Proton pressure: a simplified model

Abstract

In this work, the authors present a simplified method for determining proton pressure, based on concepts from statistical mechanics, quantum chromodynamics – specifically asymptotic freedom of quarks – and other well-established physical concepts. The result obtained for the proton pressure was $0.3570\times {10}^{\text{35}}$ Pa, which is consistent with recently reported experimental data.

Keywords:
Proton pressure; MIT bag model; fermion gas

1. Introduction

Recently, the first measurement of the proton’s internal pressure distribution was published [¹], showing it to be the highest known pressure to date, on the order of ${10}^{\text{35}}$ Pa. This initially surprising result can have its order of magnitude calculated from well-established concepts in statistical mechanics and quark asymptotic freedom [², ³].

More recently, a theoretical article presented a calculation made using Lattice-QCD [⁴] that shows the dependency of the proton’s pressure with respect to the distance to the center of the proton $r$, asserting that the maximum pressure obtained is consistent with the experimental result mentioned above..

We will use a very simplified model, which has more of a pedagogical motivation than a theoretical precision. Within the model’s limitations, the theoretical foundations are accurate. In this model, the proton is viewed as a spherical cavity in which constituent quarks are confined but free to move. The hypothesis of quark mobility is based on the fact that at short distances, QCD interaction is weak, so quarks behave almost as free particles. However, as the distance between quarks increases, the interaction rapidly strengthens, leading to confinement. Thus, an approximation would be that within the proton volume, quarks are free particles, but the proton boundaries act as a potential barrier, forcing confinement.

Next, using Fermi-Dirac statistics, we will assume that confined quarks behave as a Fermi gas and derive an expression for calculating proton pressure. We will present the results and discuss the assumed conditions.

2. Proton Modeling: Asymptotic Freedom and the Ideal Gas Assumption

According to the Standard Model, a proton is a baryon composed of 3 quarks – two up and one down. The predominant force between quarks is the strong force, described by Quantum Chromodynamics (QCD). Both quarks and gluons, the mediating particles of the strong force, are assigned charges called color.

To date, no free particle with color charge has been detected, leading to the belief that they are confined within baryons and mesons.

The strong force is powerful enough to keep quarks confined, but at short distances, the phenomenon of asymptotic freedom occurs. The closer the quarks are, the weaker the coupling becomes, and consequently, quarks interact much more weakly.

The QCD coupling constant ${\alpha }_s$, shown in Fig. 1, varies with distance $r$, where $r=1/Q$, and $Q$ is the momentum scale felt in the interaction, displayed on a logarithmic scale. At short distances, the coupling is weak – asymptotic freedom. At large distances ($r>1$ fm), the coupling becomes strong, leading to confinement.

Figure 1
Plot of the coupling constant ${\alpha }_s$ as a function of distance $r$, where $r=1/Q$, with $Q$ being the momentum scale probed in the interaction – shown here on a logarithmic scale. At short distances, the coupling is weaker, a property known as asymptotic freedom. The color points correspond to data from $\tau$ decays, jets in deep inelastic scattering, heavy quarkonia, jet shapes in $e^{+}{e}^{-}$ annihilation, fits to the Z-pole, and jet production in proton-proton and proton-antiproton collisions, while the black line is a fit based on Lattice QCD results [⁵].

Thus, we can consider the approximation that the quarks constituting the proton move as free particles,

The asymptotic freedom regime allows calculations using perturbation theory, and its discovery enabled QCD theory to develop similarly to QED. For the confinement regime, perturbation theory is not applicable, making its theoretical demonstration difficult.

Given this difficulty, several confinement models emerged, including the MIT Bag Model [⁶]. In this model, the fields describing quarks and gluons are restricted to a specific region called a “bag”. A surface is specified, and it is required that color current does not pass through it. Since both quarks and gluons carry color charge, neither can cross this barrier.

Therefore, inspired in the MIT bag model, we will assume that the proton is a hollow sphere with a hard (impenetrable) surface and inside it there is an ideal gas of free quarks, but we don’t make any claim regarding external QCD vacuum pressure or gluons.

