Vector Calculus: Gradient of separation distance

[WWCY]

Homework Statement

Could someone explain how the property,
$$\nabla (\frac{1}{R}) = -\frac{\hat{R}}{R^2}$$
where $R$ is the separation distance $|\vec{r} - \vec{r'}|$, comes about?

What does the expression $\nabla (\frac{1}{R})$ even mean?

Homework Equations

The Attempt at a Solution

I know this attempt misses the mark completely, but I'd like to know what I'm getting wrong:

The separation distance $R$ is a function of $x,y,z$ since $r$ is too. Thus
$$ \nabla ( \frac{1}{R} )= \frac{-1}{R^2} \frac{dR}{dx} \hat{x} + \frac{-1}{R^2} \frac{dR}{dy} \hat{y} + \frac{-1}{R^2} \frac{dR}{dz} \hat{z} $$
which doesn't resemble what I wrote at the beginning.

Thanks in advance!

 

[PeroK]

$\frac{1}{R}$ is a scalar function of position. Try expressing this in terms of Cartesian coordinates $(x, y, z)$.

$\nabla$ is an operator:

$\nabla f(x, y, z) = (f_x, f_y, f_z)$

where I've used $f_x = \frac{\partial f}{\partial x}$ etc. But, you seem to be using this correctly in any case.

Now, just keep going.

 

[WWCY]

Thank you for your response.

$\frac{1}{R}$ is a scalar function of position. Try expressing this in terms of Cartesian coordinates $(x, y, z)$.

So I let $\frac{1}{R} = f(x,y,z)$. So,

$\nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R$

Is this what you meant?

 

[PeroK]

Thank you for your response.
So I let $\frac{1}{R} = f(x,y,z)$. So,

$\nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R$

Is this what you meant?

Why don't you express $R$ as a function of $(x, y, x)$? Then you can differentiate it. And all will be revealed!

 

[Ray Vickson]

Thank you for your response.
So I let $\frac{1}{R} = f(x,y,z)$. So,

$\nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R$

Is this what you meant?

Yes, that is exactly what the notation $\nabla (1/R)$ means. Now just finish the job by computing $\partial R/\partial x,$ etc.

 

[PeroK]

$\nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R$

Note that, as in all your posts, these should be partial derivatives.

 

[WWCY]

Thank you both for your input, here's what I came up with.
$$R = \sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }$$
$$\partial _x R = \frac{x - x'}{\sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }} = \frac{R_x}{R}$$
Doing this for all components of the gradient vector, I get
$$\frac{1}{R} (R_x , R_y, R_z)$$
which is the unit vector pointing in the direction of separation, $\hat{R}$

Is this right?

 

[PeroK]

Thank you both for your input, here's what I came up with.
$$R = \sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }$$
$$\partial _x R = \frac{x - x'}{\sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }} = \frac{R_x}{R}$$
Doing this for all components of the gradient vector, I get
$$\frac{1}{R} (R_x , R_y, R_z)$$
which is the unit vector pointing in the direction of separation, $\hat{R}$

Is this right?

Yes. In other words $\nabla R = \hat{R}$.

 

[WWCY]

Yes. In other words $\nabla R = \hat{R}$.

Thank you and @Ray Vickson for the assistance!