Flattening equation

  • Thread starter stevebd1
  • Start date
  • #1
749
41
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 separate 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)
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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 a3 gives A+ Ba= a3 or
a3- 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!
 

Suggested for: Flattening equation

Replies
1
Views
310
  • Last Post
Replies
2
Views
239
  • Last Post
Replies
7
Views
382
  • Last Post
Replies
5
Views
314
  • Last Post
Replies
4
Views
411
  • Last Post
Replies
4
Views
494
  • Last Post
Replies
0
Views
665
  • Last Post
Replies
7
Views
533
  • Last Post
Replies
6
Views
615
  • Last Post
Replies
3
Views
280
Top