Understanding the Position Vector in Calculus Problems

[member 428835]

Homework Statement

With $\vec{r}$ the position vector and $r$ its norm, we define
$$ \vec{f} = \frac{\vec{r}}{r^n}.$$
Show that
$$ \nabla^2\vec{f} = n(n-3)\frac{\vec{r}}{r^{n+2}}.$$

Homework Equations

Basic rules of calculus.

The Attempt at a Solution

From the definition of the Laplacian
$$
\nabla^2 \vec{f} = \nabla \cdot \nabla \vec{f}\\
= \left( \partial_i \partial_j \delta_{ij} \right) \frac{\vec{r}}{r^n}\\
= \vec{r}\left( \partial_i \partial_i \right) r^{-n}\\
= \vec{r}(-n)(-n-1) r^{-n-2}\\
= n(n+1) \frac{\vec{r}}{r^{n+2}}.$$

Clearly this does not agree with the proposed solution. I think my error is assuming $vec{r}$ is constant, but we know this is not always true in other coordinate systems. Any ideas?

 

[Dr Transport]

Laplacian in spherical coordinates is not simply the second derivative wrt r and you have ignored the \vec{r}.

 

[Ray Vickson]

Homework Statement

With $\vec{r}$ the position vector and $r$ its norm, we define
$$ \vec{f} = \frac{\vec{r}}{r^n}.$$
Show that
$$ \nabla^2\vec{f} = n(n-3)\frac{\vec{r}}{r^{n+2}}.$$

Homework Equations

Basic rules of calculus.

The Attempt at a Solution

From the definition of the Laplacian
$$
\nabla^2 \vec{f} = \nabla \cdot \nabla \vec{f}\\
= \left( \partial_i \partial_j \delta_{ij} \right) \frac{\vec{r}}{r^n}\\
= \vec{r}\left( \partial_i \partial_i \right) r^{-n}\\
= \vec{r}(-n)(-n-1) r^{-n-2}\\
= n(n+1) \frac{\vec{r}}{r^{n+2}}.$$

Clearly this does not agree with the proposed solution. I think my error is assuming $vec{r}$ is constant, but we know this is not always true in other coordinate systems. Any ideas?

Your $\vec{f}$ is a vector whose $i$th component is $f_i = x_i/r^n$. What is $\nabla^2 f_i$?

 

[member 428835]

Your $\vec{f}$ is a vector whose $i$th component is $f_i = x_i/r^n$. What is $\nabla^2 f_i$?

$\nabla^2 f_i = \partial_j \partial_j f_i = \partial_j \partial_j (x_i/r^n)$ Does this look correct?

 

[Ray Vickson]

$\nabla^2 f_i = \partial_j \partial_j f_i = \partial_j \partial_j (x_i/r^n)$ Does this look correct?

Why not finish the computations? When you have done that, if your final result agrees with what you were asked to show, then you know you have done it correctly---at least if the dimensionality is 3 (that is, vectors are $(x,y,z)$ or $(x_1,x_2,x_3)$). For dimensions other than 3 the results ought to be a bit different.

 

[member 428835]

Why not finish the computations?

Ok, so I think we continue by saying $r^{-n}\partial_j \partial_j(x_i) + x_i\partial_j \partial_j(r^{-n})$ by product rule, right? However, I really don't know how to proceed from here. Partial derivatives are with respect to x,y,z and $x_i$ is technically not the same thing. Any help in the right direction here?

 

[Ray Vickson]

Ok, so I think we continue by saying $r^{-n}\partial_j \partial_j(x_i) + x_i\partial_j \partial_j(r^{-n})$ by product rule, right? However, I really don't know how to proceed from here. Partial derivatives are with respect to x,y,z and $x_i$ is technically not the same thing. Any help in the right direction here?

No, not right. First compute $f_{ij} = \partial_j (x_i/r^n)$, which will certainly involve using the product rule. Then, compute $\partial_j f_{ij}$, first without summing over $j$, then doing $j$-sum; those derivatives will involve the product rule again. So, when you differentiate twice, you cannot just use the product rule on $\partial^2$; it is a bit more complicated than that:
$$ \partial^2 (fg) =(\partial^2 f) g + 2 (\partial f) (\partial g) + f (\partial^2 g) $$
However, using this formula is not much quicker than just going ahead and differentiating twice, re-using the product rule each time.

