From reading about general relativity, the event horizon and the cosmological radius are the radius when $f(r)=0$, in the metric $$ ds^{2}=-f(r)dt^{2}+\frac{dr^{2}}{f(r)}+r^{2}d\Omega^{2} $$ However, how to distinguish between these two quantities and what happens if $f(R)=0$ has only one solution $r=2M$ or when it has two solution $r=2M$ and $r=R$ where $R>2M$.
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For a spherically symmetric system there are only two possibilities for the function $f(r)$:
$$ f(r) = 1 - \frac{2a}{r} $$
$$ f(r) = 1 - br^2 $$
Though you can combine the two to get:
$$ f(r) = 1 - \frac{2a}{r} - br^2 $$
Any of these forms is a perfectly valid solution of the Einstein equations, and the exact form of $f(r)$ is determined by specifying the initial conditions. For example if we have a non-zero mass but no cosmlogical constant then $a = M$ and $b = 0$ and we get the Schwarzschild metric. If we have no mass but a non-zero cosmological constant then $a = 0$ and $b = \Lambda$ and we get a de Sitter metric. Non-zero $a$ and $b$ give the de Sitter-Schwarzschild metric.
So the existance of one or two zeros for $f(r)$ just depends on what initial conditions you specify. Specifically it requires non-zero mass and non-zero cosmological constant.
John Rennie
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To calculate event horizon...
It is the radius sphere around a black hole through which even light fails to escape
Therefore, light energy$(E)=mc²$
Energy needed to escape
i.e potential energy=$\frac{GMm}{R}$Therefore, equating $$mc²=\frac{GMm}{R}$$ $$C²=\frac{GM}{R}$$ $$R=\frac{GM}{c^2}$$ Where, $M$ = mass of the black hole notsure if my logic is correct
SchrodingersCat
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