According to SR...

Discussion in 'Physics & Math' started by Motor Daddy, Mar 26, 2012.

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  1. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    ...what is the total distance one twin travels in this scenario:

    Initially there are two twins next to each other at t=0. One twin accelerates for 3.2 seconds at a constant acceleration rate of 74,948,114.5 \(m/s^2\), then travels at a constant velocity for 9.3 seconds, then decelerates to a zero velocity at a constant acceleration rate for 3.2 seconds.

    At exactly t=44.3 the twin starts his return journey home to his twin brother. He accelerates at a constant acceleration rate to a velocity of .8c, and it takes 3.2 seconds to get to that velocity, at which time he travels at that .8c velocity for 2,230,455,887.52 meters, and then decelerates at a constant acceleration rate for 3.2 seconds, to a zero velocity, where he meets his twin brother once again.

    What is the total distance of the journey?
    What is the total time of the journey?
     
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  3. Tach Banned Banned

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    Plug your numbers here and you'll get your answers.
     
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  5. rpenner Fully Wired Staff Member

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    The answer depends largely if we are defining constant acceleration in the initial inertial frame or in the instantaneous frame of the accelerating twin, and largely in what frame we measure distance. The question's lack of context does not distinguish if distance is net displacement or a measure of accumulated unsigned change of position.

    It's trivial to work in the inertial frame of the non-accelerating twin if we assume constant acceleration in defined in the non-accelerating twin's frame.
    \(a = 0.8 c / 3.2 \, \textrm{s} \, = c / 4.0 \, \textrm{s} \, = 74948114.5 \, \textrm{m} \, \textrm{s}^{\tiny -2} d_0 = (7.44 \, \textrm{s} \,)c = (9.3 \, \textrm{s})(0.8 c) \, = \, 2230455887.52 \, \textrm{m} v_{\tiny \textrm{max}} = 0.8 c = 239833966.4 \, \textrm{m} \, \textrm{s}^{\tiny -1} t_1 = 3.2 \, \textrm{s} t_2 = 9.3 \, \textrm{s} t_3 = 44.3 \, \textrm{s} \, - ( t_1 + t_2 + t_1 ) = 28.6 \, \textrm{s} t_{\tiny \textrm{final}} = t_1 + t_2 + t_1 + t_3 + t_1 + t_2 + t_1 = 4 t_1 + 2 t_2 + t_3 = 60 \, \textrm{s} s = \int_0^{60} v dt \, = \frac{1}{2} a t_1^2 + a t_1 t_2 + \frac{1}{2} a t_1^2 + 0 t_3 - \frac{1}{2} a t_1^2 - a t_1 t_2 - \frac{1}{2} a t_1^2 = 0 s_{\tiny \textrm{cum.elapsed}} = \int_0^{60} \left| v \right| dt \, = \frac{1}{2} a t_1^2 + a t_1 t_2 + \frac{1}{2} a t_1^2 + 0 t_3 + \frac{1}{2} a t_1^2 + a t_1 t_2 + \frac{1}{2} a t_1^2 = 2 a t^1 + 2 a t_1 t_2 = 2 a t_1 ( t_1 + t_2 ) = ( 20 \, \textrm{s} \, ) c = 20 \, \textrm{light-seconds} = 5995849160 \textrm{m} v_{\tiny \textrm{average velocity}} = s/t_{\tiny \textrm{final}} = 0 v_{\tiny \textrm{average speed}} = s_{\tiny \textrm{cum.elapsed}}/t_{\tiny \textrm{final}} = 20 \, \textrm{light-seconds} / 60 \, \textrm{s} = c/3 = 99930819 \frac{1}{3} \, \textrm{m} \, \textrm{s}^{\tiny -1} \tau = \int_0^{60} \sqrt{1 - v^2/c^2} dt \, = 4 \int_{0}^{t_1} \sqrt{1 - a^2 t^2/c^2} \, dt + 2 t_2 \sqrt{1 - a^2 t_1^2/c^2} + t_3 \sqrt{1 - 0^2} = 3.84 \, \textrm{s} \, + (8 \, \textrm{s}) \sin^{\tiny -1} \frac{4}{5}+ 6 t_2 /5 + t_3 \, \approx \, 51.0184 \, \textrm{s} \)

    Only the last calculation, of proper time for the accelerated twin involves special relativity, because that's the only step that requires expressing one observer's clocks or rulers in terms of the clocks and rulers of another observer.
     
