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Griffiths equation 5.32

  1. Apr 3, 2008 #1
    1. The problem statement, all variables and given/known data
    This question refers to Griffiths E and M book.

    The first equality makes perfect sense to me. It is a standard path-integral along the wire FOR EACH COMPONENT OF THE INTEGRAND. It is defined just like this (for each component)

    [tex]\int_C f\, ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)|\, dt[/tex]

    I do not understand the second equality at all. What exactly does

    [tex]\int \frac{d\mathbf{l'}\times \hat{\mathbf{r}}}{r^2}[/tex]

    mean?

    I have never seen an integral like. What is the formal definition of that integral? So, you parametrize your path and then do what? I just don't understand!?!?!

    BTW: how do you get that the script r that Griffiths uses in latex?

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Apr 3, 2008
  2. jcsd
  3. Apr 3, 2008 #2
    The expression in the integrand is part of the Biot-Savart Law, the part that shows the magnetic field is inverse-square and that it is always orthogonal to the direction of the current element Idl and also to the radial component of the distance from the current. Because of this double orthogonality, you need a cross product notation. The permeability and current are taken out of the integrand because in what Griffiths calls magnetostatics, current is steady. That was an experimentally determined "law" based on observational data.

    Some introductory textbooks have examples of how to actually perform this integration; for example, you can use it to find the field from a wire segment, and then expand the wire to infinity and get Ampere's law for a long wire.
     
  4. Apr 3, 2008 #3

    Dick

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    It's three path integrals in wrapped up in one vector package. Define the path l' and treat each of the x,y and z components as a separate path integral.
     
  5. Apr 20, 2008 #4
    Could you just write out exactly what this means:

    [tex]\int_{\gamma} d\mathbf{l} \times \mathbf{v}[/tex]

    where [itex]\gamma: I \to R^3 [/tex] is a rectifiable path and [itex]\mathbf{v}[/itex] is a continuous vector field? The object above is a vector, so can you write out the x-coordinate for me please?
     
  6. Apr 20, 2008 #5

    malawi_glenn

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    How would you do a cross product?

    What is the definition of : [tex] d\mathbf{l} [/itex] ?
     
  7. Apr 20, 2008 #6
    Thats what I want to know.
     
  8. Apr 20, 2008 #7
    If you introduce an explicit Cartesian basis, then [tex]d\mathbf{l} = dx \hat{\mathbf{x}} + dy \hat{\mathbf{y}} + dz \hat{\mathbf{z}}[/tex]
     
  9. Apr 20, 2008 #8

    malawi_glenn

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    and in general (3-dimensions)

    [tex] d\mathbf{r} = h_1 du_1 \mathbf{e}_{u_1} + h_2 du_2 \mathbf{e}_{u_2} + h_3 du_3 \mathbf{e}_{u_3} [/tex]

    where:
    [tex] h_i = \vert \partial \mathbf{r} / \partial u_i \vert [/tex]

    [tex] \mathbf{e}_{u_i} = \frac{\partial \mathbf{r}}{ \partial u_i} / h_i [/tex]

    I thought Griffiths had these defenitions?
     
    Last edited: Apr 20, 2008
  10. Apr 20, 2008 #9
    What malawi and genneth wrote is different.
     
  11. Apr 20, 2008 #10

    malawi_glenn

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    why? The thing i wrote is the general definition, it reduces to what genneth wrote if you have a cartesian coordinate system. I can give the [tex] d\matbf{r} [/tex] in spherical and/or cylindrical coordinates aswell if you want?

    In cartesian coordinates, all [itex] h_i = 1[/itex]

    [itex] u_1 = x, u_2 = y, u_3 = z [/itex] and [itex] \hat{\mathbf{x}} = \mathbf{e}_{u_1} [/itex] is just different notations for the same things, just as i call it [itex] d\mathbf{r}} [/itex], gennet calls it [itex] d\mathbf{l}} [/itex],

    I mean it just follows from the chain rule and the definition of differentials ( I know you have differentials and infenitesimals from another thread you created some weeks ago) :


    [tex] d\mathbf{r} = \frac{\partial \mathbf{r}}{ \partial u_i}du_i [/tex] (using einstein summation)

    So by just saying that we says different things and not even trying to give it substance or showing that you have at least tried what I and gennet posted to you, makes me sad. I think that was a quit ignorant posting of you.

    You should be glad that me and genneth, who 'works' here for free, helps you by posting here and in the other thread that you create here.
     
    Last edited: Apr 20, 2008
  12. Apr 20, 2008 #11
    Oops, I missed the part where you said "and in general (3-dimensions)" and the part where genneth said "If you introduce an explicit Cartesian basis", sorry.

    Also, sorry malawi that you are not supplied with a salary. In fact you have even given money to PF ironically. :)
     
    Last edited: Apr 20, 2008
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