## 2 Review of Cryogenic-Cooling Benefits
### Thermal Parameters
#### 2.2.1 Thermal Conductivity
For any finite value of \(Q_{0}\), the temperature, and thus the thermal aberrations are not quadratic with \(y\), and thus not correctable using a simple lens. From a practical standpoint, for lower values of \(Q_{0}\), this effect can be ignored. Equation (19) may be differentiated, to find the temperature gradient  
\[\frac{dT}{dy}=-\left[\frac{T_{0}}{T_{A}}\right]\cdot\left[\frac{Q_{0}}{k_{0}} \right]y\cdot Exp\left(-\frac{Q_{0}y^{2}}{2k_{0}T_{A}}\right) \tag{22}\]  
Here we see that reducing the ratio (\(Q_{0}/k_{0}\)) is required in this case as well to reduce the thermal profile, except that in this more realistic case, the decrease in the thermal gradient is faster that in the standard case. Figure 1 shows calculation results from the standard Equations (15) and (19) derived here. Temperature as a function of the slab thickness coordinate \(y\) is shown for three cases of heat density, 500, 1500, and 2500 W/cm\({}^{3}\), and for a slab edge temperature of 300 K. Note that we assume that Equation (18) is valid above room temperature as well. The same plot is shown in Figure 2 for an edge temperature of 80 K. The thermal conductivity is scaled according to Equation (18), and was assumed to be 0.112 W/(cm\(\cdot\) K) at 300 K, and 0.424 W/(cm\(\cdot\) K) at 80 K. The difference between the standard model and ours is small at 500 W/cm\({}^{3}\) and below, whereas at the largest heat densities of 1500 and 2500 W/cm\({}^{3}\), the difference strongly increases. For 300 K, the difference between the two models for 2500 W/cm\({}^{3}\) amounts to 23.2 K; for 80 K the difference is 6.10 degrees. The fundamental reason for the temperature increase using our model is the decrease in the thermal conductivity that occurs as one moves towards the slab center where the temperature is maximum. This effect is ignored in the standard model. As the ambient or cooling temperature decreases to 80 K, the enhanced thermal conductivity for all temperatures reduces the difference between the two models. We have solved the CW heat equation, ignoring the \(z\)-dependence, for rod elements as well, and find the following similar equation:  
\[T\left(r\right)=T_{0}\cdot\text{Exp}\left(-\frac{Q_{0}r^{2}}{4T_{A}}\right) \tag{23}\]  
For cases in which the thermal conductivity does not vary as \((1/T)\), but with a power of \(T\left(1/T^{b}\right)\) for example, we can still obtain an analytical solution, albeit a transcendental one. The solution for a slab with \(b>1\) is  
\[T\left(y\right)=\frac{T_{0}}{\left[\frac{Q_{0}\left(b-1\right)y^{2}}{2\,k} \cdot\frac{T_{0}^{1-b}}{T_{A}}\right]\frac{1}{1-b}} \tag{24}\]whereas before \(T_{0}\) is the temperature for \(y=0\), and \(T_{A}\) is the ambient temperature. For \(y=t/2\), we can find the edge temperaure \(T_{e}\), as  
\[\frac{dT}{dy}=-\frac{Q_{0}}{k}y \tag{16}\]0  
This equation is transcendental in \(T_{0}\), and may be solved numerically for any edge temperature for \(T_{0}\) after \(Q_{0}\), \(b\), \(k\), and \(T_{A}\) are specified. \[\frac{dT}{dy}=-\frac{Q_{0}}{k}y \tag{16}\]1 Temperature as a function of thickness coordinate \(y\), for the standard model (blue) and the new model described here, for heat density values of 500, 1500, and 2500 W/cm\({}^{3}\) (bottom to top) and an ambient cooling temperature of 300 K.  
\[\frac{dT}{dy}=-\frac{Q_{0}}{k}y \tag{16}\]2 Temperature as a function of thickness coordinate \(y\), for the standard model (blue) and the new model described here, for heat density values of 500, 1500, and 2500 W/cm\({}^{3}\) (bottom to top) and an ambient cooling temperature of 80 K.