INTO THE FUTURE A Planning Tool for Optimists and Hopeful Travelers © David Allan Wright 19962016 
t ↔ τ 

Leave Earth for a few years and escape into its future. Cruise an interstellar luxury liner, taking a round trip journey to a nearby star and back. Accelerating at a constant rate, you wouldn’t be weightless. Accelerating at a rate of 1g (9.81 m/s2) you’d weigh the same as you do on Earth. If you’re an optimist and willing to take on double your weight while possibly losing some weight at the same time (think health cruise), traveling a dozen years at twice your weight will get you about five hundred years away from the present.
The technical and theoretical problems of accelerating any object, the size of spacecraft and in a straight line, at a constant rate for long periods of time (more than a few seconds or minutes) are not trivial. Serious consideration of these obstacles can lead to some very curious physics and technology and this is a subject left for later review (consider the problem of accelerating a few protons close to the speed of light: http://home.web.cern.ch/about/engineering/restartinglhcwhy13tev). While engineers work out the details of luxury starship design and construction, you can plan your escape into the future. The Starship Planning Tool below allows you to enter fractions or multiples of your weight (gforces) and the time of your roundtrip. You can travel at one “g” to feel your normal weight, or choose to adapt to higher “g’s” to increase your time travel time. The longer you take on your roundtrip journey, the more you travel into the future. For example, you might choose a one g comfort cruise together with a round trip time of 10 years – the Earth would have aged 25 years to your 10. Everyone and everything you left behind aged 15 years more than the 10 years you spent on the luxury liner. If 15 years difference isn’t your idea of time travel, try traveling with a less comfortable acceleration of 1.5 g’s for 15 years and you’ll travel over 400 years into the future. Travel at 2 g’s for 20 years and travel over 29,000 years into the future! 
CHOOSE YOUR FRAME:
STARSHIP TIME (τ) TO EARTH TIME (t) EARTH TIME (t) TO STARSHIP TIME (τ) FRAME TIMES DIFF., MAX SPEED, & DISTANCE:EARTH TIME (t) MINUS STARSHIP TIME (τ) 
Starship Time as a Function of Earth Time  This form returns the time that would pass in the moving frame of your starship for a given amount of time that passes on Earth while you're gone. Determine your ship's roundtrip travel time by entering the amount of time that would pass on Earth during your journey.  
Choose your gForce:  Standard Gravity is used in this form = 9.80665 m/s^{2}. The speed of light, c = 299792458 m/s. 

Your Spaceship's acceleration will be:  m/s^{2}  
Years on Earth you'd like to pass while you're gone:  365.242 days = 3.15569 x 10^{7}seconds is used in this form.
The Ship's roundtrip time as a function of Earth time: τ = 4c/a sinh^{1}(at/4c), where τ is the accelerated frame's time, a is the acceleration, t is the rest frame (Earth) time, and c is the speed of light. BACK TO TOP (CHOOSE YOUR FRAME) 

Years passing on ship (τ) during your roundtrip:  years  
Days that will pass on your ship's roundtrip:  days  
Seconds that will pass on your ship's roundtrip:  seconds 
MAXIMUM SPEED AND FARTHEST DISTANCE