3. Ideal Fermi Gas

We know that quarks are particles governed by Fermi-Dirac statistics, as they have half-integer spin, specifically spin $1/2$. Thus, in our model, we assume, initially, that the proton consists of a non-relativistic fermion gas. To describe it, we need a partition function as well as the energy for each quantum state. We start with the quantum states of the system.

3.1. Quantum states in a solid sphere

For a particle of mass $m$ in a sphere, the eigenfunctions are given by [⁷]:

(1) $${\psi }_{nlm}(r,\theta ,\varphi )=A{j}_l\left(k_{nl}r\right)Y_l^m(\theta ,\varphi )$$

where $A$ is the normalization constant, $j_l$ are the first order Bessel functions, and $k_{nl}={\alpha }_{nl}/R$ with ${\alpha }_{nl}$ being the $n$-th zero of the Bessel function $j_l$.

The energy eigenvalues are given by:

(2) $$E_{nl}\left(m\right)=\frac{{\hslash }^2{\alpha }_{nl}^2}{2m{R}^2}$$

where $n=1,2,\dots$ is the radial number and $l=0,1,\dots$ is the orbital angular momentum.

Defining the single particle states as $\lambda \equiv (m,n,l,m_l,s)$ the energy of those states are given by:

(3) $${\epsilon }_{\lambda }^{\left(u\right)}=\frac{{\hslash }^2{\alpha }_{nl}^2}{2m_u{R}^2},{\epsilon }_{\lambda }^{\left(d\right)}=\frac{{\hslash }^2{\alpha }_{nl}^2}{2m_d{R}^2}$$

Note that the numbers ${\alpha }_{nl}$ are the same for both flavors, $u$ and $d$, changing only their masses. Given that $m_d>m_u$, then ${\epsilon }_{\lambda }^{\left(d\right)}<{\epsilon }_{\lambda }^{\left(u\right)}$, for all $\lambda$.

The way this model is proposed, the space of states is defined for three fermions, consisting of two identical $u$ quarks, which are subject to Pauli’s exclusion principle, and one $d$ quark. So, we can set one microscopic state of the system as $|{\lambda }_i^{\left(u\right)},{\lambda }_j^{\left(u\right)},{\lambda }_k^{\left(d\right)}⟩$, with Pauli’s constraint that ${\lambda }_i^{\left(u\right)}\ne {\lambda }_j^{\left(u\right)}$. Then, the energy of this microstate will be

(4) $$E_{ijk}={\epsilon }_{{\lambda }_i}^{\left(u\right)}+{\epsilon }_{{\lambda }_j}^{\left(u\right)}+{\epsilon }_{{\lambda }_k}^{\left(d\right)}$$

For the next subsection, we shall simplify the notation, dropping the $\lambda$’s, as in ${\epsilon }_{{\lambda }_i}^{\left(u\right)}={\epsilon }_i^{\left(u\right)}$.

3.2. Partition function

For this model we should use the canonical ensemble formalism [⁷]. Its partition function $𝒵$ is given by

(5) $$𝒵(R,T)=\sum _{\begin{array}{c}i,j,k\\ i\ne j\end{array}}^{\infty }\exp \left[-\beta \left({\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}+{\epsilon }_k^{\left(d\right)}\right)\right]$$

where $R$ is the sphere radius, $T$ is the temperature, and ${\epsilon }_n^{\left(f\right)}$ is the energy of state $n=i,j,k$ for a quark of flavor $f$. Note that for the two quarks of same flavor ($f=u$) the Pauli exclusion principle requires that $i\ne j$. As usual, $\beta$ is the reciprocal of the product of the temperature and the Boltzmann constant, $\beta =1/k_{B}T$.

We can partially factor the summation in equation 5

(6) $$𝒵(R,T)=\sum _k{e}^{-\beta {\epsilon }_k^{\left(d\right)}}\sum _{\begin{array}{c}i,j,k\\ i\ne j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}$$

The second sum in equation 6 can be re-written in a more convenient way. First, analyzing the condition $i\ne j$

(7) $$\sum _{\begin{array}{c}i,j,k\\ i\ne j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}=\sum _{\begin{array}{c}i,j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}-\sum _{\begin{array}{c}i=j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}$$

