7
$\begingroup$

Prologue

In my current worldbuilding project, I decided to put a ban for faster than lightspeed travel, and decided that most space flights be done in slower than lightspeed travels. Assuming efficient engines had been invented in this universe (say, some sort of 80% to 90% energy efficiency), and antimatter are not that hard to farm (just say a new technology (or some) had been invented to increase antimatter productions and storages), relativistic travel is common to travel from stars to stars. So far the setting looks nice, and ready to be filled with stories. As I start writing, I realized that onboard clock and outside observer (assuming it as a reference frame) would experience different passage of time, and that accelerating (and decelerating) to relativistic speed would also contribute to variable time passage felt onboard the ship relative to outside observer.

Common Space Travel Settings

  1. Accelerate-ballistic-decelerate settings, where ships were accelerated to desired speed, then follows ballistic path, and then decelerates toward rest near its destination. Commonly used, as the ship could be mainly dormant during ballistic path (cruise speed), saves the most power for considerable travel time. Cruise speed varies between common 0.3-0.5 times c to high priority travel between 0.7-0.9 times c.
  2. Constant acceleration-turnaround-constant deceleration toward rest, very energy consuming, usually used for urgent short-ranged distance travel (for example, interplanetary travel). It differs from the first setting only on turnaround phase, that costs nearly negligible amount of time compared to time elapsed on the whole voyages (while at the first setting most of the voyages were spent on cruise speed).

Problems

Mainly the question revolves around relativistic effect felt onboard the ship and an observer at a planet (making a huge assumption that a planet is generally 'static').

In case of first setting

  1. How to calculate time (both from reference frame of a static observer, and the ship's reference frame) required for the ship to accelerate from rest toward desired speed (or from desired speed toward rest) given x gee of acceleration (ship's reference frame)?

In case of second setting

  1. How to calculate total time required (both from reference frame of a static observer, and the ship's reference frame) for the ship to cross the distance of n kilometers (or AU) given n gee of acceleration?

Considerations

Given wide range of possible readers in term of math proficiency, I would encourage easy-to-understand formula, with terms that high-schoolers could understand (high-school education be the baseline for difficulty level). Explanations may follow after the given formula if the answerer desires (and would be appreciated when they do). That said, square root functions, divisions, multiplications (additions and substractions too, obviously), square functions, are allowed.

Closing Statements

The purpose of the aforementioned considerations are that everybody (with minimum of high-school education) could read and take advantage of the answers, basically any worldbuilder that requires the formula to calculate similar problem I am facing. And I have to admit that my physics were quite rusty (it had been two years since my high school), and quite lazy to figure it out on the net. I wrote this question mainly to benefit me (let's be honest), but it could also benefits other (mainly) hard sci-fi worldbuilder or any worldbuilder facing similar problem as of me.

Thanks in advance.

$\endgroup$
6
$\begingroup$

The term you'll want to google is the "relativistic rocket"--there's a good page on the idea here. The equations there all deal with a rocket that accelerates at a constant rate "a" (a constant proper acceleration as measured by an accelerometer on board the ship) starting from rest in some inertial observer's rest frame (say, an observer at rest relative to a star), with distance "d" and time "t" and velocity "v" measured in terms of that frame, while "T" is the time on board the ship (which is different from t due to time dilation) For problem #1, you can use the equation for velocity "v" as a function of acceleration "a" along with either the onboard time "T" or the observer's time "t":

v = c th(aT/c) = at / sqrt[1 + (at/c)^2]

Note that the symbol "th" is intended to refer to the hyperbolic tangent function, likewise some other equations use "sh" which is hyperbolic sine and "ch" which is hyperbolic cosine, all three functions are available on the online calculator here, where they are written tanh, sinh, and cosh, and the calculator also includes their inverse functions atanh, asinh, and acosh (which are needed if you want to solve one of these equations for T). If you take the first half of the above equation, v = c th(aT/c), and solve for T in terms of v and a, then you get the following equation for T (I've put in a syntax that the online calculator will understand):

(c/a)*atanh(v/c)

So, you can just substitute the speed of light for c (in whatever units you're using, might be easiest to use something like light-years for distance and years for time so that c=1), the acceleration for a (in units of light-years and years, a one-gee acceleration has a value of 1.03 as mentioned on the relativistic rocket page), and your desired final acceleration for v, and plug it into the calculator (note to other editors: this is why I'm not using the nicer-looking "MathJax" formatting for equations, I want people to be able to cut and paste into the calculator, so please don't change them). For example, with c=1, a=1.03 and v=0.5, I can plug (1/1.03)*atanh(0.5/1) into the calculator and conclude it would take about 0.5333 years of onboard time to reach 0.5c relative to the reference frame I initially started from rest in.