Given the Earth time that passes during the starship's roundtrip, what was the ship's maximum speed and how far out did it travel? The outward bound half of your roundtrip consists of two equal blocks of time (half of four equal blocks that makeup the roundtrip). At the launch of the starship, the Earth time t and starship time τ equal zero (t_{0} = 0), the ship's initial speed equals zero (v_{0} = 0), its acceleration a is constant, and it will reach its maximum speed v_{max} just before the engines are turned around midway to its destination (at the end of the first of the four equal blocks of time the make up the total roundtrip time; again, the engines are turned around so as to decelerate the ship at a constant rate and bring it to rest at its destination). From the definition of relativistic uniform acceleration, the speed v of the starship in the frame of the Earth (see below) is found to be v = dx/dt = at/√(1 + a^{2}t^{2}/c^{2}), where dx/dt is the instantaneous speed of the ship and t is the time that passes in the frame of the Earth; dx and dt are the infinitesimal change in position and time in the frame of the Earth, respectively. However, the contant acceleration a of the ship will not be the acceleration of ship as "seen" within the rest frame of the Earth; this acceleration, dv/dt = a(1  v^{2}/c^{2})^{3/2}, approaches zero as the speed of the ship v approaches the speed of light c.From the definition of relativistic uniform acceleration a_{R}, the ship's maximum speed and farthest distance reached during its roundtrip can be expressed as function of Earth time t: a_{R} ≡ dv/dt = a(1  v^{2}/c^{2})^{3/2}.Rearranging and integrating (at x = 0, t = 0), at = v/√(1  v^{2}/c^{2}) → v = dx/dt = at/√(1 + a^{2}t^{2}/c^{2}) → x = (c^{2}/a)[√(1 + a^{2}t^{2}/c^{2})  1]⇒ farthest distance = 2(c^{2}/a)[√(1 + a^{2}t^{2}/c^{2})  1)] ⇒ maximum speed = at/√(1 + a^{2}t^{2}/c^{2})  
Choose your gForce:  Standard Gravity is used in this form = 9.80665 m/s^{2}  
Your Spaceship's acceleration will be:  m/s^{2}  
Years on Earth you'd like to pass while you're gone:  This form uses 1 year = 3.15569 x 10^{7} seconds.
Ship time τ as function of Earth time t:


Earth roundtrip time in seconds (t):  seconds  
Ship's roundtrip time in years (τ):  years  
Ship's roundtrip time in seconds (τ):  seconds  
Farthest distance in meters (m):  meters 
Farthest distance = x_{max} = 2(c^{2}/a)[√(1 + a^{2}t^{2}/c^{2})  1)]
Maximum speed = v_{max} = at/√(1 + a^{2}t^{2}/c^{2}) 1 lightyear = 9.4605284 x 10^{15} meters 1 Parsec = 3.08567758 x 10^{16} meters BACK TO TOP 
Farthest distance in lightyears (ly):  lightyears  
Maximum speed reached in meters per second:  m/s  
Maximum speed in fraction of light speed:  c  
Distance reached in parsecs:  pc 
TIME TO DESTINATION
Are we there yet? Are you there yet? The time it takes to reach your favorite star and return depends on whether you’re on the starship or waiting back on Earth for good picture of your favorite solar system. This form allows to enter the distance (in lightyears) to a star, nebula, galaxy etc. and returns the time it takes for the ship to get there as viewed from the Earth or perhaps more importantly, how long it takes to get there for passengers in the starship. Though our closest star (Proxima Centauri) is about 4.2 lightyears and our galaxy is about a 100,000 lightyears across, you don't have to travel faster than light to reach your favorite star (http://www.space.com/3380constellations.html). Traveling to any star in our galaxy can be achieved within an average lifetime. The following forms allow you to determine how much time it would take, within a starship moving with constant acceleration, to travel a given distance. For example, travelling with a constant acceleration of 1g to the Andromeda Galaxy (2,538,000 light years) can take 2.5 million Earth years but will take less than 30 years for the passengers on our ideal starship.The outward bound part of your roundtrip consists of two equal blocks of time. At the launch of the starship, the time t equals zero (t_{0} = 0), the initial speed equals zero (v_{0} = 0), the acceleration a is constant, and the ship will reach it maximum speed (v_{max}) just before its engines are turned around to decelerate halfway to its destination. Slowing down with constant acceleration (deceleration) this way will bring to ship to rest (v_{f} = 0) at its chosen destination/distance.
