Thanks to Aron Yoffe for suggesting the the newtonian comparison.

Thanks to Tom Fuchs for suggesting the analogy to the stone thrown upwards.

Fuel numbers added by Don Koks 2004.

Updated by Phil Gibbs 1998.

Thanks to Bill Woods for correcting the fuel equation.

Original by Philip Gibbs 1996.

The theory of relativity sets a severe limit to our ability to explore the galaxy in space ships. As an object approaches the speed of light, more and more energy is needed to maintain its acceleration, with the result that to reach the speed of light, an infinite amount of energy would be required. It seems that the speed of light is an absolute barrier which cannot be reached or surpassed by massive objects (see the FAQ article on faster than light travel). Given that our galaxy is about 80,000 light years across, there seems little hope for us to get very far in galactic terms.

Science fiction writers might make use of worm holes or warp drives to overcome this restriction, but it is not
clear that such things can ever be made to work in reality. Another way to get around the problem may be to
use the relativistic effects of time dilation and length contraction to cover large distances within a reasonable
time span for those aboard a space ship. When a rocket accelerates at 1 *g* (9.81 m/s^{2}),
its crew experiences the equivalent of a gravitational field with the same strength as that on Earth. If this
acceleration could be maintained for long enough, the crew would eventually reap the benefits of the relativistic
effects that increase the effective rate of travel.

What then, are the appropriate equations for the relativistic rocket?

First of all, we need to be clear what we mean by continuous acceleration at 1 *g*. The
acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference
travelling at the same instantaneous speed as the rocket (see the FAQ article
on accelerating clocks), because this is the acceleration that its occupants would
physically feel—and we want them to accelerate at a comfortable rate that has the effect of mimicking their
weight on Earth. We'll call this acceleration *a*. The proper time measured by the crew of the
rocket (i.e. how much they age) is called *T*, and the time measured in the non-accelerating frame of
reference in which they started (e.g. Earth) is *t*. We assume that the stars are essentially at rest
in this Earth frame. The distance covered by the rocket as measured in this frame of reference is *d*,
and the rocket's velocity is *v*. The time dilation or length-contraction factor at any instant is the
gamma factor γ.

The relativistic equations for a rocket with constant positive acceleration *a > 0* are the
following. First, define the hyperbolic trigonometric functions sh, ch, and th (also known as sinh, cosh, and
tanh):

sh x = (e^{x}− e^{−x})/2 , ch x = (e^{x}+ e^{−x})/2 , th x = sh x/ch x .

Using these, the rocket equations are

t = (c/a) sh(aT/c) = sqrt[(d/c)^{2}+ 2d/a] , T = (c/a) sh^{−1}(at/c) = (c/a) ch^{−1}[ad/c^{2}+ 1] , d = (c^{2}/a) [ch(aT/c) − 1] = (c^{2}/a) (sqrt[1 + (at/c)^{2}] − 1) , v = c th(aT/c) = at / sqrt[1 + (at/c)^{2}] , γ = ch(aT/c) = sqrt[1 + (at/c)^{2}] = ad/c^{2}+ 1 .

These equations are valid in any consistent system of units, such as seconds for time, metres for distance,
metres per second for speeds and metres per second squared for accelerations. In these units *c*
≃ 3 × 10^{8} m/s. To do some example calculations, it's easier to use units of years for
time and light years ("ly") for distance. Then *c = *1 ly/yr and *g ≃ *1.03
ly/yr^{2}. Here are some typical values of the various parameters for *a = *1 *g*.

Ttdvγ----------------------------------------------------------------------- 1 year 1.19 years 0.56 ly 0.77c 1.58 2 3.75 2.90 0.97 3.99 5 83.7 82.7 0.99993 86.2 8 1,840 1,839 0.9999998 1,895 12 113,243 113,242 0.99999999996 116,641

So in theory you can travel across the galaxy in just 12 years of your own time. If you want to arrive at
your destination and stop, then let's say you turn your rocket around at the halfway point and decelerate at
1 *g*. In that case it will take nearly twice as long in terms of proper time *T* for the
longer journeys; the elapsed time *t* on Earth will be only a little longer, since in both cases the rocket
is spending most of its time at a speed near that of light. We can still use the above equations to work
this out, since although the acceleration is now negative, we can "run the film backwards" to reason that they must
still apply. Here's the general formula:

T = 2c/g ch^{−1}[gd/(2c^{2}) + 1] .

