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Good evening, I'm reading the paper Prehawking radiation by William G. Unruh where it says:

...a time scale of order of $m^{3}$ in Planck units, or $10^{53}$ ages of the current universe for a solar mass black hole"

How do I perform the conversion from Planck units to seconds?

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  • $\begingroup$ Technically it gave you a time by reading the age of the universe which if I recall is about 13.8 billion years (this can be looked up) all you then have to do is convert years to seconds $\endgroup$ – Triatticus Nov 14 '18 at 16:11
  • $\begingroup$ Yes, but I don't know how to manage that $m^{3}$. in Planck unit $t \approx m^{3}$, so what do I have to do to have the result? for a solar mass black hole? $\endgroup$ – Lucap Nov 14 '18 at 16:17
  • $\begingroup$ Another thing is when citing that you are reading a paper, generally a link to the paper should be provided so we can get more context. $\endgroup$ – Triatticus Nov 14 '18 at 16:18
  • $\begingroup$ Thank you, i'm new i didn't know that $\endgroup$ – Lucap Nov 14 '18 at 16:21
  • $\begingroup$ In Planck units, mass, length and time become dimensionless by setting $c$, $\hbar$ and $G$ to 1, so not only time and mass cubed have the same dimension, all combinations of these quantities do. To get from $m^3$ to seconds, insert the unique combination of factors of $\hbar$, $G$ and $c$ that converts the unit in which you are expression $m^3$ into seconds. $\endgroup$ – doetoe Nov 14 '18 at 17:10
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Another way to say what John Rennie said is:

A formula such as $$t=m^3$$ written "in Planck units" means $$\frac{t}{t_P}=\left(\frac{m}{m_P}\right)^3$$ in non-Planck units, where $$t_P=\sqrt{\frac{\hbar G}{c^5}}$$ is the Planck time and $$m_P=\sqrt{\frac{\hbar c}{G}}$$ is the Planck mass.

So in SI units the formula would be

$$t=\frac{t_P}{m_P^3}m^3=\frac{G^2 m^3}{\hbar c^4}.$$

In general, to convert any equation in Planck units to SI units, just divide every mass by the Planck mass, every time by the Planck time, and every length by the Planck length.

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The paper is saying that the time in Planck units is of order $m^3$ where $m$ is the mass of the black hole in Planck units.

The paper uses an example of a Solar mass black hole so $ m = 2 \times 10^{30}$ kg. One Planck mass is $2.18 \times 10^{-8}$ kg, so the mass of the black hole in Planck units is:

$$ m = 9.19 \times 10^{37} $$

and therefore:

$$ m^3 = 7.76 \times 10^{113} $$

So the timescale is of order $7.76 \times 10^{113}$ Planck times. One Planck time is $5.39 \times 10^{-44}$ seconds, so the time in seconds is:

$$\begin{align} t &= 4.18 \times 10^{70} \, \mathrm{seconds} \\ &= 1.33 \times 10^{63} \, \mathrm{years} \end{align}$$

And since the age of the universe is $1.4 \times 10^{10}$ years the timescale is about $10^{53}$ times greater than the age of the universe, just as it says in the paper.

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