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Yet another relativistic acceleration question

  1. Apr 3, 2010 #1
    Hello. I have read many of the posts on this forum concerning relativistic accelerations, constant accelerations to exceed the speed of light, etc. and one thing is still bothering me about the concept: why can one not accelerate constantly to achieve any arbitrary velocity as calculated in their reference frame? I have been searching for a definitive answer to this question for a long time and have yet to find one. Allow me to construct the question using the following thought experiment.

    I am traveling on a rocket that is powered by a nuclear reactor that works in such a way that it converts matter into energy at some efficiency. On board I have a normal bathroom scale, a clock, and a calculator. The engine is constantly emitting energy in such a way that the ship is constantly accelerated at about 10 m/s2 (1 g). From my perspective aboard the ship, physics should be working exactly as they would on earth due to the general equivalence principle. Since I know that I am in a rocket ship, however, and I can calculate my acceleration by dividing my weight by my mass (which I assume is remaining constant in my reference frame). I can also record the time that has elapsed since I began to accelerate using my clock (the proper time of my reference frame). Though I have no actual velocity in my constantly accelerating frame, using my calculator I should be able to calculate an "extrapolated instantaneous velocity", v, that some distant observer would see, were I not warping spacetime by accelerating. After accelerating at 10 m/s2 for about 350 days I should finally reach an extrapolated velocity of 300,000,000 m/s (speed of light) and can even exceed this to reach any arbitrary velocity, provided I have enough fuel.

    Since we assume the laws of relativity work on Earth they should also hold in this spaceship, so how can I have apparently exceeded the speed of light? In my reference frame my mass remains constant so it would never take an infinite amount of energy to cause me to accelerate from my point of view. In fact, even from a distant observer's point of view, even if my mass goes to infinity as my velocity approaches c, the mass of my fuel is also going to infinity. Because the ship converts the fuel's mass into energy to provide acceleration, I would thus have an infinite amount of energy to accelerate my infinitely massive ship. So, regardless of who's reference frame you're in, there does not seem to be any reason why I cannot reach the speed of light, or any velocity for that matter, as long as I am bringing enough fuel with me. How is this possible?

    Thanks,
    Rexx
     
  2. jcsd
  3. Apr 3, 2010 #2
    You are asking what happens to a rocket with constant trust. Here is a breakdown of this:
    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    If we had a rocket that could turn the CMBR into an accelerate of 1g we could reach the Andromeda Galaxy in 28 years, 14 years if we don't plan to stop there.

    (Maybe reflecting light from the brightest stars in our neighborhood might be a better option, it might be hard to find a thermal sink cooler then the CMBR :)
     
    Last edited by a moderator: May 4, 2017
  4. Apr 3, 2010 #3
    This explanation is still based on the perspective of a distant observer, stationary with respect to the spaceship. As I stated before, from his point of view the rocket would indeed constantly become more massive as it's velocity approached c, and would thus need a constantly increasing energy source in order to propel it further. However, since the fuel's energy is based on its mass, which is also increasing, the ship would have a constantly increasing energy source, and thus the ship would continue to be able to accelerate indefinitely. So why can we not do this?
     
  5. Apr 3, 2010 #4

    Dale

    Staff: Mentor

    Your "extrapolated instantaneous velocity" is not your velocity in any reference frame. It is not limited to c as velocity is. Rapidity is a similar quantity which has dimensions of L/T but is not speed and is not limited to c.
     
  6. Apr 4, 2010 #5
    Actually rapidity is arctanh(v/c), which is still limited to velocities below c. The extrapolated velocity is not a real velocity in any reference frame because in this situation we are accelerating, so there is no inertial reference frame. This quantity represents an assumed velocity that one would have in their own reference frame were they to stop accelerating at that instant. In any accelerating reference frame you would calculate your instantaneous velocity in the exact same way. Regardless of whether this velocity is real or not, the real point of this question is what would a distant stationary observer see at any point in time after the spaceship's proper time has indicated that the ship's "velocity" has exceeded the speed of light?
     
  7. Apr 4, 2010 #6

    Dale

    Staff: Mentor

    An inertial observer, regardless of distance, would not see anything unusual. They would see the rocket going at some v<c as indicated by the relativistic rocket equation that utesfan100 linked to earlier.
     
  8. Apr 4, 2010 #7

    Cleonis

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    In relativistic physics, when dealing with acceleration, it's very unfruitful to try and think in terms of the instantaneously co-moving frame.

    How are you going to express the current velocity? If I understand you correctly you are proposing to use what amounts to newtonian dead reckoning: "If I have sustained acceleration so-and-so for such-and-such duration as measured by my own clocks, then my accumulated velocity must exceed c."

    Now, your point of departure is some solar system, and you have a velocity relative to your point of departure. Conversely, your point of departure has a velocity relative to you. Given the current relative velocity the distance between you and your point of departure must be corrected for relativistic length contraction, and there is a corresponding correction for the relative velocity.

