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Relitivistic rocket

  1. Jun 2, 2008 #1
    what is to top speed that this rocket can achieve as seen by a stationary observer? Also if you look at the distance covered by the rocket, lets say 10 ly in its own frame is covered in 5 years.... then anyone in the rockets frame has acually gone faster than the speed of light.?right.? one more question. the horizon would not be at a fixed point at the back of the ship would it.. i would think that it would be at an angle of 90 degrees to the viewer.. so if you had a glass wall on the ship and u walked from the back of the ship to the front would it not follow you
  2. jcsd
  3. Jun 2, 2008 #2


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    It can (in principle) reach any speed below c.

    No. They never had a speed greater than c in any frame, but I see what you mean. When they get there, they will have aged 5 years and they are 10 light-years from home, exactly as if they had moved at twice the speed of light. But what really happened from their point of view is not that they moved twice as fast. They just moved half the distance.

    I don't understand this question.
  4. Jun 2, 2008 #3
    In the rocket, you can make measurements of the world around you. One thing you might do is ask how the distance to an interesting star you are headed towards changes with T, the time on your clock. At blast-off (t=T=0) the rocket is at rest, so this distance initially equals the distance D to the star in the non-accelerating frame. But once you are moving, however you choose to measure this distance, it will be reduced by your current distance d travelled in the non-accelerating frame, as well as the whole lot contracted by a factor of γ, your Lorentz factor at time T. Eventually you will pass the star and it will recede behind you. The distance you measure to it at time T is

    (D - d)/γ = (D + c2/a)/ch(aT/c) - c2/a

    A plot of this distance as a function of T shows that, as expected, it starts at D, then reduces to zero as you pass the star. Then it becomes negative as the star moves behind you. As T goes to infinity, the distance asymptotes to a value of -c2/a. That means that everything in the universe is falling "below" the rocket, but never receding any further than a distance of -c2/a as measured by you. It all piles up just short of this distance, asymptoting to a plane called a horizon. You see this horizon actually form as the rocket accelerates, because there comes a time when no signal emitted from "below" the horizon can ever reach you. Everything falls toward that plane, and as it does so it begins to redden, due to the increasing red shift of its light, because you are accelerating. Finally it fades out of visibility. In fact, as anything gets closer to the horizon, it ages more and more slowly; time comes to a complete halt there. The horizon is a dark plane that appears to be swallowing everything in the universe! But of course, nothing strange is noticed by the non-accelerating Earth observers. There is no horizon anywhere for them.
  5. Jun 2, 2008 #4
    it also says that at some point no signal from behind you can ever reach you.. This makes it sound as if you are out running the signal. however from the stationary frame (earth) they will see the signal catch you correct?
  6. Jun 2, 2008 #5

    D H

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    You did not write this, so why did you not give proper attribution?

    The source of this is the Usenet Physics FAQ article on the relativistic rocket, http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html.

    neh4pres, I am getting a strong impression from this thread and from others that you are quite young, and, to be blunt, are not as well educated in physics as you pretend to be. Before you go delving into relativity I think it would be best if you delved into the underpinnings of relativity first.
  7. Jun 2, 2008 #6
    hmm... well sorry about that.. the guy above me did not understand my initial question so i pasted the column where my question derived from. Also ill have you know that i do not pretend to be knowledgeable or educated in special relativity. if you have read my earlier postings i do admit that i have no education. However i do have a big interest in cosmology and GR. And to be quite blunt i dont think my age or lack of proper education prevents me from understanding physics or GR nor should it prevent me from putting my questions out there and searching for answers.. thank you
  8. Jun 3, 2008 #7
    Nope. If the signal is emitted too late or from too far behind it will never catch up with rocket in either the rocket or Earth frame. Have a look at the attached drawing. the curved green line is the wordline of a rocket with constant proper acceleration. The black arrow represents a signal sent at the same time as the rocket launches but from too far back. The blue arrow represents a signal sent from the same position as the rocket launched but at a later time. neither signal will catch up with the accelerating rocket. In fact any signal sent from above or to the left of the dotted line will catch up. The the observers on the rocket the dotted line is effectively an event horizon. The red arrow represents an observer that has fallen off the accelerating rocket. He continues with the velocity he had when he was last on the rocket. When he passes the dashed line the observers on the rocket can no longer see him or receive radio signals from him, so to the observers on the rocket it looks like the fallen observer has fallen into a black hole. If you are not familiar with these type of diagrams, the vertical axis represents time and the horizontal axis represents distance. (I should have labeled them)

