From a traveler's point of view, what prevents him from reaching the speed faster than light?

Question

I have read the answer to this question from Eric. That answer is still from the point of view of a standing observer that sees the traveler spending infinite amount of energy. My question is asking for what happens from a traveller's point of view.

I know from standard traveller's point of view, the rest of the universe would finally reach the speed just below the speed of light moving away from him, that even though he keeps accelerating, he would never see himself moving faster than light relative to the rest of the universe. He would also see distances shrink to almost zero and time is slowing down to almost frozen, but none would eventually reach exactly zero and frozen, let alone be negative and moving backward in time. In short, please note that I don't at all suggest there would be two objects moving relative to each other faster than light.

However, also from his point of view, everything looks normal as according to him his mass is not increasing at all, he doesn't need infinite amount of energy to keep accelerating. The result of his acceleration (or energy spent) could not be just he is moving in a constant speed (or for nothing). For when if it does happen, we could not just maintain the contracted length would be just very close to zero, or time would be just very close to frozen, because e.g. the SR equation itself would not be valid. The only choice I could think of is to let length becomes negative and time is moving backward. So, he should see outside distances finally shrink to exactly zero and eventually points in front of him would become his behind. In the same manner, outside time should finally look frozen to him and eventually moving backward.

Note that he could still see light travelling at the speed exactly $c$, but the rest of the universe is starting to move closer to him (as points at his behind become his front and time is moving backward), thus neither principle of relativity nor relativistic speed limit is broken.

So, what prevents him from seeing all those phenomena happen?

Following, I would like to try to describe it in a more step-by-step manner to avoid confusion such I don't use / understand relativistic speed addition correctly :D)

1. Please note that I don't at all suggest that the traveler would see earth moving away from him faster than light.

2. Let we just stick to the point of view of the traveler. To him, as he is accelerating, he would continuosly see outside distances contract and outside time slows down.

3. Now he reaches the speed of $0.9999.... c$, he should see outside distances are very close to zero and outside time are almost frozen.

4. Of course to him, nothing to do with other frame of references, he calculates his speed reaches c -- again nothing to do with other frame of references, and continuously so, so according to his calculation which is simply $s = tg > c$, his speed should be faster than light. There should be nothing wrong with this.

5. In fact if he doesn't calculate it to $s = tg > c$, his energy would not be conserved. Because otherwise where does the energy he spent go if it doesn't accelerate his speed?

6. Now, he plugs in his states to SR equation and of course because he is using SR equation, he MUST still see light traveling at the speed exactly $c$, but then he will get the lenght and time become negative. It's weird obviously, but he still doesn't see earth moving away from him faster than light, so no violation here.

7. Instead speculatively we could guess that he would see earth now is coming to him while in the same time outside time is running backward to him.

8. Other than that, please note again that I don't at all suggest he is moving faster than light relative to others, so it doesn't violate relativistic speed addition, but just (relative) speed faster than light according to him, WITHOUT reference to any other frame of references at all.

Answer 1

If I'm understanding your question correctly, and after having read the comments, I'll just try to put my two cents in.

First, you're asking about something a traveler sees from the traveler's point of view. I take that to mean in the traveler's rest frame, because by "point of view" we usually mean this. In the traveler's rest frame, the traveler is always at rest. By definition.

What prevents him from seeing things traveling faster than the speed of light in a vacuum? You probably know these standard answers—something about the equations of relativity, or that you need infinite energy, or something like that. Those are correct. That is the answer.

Those equations are, as you've said, with respect to a "standing observer." That means "an observer at rest." And in the rest frame of a traveler, the traveler is `standing.' Therefore all of those equations apply to the traveler in his rest frame.

This is to say that, in the traveler's rest frame, (1) he is always at rest, and (2) everything else can move no faster than $c$, because it requires infinite energy to do so. Therefore there is no way for anything to go faster than $c$.

In someone else's rest frame, for example yours as you watch the traveler go by, you are at rest, and the traveler (along with everything else in the universe) cannot go faster than $c$. Therefore there is no way for anything to go faster than $c$.

Feel free to ask more questions in the comments; I think I see your misunderstanding, but I may not be explaining well.

Answer 2

Though I also learned the energy requirement argument from pop-sci source first, once I studied relativity I've come to feel that it skips past a more fundamental reasoning.

Let's define a minimal experiment with which to discuss the problem.

