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Can I use generalized Stokes' theorem in Faraday's law?

  1. Sep 11, 2014 #1
    It is a nonsense use the generalized Stokes' theorem in right side of Faraday's Law?

    we know this is true...


    [itex]\displaystyle \oint_{\partial \Sigma}\vec{E}\cdot dl=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}\Rightarrow \int_{\Sigma}\vec{\nabla}\times\vec{E}\cdot d\vec{A}=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}[/itex]

    i was thinking if we consider [itex] \Sigma[/itex] a smooth manifold (i really dont know if i can do that)... can i do this?

    [itex]\displaystyle \oint_{\partial \Sigma}\vec{E}\cdot dl=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}\Leftrightarrow \int_{\partial \Sigma}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} \Rightarrow \int_{\Sigma}\underline{d}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} [/itex]

    where [itex]\int_{\partial \Sigma}E[/itex] is the integration in boundary of [itex] \Sigma[/itex] manifold and [itex] \underline{d}[/itex] the exterior derivative

    i'm afraid i have done a mess with the concepts, but the doubt is in my head and it dont let me sleep...
     
    Last edited: Sep 11, 2014
  2. jcsd
  3. Sep 11, 2014 #2
    So, your problem is actually that you don't know whether you can assume Sigma to be a manifold? I would not worry too much about that, but maybe it will give your head some peace that Stokes' theorem can also be formulated for chains on manifolds (sadly, the only book that I know that proves this for chains on manifolds is the classical mechanics book by V. Arnold).

    In fact, if E is the 1-form associated with E (vector field, this time, to be integrated over a line), dE is just the 2-form associated with rot(E) (vector field to be integrated over a surface).
     
  4. Sep 11, 2014 #3
    sorry, I meant curl(E) (rot is the German name).
     
  5. Sep 11, 2014 #4
    So, once I make sure Sigma is a manifold the expression [itex] \int_{\Sigma}\underline{d}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} [/itex] is valid?
     
  6. Sep 11, 2014 #5
    Yes, if E is the 1-form associated with the electric field.

    And you don't really have to make sure that Sigma is a manifold (a manifold with boundary, actually). As I said, it is possible to take chains instead, which is more general and fits more to the needs of physics (sometimes you want to integrate over something with corners or edges, like a cylinder or, in this case, a rectangle - strictly, that's not a manifold then).
     
  7. Sep 11, 2014 #6
    THX A LOT, confirm this is considerably important for me. Once i resolve [itex] \int_{\Sigma}\underline{d}E[/itex] i have to take hodge dual of the result to get the vector field resultant right?
     
  8. Sep 11, 2014 #7
    I don't get it... Why should you? The Hodge dual would turn your 2-form into a 1-form. I mean, you could do this and then apply the inverse of the translation isomorphism that you used to turn E into a 1-form. In fact:

    [itex]\omega^1_{\vec E} := \langle \vec E , \cdot \rangle = \star \omega^2_{\vec E } = \langle \vec E , \cdot \times \cdot \rangle [/itex]

    (I hope, I didn't struggle with the signs)

    As I predicted, the resultant vector field will be the curl of E.
     
  9. Sep 12, 2014 #8
    sorry but i dont familiar with that notation, [itex] \omega^{1}_{\vec{e}}[/itex] are the one-forme associate with E?

    [itex]\langle \vec E , \cdot \rangle[/itex] i really dont understand, i only see that notation in inner product.

    [itex]\star \omega^2_{\vec{E} }[/itex] is the star operator apply in 2-form associate with E?
     
  10. Sep 12, 2014 #9
    Yes.

    It's a shortcut notation for the 1-form defined as follows:

    [itex]\langle \vec E , \cdot \rangle : x \mapsto \langle \vec E, x \rangle[/itex]

    Some people (my professor in linear algebra 2, for example) prefer a dash instead: [itex]\langle \vec E , - \rangle : x \mapsto \langle \vec E, x \rangle[/itex]. The same here:

    [itex]\langle \vec E , \cdot \times \cdot \rangle : (x, y) \mapsto \langle \vec E, x \times y \rangle[/itex]

    Yes, you got it right, but I goofed there.^^ Of course, the star operator needs to be applied to the right term as well in order for the statement to make sense: It shall be an identity between 1-forms, but the right term (without the star operator applied) is a 2-form (the star operator turns it into a 1-form).
     
  11. Sep 12, 2014 #10
    the things are better for me now, but still have some ideias i dont get. The only way to use the generalized stokes theorem in a ordinary vector field is take it one-form right? once i do that i use the theorem and do exterior derivative of this one-form and integrate that in a manifold, the result of that is a 2-form right? so i apply the star operator and get a one-form... that statements are all right?
     
  12. Sep 12, 2014 #11
    The statements are right, more or less:
    The (general) Stokes' theorem is a theorem on forms. Consequently, if you want to deal with vector fields, you have to make forms out of them. However, you can also make a 2-form out of a vector field. Forms are just integrated (a k-form can be integrated over a k-dimensional manifold), but vector fields are not integrated "just the way they are", but you can integrate them along lines or over surfaces. Whether you treat them as a 1-form or a 2-form depends on what you want to do with them. I got a bit irritated when you mentioned the star operator because you don't really need it to understand what follows. You can use it, however. When I read your statements, it comes to me the apprehension that you have never seen how the classical integral theorems of Stokes and Gauss are derived from the general Stokes' theorem, am I right?

