[SOLVED] differentiation integration with respect to a vector

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December 21st 2008, 08:37 AM
fobos3

[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.
December 21st 2008, 08:40 AM
Mush

Quote:

Originally Posted by fobos3

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.
December 21st 2008, 10:55 AM
fobos3

Quote:

Originally Posted by Mush

$\nabla f$ 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.
December 22nd 2008, 09:49 PM
fobos3

Anyone? Come on guys I'm really stuck here.