Here are some of the times you will age when journeying to a few well known space marks, arriving at low speed:

dStopping at:T--------------------------------------------------- 4.3 ly Nearest star 3.6 years 27 ly Vega 6.6 years 30,000 ly Centre of our galaxy 20 years 2,000,000 ly Andromeda Galaxy 28 yearsnly Anywhere, but see 1.94 cosh^{−1}(n/1.94 + 1) years next paragraph

For distances greater than about a thousand million light years, the formulae given here are inadequate because the universe is expanding. General Relativity would have to be used to work out those cases.

If you wish to pass by a distant star and return to Earth, but you don't need to stop there, then a looping route is better than a straight-out-and-back route. A good course might be to head out at constant acceleration in a direction at about 45° to that of your destination. At the appropriate point, you start a long arc such that the centrifugal acceleration you experience is also equivalent to Earth's gravity. After 3/4 of a circle, you decelerate in a straight line until you arrive home.

To experiment with the quantities above, see David Wright's INTO THE FUTURE web page.

As a comparison to the above values of *T*, what if relativity didn't apply: that is, how
would *T* depend on *d* in a fully newtonian universe, so that the rocket could accelerate at
1 *g* forever? We'll call the elapsed time *T'* here so as not to confuse it with
the *T* above that we are going to compare it with. Begin with the usual equation for constant
acceleration from a standing start, "displacement = 1/2 × acceleration × time^{2}".
Consider that, by symmetry, *T'/2* is the time to get to the half-way point, so

which means thatd/2 = 1/2 a (T'/2)^{2},

WithT' = 2 sqrt(d/a) .

T' = 1.97 sqrt(n)years.

Here is the above table amended to include comparison values of *T'*:

You can see that in our relativistic universe, even though the rocket's speed is limited to the speed of light (unlike in the newtonian universe, which has no speed limit), the very high values attained by the time-dilation factor γ on long trips ensure that the relativistic rocket passengers still age far, far less than do the newtonian passengers undergoing the same trip.dStopping at:T(relativistic universe)T'(newtonian universe) ------------------------------------------------------------------------------------------------------------ 4.3 ly Nearest star 3.6 years 4.1 years 27 ly Vega 6.6 years 10.2 years 30,000 ly Centre of our galaxy 20 years 341 years 2,000,000 ly Andromeda Galaxy 28 years 2786 yearsnly Anywhere 1.94 cosh^{−1}(n/1.94 + 1) years 1.97 sqrt(n) years

But now, back to the relativistic rocket...

In the rocket, you can make measurements of the world around you. One thing you might do is ask how the
distance to an interesting star you are headed towards changes with *T*, the time on your clock. At
blast-off (*t = T = 0*) the rocket is at rest, so this distance initially equals the distance *D* to
the star in the non-accelerating frame. But once you are moving, however you choose to measure this distance,
at each moment it will be reduced by your current distance *d* travelled in the non-accelerating frame, as
well as the whole lot contracted by the current factor of γ. Eventually you will pass the star and it
will recede behind you. The distance you measure to it at time *T* is

(D − d)/γ = (D + c^{2}/a)/ch(aT/c) − c^{2}/a .

A plot of this distance as a function of *T* shows that, as expected, it starts at *D*, then
reduces to zero as you pass the star. Then it becomes negative as the star moves behind you.
As *T* goes to infinity, the distance asymptotes to a value of −*c ^{2}/a*. That
means that from your vantage point in the rocket, everything in the universe is falling from "above" to "below" the
rocket, but never receding any farther than a distance of −

Whereas time slows to a stop a certain distance below the rocket, it speeds up above the rocket (that is, in the direction in which it's travelling from Earth's perspective). This effect could, in principle, be measured inside the rocket too: a clock attached to the rocket's ceiling (i.e. in the rocket's "nose") ages faster than a clock attached to its floor.

For a standard-sized rocket with a survivable acceleration, this difference in how fast things age within its
cabin is very small. Even so, it tells us something fundamental about gravity, via Einstein's *Equivalence
Principle*. Einstein postulated that any experiment done in a real gravitational field—provided
that experiment has a "small" extent in space and time—will give a result indistinguishable from the same
experiment done in the above "uniformly accelerating" rocket. These space and time extents must be small so
that no "tidal" effect due to the inhomogeneous nature of the real gravity field will be apparent.
Analogously, even though Earth is not flat, we can treat it as being flat on a small patch of ground: by this we
mean that as we consider smaller and smaller patches of ground, the error we make by treating each patch as flat
grows smaller at a faster rate than the size of the patch decreases. In a similar way, the errors we make by
treating the gravity in a sequence of ever-smaller regions of spacetime as equivalent to the interior of a
uniformly accelerated rocket in the absence of gravity decrease at a faster rate than do the sizes of those
spacetime regions. So the idea that the rocket's ceiling ages faster than its floor (and that includes the
ageing of any bugs sitting on these) transfers to gravity: the ceiling of the room in which you now sit is ageing
faster than its floor; and your head is ageing faster than your feet. Earth's rotation complicates this
effect, but doesn't dominate it.

