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Divergence theorem

  1. Oct 5, 2009 #1
    In h.m. schey, div grad curl and all that, II-25:
    Use the divergence theorem to show that
    [tex]\int\int_S \hat{\mathbf{n}}\,dS=0,[/tex]
    where [tex]S[/tex] is a closed surface and
    [tex]\hat{\mathbf{n}}[/tex] the unit vector
    normal to the surface [tex]S[/tex].
    How should I understand the l.h.s. ?
    Coordinatewise? The r.h.s. is not 0, but zero vector?
     
  2. jcsd
  3. Oct 5, 2009 #2
    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.
     
  4. Oct 6, 2009 #3
    Right. However in the l.h.s. we have a vector [tex]\hat{\mathbf{n}}[/tex] 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?
     
  5. Oct 6, 2009 #4
    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.
     
  6. Oct 6, 2009 #5
    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.
     
  7. Oct 7, 2009 #6
    I see. Yes, indeed, the answer -intuitively- is the 0 vector.
    It means
    [tex]
    \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),
    [/tex]
    where [tex]\hat{\mathbf{n}}=(n_1,n_2,n_3)[/tex].
    Now [tex]n_1=\left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle[/tex] and here we can use the divergence theorem to obtain
    [tex]\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,
    [/tex]
    because [tex]div\, \mathbf{i}=0[/tex]. Nice result.
    Thanks for your helps.
     
    Last edited: Oct 7, 2009
  8. Oct 7, 2009 #7
    no. your your three integrals are integrals of vectors not inner products.
     
  9. Oct 7, 2009 #8
    Sorry, I absolutely don't understand what you think. Please, give more details, for example point out the wrong step in the calculations. Thanks.
     
  10. Oct 7, 2009 #9
    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.
     
  11. Oct 11, 2009 #10
    Agree.
    Agree.
    Of course.
    Certainly a similar calculation gives
    [tex]
    \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,
    [/tex]
    and
    [tex]
    \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
    [/tex]
    The final result, the value of the original integral is [tex]\mathbf{0}[/tex].
     
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