# Spontaneous Brillouin Scattering Spectrum and Coherent Brillouin Gain in Optical Fibers
## 2 Results
### Elastodynamic Equation and Coherent Response  
As we outlined in the introduction, the material density \(\rho(t)\) cannot physically describe acoustic phonons nor elastic waves in solids, because these have a polarization. Instead, we should consider the elastodynamic equation as a replacement for (7). We write this equation [25; 26]  
\[-\
abla\cdot(c^{\prime}\
abla\mathbf{u})+\rho_{0}\frac{\partial^{2}\mathbf{u }}{\partial t^{2}}=\mathbf{f}_{\text{es}}+\mathbf{f}_{a}, \tag{14}\]  
where \(\mathbf{u}\) is the displacement vector, \(c^{\prime}\) is the elastic tensor including a viscoelastic contribution describing material loss, and \(\mathbf{f}_{\text{es}}\) is the deterministic force resulting from electrostriction. We consider only the coherent response in this section, i.e., we first take \(\mathbf{f}_{a}=0\). In a Brillouin scattering experiment, the acoustic wavenumber \(q\) is imposed by phase matching along a given direction \(z\), e.g., the axis of an optical fiber. The acoustic or phonon frequency, however, remains free. We consider the temporal Fourier transform of (14)  
\[-\
abla\cdot\left((c+i\omega\mu)\
abla\mathbf{\hat{u}}\right)-\rho_{0}\omega^ {2}\mathbf{\hat{u}}=\mathbf{\hat{f}}_{\text{es}}. \tag{15}\]  
Implicitly, all terms with a tilde (\(\hat{\tau}\)) in (15) have an \(\exp(i(\omega t-qz))\) dependence and the nabla operator \(\
abla=\
abla_{t}-iq\mathbf{z}\) with \(\
abla_{t}\) the transverse gradient and \(\mathbf{z}\) a unit vector along the axis. The phononviscosity tensor \(\mu\) accounts for polarization dependent propagation loss. Next we assume the vicinity of a mode \(\mathbf{u}_{0}(\mathbf{r})\) of the homogeneous part of (15) as defined by  
\[-\
abla\cdot(c\
abla\mathbf{u}_{0})=\rho_{0}\omega_{0}^{2}\mathbf{u}_{0}. \tag{16}\]  
For frequencies close to the mode frequency, we assume \(\hat{\mathbf{u}}=\hat{a}\mathbf{u}_{0}\) with \(a(t)\) a scalar amplitude. Writing \(\omega=\omega_{0}+\delta\omega\), we have \(\omega^{2}\approx\omega_{0}^{2}+2\omega_{0}\delta\omega\) and then  
\[-\hat{a}(\
abla\cdot(i\omega_{0}\mu\
abla\mathbf{u}_{0})+2\omega_{0}\delta \omega\rho_{0}\mathbf{u}_{0})=\hat{\mathbf{f}}_{\text{es}}. \tag{17}\]  
Taking the scalar product with \(\mathbf{u}_{0}\) and integrating over a transverse cross-section  
\[\hat{a}\left(\int_{\Omega}\
abla\mathbf{u}_{0}\mu\
abla\mathbf{u}_{0}+i\delta \omega\int_{\Omega}\rho_{0}|\mathbf{u}_{0}|^{2}\right)=\frac{i}{2\omega_{0}} \int_{\Omega}\hat{\mathbf{f}}_{\text{es}}\cdot\mathbf{u}_{0}. \tag{18}\]  
We note that \(E=\frac{1}{2}\int_{\Omega}\rho_{0}\omega^{2}|\mathbf{u}|^{2}+\frac{1}{2}\int_{ \Omega}\
abla\mathbf{u}c\
abla\mathbf{u}\) is the energy density carried by an elastic wave with displacements \(\mathbf{u}\) at frequency \(\omega\). The kinetic and potential energy terms are actually equal for a mode of propagation and we thus note \(E_{0}=\int_{\Omega}\rho_{0}\omega_{0}^{2}|\mathbf{u}_{0}|^{2}\). We then recover the result (12) with the identifications  
\[i\delta\omega=2i\pi\xi,\ \ \frac{\Gamma}{2}=\omega_{0}\frac{\int_{\Omega} \
abla\mathbf{u}_{0}\omega_{0}\mu\
abla\mathbf{u}_{0}}{E_{0}},\ \ \hat{f}=\frac{i\omega_{0}}{2E_{0}}\int_{\Omega}\hat{\mathbf{f}}_{\text{es}} \cdot\mathbf{u}_{0}. \tag{19}\]  
In particular, we see that there was no reason to neglect the spatial propagation of the elastic wave but also that propagation losses effectively depend on the particular phonon that is excited.