Solving a differential to show the homogeneity of space.

[stephen cripps]

Homework Statement

The final part of the problem I am trying to solve requires the proof of the following equation:
$$\frac{d}{dr}(\frac{rf'(r)-f(r)+f^2(r)}{r^2 f^2(r)})=0$$[/B]

Homework Equations

I've been given the ansatz:
$$f(r)=(1-kr^2)^{-1}$$
leading to
$$f'(r)=2krf^2(r)$$
$$f''(r)=2kf^2(r)+8(kr)^2f^3(r)$$

The Attempt at a Solution

Using the quotient rule on the first equation and cancelling some terms and the denominator gets me to:
$$(rf)^2(rf''+2ff')-(rf'-f+f^2)(2rf^2+2r^2ff')=0$$
Expanding, cancelling and subbing in for f' & f'' leads me to:
$$2kr^3f^4+8k^2r^5f^5+2rf^3-2rf^4-4kr^4f^3=0$$

I have tried subbing in the ansatz value for f, but I still can't get the terms to cancel out. Would anybody be able to point out where I've made a mistake/ what I've missed?

 

[RUber]

If you were to start with your ansatz, then the problem looks like:
$\frac{d}{dr}\left( \frac{ r(2kr)f^2 - f + f^2}{r^2 f^2 }\right) =0$
Which could just as well be written as:
$\left( \frac{ r(2kr)f^2 - f + f^2}{r^2 f^2 }\right) =C$
Separate the fractions and you have: *edited to correct sign error*
$\left( \frac{ r(2kr)f^2 }{r^2 f^2 } + \frac{ -f + f^2}{r^2 f^2 }\right) =C$
That first fraction is clearly a constant...do some algebra on the second...and I think you might find your solution.

 

[stephen cripps]

Ah thanks! I've got there now. On the solution sheet they seem to suggest a different method using f''(r) but don't show it explicitly but this definitely seems valid. (Though I think the + should be a - in your third equation after splitting the fraction.)

 

[RUber]

I agree on both points. I was hasty in splitting the fractions and it did sound like the problem was enticing you to take the complicated derivative.