1. We start by assuming a heavily fueled and very efficient spacecraft in free-fall in an otherwise empty universe.

2. It does not make sense to discuss the "speed" of this thing as only relative speed has meaning and there is nothing to compare against. We fix that by gently setting a reference buoy out the air-lock to mark the original rest frame of the craft. All speed measurements will be made relative this reference.

3. We turn on the engine. For simplicity's sake I will assume it generate a constant acceleration of $3\,\mathrm{m/s^2}$ as measured by on-board accelerometers. A thousand seconds later (as measured by our on-board clocks) we will measure a relative speed of $3000 \,\mathrm{m/s} \approx 10^{-5} c$ (after suitable correction for the retardation of the signal) and a thousand seconds after that a relative speed of $6000\,\mathrm{m/s}$. All is as we expect.

4. Now we wait a while. At some point we notice that we are registering a relative speed of $2.7 \times 10^8\,\mathrm{m/s} \approx 0.9c$.

5. A thousand seconds after that we don't register a speed of $(2.7 \times 10^8 + 3000) \,\mathrm{m/s}$ but only $(2.7 \times 10^8 + 1900) \,\mathrm{m/s}$ because of the velocity composition formula $$ u = \frac{v + u'}{1 + \frac{vu'}{c^2}} \;.$$

6. The closer we get to $c$ relative that buoy, the less effect our acceleration has on our relative speed, and it is asymptotic; so we can never obtain a relative velocity at or above $c$.

We don't depend on any notion of how you accelerate, we just see that however you manage to accelerate no on-going acceleration will get you there.

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I've cheated a little here by using a finite time for calculating the speed increment. That way I was able to rig the numbers to be tractable within the precision of a hand calculator. To make the argument rigorous we would have to use infinitesimal speed increments (that is work it as a calculus problem).

Answer 3

2. Let we just stick to the point of view of the traveler. To him, as he is accelerating, he would continuosly see outside distances contract and outside time slows down.

The correct frame to represent the point of view of a uniformly accelerating traveler is called the Rindler frame. This frame does not behave as you suggest. As other things move faster they do show time dilation, however, more importantly there is in this frame a uniform "gravitational field" and an associated gravitational time dilation. This frame also has an "event horizon" called the Rindler horizon.

https://en.wikipedia.org/wiki/Rindler_coordinates

3. Now he reaches the speed of $0.9999.... c$, he should see outside distances are very close to zero and outside time are almost frozen.

His speed is always 0 in his point of view, by definition. The Rindler frame is different from standard inertial frames, but there is time dilation and length contraction. In particular, objects near the Rindler horizon will be nearly infinitely time dilated.

4. Of course to him, nothing to do with other frame of references, he calculates his speed reaches c -- again nothing to do with other frame of references, and continuously so, so according to his calculation which is simply $s = tg > c$, his speed should be faster than light. There should be nothing wrong with this.

Again, by definition his speed is always 0 in his own frame. The calculation $s=tg$ has nothing to do with his speed in his frame.

5. In fact if he doesn't calculate it to $s = tg > c$, his energy would not be conserved. Because otherwise where does the energy he spent go if it doesn't accelerate his speed?

The energy goes into accelerating the exhaust. This is always the case for a rocket at rest.

6. Now, he plugs in his states to SR equation and of course because he is using SR equation, he MUST still see light traveling at the speed exactly $c$, but then he will get the lenght and time become negative. It's weird obviously, but he still doesn't see earth moving away from him faster than light, so no violation here.

This is simply incorrect. As an object falls under the influence of the "gravitational field" they will be progressively time dilated both due to their increasing velocity and also due to their increasing depth in the "gravitational field". Rather than having anything become negative, the object will simply cross the Rindler horizon.

7. Instead speculatively we could guess that he would see earth now is coming to him while in the same time outside time is running backward to him.

There is no need to speculate. This is standard material and is well understood.

However, also from his point of view, everything looks normal as according to him his mass is not increasing at all, he doesn't need infinite amount of energy to keep accelerating. The result of his acceleration (or energy spent) could not be just he is moving in a constant speed (or for nothing).

In fact, from his point of view, by definition, his velocity is a constant speed of 0. He does not accelerate. The energy spent by the engine goes entirely into accelerating the rocket exhaust. In his point of view, the rocket's force does not produce any acceleration and simply is used to counteract the "gravitational acceleration". Thus, he experiences proper acceleration without changing speed.

Answer 4

I think I know what you mean. You are asking "how does the traveler account for the fact that she is unable to reach or surpass the speed of light relative to the rest of the Universe, in terms of what she is observing happening in her rest frame?"

