6
$\begingroup$

A question has recently come up that goes beyond my knowledge of special relativity. Suppose a pilot has his foot on the gas pedal of a rocket ship and keeps it applied to achieve a constant acceleration, and he has a magical engine that can maintain the thrust needed to accomplish such acceleration even as the vehicle acquires mass. He measures his acceleration at periodic intervals and sums it up as he goes.

Observers perceive the rocket to increase in speed and gradually approach the speed of light (but never reach it). Two possibilities are being debated for what the pilot observes, and my preference is for the first:

  1. Time dilation gives him the impression that the summation of his readings are approaching light speed at a constant rate, but there is an asymptote involved here even in the inertial frame such that he never actually perceives himself to reach, let alone exceed, $c$, because time slows down to a crawl (as seen from the reference frame) before this can happen.
  2. Although to observers he is only approaching $c$, it is possible for the summation of the measurements on his accelerometer to indicate that he has surpassed $c$ and is now traveling faster than light. From the reference frame he is approaching $c$ and his instruments in the inertial frame tell him that he has surpassed it.

I assume (1) is true beyond a doubt? If so, is there a straightforward explanation for this?

$\endgroup$
  • 1
    $\begingroup$ Hint: don't jump from one reference frame to another unexpectedly. First select most simple reference frame, then calculate task it it, then transform to another reference frame correctly. The fact is: in principle, pilot can reach Alpha Centauri for one second of personal time. This means, that his impression can be that he moved at a speed of 4 ly per second -- much faster than light. But this is not what is regarded as physical speed. $\endgroup$ – Suzan Cioc Nov 19 '13 at 7:58
  • 1
    $\begingroup$ But the pilot measures the distance moved as less than 4 lyrs, hence they don't measure their velocity as 4 lyrs/sec. $\endgroup$ – John Rennie Nov 19 '13 at 8:00
  • $\begingroup$ My text is self-containing. I don't claim my measure is correct, I just describe how I did it. Moreover, I can predict, that my form of measure will prevail in future, when interstellar travels arise. Because it is natural to measure distance relative to always "stationary" ground. Later, the measure relative to microwave background will arise (when intergalactic travel) :) Both will show superluminal speeds sometimes. $\endgroup$ – Suzan Cioc Nov 19 '13 at 8:40
  • 2
    $\begingroup$ I think you're making a bit of an error here in saying "suppose ... he has a magical engine that can maintain the thrust needed to accomplish such acceleration even as the vehicle acquires mass" and then saying "observers perceive the rocket to increase in speed and gradually approach the speed of light (but never reach it)". It's only from the observers' frame that the rocket acquires mass - from the pilot's point of view nothing about the ship is changing (but the universe outside is changing radically). $\endgroup$ – Nathaniel Nov 19 '13 at 8:52
7
$\begingroup$

The equations you need are given in John Baez's article on the relativistic rocket. If $T$ is the time measured on the rocket, and $V$ is the velocity measured on the rocket, then the equation you need is:

$$ V = c \tanh \left( \frac{aT}{c} \right ) $$

For a rocket accelerating at $g$ the velocity time looks like:

Rocket

So, as you guessed, option (1) is the correct answer.

Bear in mind that the crew of the rocket assume they are stationary and that the rest of the universe is accelerating towards them. Their situation is as if they were stationary in a constant gravitational field of $g$ and were observing everything else falling towards them in that field. It should come as no surprise that they do not observe any superluminal velocities.

If you're interested in taking this further, the coordinate system for a uniformly accelerating observer is called the Rindler coordinates.

Incidentally, one of the few things observers in two frames will agree on is their relative speed (measured as the two observers meet i.e. when they're at the same point in space). You'll find that the time dilation and length contraction balance each other out to give both frames the same relative speed.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your plot describes situation from the point of view of stationary observer. The question as I understood it, asks about the impressions of a pilot. Pilot will feel $g=9.8\approx10 m/s^2$ all the time. This implies, that after 300000000/10/60/60/24 = 347 days he will come to conclusion, that he is superluminal. Of course he will be wrong, but the question is about impressions. $\endgroup$ – Suzan Cioc Nov 19 '13 at 7:48
  • 2
    $\begingroup$ @SuzanCioc: no, this is the velocity of the rest of the universe measured by the pilot of the rocket (the pilot thinks they are stationary). The velocity of the pilot measured by the stationary observer is given by $v = at/\sqrt{1 + (at/c)^2}$. $\endgroup$ – John Rennie Nov 19 '13 at 7:57
  • $\begingroup$ Ah yes, sorry you are right at this point. But don't think this is what was asked. $\endgroup$ – Suzan Cioc Nov 19 '13 at 8:03
  • $\begingroup$ Good answer. See also a nice chart here: physics.stackexchange.com/questions/47893/… $\endgroup$ – Murod Abdukhakimov Nov 19 '13 at 15:41
  • $\begingroup$ @JohnRennie -- how does the following section in the Wikipedia article on FTL relate to my question and your answer? en.wikipedia.org/wiki/Faster-than-light#Proper_speeds Is it that the pilot does not perceive FTL, but when he does the calculation across different inertial frames, he gets a proper velocity that is faster than light? $\endgroup$ – Eric Walker Feb 16 '14 at 1:44
0
$\begingroup$