Anyway, I do not understand the rest of your difficulties: of course $x_i$ is the same thing as one of the $x$, $y$ or $z$; in fact, when YOU resorted to the use of the Einstein summation convention in post #1 you must have been assuming that your coordinates are $x_1, x_2, x_3$. These are just names; instead of using $x_i, i=1,2,3$ or $x,y,z$ I could call the coordinates red, white and blue, or Tom, Dick and Harry. The only thing about them that matters is how they appear in formulas and how they can be manipulated algebraically.

 

[member 428835]

No, not right. First compute $f_{ij} = \partial_j (x_i/r^n)$, which will certainly involve using the product rule. Then, compute $\partial_j f_{ij}$, first without summing over $j$, then doing $j$-sum; those derivatives will involve the product rule again. So, when you differentiate twice, you cannot just use the product rule on $\partial^2$; it is a bit more complicated than that:
$$ \partial^2 (fg) =(\partial^2 f) g + 2 (\partial f) (\partial g) + f (\partial^2 g) $$
However, using this formula is not much quicker than just going ahead and differentiating twice, re-using the product rule each time.

Yea, I spaced it big time! So just to be clear, how about we call $\vec{r} = r_i \vec{e_i}$. Then we have $$\partial_j \partial_j (r_i \vec{e_i} r^{-n}) = \partial_j (\partial_j (r_i \vec{e_i} r^{-n}))\\
= \vec{e_i} \partial_j(\partial_j (r_i r^{-n})\\
= \vec{e_i} \partial_j(r^{-n} \partial_j r_i + r_i \partial_j r^{-n})$$
But how do you simplify the term $r^{-n} \partial_j r_i + r_i \partial_j r^{-n}$?

Anyway, I do not understand the rest of your difficulties: of course $x_i$ is the same thing as one of the $x$, $y$ or $z$; in fact, when YOU resorted to the use of the Einstein summation convention in post #1 you must have been assuming that your coordinates are $x_1, x_2, x_3$. These are just names; instead of using $x_i, i=1,2,3$ or $x,y,z$ I could call the coordinates red, white and blue, or Tom, Dick and Harry. The only thing about them that matters is how they appear in formulas and how they can be manipulated algebraically.

I was referring to the fact that if we define the components of $\vec{r}$ to be $x_i$ then $x_i$ is not the same as x,y,z since we don't know that $\vec{r} = <x,y,z>$.

 

[Ray Vickson]

No, not right. First compute $f_{ij} = \partial_j (x_i/r^n)$, which will certainly involve using the product rule. Then, compute $\partial_j f_{ij}$, first without summing over $j$, then doing $j$-sum; those derivatives will involve the product rule again. So, when you differentiate twice, you cannot just use the product rule on $\partial^2$; it is a bit more complicated than that:
$$ \partial^2 (fg) =(\partial^2 f) g + 2 (\partial f) (\partial g) + f (\partial^2 g) $$
However, using this formula is not really much easier than just going ahead and differentiating twice, re-using the product rule each time.

Anyway, I do not understand the rest of your difficulties: of course $x_i$ is the same thing as one of the $x$, $y$ or $z$; in fact, when YOU resorted to the use of the Einstein summation convention in post #1 you must have been assuming that your coordinates are $x_1, x_2, x_3$. These are just names; instead of using $x_i, i=1,2,3$ or $x,y,z$ I could call the coordinates red, white and blue, or Tom, Dick and Harry. The only thing about them that matters is how they appear in formulas and how they can be manipulated algebraically.

Yea, I spaced it big time! So just to be clear, how about we call $\vec{r} = r_i \vec{e_i}$. Then we have $$\partial_j \partial_j (r_i \vec{e_i} r^{-n}) = \partial_j (\partial_j (r_i \vec{e_i} r^{-n}))\\
= \vec{e_i} \partial_j(\partial_j (r_i r^{-n})\\
= \vec{e_i} \partial_j(r^{-n} \partial_j r_i + r_i \partial_j r^{-n})$$
But how do you simplify the term $r^{-n} \partial_j r_i + r_i \partial_j r^{-n}$?

I was referring to the fact that if we define the components of $\vec{r}$ to be $x_i$ then $x_i$ is not the same as x,y,z since we don't know that $\vec{r} = <x,y,z>$.

