B Does gravitational force act on system or its components?

[jovi2k]

TL;DR Summary

Example: When the Earth pulls a proton, the proton also pulls Earth with force that is proportional to mass of the proton. That force is not the same as the force three quarks would exert on Earth. How the Earth "chooses" to interact with whole proton instead of three quarks?

I want to emphasize that I am not familiar with general relativity.
Consider a system of particles with masses $m_i$ in gravitational field of another body $A$. Total gravitational force exerted on $A$ will not be $$\sum G \frac{m_i m_A}{r^2}$$ where $r$ is distance between system and $A$ (I am assuming the system has small dimensions compared to distances at which gravitational force changes significantly). Instead force exerted on $A$ will be $$ G \frac{m_s m_A}{r^2}$$ where $m_s$ is mass of the system and it is not equal to the sum of masses $m_i$, instead it also includes kinetic and potential energy of the system in center of momentum frame divided by $c^2$. And we know this happens, because if we put a proton on a scale (theoretically) it will give a greater value than it would for three separate quarks. Why is this the case? Why $A$ sees a system, and not its components? Could we even say gravity acts on single quarks in the above example? How the Earth "chooses" to interact with whole proton instead of three quarks?

 

[Nugatory]

TL;DR Summary: Example: When the Earth pulls a proton, the proton also pulls Earth with force that is proportional to mass of the proton. That force is not the same as the force three quarks would exert on Earth. How the Earth "chooses" to interact with whole proton instead of three quarks?

The way GR works, gravity is not a force. There is no pulling going on. Instead, both the sun and the proton are (aside from electromagnetic forces on the proton, which likely swamp any gravitational effects) are in free fall, just drifting happily along in a straight line the way Newton's first law says they will.

The appearance of force comes because they are moving through curved spacetime, which causes their paths to converge or diverge. For an analogy, you could think about two people standing at the North pole, holding hands, and then walking south, both in a straight line but on slightly different lines of longitude - they will perceive a force that seems to be pulling them apart until eventually they have to let go of each others' hands and will end up many kilometers apart at the equator. Even better... search this forum for the video made by our member @A.T. showing the "hang" and "fall" path of an apple falling from a tree.

The curvature that makes all of this happens is calculated from a thing called the "stress energy tensor", which includes both masses but also energies and pressure. That includes the binding energy of the proton (although that is pretty much negligible compared with the binding energy of all the protons and neutrons in the sun).

 

[Dale]

Gravity interacts with anything that has stress-energy. Quarks have stress-energy and so do the electromagnetic fields as well as the strong and weak fields. Gravity doesn’t make any choice, it interacts with the whole system because the system is more than just the quarks.

 

[jovi2k]

Thank you @Nugatory and @Dale for your answers. I think I now understand better how gravity works.

 

[PeroK]

TL;DR Summary: Example: When the Earth pulls a proton, the proton also pulls Earth with force that is proportional to mass of the proton. That force is not the same as the force three quarks would exert on Earth. How the Earth "chooses" to interact with whole proton instead of three quarks?

I want to emphasize that I am not familiar with general relativity.
Consider a system of particles with masses $m_i$ in gravitational field of another body $A$. Total gravitational force exerted on $A$ will not be $$\sum G \frac{m_i m_A}{r^2}$$ where $r$ is distance between system and $A$ (I am assuming the system has small dimensions compared to distances at which gravitational force changes significantly). Instead force exerted on $A$ will be $$ G \frac{m_s m_A}{r^2}$$ where $m_s$ is mass of the system and it is not equal to the sum of masses $m_i$, instead it also includes kinetic and potential energy of the system in center of momentum frame divided by $c^2$. And we know this happens, because if we put a proton on a scale (theoretically) it will give a greater value than it would for three separate quarks. Why is this the case? Why $A$ sees a system, and not its components? Could we even say gravity acts on single quarks in the above example? How the Earth "chooses" to interact with whole proton instead of three quarks?

It is not known how gravity works on the scale of protons and quarks. The mass of a quark is not defined by how much gravitational force acts on it. In fact, at the quark scale, mass is a slippery concept.

What's needed is a quantum theory of gravity. The search for which is one of the big challenges of modern physics.

The equation you give is Newton's law of gravity, which is not compatible with the special theory of relativity. Mixing Newton and $E = mc^2$ is likely to lead to conceptual difficulties.

 

[Vanadium 50]

It is not known how gravity works on the scale of protons and quarks.

It's not as bad as that. It is known that individual neutrons fall down, and individual (neutral) atoms and molecules fall down. Proton-proton gravitational attraction is hard to see, as it is a correction in something like the 35th digit of their electrostatic repulsion.