This difference in ageings on Earth has been verified experimentally. In fact, it was absolutely necessary to take into account when the GPS satellite system was assembled: because the satellites that broadcast data to your GPS receiver are ageing slowly compared to clocks on Earth, they must be set to run slightly quickly in the factory by the same factor, before they are sent up to orbit.

We can use the above equation for the distance *(D − d)/γ* that you measure to your
destination star to illustrate the Equivalence Principle. As you accelerated from Earth and headed towards
that star, it was as if the star began falling towards you with an acceleration of *a*. After a time
measured by you as *T*, the distance you measure to the star is the above

Suppose that either your acceleration(D − d)/γ = (D + c^{2}/a)/ch(aT/c) − c^{2}/a .

Setch x = 1 + x^{2}/2! + x^{4}/4! + x^{6}/6! + ...

so that1/(1+x) = 1 − x + x^{2}− x^{3}+ ... ,

In that case we can approximate the distance you measure to the star as1/ch x ≃ 1 − xfor |^{2}/2x|<1.

The second term in the last line divided by the third term equals(D − d)/γ = (D + c^{2}/a)/ch(aT/c) − c^{2}/a ≃ (D + c^{2}/a) [1 − a^{2}T^{2}/(2c^{2})] − c^{2}/a = D + c^{2}/a − D a^{2}T^{2}/(2c^{2}) − aT^{2}/2 − c^{2}/a = D − D a^{2}T^{2}/(2c^{2}) − aT^{2}/2 .

distance to star ≃Now, what would Newton say? He'd say that the star had been falling towards you for a timeD − aT.^{2}/2

Sadly, there are a few technical difficulties you will have to overcome before you can head off into space. One of these difficulties is creating your propulsion system and generating fuel. The most efficient theoretical way to propel the rocket is to use a "photon drive". This would convert mass to photons or other massless particles which shoot out the back of the rocket. Perhaps this may even be technically feasible if we ever produce an antimatter-driven "graser" (gamma-ray laser).

Remember that energy is equivalent to mass, so provided mass can be converted to 100% radiation by means of
matter–antimatter annihilation, we just want to find the mass *M* of the fuel required to accelerate the
payload *m*. The answer is most easily worked out by conservation of energy and momentum.

At the end of the trip the fuel has all been converted to light with energyE_{initial}= (M+m)c^{2}.

By conservation of energy these must be equal, so here is our first conservation equation:E_{final}= γmc^{2}+ E_{L}.

(M+m)c........ (1)^{2}= γmc^{2}+ E_{L}

At the trip's end the fuel has all been converted to light with momentum of magnitudep_{initial}= 0 .

By conservation of momentum these must be equal, so our second conservation equation is:p_{final}= γ mv − E_{L}/c .

Eliminating0 = γ mv − E........ (2)_{L}/c

so that the fuel-to-payload ratio is(M+m)c^{2}− γmc^{2}= γmvc ,

M/m = γ(1 + v/c) − 1 .

This equation is true irrespective of how the ship accelerates to velocity *v*, but if it accelerates at
constant rate *a* then

M/m = γ(1 + v/c) − 1 = cosh(aT/c)[ 1 + tanh(aT/c) ] − 1 = exp(aT/c) − 1 .

How much fuel is this? The next chart shows the amount of fuel needed (*M*) for every kilogramme of
payload (*m = *1 kg).

dNot stopping, sailing past:M------------------------------------------------------- 4.3 ly Nearest star 10 kg 27 ly Vega 57 kg 30,000 ly Centre of our galaxy 62 tonnes 2,000,000 ly Andromeda Galaxy 4,100 tonnes

This is a lot of fuel—and remember, we are using a motor that is 100% efficient!

