In an Tolman Oppenheimer and Volkoff (TOV) equation of state for static equilibrium stars, if any one want to calculate pressure and density in $\mathrm{MeV/fm^3}$, then what is the appropriate choice for the value and the units of constant $G$ and $c$. Are they both remain same as $$G = 6.6743\times10^{-8} \quad\text{and}\quad c = 2.99792458\times10^{10}$$ as in CGS or $c=G=1$ as in geometrical units, or any other suitable value with their appropriate unit when Pressure and density is taken in $\mathrm{MeV/fm^3}$. (I think $\mathrm{MeV/fm^3}$ is a unit of pressure and density in Planck units.)


$\mathrm{MeV/fm^3}$ is a unit of energy density and a unit of pressure in SI-compatible units. In its full glory, it goes \begin{align} 1\:\mathrm{MeV/fm^3} & = 10^6\:e \:\mathrm V/(10^{-15}\:\mathrm m)^3 \\ & = 10^{51}\times1.602 \times10^{-19} \:\mathrm {C\:V/m^3} \\ & = 1.602 \times10^{32} \:\mathrm {J/m^3} \\ & = 1.602 \times10^{32} \:\mathrm {N/m^2} \end{align}

If you want to calculate those quantities in that SI unit, and the rest of the quantities involved are in SI units, then you should use the SI values of those constants, $$ G=6.674\:08\times10^{−11}\mathrm{ m^3\:kg^{−1}\:s^{−2}} \quad\text{and}\quad c=299\:792\:458 \:\mathrm{m \: s^{-1}}. $$ If you want to start with CGS values and end up with an SI final quantity, then you should really convert the CGS values to SI and then work in SI, but you can also use $G$ and $c$ in mixed units (which is discouraged unless you know what you're doing).

  • $\begingroup$ If MeV/fm^3 is a unit of energy density and a unit of pressure in SI units then what we measure pressure in N/m^2 or kg/m*s^2. $\endgroup$ – Umar Khan Apr 16 '17 at 16:32
  • $\begingroup$ @UmarKhan SI unit was perhaps a slight mischaracterization - the electronvolt is SI-compatible but not core SI. Note that $\mathrm {eV/m^3}$ is equivalent to $\mathrm {J/m^3}$ up to a numerical factor, and because $1\:\mathrm {J}=1\:\mathrm {N\:m}$, the units for energy density are identical to the units for pressure, $\mathrm {N/m^2}$. $\endgroup$ – Emilio Pisanty Apr 16 '17 at 16:59

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