Do protons and neutrons have different masses based on what kind of atom they are a part of? When atoms undergo fission or fusion, they release energy according to the famous e=mc2 equation. My understanding of the process is that mass is somehow being lost when 4 hydrogen atoms fuse into a helium atom or when a uranium atom splits into krypton and barium or whatever else it splits into. Does this mean that if we had a method to precisely measure the mass of such small particles, that we would find that a proton in a hydrogen or uranium atom has more mass than a proton in an iron atom? Or the neutrons in a uranium atom compared to a neutron in iron?
Short answer is yes.
Long answer is also yes, but it’s complicated.
You can’t actually pick out and measure a nucleon while it is part of a nucleus, you weigh the nucleus, count up how many nucleons there are, and divide by that number to get the average mass per. That mass will be less than that of a free proton or neutron.
What’s happening is that the binding energy is being converted into kinetic or electromagnetic energy. Binding energy makes up 99% of a proton or neutron’s rest mass, and they have less of it as you build atoms up to iron/nickel. A helium nucleus has about 4% less binding energy per nucleon than the protons that went into making it.
Longer answers go into nuclear shell models and maths that are above my pay grade.
Interaction itself can have mass (i.e., binding energy), so no mass is “lost”. Other than that, what do you mean (why would you divide the energy by the number of nucleons)? But it does seem that to define something like the value of quark masses we need to go into “nuclear shell models and maths that are above [our] pay grade” like some kind of lattice QCD calculations, but we can, in principle, do it, even though we do not have the free particles to work with directly.
A helium nucleus has ~96% the mass of 4 protons. Mass is lost and converted to energy in the fusion reaction.
The question in the OP was whether nucleons massed less when in a nucleus, and the answer is a qualified yes. In as much as it means anything to talk about an individual nucleon’s mass when it is part of a nucleus, they do in fact mass less than they did when they were free particles. Which is why you would divide the mass of the new nucleus by the number of nucleons, and see that the average mass of those nucleons is less than it was before the fusion.
We aren’t really defining quark masses here, and it’s likely that they are unchanged, but the mass of the quarks only makes up ~1% of the mass of their respective nucleons. It is the binding energy in the strong force that makes up the bulk of the mass, and that is what is lost as it goes towards iron/nickel on the periodic table.
If we are trying to explain why, or trying to predict the mass of various nucleons rather than determine them empirically, that’s gonna require fairly complicated math.
I’d argue the appropriate interpretation is that the nucleons in the nucleus do in fact have the exact same mass as free nucleons, but the mass of a system is not simply the sum of the masses of its constituent parts. I.e. the idea that the average mass to nucleons in a helium nucleus is just “mass of helium nucleus divided by four” is wrong.
And that’s why I say it doesn’t really mean anything to talk about an individual nucleon’s mass when it is part of a nucleus.
At the same time, they don’t have the same mass as they would when they were free, or you’d have to be adding negative mass to the helium nucleus to make it balance out.
I specifically said that the mass comes from the binding energy, not from the mass of the quarks, and the amount of binding energy does in fact change when you fuse or fission an atom.
Nothing about simple summing.
I mean, even though that’s not what I said, that is actually how you would get an average, so I’m not sure how you say it is wrong.
Meaningless, probably. But the OP’s question was not whether measuring the mass of an individual nucleon was useful, it was whether the mass of an individual nucleon in a helium nucleus is less than that of that same nucleon while free. And the answer to that, is “yes”.
Otherwise, where do you think the mass is hiding?
How would you explain the different masses of Pions and Kaons, if the mass doesn’t change based on binding energy?
Mass = energy, but I think it has to add up: in the case of helium the binding energy is -28.3 MeV, the mass of a proton is 938.3 MeV, and the mass of a neutron is 939.6 MeV. Alpha particle mass is 3727.4 MeV, and 2(938.3)+2(939.6)−28.3 = 3727.5 (close enough, considering all the rounding). That is, at this level of approximation, we might as well think of the total mass as the sum of the particle masses and their (negative) binding energy.
It does seem to be very very roughly true that if you compute the binding energy of a nucleus divided by the number of nucleons, then the largest contribution is roughly constant; see Semi-empirical mass formula - Wikipedia
It does add up, but that - sign in front of the binding energy is the amount of mass that is lost in the process. That’s where the energy we hope to harness from fusion comes from.
Does it really make sense to think of binding energy as negative energy?
