User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase

T3 Coordinates (continued)

On one accompanying wiki page we have introduced T3 Coordinates and on another we have described how Jay Call's Characteristic Vector applies to T3 Coordinates. Here we investigate the properties of our T3 Coordinate system in the special case when <math>q^2 = 2</math>; Jay Call's independent analysis is recorded on a separate page.

Special Case (Quadratic)

Coordinate Relations

When <math>q^2=2</math>, the two key coordinates are:

<math> \lambda_1 </math>		<math> = </math>	<math> \varpi \cosh\Zeta </math>
<math> \lambda_2 </math>		<math> = </math>	<math> \frac{\varpi}{\sinh\Zeta} </math>
Note also:	<math> \Chi \equiv 2\frac{\lambda_1}{\lambda_2} </math>	<math> = </math>	<math> 2 \sinh\Zeta \cosh\Zeta = \sinh(2\Zeta) , </math>

where, in this case,

<math> \Zeta \equiv \sinh^{-1} \biggl( \frac{\sqrt{2}z}{\varpi} \biggr) . </math>

For this special case, we can invert these coordinate relations to obtain analytic expressions for both <math>\varpi</math> and <math>z</math> in terms of <math>\lambda_1</math> and <math>\lambda_2</math>. Specifically, the relation,

<math> 1 = \cosh^2\Zeta - \sinh^2\Zeta = \biggl(\frac{\lambda_1}{\varpi}\biggr)^2 - \biggl(\frac{\varpi}{\lambda_2}\biggr)^2 </math>

implies that the function <math>\varpi(\lambda_1,\lambda_2)</math> can be obtained from the physically relevant root of the following equation:

<math> \varpi^4 \lambda_2^{-2} + \varpi^2 - \lambda_1^2 = 0. </math>

The relevant root gives,

<math> \varpi^2 = \frac{\lambda_2^{2}}{2} \biggl[ \sqrt{1 + (2\lambda_1/\lambda_2)^2} - 1 \biggr] = \frac{\lambda_2^{2}}{2} \biggl[ \cosh(2\Zeta) - 1 \biggr] </math>

<math> \Rightarrow ~~~~~\varpi = \frac{\lambda_2}{\sqrt{2}}\biggl[ \cosh(2\Zeta) - 1 \biggr]^{1/2} . </math>

The desired function <math>z(\lambda_1,\lambda_2)</math> is therefore,

<math> z = \frac{\varpi}{\sqrt{2}} \sinh\Zeta = \frac{\varpi^2}{\sqrt{2}~\lambda_2} = \frac{\lambda_2 }{2\sqrt{2}} \biggl[ \sqrt{1 + (2\lambda_1/\lambda_2)^2} - 1 \biggr] </math>

<math> \Rightarrow ~~~~~ z = \frac{1}{2}~\frac{\lambda_2 }{\sqrt{2}} \biggl[ \cosh(2\Zeta) - 1 \biggr] . </math>

In an effort to simplify the appearance of these and future expressions, we will henceforth adopt the notation,

<math> \Lambda \equiv \cosh(2\Zeta) ~~~~~ \Rightarrow ~~~~~ \Lambda = \sqrt{1 + \Chi^2} . </math>

In terms of <math>\Lambda</math>, then, we have,

<math> \varpi </math>	<math> = </math>	<math> \frac{\lambda_2 }{\sqrt{2}} \biggl[ \Lambda - 1 \biggr]^{1/2} ; </math>
<math> z </math>	<math> = </math>	<math> \frac{1}{2}~\frac{\lambda_2 }{\sqrt{2}} \biggl[ \Lambda - 1 \biggr] . </math>

Scale Factor Expressions

We are now in a position to express the two key scale factors purely in terms of the two key T3 coordinates. First, we note that,

<math> \ell^{2} = [\varpi^2 + 4z^2]^{-1} = \frac{2}{\lambda_2^2(\Lambda - 1)\Lambda} , </math>

and,

<math> \biggl(\frac{\lambda_1}{\lambda_2}\biggr)^2 = \frac{1}{4}\Chi^2 = \frac{1}{4}(\Lambda^2 -1) = \frac{1}{4}(\Lambda -1)(\Lambda +1) . </math>

