Convert units of a $N$-Body simulation where $G$=$M$=$L$=1

I think that it can not be so hard, but I don't get it. I have a $N$-Body simulation where total mass $M$ and the gravitational constant $G$ and the length $L$ are equal to one: $M$ = $G$ =$L$ = 1. I just don't see how I can convert the units of my simulations into SI units. If I have a time step of $t=0.001$, how many years represents such a step? And if I have a particle with mass $m_i = 1/N$, what would it be in kilograms?

I hope someone can give me a hint.

• $G$ units are $m^3 kg^{-1} s^{-2}$, so $M = G = 1$ to not fully characterize your units, you need a length, a time or something set to 1 as well. Only the particle mass can be worked out, the unit of mass in your simulation is $M$, so $m_i = (1/N) M$, where expressing $M$ in SI units give $m_i$ in SI units.
– Tony
Feb 23, 2017 at 14:32
• @Tony If I have a star cluster where each star has the same weight and I want to study the evolution of such a system for a given period of time, say 1 million years, how do I now how many time step I need? Feb 23, 2017 at 20:06
• This is a difficult question. I do not think a general answer can be given, it depends on the specificities of the problem at hand and the algorithm used to perform the evolution.
– Tony
Feb 24, 2017 at 10:21

Taking $M = G =L = 1$ basically means that you are using the following set of units: masses are expressed in units of $M$, lengths in units of $L$, and times in units of $\sqrt{L^3 / (M G )}$ (convince yourself that this is indeed homogeneous to a time).

Therefore, $m_i=1/N$ really means $m_i= (1/N) M$ and $t=0.001$ really means $t=0.001 \sqrt{L^3 / (M G )}$. Expressing $M$, $G$ and $L$ in SI units give you $m_i$ and $t$ in SI units.

• Ok, if I want to simulate a cluster of 1000 sun like stars, so mass would be 1000 sun masses, uniformly distributed inside a unit sphere (Radius =1) for 100 time steps $dt=0.001$, how exactly would I compute the real diameter of that cluster. And how do I compute the mass to SI units if I multiply it by $M=1$? Feb 24, 2017 at 12:32
• Masses are then expressed in units of 1000 sun mass. I.e. $M = 1$ means $M = 1000$ sun mass, and $m_i=1/N$ means $m_i = (1/N) 1000$ sun mass. Just put the sum mass in $kg$ in the formula for $m_i$ and $m_i$ is in $kg$. The real diameter is set by the physical situation you are considering, e.g. the diameter of some galaxy.
– Tony
Feb 24, 2017 at 14:20
• Ok, I think I understand it now better! So let me try: If I want to get the velocity in SI units I multiply the velocity of my simulation, lets say $v=0.1$ with $V=\frac{L}{T}=\frac{GMT}{L^2}$. So I compute $v = 0.1\frac{GMT}{L^2}$, right? Feb 24, 2017 at 16:35
• I think you got the point, but you mixed up the units. What is $T$ is your formula ? $T$ should be the time scale you set to 1, which is $\sqrt{L^3/(MG)}$ as we have seen, so $V = L/T = L/\sqrt{L^3/(MG)}$. Then $v=0.1 L/\sqrt{L^3/(MG)}$.
– Tony
Feb 25, 2017 at 15:40
• I just wondered how to compute the energy of a system using $M=G=L=1$. I know that $J = \frac{kg m^2}{s^2} = kg (\frac{m}{s})^2$. Then I can use the units from above and get for the energy $J = \frac{ML^2}{T^2} = MV^2 = \frac{M^2G}{L}$. Does that mean, that I can compute the energy like: $E_{real} = E_{simulation}*J = E_{simulation}*\frac{M^2G}{L}$? Apr 13, 2017 at 7:14

Setting just $G$ and $M$ to 1 is not enough to establish the time and length scaling as well. However, if you use Planck units, where $G=1$ and say your mass is 1 plank mass ($2.17\ 10^{-8} kg$), then you will know what your times and masses mean in MKS.

• What else do I have to consider if I want to work with SI units? Feb 23, 2017 at 14:43
• I'm not sure what you mean
– milo
Feb 23, 2017 at 14:45
• If you set G = 1, then you switch to a different set of units, and to get SI you need to switch back. The most common units where G = 1 are the Planck units. In the linked Wikipedia article there are conversion tables from Planck units to SI and back.
– milo
Feb 23, 2017 at 14:48