Since our technology (in particular, our maser design) is designed to function in space, we will be testing it in vacuum chambers to replicate the near-vacuum conditions of space. However, as we will need to evacuate the vacuum chambers of air (and thus de-pressurize the chamber), the drop in pressure must be sufficiently slow as to prevent structural damage to the maser body. Thus, we show a calculation here to determine the predicted stress load on the maser during the de-pressurization process, which can be then used to determine the optimal rate at which air should be evacuated and set structural requirements for material strength. We will model the maser using a considerably-simplified approximation: a uniform spherical shell of nonzero thickness made of a given material, with a gap to allow pressure in its interior and exterior to be equalized. Since a spherical shell is symmetric, the pressure on it is equal in all directions, and thus independent of position, simplifying our calculations. In our hypothetical scenario, the shell is placed within a de-pressurization chamber. Now, let $P$ denote the fluid pressure within the chamber (in our case, the fluid is simply air). The total force is given by: $ \mathbf{F} = P A \hat n $ Where $A$ is the (uniform) surface area of the object and $\hat n$ is the normal to the object's surface. If we differentiate with respect to time, we find that: $ \dfrac{d\mathbf{F}}{dt} = A \dfrac{dP}{dt} \hat n $ As the pressure $P = P(t)$ changes with time, there is a change in the force exerted on the object by the surrounding fluid (in our case, air). Now, if we divide by $A$ on both sides, we have: $ \frac{1}{A} \dfrac{d\mathbf{F}}{dt} = \dfrac{dP}{dt} \hat n $ The left-hand side of the equation is equal to the time-derivative of the stress $\sigma = \mathbf{F}/A$, so this can be rewritten as: $ \dfrac{d\sigma}{dt} = \dfrac{dP}{dt} \hat n $ This differential equation tells us that the rate of change of the stress $\sigma = \sigma(t)$ is equal to the rate of change of the pressure. Importantly, while we want to decrease the air pressure in the chamber to achieve a good vacuum, we also do not want the *change* in pressure to be too rapid. This is because our uniform spherical shell is only an approximation; in reality, different parts of the maser will experience the drop in pressure at different rates, so a rapid change in stress will cause a stress differential across the maser and might cause structural damage to the maser body. To show this, let us drop the uniform area assumption, so we would have: $ \dfrac{d}{dt}\int \sigma \cdot d\mathbf{A} = \dfrac{dP}{dt} (\hat n \cdot \hat r) $ Using the divergence theorem, we have: $ \dfrac{d}{dt} \int \sigma \cdot d\mathbf{A} = \dfrac{\partial}{\partial t} \int (\nabla \cdot \sigma) dv = \dfrac{dP}{dt} (\hat n \cdot \hat r) $ Thus, integrating w.r.t. time and differentiating w.r.t. volume we have: $ \nabla \cdot \sigma = \dfrac{dP}{dv} (\hat n \cdot \hat r) $ This tells us that the divergence of the stress is equal to the pressure per unit volume in the component of the surface normal along $\hat r$. This means that even if the pressure is constant in time, if it is not uniform in space, there will be unequal stress on different parts of the maser body. Since we may express $dv = dx\, dy\, dz$ then we have: $ dP = \dfrac{dP}{dx} dx + \dfrac{dP}{dy} dy + \dfrac{dP}{dz} dz $ These three derivatives are almost certainty not the same simply because $dx, dy, dz$ are non-uniform on the surface of the non-symmetric maser body, so the stress would (almost, though not necessarily certainly) have a nonzero divergence. The resulting variation in stress is what causes the structural damage, but if the pressure is changed slowly when the maser is under high pressure, this variation can be minimized. We may, for instance, choose to allow the pressure to decrease slowly at first, and then at low vacuum to decrease more quickly. In the mathematical terms, $\dot P$ will need to be negative for all $t$ (since we are modelling a *decreasing pressure*), where $|\dot P| \ll 1$ for small $t$ and $|\dot P| \ll 1$ for large $t$. A possible choice for $\dot P$ is given by: $ \dfrac{dP}{dt} = -k\ln(t + 1), \quad t \geq 0 $ Where $k$ is some free parameter that we can use to tune the depressurization rate (higher $k$ results in a more rapid depressurization rate; $k = 0$ is constant pressure). Now, if we let the initial pressure $P_0 = P(0)$ equal standard atmospheric pressure under room temperature conditions ($P_0 = \pu{1 atm} = \pu{101.325 kPa}$), then we have an initial-value problem (IVP) that we can then solve. The particular solution to this IVP is given by: $ P(t) = P_0 + k [t - (t + 1)\ln(t + 1)] $