Can A Vector Have Components Of Say Y or Z in the X-Hat Direction?

[Ascendant0]

Homework Statement

It was a quiz problem, asking to solve $( \vec{r} * \Delta ) * \vec{r}$

Relevant Equations

$\vec{r} = (x^2 + 4y) \widehat{x}$ (This wasn't part of the problem. Just giving an example of where I felt my answer could be the correct one (*If a vector like this is possible))

While I only have a minute to touch on this right now, when I solved this, according to my instructor, I did everything right except at the very end, I didn't drop the partial vectors for each direction. The reason I didn't do so is because up to this point, my understanding was that a vector can be defined by any variables, and as such, could very well use a partial in any direction.

For example, $\vec{r} = (x^2 + 4y) \widehat{x}$ is something like this possible, or as far as a vector, can you not have y or z components in the $\widehat{x}$ direction like that? From what I gathered from him, it seems like he thinks as soon as you add a component of y like that into a direction other than $\widehat{y}$, it is no longer a vector, it becomes a function.

If I am correct and that is possible, is there an example of where it's used that I could present to my professor as proof? According to him, a vector can't have partial derivatives like that y value there in it, but I thought this was possible. Just trying to confirm, and if I was right, get some type of proof to present to him. Thanks

 

[BvU]

Homework Statement: It was a quiz problem, asking to solve $( \vec{r} * \Delta ) * \vec{r}$

I don't understand the word 'solve' : I don't see an equation to solve. I only see an expression that I also do not understand. What do the asterisks represent (*) ?
So what do you get when the expression is written out ?

(*) Inner products are ususally rendered with a \cdot

 

[kuruman]

In your example you have a vector $\mathbf {r}$ that is entirely in the $~\mathbf{\hat x}~$ direction. The fact that its $x-$component depends on both $x$ and $y$ coordinates does not mean that it has a $y-$component.

Here are some specific examples.
On the x-axis ($y=0$) the vector is $\mathbf r=x^2~\mathbf{\hat x}~$ and varies quadratically with $x.$
On the y-axis ($x=0$) the vector is $\mathbf r=4y~\mathbf{\hat x}~$ and varies linearly with $y.$
At point $(2,1)$ the vector is $\mathbf r=8~\mathbf{\hat x}$ and so on.

Note that in all cases the vector points in the $x-$direction only. This is how vector fields are defined.

Your confusion is not uncommon and has to do with the conventional way you were taught vectors. You started with 1-d kinematics in which case everything depends on one variable, say $x$, and depends on that variable only. Then you saw 2-d projectile motion where the motion in the x-direction is independent of the motion in the y-direction. This reinforced the preconception that vector components can only depend on one direction. This is simply incorrect.

 

[Steve4Physics]

Homework Statement: It was a quiz problem, asking to solve $( \vec{r} * \Delta ) * \vec{r}$

In addition to what has already been said...

$\vec r$ is the usual symbol for the position vector. Is that what it's supposed to represent in your Post? Or is it some general vector (if so, it would be better-named as $\vec v$)?

$\Delta$ in the context of vector calculus means the Lapalacian (the divergence of the gradient of a scalar function). $\Delta f =\nabla \cdot \nabla f$. Is that what you intended?

 

[Ascendant0]

In addition to what has already been said...

$\vec r$ is the usual symbol for the position vector. Is that what it's supposed to represent in your Post? Or is it some general vector (if so, it would be better-named as $\vec v$)?

$\Delta$ in the context of vector calculus means the Lapalacian (the divergence of the gradient of a scalar function). $\Delta f =\nabla \cdot \nabla f$. Is that what you intended?

Apologies, I used "*" instead of $\cdot$.

