An extension of Dot and Cross Product

[Jhenrique]

I was thinking, if exist a product (cross) between vectors defined as:
$$\vec{a}\times\vec{b}=a\;b\;sin(\theta)\;\hat{c}$$
and a product (dot) such that:
$$\vec{a}\cdot\vec{b}=a\;b\;cos(\theta)$$
Why not define more 2 products that result:
$$\\a\;b\;sin(\theta) \\a\;b\;cos(\theta)\;\hat{d}$$

So, for my proper use and consume, I thought to create the follows definitions:
$$\begin{matrix} \cdot & \hat{i} & \hat{j} & \hat{k} \\ \hat{i} & 1 & 0 & 0 \\ \hat{j} & 0 & 1 & 0 \\ \hat{k} & 0 & 0 & 1 \\ \end{matrix}$$$$\begin{matrix} \times & \hat{i} & \hat{j} & \hat{k} \\ \hat{i} & 0 & 1 & -1 \\ \hat{j} & -1 & 0 & 1 \\ \hat{k} & 1 & -1 & 0 \\ \end{matrix}$$$$\begin{matrix} \ \odot & \hat{i} & \hat{j} & \hat{k} \\ \hat{i} & \hat{i} & \vec{0} & \vec{0} \\ \hat{j} & \vec{0} & \hat{j} & \vec{0} \\ \hat{k} & \vec{0} & \vec{0} & \hat{k} \\ \end{matrix}$$$$\begin{matrix} \ \otimes & \hat{i} & \hat{j} & \hat{k} \\ \hat{i} & \vec{0} & \hat{k} & -\hat{j} \\ \hat{j} & -\hat{k} & \vec{0} & \hat{i} \\ \hat{k} & \hat{j} & -\hat{i} & \vec{0} \\ \end{matrix}$$

I think that this definitions to generate new possibilities and facilitate some notations. For example:
$$\frac{\partial^2 f}{\partial x^2}\frac{dx^2}{dt^2}+\frac{\partial^2 f}{\partial y^2}\frac{dy^2}{dt^2}=\bigtriangledown^2f\cdot \frac{d\vec{r}}{dt}\odot \frac{d\vec{r}}{dt}$$

This is only a ideia that I'd like to share, is not a doubt. What do you think? It seems useful and applicable?
BTW, this definitions extends and generates some interesting questions:
If I can apply a scalar field f in:
$$\frac{\partial }{\partial x}\hat{x}+\frac{\partial }{\partial y}\hat{y}$$
Can I apply a scalar field f in this version of Del operator too:
$$\left ( \frac{\partial }{\partial x}-\frac{\partial }{\partial y}\right )dxdy\;\hat{k}$$
?

 

[jedishrfu]

Sounds interesting, but in order to become commonplace your extensions must really show usefulness in applications where the current notation complicates or muddles things. So showing more examples would help.

For me the two vector products suffice and make sense.

You might consider looking at the parallels of quaternions and vectors. Hamilton tried to remake physics using quaternion math an extension of complex numbers to 3D space but others extracted what was useful and created vector math. However, more recently some physicists were revisiting quaternions because of their added rotational properties.

http://en.wikipedia.org/wiki/Quaternions

 

[Jhenrique]

Yeah, I know the Quaternions, it's very interestering and full of complicated details too!
I don't know other aplications to this 'new' products, I only filled a lacuna that I found albrebraically.

However, I really have a doubt:
I noticed that exist a analogy between the operations with scalar and vector fields.
Note that:
$$\bigtriangledown \cdot \vec{f}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}$$
is a analogous to:
$$\bigtriangledown f=\frac{\partial f}{\partial x}\hat{x}+\frac{\partial f}{\partial y}\hat{y}+\frac{\partial f}{\partial z}\hat{z}$$
and is more analogous if you consider this operation:
$$\bigtriangledown \odot \vec{f}=\frac{\partial f_1}{\partial x}\hat{x}+\frac{\partial f_2}{\partial y}\hat{y}+\frac{\partial f_3}{\partial z}\hat{z}$$

So, we could define all possible operations (using the vector ∇) to a vector field, that are:
$$\bigtriangledown \cdot \vec{f}=\frac{\partial f_1}{\partial x}+\frac{\partial f_1}{\partial y}+\frac{\partial f_1}{\partial z}$$
$$\bigtriangledown \odot \vec{f}=\frac{\partial f_1}{\partial x}\hat{x}+\frac{\partial f_1}{\partial y}\hat{y}+\frac{\partial f_1}{\partial z}\hat{z}$$
$$\bigtriangledown \times \vec{f}= \left (\frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} \right )+ \left (\frac{\partial f_1}{\partial y} - \frac{\partial f_3}{\partial z} \right )+ \left (\frac{\partial f_2}{\partial y} - \frac{\partial f_1}{\partial z} \right )$$
$$\bigtriangledown \otimes \vec{f}= \left (\frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} \right )\hat{x}+ \left (\frac{\partial f_1}{\partial y} - \frac{\partial f_3}{\partial z} \right )\hat{y}+ \left (\frac{\partial f_2}{\partial y} - \frac{\partial f_1}{\partial z} \right )\hat{z}$$

