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

1. Jul 27, 2009

### symplectic

Hi

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

3. The attempt at a solution

No idea!!!

Last edited: Jul 27, 2009
2. Jul 27, 2009

### turin

Consider this kind of object:

nνναβγaαbβcγ

where n is perpendicular to the hyperplane that contains a,b,c.

3. Jul 28, 2009

### symplectic

Okey, and how can I see it as a 1-form?!!!

4. Jul 28, 2009

### turin

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.

5. Jul 28, 2009

### symplectic

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

6. Jul 28, 2009

### Ben Niehoff

Writing it out in full, you would have

$$\mathbf{n} = n_\mu \mathrm{dx}^\mu = \epsilon_{\mu\nu\lambda\sigma} a^\nu b^\lambda c^\sigma \mathrm{dx}^\mu$$

where $\mathrm{dx}^\mu$ are the basis 1-forms. Beyond that, I also don't know what you mean by "see it as a 1-form".

7. Jul 29, 2009

### symplectic

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 !!!

8. Jul 29, 2009

### turin

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.

ε(⋅,⋅,⋅,⋅)

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

ε(⋅,dx,dy,dz)=~dV

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.

η(⋅,⋅)

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

η(⋅,dx)=~dx

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.

9. Jul 29, 2009

### symplectic

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 !!!