# Spontaneous Brillouin Scattering Spectrum and Coherent Brillouin Gain in Optical Fibers
## 2 Results
### Langevin Noise Initiation  
It is customary to neglect acoustic wave propagation compared to optical wave propagation, since \(v\ll c\). Hence, the explicit \(z\) dependence of the acoustic amplitude and of the random function \(f\) can be dropped. In this case, Equation (8) limited to stochastic excitation resembles the 1D Langevin equation describing a random walk [13]  
\[\frac{\partial A_{a}(t)}{\partial t}+\frac{1}{2}\Gamma A_{a}(t)=f(t). \tag{9}\]  
Assuming the stochastic force has a Gaussian white noise probability distribution we have  
\[\langle f(t)\rangle=0\,\ \ \langle f(t)f^{*}(t^{\prime})\rangle=Q\delta(t-t^{ \prime}). \tag{10}\]  
[PERSON] et al. enigmatically state that: \"By introducing the formal solution of Equation (9) into the left-hand side of Equation (10), we find that  
\[\langle A_{a}(t)A_{a}^{*}(t)\rangle=\frac{Q}{\Gamma}.\text{''} \tag{11}\]  
(Notations have been slightly altered to serve our purpose without changing the original meaning.) While the result is certainly correct, the derivation is not obvious in our opinion and merits some further scrutiny.  
Let us first examine the coherent response, i.e., when \(f(t)\) is a deterministic force and not a random noise. Assuming \(f(t)\) is sufficiently regular to have a Fourier transform \(\hat{f}(\xi)\), it immediately follows from (9) that  
\[\hat{A}_{a}(\xi)=\frac{\hat{f}(\xi)}{2i\pi\xi+\frac{1}{2}\Gamma}. \tag{12}\]  
For an impulse (\(f(t)=\delta(t)\), \(\hat{f}(\xi)=1\)), the coherent response is \(A_{a}(t)=H(t)\exp(-\frac{\Gamma}{2}t)\), with \(H(t)\) the Heaviside distribution. Hence the intensity of the response would instantly jump to 1 at \(t=0\) and then decrease with positive times with an \(\exp(-\Gamma t)\) dependence. Alternatively, for an applied force \(f(t)=1-H(t)\) (a force held constant at value 1 until it is instantly removed at \(t=0\)), we have \(\hat{f}(\xi)=\frac{1}{2}\delta(\xi)-\frac{1}{2i\pi\xi}\) and then \(A_{a}(t)=\frac{2}{\Gamma}(1-H(t)+H(t)\exp(-\frac{\Gamma}{2}t))\). Hence, the coherent response would be constant (=2/\(\Gamma\)) for \(t\leq 0\) and would then decrease exponentially with positive times. The intensity would again decrease with positive times with an \(\exp(-\Gamma t)\) dependence.  
Obviously, the previous deterministic analysis does not hold for the noise driven equation. There is no reason that we can select an initial value at \(t=0\) nor any way to abruptly switch on and off the driving noise. Furthermore, there is no mathematical guarantee that the Fourier transform of the random variable \(f(t)\) can even be defined. We can instead rely on the Wiener-Khinchin theorem: the Fourier transform (FT) of the autocorrelation function is the power spectral density (PSD). The autocorrelation function of the noise source is \(\langle f(t)f^{*}(t+\tau)\rangle\) whose Fourier transform is \(S_{f}(\xi)=Q\) according to Equation (10). The power spectral density and autocorrelation function of the stochastic response are then  
\[S_{A_{a}}(\xi)=\frac{Q}{|2i\pi\xi+\frac{1}{2}\Gamma|^{2}},\ \ \langle A_{a}(t)A_{a}^{*}(t+\tau) \rangle=\frac{Q}{\Gamma}\exp\left(-\frac{\Gamma}{2}|\tau|\right)\text{.} \tag{13}\]  
Note the appearance of the absolute value of the delay time \(\tau\), implying that the stochastic response is as much correlated to its past as to its future values. Note also that the power spectral density is the quantity that can be measured by a spectrum analyzer. The square modulus of the coherent response, \(|\hat{A}_{a}(\xi)|^{2}=|\hat{f}(\xi)|^{2}|2i\pi\xi+\frac{1}{2}\Gamma|^{-2}\), has the same symmetric Lorentzian spectral shape as the power spectral density when \(|\hat{f}(\xi)|^{2}\) is a sufficiently smooth function.