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I initially thought it was inversely proportional to the mass, but I think that's wrong because temperature is inversely proportional to mass. If someone could give the formula(s) for finding this that would be fantastic. I've searched around and just found stuff about Swartzchild radius etc.

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4 Answers 4

up vote 6 down vote accepted

I have wondered about the same question, and I have a short attention span, so I'm going to steal the answer given by other people and put it in direct equations. Stan Liou gave the answer generally for a black hole with mass $M$, angular momentum $J$ and charge $Q$. These values being nonzero implies that it is a Kerr-Newman black hole. To get from his answer to an explicit form, I had to use $r^+$, $r^-$, $a$, $\sigma$ (constant defined in terms of other constants), $\kappa$, $T$, $A$, and finally, the luminosity. This is what I found.

$$P = \frac{1}{240} \frac{\hbar c^6 \left( 1 -\frac{Q^2 }{ 4 \pi \epsilon_0 G M^2} -\left( \frac{J c}{M^2 G} \right)^2\right)^2 }{ \pi G^2 M^2 \left( 2 +2 \sqrt{ 1 - \frac{Q^2 }{ 4 \pi \epsilon_0 G M^2} - \left( \frac{J c}{M^2 G} \right)^2 } -\frac{Q^2 }{ 4 \pi \epsilon_0 G M} \right)^3} $$

Also, I should note that not all combinations of these values are physical. Any BH that violates the inequality $ Q^2+\left ( J/M \right )^2\le M^2\, $ is unphysical, with that equation taken to be in the special universal units. This says what is already obvious from my above equation. We expect that if the quantity in the radical is negative, it won't be an allowed combination of values. So the correct qualifier on my above equation in arbitrary units is:

$$ M^2 - \frac{Q^2 }{ 4 \pi \epsilon_0 G } - \left( \frac{J c}{M G} \right)^2 \ge 0 $$

Now, let's say that it is just a Kerr black hole, implying that $Q=0$. We substitute this in to get a more compact equation.

$$ P = \frac{1}{1920} \frac{ c^6 \hbar \left(1-\left( \frac{J c}{M^2 G} \right)^2 \right)^2}{ \pi G^2 M^2 \left(1+\sqrt{ 1 - \left( \frac{J c}{M^2 G} \right)^2 }\right)^3}$$

Limiting the discussion further, let's narrow it to a Schwarzschild black hole, meaning $Q=0$ and $J=0$. That reduces the above equation to:

$$ P = \frac{\hbar c^6}{15360 \pi G^2 M^2} $$

This matches the equations you can find in Wikipedia.

Naturally, if the black hole is radiating away its mass through Hawking Radiation, mass loss and power output is connected by $E=m c^2$. So $dM/dt=P/c^2$. You could use this differential equation to find life of the black hole. I don't know what rate it would lose charge and angular momentum at.

I also wanted to verify (or refute) the claim that an angular momentum causes the black hole to radiate more slowly. I graphed the part of the equation for the Kerr-Newman black hole with the special units for charge and angular momentum. Neither of these can be zero, so I graphed from 0 to 1 for the range of both. The maximum value of the plot is exactly $240/15360=0.015625$.

Google graph

So yes, any amount of charge and/or angular momentum decreases the rate of Hawking radiation.

