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I know how to do a to d. For e, i let $\phi = \frac{1}{|r|^3}$ and $A = a \times r$, tried to simplify but did not reach the answer. Anything can help. THank you

Xstuds
  • 141
  • You might find this thread useful: https://math.stackexchange.com/questions/1275521/the-leibniz-rule-for-the-curl-of-the-product-of-a-scalar-field-and-a-vector-fiel?rq=1 – Jason Born Jun 19 '17 at 11:05

1 Answers1

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rewrite $d$ with brackets: $$\vec{\triangledown }\times (\varphi \vec{A}) =[(\vec{\triangledown }\varphi)\times\vec{A})]+\varphi[\vec{\triangledown }\times\vec{A}]$$ Let's work with first addend: $[(\vec{\triangledown }\varphi)\times\vec{A})]=[\tfrac{-3\vec{r}}{\left | \vec {r}\right |^5}\times(\vec{a}\times\vec{r})]=\tfrac{-3}{\left | \vec {r}\right |^5}[\vec {r} \times(\vec{a}\times\vec{r})]=\tfrac{-3}{\left | \vec {r}\right |^5}(\vec{a}\cdot\left | \vec {r}\right |^2-\vec{r}(\vec{a}\cdot\vec{r}))$ ${\color{Red}(\color{Red}1\color{Red})}$

Remark: we use the rule: $[\vec {a} \times(\vec{b}\times\vec{c})]=\vec{b}\cdot(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})$

For the second addend using $(b)$ statement we get: $\varphi[\vec{\triangledown }\times\vec{A}]=\tfrac{1}{\left | \vec {r}\right |^3}\cdot2\vec{a}$ $\;\;\;\;\;\;\;\;\;\;$${\color{Red}(\color{Red}2\color{Red})}$

Now, we add $(1)+(2)$ and get your result $$\vec{\triangledown }\times (\varphi \vec{A}) = \tfrac{3}{\left | \vec {r}\right |^5}\vec{r}(\vec{a}\cdot\vec{r})-\tfrac{1}{\left | \vec {r}\right |^3}\cdot\vec{a}$$

serg_1
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  • Hi, Thank you for your answer. I have never seen that rule before. Do you know what's it called, so i can search for the prove? Or is it possible for you to provide it if its short? – Xstuds Jun 19 '17 at 12:30
  • it's called vector triple product [link] (https://en.wikipedia.org/wiki/Triple_product#Vector_triple_product) – serg_1 Jun 19 '17 at 12:37