature, i.e. all the components of the curvature tensor will vanish.

    By this procedure, an accelerating observer using Rindler coordinates and an inertial observer using some Minkowski coordinates will have different coordinates for the same points, and also will have different metrics.

    Both will, however, compute a zero curvature tensor.

    I hope this helps some.
    Last edited: Apr 19, 2007
  6. Apr 20, 2007 #5
    Sorry if I keep asking, even if you have already answered; I have understood something but not everything.
    My question is: having all the necessary events for the twins, can I avoid using GR and use SR only if I find an inertial frame of ref. with respect to which I draw the twin's world lines?
  7. Apr 23, 2007 #6
    An observer in Minkowski space has no choice but to use the Minkowski metric, but he is free to choose any suitable coordinate chart.

    You are really describing coordinate charts here not metrics.

    The difference is not very important in flat spaces but in curved spaces the difference is essential.
    Last edited: Apr 23, 2007
  8. Apr 23, 2007 #7


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    I tend to use "metric" and "coordinate chart" interchangeably.

    Most people who make a big deal of the distinction are mathematicians rather than physicists.

    Looking at http://mathworld.wolfram.com/MinkowskiMetric.html

    it appears to me that the Minkowski metric is defined in terms of a particular coordinate chart, and that metrics with a vanishing Riemann may be equivalent to the Minkowski metric but are not identical to it.

    This URL may only prove that other people are as sloppy as I am about the distinction between the two.
  9. Apr 23, 2007 #8


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    If you can find some inertial frame of reference, then you can use SR to draw the twins worldlines on a flat sheet of paper, using the coordinates of that inertial frame.

    If you can't find some inertial frame of reference (for instance if you have gravity), then you'd have to draw your SR diagram on some non-flat sheet of paper to get the right answer.

    Often times the gravitational effects are small, and the error involved in using a flat sheet of paper rather than the GR approach is small. For instance, the gravitational time dilation causes a difference of .7 parts per billion between the surface of the Earth and a point infinitely far away.

    (see for instance)

    So if you don't mind a part per billion error, you can ignore the gravity of the Earth in a "twin paradox" type example.
    Last edited: Apr 23, 2007
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