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Math Help - [SOLVED] differentiation integration with respect to a vector

  1. #1
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    [SOLVED] differentiation integration with respect to a vector

    Let r=xi+yj+zk be a displacement vector. We have:

    df=r.\nabla f

    Where f is a scalar field

    What if f is a vector field. Does that mean that:
    df=r.\nabla f

    What about integration of scalar/vector fields with respect to vectors. I am interested in how you solve these:

    \int f \, dr Where f is a scalar field and

    \int g \, dr Where g is a vector field.

    Any input is appreciated.
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  2. #2
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    Quote Originally Posted by fobos3 View Post
    Let r=xi+yj+zk be a displacement vector. We have:

    df=r.\nabla f

    Where f is a scalar field

    What if f is a vector field. Does that mean that:
    df=r.\nabla f

    What about integration of scalar/vector fields with respect to vectors. I am interested in how you solve these:

    \int f \, dr Where f is a scalar field and

    \int g \, dr Where g is a vector field.

    Any input is appreciated.
     \nabla f is the gradient of scalar field f. A vector does not have a gradient in this sense. A vector can have curl and divergence, but not gradient. The gradient of a scalar field is a vector field.

    Nabla, is defined as :  \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})

    If you have a vector \vec{f}=(F_x,F_y,F_z), then you can have only two multiplicative operations with this vector and the nabla vector. Those are the dot product ( \nabla . \vec{f} )and the cross product (  \nabla \times \vec{f} ). And these represent divergence and curl respectively. It does not make sense to have scalar multiplcation between two vectors, which is what you are proposing with  \nabla \vec{f}.

    You may have scalar multiplication between two scalars, and scalar multiplcation between a scalar and a vector (a lá,  \vec{grad}(f), but you may not have scalar multiplication between two vectors.
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  3. #3
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    Quote Originally Posted by Mush View Post
     \nabla f is the gradient of scalar field f. A vector does not have a gradient in this sense. A vector can have curl and divergence, but not gradient. The gradient of a scalar field is a vector field.

    Nabla, is defined as :  \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})

    If you have a vector \vec{f}=(F_x,F_y,F_z), then you can have only two multiplicative operations with this vector and the nabla vector. Those are the dot product ( \nabla . \vec{f} )and the cross product (  \nabla \times \vec{f} ). And these represent divergence and curl respectively. It does not make sense to have scalar multiplcation between two vectors, which is what you are proposing with  \nabla \vec{f}.

    You may have scalar multiplication between two scalars, and scalar multiplcation between a scalar and a vector (a lá,  \vec{grad}(f), but you may not have scalar multiplication between two vectors.
    Huh? The gradient of a vector field is a second order tensor.

    v= v_1e_1+v_2e_2+v_3e_3

    Then \nabla v=\sum _{i=1}^3 \sum _{j=1}^3 \frac{\partial v_j}{\partial x_i}e_ie_j

    Where e_1,e_2,e_3 are vectors.

    Check this out.
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  4. #4
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    Anyone? Come on guys I'm really stuck here.
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