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Consider the force $F_\Omega$: $$ F_\Omega= \Omega^2\left(x -\frac{\beta\left(x+\alpha R\right)R^3}{\left(\left(x+\alpha R\right)^2\right)^{3/2}} -\frac{\alpha\left(x-\beta R\right)R^3}{\left(\left(x-\beta R\right)^2\right)^{3/2}} \right) $$ Where $\alpha+\beta=1$

(also to the attentive reader; the unit of mass, which is missing, has been set equal to unity)

Set $F_\Omega=0$

Now, make the substitution $x=R(u+\beta)$

The following three quintics can be obtained $$ u^2\left(\left(1-s_1\right)+3u+3u^2+u^3\right) =\alpha\left(s_0+2s_0u+\left(1+s_0-s_1\right)u^2+2u^3+u^4\right) $$ Here $s_0=\mbox{sign}(u)$ and $s_1=\mbox{sign}(u+1)$

Now to my question: How can we arrive at the quintic equation?

The equations come from a paper on Lagrange-points: http://www.physics.montana.edu/faculty/cornish/lagrange.pdf

It has been suggested to me (from the author) that I should

1) Make the substitution of $x$

2) Put everything on a common denominator

3) Make the numerator vanish by taking the term to the $\frac{3}{2}$-power to the left side of the equation

I have tried of course done the substitution, but fail to put it all on a common demonitor, and honestly can't see for myself where all these $u$'s come into the picture.

Feel free to put on some relevant tags. I couldn't think of any.

mvw
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user147163
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  • Are you missing a closing bracket above? Also are you looking for the root of the above expression ? – Chinny84 Nov 27 '14 at 15:15
  • I was missing a bracket, yes. Thank you. I rewrote the question. I am not looking for the roots I am looking for the way to rewrite the expression of the force to the quintic equation. I have also added the (of course crucial) detail that the force is 0. – user147163 Nov 27 '14 at 15:19

1 Answers1

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Presumably the variable $R$ is understood to be positive. Assuming this, let's see what the substitution $x=uR+\beta R$, with $\alpha+\beta=1$, does to the three pieces of the expression inside the brackets:

$${\alpha(x-\beta R)R^3\over((x-\beta R)^2)^{3/2}}={\alpha uR^4\over|uR|^3}={\alpha Rs_0\over u^2}$$

where $s_0$ is the sign of $uR$, which is presumably the same as the sign of $u$.

$${\beta(x+\alpha R)R^3\over((x+\alpha R)^2)^{3/2}}={\beta(u+\beta+\alpha)R^4\over|(u+\beta+\alpha)R|^3}={(1-\alpha)Rs_1\over(u+1)^2}$$

where $s_1$ is the sign of $(u+1)R$, which is presumably the same as the sign of $u+1$.

$$x=uR+\beta R=(u+1-\alpha)R$$

Putting these together and cancelling out the common $R$ gives

$$u+1-\alpha={\alpha s_0\over u^2}+{(1-\alpha)s_1\over(u+1)^2}$$

Can you take it from there?

Barry Cipra
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  • I am sorry for not having added this information. R is a distance and is positive. I had never given it any thought that it would mean anything, but I see know, of course. Thank you very much. – user147163 Nov 27 '14 at 19:02
  • I am almost there, but I might have done something wrong? $$u+1- \alpha={ \alpha s_0\over u^2}+{(1- \alpha)s_1\over(u+1)^2}$$ $$u^3+3u-2u \alpha+3u^2-u^2 \alpha- \alpha+1=\alpha (2s_0u+s_0)+s_1-s_1 \alpha$$ $$(1-s_0)+u^3+3u^2+3u=\alpha (s_0+2s_0u+(1-s_1)+2u+u^2)$$

    I have of course tried to make it look like the actual equations as much as possible. Can I multiply by $|u|/u$? That is of course to get the last $s_0$ on the right side.

    I though about multplying with u^2 on both sides, but isn't quite it.

     – user147163 Nov 27 '14 at 19:13
  • @user147163, yes, you need to do the algebra more carefully. When you multiply both sides by $u^2(u+1)^2$ to clear out the denominator, you should get $$u^2(u+1)^2(u+1-\alpha)=(u+1)^2\alpha s_0+u^2(1-\alpha)s_1$$ Try taking it from there, moving all the $\alpha$'s over to the right. – Barry Cipra Nov 27 '14 at 19:59
  • I see, I was just multiplying by $(u+1)^2$. But I will try to move om from your's. Tank you. – user147163 Nov 27 '14 at 20:17