# Why is (rest) mass not discrete? [duplicate]

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We say charge is discrete because it always has the form $q = n e$, where $n \in \mathbb Z$.

Similarly, I can say that $m = n N_a \mu$, where $n$ is number of moles, $N_a$ is avogadro's constant and $\mu$ is the mass of one atom/molecule of that element. Since $nN_a$ is the number of atoms/molecules, it will not make sense to have $\pi$ atoms or $1.5$ atoms, so $nN_a$ is a natural number.

By this mass is clearly discrete like charge. However, my textbook says the following

One of the main differences between charge and mass is that mass is continuous and charge is not.

## marked as duplicate by DilithiumMatrix, John Rennie, sammy gerbil, ZeroTheHero, YashasApr 10 '17 at 2:48

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## 1 Answer

You can decompose an atom into its sub-particles and observe masses which are not integer multiples of, in your terminology, $\mu$. Even without decomposing, you can still change the mass of an atom by giving it energy in different forms like heat, which will increase its overall energy hence mass, à la $E=mc^2$. This is not the case for charge.

• But charge can also be in fractions. – A---B Apr 9 '17 at 6:19
• @A---B Only as small as $e/3$, of which charges are integer multiples, whereas masses are more like real numbers. – J.G. Apr 9 '17 at 7:50
• @J.G. But you can only divide mass to the level of protons, neutrons and electrons right ? and there is no half proton or 0.1 electron. So how mass is real number ? – A---B Apr 9 '17 at 11:31
• @A---B It's partly because there isn't a neat ratio for those particles' masses (not to mention some others we won't get into), but the other issue is binding energy reduces slightly the total mass of a nucleus, atom or molecule because $E=mc^2$. – J.G. Apr 9 '17 at 11:34
• @J.G. I can't argue much because I don't know much. But the obvious conclusion is that mass was discrete before Einstein was born. – A---B Apr 9 '17 at 11:41