# Energy of the nucleon bag (MIT bag model)

I am studying MIT bag model.

The Lagrangian density for this model is $$\mathcal{L} = (i\bar{\psi}\gamma^\mu\partial_\mu\psi - B)\theta_V - \frac{1}{2}\bar{\psi}\psi\delta_S .$$

The nonlinear boundary condition of the MIT bag model is $$B = -\frac{1}{2}n\cdot\partial[\bar{\psi}\psi]_{r = R}$$, where $$B$$ is phenomenological energy density term added to Bogolioubov model for conservation law of energy-momentum tensor.

We can get the energy of the nucleon by calculating $$P^0 = \int d^{3}x T^{00}(x)$$ and the energy is

$$E(R) = \frac{\sum{\omega_i}}{R}+\frac{4\pi}{3}R^3B.$$

This model implies that $$\frac{\partial E}{\partial R} = 0$$ but, I can not understand how this result came out because here is an assumption of the model that $$B$$ is constant.

So I would like to show that

$$\frac{\partial E}{\partial R} = -\frac{\sum{\omega_i}}{R^2}+4\pi R^2B = 0.$$

Can I get some advice for it?

I am studying this model with 'Chiral symmetry and the bag model : A new starting point for nuclear physics, A. W. Thomas, 1984'. In this text, author said that we can explicitly show this result but I failed.

• So, what, exactly, is your problem? That energy is energy density integrated over volume? Commented Oct 18, 2019 at 19:58
• Yes, $B$ is the vacuum energy density. What I asked is "How partial derivative of energy with bag radius becomes zero?". Is this stability equation obvious?
– Seal
Commented Oct 19, 2019 at 19:07
• ?? You mean how is a minimum-energy configuration for a bag of labile size most stable? Commented Oct 19, 2019 at 20:00
• Because it has radiated away all excess energy and cannot lose more. Commented Oct 19, 2019 at 21:37
• Well, I'm not clear yet. But I can keep thinking about it. Thank you very much for your precious comment :)
– Seal
Commented Oct 19, 2019 at 21:57

Consider the function $$E(R)$$ rather than its derivative. For $$R<<$$ the dominating term is the $$1/R$$ and at $$R \rightarrow 0$$ we have $$E(R \rightarrow 0) \rightarrow \infty$$ (assuming that the $$\omega$$ are positive). On the other hand, as we approach infinity the $$1/R^3$$ dominates the description of the function and we have $$E(R \rightarrow \infty) \rightarrow \infty$$.