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Math Help - Is A small?

  1. #1
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    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?
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  2. #2
    Super Member Deadstar's Avatar
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    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...
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  3. #3
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    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!
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  4. #4
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    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.
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