divergence theorem

[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?

[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.

[sszabo]

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.
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?

[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.

[wofsy]

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?

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.

[sszabo]

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.
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.

[wofsy]

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.

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

[sszabo]

no. your three integrals are integrals of vectors not inner products.
Sorry, I absolutely don't understand what you think. Please, give more details, for example point out the wrong step in the calculations. Thanks.

[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.

[sszabo]

I'm just saying that n1 n2 and n3 are inner products and thus are scalars.
Agree.

Originally you wanted to integrate a vector.
Agree.
Splitting into components still requires keeping the basis vectors in the integral.

Of course.
I don't think you did this.
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

Instead I thing you substituted scalars.Unless I don't understand your post.
The final result, the value of the original integral is \mathbf{0}.