Qualitatively, you have the right idea regarding that as she accelerates, the "size" of objects in the rest frame will be compressed more and more in the direction of travel while they continue to pass at speeds near the speed of light, but you're then asking "why then don't they just shrink to zero?"

And the answer to that really is that while qualitatively correct, it is not quantitatively so, and we have to calibrate and sharpen this intuitive notion with mathematical rules. The precise mathematical description of the scale of the effects you describe is afforded by a quantity known as the rapidity, which is given by:

$$\phi(v) := \tanh^{-1}(v)$$

where we work in natural units ($c = 1$, so $v$ is the speed as a fraction of the speed of light, from $0$ to $1$). Here $v$ is the speed of the traveler seen by the ground frame. What rapidity represents in the traveler's rest frame is basically this: if we imagine posts (or perhaps, asteroids, to be more realistic) to have been studded all along her route, at rest in the "ground frame", each set and marked with regular distances (say 1 Gm) between them - think about a cosmic version of the mile/kilometer markers by the side of the road - then the number of posts which pass her, which is determined by a conspiracy between the length contraction and the increase in apparent speed, is proportional to this rapidity. Since $c$ is about $0.3\ \mathrm{Gm/s}$, then at a rapidity of $1$, the 1 Gm posts pass at about 3-second intervals, and at a rapidity of 2, they pass at 1.5-second intervals, and so forth.

Now, there are two relevant features we need here. The first one is that a ground speed of $c$ corresponds to an infinite rapidity ($\phi(1) = +\infty$): the inverse hyperbolic tangent has a singularity there. This agrees with what you said that at the speed of light distances should shrink to zero - in effect, the traveler should be everywhere along the route at once and hence passing the gigameter-asteroids "infinitely fast".

But the one that answers your question is this in combination with the second: when you "accelerate at a constant rate" (e.g. "one gee", $9.8\ \mathrm{(m/s)/s}$ or, in perhaps better units for our purposes, $9.8\ \mathrm{(Gm/s)/Gs}$), what actually increases at this "constant rate" is your rapidity, not your ground velocity.

Hence, the traveler will see the gigameter-asteroids passing them as though they were accelerating steadily in a Newtonian-like fashion, even seeming to surpass $c$, but they will attribute it to a different set of physical effects: instead of a relative velocity between them and the asteroids that is greater than $c$, in line with what is mentioned earlier they will treat this as resulting at first from increasing relative speed (in the early part of the acceleration when "ground speed" is still well below $c$ - i.e. before the first 0.03 Gs of travel in the one-gee scenario), and then as their ground speed approaches $c$, they will treat further gains as increasingly resulting from the escalating length contraction of the distances between the asteroids, so they are steadily becoming much shorter than the marked 1 Gm, but still not ever actually reaching zero any more than the traveler's ground speed reaches $c$.

But here's another aspect to look at, too: suppose that, say, at the beginning of her trip, someone sends out a photon in the traveler's direction of accelerating motion. This photon represents the "true" speed of light, $c$, in every frame. No matter how much she accelerates, the traveler will never be able to "catch" this or even gain on it, because she will consider it as always receding at $c$ relative to her, and thus in effect at greater and greater "gigameter-asteroid speeds". Much like the story of Tantalus, from whence derives our word "tantalize", being lured by water that would always stand off as he tried to approach it, the speed of light always stays "forever in the distance", no matter how much you try to approach. Thus, if it cannot be even reached, it surely also cannot be surpassed.

ADD: apparently this isn't quite the right profile, it's not literally a Newtonian profile in the rapidity, but the point is the increasing apparent "speedup" past $c$, and how it occurs, without becoming infinite.

Answer 5

Note that special relativity is all about flat spacetime, so bringing the rest of the universe into these types of problems can lead to a lack of clarity.

Regarding your inertial observer who has accelerated to $c-\epsilon$ for some time, the magic of special relativity is that he is still at rest. Locally, his spacetime looks exactly the same as before he started: he feels neither Lorentz contracted nor time dilated, and light propagates at $c$ in all directions. As far as he is concerned, he is a long way from $c$. Moreover, if he repeated the process and accelerated to $c-\epsilon^2$, nothing would change.

You argument that $v = at$ is just not applicable. This is testable. Next time you are at the LHC, and the proton beam is at full energy ($E=7$ TeV, $\gamma = E/M_p = 7462$), the protons are moving at:

$$ v \approx 1 -\frac 1 {2\gamma^2} = 0.999999991022c$$

So all you have to do is hop on a bike and accelerate to a few meters per second up the beam line and the protons would be superluminal by your methods: they are not.