This problem deals with Jackson's problem 11.4. A set of twins are 20 years old - earth time. One of the twins (twin A) leaves the earth (year: 2000) on a rocketship. First twin A accelerates (a=g) in a straight line with respect to its own frame for five years, and then decelerates for five years. Then twin A returns the same way: accelerates for five years and decelerates for five years. Now twin A believes it is 40 years old. How old is twin B? Twin B is 267.84! (Wow! I wonder if any of us will live to be that old!)
Complete Statement of the Assignment
Solution to Problem
I began by coping the Bland Starter Program in directory:
to my own directory. This program calculates how much time it takes twin A to journey some distance out in space (before deciding to come back) according to: twin A (time=tau) and twin B (time=t). Throughout the calculation of time, various other variables are being calculated such as the distance that the earth (twin B) thinks the rocketship has traveled (x in meters and in light-years).
An adjustment of the program is necessary inorder that the rocketship may come back to the earth from alpha-Centauri. Yes! The twin is now able to come home!
Upon changing the parameters, I realized that the numbers do not match the homework solutions nor are the numbers close enough to call them good enough. First I thought it might be significant figures but this proved wrong. Then I realized that changing n1 to 1826.25 (5 yrs) and dtau to 1 was creating a problem. So I changed it to 1826 and then after each loop and have the program recompute variables using dtau set to 0.25. Then I reset dtau to 1 before begining the next loop. This does not help much. I wonder if the problem has to do with beta. For example, beta increases to 0.999939 at the end of the first acceleration, but it decreases only to 0.047069. The number should be zero here or atleast something smaller than 0.047069! Well, I have tried various ideas to correct for this. For the first acceleration, the distance according to the earth (x) is 83.9 light years compared to the solutions where x = 83.0. This is extremely close - 1.1%. It is after the second acceleration where x = 171.0 (compared to x = 171.0) the percent difference is 0.0% After the third acceleration, x = 259.148 compared to 249.135 - a 4.0% difference. After the fourth and final acceleration, x = 350.595 compared to x = 332.18 - a 5.5% difference!
Looking at the earth time, after the first acceleration, t is 30995.447 days compared to the solution of 30684.653. That is a 1.0% difference. After the second acceleration, t = 63156.414 compared to t = 61369.305. That is a 2.9& difference. After the third acceleration, t = 95678.492 compared to t = 92053.958 - a 3.9% difference. After the fourth and final acceleration, t = 129400.000 compared to t = 122745.915 - a 5.4% difference!
Overall, as tau increases so does the error in the programs value for t and x. This is due to beta not functioning properly. If I had more time, I would investigate more thoroughly as to why this is true. Notice that tau has been perfectly correct the entire time.
Okay, wait. Changing the code so that the number of days for each acceleration, the acceleration, earth time (t), tau, beta and the earth distance (x) are doubles, results in a properly behaving beta. (It goes to zero when it should.) Now the values of x, and t after the total trip are very close to the written solutions - 0.01% and .02% respectively.