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- Thread starter neh4pres
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Fredrik

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It can (in principle) reach any speed below c.what is to top speed that this rocket can achieve as seen by a stationary observer?

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, exactlyAlso 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.?

I don't understand this question.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

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(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.

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You did not write this, so why did you not give proper attribution?In the rocket, you can make measurements of the world around you. ...

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" [Broken].

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.

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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)

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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 dont understand what you mean by "they moved half the distance"

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Fredrik

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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.I dont understand what you mean by "they moved half the distance"

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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.

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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 traveling 100 mph would eventually catch something traveling 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.

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.

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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.

Thank you. that is exactly my point. If anyone figures this out i would like to know the answer.

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DrGreg

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The equation of the rocket (the green line) isPlease 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.

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

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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.

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:

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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.Thank you. that is exactly my point. If anyone figures this out i would like to know the answer.

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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.

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No. The signal is sent away from the rocket, and the rocket always travels with some v<c, so the distance between the rocket and the signal is always strictly increasing.I added a brown signal. Will this signal ever reach the rocket?

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The signals sent from the rocket (curved green wordline) to the Earth (vertical brown worldline in the diagram) are represented by the blue arrows in the diagram and always arrive, so the rocket is always visable from the Earth. Signals sent from Earth before time t2 in the diagram (black arrows) catch up with the rocket. Signals sent after time t2 (red arrows) never catch up with the rocket as long as the rocket maintains constant proper acceleration.

[EDIT] I just noticed you mentioned the rocket is not emitting light, so if the only illumination of the rocket is light reflected from the Sun then it would not be visible after time t2. The rocket would also be invisible to radar signals from Earth after time t2 but the rocket would also have to maintain radio silence if it wants to stay in stealth mode ;)

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very nice graph kev

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