# Acceleration up to lightspeed (From the inside)

1. Sep 12, 2008

### Korgeghi

Hi, I have a question that has been bugging me for ages now. It concerns acceleration up to the speed of light. I know that if an attempt is made to accelerate a spaceship/particle up to the speed of light from the outside, it is not possible to get it to the speed of light because the energy requirement to accelerate the spaceship/particle gets exponentially larger as the speed of light is approached, and tends to infinity at the speed of light.

However, what if you are propelling the spaceship/particle from the inside instead? For example, what if you are inside a spaceship that has its own propulsion system, and is accelerating towards the speed of light? You would experience that the spaceship remains at its rest mass energy, as you are in the inertial frame of the spaceship at all times. Therefore there should be no exponential increase of energy required to accelerate the spaceship as it approaches the speed of light, right? Or am I missing something here? I figure I must be missing something because otherwise it appears to me that a craft can accelerate up to the speed of light under its own propulsion, which surely is impossible?

2. Sep 12, 2008

### JesseM

If your ship is experiencing constant proper acceleration--constant G-force as experienced on the ship--then from the perspective of a given inertial frame, your rate of coordinate acceleration would be continually decreasing as you approached c, so you'd never reach it. In order for your coordinate acceleration in an inertial frame to be constant, your proper acceleration would have to be growing without bound, reaching infinity at the moment you reach c in the inertial frame (which is impossible, of course).

For some formulas relating to constant proper acceleration, check out the "relativistic rocket" page here:

http://www.math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

3. Sep 13, 2008

### Korgeghi

I still have some questions to help me clarify this situation. OK, am I right in saying that the coordinate acceleration of the spaceship is the acceleration as observed by someone in an inertial frame outside the ship (as the ship's frame of reference is an accelerating one)? If so, then I can understand that indeed that observer could never see the ship reach c, as from their point of view, all the ship's processes are slowing down due to time dilation as it accelerates, thus slowing the rate of coordinate acceleration.

However, what would an observer in the ship see? If this observer were to keep track of the ship's velocity with respect to an inertial frame (such as Earth's, assuming it is in a constant 'stationary' inertial frame and ignoring centripetal accelerations), what would that velocity be measured as while the ship accelerates with a constant proper acceleration? For an observer inside the ship, the ship's processes are not slowing. Would the ship not appear to be travelling faster and faster, with its constant thrust, or does that thrust lose its effectiveness at relativistic speeds? If so, why, if the spaceship's mass energy is not increasing with respect to its internal passengers?

4. Sep 13, 2008

### JesseM

Yes, that's right.
Time dilation has something to do with it, but it's not the only thing that's relevant. Constant proper acceleration means that at any given moment, if you take the inertial reference frame where the accelerating ship is momentarily at rest (the 'comoving frame'), and look at the amount the coordinate velocity increases in that frame for a brief period of time, then if you later switch to the new comoving inertial frame where the ship is momentarily at rest and repeat the process, the amount the coordinate velocity increases for a brief time-interval in the new comoving frame will be the same as it was in the previous comoving frame.

