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What is a null coordinate?

  1. Apr 28, 2008 #1

    DrGreg

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    Coordinates are sometimes described as "null coordinates". An example in SR is the coordinate [itex]u = x - ct[/itex]. Another example is one of the coordinates in the Eddington-Finkelstein metric. But I've never seen an explicit rigorous definition of a null coordinate. The defining property seems to be that any hypersurface of constant coordinate should possess a null tangent vector at every event within it.

    My first thought was to say that [itex]x^0[/itex] is a null coordinate if the tangent vector [itex](dx^0, 0, 0, 0)[/itex] is null. But it didn't take long, considering the above examples, to realise this was wrong. Should I, instead, be considering the cotangent covector [itex](dx_0, 0, 0, 0)[/itex]? If I haven't made a mistake in my maths, this seems to work for the examples above, but I can't convince myself it will always work.

    And then what about "timelike" coords and "spacelike" coords? Can they be defined in a similar way?

    If I have an explicit equation for a metric, e.g.

    [tex]ds^2 = -\left(1-\frac{2M}{r} \right) du^2 - 2 du dr + r^2 \left(d\theta^2 + \sin^2 \theta\, d\phi^2 \right) [/tex]​

    is there a quick way of inspecting this to classify each of the four coordinates as timelike, spacelike, or null?
     
  2. jcsd
  3. Apr 28, 2008 #2
    You are correct. A null coordinate [tex]f[/tex] is one for which the 1-form [tex]df[/tex] has zero norm. In other words, the hypersurface of constant [tex]f[/tex] posesses a null "normal 1-form". The 1-form is what you might call a covariant vector. Thus [tex]df[/tex] has components [tex] (\tfrac{\partial f}{\partial x^0}, \tfrac{\partial f}{\partial x^1}, \tfrac{\partial f}{\partial x^1}, ...)[/tex]

    To quickly classify coordinates, one usually does it by inspection. Otherwise, of course, you could first diagonalize the metric and work from there.

    The rough 'n' ready way of finding null/spacelike/timelike vectors is simply to look at tangent vectors to curves that are perpendicular to the hypersurface. Hence, you want to solve, say,
    [tex]0 = ds^2 = -\left(1-\frac{2M}{r} \right) \frac{du}{d\lambda}^2 - 2 \frac{du}{d\lambda} \frac{dr}{d\lambda}[/tex],
    hence

    [tex]-\left(1-\frac{2M}{r} \right) \frac{du}{d\lambda} - 2 \frac{dr}{d\lambda} = 0[/tex]
    fits the bill.
    So take the vector to be
    [tex]\vec{v} = (-2, \left(1-\frac{2M}{r} \right) , 0, 0)[/tex]
    and find the corresponding covariant vector [tex]v_\alpha[/tex].

    I suspect that it'll have components [tex]v_\alpha = (?, 0, 0, 0)[/tex] since [tex]u[/tex] looks to me like a null coordinate.

    ** The only caveat of course is that null vectors are orthogonal to other null vectors. But looking back at the problem, it is obvious that the 1-form [tex]du[/tex] is null, while the 1-form [tex]dr[/tex] is spacelike. I tried to motivate my explanation without actually making use of the geometrical interpretation of a 1-form, which is difficult. I therefore invite you to read up on vectors and 1-forms in differential geometry.
     
    Last edited: Apr 28, 2008
  4. Apr 29, 2008 #3

    DrGreg

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    Thanks very much. Your reply was very helpful and gave me the right terminology to search further. I dug out my quarter-read copy of Roger Penrose's Road to Reality and was able to make some sense out of 1-forms from that. Though I've always known the coordinate definition of a covector as [itex]v_a = g_{ab} v^b[/itex], I've never had much geometrical insight into what this meant, and now I understand better.

    In the case of an "almost diagonal" metric like the one I quoted, it's not very difficult to calculate the "inverse metric" [itex]g^{ab}[/itex] and then you can just read off the diagonal entries to determine timelike/null/spacelike coordinates.

    I'm assuming that the definition of (locally) timelike/null/spacelike works for any scalar field [itex]f[/itex], not just a "coordinate".

    After reading all this, it occurred to me that the "dual" of the usual metric "definition" of [itex]ds^2 = g_{ab} dx^a dx^b[/itex] must be the operator

    [tex]g^{ab}\frac{\partial}{\partial x^a} \frac{\partial}{\partial x^b}[/tex]​

    which, if we had a Euclidean metric, I would call the Laplacian [itex]\nabla^2[/itex] or [itex]\triangle[/itex]. Wikipedia calls this the d'Alembertian in the case of (flat) Minkowski space but makes no mention of any other metric. Is that terminology used in GR?
     