What if we prefer to stop at the destination? We accelerate to the half-way point at 1 *g* and then
immediately switch the direction of our rocket so that we now decelerate at 1 *g* for the second half of the
trip. The calculations here are just a little more involved since the trip is now in two distinct halves (and
the equations at the top assume a positive acceleration only). Even so, the answer turns out to have exactly
the same form: *M/m = *exp*(aT/c) − 1*, except that the proper time *T* is now almost
twice as large as for the non-stop case, since the slowing-down rocket is losing the ageing benefits of
relativistic speed. This dramatically increases the amount of fuel needed:

dStopping at:M------------------------------------------------ 4.3 ly Nearest star 38 kg 27 ly Vega 886 kg 30,000 ly Centre of our galaxy 955,000 tonnes 2,000,000 ly Andromeda Galaxy 4.2 thousand million tonnes

Compare these numbers to the previous case: they are hugely different! Why should that be? Let's take the case of Laurel and Hardy, two astronauts travelling to Vega. Laurel speeds past without stopping, and so only needs 57 kg of fuel for every 1 kg of payload. Hardy wishes to stop at Vega, and so needs 886 kg of fuel for every 1 kg of payload. Laurel takes almost 28 Earth years for the trip, while Hardy takes 29 Earth years. (They both take roughly the same amount of Earth time because they are both travelling close to speed c for most of the journey.) They travel neck and neck until they've both gone half way to Vega, at which point Hardy begins to decelerate.

It's useful to think of the problem in terms of relativistic mass, since this is what each rocket motor "feels"
as it strives to maintain a 1 *g* acceleration or deceleration. The relativistic mass of each
traveller's rocket is continually decreasing throughout their trip (since it's being converted to exhaust
energy). It turns out that at the half-way point, Laurel's total relativistic mass (for fuel plus payload) is
about 28*m*, and from here until the trip's end, this relativistic mass only decreases by a tiny amount, so
that Laurel's rocket needs to do very little work. So at the halfway point his fuel-to-payload ratio turns
out to be about 1.

For Hardy, things are different. He needs to decrease his relativistic mass to *m* at the end where
he is to stop. If his rocket's total relativistic mass at the halfway point were the same as Laurel's
(28*m*), with a fuel-to-payload ratio of 1, Hardy would need to decrease the relativistic mass all the way
down to *m* at the end, which would require more fuel than Laurel had needed. But Hardy wouldn't have
this much fuel on board—unless he ensures that he takes it with him initially. This extra fuel that he
must carry from the start becomes more payload (a lot more), which needs yet more fuel again to carry that.
So suddenly his fuel requirement has increased enormously. It turns out that at the half-way point, all this
extra fuel gives Hardy's rocket a total relativistic mass of about 442*m*, and his fuel-to-payload ratio
turns out to be about 29.

Another way of looking at this odd situation is that both travellers know that they must take fuel on board
initially to push them at 1 *g* for the total trip time. They don't care about what's happening
outside. In that case, Laurel travels for 28 Earth years but ages just 3.9 years, while Hardy travels for 29
Earth years but ages 6.6 years. So Hardy has had to sit at his controls and burn his rocket for almost twice
as long as Laurel, and that has required more fuel, with even more fuel required because of the
fuel-becomes-payload situation that we mentioned above.

This fuel-becomes-payload problem is well known in the space programme: part of the reason the Saturn V moon rocket was so big was that it needed yet more fuel just to carry the fuel that it was already carrying.

Well, this is probably all just too much fuel to contemplate. There are a limited number of solutions that don't violate energy–momentum conservation or require hypothetical entities such as tachyons or worm holes.

It may be possible to scoop up hydrogen as the rocket goes through space, using fusion to drive the rocket. This would have big benefits because the fuel would not have to be carried along from the start. Another possibility would be to push the rocket away using an Earth-bound graser directed onto the back of the rocket. There are a few extra technical difficulties but expect NASA to start looking at the possibilities soon :-).

You might also consider using a large rotating black hole as a gravitational catapult, but aside from whether
such a thing actually exists, it would have to be *very* big to avoid the rocket being torn apart by tidal
forces or spun at high angular speed. If there is a black hole at the centre of the Milky Way, as some
astronomers think, then perhaps if you can get that far, you can use this effect to shoot you off to the next
galaxy.

One major problem you would have to solve is the need for shielding. As you approach the speed of light
you will be heading into an increasingly energetic and intense bombardment of cosmic rays and other
particles. After only a few years of 1 *g* acceleration, even the cosmic background radiation is
Doppler shifted into a lethal heat bath that's hot enough to melt all known materials.

(You can prove that easily using the standard textbook equations for constant acceleration, "h(t) = sqrt(2gH) t − gt^{2}/2 .

The Equivalence Principle suggests that provided we don't throw the star "too far" from a region that can be
considered to have uniform gravity, we can treat the motion of this stone moving up and coming back down as
identical to that of the above star. So let's use the rocket equations, but now think of the star as a stone
and the rocket as a human observer, and alter the equations so that the observer moves "downwards" through the
stone at time *t = 0* (but always accelerating upwards with some *g > 0*), then slows to a stop at a
distance *H* "below" the stone, then moves back up towards the stone. A plot of the observer's motion
in spacetime is still the same hyperbola that the rocket has at the top of this page, so we can mostly use the
rocket equations from there; but in this new scenario the hyperbola has been shifted somewhat on the spacetime
diagram relative to its origin, so we'll need to tweak the equations to produce that shift.