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As Lubos mentioned, do in your power equation do we also equate G and c to 1? Also, is my $dM/dt = (c^2(dR/dt))/(c^4)$ wrong and why? Also, how can you defend your rate of mass loss? No offense, but a few answers have emerged and they all seem promising. – Stumbleine75 Nov 21 '12 at 23:49
@ThroatOfWinter57 I edited my answer to make my unit system more clear. I'm now only reporting equations in the most literal form possible, that means that the units of the equation can be anything, as long as they're consistent, like most physics equations you're used to. It contains physical constants, so those must match units of the variables you put into it. I do NOT use $G=1$ units except for the one in-line equation now. For a non-rotating BH $R=GM/c^2$ and your $dM/dt$ equation lacks a $G$, so no, it can't be right. – Alan Rominger Nov 25 '12 at 2:21
That makes sense. However, I went to your wikipedia link, and my doubt was rekindled. You had $dM/dt = Pc^2$ due to energy-mass equivalence. Wikipedia has $P = -(dE/dt) = -c^2(dM/dt)$. Therefore, according to wikipedia, $dM/dt = P/-c^2$ – Stumbleine75 Nov 25 '12 at 3:16
@ThroatOfWinter57 Oh that was really a careless mistake on my part. Thanks for finding it. I agree with your last equation. – Alan Rominger Nov 25 '12 at 14:01
hi AlanSE, can you show me what source are you using for that formula? can you point me where did you find it, or from where are you deriving it? – lurscher Dec 19 '12 at 17:53

This site seems to have the kinds of things you're looking for: It has a formula for the temperature of a Kerr-Newman BH about 3/4 of the way down. Ignore 'e' if you don't care about charge.

Another web page offers a formula, marked (11), just short of halfway down, but I don't see Boltzmann's constant, and somehow it doesn't look right.

If I didn't botch up the algebra, for a chargeless BH with angular momentum the temperature is

$T_{BH}={\hbar c \over 4\pi k_B}{c^2\over{GM}}{{\sqrt{1-({a\over{M}})^2}}\over{{1+\sqrt{1-({a\over{M}})^2}}}}$

The rotational parameter $a$ is related to angular momentum by

$a = {cJ\over GM}$

Given a temperature, rate of energy loss (luminosity, or power) is given by standard thermodynamics

$P = \sigma A T^4$

A bit of math, integrating this and equating to the total mass of the BH (time c^2) gives the lifetime. Ignoring spin, this goes up with the mass cubed. With increasing spin, the temperature goes down, and the lifetime longer.

To get a feel for typical physical values, but alas only for a non-spinning BH, one can play with the online calculator at

I don't like listing wikipedia as a reference (but love it for browsing) but still, in case someone stumbling upon this question and answer needs more background, see

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In $D$ spacetime dimensions, the temperature goes like $$ T\sim 1/R $$ and the non-degenerate (e.g. spherical) black hole has a horizon area scaling like $$ A\sim R^{D-2}.$$ Multiply this by the Stefan-Boltzmann law for the energy radiated per unit area $$ \frac{dE}{dt} \sim -\sigma T^D\cdot A $$ where I included $\sigma$ just to remind you it is the Stefan-Boltzmann law and you get $$ \frac{dE}{dt} \sim - T^{2}\sim - \frac{1}{R^2} $$ independently of the dimension. Because the mass of the black hole (energy that will be radiated away) is $$ E\sim R^{D-3}, $$ you may also estimate the lifetime of the black hole as $$ t_{\rm life} \sim \frac{R^{D-3}}{1/R^2}\sim R^{D-1}$$ In all the formulae above, substitute $D=4$ for our spacetime.