    I think it's interesting to focus on the difference between newtonian physics and relativistic physics here.
    The common element is that if you build up enough velocity you can travel from one end of the Galaxy to the other in decades, even though the same journey takes light thousands of years.
    - In terms of newtonian physics you cut the cross-Galaxy journey time down by exceeding the speed of light.
    - In terms of relativistic physic the time it takes to travel to the other side of the Galaxy is to be accounted for by time dilation and length contraction.

    It's not that you cannot apply the counterpart of dead reckoning: you can. But it has to be relativistically correct dead reckoning. You must extrapolate the location of your point of departure, including correction for length contraction.
     
  9. Apr 5, 2010 #8
    Hi Cleonis, thanks for your response. I couldn't agree more about the serious lack of fruit present when it comes to using instantaneous comoving frames in relativistic acceleration, which is where my whole problem with this problem lies. According to the link that was previously put in this post concerning the equations of a relativistic rocket, acceleration is defined exactly in this fashion, i.e. a combination of velocities as measured by a stationary observer (which will suffer from the gamma factor) and times measured by an inhabitant of the relativistic rocket (which will not). In my proposed thought experiment acceleration is measured directly within the reference frame of the rocket, by using a=F/m, not a = dv/dt, as v is essentially meaningless in one's own reference frame. The problem that I have with the problem of the rocket not really starting out at rest because whatever solar system it's starting out in is already moving is that the rocket is not starting with a velocity of 0, just some arbitrary velocity, and is accelerating constantly to achieve another arbitrary velocity that is c more m/s than the original one was. In this way, the important value is not your initial or final "absolute velocity" as compared with some "absolute reference frame" (which Einstein showed could not exist), but it is the amount of (proper) time one has been accelerating for that is important. Once this value reaches the speed of light (which is constant in any reference frame), I would think that something unusual would happen, because one would calculate themselves to have exceeded this fundamental velocity in their own reference frame.
     
  10. Apr 5, 2010 #9

    Dale

    Staff: Mentor

    Nothing unusual would happen at this point in any reference frame.
     
  11. Apr 5, 2010 #10
    Hi,

    Here are all the equations you are looking for. Nothing "unusual" happens when v->c as long as v never equals c. No massive object can reach the speed of light. It is easy to show that the mechanical work required to do that would be infinite.
     
  12. Apr 5, 2010 #11
    These are a more general form of the same equations posted before, which still define acceleration in a way differently than I have. In this set of equations acceleration is being measured by a distant observer which can then not be used in conjunction with the proper time of the traveller, as they do not refer to the same reference frame. If I were to look at the clock on the ship and then multiply it by the acceleration I directly observe the ship to have, these equations would be correct. However, in order to actually calculate the ships "velocity" in the same way that a traveller in the ship would, it would have to be based on the mass of the traveller (remember, this is how the traveller determined his ship's acceleration). Because the traveller's relativistic mass would be multiplied by a factor of gamma from my point of view, the relativistic acceleration would be multiplied by a factor of 1/gamma because we are using a = F/m. This is the correct acceleration. We can would multiply this by the proper time (which is again multiplied by a factor of gamma from our stationary POV) to find that the gammas cancel, and we have a velocity of simply a*tau. Also, how does one easily show that the mechanical work required for an object to accelerate itself to the speed of light is infinite?
     
  13. Apr 5, 2010 #12
    Why not? In my own reference frame I have just broken the light barrier, something that is supposed to be impossible. According to Einstein, all reference frames are equally valid, so I have definitely just gone faster than the speed of light in some way. Are you telling me that at the very least some hatch built into the universe filled with balloons and confetti wouldn't open up to congratulate me for achieving the physically impossible? But seriously, because of the gamma factor, if I have exceeded c in my reference frame, then this gives imaginary results to relativistic equations of motion in all other reference frames with a velocity less than mine. How is this not something unusual? What does a spaceship moving through imaginary space or time look like?
     
  14. Apr 5, 2010 #13
    No, they don't. a=F/m. I can derive them for you in a few lines.

    F=m*d/dt(v/sqrt(1-(v/c)^2)

    so,

    F/m=d/dt(v/sqrt(1-(v/c)^2)


    Since F/m is constant, the above becomes a very simple differential equation with the solution:

    v=at/sqrt(1+(at/c)^2)

    For at<<c, you recover the Newtonian equation v=at

    If you integrate one more time, you will get x as a function of a and t. Indeed:

    dx/dt=at/sqrt(1+(at/c)^2)


    x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)

    Again, for at<<c, you recover the Newtonian formula x(t)=at^2/2

    Work=Integral{Fdx}=Integral_0_to_c{m*d/dt(v/sqrt(1-(v/c)^2)*dx}=
    =m*Integral_0_to_c{d/dt(v/sqrt(1-(v/c)^2)*vdt}=
    =m*Integral_0_to_c{v/(sqrt(1-(v/c)^2))^3*dv}=oo


    No, you didn't. You need infinite energy to even reach light speed (see above).
     
