Minguzzi Invariant Inertial Frame Time Here I have chosen to let the rocket pilot's proper time schedule be the driving factor for the plots. I have also divided the trip into 6 acceleration phases. These are, consecutively {a,0,-a,-a,0,a} and the times for each leg are {ta,tc,ta,ta,tc,ta}. As you can see, these allow the rocket, if started at zero speed, to return to the start at zero speed. I feel that this acceleration program should provide enough detail for most people. For that reason I have plotted the rocket times linearly on the right hand vertical axis. I have placed this axis at an x value of xMax=c*tTotal/2. The reason for this axis placement is that the farthrest that the rocket can go in time tTotal is xMax=c*tTotal/2 even if the entire trip proceeds at the speed of light. The left hand vertical axis contains the Earth frame times that correspond to the set of linearly spaced rocket proper times. On the bottom horizontal axis I show the rocket distance as viewed in the earth frame. On the top horizontal axis I show the rocket distance from start as viewed (as contracted) by the rocket. I think this method of showing the results of a space trip represents the best way that we can present them. The Earth times presented are from the paper by Minguzzi. These are chosen to be invariant times T(tau) where T(tau)^2=[t(tau)-t(0)]^2-(x(tau)-x(0))^2 and lower case t and lower case x represent the standard transformed expressions for time and displacement respectively t(tau)-t(0)=c*Integral[cosh(Integral(a(t'') dt'' {0,t'}]dt' {0,tau}] and x(tau)-x(0)=c*Integral[sinh(Integral(a(t'') dt'' {0,t'})dt' {0,tau}] where a(t) is the acceleration profile and tau is the rocket proper time. T(tau) is the age seen by an inertial observer who has moved along a geodesic to the point x(tau) so that, for him, x(tau)-x(0)=0 and therefore, for him, T(tau)=t(tau)-t(0). The learner will find that most of the time difference between the inertial observer clock and the rocket clock accrues during the second half of the journey. The reason for this asymmetry is that, during the outward trip, T(tau)^2 is reduced by an increasing [x(tau)-x(0)]^2 while during the trip back, the dx^2 term decreases. I have also included a color coded shading of the ratio of the invariant observer time to the rocket proper time as a function of the rocket proper time.
Minguzzi Paper Minguzzi Equation Derivations
a TT TC/TT Ratio