And if you want to calculate the time in the observer's frame rather than the onboard time, you solve the equation v = at / sqrt[1 + (at/c)^2] for t, which gives the following equation for t:

(v/a)/(sqrt(1 - (v/c)^2))

So for the same numbers of c=1, a=1.03 and v=0.5, you'd plug in (0.5/1.03)/(sqrt(1 - (0.5/1)^2)) which gives an answer of t = 0.5605 years for the observer in the frame where the ship started at rest.

For problem #2, you'll want to start with these equations from the relativistic rocket page:

d = (c^2/a) [ch(aT/c) − 1] = (c^2/a) (sqrt[1 + (at/c)^2] − 1)

...and solve the first part for T in terms of d and a, then solve the second part for t in terms of d and a. Solving the first part for T gives:

(c/a) * acosh((d*a/c^2) + 1)

And solving the second part for t gives:

sqrt(d^2 + (2*d/a))

So for example if you want to figure out the onboard time T and observer's time t for the ship to travel a distance of 10 light years at a one-gee acceleration, you'd use c=1, a=1.03 and d=10, so for T you'd plug (1/1.03) * acosh((10*1.03/1^2) + 1) into the calculator which gives an answer of 3.0252 years of onboard time, and for t you'd plug sqrt(10^2 + (2*10/1.03)) into the calculator which gives 10.9278 years of observer's time.

Also, note these distance formulas are for a ship that accelerates at a constant rate, continually gaining speed relative to the inertial frame where it started at rest. If you like you can easily adapt the formulas to a scenario where you accelerate for the first half of the journey and then decelerate for the second half, coming to rest at your destination. The way to do this is just to pick the distance to the halfway point and plug that into the formulas, giving the time to get to the halfway point. Then since the acceleration and deceleration phases are totally symmetrical, you can just double the time to the halfway point to get the total time to arrive and come to rest at the destination.

Summing up:

If you have the acceleration A and the final velocity V, plug them into the following along with the value of the speed of light c in your preferred units, and evaluate in the online calculator to get the onboard time T:

(c/A)*atanh(V/c)

If you have the acceleration A and the final velocity V, plug them into the following along with the value of the speed of light c in your preferred units, and evaluate in the online calculator to get the time t in the inertial frame of an observer who saw the ship initially at rest:

(V/A)/(sqrt(1 - (V/c)^2))

If you have the acceleration A and the final distance D, plug them into the following along with the value of the speed of light c in your preferred units, and evaluate in the online calculator to get the onboard time T:

(c/A) * acosh((D*A/c^2) + 1)

If you have the acceleration A and the final distance D, plug them into the following along with the value of the speed of light c in your preferred units, and evaluate in the online calculator to get the time t in the inertial frame of an observer who saw the ship initially at rest:

sqrt(D^2 + (2*D/A))

One last thing:

Another issue you may want to consider for worldbuilding purposes is the ratio of the initial total mass of the rocket, including both fuel and payload, to the final payload mass at the end of the acceleration when all the fuel has been expended (if it ends up being something ridiculous, like millions of tons of fuel per kilogram of payload mass, you may have to rethink your acceleration and time somewhat). I covered the equations needed to calculate this in this post, if anyone's interested. The answer depends on the type of rocket and the velocity of its exhaust, with the most efficient possible case being a rocket that totally annihilates the mass of the fuel and converts it into high-energy photons that get shot out the back, in which case the exhaust velocity would be c. A bunch of exhaust velocities for rockets both existing and theoretical can be found on this page, see the column for velocity in meters/second (which you need to convert to whatever units you're using for the other quantities in your equations, like light-seconds and seconds, light-years and years etc.).

$\endgroup$
  • $\begingroup$ That is exactly what I am asking for! Thanks! $\endgroup$ – Hendrik Lie Mar 10 '16 at 7:06
  • $\begingroup$ @HDE 226868 - I think you missed it, but my answer included a comment about this after I pointed out that people could copy and paste the equations into this online calculator and plug in their own desired values, after which I said: (note to other editors: this is why I'm not using the nicer-looking "MathJax" formatting for equations, I want people to be able to cut and paste into the calculator, so please don't change them) $\endgroup$ – Hypnosifl Apr 5 '16 at 4:41
  • $\begingroup$ @Hypnosifl My bad; I totally missed that. $\endgroup$ – HDE 226868 Apr 5 '16 at 21:13
  • $\begingroup$ One last thing, in solving the first scenario, we must sum up total time (say in observer time frame) of first acceleration, then normal time dilation for cruise speed, then deceleration phase. The equation to calculate t required to obtain speed v and acceleration a is given, but how would we know its d for distance traveled during acceleration (or deceleration) phase? $\endgroup$ – Hendrik Lie Apr 1 '17 at 10:33
  • $\begingroup$ Oh wait, I missed it previously, but we could use d = (c^2/a) (sqrt[1 + (at/c)^2] − 1) to obtain d, right? $\endgroup$ – Hendrik Lie Apr 1 '17 at 10:55
5
$\begingroup$

There are two different causes for time dilation. For the most part, you'll ignore General Relativity and just use Special Relativity, but I've laid out both as best as I know.

