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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 `x_(Max)=c*t_(Total)/2`. The reason for this axis placement is that the farthrest that the rocket can go in time `t_(Total)` is `x_(Max)=c*t_(Total)/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 Minguzzi Paper. 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)=cint_0^taucosh(int_0^(t')(a(t'') dt'')dt'`

`x(tau)-x(0)=cint_0^tausinh(int_0^(t')(a(t'') dt'')dt'`

Minguzzi considered what he called the reconstruction problem during acceleration in special relativity. For 1+1 dimensions this requires that the accelerated observer O have an accelerometer that records his acceleration Vs time along his spatial direction. His ansatz is that the acceleration, `a`, can be expressed as

`a =d/dt(gamma v)=d/dt(v/sqrt(1/sqrt(1-v^2)))\ \ \ \ \ (1.1)`

where v is speed and it has been assumed that c=1Instead of using this equation directly to compute the speed Vs time he reverts to the Lorentz invariant value a2 by taking the square of the time-like component and subtracting the square of the space-like component and obtains the following result

`−a^2=((dgamma)/dt)^2−((d(γv))/dt)^2\ \ \ \ (1.2)`

Taking the differentials we get:

`a^2=1/(1-v^2)^2((dv)/dt)^2\ \ \ \ (1.3)`

which is quite different from the result that we would get using equation (1.1) which is

`a^2=1/(1-v^2)^3((dv)/dt)^2\ \ \ \ (1.4)`

Solving equation (1.3) for v we obtain:

`int_0^v(dv')/(1-v'^2)=tanh^-1v=int_0^ta(t')dt'`

and therefore:

`v(t)=tanh[int_0^ta(t')dt'+tanh^-1(v_0)]=tanh(theta)`

Then the change in the time like component is just the integral `δτ=int_0^tcosh[θ(t')]dt'` and the change in space like component is `δx=int_0^tsinh[θ(t')]dt'`.

In order to avoid problems of simultaneity at a large distance, Minguzzi chooses to compute the invariant `T^2(t)` which is the difference between the squares of `deltatau` and `deltax` or

`T^2=deltatau^2-deltax^2`

This expression turns out to be very simple since `deltatau^2−deltax^2=(deltatau−deltax)(deltatau+deltax)` and we may form the components of this product inside the integrals:

`deltatau-deltax=int_0^t{coshtheta(t')-sinhtheta(t')}dt'=int_0^texp[-theta(t')]dt'`

`deltatau+deltax=int_0^t{coshtheta(t')-sinhtheta(t')}dt'=int_0^texp[theta(t')]dt'`

`T^2=int_0^texp[theta(t')]int_0^texp[-theta(t')]`

We would like to derive an expression for dT/dt from our expression for `T^2`.

`2T(dT)/dT=d/dt{int_0^texp[theta(t')]int_0^texp[-theta(t')]}`

`(dT)/(dT)=1/(2T)d/dt{int_0^texp[theta(t')]int_0^texp[-theta(t')]}`

`(dT)/(dT)=1/(2T){exp[theta(t)]int_0^texp[theta(t')]+exp[-theta(t)]int_0^texp[-theta(t')]}`

Using the expression for `T` we have:`(dT)/(dT)=1/2{exp[theta(t)]int_0^texp[theta(t')]+exp[-theta(t)]int_0^texp[-theta(t')]}/sqrt(int_0^texp[theta(t')]int_0^texp[-theta(t')]`

To simplify this expression let:`b=sqrt(int_0^texp[-theta(t')])/sqrt(int_0^texp[theta(t')])`

Note that `b=exp[ln(b)]` then we can re-write:`(dT)/(dt)=1/2{bexp[theta(t)]+exp[-theta(t)]/b}=cosh[theta(t)+ln(b)]`

Finally:`(dT)/(dt)=cosh[int_0^ta(t')dt'+1/2ln({(int_0^texp[-int_0^va(t"''"")dt"''""]dt')])/(int_0^texp[-int_0^va(t"''"")dt"''""]dt')]))}`