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I A spaceship traveling close to the speed of light sending some data...

  1. Dec 28, 2016 #1
    A spaceship travelling at speed of light close to speed of light (wrt inertial reference frame) sending some data every second on their clock to people who are stationary (wrt inertial reference frame). At what time these people would receive this data on their own clock?

    Let's say for a second passed in the spaceship, 10 seconds passes for the stationary people.

    Sorry if this makes no sense, I don't have a good understanding of relativity, I only know that at high speeds, the time is slower than the people at low speeds.
     
    Last edited: Dec 28, 2016
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  3. Dec 28, 2016 #2

    Orodruin

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    Assuming your space ship has mass, it cannot do this.

    If you exchange "at the speed of light" for "close to the speed of light", it will depend on whether the space ship is moving towards or away from the people. The relation will be given by the relativistic Doppler formula.
     
  4. Dec 28, 2016 #3

    Ibix

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    It's impossible to describe something with mass moving at the speed of light in relativity. However, you specified a tick rate ratio of 10:1, which implies a specific speed. If I see you travelling at speed ##v## then I will determine your clocks to be ticking once every ##1/\sqrt {1-v^2/c^2}## seconds.

    As Orodruin says, when the people who are stationary in the inertial reference frame receive the pulses depends on where they are. The emitting ship is moving - so if it is coming towards you the second pulse does not have as far to travel as the first, so pulses come in squashed together. If it is moving away then the pulses come in spread apart. The relevant formula is that if the pulses are emitted with frequency ##f## (once per second, for example) then they are received with frequency ##f'=f\sqrt {(c+v)/(c-v)}##. Note that ##v## can be positive ot negative and that I'm assuming that the ship is coming straight towards or straight away from the observer.

    If the receiver corrects for the travel time of light then they will calculate that the pulses were emitted once every ten seconds.

    Note that the situation is symmetric. The "moving" ship may regard itself as stationary and the "stationary" ship as moving, and will be able to carry out the same measurements and calculations.
     
  5. Dec 28, 2016 #4
    what if the ship is moving in circles around the receiver and keeping the radius with the receiver always constant, can we forget about doppler effect? would the constant velocity change of the ship affect the calculations?
     
  6. Dec 28, 2016 #5

    Ibix

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    In that case, yes, the receiver would receive pulses every ten seconds. Note the ##v^2## in the time dilation formula - it's only the magnitude of the velocity that matters.

    Note that this situation is no longer symmetric. The ship moving in circles would receive ten pulses per (its) second from the stationary ship was emitting once per second by its clock.

    Note that the thrust or time requirements to do this experiment are likely to be absurd.
     
  7. Dec 28, 2016 #6

    BvU

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    Note that you have to follow an orbit e.g. around the solar system with a huge radius:
    Orbiting the solar system (never mind the earth) at a sensible radius would also be impossible without crushing every bone of the space travellers: ##v^2/r=10## g at ##10^{12}## km ! That's 6000 astronomical units, less than an hour per round trip (pluto has 40 AU and 248 year) -- all calculated nonrelativistically, of course :smile:.
     
  8. Dec 28, 2016 #7
    yes I see that but we are just doing some mental experiments and we happen to have all the tools and technology to achieve that :) also keep in mind that if you accelerate every atom of your body at the same rate no acceleration can crush your bones, it is the difference of acceleration throughout the body that crushes bones.
     
  9. Dec 28, 2016 #8

    Ibix

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    True as far as it goes. But we have no way to do that and you will be crushed. Also, accelerating all parts of your body at the same rate is a problematic statement in relativity. The different parts of your body don't agree on what "the same rate" means due to time dilation and the relativity of simultaneity. The end result is painful.
     
  10. Dec 28, 2016 #9

    BvU

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    Care to hint at a way to accelerate every atom of an object in a practical manner ?
     
  11. Dec 28, 2016 #10
    I surely would care if I had any idea how. Again, we are not talking in terms of today's technology it's just mental experiments.
     
  12. Dec 28, 2016 #11

    Ibix

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    As noted above, you can't do so at the same rate, since "at the same rate" is an incomplete sentence.
    It's fine to think about what happens in extreme circumstances and leave the details to the engineers, so to speak. It's important, though, to be aware that there can be complications that you can't handwave away.
     
  13. Dec 28, 2016 #12
    What about if we drilled a hole to the center of a planet, having the right size and mass, and then dropped an astronaut in there.

    His legs would be subject to less acceleration than his head, seen from the perspective of an observer at rest to the planet. He would be shrinking as predicted by SR (length contraction).
    If the size and mass of the planet was chosen properly, one might be able to calculate that the astronaut won't be able to register any acceleration at all, as in his body registering a change of structure due to pressure/forces he would be subject to - or "feeling" acceleration as some loosely describe it.

