Vector product - operator's a vector?

In summary, the conversation discusses the use of the del operator in determining the curl of a force, specifically in the context of classical mechanics. The del operator, though not a vector, can be treated as one in calculations. However, it is important to remember that it is not a true vector and should not be used as such. The correct application of the del operator involves applying it before making cross-products between vectors.
  • #1
Oblio
398
0
Greetings all!
So,

[tex]\nabla X \vec{F}[/tex] is confusing me.

I understand that it can be used to tell whether a force is conservative in that, if the curl is 0 then the work done all all paths are the same... that's fine.

However,
I was looking at it, for example, in the context of the gravitational field. When drawing it out, one can see that the curl is indeed 0, and I've been told that it is proven by [tex]\nabla X \vec{F}[/tex], BUT from what I understand, and from what I've been told, the operator [tex] \nabla [/tex] has no direction, and thereby not a vector...

yet used in a vector product?

How can it even by setup in the first place if the operator isn't a vector?
 
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  • #2
This got moved I see? Yet its from my classical mechanics class. lol
 
  • #3
this post can probably be disregarded unless people with a classical mechanics context can answer my question, and not a purely mathematical one; which would probably mean very little to me.
 
  • #4
The operator is defined in such a clever way that you MAY regard it as a vector, so that the curl comes out properly.

Of course, it is not really a vector, but how we define it enables us to treat it as a vector in our calculations.
 
  • #5
Actually, the curl operator is curl, rot or [itex]\vec\nabla\times[/itex]. You can apply it to a vector [itex]\vec V = (V_1, V_2, V_3)[/itex] so you get [itex]\operatorname{curl} \vec V[/itex], [itex]\operatorname{rot} \vec V[/itex] or [itex]\vec\nabla \times \vec V[/itex], which is defined by

[tex]\operatorname{curl} \vec V = \operatorname{rot} \vec V = \nabla \times V \stackrel{\mathrm{def}}{=} \begin{pmatrix}
\frac{\partial V_3}{\partial y} - \frac{\partial V_2}{\partial z} \\
\frac{\partial V_1}{\partial z} - \frac{\partial V_3}{\partial x} \\
\frac{\partial V_2}{\partial x} - \frac{\partial V_1}{\partial y} \\
\end{pmatrix}
[/tex].

By way of mnemonic, you can remember
[tex]\nabla = \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{pmatrix} [/tex]
and you can multiply it in the cross product like it is an actual vector and it will give you the correct result.
But you should consider this as a coincidence, and keep in mind that although [itex]\nabla[/itex] behaves in most ways like a vector, the operators that work on the vector are really things like [itex]\nabla\cdot, \nabla\times[/itex], etc. and it is not really a vector product in the sense of [itex]\vec V \times \vec W[/itex] or [itex]\vec V \cdot \vec W[/itex]. If you want, also consider it a coincidence that the same symbol [itex]\nabla[/itex] occurs in all these notations, though of course this is because we think of [itex]\nabla[/itex] as something with partial derivatives, which is in all these operators.

Sorry, it became a mathematical explanation after all. But for most practical purposes (actually, all I have encountered in classical mechanics) you can pretend that it is a vector (but again, it is not).
 
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  • #6
[tex]\nabla[/tex] can be thought of like a vector only in cartesian coordinates.
Note the warning at the bottom of http://users.aber.ac.uk/ruw/teach/260/260del.html [Broken] .

So, in starting out, it might be a useful mnemonic... but one should quickly free oneself from it. (Similarly, since evaluating determinants with the diagonal lines only work with 2x2 and 3x3, one should quickly learn the more general method and its interpretation.)
 
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  • #7
robphy said:
[tex]\nabla[/tex] can be thought of like a vector only in cartesian coordinates.
Del is a vector differential operator.
It's components only have that simple form in Cartesian coordinates, but it is still a vector.
All its operations can be carried out independently of any coordinate system.
 
  • #8
robphy said:
[tex]\nabla[/tex] can be thought of like a vector only in cartesian coordinates.
Note the warning at the bottom of http://users.aber.ac.uk/ruw/teach/260/260del.html [Broken] .

So, in starting out, it might be a useful mnemonic... but one should quickly free oneself from it. (Similarly, since evaluating determinants with the diagonal lines only work with 2x2 and 3x3, one should quickly learn the more general method and its interpretation.)
The correct application of the del operator will always work out properly, irrespective of your choice of coordinates.

Here's how the correct application goes, using cylindrical coordinates for convenience, calculating the curl.
The trick lies in applying the differential operator BEFORE making cross-products between vectors proper:
[tex]\nabla\times\vec{F}=\vec{i}_{r}\times\frac{\partial\vec{F}}{\partial{r}}+\vec{i}_{\theta}\times\frac{\partial\vec{F}}{r\partial\theta}+\vec{i}_{z}\times\frac{\partial\vec{F}}{\partial{z}},\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{i}_{z}\frac{\partial}{\partial{z}}, \vec{F}=F_{r}\vec{i}_{r}+{F}_{\theta}\vec{i}_{\theta}+F_{z}\vec{i}_{z}[/tex]

If you calculate this, you'll end up with the correct expression for the curl of F

So, there is no need to free yourself from that mnemonic, as long as you remember it IS a mnemonic..
 
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1. What is a vector product?

A vector product, also known as a cross product, is a mathematical operation that takes two vectors as input and produces a third vector as output. It is denoted by the symbol "x" and is often used in physics and engineering to calculate torque, angular momentum, and other quantities.

2. How is the vector product different from the dot product?

The vector product is different from the dot product in that it produces a vector as output, whereas the dot product produces a scalar. Additionally, the vector product is not commutative, meaning the order in which the vectors are multiplied matters, while the dot product is commutative.

3. What are some properties of the vector product?

Some properties of the vector product include: it is distributive, meaning a x (b + c) = (a x b) + (a x c); it follows the right-hand rule, meaning the resulting vector is perpendicular to the two input vectors; and its magnitude is equal to the product of the magnitudes of the two input vectors multiplied by the sine of the angle between them.

4. How is the vector product calculated?

The vector product is calculated by taking the determinant of a 3x3 matrix. The first row contains the unit vectors i, j, and k, the second row contains the components of the first vector, and the third row contains the components of the second vector. The resulting vector is the third row of the resulting matrix.

5. What are some applications of the vector product?

The vector product has many applications in physics and engineering, including calculating torque, angular momentum, and magnetic force. It is also used in computer graphics to rotate objects in three-dimensional space and in navigation to calculate the direction of a plane's heading.

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