    Last edited: Mar 26, 2012
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  7. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    So you are saying that the total distance traveled is 20 light seconds, and the total time of the trip is 60 seconds, according to the stay at home twin.

    You are saying the traveling twin says the trip only took approximately 51.0184 seconds.

    If a light sphere was emitted at t=0 from the coordinate (0,0,0) when and where the traveling twin started traveling away from his stay at home brother, what is the radius of the light sphere when the traveling twin returns back from his journey and meets back up with his brother?

    The sphere can have only one radius at that point in time, correct?
     
  8. arfa brane call me arf Valued Senior Member

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    You might think this is a deal breaker for the twin paradox, but it isn't.

    The twins start in the same place and agree that t = 0. When the travelling twin returns, the two agree what time it is at their present location. Therefore a light sphere emitted at t = 0 will have the same radius for both twins (if they somehow measure the radius).

    What they don't agree on is how long the travelling twin's journey-time was.
     
  9. rpenner Fully Wired Staff Member

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    Since the accelerating twin was accelerating, he wasn't continuously in a single inertial frame and so has no special relativistic reason to assume the light sphere has a radius of 51.0184 light-seconds. Special Relativity distinguishes inertial frames of reference as those where the laws of physics (and thus the speed of light) look the same.

    There are four inertial coordinate frames involved here, and you have to use the Poincaré transform (inhomogeneous Lorentz transform) since t=0, x=0 is not the same event for all.

    Seen from the stay-at-home twin, both twins start at time=0 in frame \(S\), and the accelerating twin immediately begins accelerating. Between t = 3.2 s and t = 12.5 s the accelerating twin is drifiting at +0.8 c in frame \(S'\). Likewise, between t=15.7 s and t=44.3 s, the accelerating twin is drifting at 0 in frame \(S''\). Likewise, between t=47.5 s and t=56.8 s, the accelerating twin is drifting at -0.8 c in frame \(S'''\). Finally, at and after t=60 s, the accelerating twin is drifting at 0 c in frame \(S''''\).

    How does frame \(S''''\) relate to frame \(S\)?
    \(x'''' = x \\ t'''' = t - 8.9816 \, \textrm{s} \\ \Delta x'''' = \Delta x \\ \Delta t'''' = \Delta t\)
     
    Last edited: Mar 27, 2012
  10. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    The journey time was the duration of time from the point in time when the two twins were together at coordinate (0,0,0) until the two twins were together once again at coordinate (0,0,0). The elapsed time is equal to the radius of the light sphere that was emitted at t=0, divided by c, the speed of light.

    The radius of the light sphere has one radius, and that radius is a light clock, which measures the duration of the trip, which each twin has to agree to, since they were both at coordinate (0,0,0) when the light sphere was emitted, and they are both at coordinate (0,0,0) when the radius is divided by c to determine the elapsed time.

    At ANY time t I can tell you the coordinate of the traveling twin, and I can tell you the radius of the light sphere, and they will match according to the accelerations and velocities he has undergone up to that time t. There is no debate about it, the time of travel at any time t is equal to r/c for both twins!
     
  11. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    So what you are saying is that SR is not capable of dealing with the twin paradox, since in the real world the twin MUST accelerate at least 4 times if traveling to and from a distant point along the same path, and that means the twin doesn't have a constant inertial frame of reference.

    There is one radius of the light sphere, and that is ct, and that radius is valid for ALL observers in the universe. There is not two or more realities, there is ONE reality, and the radius of the light sphere defines distance according to light travel time. It's non-negotiable.
     
  12. Tach Banned Banned

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    Contrary to one of the classical misconceptions amongst antirelativistic crackpots SR deals with acceleration just fine. I knew I wasted my time when I pointed out the link that explains how SR deals with acceleration.