To do that, first consider the stone to be at a distance *H* from the uniformly accelerated observer,
who is momentarily at rest, then starts to move towards the stone (while always accelerating towards it). The
stone is reached by the observer after a time *t _{0} = sqrt(H^{2}/c^{2} +
2H/g)*. We'll offset the graph of

The "height"d = (c^{2}/g) (sqrt[1 + g^{2}(t − t_{0})^{2}/c^{2}] − 1) − H .

So the height of the stone above the observer at his proper timeT = (c/g) sh^{−1}(g(t − t_{0})/c) − (c/g) sh^{−1}(−gt_{0}/c) = (c/g) sh^{−1}(g(t − t_{0})/c) + (c/g) sh^{−1}(gt_{0}/c) .

whereh = −[(c^{2}/g) (sqrt[1 + g^{2}(t − t_{0})^{2}/c^{2}] − 1) − H]/γ ,

So to plott_{0}= sqrt(H^{2}/c^{2}+ 2H/g) , γ = sqrt(1 + g^{2}(t − t_{0})^{2}/c^{2}) .

then insert thisg(t − t_{0})/c = sh(gT/c - sh^{-1}(gt_{0}/c)) ,

You can show that this height *h(T)* reduces to the newtonian limit by supposing that (1) gravity is weak:
*gt _{0} /c ≪ 1*, and (2) we don't consider times too far outside the scenario:

Use the fact that shγ ≃ 1 + g^{2}(t − t_{0})^{2}/(2c^{2}) .

Now the point here is to plot *h* versus *T*, and compare it to the usual newtonian expression
for the stone's motion in space and time—which is of course a parabola. The figure shows a plot
of *h* versus *T* for *g =* 1 light year/year^{2}, *c =* 1 light year/year (of
course), and *H =* 10 light years.

Why this apparently anomalous acceleration and deceleration? Remember that the stone is not really
accelerating in an inertial frame; instead, the observer is accelerating in an inertial frame. As the
observer accelerates, his standard of simultaneity changes in a non-trivial way (see the FAQ article on
Do moving clocks *always* run slowly?). As he moves down past the
stone, his "line of simultaneity" coincides with the recent past of the stone's world line and mostly translates
through spacetime; then as he slows to a stop, his line of simultaneity begins to rotate in spacetime, sweeping
quickly along the stone's world line until it intersects that world line in his future. Then as he picks up
speed, his line of simultaneity mostly stops rotating in spacetime and simply translates again. The effect of
this is that he measures the stone initially to *accelerate* upwards, then slow for a time, stop, accelerate
back down, and then slow down shortly before it hits him. This motion might be non-intuitive, but most things
in relativity are non-intuitive!

We can find the values of the parameters that will make this non-intuitive behaviour appear: just demand that
the curve of *h* versus *T* be concave up at *T = 0*. So calculate the second
derivative *h''(T)* and demand it be positive at *T = 0*. The requirement for this to happen
turns out to be *g > (*sqrt*(2) − 1) c ^{2}/H*. This means that if you want to
perform such an experiment, either

This motion of objects is suggestive of current ideas in cosmology. Cosmology is built on the simplest
picture of why galaxies are observed to be rushing away from us: rather than assume they really are rushing away
from us, it is perhaps simpler to posit that they are all in a sense "at rest" in a universe whose very fabric
(spacetime) is itself expanding, as dictated by Einstein's equations of General Relativity. In other words,
if the galaxies are represented by raisins in a pudding, then we needn't imagine that the raisins are somehow
racing outwards through the pudding; instead, they can be "locally at rest" in the pudding, while the entire
pudding expands as it bakes. Recent observations of galactic recession have been interpreted as a sign that
the universe's rate of expansion is increasing, but the above description of free fall in which an object thrown up
can actually accelerate upwards demonstrates that what might at first be seen as an anomalous acceleration is
actually fully in keeping with relativity. The catch, of course, is that cosmology does *not* treat
galaxies as "objects thrown upward [or outward]". Even so, perhaps we don't need to invoke new ideas such as
dark energy to explain the universe's expansion: the above scenario shows that in certain situations, an object
thrown upwards can accelerate up. Our ideas of simultaneity are not always up to the job of finding these
behaviours to be intuitive.

*For a derivation of the rocket equations, see "Gravitation" by Misner, Thorne, and Wheeler, Section 6.2.*