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Physical significance of dot and cross products in electrodynamics

  1. Aug 19, 2011 #1
    i know that del.v=divergence of vector v
    and del x v=curl of vector v

    can anyone justify for the same? how dot product is physically connected to divergence property?
  2. jcsd
  3. Aug 19, 2011 #2


    Staff: Mentor

    It is just notational convenience. The formula for divergence looks like a dot product.
  4. Aug 19, 2011 #3


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    Science Advisor
    Gold Member

    If you have an Euclidean space (e.g., [itex]\mathbb{R}^3[/itex] that describes the spatial hypersurface of an observer in an inertial frame in both Newtonian and special-relativistic physics), the derivatives

    [tex]\partial_i=\frac{\partial}{\partial x^i}[/tex]

    act formaly like components of a co-vector (covariant components).

    It's convenient to write this in vector notation as the nabla symbol [itex]\vec{\nabla}[/itex]. In Cartesian coordinates you can now easily play with that symbol to create new tensor fields out of given tensors by operating on their corresponding components wrt. to the Cartesian basis (and co-basis). The most important examples, all appearing in electromagnetism as well as fluid dynamics are

    The Gradient

    The gradient of a scalar field is a vector field (more precisely a one-form) and defined by

    [tex]\mathrm{grad} f=\vec{\nabla} f=\vec{e}^j \partial_j f.[/tex]

    The Divergence

    The divergence of a vector field is a scalar field defined by

    [tex]\mathrm{div} \vec{V}=\vec{\nabla} \cdot \vec{V}=\partial_i V^i.[/tex]

    The Curl

    The curl of a vector field is given by

    [tex]\mathrm{curl} \vec{V}=\vec{\nabla} \times \vec{V}.[/tex]

    The structure of this becomes more clear by looking at it in terms of alternating differential forms (Cartan calculus). Given a 1-form (co-vector) [itex]\omega = \mathrm{d}x^j \omega_j[/itex], you get an alternating 2-form by

    [tex]\mathrm{d} \omega = \mathrm{d} x^j \wedge \mathrm{d} x^k \partial_j \omega_k = \frac{1}{2} \mathrm{d} x^j \wedge \mathrm{d} x^k (\partial_j \omega_k-\partial_k \omega_j).[/tex]

    In more conventional terms, you have a antisymmetric rank-2 tensor, given by its covariant components [itex]\partial_j \omega_k-\partial_k \omega_j[/itex]. In 3 dimensions, you can map this uniquely to a vector field by hodge dualization:

    [tex][\mathrm{curl} \vec{V}]^j=\epsilon^{jkl} (\partial_k V_l-\partial_j V_k).[/tex]

    Further, since in Euclidean spaced the metric has components [tex]\delta_{jk}[/tex] wrt. Cartesian Coordinates, you have [tex]V^j=V_j[/tex].
  5. Aug 19, 2011 #4
    can u justify with an example?
  6. Aug 19, 2011 #5


    Staff: Mentor

    [tex](a,b,c)\cdot(d,e,f) = ad+be+cf[/tex]
    [tex]\text{Divergence}(f)=\frac{\partial}{\partial x}f_x+\frac{\partial}{\partial y}f_y+\frac{\partial}{\partial z}f_z=\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot \left( f_x, f_y, f_z \right) = \nabla \cdot f[/tex]
  7. Aug 20, 2011 #6
    how is the left hand side of your second equation, a divergence?
  8. Aug 20, 2011 #7
    [tex]\nabla.\phi = \frac{\partial \phi_x}{\partial x} + \frac{\partial \phi_y}{\partial y}+\frac{\partial \phi_z}{\partial z}[/tex]

    It looks like the dot product of [tex]\frac{\partial}{\partial x} +\frac{\partial}{\partial y} + \frac{\partial}{\partial z}[/tex] and [tex](\phi_x,\phi_y,\phi_z)[/tex].

    It's an abuse of notation.
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