    Nothing to do with your fixation with "light sphere".
     
  13. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    It has everything to do with the light sphere, because the light sphere is the clock for the universe, whether you realize it or not.
     
  14. AlphaNumeric Fully ionized Moderator

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    MD, you seem to have a conceptual block when it comes to this stuff. You aren't going to construct a thought experiment which disproves special relativity. It's a coherent mathematical construct, within which the behaviour of light is well understood and consistent.

    You repeatedly try to come up with thought experiments but all you do is invent thought experiments which either don't conclude what you think they do or are not actually special relativity. It's easy to come up with a thought experiment which doesn't do what SR says, if you assume the universe works like Newtonian mechanics thinks. But you haven't shown that is actually how the universe behaves.

    As such you need to provide experimental evidence to refute special relativity. Attempting to show it's mathematically flawed is doomed to failure, the only avenue is to show the universe doesn't work like that. You've been at this for years, when are you going to understand this? It's your own time you're wasting.
     
  15. mikelizzi Registered Senior Member

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    Despite what Motor Daddy has posted before, this latest post does not appear to me to be crack pot. It caught me for a minute.

    The issue, as I understand it is this. The radius of a light sphere is directly proportional to the elapsed time since it was created. If you and I are standing together at rest with respect to each other and at the source of the light sphere, we should agree on the radius of the light sphere. But if our clocks read different elapsed times since the emission of the light how can we do that?

    Consider the Twins example. The light is emitted when the twins are together, noted by each as t=t’=0. One twin goes on a round trip. Upon return, the twins calculate the radius of the light sphere as c * t. So, wouldn’t one twin believe the radius was ct and the other ct’ where t and t’ are no longer equal?

    The resolution again has to do with the acceleration required for the twins to meet up, but I'm struggling with it. I think it’s a neat problem to analyze and I’m going to include it in my set of problems for my web site.
     
  16. Tach Banned Banned

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    The light wavefront is spherical in the inertial frame of the "at home" twin (and, in any other inertial frame). It is not spherical for the Rindler frames of the accelerated twin.
     
    Last edited: Mar 27, 2012
  17. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    At t=0 when the light sphere was emitted, and at t=60 when the twin is back home, the twins are in the same frame, and they MUST agree on the radius of the light sphere that was emitted at t=0 when they were together in the same frame, as the radius of the light sphere is exactly the elapsed time of travel times c. It's non-negotiable, Tach.
     
  18. Tach Banned Banned

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    They do? Because your "senses tell you so"?
     
  19. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    It's got nothing to do with my senses. The light sphere never lies! Should I incorporate a mirror, which the twins place 1 light minute away from the coordinate (0,0,0) which they both are at t=0 and t=60? That way after the twin returns home after his journey, the twins are once again together in the same frame, and the light returns to them at t=120, which is 60 seconds after the twin returns home. They BOTH agree to the light sphere, Tach! The traveling twin therefore must conclude that his watch was full of it!
     
  20. Tach Banned Banned

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    Sure it does, there is absolutely no reason for the two observers to agree on the radius. The traveling twin does not even agree that the wavefront is a sphere. You missed that, didn't you?
     
  21. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    The traveling twin isn't traveling at t=60. At t=60 the twins are side by side, in the same frame, motionless.

    Prior to traveling the twins agreed that the radius of the light sphere is always ct, so they agreed that since the light sphere never lies, and that they will each be standing side by side after the twin returns home, that the radius of the light sphere divided by c is the official elapsed time
    of travel. Period!
     
  22. waitedavid137 Registered Senior Member

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    As noticed by someone mentioning that it isn't spherical for an accelerated frame you already know this part of your argument to be wrong so why do you continue?
     
  23. mikelizzi Registered Senior Member

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    True, the wavefront is not spherical to the astronaut when the astronaut is accelerating. But that is not the issue. The issue is what is the shape and size of the light sphere when the twins join up. At that point they are both in the same inertial reference frame. And, even though they have recorded different elapsed times since the light was emitted, they must both agree on the shape, size and location of the sphere.
     
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