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So, regarding your second to last formula, if Energy to be radiated away is given by $R$ when $D = 4$ does that mean that the rate of mass loss is given by $dM/dt = (c^2(dR/dt))/(c^4)$ when you factor in Energy - Mass equivalence? Also, is the $R$ in your formula radius or redshift factor? I have no background in upper level physics, so this is intense speculation. – Stumbleine75 Nov 16 '12 at 13:45
$R$ is the radius, as measured from the angular components of the metric i.e. from the proper area of the horizon. Yes, in $D=4$, $dM/dt$ is proportional to $dR/dt$ simply because $M\sim R$. The Schwarzschild radius, for example, is $R=2GM/c^2$ in normal units. You put $G=1$ right? – Luboš Motl Nov 16 '12 at 16:25
I actually didn't use $G$ to get $R$. I first heard that $R = 2GM/c^2$, but I later read that it was equal to $2M$. I'm not so sure that $R = 2M$ is correct. I'll re-run the necessary parts of my work with the other formula for $R$. Anyway, if I what should I use $G = 1$ in and where? All I did was use $R$ to find $dM/dt$ in your mass loss rate formula. My big problem now is finding $dR/dt$ so that I can find the rate the volume decreases in relation to mass loss. – Stumbleine75 Nov 16 '12 at 23:27
Hi, $R=2M$ is clearly the same formula as $R=2GM/c^2$, expressed in "general relativistic" units with $G=c=1$. You may check that e.g. in the SI units, $R=2M$ is a wrong equation even dimensionally, can't you? If you can't, you shouldn't try to study evaporation of black holes because it is a problem of difficulty higher by about 7 levels of education relatively to the problem of distinguishing meters from kilograms. – Luboš Motl Nov 17 '12 at 6:10

A Kerr or Kerr-Newman black hole with mass $M$, charge $Q$, and angular momentum $J$ has the horizons: $$r_\pm = \frac{G}{c^2}\left[M\pm\sqrt{M^2-\frac{1}{4\pi\epsilon_0G}Q^2-\frac{c^2}{G^2}\frac{J^2}{M^2}}\right]$$ and surface gravity relative to infinity: $$\kappa = c^2\frac{r_+-r_-}{2(r_+^2+a^2)},$$ where $a = J/(Mc)$. The surface area is: $$A = 4\pi(r_+^2 + a^2)$$ And finally the temperature: $$T = \frac{\hbar}{c k_B}\frac{\kappa}{2\pi}.$$ Semiclassically a black hole is a perfect blackbody, so its luminosity is given by the usual Stefan-Boltzmann law: $$L = \sigma AT^4 \longrightarrow \frac{\eta_{\text{eff}}}{2}\sigma AT^4$$ I put in some factors of physical consntants that physicists usually ignore by working in natural units ($c = G = 4\pi\epsilon_0 = k_B = \hbar = 1$). Ignore them if you do as well. The mass loss rate is naturally related to this by $c^2$.

One complication is that when the temperature is high enough compared to rest mass of massive particles in units of Boltzmann constant $k_B = 1$, the luminosity is higher as the number of effectively accessible particle species increases. For a cold black holes, as astrophysical ones should be, $\eta_{\text{eff}} = 2$, counting both polarization types of a photon. If a black hole is hot enough to emit charged particles, it will do so in a biased way, thus changing its charge $Q$. There is probably a similar effect to change the angular momentum through Hawking radiation, but I'm uncertain of the details.

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I don't need all of that to find the mass loss rate do I? Bottom line, what is the formula? Call me lazy, but I'd really like a formula that can model the mass loss rate itself. I don't want to derive your horizon formula to get dm/dt. There has to be something easier. At least I hope there is. – Stumbleine75 Nov 16 '12 at 4:15
Lazy. $$\frac{\eta_{\text{eff}}}{2}\frac{\hbar}{240\pi}\frac{(r_+ - r_-)^4}{(r_+^2+a^2)^3}$$ – Stan Liou Nov 16 '12 at 4:43
@StanLiou I'm mostly able to match your formula with that of DarenW with a notable exception of the exponent of 3 in your denominator. The equation for luminosity and power has a $T^4$ term, and I don't see how this evolves into what you got. – Alan Rominger Nov 19 '12 at 17:03
@StanLiou Never mind. I was able to derive the non-rotating expression for luminosity from both your equation and DarrenW's. The difference in the factor came from the definition of $A$, where I used the Wikipedia equation that was for non-rotating. Your equation for area is accurate for rotating as well as non-rotating, although they both (predictably) converge to the right expression for non-rotating. – Alan Rominger Nov 19 '12 at 17:24
Hi Stan, can you point me to a source where those equations are shown? – lurscher Dec 19 '12 at 18:18

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