In the case of helium the total amount of binding energy is actually closer to 3690 MeV, when you account for the binding energy in the nucleons themselves (remember, quarks only make up ~1% of the mass, the rest is binding energy.) I don’t know that it makes sense to have two different entries for binding energy, one positive and one negative.
So, binding energy in 4 protons=~3,715MeV. Binding energy in helium nucleus=~3690Mev. There is less binding energy in the end result than in the parts that went in. The binding energy that is lost from the system is the energy that comes out. It seems you are saying that it should be that the binding energy of a helium nucleus should be ~3715MeV - ~28.3MeV. I agree with that, but what is the purpose of leaving the equation unsolved?
It’s not like you are subtracting oranges from apples here, you are subtracting apples from apples.
Sorry if it wasn’t clear; the 28.3 MeV is from the residual nuclear force (binding the protons/neutrons together); as you say it does not count forces between quarks.
As for the minus sign, I was thinking of this kind of diagram:
(the bound state having lower energy than the unbound state)—to blast apart the nucleus you would have to put that energy back in, or, conversely, you can get it out by fusing the nucleus together
OK, sure, so add negative mass to the helium nucleus. Why not?
And while we’re at it, a proton or neutron masses much more than the sum of the masses of its constituent quarks.
Because if you add negative mass, then you are subtracting mass. That’s fine, it just means you end up with less mass. That was my point.
I thought I had already pointed that out.
Given the binding force inside a nucleus is mediated by the exchange of pions, which themselves are composed of two quarks, I wonder if it makes sense to talk about “the” mass of nucleons in a nucleus. The nucleus is a dynamic thing, pion exchange is occurring constantly in order to keep the thing together. At any instant one might say that the set of nucleons have a range of masses depending upon what exchange they are participating in.
Also, way above my pay grade.
I don’t think it really does. There is no way, even in principle, to measure the mass of an individual nucleon inside the nucleus. You’d have to remove it, which ruins the whole experiment.
But the OP didn’t ask if it made sense, they asked, in a way, if it did make sense, would the mass be reduced.
And while the question itself doesn’t have a great answer that makes sense, it still brings up the point that most of the mass of a proton is not intrinsic. A proton in free space needs to be a certain mass to be stable, but that doesn’t mean that a proton is defined by that mass.
Right, there is a quark field for each quark. A quark has an intrinsic mass that must be the right mass for to exist as a real particle. There is no proton field, a proton is a composite particle, it has no intrinsic mass, it has the mass of its quarks plus the binding energy.
If you could somehow measure the mass of a proton in a nucleus of helium, it would mass less than a proton by itself. But, you’ve brought up a good point that it would mass even less, as some of that mass/energy is no longer in any of the nucleons, but is instead in the process of moving between them.
At some level, “mass” and “energy” are just accounting buckets we humans use to understand what’s going on. Where we draw the lines is more a matter of convenience than intrinsic to nature.
The basic answer to the OP is indeed ‘yes’, but the proton’s (or, generally, nucleon’s) “effective mass” is not uniquely defined. The most useful effective mass definition for a given context will vary. One perfectly valid (but not actually that useful) effective mass definition would be just dividing the total nucleus’s mass by the number of nucleons (or something slightly more clever given neutron-proton differences). Nothing wrong with that.
But more commonly of interest is some effective mass that allows one to approximately treat a nucleon as a quasi-independent particle inside a potential (i.e., experiencing forces from the rest of the nucleus). This in general leads to dynamic effective masses.
You can. The most effective way to probe nuclear structure at that scale is with electron scattering. You shoot electrons with suitable momenta at a nucleus and watch them collide with / scatter off the nucleonic substructure, thereby measuring properties of those nucleons.
But many other experimental phenomena allow access to the distinct* nucleons within a nucleus.
*How distinct they are depends heavily on the momentum transferred from the broader system (e.g., the incoming colliding particle). For the lowest momentum transfers, the nucleus as a whole is all that is resolved; individual nucleons are not noticed. At much higher momentum transfers, the nucleon substructure becomes the most relevant, and the nucleons themselves become the unseen forest for the quark/gluon trees inside. These two regimes (called “coherent” and “deep inelastic”, respectively) have lying between them the regime of interest here.
Indeed, nucleon effective masses are generally dynamic and can depend on the nucleon’s instantaneous momentum or energy, or on the kinematic specifics of whatever collision / scattering process is being used to probe the system.
Sort of? Quarks were an unlucky choice for your example here as their masses are a deep topic, and when a quark’s mass is listed, the specific definitional scheme must also be listed. This stems from their confined nature. (But this is just a pedantic and not overly relevant point of interest for the thread at hand.)
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