Call the components whatever you want. If you like $\vec{r}=(x_1, x_2, x_3)$ then go ahead and use those; if you prefer $\vec{r}=(x,y, z)$ then go with those; if you prefer $\vec{r}=(\psi, \lambda, \sigma)$ then use those instead. All that matters is how you use them.

However, if you want to use summation notation (and, possibly, the Einstein summation convention) then you had better use something like $\vec{r}=(x_1, x_2, x_3)$ or $\vec{r}=(\xi_1,\xi_2, \xi_3)$ or $\vec{r}=(p_1, p_2, p_3)$, or even $\vec{r} = (r_1,r_2,r_3)$ as you seem to like best. It really and truly does not matter, but books and papers tend to use $x,y,x$ or $x_1,x_2,x_3$ most commonly, and the latter is what I was referring to.

As to the rest of your question: you seem to be saying that you do not know how to find derivatives like $\partial x_i /\partial x_j$ (of course, for $i = j$ and for $i \neq j$), or derivatives like $(\partial /\partial x_j) (x_1^2+x_2^2+x_3^2)^{-n/2}$, and I find that hard to believe.

 

[member 428835]

All of your post makes sense. I think what I'm having trouble with is thinking about $\vec{r}$. Let's say $\vec{r} = r_i \vec{e_i} = <r_1,r_2,r_3>$. Now let's get concrete and define $<r_1,r_2,r_3> \equiv <2x,z,y+2>$. Then $\partial_1 r_1 = \partial (r_1) / \partial x_1 = 2 \neq 1$. However, if we say $\vec{r} = x_i \vec{e_i} = <x_1,x_2,x_3>\equiv <2x,z,y+2>$ then $\partial_1 x_1 = \partial (x_1) / \partial x_1 = 1 \neq 2$. Can you see what I'm confused about?

 

[Ray Vickson]

All of your post makes sense. I think what I'm having trouble with is thinking about $\vec{r}$. Let's say $\vec{r} = r_i \vec{e_i} = <r_1,r_2,r_3>$. Now let's get concrete and define $<r_1,r_2,r_3> \equiv <2x,z,y+2>$. Then $\partial_1 r_1 = \partial (r_1) / \partial x_1 = 2 \neq 1$. However, if we say $\vec{r} = x_i \vec{e_i} = <x_1,x_2,x_3>\equiv <2x,z,y+2>$ then $\partial_1 x_1 = \partial (x_1) / \partial x_1 = 1 \neq 2$. Can you see what I'm confused about?

No. The $x,y,z$ (or $r_1, r_2, r_3$ if you prefer) are not functions of anything in this problem. Something else is a function of THEM, though, and that is what you have. That is, you are given a vector function
$$\vec{f} = \vec{i} \frac{x}{(x^2+y^2+z^2)^{n/2}}+ \vec{j} \frac{y}{(x^2+y^2+z^2)^{n/2}}+ \vec{k} \frac{z}{(x^2+y^2+z^2)^{n/2}}, $$
where $\vec{i}$, $\vec{j}$ and $\vec{k}$ are the unit vectors along the x-, y- and z-axes. You want to compute
$$\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \vec{f}, $$
and this can be done by doing it separately for the x-, y- and z-components of $\vec{f}$ (because the unit coordinate vectors $\vec{i}$, $\vec{j}$ and $\vec{k}$ are constants).

Just to simplify the work---not for any other reason at all---I prefer to use the symbols $x_1, x_2, x_3$ instead of the symbols $x,y,z$ and to speak of components 1,2, and 3 of $\vec{f}$ instead of the x-, y- and z-components. That's all, nothing mysterious about it. Anyway, that is what YOU did in post #1 when you, yourself used the notation $\partial_i$, etc. I am just following YOUR lead!

 

[member 428835]

No. The $x,y,z$ (or $r_1, r_2, r_3$ if you prefer) are not functions of anything in this problem. Something else is a function of THEM, though, and that is what you have.

This is exactly what I was confused about. So "position vector" implies the vector $<x,y,z>$, or yea, any other name you want. It never (unless other wise stated) would imply something that I wrote, like $<x^2,z,4>$, right? This makes sense since that must be what your position is. Sorry for all the other stuff; at the end of the day this was my real problem. Thanks!

In fact, that's why I seemingly didn't know derivatives or understand the dummy-variables behind the symbols $x_i$ and $r_i$, etc. To me, it seemed like taking $\partial r$ was as arbitrary as taking $\partial f$.