# What does using these units (c=G=M=1) in MHD simulations mean?

I was reading some papers on numerical HD/MHD simulations of black hole accretion disks, and most of them scale their physical quantities to this form of units: c = G = M = 1. I know c = 1 and G = 1 correspond to geometrisized units which rely on length as a standard unit, but I wasn't sure what M=1 corresponds to in SI/cgs units (does it mean their standard unit is mass) and how to convert other units with this scaling?

For reference, this is one of the papers I was looking at: https://ui.adsabs.harvard.edu/abs/2012ApJ...761..130Y/abstract

• As to the simulation aspect, I suspect this will help the accuracy. c in SI units would be of order 10^8, while G would be of order 10^-11. Add in a square c or so, and the range of values is so large, double precision accuracy is not good enough to properly add or multiply values. Scaling everything to one will help with the accuracy of the simulations. Commented Jun 25, 2022 at 8:54
• @9769953: double precision is perfectly capable of handling numbers of this magnitude -- it can represent numbers higher than 10^308 or lower than 10^-308 without any problem. It is simply a question of keeping the equations uncluttered. Commented Jun 25, 2022 at 18:20
• @TonyK Add 1e-11 to 1e8; the result will remain 1e8. Of course, that is not a problem here, but there are times where that will matter, and you're better off normalizing your equations & numbers. It is not so much whether the number fits within the accuracy, but whether calculations with the numbers are precise. Commented Jun 25, 2022 at 19:15
• @9769953: You don't add lengths to masses. You are seeing a problem where none exists! No precision is lost or gained by the choice of units. Commented Jun 25, 2022 at 20:27
• @TonyK Of course, there would be proper factors to account for the different units. I'm purely talking about numerical accuracy, leaving aside the physical units etc. Besides, I never said anything about the units of those numbers. They are certainly not length nor mass. As for another example, that could be matrix inversion to solve a set of linear equations, where normalizing (and orthogonalizing) the matrix before inversion would be a good thing to do for numerical accuracy. You can generalize that principle to other equation solving methods. Commented Jun 25, 2022 at 22:49

It just means you choose a set of units where $$M=1$$. What those units are would then depend on what the mass of the object is.
For example, in geometrised units, the Schwarzschild radius is $$2M$$. In $$G=c=M=1$$ units, the Schwarzschild radius is $$2$$.
To convert to SI, you would multiply a length by $$GM/c^2$$. The conversions for other quantities would be similar to those for geometrised units, but anywhere a quantity featured the mass $$M$$ (energy, angular momentum), you would need to include mass in the conversion as well as the relevant combination of $$G$$ and $$c$$.
• @Ruslan the OP implied they understood geometrised units. If $r_s = 2GM/c^2$ and you set $G=c=1$, then $r_s = 2M$. Commented Jun 26, 2022 at 6:37