    What you are trying to do is more or less exactly proving the classical Stokes' theorem. You might know that there are (in Euclidean R³) the following isomorphisms (X is an open subset):

    [itex] C^\infty (X) = \Omega^0 X[/itex]
    [itex]V(X) \cong \Omega^1 X : F \mapsto \omega^1_F := ( r \mapsto \sum_i F_i(r) dx^i )[/itex]
    [itex]V(X) \cong \Omega^2 X : F \mapsto \omega^2_F := (r \mapsto \tfrac 1 2 \sum_{i, j, k} \epsilon^i{}_{j k} F_i (r) dx^j \wedge dx^k ) = \star \omega^1_F [/itex]
    [itex]C^\infty (X) \cong \Omega^3 X : \Phi \mapsto \omega^3_\Phi := (r \mapsto \Phi (r) \bigwedge_i dx^i) = \star \Phi[/itex]

    These have the following geometrical meaning:

    [itex] \omega^1_F (\xi ) = \sum_i F_i \xi^i = \langle F, \xi \rangle [/itex] (scalar product)
    [itex] \omega^2_F (\xi, \eta ) = \sum_{i, j, k} \epsilon^i{}_{ik} F_i \xi^j \eta^k = \langle F, \xi \times \eta \rangle [/itex]
    [itex] \omega^3_\Phi (\xi, \eta, \zeta ) = \Phi \cdot \langle \xi , \eta \times \zeta \rangle [/itex]

    You are familiar with that?

    Now, assume any vector field [itex]F = \sum_i F_i e_i [/itex] (I know, the notation sucks, because one cannot decide whether one wants to write super- or subscript indices for vector field components, once you want to use them for forms as well...). Let [itex]\omega = \omega^1_F = \sum_i F_i dx^i[/itex]. Then [itex]d \omega = \sum_i d F_i \wedge dx^i = \sum_i (\sum_j \partial_j F_i dx^j ) \wedge dx^i = \sum_{i, j, k, l} \tfrac 1 2 (\delta_k^j \delta_l^i - \delta_l^j \delta_k^i) \partial_j F_i dx^k \wedge dx^l [/itex]
    [itex]= \sum_{i, j, k, l, m} \tfrac 1 2 ( \epsilon_m{}^{ji} \epsilon^{m}{}_{kl} ) \partial_j F_i dx^k \wedge dx^l = \sum_{m, k, l} \tfrac 1 2 (\nabla \times F)_m \epsilon^m{}_{k l} dx^k \wedge dx^l = \star \omega^1_{\nabla \times F} = \omega^2_{\nabla \times F} [/itex]

    So, you get for a 2-dimensional manifold (a surface) S

    [itex] \oint_{\partial S} \langle F, \tau \rangle d r = \oint_{\partial S} \omega = \int_S d \omega = \int_S \omega^2_{\nabla \times F} = \int_S \langle \nabla \times F , \nu \rangle d \sigma [/itex]

    This proves the classical Stokes' theorem. This is what I told you. ;)

    Similarly, the Cartan derivative of a 0-form corresponds to the gradient (the result is a 1-form = a vector field from the isomorphism above), of a 1-form corresponds to the curl (as shown), of a 2-form corresponds to the divergence (the result is a 3-form = a scalar field).

    Hope this helped a little bit...
     
    Last edited by a moderator: Sep 13, 2014
  13. Sep 13, 2014 #12
    I hope, I didn't confuse you. Please ask, if there is anything unclear left.
    The main thing that you must understand is that integrating forms differs from integrating vector fields. Suppose you have a line element [itex]d \vec r[/itex]. What you do if you want to integrate a vector field along a line is to integrate the scalar product [itex]\langle \vec F , d \vec r \rangle[/itex] (don't show to a mathematician^^). There is an important theorem from linear algebra (I suppose, it is called Riesz' theorem) stating that there is a 1-form doing this: [itex]\langle \vec F , d \vec r \rangle = \omega (d \vec r )[/itex]. Vector fields are not necessarily made for computing scalar products, but 1-forms are just made for applying to vectors, so it is more natural to integrate forms. The fact that vector fields can be used in several ways is why you always need to think about whether you want it to act as a 1-form (integrate along lines) or a 2-form (integrate over surfaces).
     
  14. Sep 14, 2014 #13
    Yes you are completely right, in fact this same time in last saturday i did not know there something called exterior calculus, i already finished my calculus disciplines but all of then here in brazil its a little superficial in undergraduate level, because of that many concepts still unclear in my mind.

    hahaha, i really sorry, im afraid since the start of topic of say any math aberration, but i take the risk, make math more clear for me worth spending a bit of shame rsrs.

    My motivation of think in star operator is to get a vector field in end of the process, because i want see how can i (or if can i) take the Electromagnetic four-potential in this way, i know we have another ways to do that, but i think if i maintain mathematically coherence maybe i get something interesting. So i want get the hexa-potential (so to speak) if i considerate (5+1)D, thats why i think in star operator, to reduce my result to a vector field. Sorry if i said a lot of nonsense rs

    but now i want ask you some patience in my doubts

    that means V(X) is isomorph to [itex]\Omega^1 X[/itex] through isomorphism [itex]\omega^1_F := ( r \mapsto \sum_i F_i(r) dx^i )[/itex] thats right?

    thats a "bit" confusing for me, if the star operator give me (k-1)-form, how can star operator applied in 1-form in right side of the equation equal to 2-form?

    the rest of my doubts i will say after i solve above doubts in my head, thus i avoid talking nonsense, and meanwhile im going read more about this subject

    thx a lot for patience
     
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