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Two Unrelated Questions about GR: Light Cones and VS Dust

  1. Oct 15, 2012 #1
    Question 1: How does the metric relate to the light cone at a particular point? Obviously, the metric should determine the shape of the light cone, but is the converse true? Does the orientation and width of the light cone tell you everything about the metric? I'm guessing not due to the different degrees of freedom, but can anyone articulate a concise relationship between the two?

    Question 2: I've looked at these two sources about Van Stockum Dust and seem to have encountered an internal contradiction. Is anything wrong here?

    http://rampke.de/data/2009/12/ba.pdf (pages 7 and 8)
    http://en.wikipedia.org/wiki/Van_Stockum_dust

    The frame field transformation defined from basis vectors to frame vectors does not seem to be the inverse of the frame field transformation from basis 1-forms to frame 1-forms. If you write out the transformations

    a1714c4cb799639e4de100c34879f07b.png
    59a58938bf57d769d13cb2d3cf3eb7ee.png

    as matrices, you get, respectively:

    \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & f(r) & 0 & 0 \\ 0 & 0 & f(r) & 0 \\ -h(r) & 0 & 0 & \frac{1}{r} \end{pmatrix}

    \begin{pmatrix}-1 & 0 & 0 & h(r)r \\ 0 & \frac{1}{f(r)} & 0 & 0 \\ 0 & 0 & \frac{1}{f(r)} & 0 \\ 0 & 0 & 0 & r \end{pmatrix}

    Their product is clearly not the identity, but isn't it always true that any change of coordinates does the opposite thing to the basis covectors as it does to the basis vectors? What's going on here?

    Thanks in advance!
     
  2. jcsd
  3. Oct 15, 2012 #2

    bcrowell

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    I think something similar to this is true, but not quite this.

    Knowing all the light cones definitely doesn't determine the metric, because the light cones don't change under a conformal transformation. Essentially this is the idea that you can't make a clock out of light.

    I think what is true is that if you know the inner product of arbitrary lightlike vectors at a given point, you can infer the metric. I think this follows because you can find a basis consisting of four lightlike vectors. IIRC there's a discussion of this in Hawking and Ellis.
     
  4. Oct 15, 2012 #3

    Mentz114

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    For the VS dust,

    [tex]
    ds^2=-{dt}^{2}+
    \,\left( r^2-h^2\right) {d\phi}^{2} +2\,h\,d\phi\,dt+{dz}^{2}\,{f}^{2}+{dr}^{2}\,{f}^{2}[/tex]
    I'm using this for the co-frame field and frame field respectively
    [tex]
    C=\begin{pmatrix}-1 & 0 & 0 & h \\ 0 & f & 0 & 0 \\ 0 & 0 & f & 0 \\ 0 & 0 & 0 & r \end{pmatrix},\ \ F=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & \frac{1}{f} & 0 & 0\\ 0 & \frac{1}{f} & 0 & 0 \\\frac{h}{r} & 0 & 0 & \frac{1}{r} \end{pmatrix}
    [/tex]
    and [itex]C^{T}F = \delta[/itex].

    Unfortunately my f and h are not the same as the Wiki, so this might sow confusion.

    Obviously [itex]C^{T}.\eta.C=g[/itex]
     
    Last edited: Oct 15, 2012
  5. Oct 15, 2012 #4

    PeterDonis

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    Four lightlike, or two lightlike and two spacelike? The set of lightlike vectors at a given point is only a two-parameter family, isn't it?
     
  6. Oct 15, 2012 #5

    bcrowell

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    Here's a set of four linearly independent, lightlike basis vectors (with components in t,x,y,z order): [1,1,0,0],[1,0,1,0],[1,0,0,1],[1,-1,0,0]

    I checked that they were linearly independent with the following Maxima code:

    Code (Text):
    m:matrix([1,1,0,0],[1,0,1,0],[1,0,0,1],[1,-1,0,0]);
    determinant(m);
    I think it's probably enough to know, at any given point, the light cone and one nonvanishing inner product (which fixes the conformal factor). I.e., basically an observer can determine the metric using a single clock.
     
  7. Oct 15, 2012 #6

    PeterDonis

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    I get the same Maxima result that you do, but do these four vectors span the entire space of vectors? I'm having trouble expressing a vector like [0,0,1,0] or [0,0,0,1] as a linear combination of these vectors, and I think it's because it can't be done, though I haven't had time to work up a proof.
     
  8. Oct 15, 2012 #7

    bcrowell

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    If you have four linearly independent vectors in a four-dimensional vector space, they span the whole space.
     