We can imagine the ship's acceleration as not being totally continuous, but as being a series of jumps in velocity such that the velocity increase is always the same in the comoving frame before the jump, and the time between jumps is always the same in terms of clocks on board the ship (this discontinous acceleration can be seen as an approximation, since in the limit as the time between jumps goes to zero this will get arbitrarily close to looking like perfectly smooth acceleration). So time dilation is relevant in the sense that an observer in some fixed inertial frame will see the time between jumps getting longer and longer since the ship's clock is ticking slower and slower as its speed increases. But the observer in the fixed frame will also see the coordinate velocity increase in each jump get smaller and smaller because of the way relativistic velocity addition works--if I'm moving at some velocity u relative to you, and I shoot out a rocket which goes at velocity v in my frame (or I make an increase in velocity such that my own velocity increases by v in the inertial frame where I was previously at rest before the increase), then instead of your seeing the rocket (or my ship) now moving at (u + v) as you'd expect in Newtonian physics, in relativity you'll only see it moving at (u + v)/(1 + uv/c^2). For any fixed increase in velocity v, the closer u already is to c, the smaller the difference will be between u and (u + v)/(1 + uv/c^2). And the fact that relativistic velocity addition works this way is not solely a matter of time dilation, it has to do with the fact that each frame measures the velocity of objects in terms of distance/time according to rulers with synchronized clocks placed at various points along them (so if I measure you to pass by the 10 light-second mark on my ruler when the clock there reads 0 seconds, and later you pass by the 18 light-second mark on my ruler when the clock there reads 10 seconds, then I define you to be moving 8 l.s./10 s = 0.8c in my frame). So, not just time dilation but also length contraction and the relativity of simultaneity (the fact that clocks which are synchronized in one inertial frame appear out-of-sync in another) come into play.
By "see" do you mean what the ship would see visually, or do you mean what the coordinate velocity of outside objects would be in the ship's instantaneous comoving inertial rest frame from one moment to another, or do you mean what the coordinate velocity of outside objects would be in a single accelerating reference frame where the ship was at rest? Remember that only in inertial frames are massive objects restricted to move at less than c (and only in inertial frames does light itself always move at c), in non-inertial frames there is no such restriction, and there is no single "correct" way to construct a coordinate system where a non-inertial object is at rest, so it doesn't really make sense to talk about "the" non-inertial rest frame of an accelerating observer since there are a lot of different coordinate systems where that observer would be at rest.

5. Sep 13, 2008

### ZikZak

While what you say about energy is true, it is not the fundamental reason that you can't accelerate a body to c, and thinking this way (that the inability to get to c has something to do with energy) really prevents you from appreciating the root character of Relativity, which is geometry.

The reason bodies cannot accelerate to c is not dynamic. It's not about energy. The reason massive bodies cannot reach c in essence is very simple: c is an invariant speed. Suppose someone shines a light beam past you, and you start running towards it in an attempt to catch up with it. No matter how fast you run, not only can you never catch up with the light ray, you never make even the least amount of headway. No matter how fast you run, the light beam continues to recede from you at c. Running at 0.99c (relative to the road) is exactly as far from the speed of light as being at rest. If you are slower than the light ray in your frame, you are slower than the light ray in every frame. Thus you can never reach c in any frame. No mention of energy is required to derive this, so it must not be an energy issue.

At slow speeds, it's easy to talk about the body's "velocity." An observer on the side of the road can (1) time how long it takes the rocket to get from one mile post to the next. (2) Or the rocket pilot can time how long it takes to get from one mile post to the next. (3) Or the roadside observer can observe the time between when the front of the rocket (of pre-measured rest length) reaches a milepost and when the back of it gets there. (4) Or the rocket pilot can integrate his acceleration with respect to time. In Newtonian physics, all these definitions of velocity are the same and it doesn't matter which one you use. But in Relativity, they are not. We call definition (1) the "coordinate velocity." This is the velocity that we say cannot reach c. But definitions (2) and (3) are equivalent and called the "celerity," and these increase without bound. Definition (4) is called the "rapidity" and also increases without bound.

So really, the reason a body can't reach c is that when getting close to c, we use a dumb definition of "velocity" to describe it. It turns out that acceleration is change in rapidity with time, NOT the change of velocity. At low speeds, this makes no difference (because the velocity, celerity, and rapidity are all the same), but at nearly c, it makes all the difference. As Jesse says, you cannot add velocities together. So since total velocity is not v+dv, then acceleration cannot be dv/dt. It's only rapidities that add properly. As the rocket accelerates, the rapidity of the rocket happily increases uniformly, at the same time as the velocity slows down its rate of increase, never reaching c.

So as the rocket accelerates, it can maintain the same feeling of acceleration indefinitely, as its rapidity increases without bound. So the rocket pilot always feels the same acceleration. His change of velocity slows down, so you might think he feels less acceleration, but proper acceleration is change of rapidity, not change of velocity.

