Is This Special Case of Stokes Theorem True?

[paweld]

Do you agree that the following identity is true:
$$\int_S (\nabla_\mu X^\mu) \Omega = \int_{\partial S} X \invneg \lrcorner \Omega$$
where $$\Omega$$ is volume form and $$X\invneg \lrcorner \Omega$$
is contraction of volume form with vector X.

 

[arkajad]

From Loomis, Sternberg, "Advanced Calculus", p. 447, where $$\Omega$$ is the volume form corresponding to density $$\rho$$:

 

[jasomill]

Under the appropriate hypotheses, Stokes' theorem says that

$$\int_S d\omega = \int_{\partial S} \omega.$$

By definition,

$$(\mathrm{div}\ X) \Omega = d(X \lrcorner \Omega),$$

so

$$\int_S (\mathrm{div}\ X) \Omega = \int_{\partial S} X \lrcorner \Omega.$$

Assuming your $$\nabla_\mu X^\mu$$ is the divergence of $$X$$, isn't this just the Divergence Theorem?

Sternberg's $$\iota^*$$ is just pullback along the inclusion map, i.e., restriction of a form on $$M$$ to the n-1 dimensional submanifold $$\partial D$$; instead of a manifold-with-boundary, he's defined $$D \subset M$$ as a domain in an n-dimensional manifold $$M$$. In case you're interested, on p. 109 of his Lectures on Differential Geometry, he gives Stokes' Theorem in the usual way, and on p.119, he gives this same "Stokes' theorem for domains."

Cheers,
Jason

 

[paweld]

Thanks for your replays. Nabla symbol which I used means covariant derivative
and its trace with vector happens to be just divergence of this vector, so my
identity is true.

 

[arkajad]

Thanks for your replays. Nabla symbol which I used means covariant derivative
and its trace with vector happens to be just divergence of this vector, so my
identity is true.

Then Omega must be the volume form of the Riemannian metric for which Nabla is the covariant derivative. Without specifying these data your formula may be confusing.

 

[paweld]

Then Omega must be the volume form of the Riemannian metric for which Nabla is the covariant derivative. Without specifying these data your formula may be confusing.

Yes, you are right.

 

[Phrak]

Under the appropriate hypotheses, Stokes' theorem says that

$$\int_S d\omega = \int_{\partial S} \omega.$$

In my understanding, omega must be an (n-1)-form on an n-dimensional differential manifold, and so Stokes theorem does not apply to the question involving integrals of a vector and scalar on, say, manifolds of dimension 4. Is this incorrect?

 

[jasomill]

In my understanding, omega must be an (n-1)-form on an n-dimensional differential manifold, and so Stokes theorem does not apply to the question involving integrals of a vector and scalar on, say, manifolds of dimension 4. Is this incorrect?

A volume form on an n-dimensional manifold is, by definition, a nonvanishing n form. Thus $$X\lrcorner\Omega$$, as the contraction of a vector field with an n form, is an n-1 form.

In other words, having fed it one vector field, it still expects n-1 more.

The exterior derivative takes k forms to k+1 forms, so $$d(X\lrcorner\Omega)$$ is another n-1+1=n form. The equation

$$d(X\lrcorner\Omega) = (\mathrm{div}\ X)\Omega$$

was not an identity, but my (perfectly sensible) definition of the divergence with respect to this volume form. In other words, since the top exterior power is 1-dimensional, $$d(X\lrcorner\Omega)$$ must be a scalar multiple of $$\Omega$$. I defined the divergence to be this scalar (I didn't write something like $$\mathrm{div}_\Omega\ X$$ because there was only one volume form in sight).

It's not hard to verify that this agrees with both the classical definition in $$\mathbb{R}^n$$, and with the original poster's definition in terms of the metric covariant derivative on a Riemannian manifold.

Perhaps it will clarify things to point out that, in the Riemannian case, if we take $$\omega$$ to be the induced volume form on the boundary and $$N$$ to be the outward-pointing unit normal vector field, we also have

$$\int_S (\mathrm{div}\ X) \Omega = \int_{\partial S} \langle X, N\rangle \omega.$$

Cheers,
Jason

 

[arkajad]

Perhaps it is also useful to notice that, looking at

$$d(X\lrcorner\Omega) = (\mathrm{div}\ X)\Omega$$

on the left there is a "d" that will also differentiate the coefficients of $$\Omega$$ in a coordinate basis.
Then one may wonder where are these derivatives on the RHS? The tricky part is that they hide themselves in the Christoffel symbols entering the definition of the covariant divergence!