Divergence theorem

1. Oct 5, 2009

sszabo

In h.m. schey, div grad curl and all that, II-25:
Use the divergence theorem to show that
$$\int\int_S \hat{\mathbf{n}}\,dS=0,$$
where $$S$$ is a closed surface and
$$\hat{\mathbf{n}}$$ the unit vector
normal to the surface $$S$$.
How should I understand the l.h.s. ?
Coordinatewise? The r.h.s. is not 0, but zero vector?

2. Oct 5, 2009

wofsy

the divergence is the total flux through the region. it is not a vector.
your calculation of divergence integrates the inner product of the flow with the unit normal to the surface. this inner product is a scalar.

3. Oct 6, 2009

sszabo

Right. However in the l.h.s. we have a vector $$\hat{\mathbf{n}}$$ and not an inner product which would be a scalar. I can integrate only vector-variable scalar-valued functions on a surface and not a vector-valued function.
So what does it mean?

4. Oct 6, 2009

slider142

If the surface is closed, intuitively, the unit normals should "sum" to the 0 vector (generalizing the fact that on a "nice" closed curve, summing the normals should lead to no displacement). This should then be proven properly.

5. Oct 6, 2009

wofsy

The integral of the unit normal is just a vector, the result of a Riemann sum of vectors.
But this is not the divergence theorem.

6. Oct 7, 2009

sszabo

I see. Yes, indeed, the answer -intuitively- is the 0 vector.
It means
$$\int\int_S \hat{\mathbf{n}}\,dS=\left( \int\int_S n_1\,dS,\int\int_S n_2\,dS,\int\int_S n_3\,dS\right),$$
where $$\hat{\mathbf{n}}=(n_1,n_2,n_3)$$.
Now $$n_1=\left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle$$ and here we can use the divergence theorem to obtain
$$\int\int_S n_1\,dS=\int\int_S \left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{i}\, dV=0,$$
because $$div\, \mathbf{i}=0$$. Nice result.
Thanks for your helps.

Last edited: Oct 7, 2009
7. Oct 7, 2009

wofsy

no. your your three integrals are integrals of vectors not inner products.

8. Oct 7, 2009

sszabo

Sorry, I absolutely don't understand what you think. Please, give more details, for example point out the wrong step in the calculations. Thanks.

9. Oct 7, 2009

wofsy

I'm just saying that n1 n2 and n3 are inner products and thus are scalars. Originally you wanted to integrate a vector. Splitting into components still requires keeping the basis vectors in the integral. I don't think you did this. Instead I thing you substituted scalars.Unless I don't understand your post.

10. Oct 11, 2009

sszabo

Agree.
Agree.
Of course.
Certainly a similar calculation gives
$$\int\int_S n_2\,dS=\int\int_S \left\langle \mathbf{j},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{j}\, dV=0,$$
and
$$\int\int_S n_3\,dS=\int\int_S \left\langle \mathbf{k},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{k}\, dV=0$$
The final result, the value of the original integral is $$\mathbf{0}$$.

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