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Flattening equation 
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#1
Apr2708, 11:32 AM

P: 608

The equation is rooted in astrophysics but the question is related to the mathematical modification of two equations.
Basically the following are two equations which provide the flattening ratio for a rapidly spinning neutron star [tex] \frac{3\pi}{2GT^{2}\rho} = \textit{f} = 1\frac{b}{a} [/tex] G gravitational constant, T rotation period (in seconds) and [tex]\rho[/tex] is density, [tex]\textit{f}[/tex] flattening ratio, a long radius (equatorial), b short radius (polar) The LHS equation is relevant to rotation and density (which is relevant to length a, the equatorial radius and length b, the polar radius). The RHS equation is relevant to the relationship between the 2 radii, a and b. I haven't find a method which simply produces an answer for the equatorial radius a, instead it's a process of trial and error using both equations until they agree on an answer. In an attempt to create just one equation, I substituted the formulas for time and density and based on the apparent fact that during flattening, the volume of the neutron star changes but the actual cross sectional area of a compact star ~ stays the same, b = [tex]R^{2}[/tex]/a, which produces the equation below [tex]\frac{3.125 R^{2}\alpha^{2}Gm}{a^{3}c^{2}}+\frac{R^{2}}{a^{2}}=1[/tex] R radius of neutron star at rest, [tex]\alpha[/tex] spin parameter (0 1), G gravitational constant, m mass, a long radius (equatorial), c speed of light which working out the constants provides [tex]\frac{2.32e10^{27}R^{2}\alpha^{2}m}{a^{3}}+\frac{R^{2}}{a^{2}}=1[/tex] which makes the trial and error process for calculating a (equatorial radius) much quicker. My question is, how do I combine the two seperate fractions into one (removing the plus sign) and if possible, how do I move 'a' over to one side of the equation in order to simply calculate the equatorial radius. Any feedback is welcome. Steve Equations [tex]T=\frac{1}{f_{Hz}}[/tex] T rotation period (in seconds), [tex]f_{Hz}[/tex] frequency (in Hz) where [tex]f_{Hz}=\frac{v}{2\pi a}[/tex] v rotational velocity at equator edge, a long radius (equatorial) [tex]v=\frac{J}{ma(2/5)}[/tex] J angular momentum, m mass, a long radius (equatorial) [tex]J=\frac{\alpha Gm^{2}}{c}[/tex] J angular momentum, [tex]\alpha[/tex] spin parameter (0  1), G gravitational radius, m mass, c speed of light and [tex]\rho=\frac{m}{V}[/tex] m mass, V volume where [tex]V=\frac{4}{3}\pi a^{2}b[/tex] a long radius (equatorial), b short radius (polar) 


#2
Apr2708, 12:32 PM

Math
Emeritus
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Thanks
PF Gold
P: 39,300

Ignoring all but the parts relevant to your specific question, you have
[tex]\frac{A}{a^3}+ \frac{B}{a^2}= 1[/tex] Multiplying both sides of the equation by a^{3} gives A+ Ba= a^{3} or a^{3} Ba A= 0. There isn't going to be any simple solution. You can try using Cardano's cubic formula: http://www.math.vanderbilt.edu/~schectex/courses/cubic/ but it isn't going to be nice! 


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