The exact problem was:

"Calculate $(\vec{r} \cdot \Delta) \vec{r}$ in cartesian coordinates"

With my final answer, I kept the partials (taking into consideration the possibility of a vector like I mentioned previously), so I wrote it out like this:

$[x(\partial x_x / \partial x)] + y( \partial x/ \partial y) + z \partial x/ \partial z] \widehat{x}$

And of course similar for the y and z directions. I used the $x_x$ for the first one because I wasn't sure how to clarify it's not going to just be $\partial x / \partial x = 1$, so I clarified that it was so the partial was taken, and not just making it equal 1.

But, after clarifying this, I just realized I had a complete brain-fart on this problem. For the first $\vec{r}$ on the left side, I multiplied them as $x(\partial / \partial x) + y(\partial / \partial y) + z(\partial / \partial z)$, which was me assuming it to be the basic vector $x \widehat{x} + y \widehat{y} + z \widehat{z}$. Wow, I don't know what I was thinking here. I overthought the final solution I had, not even realizing that. So yea, this is pointless, lol. Sorry to have wasted your time guys.

 

[kuruman]

So what is your answer to this?

 

[PhDeezNutz]

In post#5 OP seems to be using the gradient instead of the Laplacian. I say this because there are no mention of second partial derivatives. Assuming we are to use the gradient instead of the laplacian

Is the problem supposed to be $\left(\vec{r} \cdot \nabla \right) \vec{r}$?

Just for additional context this kind of calculation is useful in E&M when you want to find the force that an electric field exerts on a point dipole where obviously Coulomb’s law doesn’t work because a dipole has no net charge.

$\vec{F} = \left( \vec{p} \cdot \nabla \right) \vec{E}$

 

[Ascendant0]

So what is your answer to this?

All the partials from the del function cancel, just breaks down to $x \hat{x} + y \hat{y} + z \hat{z}$

The del function only operates on vectors to the right of it. So, with the first $\vec{r}$, when you do the parenthesis, it just comes out to:

$x ( \partial / \partial x ) + y ( \partial / \partial y ) + z ( \partial / \partial z )$

Each individual vector component of the first vector to the left of the del operator is multiplied by the del component of that directional vector. It doesn't get applied to that vector, because it's to the left of the del operator.

That completes what's in the parenthesis. Then, keeping in mind that's now a scalar with no specific vector components, it gets multiplied by the $\vec{r}$ to the right of the del operator. I don't have time right now to write it all out (plus, with how OCD people are on here about not handing people answers, I doubt they'd want me to fully write every single step out anyway), but basically the scalar above is applied to each vector component of $\vec{r}$, so x, y, and z, in each respective direction.

Now, say for example with the $\hat{x}$ direction, you get:

$[x(\partial x / \partial x) + y( \partial x/ \partial y) + z (\partial x/ \partial z) ] \hat{x}$

The three x that are being operated on by the del functions is because this full scalar is getting multiplied to each individual vector component of the final vector $\vec{r}$ on the right. So, the same scalar is multiplied to all components of $\vec{r}$, so for example with the $\hat{y}$ direction, the y component of the vector will be multiplied by the scalar, so the "x" in the numerator of the del operator would all be "y", but the rest would remain the same. Of course same for z.

Looking at the partial, you can see the second and third partials become zero, since there is no y or z components in the $\hat{x}$ direction to take the partials of. And of course $(\partial x / \partial x) = 1$. So, this leaves you with $x \hat{x}$ for the portion cited above. Applying the same to the y and z directions, each time you're only left with the value y and z for each respective direction as well.

So, the final result just gives you the original vector $\vec{r}$ back:

$x \hat{x} + y \hat{y} + z \hat{z}$ which is simply $\vec{r}$

 

[Steve4Physics]

At the risk of excess repetition (see various previous posts), you wrote:

The exact problem was:

"Calculate $(\vec{r} \cdot \Delta) \vec{r}$ in cartesian coordinates"

Are you sure? It is not the same as asking:
"Calculate $(\vec{r} \cdot \nabla) \vec{r}$ in cartesian coordinates"

$\nabla$ and $\Delta$ are different operators.