This question intrigued me enough to spend a few hours on it. I like to address special relativity problems in more than one way, so this gets to the same answer as John's but with different logic.

The problem with the initial concept is that we are assuming v = at, as Newton would say. If the acceleration is 9.81 m/s2, then sure enough, after 1 year (31e6 seconds), we would calculate a velocity greater than light speed. Why does this not work?

One way to demonstrate it is to use a conservation of energy approach. As you know, Newton would say kinetic energy is 1/2mv^2 while in relativity, we show that is only true for velocities well below light speed. We would say if we use enough force to create an apparent acceleration of 9.8 m/s2 that:

\begin{alignat}{7} \int{ F ds } & ~=~ (\gamma - 1)mc^2 \end{alignat}

Plug in that \begin{alignat}{7} F & ~=~ \frac{d(mv)}{dt} \end{alignat}

and acceleration a is dv/dt, you can eventually show that \begin{alignat}{7} v & ~=~ \frac{1}{\gamma^3}at \end{alignat} which reduces to the Newtonian result when velocity is modest. I am not sure if that is a well known equation, but it solves this problem nicely. Plus you see that the underlying cause is that the acceleration (which is energy addition) is not going to velocity increase, but rather to momentum increase. And momentum can go to infinity without the velocity reaching lightspeed.

The formula also explains why the pilot doesn't calculate greater than light speed by "the summation of the measurements on his accelerometer." He should not increment his speed by at. Instead, his calculation will be, solving numerically, the previous total calculated velocity plus over the next time period, a reduced acceleration, so that light speed is approached asymptotically. I did a spreadsheet and it looks like John's result: enter image description here

| cite | improve this answer | |
$\endgroup$
-2
$\begingroup$

The most close to correct is (2).

To measure acceleration a pilot should just measure G-force (over weight). If he assumes some overweight as some acceleration and will sum it linearly, he will obtain more-than-c values at some time.

Moreover, a pilot will see, how surrounding Universe passing by faster than C. For example, if some mileposts are installed across Universe, he will see how they counting miles faster than C.

This is because Universe is contracted along pilot's trajectory.

Another point of view says that pilot's time is slowing, but he doesn't notice this and hence feel that mileposts passing faster.

Simultaneously, if the pilot will mark some star or planet and will measure the speed of this body, the result of measurement will say a value slower-than-c.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Can you show the calculations to support this conclusion? $\endgroup$ – John Rennie Nov 19 '13 at 7:25
  • 1
    $\begingroup$ What to calculate? What are you doubt in? That summing of 10 will supersede 300000000 after some number of operations? $\endgroup$ – Suzan Cioc Nov 19 '13 at 7:42
  • 2
    $\begingroup$ The universe does not actually add velocities by simple summation. It uses a more complicated rule in the form $v' = (u+v)(1 + \frac{uv}{c^2})^{-1/2}$. This is not noticeable in day to day life because $u,v \ll c$ and the correction is negligible, but it is experimentally verified to extremely high precision in particle accelerators every day. $\endgroup$ – dmckee --- ex-moderator kitten Nov 19 '13 at 7:47
  • 1
    $\begingroup$ There would be no velocity summing by Universe. From the point of view of pilot, he will be stationary. But he will feel G-force. Assuming it is 10m/s^2, pilot can sum this value each second and at some time will get a value greater than C. $\endgroup$ – Suzan Cioc Nov 19 '13 at 7:51
  • 6
    $\begingroup$ You comment to the original question assumes that you can use the timing of one frame and the distance of another (that is you have assumed a preferred frame for distance). You can't you have to make the velocity measurement in one frame. The pilot will never measure a speed that exceeds $c$. $\endgroup$ – dmckee --- ex-moderator kitten Nov 19 '13 at 8:30

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.