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Special relativity tells us that interactions with forces contributes to the masses of objects. In the case of electromagnetic interactions this causes the slight difference in mass between protons which are charged and neutrons which are neutral, for instance. As far as I have understood it in a classical picture, this can be explained via the emission of radiation. Accelerating charged particles emit radiation in the form of photons which slows them down (either by self force arguments or energy conservation arguments), this slowing down can be thought of as an increased mass compared to what would be expected for a neutral particle.

I have two questions,

  1. Radiation has a dependence on acceleration and velocity. Therefore one would expect this perceived mass to be dependent on velocity or its derivatives. Thankfully special relativity also tells us that mass has a velocity dependence in the form of m$\gamma$. Are these velocity dependencies consistent with each other?

  2. I know that other forces like the strong force also contribute to the mass of particles, and I am wondering if the mechanism behind them is the same? By that what I mean is, are particles also radiating through the strong force which causes them to slow down which is perceived as an additional mass? The problem that seems to have is that they would have to emit gluons instead of photons but gluons cannot be isolated as far as I know.

I understand that the intuition I am presenting here is completely classical so it may not translate well to the strong force which really needs a quantum mechanical description.

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    $\begingroup$ The correct mass difference between proton and neutron comes from the whole standard model of particle physics (and possibly beyond). See physics.stackexchange.com/q/34245 $\endgroup$ Commented 22 hours ago
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    $\begingroup$ What Jeanbaptise Roux said, and on top of that, if the mass difference of proton v.s. neutrons are due solely to electromagnetism, then the proton should be heavier than neutron, but it is manifestly the opposite. $\endgroup$ Commented 22 hours ago
  • $\begingroup$ Related: physics.stackexchange.com/q/721760 $\endgroup$
    – Hyperon
    Commented 21 hours ago

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