  9. Oct 15, 2012 #8
    At some point I'll check to see if I can make your functions equivalent to Wikipedia's, but in the meantime, do you have any idea what's up with Wikipedia? It seems to clearly be wrong.

    Also, I thought the matrices had to be actual inverses, as in their product is the Kronecker Delta, not the flat metric. I mean, if you contract two linear transformations with each other (one index each), you get a linear transformation, not a (0,2) tensor, so it wouldn't make any sense for their product to equal the flat metric.
     
  10. Oct 15, 2012 #9

    Mentz114

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    Try this,

    Code (Text):

    m1:matrix([1,1,0,0]);
    m2:matrix([1,0,1,0]);
    m3:matrix([1,0,0,1]);
    m4:matrix([1,-1,0,0]);
    a*m1+b*m2+c*m3+d*m4;
    /** solution for [0,1,0,0] ***/
    algsys([d+c+b+a,a-d-1],[a,b,c,d]);
     
    and it seems [itex]a=1,b=1,c=-2,d=0[/itex]

    is a solution for [0,1,0,0].
     
  11. Oct 15, 2012 #10

    Mentz114

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    Yes, you're probably right about the product being the delta. I'll have to think about it and check a script but it's rather late ... Same goes for the Wiki page.

    [Edit] The sign of F[0,0] is (was) wrong. If F[0,0] = -1 then the product is the Kronecker delta. Apologies. I've edited the post.
     
    Last edited: Oct 15, 2012
  12. Oct 15, 2012 #11

    PeterDonis

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    That's actually not one of the vectors I was concerned about; I was concerned about [0,0,1,0] and [0,0,0,1]. However, I see now that, for example,

    - 1/2 * ([1,1,0,0] + [1,-1,0,0]) + [1,0,1,0] = [0,0,1,0]

    and a similar solution obviously exists for [0,0,0,1], so I now agree that the vectors bcrowell gave span the entire spacetime (since it is obvious how to make [1,0,0,0] and [0,1,0,0] in the basis bcrowell gave, so we have a complete expression for an ordinary basis, with one timelike and three spacelike vectors, in terms of bcrowell's basis).

    This is very interesting to me because I had always thought that a basis could have at most two lightlike vectors (with the other two spacelike). The intuition underlying that belief basically went like this: in ordinary light-cone coordinates (u, v, y, z), where u = t - x and v = t + x, (u, v) obviously span the same subspace as (t, x), and that subspace is orthogonal to the (y, z) subspace. The standard physical interpretation of this is that the (t,x) / (u, v) subspace describes the possible directions of light rays emitted at the origin, and the (y,z) subspace describes the possible polarizations of those light rays (this is probably a somewhat loose way of stating it).

    I had thought that *any* basis would have to be divisible into orthogonal subspaces by pairing up coordinates in this way, but obviously that's not the case. In bcrowell's basis, we have (u, v) as above, and two new null coordinates which I'll call q = t + y and r = t + z. The key point is that the (q, r) subspace is *not* orthogonal to the (u, v) subspace, and in fact there does not appear to be *any* way of splitting up the spacetime into orthogonal subspaces by pairing up coordinates from (u, v, q, r).
     
  13. Oct 16, 2012 #12

    PeterDonis

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    On looking into bcrowell's 4-null-vector basis further, it gets even more interesting. First, just to be explicit, some definitions. The new basis vectors are:

    [tex]\partial_u = \partial_t - \partial_x[/tex]

    [tex]\partial_v = \partial_t + \partial_x[/tex]

    [tex]\partial_q = \partial_t + \partial_y[/tex]

    [tex]\partial_r = \partial_t + \partial_z[/tex]

    The corresponding basis 1-forms are (I'm using a +--- metric signature; see the line element below):

    [tex]du = dt + dx[/tex]

    [tex]dv = dt - dx[/tex]

    [tex]dq = dt - dy[/tex]

    [tex]dr = dt - dz[/tex]

    (Note the sign changes when we lower indexes.)

    The Minkowski line element in the standard (t,x,y,z) coordinates is:

    [tex]ds^2 = dt^2 - dx^2 - dy^2 - dz^2[/tex]

    This line element transforms into the (u,v,q,r) coordinates (the simplest way is to invert the basis 1-forms given above, to find dt, dx, dy, dz in terms of du, dv, dq, dr and then substitute into the above line element) as:

    [tex]ds^2 = \left( du + dv \right) \left( dq + dr \right) - \frac{1}{2} du^2 - \frac{1}{2} dv^2 - dq^2 - dr^2[/tex]

    This line element is interesting because it shows that, unlike the basis vectors (all of which are null), the basis 1-forms are all spacelike; that is, a line element with just one of du, dv, dq, dr nonzero is spacelike.
     
