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I'm trying to find the mass matrix from the actions in $(2.10)$ and $(2.11)$ in this paper. The action is expanded around a classical background field $B^{i}$ and the action for fluctuations is given as

$$ S_{Y} = i \int d\tau \Big(\frac{1}{2}Y^{i}_{1}(\partial_{\tau}^2 - r^2)Y^{i}_{1} + \frac{1}{2}Y^{i}_{2}(\partial_{\tau}^2 - r^2)Y^{i}_{2}+ \frac{1}{2}Y^{i}_{3}\partial_{\tau}^2 Y^{i}_{3} -\sqrt{g} \epsilon^{a3x} \epsilon^{cbx}B^{i}_{3}Y^{j}_{a}Y^{j}_{b}Y^{j}_{c}-\frac{g}{2}\epsilon^{abx}\epsilon^{cdx}Y^{i}_{a} Y^{j}_{b} Y^{i}_{c} Y^{j}_{d} \Big) $$

Here, the $Y^{i}_{a}$s are coefficients in the expansion of the fluctuations around the background field in a $\sigma$ matrix basis and $i = 1,2,\dots ,9$.

The gauge field action is given by

$$ S_{A} = i \int d\tau \Big(\frac{1}{2}A_{1}(\partial_{\tau}^2 - r^2)A_{1} + \frac{1}{2}A_{2}(\partial_{\tau}^2 - r^2)A_{2}+ \frac{1}{2}A_{3}\partial_{\tau}^2 A_{3}+ 2 \epsilon^{ab3}\partial_{\tau}B^{i}_{3}A_{a} Y^{i}_{b} + \sqrt{g}\epsilon^{abc} \partial_{\tau} Y^{i}_{a}A_{b}Y^{i}_{c} -\sqrt{g} \epsilon^{a3x} \epsilon^{bcx}B^{i}_{3}A_{a}A_{b}Y^{i}_{c}-\frac{g}{2}\epsilon^{abx}\epsilon^{cdx}A_{a} Y^{i}_{b} A_{c} Y^{i}_{d} \Big)$$

Now I know that the mass squared matrix is given by the second derivative of the potential, evaluated at the classical minimum potential, i.e.

$$ m^{2}_{ab} = \frac{\partial^2 V(\phi)}{\partial \phi^{a} \partial \phi^{b}}\Bigg\rvert_{\phi = \phi_{0}} $$

I understand that I have to diagonalize $m^2_{ab}$ to get the masses.

I need to understand how to use this get the following particles:

$16$ bosons with $m^2 = r^2 = b^2 + (v \tau)^2 $

$2$ bosons with $m^2 = r^2 + 2v $

$2$ bosons with $m^2 = r^2 - 2v $

$10$ massless bosons

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