Why is the gradient of a vector function/field meaningless?

[flyingpig]

Homework Statement

Let's say \vec{F} = <P,Q,R>

If I take the gradient, shouldn't I get

\nabla \vec{F} = <\frac{\partial P }{\partial x}, \frac{\partial Q}{\partial y}, \frac{\partial R}{\partial z}>

Also why is grad(div f) meaningless? My book says it's because div(f) gives a scalar field. My question what's wrong with taking the gradient of a scalar field?

 

[SammyS]

...
Also why is grad(div f) meaningless? My book says it's because div(f) gives a scalar field. My question what's wrong with taking the gradient of a scalar field?

Is f, itself, a scalar field, or is it a vector field?

 

[flyingpig]

f itself is a scalar field.

 

[Matterwave]

The gradient of a vector field is generally a tensor. The divergence of a scalar field is ill defined. In general, a divergence lowers the rank of whatever it's operating on by 1, so the divergence of a rank 2 tensor is a vector, and the divergence of a vector is a scalar. As a scalar is rank 0, and there are no objects with rank -1, the divergence is ill defined. How would you propose to define the divergence of a scalar field?

 

[flyingpig]

The gradient of a vector field is generally a tensor. The divergence of a scalar field is ill defined. In general, a divergence lowers the rank of whatever it's operating on by 1, so the divergence of a rank 2 tensor is a vector, and the divergence of a vector is a scalar. As a scalar is rank 0, and there are no objects with rank -1, the divergence is ill defined. How would you propose to define the divergence of a scalar field?

Okay, let's try to use some words I can understand...

 

[Matterwave]

So, the gradient of a vector field is meaningful. It returns a tensor. You can think of a tensor, for now, as something like a matrix. The rank of a tensor is the number of indices you need to specify a component of the tensor. A scalar requires no indices, a vector requires 1 index, a rank 2 tensor (like a matrix) requires 2 indices.

The divergence of a scalar field is simply not well defined.

 

[flyingpig]

No my book says gradF is also meaningless because F is not a scalar field.

 

[Matterwave]

Well, it's meaningless in that it doesn't return an object which you have had to deal with before (a tensor), but more generally gradF returns a tensor.

 

[SammyS]

f itself is a scalar field.

Divergence is defined for a Vector field, not a scalar field.

So, Div(f) is meaningless.

Thus, Grad(Div(f)) is also meaningless.

 

[I like Serena]

Let's say \vec{F} = <P,Q,R>

If I take the gradient, shouldn't I get

\nabla \vec{F} = <\frac{\partial P }{\partial x}, \frac{\partial Q}{\partial y}, \frac{\partial R}{\partial z}>

The proper gradient of F is:

<br /> \nabla \vec{F} = &lt;\frac{\partial \vec{F} }{\partial x}, \frac{\partial \vec{F}}{\partial y}, \frac{\partial \vec{F}}{\partial z}&gt;<br />

This is a matrix.