<p>It just means you choose a set of units where <span class="math-container">$M=1$</span>. What those units are would then depend on what the mass of the object is.</p>
<p>For example, in geometrised units, the Schwarzschild radius is <span class="math-container">$2M$</span>. In <span class="math-container">$G=c=M=1$</span> units, the Schwarzschild radius is <span class="math-container">$2$</span>.</p>
<p>To convert to SI, you would multiply a length by <span class="math-container">$GM/c^2$</span>. The conversions for other quantities would be similar to those for <a href="https://en.m.wikipedia.org/wiki/Geometrized_unit_system" rel="nofollow noreferrer">geometrised units</a>, but anywhere a quantity featured the mass <span class="math-container">$M$</span> (energy, angular momentum), you would need to include mass in the conversion as well as the relevant combination of <span class="math-container">$G$</span> and <span class="math-container">$c$</span>.</p>

CC BY-SA 4.0
2022-06-25T20:17:32.350Z
This answer would be more useful if it listed the expressions to convert the relevant quantities between nondimensioned form and the dimensionful one. It doesn't seem to be straightforward when a length is equal to two masses.
Ruslan
CC BY-SA 4.0
2022-06-26T06:37:40.780Z
@Ruslan the OP implied they understood geometrised units. If $r_s = 2GM/c^2$ and you set $G=c=1$, then $r_s = 2M$.
ProfRob
CC BY-SA 4.0
2022-06-24T15:38:52.873Z
2022-06-26T06:52:19.193Z
2022-06-26T06:52:19.193Z
ProfRob
<p>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?</p>
<p>For reference, this is one of the papers I was looking at: <a href="https://ui.adsabs.harvard.edu/abs/2012ApJ...761..130Y/abstract" rel="nofollow noreferrer">https://ui.adsabs.harvard.edu/abs/2012ApJ...761..130Y/abstract</a></p>

CC BY-SA 4.0
2022-06-25T08:54:34.513Z
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.
9769953
CC BY-SA 4.0
2022-06-25T18:20:47.507Z
@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.
TonyK
CC BY-SA 4.0
2022-06-25T19:15:53.600Z
@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.
9769953
CC BY-SA 4.0
2022-06-25T20:27:17.313Z
@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.
TonyK
CC BY-SA 4.0
2022-06-25T22:49:11.403Z
@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.
9769953
CC BY-SA 4.0
2022-06-26T00:13:47.290Z
@9769953: I see you're not listening, so I won't say any more.
TonyK
CC BY-SA 4.0
2022-06-26T08:25:06.700Z
@TonyK I don't disagree with what you are saying, but I feel it misses the point about numerical stability. Using proper solvers helps that, but normalizing the equations (or even rewriting them in a log space) can help with keeping the computations stable.
9769953
<accretion-discs><units>
What does using these units (c=G=M=1) in MHD simulations mean?
Ferzem Khan
# What does using these units (c=G=M=1) in MHD simulations mean?

(Asked by: username_0 on 2022-06-24)

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>

**Tags:** accretion-discs, units

> **Comments:**
> **username_1**: 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.  
> **username_2**: @username_1: 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.  
> **username_1**: @username_2 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.  
> **username_2**: @username_1: 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.  
> **username_1**: @username_2 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.  
> **username_2**: @username_1: I see you're not listening, so I won't say any more.  
> **username_1**: @username_2 I don't disagree with what you are saying, but I feel it misses the point about numerical stability. Using proper solvers helps that, but normalizing the equations (or even rewriting them in a log space) can help with keeping the computations stable.  

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## 1 answers

### Answer by username_3 on 2022-06-24. Score: 8

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](https://en.m.wikipedia.org/wiki/Geometrized_unit_system), 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 $c$.

> **Comments:**
> **username_4**: This answer would be more useful if it listed the expressions to convert the relevant quantities between nondimensioned form and the dimensionful one. It doesn't seem to be straightforward when a length is equal to two masses.  
> **username_3**: @username_4 the OP implied they understood geometrised units. If $r_s = 2GM/c^2$ and you set $G=c=1$, then $r_s = 2M$.  

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