General Relativity

First is acceleration, whether this is a result of a gravitational field (such as sitting on a planet) or more typical acceleration (such as being in a rocket). I'm getting mixed results trying to find formulas for this, but at any acceleration a human can survive, the difference will be fairly negligible. From this Physics FAQ article, it seems the effect of acceleration is just that it's constantly changing the relative speed, as shown below. While you're in a gravity field (such as on a planet), the time dilation has to do with how it warps spacetime to make it look like you're going different speeds.

The result is when your guys are far from any planets, you can just use the relative speed (special relativity) calculations below for various points throughout their acceleration (that is, integrate). For example, break the trip up into 50 stages as they accelerate, 1 stage of constant velocity, then 50 stages of deceleration. If acceleration is the same rate as deceleration, you can just calculate one and map it to the other.

When they're hovering near a planet (or standing on the planet), you use the following formula (from this physics.SE answer):

$t'=t\sqrt{1-{2GM\over rc^2}}$

where
$G$ is the gravitational constant, $\approx 6.674\times 10^{-11}{N\cdot m^2\over kg^2}$,
$M$ is the mass of the planet / star / black hole / small moon / Death Star in question,
$r$ is the distance from the center of the planet.

Doing this calculation for Earth, we see that time on Earth passes about 0.999999999305 times as fast as it would with zero gravity around. Which means we see about 0.7 seconds less than a spaceship in deep space would over a 30 year period. So you need a really massive planet before you start seeing a notable difference. To get $t'=0.9t$, you need $M\approx 8.2\times 10^{32}kg\approx 410 M_\odot$ (about 410 times the mass of our sun), where you'd be experiencing about 136 million gees.

If they're orbiting near the same planet, you use this formula (from the same physics.SE answer):

$t'=t\sqrt{1-{3\over 2}{2GM\over rc^2}}$

I believe neither of those answers applies if you're inside a planet, but that shouldn't be an issue for the most part. Also, I presume that during a transition from hovering to orbit, you'd need to integrate between the two formulas, but I doubt your spaceships will spend enough time in this transition to really care. And about the only place it's going to matter anyways is if they're really close to a black hole or neutron star or something.

Special Relativity

Second is relative speed. As your spaceship gets close to light speed relative to some "stationary" object (your reference frame), its perceived time will be lower. The formula is given by (click the link for a calculator):

$t'=t\times\sqrt{1-{v^2\over c^2}}$

where
$t'$ is the time perceived by the moving spaceship,
$t$ is the time perceived by Earth,
$v$ is the speed of the spaceship relative to Earth,
$c$ is the speed of light.

You get a 1% difference ($t'=0.99t$) in clocks around $v=0.14c$, a 10% difference ($t'=0.9t$) around $v=0.44c$, a 50% difference ($t'=0.5t$) around $v=0.87c$, a 90% difference ($t'=0.1t$) around $v=0.995c$, and a 99% difference ($t'=0.01t$) around $v=0.99995c$.

The above calculator gives the moving object's time as a fraction of "normal" time. If you'd like to go the other way, you can either just plug $1\over t'$ into a calculator, or use this calculator that returns normal time as a (greater than 1) fraction of the moving object's time.

Technically, you could use anything, including imaginary objects, as your base reference frame, but it's easiest to use something like a star or planet so everyone can be on the same schedule. Stars do move relative to each other, and realistically your scientists would carefully analyze the differences, but for practical purposes all stars are in the same reference frame. Barnard's Star is the fastest-moving star near us, moving about 143 km/s, which translates to a relativistic factor of 1.000000114, or 0.999999886, depending on whose perspective we're looking from. If someone near Barnard's Star counted 1,000,000,000 seconds passing, someone near Earth would count 1,000,000,114 seconds. That's a difference of about 2 minutes every 32 years.

$\endgroup$
  • $\begingroup$ Actually, I was asking for time dilation during accelerations, but yes, you provided me answer I don't even think of to ask. So up vote for that. $\endgroup$ – Hendrik Lie Mar 7 '16 at 11:22
  • $\begingroup$ Dealing with acceleration doesn't require general relativity, special relativity can handle acceleration in a spacetime which isn't curved by the gravity of massive objects (see this answer on the Usenet Physics FAQ). Only when spacetime is curved is general relativity needed. $\endgroup$ – Hypnosifl Mar 10 '16 at 5:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.