Hence,

<math> h_1^2 </math>	<math> = </math>	<math> \lambda_1^2 \ell^2 </math>	<math> = </math>	<math> \frac{1}{2}\biggl[ \frac{\Lambda + 1}{\Lambda} \biggr] ; </math>
<math> h_2^2 </math>	<math> = </math>	<math> (q^2-1)\biggl( \frac{\varpi z \ell}{\lambda_2} \biggr)^2 </math>	<math> = </math>	<math> \frac{1}{8}~ \frac{(\Lambda - 1)^2}{\Lambda} . </math>

We note also that,

<math> \frac{d\Lambda}{dt} </math>

<math> = </math>

<math> \frac{d}{dt}\biggl[ 1+(2\lambda_1/\lambda_2)^2 \biggr]^{1/2} </math>

<math> = </math>

<math> \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr] \frac{d\ln(\lambda_1/\lambda_2)}{dt} . </math>

Hence, starting from the general expression for <math>d\ln h_2/dt</math> derived elsewhere, we deduce that,

<math> \frac{d\ln h_2}{dt} </math>

<math> = </math>

<math> 2 h_1^4\frac{d}{dt}\biggl[ \ln(\lambda_1/\lambda_2) \biggr] </math>

<math> = </math>

<math> \frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda} \biggr)^2 \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr]^{-1} \frac{d\Lambda}{dt} </math>

<math> = </math>

<math> \frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda-1} \biggr) \frac{d\ln\Lambda}{dt} . </math>

(This identical expression also can be derived straightforwardly from the specific expression for <math>h_2</math> given above.) From the expression given above for <math>h_1</math>, we also deduce that,

<math> \frac{d\ln h_1}{dt} = - \frac{1}{2(\Lambda + 1)} \frac{d\ln\Lambda}{dt} . </math>

As a check, we note that the relationship between <math>d\ln h_1/dt</math> and <math>d\ln h_2/dt</math> in this specific case (i.e., <math>q^2 = 2</math>) matches the general relationship between these two logarithmic time-derivatives that has been derived elsewhere, namely,

<math> (h_1 \lambda_1)^2 \frac{d\ln h_1}{dt} + (h_2 \lambda_2)^2 \frac{d\ln h_2}{dt} = 0 . </math>

Equation of Motion

According to Equation EOM.01, as derived elsewhere in the context of T3 coordinates, a general expression for the second component of the equation of motion is,

<math> \frac{d(h_2 \dot{\lambda}_2)}{dt} = \biggl(\frac{\lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) \frac{dh_2}{dt} </math>

<math> \Rightarrow ~~~~~ \frac{\ddot{\lambda}_2}{\lambda_2} = \biggl[\frac{\dot{\lambda}_1}{\lambda_1} - \frac{\dot{\lambda}_2}{\lambda_2} \biggr]\frac{d\ln h_2 }{dt} . </math>

Based on the derivations provided above, both factors that make up the term on the RHS of this expression can be written entirely in terms of the variable, <math>\Lambda</math>. This allows us to rewrite Equation EOM.01 as,

<math> \frac{\ddot{\lambda}_2}{\lambda_2} = \biggl[\frac{\Lambda}{\Lambda^2-1} \frac{d\Lambda}{dt}\biggr]\biggl[ \frac{1}{2\Lambda} \biggl( \frac{\Lambda + 1}{\Lambda - 1} \biggr) \frac{d\Lambda}{dt} \biggr] = \frac{1}{2(\Lambda - 1)^2} \biggl[\frac{d\Lambda}{dt} \biggr]^2 </math>

T3Q.01

<math> \Rightarrow ~~~~~ 2\frac{\ddot{\lambda}_2}{\lambda_2} = \biggl[\frac{d\ln(\Lambda-1)}{dt} \biggr]^2. </math>

Solution Strategy

I'm not sure whether the following strategy is fully legitimate, but let's explore it anyway. Because the LHS of Equation T3Q.01 displays an explicit dependence only on the coordinate <math>\lambda_2</math> while the RHS displays an explicit dependence only on <math>\Lambda</math> — that is, only on the ratio of the two coordinates <math>\lambda_1/\lambda_2</math> — perhaps we can use a separation of variables technique to derive a solution. Specifically, suppose the LHS and the RHS separately are set equal to the same value, call it <math>n</math>.

Then, for the LHS:

<math> \ddot{\lambda}_2 = \frac{n}{2}\lambda_2 </math>

And, for the RHS:

<math> \frac{d\ln(\Lambda-1)}{dt} = \sqrt{n} </math>

 

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation