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Calculating speed from acceleration

  1. Jun 21, 2004 #1
    [SOLVED] Calculating speed from acceleration

    I'd like to understand what could be the problem with this reasoning (apart from practical considerations):

    An imaginary rocket is flying through space. The imaginary astronaut has the engines on all the time and the rocket, being imaginary and not suffering practical limitations such as lack of fuel, is accelerating at the constant rate of 20 m/sec^2. Acceleration inside the rocket is measured by a little device with a spring (calibrated inside the rocket itself).

    After one hour, the astronaut calculates the speed he's supposed to be moving at: 20 * 60 * 60 = 72,000 m/sec. A very "Newtonian" speed.

    Now the rocket keeps on accelerating at the same rate, and after one year the astronaut calculates his speed again: 20 * 60 * 60 * 24 * 365 = 630,720,000 m/sec!

    What is wrong there? How did I find a value > 2c, in clear violation of relativity?
    Last edited by a moderator: Jun 21, 2004
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  3. Jun 21, 2004 #2


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    You're doing Newtonian math.

    I don't understand your question. You're applying newtonian mechanics, and then getting a newtonian answer.

    Let's say that the rocket exerts a constant force - then the acceleration drops off as the rocket gets faster.

    Alternatively, you can do the calculations for constant acceleration, and discover that the required force goes to infinity as the rocket's speed approaches c.
  4. Jun 21, 2004 #3
    Are there different definitions for speed and acceleration under relativity? I thought v = dx/dt and a = dv/dt were also correct definitions in relativity.

    I'm using an imaginary rocket, and I said the acceleration is constant. When you say "faster", what do you mean? Faster relative to what?

    OK, let's examine what will prevent the rocket from maintaining constant acceleration. When you say the force goes to infinity when speed approaches c, what reference frame are you using?

    What I don't understand is that, as far as I can tell, I can calculate my own increase in speed simply by multiplying the reading on the rocket's accelerometer by the reading on the rocket's clock. I cannot calculate my "actual" speed, because I would need to know my "actual" starting speed, but if I'm inside a rocket accelerating at 20m/sec^2, I know that my speed increases by 20 m/sec every second - by definition. What exactly happens inside the rocket that causes its accelerometer to read smaller and smaller values until it approaches zero? Isn't a frame of reference required for that kind of observation?
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  5. Jun 21, 2004 #4


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    Fair enough:

    Since this is a one-dimensional case, the relativistic mass is:
    [tex]m=\gamma m_0=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}m_0[/tex]

    where [tex]m_0[/tex] is the rest mass of the rocket, and [tex]v[/tex] is the velocity of the rocket relative to the observer. As you can see [tex]\gamma \rightarrow \infty[/tex] as [tex]v \rightarrow c[/tex]. So for the observer 'on the ground' the force required to achieve constant acceleration goes to infinity.

    Alternatively, if you loom at a constant force, you've got [tex]F=ma[/tex] with relativistic mass leading to a progressively slower acceleration.
  6. Jun 21, 2004 #5
    All I have is a guy inside a rocket with an accelerometer, a clock, and a calculator. What observer are you talking about?

    What you posted is basic stuff, but I'm trying to get beyond the basics here. A rocket at constant acceleration is equivalent to a gravitational field, and it's always possible to measure the acceleration the same way we measure the value of g on earth. What is really strange is that this mass increase thing involves velocity and therefore requires an observer, whereas acceleration does not. If you measure acceleration and you see no massive object around you, you can be sure you are moving in an "absolute" sense.

    A lot of people are confused by what seems a discrepancy between the concepts of movement in SR versus GR. Since every single respectable physicist guarantees the discrepancy does not exist, I think we should try and learn how to see things such that the discrepancy disappears. Which is why I'm asking the question.
    Last edited by a moderator: Jun 21, 2004
  7. Jun 21, 2004 #6


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    Why do you think that the accelerometer must read smaller and smaller values until it approaches zero? Your thought experiment is not inaccessible by the theory, but your analysis is on shakey ground. NateG said it rather well, IMO:

    "You're applying newtonian mechanics, and then getting a newtonian answer."

    What do you think is wrong with a velocity > 2c (and don't say, "because relativity ..." :smile: )? I suggest you focus on really understanding the problem with v > c before you get confused about this or that overly specific gedanken experiment.

    Some points that I did notice:
    Your definitions for v and a are, while not strictly incorrect, dangerous for the comparison. There is a distinction between local and global measurements and quantities. This distinction is somewhat trivial for newtonian velocity and acceleration. However, in SR, the distinction becomes important. Your accelerometer will read local acceleration (sometimes refered to as proper acceleration). You can go either way with velocity (local or global). For local velocity, you are probably mostly concerned with the spatial-components. The magnitude of the local velocity for any massive object is c (an invariant), and this projects onto the time and space axes, if you will. The global measurement is the more newtonain way of looking at velocity. For this, you certainly would need a defined frame of reference (inertial, i.e. not the rocket). Position doesn't have much meaning at all as a local measure, so I will just make the somewhat rude generalization that position is only a global measure (it must have a reference structure, otherwise it is meaningless).

    As the rocket accelerates, the spatial component of the velocity gradually changes in some inertial reference frame (not the rocket). This global effect is "felt" on board the rocket as what we commonly call acceleration or force (I do not mean to equate acceleration with force, but to emphasize the effect that one feels and then to give it a name without being too picky). The newtonian acceleration would be observed to behave assymptotically to zero, which is usually confounded with some consideration of relativistic mass or such. The proper acceleration (the local measure that the rocket passengers would see on the accelerometer) should remain unchanged (as specified by you). This seeming inconsistency arises from the non-Euclidean nature of space-time as a geometry. Where you might expect the lengthening of the spatial velocity to steadily and uniformly increase, the velocity is actually a 4-D vector (which we can reduce to 2-D for this discussion) in a non-Euclidean spacetime that maintains a constant length. Therefore, as the rocket accelerates, in some inertial frame from which it is being observed, its spatial component of velocity is increasing to the point that it can only approach c but never quite get there. Its time component actually also increases (instead of decreasing as you might expect in Euclidean space). This is related to the momentum which, in turn is related to the energy. In order to rotate this velocity vector (wrt some inertial frame) either into or out of the time dimension, you must therefore give or take some energy. In the case of your rocket, you are probably most interested in a qualitatively increasing speed, which entails an increasing time component, and therefore an increasing energy. The law of conservation of energy still applies, and so more energy is required the faster the rocket goes. The rate at which the energy increases is, however, local to the rocket, and is another qualitative way of explaining the constant accelerometer reading as a rate of energy increase that would be observed in some inertial frame outside the rocket.

    When considering observation inside the rocket itself, SR no longer applies, and you must go to some simple GR considerations.
  8. Jun 21, 2004 #7


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    According to my understanding the pilot of the continuously accelerating space ship would not observe any change in mass of his ship or slacking of the acceleration. Remember that in the frame of reference of the ship all of Newtons mechanics must hold. Remember also that anytime that a measurement of the speed of light was done you would still measure c as the velocity in your frame of reference. Now we cannot meaningfully speak of velocity without some second reference point. What is the reference point of the accelerating space ship? The only meaningful one would be the point at which he started his journey. Now let us suppose that the ship maintains constant radio contact with the home port. He will never be out of contact because at any time he does a measurement of the speed of radio waves he measures c, what he does observe is a continual shift in frequency of the messages from home, he will also learn that he is not as far away as his Newtonian calculations tell him he should be. The ONLY way to reconcile his observations with calculations is to use the equations of GR (since he is constantly acceleration SR cannot be directly applied).

    Simply applying Newtonian methods will not and should not yield correct position information. If the pilot of the space ship is ignorant of Relativity he will be lost, if he has knowledge of relativity his calculations will correspond to observation. Meanwhile irregardless of how long he continues to accelerate he will always measure the speed of light to be c. Therefore he can expend an infinite amount of fuel and still not be any closer to the speed of light and even though the messages from home have been shifted in frequency they still arrive.
  9. Jun 21, 2004 #8


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    Accelartion does apply to SR after all it wouldn't be a great kinematic theory if it didn't.

    In Newtonian mechanics:

    [tex]\vec{a} = \frac{d^2\vec{x}}{dt^2}[/tex]

    in special relativity:

    [tex]A^{\mu} = \frac{d^2 x^{\mu}}{d\tau^2}[/tex]

    For someone who is accelarting constantly in their (accelarted) frame, in another (inertial) frame the accelarting object will approach c asymptotically.
  10. Jun 22, 2004 #9


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    Do not use relativistic mass. Use modern relativity wherin mass is invariant. If you use [tex]M = \gamma m[/tex] in [tex]f = Ma[/tex] as someone previously suggested, you will get the wrong answer for the dynamics according to relativity. The kind of acceleration that its accelerometer measures is called the invariant acceleration or in some contexts proper acceleration [tex]|g_{\mu }_{\nu }A^{\mu }A^{\nu }|^{\frac{1}{2}}[/tex] and this is not equal to the coordinate acceleration, [tex]a^{i} = \frac{dv^{i}}{dt}[/tex] for most frames. For your case of motion, the proper acceleration is constant and equal to [tex]\alpha [/tex] and can be related to the coordinate acceleration by [tex]\alpha = \gamma ^{3}a[/tex].
    This is a differential equation that can be integrated after seperation of variables from [tex]a = \frac{dv}{dt}[/tex].
    [tex]\alpha t = \int \frac{dv}{(1 - \frac{v^2}{c^2})^{3/2}}[/tex]
    The solution of which can be symplified to
    [tex]v = \frac{\alpha t}{\sqrt{1 + (\frac{\alpha t}{c})^{2}}}[/tex]
    This never violates relativity. For small times it approaches
    [tex]v = \alpha t[/tex] and for large times it approaches
    [tex]v = c[1 - \frac{1}{2}(\frac{c}{\alpha t})^{2}][/tex] and the second term in the expansion here vanishes as time becomes very large.
    Last edited: Jul 14, 2004
  11. Jun 22, 2004 #10
    That is exactly what I thought.

    I don't believe that is strictly correct, given that the rocket is not an inertial frame. But this doesn't seem extremely relevant anyway.

    OK, I was trying to get exactly at this. I'm glad you saw it.

    Can we look at exactly the same scenario from a slightly different perspective? Suppose our astronaut wants to reach a certain star, and estimates the time it will take to get there based on his acceleration as measured from inside the rocket. Suppose also that he is ignorant of relativity and believes Newtonian mechanics will give them the right numbers.

    Now this is what I want to understand: does the astronaut reach his destination in the time he calculated, or does it take longer? It doesn't seem like an easy question to me, especially since no one so far touched on a very important issue: the clock inside the rocket!

    I believe (but I may be wrong) that the astronaut may well find his speed to be greater than c, and to reach his destination exactly at the time he calculated. That is, from his perspective, he has constantly accelerated until he reached the star, and got there at the time he thought he would, having traveled at superluminal speeds for a good portion of the trip. That is his perspective, and I can't see what's wrong with it (although as I said, I may be wrong)

    However.... the people on earth don't think so. For them clocks are running far slower, and while the astronaut measures the duration of his trip to be, say, one year, for the people on earth it took almost a century. So they won't agree with the astronaut that he traveled at the speed he says he did, even though he reached his destination at the time he thought he would.

    All this seems to be in agreement with relativity, except perhaps the fact that the astronaut won't perceive his trip as being affected by relativistic effects.

    Well, this is exactly the bit I don't get. I would agree that applying Newtonian mechanics will not yield the correct time, and therefore the astronaut's measurement of his own velocity must necessarily disagree with measurements of the rocket's velocity done by an observer on earth. But it seems to me that, given the time dilation effect, both parties will always agree as to the rocket's position, even though one calculates it using Newtonian methods ("in the frame of reference of the ship all of Newtons mechanics must hold") while the other must use GR.

    If the above is correct, I would like to ask the question I'm really, really interested in knowing the answer.
  12. Jun 22, 2004 #11


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    You are wrong, his velocity will never exceed c, he will not calculate the correct velocity or position if Newtonian mechanics is used. Remember that Newtonian Mechanics is a first order approximation to the real (relativistic) equations. Any time an approximation is used outside of its region of applicability errors will result. As long as your velocity is small wrt to c it is safe to neglect the higher order terms of the relativistic equations. This will apply to any measurement made wrt to your starting position, thus to any calculation of velocity or position. Since you are proposing velocities approaching c, you cannot use the approximation.

    And yes, you will always measure c as the speed of light. Keep in mind that the earth is not an inertial frame of reference, yet we measure c as the speed of light.
  13. Jun 22, 2004 #12
    A quote from Albert Einstein himself:

    "according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position"

    This doesn't sound as clear-cut an issue as some people seem to be implying. It's really hard to understand all the subtleties and implications of Einstein's theory; he is known to have complained that people misunderstood him in his own time. So who knows what the guy really, really had in mind?
    Last edited by a moderator: Jun 22, 2004
  14. Jun 22, 2004 #13


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    I can look at that.

    The velocity function I arrived at earlier was
    [tex]v = \frac{\alpha t}{\sqrt{1 + (\frac{\alpha t}{c})^{2}}}[/tex]

    Input into [tex]\gamma [/tex] this yields
    [tex]\gamma = \sqrt{1 + (\frac{\alpha t}{c})^{2}}[/tex]
    [tex]dt = \gamma d\tau[/tex]
    [tex]dt = \sqrt{1 + (\frac{\alpha t}{c})^{2}} d\tau [/tex]
    [tex]\tau = \int \frac{dt}{\sqrt{1 + (\frac{\alpha t}{c})^{2}}}[/tex]
    Integration and simplification yields
    [tex]t = \frac{c}{\alpha}sinh(\frac{\alpha \tau }{c})[/tex]
    Insert this and dt into the velocity function and simplify to arrive at
    [tex]dx = csinh(\frac{\alpha \tau }{c})d\tau [/tex]
    [tex]x = (\frac{c^2}{\alpha })[cosh(\frac{\alpha \tau }{c}) - 1] [/tex]

    Using Newtonian mechanics he thinks the answer should be
    [tex]x = \frac{1}{2}\alpha \tau ^2[/tex].
    Using relativity we find that the answer is actually
    [tex]x = (\frac{c^2}{\alpha })[cosh(\frac{\alpha \tau }{c}) - 1] [/tex]
    So, no Newtonian mechanics doesn't give the right time even if you use his time in the relativistic expression. Relativity predicts that he will be there even sooner according to his own watch than Newtonian mechanics does.

    This is touched on in places, just not previously in this disscussion.
    Last edited: Jun 22, 2004
  15. Jun 22, 2004 #14
    That is very interesting. I didn't think Newtonian mechanics would give the right answer, but the phenomenon I imagined happens nonetheless. Can you bear with me as I try to imagine what the astronaut might be thinking? Suppose he left earth with the goal of reaching the vicinity of some star, and suppose he knows the distance from earth to that star as measured from earth. Then he climbs on his rocket, flies to his destination at constant acceleration, and annotates the time on his watch/calendar as he passes by the star. He now has two bits of information: the distance he traveled, which he knew before he left, and the time it took him to reach his destination. So he divides one by the other and finds that his average speed during the trip is greater than the speed of light. That is, of course, because he's not taking into account relativity, but as I said, he's ignorant of it and has every reason to believe his calculations are correct.

