Recent content by symplectic

This helps me a lot, I guess I got it, in GR the problem is not to calculate the "value" of the Einstein-Hilbert action, but our aim is to do a Lagrangian formulation, i.e: write out a functional of the fields such that the field configuration which extremize the later is the solution of the...

I do not know if the problem is with me but ... I know all that! but even with that I cannot understand why the integral is not covariant if one does not use the covariant volume form !

Why should it be so?

Hi, Sorry I'm not convinced by simply saying "make sens" :frown:, what I need is a mathematical explanation that force the volume element to be covariant in order that the integral becomes well defined. Another example: Here is an explicit example: I = \int d^{d}x f(x) I' = \int d^{d}x'...

@Dick: I know that nothing will change, and that is precisely the problem for me: if nothing will change, why do we introduce a covariant volume element? @nicksauce: For the book I don't have it. For the link, they simply say that we have to multiply by the square root of the metric determinant...

Hi, We use as an integration form in Riemannian geometry the covariant \int \sqrt{g}d\Omega I understand how this is invariant under an arbitrary change of coordinates (both Jacobian and metric square root transformation coefficient will cancel each other), what I don't understand is why don't...

Okey, you succeeded ... I don't see it anymore as a vector :smile: If I understood, writing n as : nμ = εμνρσdxνdxρdxσ Tells us that it is in fact a 1-form, and not a vector !

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...

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:

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

Hi Homework Statement In the ADM formalsim, we need at some point to see the normal vector (to the 3D surfaces) as a 1-form. Homework Equations How can we (simply) see that? (Please no Frobenius theorem, otherwise, with more explanation about this theorem) The Attempt at a Solution No idea!