I am reading a paper in which the author is deducing from the following differential equation $$\dot{v} + \Big(\frac{\dot{a}}{a} + \frac{\dot{f_2}}{f_2}\Big)(1-v^2)v = 0, $$ that $$\frac{v}{\sqrt{1-v^2}} \propto (af_2)^{-1}.$$ For context, $a$ represents the scale factor in FRW spacetime and $v$ is the velocity of a $0-$brane in $1+1$ dimensions; $f_2 = f_2(R)$ is a function of the scalar curvature. I cannot see how to show this is true. Any help is appreciated. Thanks.
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Try differentiating $\ln (a f_2 v/\sqrt{1-v^2}) $. – Cosmas Zachos Jul 05 '19 at 00:03
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@CosmasZachos Thanks, that solves my problem. – erdoswiles Jul 05 '19 at 00:21
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If your problem is solved then please write an answer and accept it so the system has a way to know it's closed and other users can find the answer if they look. – StephenG - Help Ukraine Jul 05 '19 at 06:28
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For an arbitrary constant c, $$ \frac{d}{dt}\ln ~ ( c~ a~ f_2 ~ \frac{v}{\sqrt{1-v^2}} ) = \frac {\dot{a}} {a} + \frac {\dot{f_2}} {f_2} + \frac {\dot{v}} {v(1-v^2)} ~, $$ so the zero of the r.h.s. dictates $$ \frac{C}{a~ f_2} = \frac{v}{\sqrt{1-v^2}} ~~, $$ for some constant C.
Cosmas Zachos
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