# 'Density' of a proton

I was doing some exercises the other day, when I came across this question in my book:

A proton weighs about 1.66 x 10-24 g and has a diameter of about 10-15 m. What is its density in g/cm3?

As you can see, a really simple and standard question, but... does it even make sense to say that a proton has a mass? And calculate its density?! If so, do I consider it a sphere?

It seems to me, subatomic particles are so small it doesn't really make sense to talk about mass, volume and density as if we were talking about... tennis balls!

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so, where do you think the mass comes from, if not from elementary particles.. ? from the dark matter? –  mykhal Aug 30 '12 at 12:02
You need to consider it as sphere. –  BigSack Aug 30 '12 at 13:04
I bet you are not surprised a hydrogen atom to have a mass. A hydrogen atom consists of a proton and an electron. Most of its mass is due to proton. And a mass of a hydrogen molecule $\text H_2$ is almost a mass of two protons in it. –  Yrogirg Aug 30 '12 at 16:37
The "diameter" comes from the root mean square charge radius, which is basically a measure of the "average" extent of the charge from a center point. –  Snowball Aug 30 '12 at 16:41
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## 2 Answers

A proton is a bound state of three quarks. The quarks themselves are (as far as we know) pointlike, but because you have the three of them bound together the proton has a finite size. It doesn't have a sharp edge any more than an atom has a sharp edge, but an edge is conventionally defined at a radius of 0.8768 femtometres. Protons are spherical in the same way that atoms are spherical even though they're made up of discrete electrons.

The three quarks have a mass, but actually the proton is a lot heavier than the combined mass of the three quarks. That's because the binding energy of the quarks is very high, and that energy increases the mass in line with Einstein's famous equation $E = mc^2$.

So, yes, it does make sense to calculate a density for the proton just as it makes sense to calculate a density for an atom.

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The proton is not a bound state of three quarks, it has the symmetry numbers of three quarks put together. It is so much more massive than the quarks that it makes no sense to say that there are 3, since quarks and antiquarks can and do pop out of the vacuum. The proton's interaction with the quark condensate alone makes the quark number meaningless. –  Ron Maimon Aug 31 '12 at 3:21
Completely meaningless? Would it not have an average of 3 quarks at a given time even if there are occasionally more or less, the same way temperature is an average of the kinetic energy yet it fluctuates from particle to particle? –  Ehryk Sep 4 '12 at 8:38
The proton is a bound state, and it's valence content is three quarks. Calling a proton "a bound state of three quarks" may leave out a lot of information, but it is certainly sensible. Note that the radius figure that John has chosen here is define the thing. A different way to approach the problem would be to ask for the density of nuclear matter in the large $A$ limit which is similar and perhaps better defined. –  dmckee Sep 13 '12 at 16:44
@Ehryk: It's completely meaningless due to the condensates, the "quark number" is indefinite in an eigenstate. The "3 quark" business is a symmetry under global transformations, and it should not be thought of as three particles bound together weakly, because it isn't. –  Ron Maimon Sep 14 '12 at 2:55
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i would say that if $\rho (density) = \frac{m}{V}$ and the volume must be proportional to $V= m |\Psi (x,t)|^{2}$ then $\rho = \frac{1}{|\Psi(x,t)|^{2}}$

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