What the pilot "sees" depends on your definition of "sees." But the easiest thing for the pilot to "see" is his celerity. If he times the mileposts as they pass by, they pass faster and faster and faster as he accelerates, without any limit. Although no single milepost has a coordinate velocity of more than c, the distances between them length-contract. To the rocket pilot, the distance between them gets very short, and is no longer a mile. In other words, as the rocket accelerates, the mileposts at first simply appear to travel faster and faster, but as they approach coordinate velocity c, they length-contract together instead of getting faster. As a result, he can pass mileposts at an arbitrarily rapid rate, even though their velocity is never more than c.

Last edited: Sep 13, 2008
6. Sep 13, 2008

### MeJennifer

Nice, and refreshing, explanation ZikZak!

7. Sep 13, 2008

### Korgeghi

Yes, c is an invariant speed. But only in an inertial frame of reference. Not in an accelerating frame. Also, the theory implies that even if you run at c, you will be as far from the speed of light as being at rest. So comparing your speed to c does not tell you if you have reached c. In fact, I would suggest that this planet could be travelling at c relative to something in this universe right now. Is there any reason why not?

It seems to me that what you are saying is that from the pilot's point of view, he can reach the speed of light. But from an external observer's point of view, he never can. So there is nothing to stop a spacecraft from accelerating up to the speed of light, as far as I can tell from what you have said. From the pilot's POV, that is. Which is the POV I am interested in.

Last edited: Sep 13, 2008
8. Sep 13, 2008

### Korgeghi

I mean if you were to measure how quickly the Earth was receding from the spaceship as it accelerates. I'm guessing it's not symmetric to viewing the accelerating spaceship from the Earth, as the Earth remains in the same inertial frame whereas the spaceship is constantly changing its inertial frame. Initially the Earth would appear to accelerate away from the ship at the proper acceleration of the spaceship. But as the spaceship approaches the speed of light, what does the Earth appear to do then?

9. Sep 13, 2008

### Staff: Mentor

I think you may have missed this line in that quote:
Put another way, it depends on how the pilot chooses to measure his speed. Using mileposts is only one way and is not necessarily a good way because it requires mixing frames of reference (remember, there is no absolute notion of speed. It is only meaningful to measure your speed with respect to a specific reference point). Another way would be to fire laser pulses behind him at the earth to measure the change in distance with time. If he measures his speed that way, he'll find it never exceeds C.

10. Sep 13, 2008

### JesseM

I think you are implicitly using energy here, because you are assuming that when you try to run to catch up with light, you cannot run with infinite proper acceleration. Certainly we can draw curves through spacetime that have the property that, if any actual object was imagined to follow that curve, the object would start out slower than light but pass light speed in some time (as viewed from the perspective of some fixed inertial frame). But such a curve cannot describe the path of any actual object, because the energy required to accelerate it along that curve would go to infinity in a finite time, as would the proper acceleration.

11. Sep 13, 2008

### JesseM

As long as you look at the instantaneous comoving inertial rest frame of the pilot, he will always observe objects that are at rest in the Earth's frame to be moving past him at some velocity less than c in his instantaneous rest frame. Also, if the Earth was firing a steady stream of photons at the pilot, the pilot will always find that a photon which is passing next to him at some point on the trip will reach the destination before he does. What the pilot can do is reach destinations that were X light-years away in the Earth's frame in significantly less than X years; there's a table of times to reach various destinations on the relativistic rocket page. Remember, though, that once the pilot has reached a significant fraction of light speed in the frame of Earth and the destination, then in the pilot's instantaneous inertial rest frame the distance between Earth and the destination will be shrunk to significantly less than X light-years thanks to Lorentz contraction.

12. Sep 13, 2008

### ZikZak

That is true, but for accelerations that the pilot is likely to survive, the difference is negligible.

It depends what you mean by "travelling." If we're just talking about locally, such that the expansion of the universe is not a factor, then yes, there is a reason why not: the planet cannot catch up to a light ray fired in its direction of motion that for every observer travels at c. Thus the planet travels at less than c for every observer.