Representing the farthest distance reached by the ship with Χ, the Earth time to reach that distance T, and the Ship time to reach that distance T', these quantities can be expressed as functions of x:
x = (c^{2}/a)[√(1 + a^{2}t^{2}/c^{2})  1] → farthest distance = x_{max} = 2(c^{2}/a)[√(1 + a^{2}t^{2}/c^{2})  1] Rearranging and solving for the Earth time (t) and ship time (τ) as functions of x:t = √(x^{2}/c^{2} + 2x/a) → Earth time to farthest distance = T = 2√((0.5Χ)^{2}/c^{2} + Χ/a), τ = c/a sinh^{1}[√(a^{2}x^{2}/c^{4} + 2ax/c^{2})] → Ship time to farthest distance = T' = 2c/a sinh^{1}(a/c T) = 2c/a sinh^{1}[a/c √((0.5Χ)^{2}/c^{2} + Χ/a)].  
Choose your gForce:  Standard Gravity is used in this form = 9.80665 m/s^{2} The speed of light, c = 299792458 m/s. 

Your Spaceship's acceleration will be:  m/s^{2}  
Enter the distance in lightyears to your favorite destination:  365.242 days are used in this form (3.15569 x 10^{7} seconds).
Rest Frame time (before any turnaround) = t = √(x^{2}/c^{2} + 2x/a), Earth time to farthest distance = T = 2√((0.5Χ)^{2}/c^{2} + Χ/a), Ship time to farthest distance = T' = 2c/a sinh^{1}(a/c T)= 2c/a sinh^{1}[a/c √((0.5Χ)^{2}/c^{2} + Χ/a)]. 

Earth time of ship's arrival at star:  years  
Earth time of ship's arrival at star (in seconds):  seconds  
Years on starship travel needed to reach the star:  years  
Starship time needed in seconds:  seconds  
Maximum speed reached before turnaround (m/s):  m/s  Maximum speed = v_{max} = at/√(1 + a^{2}t^{2}/c^{2}) 
Maximum speed reached as fraction of lightspeed:  c  
Maximum speed reached in m/s  NOT ROUNDED:  m/s 
Maximum speed as a function of distance (x = 0.5Χ) = v_{max} = a√(x^{2}/c^{2} + 2x/a)/(1 + ax/c^{2}) = v_{max} = a√(Χ^{2}/4c^{2} + Χ/a)/(1 + aΧ/2c^{2}) 
Maximum speed reached as fraction of lightspeed  NOT ROUNDED:  c 
The forms above use the following equations, respectively:
STARSHIP TIME (τ) TO EARTH TIME (t): T = 4c[e^{aT'/4c}  e^{aT'/4c}]/2a = (4c/a)sinh(aT'/4c),
EARTH TIME (t) TO STARSHIP TIME (τ): T' = (4c/a) ln[aT/4c + sqrt( 1 + a^{2}T^{2}/16c^{2}] = 4c/a sinh^{1}(aT/4c),
EARTH TIME (t) MINUS STARSHIP TIME (τ): T  T' = T  [4c/a sinh^{1}(aT/4c)] = (4c/a)sinh(aT'/4c)  T'.
T is the time that passes within the rest frame (Earth frame), a is the acceleration, c is the speed of light, and T^{’} is accelerated frame (the starship) time. The roundtrip journey is broken into four parts (note the 4’s in the equations) allowing for constant acceleration. The ship leaves Earth with constant acceleration, briefly turns around its rockets at the halfway point on its outward journey, and decelerates until it comes to rest at its destination. The ship then turns its rockets around again and returns to Earth in the same fashion.
The following is a derivation of a relationship between the time that passes with a rest frame (Earth) and the time that passes within a moving frame (a starship) that starting from rest, moves with a predetermined constant acceleration to a predetermined destination and back.
Here's a more detailed derivation in a pdf format:
An Introduction to Acceleration in Special Relativity.