    Attached Files:

    Last edited: Jun 3, 2008
  9. Jun 3, 2008 #8
    thank you kev.. i understand the graph and everything you said. However i find it very hard to fathom. It seems that if something is traveling at 100mph it would someday catch up to something traveling 99.9 mph. It would also seem that the green line could never be parallel with that of the signals because it would need to cover the same distance over time. This means it would have to be traveling AT the speed of light correct?
  10. Jun 3, 2008 #9
    I dont understand what you mean by "they moved half the distance"
  11. Jun 3, 2008 #10


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    They're moving at the speed [itex]\sqrt{3}c/2[/itex] relative to the endpoints of their path in space, which means that in their rest frame, the distance is Lorentz contracted by a factor of 2, so they only have to move 5 light-years instead of 10.
  12. Jun 3, 2008 #11


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    kev is correct here. If you look at the graph you can see that the accelerating observer follows a path that looks like a hyperbola. As you know, hyperbolas have asymptotes which are lines that they get arbitrarily close to, but never cross. In the case of an accelerating observer the asymptote is a line at the speed c. Light from any event on the other side of that asymptote will never "catch" the accelerating rocket.

    In your post you mention that something travelling 100 mph would eventually catch something travelling 99.9 mph. That is correct, but does not apply here. The rocket is not going at a steady .999c (in which case there would be no event horizon), but it is constantly accelerating, .999c now, .9999c soon, .99999c after that. In that case the light on the other side of the asymptote does not catch up.

    Also, since all frames agree on c all frames agree on the asymptote, so all frames agree which events can send signals that will reach the accelerating rocket and which can not.
    Last edited: Jun 3, 2008
  13. Jun 3, 2008 #12

    Please correct me if I am wrong, but I believe neh4pres is referring to a stationary frame (ie the frame that sent the signal.) It does seem very odd that in this frame the rocket never reaches the speed C but the signal is traveling at C so it seems that from this frame it would eventually catch up before infinity (it might be a really long time though). If I had the math skills I would set up the an equation/graph to find the distance between the asymptote and the hyperbola and see if it was possible or not.
  14. Jun 3, 2008 #13
    Thank you. that is exactly my point. If anyone figures this out i would like to know the answer.
  15. Jun 3, 2008 #14


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    The equation of the rocket (the green line) is

    [tex]x_r^2 - c^2 t^2 = c^4 / a^2 [/tex]​

    The equation of the photon (the dotted pink line) is

    [tex]x_p^2 - c^2 t^2 = 0 [/tex]​

    From which it follows that

    [tex]x_r > x_p [/tex]​

    for all values of [itex]t[/itex].

    Reference: The Relativistic Rocket
  16. Jun 3, 2008 #15
    It is not too difficult to do the math and prove that the light signal will never catch the accelerating rocket if the rocket has sufficient head start, but it still does not give an intuitive sense of why that is so. It is a bit like Zeno's paradox of the Achilles and the tortoise except this time the Achilles really can not ever catch up with the tortoise. :tongue:
  17. Jun 3, 2008 #16
    Ok. so here is a new question. if no light can catch the accelerating rocket, and the rocket is not emitting any light. Then as long as the rocket continues to accelerate it will be completely inviable to the stationary frame earth. correct?
  18. Jun 3, 2008 #17


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    It has been figured out and already answered. Perhaps you should read up on asymptotes. They come up in so many different situations that you really should understand them.
  19. Jun 3, 2008 #18


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    Light can catch the rocket depending on when and where it was emitted. If it is on the "near side" of the asymptote it will catch the rocket, if it is on the far side it will not. The earth could therefore recieve reflected light from any event on the near side of the asymptote.
  20. Jun 4, 2008 #19
    Hey I am trying to make sure I understand these diagrams. I added a brown signal. Will this signal ever reach the rocket? I assume that the signal is received when the rocket line and The ship line cross.

    Attached Files:

  21. Jun 4, 2008 #20
    i wouldent believe so becouse where you put it says you sent the signal before the rocket left, therefore the rocket would have to catch up to the signal.
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