Gradient (1/r): Homework Equation Solutions

[physics12]

Homework Statement

gradient(1/r) = r(hat) / r^2

Homework Equations

r = (x-x')i + (y-y')j + (z-z')k

[Mentor Note -- Poster was reminded to always show effort on schoolwork questions]

 

[ShayanJ]

gradient(1/r) = -r(hat) / r^2

 

[physics12]

how do you prove that?

 

[ShayanJ]

In spherical coordinates, the gradient of a scalar function f is:
$\vec\nabla f(r, \theta, \phi) = \frac{\partial f}{\partial r}\hat r+ \frac{1}{r}\frac{\partial f}{\partial \theta}\hat\theta+ \frac{1}{r \sin\theta}\frac{\partial f}{\partial \phi}\hat\phi$.
And we have $\frac{d}{dr}\frac 1 r=-\frac{1}{r^2}$.

 

[physics12]

can this be done in cartesian coordinates?

 

[HallsofIvy]

Or, in Cartesian coordinates,
$$\frac{1}{r}= \frac{1}{x^2+ y^2+ z^2}= (x^2+ y^2+ z^2)^{-1/2}$$

$$\left(\frac{1}{r}\right)_x= -\frac{1}{2}(x^2+ y^2+ z^2)^{-3/2}(2x)= -\frac{rcos(\theta)sin(\phi)}{r^3}= -\frac{1}{r^3} (rcos(\theta)sin(\phi))$$

$$\left(\frac{1}{r}\right)_y= -\frac{1}{2}(x^2+ y^2+ z^2)^{-3/2}(2y)= -\frac{rsin(\theta)sin(\phi)}{r^3}= -\frac{1}{r^3} (rsin(\theta)sin(\phi)($$

$$\left(\frac{1}{r}\right)_z= -\frac{1}{2}(x^2+ y^2+ z^2)^{-3/2}(2z)= -\frac{rcos(\phi)}{r^3}= -\frac{1}{r^3} (rcos(\phi))$$

So that $$\nabla \frac{1}{r}= -\frac{1}{r^3}\vec{r}= -\frac{1}{r^2}\frac{\vec{r}}{r}= -\frac{1}{r^2}\hat{r}$$

Where $\hat{r}$ is the unit vector in the direction of $\vec{r}$.

 

[kunu]

but this negative

 

[Frabjous]

but this negative

Yes it is. 1/r decreases with increasing r.

 

[vanhees71]

Sometimes it's easier to work in Cartesian coordinates. For any function $\Phi(r)$ you have
$$\vec{\nabla} \Phi(r)=\Phi'(r) \vec{\nabla} r.$$
Now
$$\vec{\nabla} r= \begin{pmatrix} \partial_1 \\ \partial_2 \\ \partial_3 \end{pmatrix} \sqrt{x_1^2+x_2^2+x_3^2}.$$
The first component thus reads
$$\partial_1 r=2 x_1 \frac{1}{2 \sqrt{x_1^2+x_2^2+x_3^2}}=\frac{x_1}{r}.$$
In the same way you get the other two components, leading to
$$\vec{\nabla} r= \frac{\vec{r}}{r}=\hat{r}.$$
Thus
$$\vec{\nabla} \frac{1}{r} = -\frac{1}{r^2} \hat{r}.$$