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Speed of light puzzle

  1. Sep 24, 2011 #1
    There has been so much interest lately in objects going faster than the speed of light, I though I would propose a little thought experiment.

    An observer is standing stationary relative to Object A, which is 1x10E10 meters away.

    Since there is no relativistic limit on acceleration, suppose the observer accelerates to velocity 2.85 x 10E8 m/s in 0.1 seconds.

    At this velocity, v/c = .9524 and (1-(v/c)^2)^.5 = .305. Assume c = 3 x 10E8 m/s.

    The instant that the accelerating observer becomes stationary in an inertial reference frame going at velocity 2.85 x 10E8 m/s, Object A is now approximately 3.05 x 10E9 meters away (length contraction). I know Object A is actually closer than that, but I am not factoring this in because the trick I am using doesn't need the distance to be any shorter than 3.05 x 10E9.

    Object A has traveled 6.95 x 10E9 meters in 0.1 seconds.....or 6.95 x 10E10 m/s.
     
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  3. Sep 24, 2011 #2

    HallsofIvy

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    No, it hasn't because one measurement, the initial distance, and the other, the distance after the acceleration, were measured in different frames of reference. You cannot compare them.
     
  4. Sep 24, 2011 #3
    First of all, if the motion of the observer is uniformly accelerated relative to object A, then the distance covered by the moving observer in those 0.1 seconds would be:

    [tex]
    s = \frac{a t^{2}}{2} = \frac{v_{f} t}{2} = \frac{2.85 \times 10^{8} \, \frac{ \mathrm{m} }{ \mathrm{s} } \times 0.1 \mathrm{s} }{2} = 1.425 \times 10^{7} \, \mathrm{m}
    [/tex]

    This distance is smaller than the inital distance, so the observer had not flown by object A while accelerating.

    Second of all, it is true that the distance is contracted by a Lorentz factor of [itex]\frac{1}{\gamma}[/itex], but time is also contracted by the same factor [itex]\frac{1}{\gamma}[/itex], since the proper time is relative to the moving observer and that is the shortest time.

    Thus, the speed that the observer would deduce for object A, after he had started moving with constant velocity is:
    [tex]
    \frac{\frac{1}{\gamma} \, (L - s)}{\frac{1}{\gamma} \, \frac{L - s}{v}} = v
    [/tex]

    in the opposite direction of the direction motion of the observer to the object A.
     
  5. Sep 24, 2011 #4
    Notice that the proper time according to the moving observer while they are accelerating takes:

    [tex]
    \tau = \frac{t_{0}}{2} \left[ \frac{\arcsin{\beta}}{\beta} + \sqrt{1 - \beta^{2}}\right], \ \beta = v_{f}/c
    [/tex]

    For [itex]\beta = v_{f}/c = (2.85 \times 10^{8} \, \mathrm{m \, s}^{-1})/(3.00 \times 10^{8} \, \mathrm{m \, s}^{-1}) = 0.950[/itex], the factor is:

    [tex]
    \tau = 37.9 \, t_{0}
    [/tex]
     
  6. Sep 25, 2011 #5
    Yes, the measurements are made in two different frames of reference. Does this solve the puzzle?

    Suppose the observer and Object A are fastened to the end of a pole and they undergo the same acceleration as a unit. When they arrive in the moving reference frame, they will now be 1 x 10E10 meters apart. They started out 3.05 x 10E9 meters apart as seen by this moving frame and now have increase their spacing by 6.95 x 10E9 meters in a time of around .3 seconds.
     
  7. Sep 25, 2011 #6

    ghwellsjr

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    Combining measurements made in two different Frames of Reference is a good way to create puzzles, not solve them. You need to describe and analyze your entire scenario using just one FoR. The idea that each observer and object "owns" their own rest FoR is just wrong. It's OK to analyze your scenario from each of these FoRs, but you need to do them one at a time and use the Lorentz Transform, not guess-work, to get from one frame to the other.
     
  8. Sep 26, 2011 #7
    I am not familiar with this equation. In my original post above, I assume that the .1 second is the time seen by the accelerating observer. The inertial references frames will therefore see an acceleration time of .328 seconds. The way this is calculated is given in the attachment to this post.
     

    Attached Files:

  9. Sep 26, 2011 #8
    I will not download any .doc files. If you can export it in PDF, fine. Otherwise, I'm afraid I cannot open it.

    If the observer had accelerated with constant accelerating relative to a co-moving inertial reference frame at any instant, then, the acceleration relative to a fixed inertial reference frame is:

    [tex]
    a \equiv \frac{d v}{d t} = \frac{d \left( \frac{v' + V}{1 + \frac{v' V}{c^{2}}} \right)}{d \left( \frac{t' + \frac{V x'}{c^{2}}}{\sqrt{1 - \frac{V^{2}}{c^{2}}}}\right)} = \frac{d v'}{d t'} \frac{(1 - V^2/c^2)^{3/2}}{(1 + v' V / c^2)^3}
    [/tex]

    In the co-moving frame [itex]v' = 0[/itex], [itex]a' = d v'/d t' = a_{0}[/itex], and [itex]V = v[/itex], so:
    [tex]
    a = \frac{d v}{d t} = a_{0} \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}
    [/tex]

    If we want to find the derivative w.r.t. proper time, we need to remember that:
    [tex]
    dt = \frac{d \tau}{\sqrt{1 - v^2/c^2}}
    [/tex]
    so:
    [tex]
    \frac{d v}{d \tau} = \frac{d v}{d t} \frac{d t}{d \tau} = a_{0} \left( 1 - \frac{v^{2}}{c^{2}} \right)
    [/tex]

    This equation may be integrated:
    [tex]
    d \tau = \frac{d v}{a_{0} (1 - v^2/c^2)}
    [/tex]
    [tex]
    \tau = \frac{1}{a_{0}} \, \int_{0}^{v}{d \bar{v} \frac{1}{1 - \bar{v}^2/c^2}}
    [/tex]
    The intergral over velocities is most easily done if we make the hyperbolic trigonometric substitution:
    [tex]
    \frac{\bar{v}}{c} = \tanh{p} \Rightarrow d \bar{v} = c \, \mathrm{sech}^{2}{p} \, dp, 1 - \bar{v}^2/c^2 = 1 - \tanh^{2}{p} = \mathrm{sech}^{2}{p}
    [/tex]
    and we have:
    [tex]
    \tau = \frac{c}{a_{0}} \, \tanh^{-1}{ \left( \frac{v}{c} \right)} \Rightarrow v = c \tanh{\left( \frac{a_{0} \tau}{c} \right)}
    [/tex]

    The time elapsed by the stationary frame is:
    [tex]
    dt = \frac{d \tau}{\sqrt{1 - \frac{v^2}{c^2}}} = \cosh{\left( \frac{a _{0} \tau}{c} \right)} \, d\tau \Rightarrow t = \frac{c}{a_{0}} \, \sinh{\left( \frac{a_{0} \tau}{c} \right)}
    [/tex]
    [/tex]

    The distance traveled during this acceleration is:
    [tex]
    dx = v \, dt = c \tanh{\left(\frac{a_{0} \, \tau}{c} \right)} \, \cosh{\left( \frac{a_{0} \tau}{c} \right)} \, d\tau = c \, \sinh{\left( \frac{a_{0} \tau}{c} \right)} \, d\tau
    [/tex]

    [tex]
    x = \frac{c^{2}}{a_{0}} \left[\cosh{\left( \frac{a_{0} \tau}{c} \right)} - 1 \right]
    [/tex]

    Eliminating [itex]a_{0}[/itex] from the relation between velocity and proper time, we get the following expressions for the elapsed time and distance traveled relative to the fixed frame:
    [tex]
    t = \tau \frac{\beta}{\sqrt{1 - \beta^{2}} \, \tanh^{-1}{\beta}}
    [/tex]

    [tex]
    s = \frac{c \tau}{\tanh^{-1}{\beta}} \, \left[ \frac{1}{\sqrt{1 - \beta^{2}}} - 1 \right]
    [/tex]

    Taking [itex]\beta = 0.950[/itex] and [itex]\tau = 0.1 \, \mathrm{s}[/itex], we get:
    [tex]
    t = 0.166 \, \mathrm{s}
    [/tex]

    [tex]
    s = 3.61 \times 10^{7} \, \mathrm{m}
    [/tex]
     
  10. Sep 26, 2011 #9
     

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  11. Sep 26, 2011 #10

    ghwellsjr

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    The details of how the acceleration takes place have nothing to do with this problem. It won't matter whether the acceleration is constant over the 0.1 second interval or all occuring at any instant of time during that interval. All that matters is the difference in speed between the beginning of the interval and the end of the interval.

    You need to pay attention to posts #2 and #6.
     
  12. Sep 28, 2011 #11
    My thought experiments here are examples of one of my favorite types of thought experiments. For those readers who suspected reference frames are involved in the analysis, you get a gold star. I have seen some confusion in a number of posts throughout this web site on what a reference frame is. So, I would analyse the experiment the following way:

    A reference frame is nothing more than a point of view from which to do an analysis (or a series of calculations). Object A is an element in the stationary and moving reference frames - in fact, it is an element of all reference frames in the universe. Special Relativity is totally concerned with converting results from one reference frame to another, so I would disagree with saying that you can't compare calculations in two reference frames. Of course you can. So I will. Assume the distance traveled by the accelerating observer is not significant to the overall dimensions of this problem. Let's look at the problem from the stationary reference frame. The observer starts off 1 x 10E10 meters from Object A and ends up 1 X 10E10 meters away after the acceleration. Object A never breaks the speed of light. Looking at it from the moving reference frame, the observer starts off 3.05 x 10E9 meters from Object A and ends up 3.05 x 10E9 meters from Object A. There never was a puzzle. The trick I used was to word the experiment as though something happened, but that thing never did happen.

    Having said that, the discussion on reference frames shows that I completely bungled getting my real trick to work. Here's what I mean:

    The observer starts the experiment 1 x 10E10 meters from Object A and ends up 3.05 x 10E9 meters from Object A. Doesn't this imply that an accelerated reference frame can see a speed faster than the speed of light? The answer: no. Just because the observer is an accelerating reference frame doesn't mean that the laws of physics don't apply. One main foundation of Special Relativity is that the laws of physics apply in all reference frames - and I see no justification for restricting that notion to just inertial frames.

    What happened? The key idea in analyzing the experiment from the accelerated reference frame is "failure of simultaneity at a distance." The trick I was trying to use in this example is that I have never found a text that coherently addresses how Special Relativity relates to accelerated reference frames. Certainly, simultaneity would be a challenge for any analysis of acceleration. Some might say that SR only deals with inertial frames. I would say that this was true 100 years ago, but don't you think it's time we moved beyond that limitation?

    The same thought applies to my post #5. Although all the analysis in that experiment is done from the moving reference frame, the observer and Object A cannot accelerate simultaneously as seen by this frame, even if they are linked together with a pole. Once again, there's no standard way to deal with accelerating objects as view by moving reference frames.

    The two experiments I cite are pretty simple experiments. Their solution should be standard textbook stuff. It appears to me that there is a giant chunk of science missing from current SR theory.
     
  13. Sep 28, 2011 #12

    ghwellsjr

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    Accelerated reference frames are trickery. That's why you can't find any texts that deal with them.

    In your scenario, the observer has accelerated to 0.9524c in 0.1 seconds. Even if the entire acceleration occured instantly at time zero and actually went to almost c and then dropped back down to 0.9524c after 0.1 seconds, the farthest the observer could have traveled is 3 x 10E7 m and would be no closer to Object A than 1000 x 10E7 - 3 x 10E7 = 997 x 10E7. He's barely moved. I don't know where you're getting these other numbers from.

    Look at it this way: it would take light 33.3 seconds to travel 1 x 10E10 meters. In a tenth of a second, it has only gone 1/333 of the distance.
     
  14. Sep 28, 2011 #13

    PeterDonis

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    It would seem that you haven't looked very hard. Take a look at the Wikipedia page on non-inertial reference frames:

    http://en.wikipedia.org/wiki/Non-inertial_reference_frame

    Note particularly the list of references at the bottom.

    Yes. We have. See above. Also try this page on the Usenet Physics FAQ:

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

    This is just the Bell Spaceship Paradox in another guise. See here:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken]

    Or Google it. Or search this forum for previous threads on that topic.
     
    Last edited by a moderator: May 5, 2017
  15. Sep 28, 2011 #14
    PeterDonis

    Let me further explain my post #11 by reviewing the links you gave. The Usenet Physics FAQ discussion: I see that acceleration is discussed, including the twin paradox, General Relativity and rotating objects. Even some simple equations are given. But this discussion doesn't approach a solution procedure when I ask if an accelerating observer can see Object A go faster than the speed of light (my post #1). The point I was trying to make in post #11 - shouldn't such a simple question have a relatively simple answer?