Let v be the velocity of a point in S, and v' the velocity in S’.
v = (v' + V)/(1 + Vv'/c^{2})
t = (t' + Vx'/c^{2})/sqrt(1  V^{2}/c^{2}),
Taking the differentials of each,
dv = dv'/(1 + Vv'/c^{2})  (v' + V)(V/c^{2})
dv'/(1 + Vv'/c^{2})^{2} = (1 V^{2}/c^{2})
dv'/(1 + Vv'/c^{2})^{2}
dv = (1 V^{2}/c^{2})
dv'/(1 + Vv'/c^{2})^{2};
dt = (dt' + Vdx'/c^{2}) /sqrt(1  V^{2}/c^{2})=
dt'[1 + Vdx'/(dt'c^{2})]
/sqrt(1  V^{2}/c^{2})
v'= dx'/ dt'
dt = (1 + Vv'/c^{2}) dt' /sqrt(1  V^{2}/c^{2})
Dividing dv by dt, we find the acceleration in S; dv’/dt’ is the acceleration experienced in S’.
dv/dt = (1 – V^{2}/c^{2})^{3/2} dv'/[(1 + Vv'/c^{2})^{3}dt']
If S’ is the instantaneous comoving system of the point P, then v’ = 0, V = v; and if dv’/dt’ = a', then
dv/dt = (1  V^{2}/c^{2})^{ 3/2}a'
a = (1  V^{2}/c^{2})^{ 3/2}a'
Rearranging,
(1  V^{2}/c^{2})^{3/2}dv = a'dt
Integrating,
v(1  V^{2}/c^{2})^{1/2}dv = a't
Rearranging, recalling SR's Clock Hypothesis, and integrating,
T' = (c/a) ln{aT/c + sqrt( 1 + a^{2}T^{2}/c^{2}} = c/a sinh^{1}(aT/c)
This represents the time that goes by on Earth (rest frame) as function (transform) of the time the goes by within the starship (moving frame), moving with a constant acceleration (a'). At the moment of the engine's first turnaround (and its assumed that the duration of the moment is neglible), halfway to the ship's outward destination, the the ship's acceleration is revearsed so the it can slow and come to rest at the end of the second quarter of its four part roundtrip.T'/4 = (c/a) ln{aT/4c + sqrt( 1 + a^{2}T^{2}/16c^{2}} = 4c/a sinh^{1}(aT/4c)
REFERENCES
Acosta, V., Cowan, C. L., & Graham, B. J. (1973) Essentials of Modern Physics. Harper & Row: New York.
Arzeleiès, Henri, (1972) Relativistic Point Dynmaics . trans. by P. W. Hawkes, Pergamon P: Oxford.
DeWitt, Bryce (2011) Bryce DeWitt’s Lectures on Gravitation, Edited by Steven M. Christensen, Lecture Notes in Physics, 826. SpringerVerlag: BerlinHeidelberg.
Einstein, Albert (1952) The Principle of Relativity (On the Electrodynamics of Moving Bodies, trans. by W. Perrett and G. B. Jefferey from “Zur Elecktrodynamik bewegter Körper,” Annalen der Physik, 17, 1905), Dover: New York.
(Einstein, Albert, 1918, Dialogue about Objections to the Theory of Relativity/Dialog über Einwände gegen die Relativitätstheorie . Naturwissenschaften, 6, 697–702.)
Marder, L., (1971) Time and the SpaceTraveller . U of Pennsylvania P: Philadelphia.
Møller, C., (1962) The Theory of Relativity . Oxford UP: Amen House, London.
Powers, Robert M., (1981) The Coatails of God: The Ultimate Spaceflight  The Trip to the Stars. Warner Bks: New York.
Schlegel, Richard, (1968) Time and the Physical World . Dover Publications, Inc. (1961, Michigan State University Press): New York.
LINKS OF INTEREST
Susskind Lectures on Relativistic Kinematics
George Smoot on Relativity: Physics 139 Relativity, UC Berkeley
The Original Usenet Physics FAQ: The Relativistic Rocket
scienceworld.wolfram: "Proper Time"
scienceworld.wolfram: "Velocity FourVector"
Leonard Susskind on Special Relativity (YouTube)
Restarting the LHC: Why 13 Tev?
NASA Science News: Scientists Examine Using Antimatter and Fusion to Propel Future Spacecraft