Vector Calculus: Prove ∇ × (a×r/r^n) = (2-n)a/r^n + n(a.r)r/r^(n+2)

MrWarlock616
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Homework Statement



Prove that ##∇ × (\frac{\vec{a} × \vec{r}}{r^n})= \frac{2-n}{r^n}\vec{a}+\frac{n(\vec{a}.\vec{r}) \vec{r}}{r^{n+2}}##

Nothing is mentioned about ##\vec{a}## so I'm assuming it is a constant vector. Also ##\vec{r} = <x, y, z> ## and ##|r|=r##

Homework Equations





The Attempt at a Solution



What I don't get is why am I not allowed to do the following:

##L.H.S = \frac{1}{r^n} (∇ × (\vec{a} × \vec{r}) ) ##

## = \frac{1}{r^n} ((∇.\vec{r})\vec{a} - (∇.\vec{a})\vec{r} )## ---(using vector triple product formula)

And since ##∇.\vec{a} = 0## since ##\vec{a}## is a constant and also ##∇.\vec{r}=3##

##L.H.S = \frac{3\vec{a}}{r^n} ##

But then a black hole gets generated somewhere in the universe because the identity cannot be proved from here.
 
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I get it... r is not a constant. damn me
 
MrWarlock616 said:
I get it... r is not a constant. damn me

If you don't mind, can you please post the solution you have found? Thanks!
 
Easy way:

Using that for scalar valued point function ∅ and a vector valued point function ##\vec{f}##,

##∇× (∅\vec{f}) = ∇∅ × \vec{f} + ∅ (∇ × \vec{f})##

L.H.S. ## = ∇(\frac{1}{r^n}) × \vec{f} + ∅ (∇ × \vec{f}) ##

## = (-nr^{-n-2}\vec{r} ) × (\vec{a} × \vec{r}) + \frac{1}{r^n} (∇ × (\vec{a} × \vec{r}) )##

## = -nr^{-n-2} (\vec{r} × (\vec{a} × \vec{r})) + r^{-n} (2\vec{a}) ##

## = -nr^{-n-2} [(\vec{r} . \vec{r})\vec{a}-(\vec{r}.\vec{a})\vec{r}] + r^{-n} (2\vec{a}) ##

## = -nr^{-n-2} r^2\vec{a} + nr^{-n-2}(\vec{r} . \vec{a})\vec{r} + 2r^{-n}\vec{a} ##

phew..simplifying further will yield the RHS.

Note that I've used the following:

## ∇r^n=nr^{n-2} \vec{r}##

##∇ × (\vec{a} × \vec{r}) = 2\vec{a} ##

Hard way:

You can do it by finding the actual cross product on the LHS like I did, keeping in mind that r is not a constant scalar xD

Also, later on you have to make an adjustment by adding and subtracting ## \frac{n}{r^{n+2}} a_{1,2,3} x^2 ## in each component of the cross product.
 
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From my book

I've made many corrections to my last post
 

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MrWarlock616 said:
From my book

I've made many corrections to my last post

Thanks a lot MrWarlock! :)

The alternative method is much better.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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