# Endless acceleration

1. Mar 25, 2009

### neopolitan

I was going to post on another thread that a spaceship could theoretically accelerate forever without reaching the speed of light (let's say this is relative to the spaceport where it began its journey).

If I am at the spaceport, I can certainly agree. I see the spaceship getting closer and closer to the speed of light, but never reaching it. But I would be seeing the acceleration slowing down.

If I am in the spaceship, what happens? It the spaceship is accelerating, I will be able tell that, from the way things move if I let go of them. I know I am not in an inertial frame.

If I read what time has elapsed on my clock, then multiply that by the acceration that I can calculate by putting a known mass on a set of scales, then I should be able to calculate my change in velocity.

If I accelerate forever, I would have on my clock ($$\Delta t$$=infinity), multiplied by the non-zero acceleration (a) and have a velocity of infinity.

This can't work.

What I think the resolution might be, is this:

According to the spaceport, the spaceship may accelerate forever, but at a hyperbolically reducing rate. I could calculate that the acceleration would seem constant according to the spaceship.

According someone on the spaceship, there would not be enough time to accelerate to the speed of light. For example, if the spaceship accelerated at a constant 10m/s2 to make a nice simulation of Earth gravity, then it would reach the final moment of the universe after 22 years (plus a bit).

I'm not totally happy with it.

Please note, I know there are issues with accelerating something forever, we have to apply some "mind experiment magic" to give us the ability to do so. Once the engineering issues are out of the way though, is my resolution correct, or does something else happen?

cheers,

neopolitan

2. Mar 26, 2009

### Fredrik

Staff Emeritus
You're right. Constant proper acceleration means that at any point on the world line, the coordinate acceleration in the co-moving inertial frame is the same as at any other point on the world line. (Note that the co-moving inertial frame at another point on the world line is another frame). The world line calculated from this condition is a hyperbola, and its slope approaches 1 but never reaches it. DrGreg posted a nice derivation of this in a thread I started that has the word "acceleration" in the title, so you can search for that if you're interested.

Last edited: Mar 26, 2009
3. Mar 26, 2009

### DrGreg

This is the thread, specifically posts #13,14,15 (and a correction in post #28). But, to understand that post, you will need some knowledge of hyperbolic functions and of calculus.

If you multiply constant proper acceleration by proper time ("proper" means you are measuring yourself, roughly speaking) then you get, not velocity, but something called rapidity, $\phi = \tanh^{-1}(v/c)$. Note that there are no limits to rapidity and the rapidity of light is infinite. Rapidity has the useful property that you can just add two rapidities together, unlike velocities. It is this property which associates it with proper acceleration and proper time. However, it does have one serious shortcoming -- it is a scalar quantity, not a vector, so it is of use only in cases of one spatial dimension, i.e. where everything under consideration is moving along the same straight line.

Rapidity can also (equivalently) be defined as the natural logarithm of the doppler factor between inertial observer and measured object.

In neopolitan's example, by the equivalence principle, the spaceport is falling "under gravity" away from the spaceship, and as it approaches a horizon at a distance of $c^2/a$ under the spaceship, "gravitational" time dilation takes effect and it never actually reaches the horizon. So, from the spaceship's point of view, the relative velocity of the spaceport does not approach the speed of light at all, far from it, after initially increasing it eventually decreases to zero! You can explain this by saying that eventually the rate of relativistic length contraction almost exactly matches the rate of acceleration, the two cancel each other out giving a net relative acceleration approaching zero. This also shows that accelerating frames don't behave much like inertial frames.

This is true.

4. Mar 26, 2009

### neopolitan

Was my a little over 22 years correct? (or more accurately a little under 23)

It was calculated with the assumption that someone on the spaceship would never calculate that the speed of light was reached, but just under would be acceptable (ie the speed of light would be asymptotically approached). 30 billion seconds is about 22.8 years.

I am wondering if the solution here is too simplistic, or is it actually a good rough approximation?

cheers,

neopolitan

5. Mar 27, 2009

### neopolitan

Does no one have any idea?

6. Mar 27, 2009

### DrGreg

I don't really understand what you think you have done here. From the point of view of the spaceship, the local speed of light is always 299792458 m/s relative to itself, so its progress towards that target is always 0% and so it never gets even close to it.

From the point of view of the spaceport, if the ship accelerated at a constant 9.81 m/s2 relative to the spaceport's frame, it would take just under a year (299792458/9.81 seconds) of spaceport time to reach the speed of light, if that were possible. Of course it isn't possible, because to maintain that acceleration the ship's proper acceleration (i.e. as measured by its own on-board accelerometer) would have to increase, ultimately to infinity and beyond.

I don't understand where your 22.8 years came from. I think you might have a decimal point in the wrong place, or you multiplied where you should have divided.

7. Mar 27, 2009

### neopolitan

If the space ship has an acceleration of 10 m/s2, which is detectable as a pseudogravity, for a certain amount of time, then someone in the ship might be able to calculate the speed they should have reached. It's inside the ship, so the acceleration has been constant.