The term with $i=j$ can be evaluated as

(8) $$\sum _{\begin{array}{c}i=j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}=\sum _{\begin{array}{c}i\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_i^{\left(u\right)}\right]}=\sum _{\begin{array}{c}i\end{array}}{e}^{-2\beta {\epsilon }_i^{\left(u\right)}}$$

Then, we work on the first term of equation 7 by factoring it

$$\begin{array}{ll}{\displaystyle \sum }_{\begin{array}{c}i,j\end{array}}{e}^{-\beta [{\epsilon }_i^{(u)}+{\epsilon }_j^{(u)}]}\hfill & ={\displaystyle \sum }_{\begin{array}{c}i,j\end{array}}\left(e^{-\beta {\epsilon }_i^{(u)}}\cdot e^{-\beta {\epsilon }_j^{(u)}}\right)\hfill \\ \hfill & ={\displaystyle \sum }_{\begin{array}{c}i\end{array}}\left(e^{-\beta {\epsilon }_i^{(u)}}\right){\displaystyle \sum }_{\begin{array}{c}i\end{array}}\left(e^{-\beta {\epsilon }_i^{(u)}}\right)\hfill \end{array}$$

But since $i$ and $j$ are dummy indices, we obtain the final form for this term as:

(9) $$\sum _{\begin{array}{c}i,j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}={\left(\sum _{\begin{array}{c}i\end{array}}{e}^{-\beta {\epsilon }_i^{\left(u\right)}}\right)}^2$$

All of these steps make the second summation in equation 6 take this form:

(10) $$\sum _{\begin{array}{c}i,j,k\\ i\ne j\end{array}}{e}^{-\beta \left[{\epsilon }_i^{\left(u\right)}+{\epsilon }_j^{\left(u\right)}\right]}={\left(\sum _{\begin{array}{c}i\end{array}}{e}^{-\beta {\epsilon }_i^{\left(u\right)}}\right)}^2-\sum _{\begin{array}{c}i\end{array}}{e}^{-2\beta {\epsilon }_i^{\left(u\right)}}$$

Now, with the first summation of equation 6 and equation 10, we can define the individual partition functions

$$Z_1^{\left(u\right)}(R,T)=\sum _{\begin{array}{c}i\end{array}}{e}^{-\beta {\epsilon }_i^{\left(u\right)}},Z_1^{\left(u\right)}(R,T)=\sum _{\begin{array}{c}i\end{array}}{e}^{-\beta {\epsilon }_i^{\left(d\right)}}$$

And finally, the partition function is

(11) $$Z(R,T)=Z_1^{\left(d\right)}\left[{\left(Z_1^{\left(u\right)}\right)}^2-\sum _{\begin{array}{c}i\end{array}}{e}^{-2\beta {\epsilon }_i^{\left(u\right)}}\right]$$

where the term in brackets is purely a Fermi–Dirac effect (Pauli correction for the two $u$’s). Also, it is interesting to note that, as the model assumes that $T$ is constant, we can just denote the partition function as $Z\left(R\right)$, or to simplify, just $Z$.

3.3. Pressure from the partition function

From the Helmholtz free energy, related to the partition function as $F=-k_{B}T\ln Z$, the pressure can be obtained as:

(12) $$P=-{\left(\frac{\partial F}{\partial V}\right)}_T=\frac{1}{\beta }{\left(\frac{\partial \ln Z}{\partial V}\right)}_T$$

Since $V=4/3\pi R^3$, then

$$\frac{\partial }{\partial V}=\frac{1}{4\pi R^2}\frac{\partial }{\partial R}$$

and placing in equation 12:

(13) $$P=\frac{k_{B}T}{4\pi R^2}\frac{\partial \ln Z}{\partial R}$$

The $R$ dependency comes from the energies $\epsilon \propto R^{-2}$. With that, we first get the derivative of a generic term:

(14) $$\frac{\partial }{\partial R}{e}^{-\beta {\epsilon }_i}=e^{-\beta {\epsilon }_i}\left(-\beta \right)\frac{\partial {\epsilon }_i}{\partial R}$$

Now, for the remaining part of the derivative, we make ${\epsilon }_i=A_i/R$ and do:

(15) $$\frac{\partial {\epsilon }_i}{\partial R}=-\frac{2A_i}{R^2}=-2\frac{{\epsilon }_i}{R}$$

Then, substituting the result of equation 15 in equation 14:

(16) $$\frac{\partial }{\partial R}{e}^{-\beta {\epsilon }_i}=e^{-\beta {\epsilon }_i}\left(-\beta \right)\left(-2\frac{{\epsilon }_i}{R}\right)=\frac{2\beta }{R}{\epsilon }_i{e}^{-\beta {\epsilon }_i}$$

Now, generalizing the result from the generic term to the partition function $Z$. For any sumation of the form:

$$Z=\sum _i{e}^{-\beta E_i}$$

we have:

(17) $$\frac{\partial Z}{\partial R}=\frac{2\beta }{R}\sum _i{E}_i{e}^{-\beta E_i}=\frac{2\beta }{R}{⟨E⟩}_{z}Z$$

then, dividing both sides by $Z$:

(18) $$\frac{\partial \ln Z}{\partial R}=\frac{2\beta }{R}⟨E⟩$$

where $⟨E⟩$ is the mean energy, but in the canonical ensemble $⟨E⟩=U$, therefore:

(19) $$\frac{\partial \ln Z}{\partial R}=\frac{2\beta }{R}U$$

Now, recalling equation 13 and that $\beta =1/\left(k_{B}T\right)$ we get:

(20) $$P=\frac{1}{4\pi R^2\cancel{\beta }}\left(\frac{2\cancel{\beta }}{R}U\right)=\frac{2U}{4\pi R^3}$$

Since $V=4\pi R^3/3\Rightarrow 4\pi R^3=3V$ we finally have:

(21) $$P=\frac{2}{\text{3}}\frac{U}{V}.$$

Note that this result is identical to that of a classical monoatomic ideal gas.

4. Pressure Determination

In this model, the internal energy can be considered as the proton’s rest energy:

$$U=m_p{c}^2,$$

one could argue that the internal energy would be due to the kinetic energy of the quarks, but it is understood that most of the proton mass($\sim 99\%$) comes from the quarks dynamics, so this assumption is not that absurd.So by Eq. (21), the proton’s internal pressure is:

$$P=\frac{2}{\text{3}}\frac{m_p{c}^2}{V}.$$

For the volume, we assume that quarks are confined to a sphere with a radius equal to that of the proton:

$$\begin{array}{ll}P\hfill & =\frac{2}{\text{3}}\cdot \frac{3}{\text{4}}\frac{m_p{c}^2}{\pi r_p^3}\hfill \\ \hfill & =\frac{1}{\text{2}}\frac{m_p{c}^2}{\pi r_p^3}.\hfill \end{array}$$

Using the proton radius value from Ref. [⁸] (CODATA 2014) and the proton mass from the same reference:

(22) $$P=0.3570\times {10}^{\text{35}}\text{Pa}.$$

According to reference [⁴], the pressure has its maximum in the range $0.3<r<0.5$, going to zero at $r=0$. Our very simple model does not have the sophistication to give a “well behaved” pressure dependence in $r$, since from equation 4 one can easily see that the pressure diverges at the origin.

5. Results and Conclusion

We discussed a simplified model for the proton, where, due to quark confinement by the strong force, we consider the quarks to be trapped in a volume defined by a spherical surface. Due to asymptotic freedom at short distances, we treat them as free particles, approximating the proton as a quantum gas of fermions at constant volume.

This model allowed for the calculation of proton pressure, achieving the same order of magnitude as recent experimental results [¹].

We assumed a non-relativistic Fermi gas. While quarks could reach high speeds within the proton, the ultra-relativistic case would yield $P=\frac{1}{\text{3}}\frac{U}{V}$[⁹] (losing a factor of 2 compared to Eq. 21), not changing the final order of magnitude.

This result is significant because basic physical concepts yield good agreement with recent experimental data. Achieving the same order of magnitude validates the approximations and demonstrates that important information can be extracted from simple models.

Data Availability

The entire dataset supporting the results of this study is published in the article.

References

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Edited by

• Editor-in-Chief:
  Marcello Ferreira https://orcid.org/0000-0003-4945-3169

Publication Dates

• Publication in this collection
  08 May 2026
• Date of issue
  2026

History

• Received
  30 Oct 2025
• Reviewed
  22 Mar 2026
• Accepted
  30 Mar 2026