    The question I'm eager to know the answer is, would this astronaut, in his ignorance of relativity, have any reason to suspect he has not really exceed the speed of light? That is, apart from the fact that people on earth will not agree with him as to the duration of his trip, can he make any observation of the universe around him that will cause him to suspect he's actually traveled far slower than his measurements implied?

    I'm sorry, I have never seen this particular issue being discussed anywhere. If you have any references, I'd be glad to hear about them.

    Thanks for your reply.
  16. Jun 22, 2004 #15


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    But as the pilot goes past the star he will also observe that his (instantaneous) relative velocity is less than c, even though his Newtonian calculations tell him otherwise.
  17. Jun 22, 2004 #16
    Sure enough. He will be very intrigued by two facts:

    - as DW said, (according to his own clock) he will get to the star faster than his calculations told him.
    - when he measures his speed as he passes the star, he finds his moving a lot slower than his calculations told him, which leaves him wondering how he could have reached the star at all.

    It will be really hard for him to understand what is going on, but a little knowledge of relativity puts some sense back in the whole thing. I'm glad to find people who seem to agree with that, because now I can ask the really, really interesting question:

    Given the principle of equivalence between gravitational and inertial mass, and given the fact that we, here on earth, experience constant acceleration of some 9.8 m/sec^2, would it be correct to say our situation is not really different from the one my imaginary astronaut finds himself in?

    As far as I can tell, what left my astronaut puzzled was his ignorance of the effects caused on his observations by constant acceleration. His universe is distorted but he doesn't realize it until some measurement turns out a value quite different from what he expected. And that makes me think our perspective of the universe may also be distorted in ways that may not yet have been discovered.

    I don't know if that can be easily understood, or if it makes much sense. It's a question I have always wondered about ever since I learned about the principle of equivalence, because to me it seems to imply the earth is "pushing" us at a constant acceleration and we are, in a sense, "moving" across the universe at increasing velocity. It also seems to imply that only objects in free fall are not really "moving" and can see the universe from a perspective that isn't distorted, but then I don't really understand what "moving" could possibly mean in those contexts.
  18. Jun 22, 2004 #17


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    Surely you understand the difference between velocity and speed?
  19. Jun 22, 2004 #18


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    Pages 32-33 and problems on 34

    Pages 64-65

    relativistic rocket at constant proper acceleration

    proper velocity discussion-
  20. Jun 23, 2004 #19
    Thanks for the links, but the particular issue I have in mind is a bit different. As I tried to explain on my last e-mail, it seems correct to me to say that our situation, here on earth, is not different from being on a rocket traveling through space at constant acceleration. I don't know the full extent of the principle of equivalence between gravity and inertia, but in my understanding an object, on the earth's surface, that is not in free fall is constantly being accelerated at g. Isn't that right, and doesn't that make our perspective of the universe exactly the same as that of the astronaut in my example?

    If you know of links about that, I'd be interested to learn.
  21. Jun 23, 2004 #20


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    It is only right for local experiments. On large scales there is a very big difference between the two which is the curvature of the spacetime and anything which this effects. For example, the global spacetime's differential geometry as expressed by the rocket frame observer is a frame transformation of the metric of special relativity which has no Riemannian spacetime curvature -
    [tex]ds^{2} = (1 + \frac{\alpha x'}{c^2})^{2}dct'^{2} - dx'^{2} - dy'^{2} - dz'^{2}[/tex].
    Whereas for an observer at the surface of the Earth at least for the exterior of the Earth it is a frame transformation of the Schwarzschild solution which does have Riemannian spacetime curvature. See equation 10.3.3 at
    for the Schwarzschild interval expressed with Earth based observer coordinates. To write it fully in that observers coordinates is easy with 10.3.2, but complicates it.
    These very different spacetime geometries have different results for the behavior of surrounding different clocks at a distance and very different results for distant geodesic behavior etc. It is only in sufficiently local experiments that the two spacetimes yeild results that can be shown to be equivalent.
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