    Last edited: Apr 5, 2010
  15. Apr 5, 2010 #14

    Dale

    Staff: Mentor

    No, in your own reference frame you are at rest. As I said before, the number you are calculating is not your velocity in any reference frame, so it is not limited to c.

    No you have not. You have merely calculated a number that is greater than c, which is not prohibited by relativity.

    Again, you are at rest in your reference frame, and you are not at v>c in any reference frame.
     
  16. Apr 5, 2010 #15

    Janus

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    Not quite. Even if we accepted this as true, You haven't accounted for time dilation and other factors. Since from that distant observer's point of view time would be running slower and slower for you, you would be burning your fuel at a slower and slower rate.
    This and the other factors mentioned would reduce your rocket's exhaust velocity as seen by this observer. For example, if your exhaust velocity was .5c as measured by you, then when you are moving at .75c relative to this observer he would measure your exhaust velocity as being .35c. At .9c it would drop to .172c, and at .99c it drops to .0192c.

    Since it is the exhaust velocity that determines how efficient your engine is (how much change in velocity burning a given amount of fuel produces), Your outside observer will see the efficiency of your rocket falling off more and more as your velocity increases with respect to him. You'll never get that last bit of change in velocity to get you to c.
     
  17. Apr 5, 2010 #16
    There are several problems with this. First of all, whose reference frame are you talking about here? Is 'v' the velocity measured by the stationary observer? If so where did the gamma factor come from? Are you saying that the velocity of the spaceship will be observed to be v = gamma*v? Second, the equation to find the force in this situation is not correct from anyone outside of the spaceships point of view. One cannot use simply F = ma = dv/dt because this is assuming constant mass. Newton's real second law is F = dp/dt = d(mv)/dt = v(dm/dt) + m(dv/dt). Because the mass of the ship is changing (due to the expenditure of fuel) the simple F=ma equation does not work outside of the ship. This is the crux of my argument. Because the fuel being used to accelerate the ship is, itself, accelerating, it will undergo relativistic mass dilation just as the ship does. Because all fuels ultimately work by converting potential energy (in the form of mass) into kinetic energy, there will be an infinite amount of energy available to continue accelerating the ship to any velocity, provided there is enough fuel. This can be seen in your own equations by introducing gamma factors where they are missing (m = gamma*m0) and keeping track of whose reference frame you are talking about.
     
  18. Apr 5, 2010 #17
    From F=dp/dt and p=mv/sqrt(1-(v/c)^2)

    This is basic relativity, did you ever take a class?

    No, see above.


    Again, you are assuming wrong. See above. You should really consider taking a class before making all these allegations.
     
  19. Apr 6, 2010 #18

    Fredrik

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    Because the work required to accelerate to speed v goes to infinity as v goes to c.

    See this post.

    What you're describing is constant proper acceleration, not constant coordinate acceleration. You will have the same coordinate acceleration in every co-moving inertial frame, but that's a different frame at every point on your world line. Your world line, as described using an inertial coordinate system in which you started out at rest, is a hyperbola. (One of the curves in the spacetime diagram here).

    The result is your proper acceleration, not your coordinate acceleration in the inertial coordinate system in which you started out at rest (or with some specified velocity v).

    Yes, that's the proper time, a coordinate independent property of the curve that represents your motion.

    What you're doing has no effect at all on the geometry of spacetime.

    If you're going to calculate velocity as "acceleration*time", you have to use the coordinate acceleration and the coordinate time of a specific inertial frame, and what you're plugging into the calculation is the proper acceleration and the proper time. Two wrongs don't always make a right.
     
    Last edited: Apr 6, 2010
  20. Apr 6, 2010 #19

    Dale

    Staff: Mentor

    No, the crux of your argument is a simple mistake. The number you are calculating is not velocity as you claim.
     
  21. Apr 6, 2010 #20
    Before you start out on your journey, you survey the night sky and pick out some stars that appear to be at rest with respect to you to use as a reference. You accelerate away in your rocket with constant proper acceleration for a time according to your onboard clock and calculate using your dead reckoning method that your velocity is say three times the speed of light. You then look out the window and calculate your velocity relative to your chosen reference stars and discover that your velocity is not greater than c relative to them. This should tell you that there is something wrong with your dead reckoning calculations.

    Perhaps something is happening to your clock that makes you arrive at this wrong conclusion? Perhaps your sense of length when you have relative motion to something, is different to when you when you were at rest with it? Wait a sec, doesn't the theory of Special Relativity say something about the non absoluteness of space and time? Might be worth looking into ;)
     
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