Jump to content

hu??

Members
  • Content Count

    12
  • Joined

  • Last visited

Community Reputation

2 Neutral

About hu??

  • Rank
    Quark

Profile Information

  • Favorite Area of Science
    physics
  1. Very cool. It is now more than three years later. Do you have new results that confirm or contradict this?
  2. Indeed, it is completely irrelevant. I only mentioned this to show that it is possible to nullify the difference in start of acceleration for both observers by positioning the rocket motor between both observers instead of one at the level of the motor and the other one a some distance.
  3. I don't think that is relevant, because that can be compensated for by a mechanic construction (e.g. with the rocker motor is positioned in the middle of the rocket). But for the sake of the argument lets assume the rocket stays in the back, what would that entail? Let's see: The difference in start and stop is given by a constant time: For a complete rigid rocket this only depending the height: delta_t = h/c where: h: height difference between observers c : speed of light. The top observer will start later, but also stop later. The result is that it is 'pushed' as long as the bottom observer (in the reference frame of the bottom observer). Now the difference in traveltime, can be calculated as follows. The timedilation is given by (for relative small acceleration and 'height' difference): v = a * h/c T = t1/t0 = 1/sqrt( 1 - v^2/c^2 ) where: a: acceleration of rocket as observed by observer in the bottom. The result T is slightly lower than in reality, but the difference will only be become significant for very large a or h. So, the total traveltime disagreement is: dt = T * tb where: tb: total traveltime for observer in bottom of of rocket. Clearly, since delta_t is constant (and canceled out in the end, and can be compensated for) while dt depends on the total traveltime of observer in the bottom, the observers will disagree on the total traveltime at the end of the trip.
  4. Sorry to be absent for some time. They don't have to climb: the rocket just has to stop accelerating. That is the whole thing. for instance, when the rocket stops accelerating half-way the trip, then turns around and starts accelerating again. Then stops accelerating when the velocity is zero relative to the starting point. The observers will disagree on the trip-time! I think that is just weird.
  5. I realized this last night myself as well: the proper local acceleration depends on the 'height' within the rocket because the observed differences in speed of time. It even does explain the contradiction i started with: the travelers will experience different travel-time although traveling the same. I think i just have to accept relativity produces weird effects. 🤪
  6. i know it is accurate, but in this case it is not relevant. I you want to go into technical details: the deformation can also be compensated for with actuators so the rod will act as completely rigid up to measurement precision.
  7. Correct. I can be wrong, but i think not this one. If so, please show me how.
  8. I though this thought experiment would be clear. Appearently it is not. The time-dilation will be ridiculously small for any practical rocket unless you are using a start-trek like impulse drive for a very long lime, and we all know that is not available. So the though experiment is just that, and handles with an idealized 'rocket' or call it a space-hook if you like. We assume the rocket is idealized rigid. Clock synchronization is done while the rocket is not accelerating! Edit: i inadvertently posted to early.
  9. Correct, this is not a contradiction. When the travelers synchronize clocks before acceleration, then accelerate with 1 in to top and one in the bottom of the rocket, then after acceleration they compare clocks again: they will not have experienced the same time! Or do they and is my reasoning incorrect?
  10. I know this. I did not want to complicate things by mentioning this.because it is not relevant here.
  11. The gravitational time-dilation is not caused by the difference in gravitation gradient (caused by the distance from a gravitational attractor), but only by the difference in gravitational potential (i hope i use the correct terminology). Hence the equivalence principle. If that is not convincing: tt is also possible to derive the time-dilation within the accelerating rocket simply using SR.
  12. When a rocket is standing on earth the time in the top is going faster that the time at ground/rocket-engine level due to gravitational time-dilation. Using the equivalence principle with the same rocket accelerating in deep space, again an observer in the top will see the time passing by faster in the top then a traveler at rocket-engine level. To me this looks like a contradiction because they both will experience the same acceleration and duration of acceleration! I am confused: how is this possible? What am i missing?
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.