Griffiths 1.13 Let r be the separation vector from a fixed point (x', y', z') to the point (x,y,z), and let r be its length. Show that

(a) $\nabla (r^2) = 2\mathbf{r}$

(b) $\nabla (1/r) = -\mathbf{r}/r^2$

(c) What is the general formula for $\nabla (r^n)$?

Griffiths 1

Pairoaj

define

$$\mathbf{\hat{r}} = \hat{\mathbf{x}}+\hat{\mathbf{y}}+\hat{\mathbf{z}}$$

$$\mathbf{r} = (x-x^\prime)\hat{\mathbf{x}}+(y-y^\prime)\hat{\mathbf{y}}+(z-z^\prime)\hat{\mathbf{z}}$$

$$r=\sqrt{(x-x^\prime)^2+(y-y^\prime)^2+(z-z^\prime)^2}$$

**part (a)**

$$\nabla (r^2) = \frac{\partial}{\partial x}(r^2)+\frac{\partial}{\partial y}(r^2)+\frac{\partial}{\partial z}(r^2)$$

$$\nabla (r^2) = 2(x-x^\prime)\hat{\mathbf{x}}+2(y-y^\prime)\hat{\mathbf{y}}+2(z-z^\prime)\hat{\mathbf{z}}$$

$$\nabla (r^2) = 2[(x-x^\prime)\hat{\mathbf{x}}+(y-y^\prime)\hat{\mathbf{y}}+(z-z^\prime)\hat{\mathbf{z}}]$$

$$\nabla (r^2) = 2\mathbf{r}$$

**part (b)**

$$\nabla (1/r) = \nabla (r^{-1}) = \frac{\partial}{\partial x}(r^{-1})+\frac{\partial}{\partial y}(r^{-1})+\frac{\partial}{\partial z}(r^{-1})$$

I will just show the partial for the x-component -- it is the same for y and z components

$$\nabla (r^{-1})=\frac{\partial}{\partial x}[(x-x^\prime)^2+(y-y^\prime)^2+(z-z^\prime)^2]^{-1/2}+\ldots$$

$$\nabla (r^{-1}) = -\frac{1}{2}(r^2)^{-\frac{3}{2}}2(x-x^\prime)\hat{\mathbf{x}}+\ldots$$

$$\nabla (r^{-1}) = -(r^2)^{-\frac{3}{2}}(x-x^\prime)\hat{\mathbf{x}}+\ldots$$

$$\nabla (r^{-1}) = -r^{-3}(x-x^\prime)\hat{\mathbf{x}}+\ldots$$

put y and z components back in

$$\nabla (r^{-1}) = -r^{-3}(x-x^\prime)\hat{\mathbf{x}}+-r^{-3}(y-y^\prime)\hat{\mathbf{y}}+-r^{-3}(z-z^\prime)\hat{\mathbf{z}}$$

$$\nabla (r^{-1}) = -r^{-3}[(x-x^\prime)\mathbf{\hat{x}}+(y-y^\prime)\mathbf{\hat{y}}+(z-z^\prime)\mathbf{\hat{z}}]$$

$$\nabla (r^{-1}) = -\frac{1}{r^3}\mathbf{r}=-\frac{1}{r^2}\mathbf{\hat{r}}$$

**part (c)**

in general (for the x component -- again its the same for the y and z),

$$\frac{\partial}{\partial x}(r^n)=nr^{n-1} \frac{\partial r}{\partial x} = nr^{n-1}(\frac{1}{2}\frac{1}{r}2r_x)=nr^{n-1}\mathbf{\hat{r}_x}$$

so with the y and z components,

$$\nabla (r^n)=nr^{n-1}\mathbf{\hat{r}}$$

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Asked: 7 months ago