    (Clocks that were formerly synced, located at the astronaut's legs and head, before he gets dropped into the hole, would go out of sync once he would be in free fall, but devices trying to measure acceleration by measuring a change in the (space) structure, like a mass connected to a spring, should fail, or so i believe)

    Contrary to letting the astronaut drop from orbit onto the planet surface, where his legs would be subject to more acceleration than his head seen from the perspective of an observer on the planet surface, hence the astronaut's body would have counter a force trying to elongate him.

    Just a few thoughts. I did not really think this out or tried to figure out how one would calculate this from the perspective of the astronaut and what kind of mass/size the planet would have to be.
     
    Last edited: Dec 28, 2016
  14. Dec 28, 2016 #13

    Ibix

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    Well, this isn't special relativity any more.

    Given that you are relying on careully chosen tidal forces to do your work, I wouldn't bet on being able to use SR concepts like length contraction - certainly not without careful calculation beforehand.

    I think what you are trying here is a variant on Bell's spaceship paradox. In the "vanilla" version of that, two ships initially at rest and joined by a string undergo equal constant proper accelerations. According to the original rest frame of the ships, the string ought to length contract and break. But in the frame of the ships, why would the string break? It isn't length contracting. The solution is to realise that there is no "frame of the ships" - due to the relativity of simultaneity the two ships aren't at relative rest from either of their perspectives. That was the kind of thing I (and Bell, I think) was warning that you can't handwave away.

    But you could certainly set up a variant where the two ships' accelerations are carefully chosen so that the separation between them (as measured in their initial rest frame) decreases at the same rate as the string length contracts. But that isn't equal acceleration in the initial rest frame and I have my doubts that it amounts to equal acceleration in anything other than a very carefully picked non-inertial frame.

    I'd also suggest that this is far enough off topic to be another thread if you want to continue the discussion.
     
  15. Dec 28, 2016 #14

    BvU

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    With an orbit radius of 6000 au the required centripetal force is pretty much the same for all your atoms.
     
  16. Dec 28, 2016 #15

    PAllen

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    Born rigid acceleration achieves this. Indeed the proper acceleration varies continuously along the rockets and string such that all neighboring distances in MCIF frames remain constant. Obviously, in the initial inertial frame, the rockets get closer together.

    Herglotz-Noether theorem puts severe constraints on Born rigid acceleration, but it is quite generally possible if you exclude rotation.
     
  17. Dec 29, 2016 #16

    Ibix

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    Agreed. But OP was originally talking about the same acceleration being applied to the whole body. The proper acceleration clearly isn't the same along the length, and all I was noting was that I suspect that the only coordinate system where the coordinate acceleration is equal along the length is one constructed to achieve that.
     
  18. Jan 1, 2017 #17

    pervect

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    I believe you're asking about what physicists would call the transverse doppler effect. The short answer is yes, see the wiki_article for more details.

    If the space ship, considered to be small and pointlike (so we don't have to talk about what happens if it's not) is accelerating in a circular orbit and broadcasting a signal (via radio or laser, i.e. some known physical means that sends signals at the speed of light), said signal will be red-shifted when it's received by a stationary observer at the center of the space-ships circular orbit. This means that the received frequency will be lower, and if the space-ship was sending a video signal, the video will play out in slow motion.

    If the receiver isn't at the center of the circular orbit, the velocity won't be transverse anymore, and you'll need a different more complicated calculation which basically involves a full doppler effect calculation. The distance from the receiver to the space-ship will only be constant if the receiver is at the center of the circular orbit.

    Some experimental tests of this have been done (but with atoms, not spaceships), the results are in agreement with special relativity.
     
  19. Jan 2, 2017 #18
    This situation appears analogous to the one in Schwarzschild spacetime, where the far away observer (who is in free-fall) sees the shell observer close to the black hole (who accelerates in his direction) redshifted. If true and assuming constant distance during the measurement of the redshift then an observer who hasn't any additional information couldn't distinguish between the two scenarios: If the source accelerates in a circular orbit with him at the center in flat spacetime or if it has constant r-coordinate in Schwarzschild spacetime,right?
     
  20. Jan 2, 2017 #19

    pervect

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    Given Ashby's analysis of the orbiting clocks in the GPS system, I'd say no. See for instance
    Relativity in the Global Positioning System.

    I'd say more, but I'm pressed for time.
     
  21. Jan 3, 2017 #20
    Although this "spaceship thought experiment" is included in most popular books on Special Relativity (SR) it does not demonstrate time dilation due to relative speed. If the spaceship were approaching Earth the data would arrive faster then once per second due to the decreasing distance the signal has to travel. SR never predicts "time contraction".

    Time dilation is due to light's universal measured speed. (See "Einstein's light clock" for explanation.) If any thought experiment lacks the universal measured speed of light as a necessary condition -- as yours doesn't -- time dilation cannot be assumed. Many books written by "authorities" on SR contain similar thought experiments which are really unrelated to SR.

    To test whether an example of time dilation is legitimate, reverse the situation. If time intervals get shorter it's not a demonstration of light's universal measured speed, but only of light's finite speed. The two are often conflated.
     
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