Critical temperature for bubble nucleation

[spaghetti3451]

Consider that the criterion for bubble nucleation (in a field theory) is the following:

$$\exp(-S_{3}/T) \gtrsim \frac{3}{4\pi} \left(\frac{H}{T}\right)^{4} \left(\frac{2\pi T}{S_{3}}\right)^{3/2},$$

where $S_3$ is the three-dimensional action of the theory and $H$ is the Hubble scale.

1. What is meant by the critical temperature of bubble nucleation?

2. Why does taking the derivative of this criterion give us the critical temperature for bubble nucleation?

N.B. : The critical temperature $T_c$ for bubble nucleation is $T_{c} = (2/11)S_{3}$.

 

[AndyW]

1. The critical temperature of bubble nucleation is the temperature at which the criterion for bubble nucleation is satisfied. In other words, it is the temperature at which the probability of bubble formation becomes significant.

2. Taking the derivative of the criterion gives us the critical temperature because at the critical temperature, the criterion is just satisfied. This means that the left side of the inequality is equal to the right side, and taking the derivative will give us the slope of the curve at that point, which corresponds to the critical temperature. This is similar to finding the maximum or minimum of a function by taking its derivative and setting it equal to zero. In this case, the derivative will give us the temperature at which the probability of bubble formation is the highest.