  5. Apr 29, 2008 #4
    The Laplacian is different from the d'Alembertian since the later contains a temporal term. For details please see - http://en.wikibooks.org/wiki/Electrodynamics/Four-Vectors#The_d.27Alembertian

    The operator you posted is the d'Alembertian in an arbitrary coordinate system and thus holds in GR. The metric in that expression, g, represents an arbitrary metric. This is easily seen from the strong equivalence principle which requires replacing the Minkowski metric with an arbitrary metric in arbitrary (but physically meaningful) coordinate system.

    Pete
     
    Last edited: Apr 29, 2008
  6. Apr 29, 2008 #5
    Hold your horses!

    The inverse metric is the (2,0) tensor
    [tex]
    g^{ab} \frac{\partial}{\partial x^a} \otimes \frac{\partial}{\partial x^b}
    [/tex],
    meaning that the action on two 1-forms [tex]\tilde{\mu}[/tex] and [tex]\tilde{\nu}[/tex] is
    [tex]
    g^{ab} \frac{\partial}{\partial x^a} \otimes \frac{\partial}{\partial x^b} (\tilde{\mu},\tilde{\nu}) = g^{ab} \frac{\partial}{\partial x^a} \otimes \frac{\partial}{\partial x^b} (\tilde{\mu},\tilde{\nu}) = g^{ab} \left\langle \frac{\partial}{\partial x^a}, \tilde{\mu} \right\rangle \left\langle \frac{\partial}{\partial x^b}, \tilde{\nu} \right\rangle = g^{ab} \mu_a \nu_b
    [/tex].

    Unfortunately this is one place where the [tex]\otimes[/tex] is kind of important, because otherwise the thing you wrote down *looks* like a Laplacian (d'Alembertian, doesn't really matter... call it the Laplace-Beltrami if you will).

    The Laplacian is an altogether different beast, namely [tex]\nabla_a \nabla^a f = \frac{1}{\sqrt{|g|}} \partial_a \left ( \sqrt{|g|}\, \partial^a f\right)[/tex].

    Thus
    [tex]g^{ab} \frac{\partial^2 f}{\partial x^a \partial x^b}[/tex] is not a coordinate invariant expression and means something altogether different (and happens to be meaningless)

    Cheers
     
  7. May 1, 2008 #6

    DrGreg

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    Fair enough. I don't entirely follow every detail of your reply, but don't worry, I don't want you to elaborate. I'm aware that there are different ways of differentiating in GR and I'll get to grips with the technicalities of that later.

    The point of my supplementary question was this: usually when the components of a specific metric are to be specified, it's conventional to write the equation [itex]ds^2 = g_{ab} dx^a dx^b[/itex] instead of the 4 × 4 matrix [itex]g_{ab}[/itex]. Is there a convenient equation you could use instead of the 4 × 4 matrix [itex]g^{ab}[/itex]? And if there is, does anyone actually use it in practice?
     
  8. May 1, 2008 #7
    I'll try to elaborate without elaborating =) The answer is yes, and it is written
    [tex] g^{ab} \frac{\partial}{\partial x^a} \otimes \frac{\partial}{\partial x^b} [/tex]

    The point is, though, that this isn't actually a differential operator. It is a tensor formed out of the tensor product of basis vectors. In differential geometry, upon which GR is based, there is a natural one-to-one correspondence between (contravariant) vectors and directional derivatives. Similarly, there is a natural one-to-one correspondence between (covariant) vectors ([tex]dx[/tex] and the like) and gradients.

    If you'd like, I could write the inverse metric as [itex] g^{ab} \vec{e}_a \otimes \vec{e}_b [/itex]. In matrix notation, you might want to write this as
    [tex]g^{-1} = \sum_{ab}(g^{-1})^{ab} \vec{e}_a {\vec{e}_b}^{\,T}[/tex],
    where you might want to write [tex]\vec{e}_1 = (1,0,0,0)[/tex], etc.

    I don't know if any of this has helped you. Does anyone actually do this in practice? Sometimes. Most of the time you work with [tex]g_{ab}[/tex] and just compute [tex]g^{ab}[/tex] as needed. I think the best path is to understand this admittedly cumbersome notation and the notion of vectors and 1-forms first, and then when you're more comfortable with what tensors do, to just use index notation. The metric is sort of the exception because you usually want to communicate it's components, and writing it in the form [tex]ds^2 = ...[/tex] is efficient.
     
  9. Aug 4, 2010 #8
    My question about null-time-space coordinates:
    are there any restriction about the distribution of coordinates?
    I mean it is possible to use four null-coordinates to describe a spacetime
    or two time- and two space-coordinates for example?
     
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