# Is A small?

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• May 25th 2009, 03:44 PM
Kiwi_Dave
Is A small?
I am working through a derivation in relativity and find the statement that A (unit is 1/distance) is small. I don't see how we can make such a statement when A could have the unit $m^{-1}$ or equally the unit could be $light years^{-1}$.

Some background:
The equation that we are solving is:

$\frac{d^2u}{d \phi^2}+u-\frac \alpha 2 = \frac 32 \beta u^2$

We propose a solution of the form:
$u=\frac 1 \rho = A(1+\epsilon cos(\phi-\phi_0))$

Where $\rho$ is the distance between a planet and its sun, A and $\epsilon$ are constants. $\phi_0$ is a slowly varying function of $A\phi$. Clearly A could be, numerically, small if we use light years as the unit of measure.

Does the given DE tell me that u (and hence A) is small?
• May 27th 2009, 09:28 AM
Deadstar
Not sure if this is what you're after but surely since we're talking distance between planets and suns then 1/p is going to be tiny.

I think the statement that A is small is meant as a guideline so that if you get A=a billion then clearly somethings gone wrong...
• May 27th 2009, 10:27 AM
HallsofIvy
A typical distance between a star and its planets is on the order of $10^{-7}$ light years so, no, A is not small at all!
• May 28th 2009, 02:39 AM
Kiwi_Dave
The motivation for saying that A is small is to allow terms in A^3 to be crossed out (because if A is small then A^3 is very small) of a complicated equation allowing an acurate aproximation to be found.