# Natural units question

by JoePhysicsNut
Tags: natural, units
 P: 34 I'm slightly confused about natural units. Take mass as an example: I can measure something in kg's, but then decide to convert to MeV/c^2, for instance. To do that I would multiply the quantity in kg by 3*10^8 squared and divide by 10^6*1.6*1-^(-19) i.e. the SI values of the constants. If I then want to express the mass in natural units, which would be MeV instead of MeV/c^2, would I multiply by 3*10^8 squared again?
P: 34
 Quote by JoePhysicsNut I'm slightly confused about natural units. Take mass as an example: I can measure something in kg's, but then decide to convert to MeV/c^2, for instance. To do that I would multiply the quantity in kg by 3*10^8 squared and divide by 10^6*1.6*1-^(-19) i.e. the SI values of the constants. If I then want to express the mass in natural units, which would be MeV instead of MeV/c^2, would I multiply by 3*10^8 squared again?
OK, so I've figured out that the answer is no I shouldn't multply by 3*10^8 squared again. The mass would be the same number in MeV/c^2 as it is in MeV, only in the latter case it's measured in natural units.

But to rephrase my question: if the number didn't change when going from MeV/c^2 to MeV, did the definition or "size" of MeV change?
 Mentor P: 10,854 In natural units, c=1. MeV/c^2 is compatible with SI units, MeV is not. $$1kg = 1kg\frac{m^2}{s^2} \frac{s^2}{m^2} = 1J \frac{s^2}{(3\cdot 10^8 m)^2} 9\cdot 10^{16} = 1J/c^2 \cdot 9\cdot 10^{16}$$ $$1J=1C\cdot 1V = 6\cdot 10^{18} q_e \cdot 1V = 6\cdot 10^{12}MeV$$ Therefore: $$1kg = 1J/c^2 \cdot 9\cdot 10^{16} = 5.4\cdot 10^{29}\frac{MeV}{c^2}$$ As you can see, I did not multiply with anything, I just converted units. (a better value for the inverse electric charge would give 5.61 instead of 5.4)

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