My thought was that there may be some sort of "relativistic censorship" which effectively prevents anyone in the ship from calculating that the change in speed was c.

The calculation was quite simple, c/g = 3x107s. It seems that I did screw up, which shows why you should not post at work. I left out a 24.

3x107s / 3600 (s/hr) / 24 (hr/day) / 365.24 (day/yr) = 0.95 year.

I'm even less happy with the figure but what I think might happen is that a ship undergoing constant proper acceleration will experience increased slowing of clocks such that the captain of the ship can't accelerate for longer than 0.95 years at 10m/s without shipboard time slowing down so much that the lifetime of the universe passes.

Some form of element analysis or calculus might be able to show that no matter what the ship's relative starting speed was for an external observer, the calculated time passed by on board will be 0.95 years, if that ship accelerated "forever" at g. Or it might be a different number.

All it would be saying is that, from an external viewpoint, clocks on the ship slow down asymptotically and the proper time reading on the shipboard clock as time dilation approaches infinity will be dependent on the proper acceleration.

It's like the geometric series 1+0.5+0.25+0.125+0.0625 ..... while some might think if you go on forever adding more and more, it must add up to infinity, but it doesn't. It adds up to 2.

Similarly, if the clock on the ship slows down and slows down, it will approach a time past which it will not tick (or alternatively, the observers at the spaceport would have to wait infinitely long to see it tick which is the same as saying it never does).

I'd appreciate being told I'm wrong by those who have made a proper effort to understand what I am trying to understand here and can explain where I have erred.

cheers,

neopolitan

8. Mar 27, 2009

### darkhorror

First off all you can't really "use" infinity as a number, thats what limits are for. But that really isn't the problem you are having there, you seem to be wondering why you can always accelerate at 1g never go the speed of light.

First off for the calculations you are using you use the ship accelerating at 1g with respect to an outside observer. The problem with this is that the ship would not experience 1g acceleration, but would be experiencing more and more acceleration. Now if you were to keep constant 1g acceleration on the ship. To the outside observer you would not appear to have 1g acceleration the faster you go but to them it would slow down.

Also lets say you have something 100 light years away, and you leave for it on a ship and you constantly accelerate at 1g with respect to the ship. It will take you much less than 100 years to reach that place.

9. Mar 27, 2009

### DrGreg

neopolitan,

Your calculation is flawed because it is based on the false assumption that

(proper acceleration) x (proper time) = velocity ​

As I explained in post #3 the answer you get is rapidity, not velocity. You have calculated how long it would take (in ship time) to reach a rapidity numerically equal to c. But the rapidity of light is infinite, so it would take an infinite time to reach light rapidity.

If you approximate the journey by a sequence of one-second segments each at constant velocity:

After one sec, your velocity is 10 m/s (relative to the spaceport).

After two secs, your velocity (relative to the spaceport) is not 20 m/s, but (10 + 10) / (1 + 100/c2) = 19.99999999999997. But your rapidity is 20 m/s (relative to the spaceport).

After 3 secs, your velocity is (19.99999999999997 + 10) / (1 + 199.9999999999997/c2) = 29.99999999999991. But your rapidity is 30 m/s.

And so on.

After a long time, your velocity will add up (a bit like a geometric series) to something approaching c but your rapidity will just keep on increasing without limit.

There is no finite number for the answer you were looking for -- it will take the spaceship an infinite time on its own clock $\tau$ to reach infinite time on the spaceport's clock $t$.

The exact formulas, when you take the calculus limit as the segment duration drops to zero, are:

$$t = \frac{c} {2g} \left( e^{g\tau/c} - e^{-g\tau/c} \right)$$
$$x = \frac{c^2}{2g} \left( e^{g\tau/c} + e^{-g\tau/c} - 2 \right)$$
$$v = c \frac {e^{g\tau/c} - e^{-g\tau/c}} {e^{g\tau/c} + e^{-g\tau/c}}$$
$$\mbox{rapidity} = g\tau \mbox{ (in velocity units)}$$​

(all of the values on the left hand side being measured in the spaceport's inertial frame and all the values on the right hand side measured in the ship's accelerating frame).

10. Mar 27, 2009

### neopolitan

Ok, so if I have understood it right (including what I read on http://en.wikipedia.org/wiki/Rapidity" [Broken]), all proper accelerations divided by proper time give you rapidity which where v << c is close enough to v for most circumstances.

I think your equation for working out the relative velocity between the spaceship and the spaceport was the one shown http://en.wikipedia.org/wiki/Relativistic_addition_of_velocities_formula" [Broken] where v is the previously calculated velocity and u is the rapidity over the last period.

Furthermore, if I have understood it right, if the spaceship starts in empty space, with no reference (so there is no relative velocity), then the only measure the captain could sensibly use is rapidity which can rise to higher than 3x108 without any problems. Or is there a way, absent any reference, that the captain could sensibly calculate a subluminal velocity?

(I suppose that he could refer to his departure point but that seems a little like considering space as being a fabric. Relative to the ship, the spaceport can be seen to be moving away, but the same is not the case with a featureless departure point.)

Last edited by a moderator: May 4, 2017