# Conversion from Planck unit to SI [closed]

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?

## closed as off-topic by Kyle Kanos, JMac, ZeroTheHero, user191954, Bill NNov 19 '18 at 15:16

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• 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 – Triatticus Nov 14 '18 at 16:11
• 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? – Lucap Nov 14 '18 at 16:17
• 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. – Triatticus Nov 14 '18 at 16:18
• Thank you, i'm new i didn't know that – Lucap Nov 14 '18 at 16:21
• 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. – doetoe Nov 14 '18 at 17:10

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.

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.