Then I think that should exist 4 analogous operations to a scalar field:
$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f_1}{\partial z}$$
$$\frac{\partial f}{\partial x}\hat{x}+\frac{\partial f}{\partial y}\hat{y}+\frac{\partial f}{\partial z}\hat{z}$$
$$\left (\frac{\partial f}{\partial y} - \frac{\partial f}{\partial z} \right )+ \left (\frac{\partial f}{\partial y} - \frac{\partial f}{\partial z} \right )+ \left (\frac{\partial f}{\partial y} - \frac{\partial f}{\partial z} \right )$$
$$\left (\frac{\partial f}{\partial y} - \frac{\partial f}{\partial z} \right )\hat{x}+ \left (\frac{\partial f}{\partial y} - \frac{\partial f}{\partial z} \right )\hat{y}+ \left (\frac{\partial f}{\partial y} - \frac{\partial f}{\partial z} \right )\hat{z}$$
*More one times I say: I'm just filling the lacuna that I saw through of algebra. I don't know why exist theory only for 3 of this 8 hypothesis. I intend to study case by case, but, as my base isn't a interpretation geometric described algebrically, but the contrary -- a algebraic deduction -- for which I intend to attribute a geometric interpretation, before, I need to know if is possible, through of some operation, combine the vector ∇ with a scalar field f to result the last 4 equations above.
The underlined part is my question.

 

[ChrisVer]

I am trying to understand why someone would need extra products.
For example you can always change the sin to cos and cos to sin (if you really want to change them) by adding a constant phase, that is rotating one of your vectors again by a constant angle.
You wouldn't get something interesting I guess...

Also the analogy you noticed for vector and scalar fields is nothing more than the definition of the Δ operator. Otherwise they are not equivalent at all... one is vector the other is scalar (under general transformations they transform differently)

 

[nonequilibrium]

Concerning your first post, defining $ab \sin \theta$ and $ab \cos \theta \; \hat d$ is useless/ill-defined. For example:

(1) concerning $ab \cos \theta \hat d$, take the case that theta is zero. Then a and b point in the same direction so they don't define a plane, hence the vector $\hat d$ normal to that plane is not well-defined. (In the normal cross-product this problem is avoided since if the angle is zero there, the norm of the vector-product becomes zero.)

(2) concerning $ab \sin \theta$. This is well-defined (at least if you define theta as the smallest angle between a and b, hence $\leq \pi$) but just useless since it's nothing more than the norm of the cross-product. (Note that since theta < pi, $|\sin\theta| = \sin \theta$)

It's good to think about these things though.

 

[Jhenrique]

(1) concerning $ab \cos \theta \hat d$, take the case that theta is zero. Then a and b point in the same direction so they don't define a plane, hence the vector $\hat d$ normal to that plane is not well-defined. (in the normal cross-product this problem is avoided since if the angle is zero there, the norm of the vector-product becomes zero.)

Humm, the unit vector d haven't a direction that's orthogonal to vectors a and b... is hard define intuitively which is the direction of new vector d.

also the analogy you noticed for vector and scalar fields is nothing more than the definition of the Δ operator.

Wrong. The definition for ∇ vector is:

$$\vec{\bigtriangledown} = \begin{bmatrix} \frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\\ \end{bmatrix}$$

and for this I asked if is possible to define some operation (dot; cross; other...) that combine the vector ∇ with a scalar field f and that can results some of 4 last equation of my post #3. Note that for the 4 last equations exist, is necessary to define the dot, cross (or other more if necessary) product between a vector (∇) and a scalar (f). And here my question more one time: is possible to define the dot, cross (or other ) product between a vector (∇) and a scalar (f) for exist the 4 last equations of my post #3?

 

[nonequilibrium]

Humm, the unit vector d haven't a direction that's orthogonal to vectors a and b... is hard define intuitively which is the direction of new vector d.

I see. Having looked at your above definition for $\odot$ I can note two even more fundamental problems:

(1) The definition you gave in the table for $\odot$ does not give something of the form "$ab \cos \theta \; \hat d$". After all, the norm of $\mathbf a \odot \mathbf b$ is not equal to $ab \cos \theta$.

(2) Your notion of $\odot$ is basis-dependent! This means that $\mathbf a \odot \mathbf b$ takes on a different value depending on the basis you chose to represent the vectors in! Surely you see this makes the notion of $\odot$ useless.

 

[Jhenrique]

I see. Having looked at your above definition for $\odot$ I can note two even more fundamental problems:

(1) The definition you gave in the table for $\odot$ does not give something of the form "$ab \cos \theta \; \hat d$". After all, the norm of $\mathbf a \odot \mathbf b$ is not equal to $ab \cos \theta$.

(2) Your notion of $\odot$ is basis-dependent! This means that $\mathbf a \odot \mathbf b$ takes on a different value depending on the basis you chose to represent the vectors in! Surely you see this makes the notion of $\odot$ useless.

Interesting! I like your answer, I like really!

 

[ChrisVer]

what you called an analogy was the result of your ∇ operator acting on something. Of course you'd get something analogous in any way and that's what I said. That's the only similiarity between them, and the rest (the thing that is in the derivative) changes giving different results.
For example divergence and gradient are different things. The one is a scalar and thus transforms differently under transformations , than the gradient which is a vector.

∇ is still an operator so I don't get why you did it bold...

 

[R136a1]

(2) Your notion of $\odot$ is basis-dependent! This means that $\mathbf a \odot \mathbf b$ takes on a different value depending on the basis you chose to represent the vectors in! Surely you see this makes the notion of $\odot$ useless.

The gradient is basis-dependent too. I don't see how the gradient is useless.

 

[nonequilibrium]

The gradient is not basis-dependent.