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Homework Help: ADM: how to see the normal vector proportional to a 1-form?

  1. Jul 27, 2009 #1

    1. The problem statement, all variables and given/known data

    In the ADM formalsim, we need at some point to see the normal vector (to the 3D surfaces) as a 1-form.

    2. Relevant equations

    How can we (simply) see that? (Please no Frobenius theorem, otherwise, with more explanation about this theorem)

    3. The attempt at a solution

    No idea!!!
    Last edited: Jul 27, 2009
  2. jcsd
  3. Jul 27, 2009 #2


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    Consider this kind of object:


    where n is perpendicular to the hyperplane that contains a,b,c.
  4. Jul 28, 2009 #3
    Okey, and how can I see it as a 1-form?!!!
  5. Jul 28, 2009 #4


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    It has a lower index. The convention that I learned is that upper indices are n-tensor indices, and lower indices are n-form indices. You can reverse this convention if you like. The main point is that the three sets of coefficients on the RHS that contract with the Levi-Civita symbol are 1-tensors, and their indices are in the opposite position to the coefficient on the LHS, which is a 1-form (coefficient).

    Maybe I don't know what you mean by "see it as a 1-form". If so, I appologize for the distraction. In the language of MTW, when a 1-tensor is contracted with a 1-form, the result sort-of tells you how many 1-form "surfaces" are pieced by the 1-tensor. In that description, the 1-tensor (coeffients with upper index) is sort-of a line, and the 1-form (coefficients with lower index) is sort-of a series of surfaces.
  6. Jul 28, 2009 #5
    Thank you turin,

    Yes, I guess that this is a manner to see it but .... I wonder how to do if there were no convention!!!

    I still wait an answer which convince me :wink:
  7. Jul 28, 2009 #6

    Ben Niehoff

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    Writing it out in full, you would have

    [tex]\mathbf{n} = n_\mu \mathrm{dx}^\mu = \epsilon_{\mu\nu\lambda\sigma} a^\nu b^\lambda c^\sigma \mathrm{dx}^\mu[/tex]

    where [itex]\mathrm{dx}^\mu[/itex] are the basis 1-forms. Beyond that, I also don't know what you mean by "see it as a 1-form".
  8. Jul 29, 2009 #7
    Hi againe,
    OK, here is what I found in almost all papers that I read:
    We have a 3d surface ∑ embedded in a 4d space M via the embedding Xt defined by:
    Xt(x) = X(t,x) where X is a diffeomorphism X : σ x ℝ → M (x are coordinates on ∑ and X are those of M) so one can think about the surface as defined by : f(X) = t = constant so that:
    0 = limε→0 [f (Xt (x + εb) − f (Xt (x))]/ε = baXμ,a(f)|X = Xt(x) for any tangential b of σ = X(∑) on x .... so that on can see n as : nμ = F f (a 1-form)
    My problem is that I don't understand very well what we mean by all that !!!
  9. Jul 29, 2009 #8


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    symplectic, I think you missed my point. The convention does not matter. The RHS has 1-tensor coefficients, and the LHS has a 1-form coefficient. The only reason that I mentioned the convention was to clarify that the position of the index tells you that it is a 1-form.

    Here's a different (more explicit) way to look at it. A 3-surface is basically a volume. A differential "area" of the 3-surface would have the form dxdydz, right. So, generalize these differentials to 1-tensors: dxαdyβdzγ, where these differentials are some specific directions in space-time. This product has units of 3-surface area (volume), but it is a 3-tensor, not a 1-form. You convert it to a 1-form by contracting with the Levi-Civita symbol. This is similar to converting a 1-tensor to a 1-form by contracting with the metric.

    I don't know if this is standard notation, but, instead of using indices, imagine that the Levi-Civita symbol is a function that "eats" n-tensors and spits out (4-n)-forms.


    So, in this case, the Levi-Civita symbol "eats" three 1-tensors, dx, dy, and dz.


    The tilde in front is my way to identify the object as a n-form. 1 argument of the Levi-Civita symbol is left open, so the result is a 1-form.

    This is similar to the idea of the metric "eating" an n-tensor, resulting in a (2-n)-form.


    So, if the metric "eats" a vector (1-tensor), the result is a 1-form.


    In the index approach, this is called "lowering" the index.

    One more point to make: ε(dt,dx,dy,dz) is Lorentz invariant. This is true regardless of what dt, dx, dy, and dz represent, as long as they are 1-tensors.
  10. Jul 29, 2009 #9
    Okey, you succeeded ... I don't see it anymore as a vector :rofl:
    If I understood, writing n as :
    nμ = εμνρσdxνdxρdxσ
    Tells us that it is in fact a 1-form, and not a vector !!!
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