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Natural units question

by JoePhysicsNut
Tags: natural, units
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JoePhysicsNut
#1
Feb12-13, 12:25 PM
P: 35
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?
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JoePhysicsNut
#2
Feb12-13, 02:54 PM
P: 35
Quote Quote by JoePhysicsNut View Post
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?
mfb
#3
Feb12-13, 03:09 PM
Mentor
P: 11,928
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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