It is common to state that a proton is bounded state of three quarks, and that the QCD energy binding (associated to a "cloud" of gluons joining together the three quarks) is responsible for the 99% of the inertial mass of a proton. For example, Durr et al. (Ab-initio Determination of Light Hadron Masses), confidently state that: According to the mass-energy equivalence, $E = m c^2$, we experience this energy as mass. This is an informal statement, and it is not clear to me why any amount of energy in the form of massless quanta of a quantum field contribute to "inertial mass".

In addition, I have checked other articles computing the proton mass, for example, Yang et al. (2018), Proton Mass Decomposition from the QCD Energy Momentum Tensor, presents a detailed and beautiful computation for the component $\langle T_{44}\rangle$ of stress-energy tensor of a proton. Taking the "QCD-mass of a proton" as $M_p = \langle T_{44}\rangle/c^2$, Yang et al. obtain that $M_p = f(m_\pi^2)$ where $m_\pi$ is the mass of a pion, taking the experimental value of $m_\pi c^2 \approx 139$ MeV, they obtain $M_p c^2 \approx 938$ MeV a beautiful numerical agreement.

But: (1) How can we demonstrate that the bounded quark-gluon state which constitutes a proton presents a resistance to be accelerated, or ''inertial mass'', equivalent to $E_p/c^2$? (I understand the informal affirmations, but I am looking for a clear/formal mathematical proof/argument/computation).

My guess is this one: A box contains a quantity of energy, an observer at rest with respect to the center of the box will measure an stress-energyimpulse tensor given by:

$$T^\mu_\nu = \begin{pmatrix} \epsilon & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$

(assume for simplicity the space-time is Minkowskian with metric -+++). If the box is accelerated until it moves with uniform velocity (directed along the X-axis) the final stress-energy tensor will be:

$$T^\mu_\nu = \begin{pmatrix} \gamma\epsilon & \gamma\beta\epsilon & 0 & 0\\ \gamma\beta\epsilon & \gamma\beta^2\epsilon & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$

with $\gamma = 1/\sqrt{1-v^2/c^2}$ and $\beta = v/c$. During the acceleration process we need a force density $f$, and the equations of motion will be:

$$\frac{1}{c}\frac{\partial T^0_0}{\partial t} + \frac{\partial T^1_0}{\partial x} = \gamma \frac{fv}{c}$$ $$\frac{1}{c}\frac{\partial T^0_1}{\partial t} + \frac{\partial T^1_1}{\partial x} = \gamma f$$

Assuming that the force is continuous and only acts for a finite time, it is easy to see that the force will always be proportional to:

$$f\propto \frac{\text{d}}{\text{d}\tau}\left( \frac{(\epsilon/c^2)v}{\sqrt{1-v^2/c^2}} \right)$$

Thus the term $\epsilon/c^2$, behaves indistinguishably from a mass density at rest $\rho$.

  • 2
    $\begingroup$ Einstein's equivalence principle. The energy must gravitate because the Einstein tensor is proportional to the stress-energy tensor. So the energy must have a gravitational mass. Since the EP tells us the gravitational and inertial masses are the same it must have an inertial mass. This isn't a derivation, because there is no derivation of the EP, but the validity of the EP is (so far) confirmed by experiment. $\endgroup$ May 17 at 10:17
  • $\begingroup$ But a single photon also gravitates, because it has energy, but that does not imply that it contributes to the inertial mass. $\endgroup$
    – Davius
    May 17 at 10:25
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    $\begingroup$ "But a single photon also gravitates" - it's not that simple. I can't speak with any authority but I believe a single photon does not have a gravitational field in the sense a point mass does. Rather it produces a gravitational wave that moves along with it. A bound state of photons (a kugelblitz) would have a gravitational mass but a single photon, or indeed any massless particle, does not have a gravitational field in the sense we usually mean. $\endgroup$ May 17 at 10:41
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    $\begingroup$ Because in GR the curvature is coupling to the stress-energy tensor, it is clear that it is NOT mass that is gravitating, but rather ALL the energy. And since rest mass is really a form of energy, just that it dominated old physics energy, that is why Newtonian gravity wrote masses rather than energy. Similarly, relativity also shows that it is energy that has the inertia property, not mass. All the old physics references to masses are just a mislabelling. Once you understand that, you will have a leap of understanding into modern physics. $\endgroup$ May 17 at 14:34
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    $\begingroup$ The energy and momentum of photons would also contribute to the stress energy tensor. Because there is no rest frame for a photon, there is a mild difference in how a single photon behaves v.s. a single localisable massive particle. Otherwise, it is all the same. It does not matter, because when we just consider two or more non-parallel photons, then there is a rest frame, and then you can see that they would have gravitational and inertial properties, precisely because they have energy. $\endgroup$ May 17 at 14:38


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