  14. Oct 16, 2012 #13

    George Jones

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    Careful. Suppose

    [tex]\partial_u = \partial_t[/tex]
    [tex]\partial_v = \partial_x[/tex]
    [tex]\partial_q = \partial_y[/tex]
    [tex]\partial_r = \partial_z[/tex]

    Does it then follow that [itex]dv = -dx[/itex] (etc.)?

    I hope to follow with a much longer post.
     
  15. Oct 16, 2012 #14

    PeterDonis

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    I see what you mean; I wasn't being consistent in how I defined what "basis 1-forms" were. I'll have to re-think this while I await the longer post.
     
  16. Oct 16, 2012 #15

    George Jones

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    In Minkowski spacetime, define new coodinates by

    [tex]
    \begin{align}
    u & = \frac{1}{2} \left( t - x - q - r \right)\\
    v & = \frac{1}{2} \left( t + x - q - r \right)\\
    q &= y\\
    r &= z.
    \end{align}
    [/tex]
    Equivalently (I think),

    [tex]
    \begin{align}
    t & = u + v + q + r\\
    x & = -u + v\\
    y &= q\\
    z &= r.
    \end{align}
    [/tex]
    Then,

    [tex]
    \begin{align}
    \frac{\partial}{\partial u} &= \frac{\partial t}{\partial u} \frac{\partial}{\partial t} + \frac{\partial x}{\partial u} \frac{\partial}{\partial x} + \frac{\partial y}{\partial u} \frac{\partial}{\partial y} + \frac{\partial z}{\partial u} \frac{\partial}{\partial z}\\
    \frac{\partial}{\partial u} &= \frac{\partial}{\partial t} - \frac{\partial}{\partial x}
    \end{align}
    [/tex]
    Similarly, I think that the other vectors work out to what Ben and Peter wrote.

    From the second group of four equations above, it follows that

    [tex]
    \begin{align}
    dt & = du + dv + dq + dr\\
    dx & = -du + dv\\
    dy &= dq\\
    dz &= dr,
    \end{align}
    [/tex]
    and

    [tex]
    dt^2 - dx^2 - dy^2 - dz^2= 4dudv+2dudq+2dudr+2dvdq+2dvdr+2dqdr.
    [/tex]
    The coordinates u, v, q, r are all lightlike. q and r are lightlight even though y and z are spacelike, and q = y and r = z! This is because, for example, an r coordinate line involves holding u, v, and constant, and a z coordinate line involves holding t, x, and y constant.
     
  17. Oct 16, 2012 #16

    PeterDonis

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    George, thanks for your post, it helps a lot in clearing up what's needed to fully define the correspondence between one basis and another.

    A small point; strictly speaking, shouldn't y and z appear on the RHS of the first and second definitions, instead of q and r? Thus:

    [tex]
    \begin{align}
    u & = \frac{1}{2} \left( t - x - y - z \right)\\
    v & = \frac{1}{2} \left( t + x - y - z \right)\\
    q &= y\\
    r &= z.
    \end{align}
    [/tex]

    The inverse definitions would still work out the same, since y + z = q + r, so you can substitute the latter for the former in the equation for t.

    Yes, I see, this is the same sort of thing that happens with the r coordinate in the Painleve vs. the Schwarzschild chart for the region inside the horizon.
     
  18. Oct 16, 2012 #17

    Mentz114

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    The metric can be transformed using Ben's matrix, 'm'.

    [tex]
    m.\eta.m^{T} = \begin{pmatrix}0 & -1 & -1 & -2 \\ -1 & 0 & -1 & -1 \\ -1 & -1 & 0 & -1\\ -2 & -1 & -1 & 0 \end{pmatrix}
    [/tex]
    which looks like George's transformed line element except all the signs are reversed because I'm using signature (-,+,+,+).
     
    Last edited: Oct 16, 2012
  19. Oct 17, 2012 #18

    pervect

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    If we let p = a column vector of [itex]\left[ \partial_t, \partial_z, \partial_r, \partial_\phi \right] [/itex]

    and q = a column vector of [itex]\left [dt, dz, dr, d\phi \right] [/itex]

    Then we are saying

    column_vector_of e = (A) p
    column_vector_of [itex]\sigma[/itex] = (B) q

    Now p and q are already dual, because of the cylindrical coordinates are orthogonal. In matrix notion, I believe the way we express this is:

    [itex]q^T \cdot p = I[/itex]

    SImilarly for the e's and the [itex]\sigma[/itex]'s to be orthogonal, we want

    [itex] (B \sigma)^T (A e) = \sigma^T B^t A e = I[/itex]

    So we want B(transpose) A to be the identy matrix, I think...
     
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