# [SOLVED] differentiation integration with respect to a vector

• Dec 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:

$\displaystyle df=r.\nabla f$

Where f is a scalar field

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

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

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

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

Any input is appreciated.
• Dec 21st 2008, 08:40 AM
Mush
Quote:

Originally Posted by fobos3
Let r=xi+yj+zk be a displacement vector. We have:

$\displaystyle df=r.\nabla f$

Where f is a scalar field

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

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

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

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

Any input is appreciated.

$\displaystyle \nabla f$ 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 : $\displaystyle \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})$

If you have a vector $\displaystyle \vec{f}=(F_x,F_y,F_z)$, then you can have only two multiplicative operations with this vector and the nabla vector. Those are the dot product ($\displaystyle \nabla . \vec{f}$)and the cross product ($\displaystyle \nabla \times \vec{f}$). 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 $\displaystyle \nabla \vec{f}$.

You may have scalar multiplication between two scalars, and scalar multiplcation between a scalar and a vector (a lá, $\displaystyle \vec{grad}(f)$, but you may not have scalar multiplication between two vectors.
• Dec 21st 2008, 10:55 AM
fobos3
Quote:

Originally Posted by Mush
$\displaystyle \nabla f$ 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 : $\displaystyle \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})$

If you have a vector $\displaystyle \vec{f}=(F_x,F_y,F_z)$, then you can have only two multiplicative operations with this vector and the nabla vector. Those are the dot product ($\displaystyle \nabla . \vec{f}$)and the cross product ($\displaystyle \nabla \times \vec{f}$). 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 $\displaystyle \nabla \vec{f}$.

You may have scalar multiplication between two scalars, and scalar multiplcation between a scalar and a vector (a lá, $\displaystyle \vec{grad}(f)$, but you may not have scalar multiplication between two vectors.

Huh? The gradient of a vector field is a second order tensor.

v=$\displaystyle v_1e_1+v_2e_2+v_3e_3$

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

Where $\displaystyle e_1,e_2,e_3$ are vectors.

Check this out.
• Dec 22nd 2008, 09:49 PM
fobos3
Anyone? Come on guys I'm really stuck here.