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.

  • $\begingroup$ So, what, exactly, is your problem? That energy is energy density integrated over volume? $\endgroup$ Commented Oct 18, 2019 at 19:58
  • $\begingroup$ 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? $\endgroup$
    – Seal
    Commented Oct 19, 2019 at 19:07
  • $\begingroup$ ?? You mean how is a minimum-energy configuration for a bag of labile size most stable? $\endgroup$ Commented Oct 19, 2019 at 20:00
  • $\begingroup$ Because it has radiated away all excess energy and cannot lose more. $\endgroup$ Commented Oct 19, 2019 at 21:37
  • $\begingroup$ Well, I'm not clear yet. But I can keep thinking about it. Thank you very much for your precious comment :) $\endgroup$
    – Seal
    Commented Oct 19, 2019 at 21:57

1 Answer 1


Prof. Zachos gave a nice and clean argument, however since there is a comment that his reply is not yet very clear let me venture a different approach. If my answer is not satisfactory, let me know and I will delete it.

What I understand from the comments is that you need some clarification for the minimum.

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 $E(R \rightarrow \infty) \rightarrow \infty$.

You can plot the function

The Energy plot

Inspection of the plot reveals a minimum and at the minimum, the derivative will be zero.

Let me also post a link that uses the above logic. I would post directly the link to the pdf-file, however it is not preferred in the site. You want to go on page 2 and check problem 8.


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