Both the pilot and the "rest" observer will observe the same larger-than-c celerity for the other. But celerity is a "mixed" velocity where the distance and time that go into measuring it are taken from different frames. Whenever anyone measures a velocity, for which both measurements are from the same frame, the resulting coordinate velocity is always less than c.

13. Sep 13, 2008

### ZikZak

I think that is a reasonable assumption, although internally I was merely thinking of a succession of inertial frames.

I can also draw a curve on a simple x vs. t diagram that has the property that it shows a 1968 VW beetle traveling at supersonic speeds. That doesn't mean that a VW beetle can travel supersonically. If there is a reason bodies cannot travel at c, it must be a physical reason. I don't quite understand what you're getting at by pointing out that there are curves in a spacetime diagram that do not represent the actual motions of particles.

14. Sep 13, 2008

### JesseM

My point is basically that if we make a distinction between kinematics (which only deals with measurements of motion, the fact that acceleration is the derivative of velocity and so forth, but not with issues like energy and forces) and dynamics (which deals with the full laws of physics governing motion), then the fact that an object cannot accelerate past light speed (and the fact that a VW beetle won't be able to accelerate to supersonic speeds under its own power) can only be understood in terms of the dynamical laws, you can't show this without consideration of energy and forces and such. You seemed to be saying otherwise when you said "The reason bodies cannot accelerate to c is not dynamic. It's not about energy. The reason massive bodies cannot reach c in essence is very simple: c is an invariant speed." I'd say we could perfectly well imagine a hypothetical universe with different dynamical laws, where the geometry of spacetime is still the same in the sense that things moving at c in one inertial frame move at c in all inertial frames, and velocity addition between inertial frames works the same way, and the proper time along worldlines is calculated in the same way for timelike worldlines, but the laws of physics do allow objects to follow worldlines that make a transition from timelike to spacelike, accelerating past c in some finite time. So, the reason this is impossible in the real universe must have to do with the specific form of the dynamical laws.

15. Sep 15, 2008

### ZikZak

I disagree. I think you would find that "accelerating past c in some finite time" is mathematically inconsistent with the velocity addition rule, which is a direct (kinematic) consequence of invariance of c.

I'll concede that there is one dynamic assumption in my argument, and that is that velocities must be continuous. If we allow your universe to contain particles to "jump" velocities discontinuously to speeds greater than c, then you might have a point. Of course, at that point, your particles can violate causality. So I think that I can maintain my argument on kinematic + causality grounds without resorting to dynamics.

16. Sep 15, 2008

### JesseM

What do you mean by "mathematically inconsistent with the velocity addition rule", exactly? Inconsistent how? Suppose that in some inertial frame, the instantaneous coordinate velocity as a function of coordinate time is v(t) = (0.5 light years/year^2)*t. So in this frame, the object will reach v=c at t=2 years. Now transform into any other inertial frame. For any event on the worldline of the object which was before t=2 years in the first frame, the velocity was less than c in the first frame, and the velocity will be less than c in the new frame too. For the event on the worldline of the object at exactly t=2 years in the first frame, the velocity was c in the first frame, and it will be c in the new frame too. For any event on the worldline of the object which happened later than t=2 years in the first frame, the velocity is greater than c in the first frame, and the velocity will be greater in the new frame. Do you disagree? If not, how is the path any more problematic in the second frame than it was in the original frame?

Of course, if you calculate the proper acceleration as a function of coordinate time in any frame, it approaches infinity in the limit as you approach the time when the velocity approaches c. But without considerations of the dynamical laws I don't see why this should be viewed as impossible.
I see no need for any discontinuous jumps in velocities in any frame. If you take events closer and closer to the event where the instantaneous velocity is c in one frame, the velocity gets arbitrarily close to c, and this will be true in other frames too.

17. Sep 16, 2008

### DrGreg

JesseM and ZikZak, I'm curious to see where your argument is going, so let me toss in a specific example and see where it takes each of you.

