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Transformation to a local inertial Frame

  1. Dec 5, 2006 #1
    So ive been working on this problem. Im given the metric in Kruskal coordinates, so

    ds^2=32M^2exp(-r/2M)/r(-dT^2+dX^2)+r^2(dθ^2+sin^2(θ)dΦ^2)

    And the path of a particle is

    X=0 T=λ θ=π/2 Φ=0

    And the path of the observer is

    X=-1/2*T+1/2 θ=π/2 Φ=0

    And im asked to find the 3 velocity of the particle as seen by the observer when the two intersect. So far this is what i have

    The 4-velocity of the observer in Kruskal Coordinates is

    u=√[4/3*r/(32M^2exp(-r/2M)](1,-1/2,0,0)

    So, I think i need to find a local intertial frame, but im unsure how to do that and im not sure what to do with it once ive found it! Thanks in advance!
     
  2. jcsd
  3. Dec 5, 2006 #2

    pervect

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    Well, the first thing I would do is to write the Kruskal metric as an orthonormal basis of one forms.

    This is actually easy to do once you know how. You basically have a diagonal metric

    ds^2 = -A dT^2 + B dx^2 + C d theta^2 + D d phi^2

    so w0=sqrt(A) dT, w1= sqrt(B) dX, w2=sqrt(C) d theta, and w3=sqrt(D) d phi is such an orthonormal basis.

    Do you know about one-forms (also called cotangent vectors?), and do you know how to take the product of a one-form and a vector to get a scalar? Do you see why dT and dX are one-forms, and why w0, w1, w2, and w3 are othornormal?

    (hint: the length of a vector is g_ij x^i x^j, the length of a one-form is g^ij x_i x_j). w0 has components (w0)_i, i=0,1,2,3

    Do you see why the duals to w0, w1, w2, and w3 (which we can call e0, e1, e2, and e3) are unit vectors i.e. e0 = [itex]\hat{T}[/tex], e1 = [itex]\hat{X}[/itex], e2 = [itex]\hat{\theta}[/itex], e3 = [itex]\hat{\phi}[/itex]

    Finally, do you see how
    [tex]
    v = (w0 v) \hat{T} + (w1 v) \hat{X} + (w2 v) \hat{\theta} + (w3 v) \hat{phi}
    [/tex]

    Lastly, do you see how this solves your problem? :-)
     
  4. Dec 5, 2006 #3
    So, i really havn't done anything with one forms....ive looked them up, but they dont really make any sence. Im not sure how this helps :cry:Is there a way to do this without 1 forms?....we havn't been taught those in class......so there should be a way to do the problem without that....
     
    Last edited: Dec 5, 2006
  5. Dec 5, 2006 #4

    pervect

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    Hmmm. Well, while you should really learn about one-forms (or cotangent vectors or dual vectors) there is a work-around that gets the same answer, but you'll have to do coordinate transformations.

    How do you deal with tensors at all without knowing about duals, though?

    What you can do instead is to introduce new scaled coordinates, say x1, t1, theta1, phi1 in such a way that you have a locally Minkowskian metric (i.e. in this case so that you normalize the metric coefficients to unity) in terms of your new coordinates. Here x1= alpha x, t1 = beta t, etc.

    Then dx1/dt1 = (dx1/dtau) / (dt1 / dtau) will be your desired velocity. See for instance https://www.physicsforums.com/showpost.php?p=602558&postcount=29 where I get the right answer after a few false starts for the velocity of an object falling into a black hole relative to a "hovering" observer. You'll probably need to read the entire thread to get the background of the problem, however, earlier attempts to solve the problem got the wrong answer. At the end of the theread, George Jones also works the problem via a slightly different means to get the same answer.

    If you use something like GrTensorII, it's really easy to specify a metric by specifiying the ONB of one forms rather than the metric itself. You can then have a choice of working with either coordinate basis vectors, or a local orthonormal tetrad of unit vectors.

    You can also try reading http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity which uses slightly different language to take the same approach I did.
     
    Last edited: Dec 5, 2006
  6. Dec 5, 2006 #5
    Thanks for the help, i was able to find a transformation that made the metric locally minkowski, and everything just fell out from there! Thanks!
     
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