    As I look at the Bell Spaceship Paradox, my feelings are only amplified. Is this really the best solution to such a simple example? This is just more complicated than I feel should be necessary. How would this technique answer my main question from post #1?

    I understand that you may prefer the solution techniques presented in the links you gave. This is fine. I am always hoping for a more visual, more intuitive solution technique (dare I say it? - even a more Newtonian feeling solution technique). This is just my personal preference.
     
  16. Sep 28, 2011 #15

    PeterDonis

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    It does: "no". :wink:

    Bear in mind that my links were not directed at the specific problem you posed, because you admitted that you posed it not to seek an answer but to make a claim that SR somehow does not deal with acceleration. (In post #11 you admit as much, and you give the correct answer to your own question, which is the one I just gave above.) My post was addressed to that claim, which is a very general one, and which I was arguing was false: SR has plenty of methods for dealing with acceleration, but apparently you are not familiar with them. That's ok, but there's no need for trickery to elicit information. :smile:

    Have you tried drawing a spacetime diagram of your proposed scenario? That's a pretty basic visual technique, which I use a lot. Plus, it forces you to confront and resolve any ambiguities or vagueness in your formulation of the problem. If you're not familiar with them, I would recommend a basic relativity text, for example Taylor-Wheeler's Spacetime Physics.
     
  17. Sep 30, 2011 #16
    An exact calculation of the speed of Object A relative to the accelerating observer (as defined in my Post #1 above). See attached. I have changed the acceleration period in the attachment to mean that the acceleration of 0.1 seconds is measured relative to the stationary reference frame. This gives a more extreme example of Post #1 experiment and provides a better representation of what is going on.
     

    Attached Files:

  18. Sep 30, 2011 #17

    PeterDonis

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    A couple of questions/comments:

    (1) You say "the accelerated reference frame", and you talk about an observer "at xo in the accelerated reference frame". What do you mean by "accelerated reference frame" and "xo"? Do you mean "the inertial frame moving at velocity [itex]\beta[/itex], in which the accelerated observer is at rest once he stops accelerating"? In which case the location xo would be defined as the spatial location of object A at the instant, in that frame, that the acceleration stops? Or do you mean something else? (As I noted before, drawing a spacetime diagram of the scenario, instead of just the spatial diagrams you drew, forces you to confront and resolve these types of ambiguities in your formulation of the problem.)

    (2) You seem to be implying that, if you can calculate a velocity greater than c using anyone's "frame", it's a problem. That's not the case. The rule in SR that "nothing can go faster than light", applied as you're applying it, only holds in inertial frames. The more general form of the rule, which holds even in curved spacetime and for objects in arbitrary states of motion, is simply that all objects with non-zero rest mass must travel on timelike worldlines (i.e., their worldlines must lie within the light cones at each event they pass through). In your example, that appears to be true by hypothesis, so there's no problem.
     
  19. Sep 30, 2011 #18

    PeterDonis

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    You might also want to check out the following page on the relativistic rocket equation:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    Looking at the formulas there, I'm not sure all of them in your posted calculation match up, but some of your terminology is a little different. Anyway, it's a useful check.
     
    Last edited by a moderator: May 5, 2017
  20. Sep 30, 2011 #19
    Assume the accelerated observer is at the origin of a massless, infinitely stiff framework (which still is subject to length contraction, naturally). When he accelerates, the framework goes with him.

    I'm not sure I understand your second question. My calculation shows that even for the extreme acceleration shown, Object A is not near the speed of light. I am (vaguely) aware of the hypothesis you state and agree with it.
     
  21. Sep 30, 2011 #20

    PeterDonis

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    There's no such thing. The stiffness of materials is limited by special relativity: one way of stating the limit is that the speed of sound in the material can't exceed the speed of light. Your prescription would require an infinite speed of sound, so the information that a certain part of the framework has started to move could propagate infinitely fast; that's the only way the framework could maintain its structure as it accelerates. That's not physically possible, even in principle, so it's not allowed even in a thought experiment. So this definition of "accelerated frame" won't work.

    It looked to me like at least one of the calculated velocities you quoted in your PDF was greater than 3 x 10^8 meters per second. I may have misread. If you agree with what I stated about worldlines staying within the light cones, that's good.
     
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