Oops! After writing this, I've just re-read JesseM's post #16 in more detail and realised my example below is the same as his! But I'll leave it as I wrote it...

In flat Minkowski spacetime, consider the curve whose equation is x = 1/2at2. Mathematically, there's nothing wrong with that curve, nor physically for shorter times. The problem is that the curve is timelike for t < c/a, null for t = c/a, and spacelike for t > c/a, so we know it can't be the worldline of a particle in our real universe (for larger values of t). The proper acceleration $\gamma^3\,a$ diverges to infinity as $t \rightarrow c/a$ and is undefined beyond that point.

The curve is continuous regarding x as a function of coordinate time t. However, regarded as a function of proper time $\tau$ there is a singularity at t = c/a.

Does this example help either of you to make your point?

18. Sep 17, 2008

### ZikZak

First, I want to point out that I don't want to be overly dogmatic about the speed limit as being purely kinematic. Jesse, if you're willing to concede that all finite acceleration cases can be treated purely kinematically, then I'm pretty OK with just that. My main concern is the way that Relativity tends to be taught in schools as a purely dynamic affair totally immersed in masses and energies and E=mc^2's, when such things are about the last thing one should be getting to in a course on SR. The result is usually a student's very semi-Newtonian semi-understanding of what is really a beautiful and elegant and, at root level, understandable theory.

I think that when most people have the idea that energy is the factor that prevents travel through c, they are basically thinking in terms of firing a rocket at constant thrust, or a finite constant proper acceleration. If you are willing to allow me to treat those cases as being entirely kinematic, then I am happy.

With that said, I think there are several potential responses that I could make to the example of a particle with coordinate velocity $v(t) = 1/2\, a t^2$, although I haven't had much time to seriously think about it. First, I think that this is one of a category of unphysical worldlines that one can draw on a spacetime diagram, including the worldline corresponding to $v(t) = \left\{ \stackrel{1/2\, a t^2,\;\;\;t<c/a}{c,\;\;\;\;\;\;t\geq c/a} \right.$. I think that we all can agree that this second worldline is unphysical kinematically since it places the particle in an inertial frame where v=c, violating the second postulate of relativity. So it is demonstrated that at least some worldlines drawable on a spacetime diagram are indeed exludable on purely kinematic grounds.

(and as an aside, it seems to me that if he's not allowed to stop there, kinematically, he probably shouldn't even be allowed to visit.)

The only remaining case is that of infinite proper acceleration through v=c. That's a tough one, and as I said, I'm more or less willing to let you have it. I view a restriction on infinite proper acceleration as being unphysical on its face. This is sort of "unstoppable force hits an unmovable object" territory. But you're probably right... that intuition is probably dynamical in nature. So I would say in this case that you can kinematically derive that for an accelerating observer, an event horizon forms at a distance 1/a behind him. At a=Infinity, he is inside this horizon, and he cannot, even in kinematic principle, construct any local Lorentz frame in which to do any physics at all, and thus he is torn apart by the acceleration.

19. Nov 24, 2008

### Ich

That's a very interesting question.
Let me say that I fully agree with ZikZak, especially with what he says in the first paragraph of #18. But JesseM certainly has a point when he says that there is dynamics involved. Maybe we can reconcile both viewpoints:
When someone asks "why is it impossible to reach light speed", he/she means normally "... as opposed to arbitrary (but finite) speeds in the Newtonian framework". I'd say that in most cases the question is definitely not why infinite speeds can't be reached in Newton's theory.
That is the only point where dynamics is needed, to explain that speeds are finite rather than infinite, with inertia as a reason (or as a consequence, whatever). But that concerns equally Newtonian physics and SR. The new thing that is asked about is "why are not arbitrary (finite) speeds allowed, why is there a speed limit?". And this new feature of SR is definitely a kinematical one, and well described the way ZiKZak does. It's rather a mapping of all conceivable speeds (with rapidity as the natural measure) to the region +-c, a geometric property of spacetime. You can't reach c because it "plays the part of an infinitely great velocity", as Einstein put it.