1. Model 5:Telescope mirrors Reflecting telescopes use large mirrors to focus light. A standard Newtonian telescope design is depicted in figure 1: Light from star Eyepiece Inside surface of mirror x=0 outside surface; x=L Figure 1: Newtonian telescope 2003/ 1

2. Some can be quite big! The 200 inch telescope at Mt. Palomar Thermal problems during casting of the mirror were solved in a novel way.

3. The honeycomb structure below the reflecting surface solved thermal conductivity- thermal expansion/contraction and cooling problems. Be aware that thermal problems of a different nature exist – even for small amateur telescopes!

4. Thermal problems As night approaches small temperature differences between the mirror and its surroundings can result in optical problems due to 1) thermal expansion of the glass and 2) disturbances in the air above the mirror. A thick mirror will take longer to equilibrate than a thin one. An hour is a long time to wait for a mirror to equilibrate. Can a mirror be too thick? The diameter of a telescope mirror is usually much greater than its thickness so we can treat this as a 1 dimensional problem. Away from the edges the temperature (as it cools down) will vary principally across the thickness and not radially. Let us calculate the temperature distribution through the thickness 0 to l at or near the centre of the disk.

5. 1)Model and equation Use the 1 -dimensional heat equation x=0 T= 0 x=L T= 0 (27) (i.e. α 2 = K/σρ) L D Assume D >> L Inside surface Outside surface

6. Step 2: General solution Instead of guessing we again use the method of “Separation of Variables”, making the assumption that the solution will be some function of x only multiplied by some function of t only: u = X(x)T(t) (28) Equation 28 is a solution:

7. Step 3: Apply the boundary conditions Both ends were fixed at 0ºC. The solution is the same one we obtained for the copper conductor: (29)

8. Step 4: Satisfy the initial conditions If we know the initial temperature, we can find values for the constants D n. Let’s assume that the mirror temperature was the same across the entire thickness of glass. At t=0 we carry the telescope outside where the temperature is lower. We can say that the initial temperature u(x,0)=T 0 where T 0 is the difference between the temperature inside and outside. (30) T0T0 x=0 x=L Temperature at time t=0:

9. The solution will be: (31) If the temperature outside is 1°C lower then T 0 = 1 °C and the solution is:

10. Sometimes the bottom surface of the mirror cannot equilibrate as easily. For this case we can assume a worst case scenario - that the bottom boundary is fixed at inside temperature T 0. 1) Model and equation (Note new b.c.) x=0 T= T 0 x=L T= 0 l D Inside surface Outside surface (27) (i.e. α 2 = K/σρ)

11. For problems involving non-zero boundary conditions we solve for the steady state (u s (x) ) and transient (U(x,t)) parts of the solution and then add the two together for the complete solution: u(x,t) = u s (x) + U(x,t) steady statetransient (32)

12. Dealing with non-zero (non-homogeneous) boundary conditions. 1) Model and equation 2) General solution 2s) Steady state general solution2t) Transient general solution 3s) Fit steady state general solution to non-zero boundary conditions 3t) Fit transient general solution to zero (homogeneous) boundary conditions 4) Satisfy the initial conditions: Adjust the initial conditions by subtracting the steady state solution from them. Fit transient solution to the ‘adjusted’ initial conditions. Final: Add the two solutions together

13. Step 2s: Find a general form for the steady state solution At the steady state the left hand side of equation 27 is zero: 0 A solution for u is: where F and G are constants. (33)

14. Step 3s: Fit the steady state solution to the non- zero boundary conditions At x=0 u s = T 0. G= T 0. At x=L u s = 0. (33)

15. A plot of the steady state solution: 0 L T 0 0º

16. How long will the mirror take to reach a steady state? To estimate this we must solve the full equation. This solution will include both the steady state solution u s (x) and the transient solution (U(x,t)).

17. Step 2t (t for transient): Find the general transient solution We found the steady state solution earlier: This is a solution to the heat equation: The transient solution U(x,t) must also be a solution to equation 27: The form of U(x,t) will be (similar to eqn. 28): (27)

18. Step 3t: Fit boundary conditions to transient solution What boundary conditions should we use? Consider the complete solution: I will demonstrate that the transient solution (U(x,t)) has zero (homogeneous) boundary conditions!

19. Step 3t continued: 2 nd boundary The transient solution has zero (homogeneous) boundary conditions! Therefore U(0,t) = 0º 1 st boundary Therefore U(L,t) = 0º

20. Thus the solution for U(x,t) will be of the same form as eqn. 29: (29*)

21. Step 4: Satisfy the initial conditions What are the initial conditions? 0 L T 0 0º At t<0 (when the telescope is still inside your house) the temperature is T 0 across the entire thickness of glass.

22. Compare the initial conditions with the steady state: 0L T 0 ºC 0ºC 0L T 0 ºC 0ºC t=0 t=∞ (steady state) The total solution is the sum of the steady state and transient solutions. At t=0 these must match the initial conditions (when t=0):

23. The total solution at t=0 and all times is the sum of the steady state and transient solutions: Put in values from model: Rearrange and solve for U(x,0): We see that the transient solution at t=0 is the difference between the initial conditions and the steady state solution.

24. We proceed to solve for D* n using the following equation (similar to eqn. 15): The solution to U(x,t) is: using

25. Evaluate D * m : 0 L Substitute Rearrange: Calculate the integral: Evaluate the integral between 0 and L: Simplify: = D * m

26. The solution to U(x,t) is:

27. Some materials and their K, σ and ρ values MaterialThermal conductivity (K) W/mºC Specific heat J/gmºC Density gm/cm 3 Steel (AISI 304) 16.270.468.01 Polyurethane foam 0.0261.140.029 Pyrex**** 1.350.775 2.23 Use constants for Pyrex: Calculate the diffusivity of Pyrex in units of m 2 /s: (  2 = 7.81  10 -7 m 2 /s)

28. End and review: non-zero b.c.’s 1) Model and equation 2) General solution 2s) Steady state general solution2t) Transient general solution 3s) Fit steady state general solution to non-zero boundary conditions 3t) Fit transient general solution to zero (homogeneous) boundary conditions 4) Satisfy the initial conditions: Adjust the initial conditions by subtracting the steady state solution from them. Fit transient solution to the ‘adjusted’ initial conditions. Final: Add the two solutions together