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- Thread starter Dave Apsey
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A spaceship cannot travel at the speed of light. See the FAQ: [thread]511170[/thread].

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If the speed of light is fixed for both observers then to the observer on Earth it would appear that the ship took say 5 years to reach it's destination. Would it also seem to the traveller that it took 5 years or would it seem to them that the distance had shrunk to zero and so it took no time?

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ghwellsjr

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Although a spacecraft cannot quite achieve the speed of light, it can, forgetting the difficulties, as you say, get as close to it as you want. So here is a formula that you can use to determine what speed you have to travel at to traverse any distance, d, in any time, t:

v = 1 / √[1 + (t/d)²]

Note that the velocity is a fraction of the speed of light and that d and t are in compatible units like light-years and years. Also, be aware that d is the distance for the observer on Earth and t is the time for the spacecraft and it must be greater than zero. No matter how large d is and how small t is (as long as it is not zero), the velocity will always be less than 1.

So although the trip cannot be done instantly, it can be as short as you would like, say a nanosecond. You need to have a calculator with a lot of precision if you want to see what velocity that will take.

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PAllen

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1) You could travel to a star while aging only a minute.

2) While going that fast, the distance to the star would appear to be a little less than one light minute, so you still see the star moving less than c (of course, you consider yourself stationary relative to yourself).

This scenario is verified zillions of times a day: muons produced by cosmic rays high in the atmosphere should decay within some hundreds of meters. Virtually none should reach the ground. In fact, almost all reach the ground.

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Does this mean that when our traveller disembarked at their destination, time for them would pass at the same rate as the observer on Earth? Let's not worry about stars moving apart or the universe expanding for now. Their twin would be 5 years older but they would now both age at the same rate.

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PAllen

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Once they are co-located and not moving fast relative to each other, they age at the same rate. If you ignore gravity (general relativity), you don't even need to worry about co-location, only relative speed. So yes, after disembarking, the traveler would age at the same rate as their earth twin - remaining 5 years younger (for example).

Does this mean that when our traveller disembarked at their destination, time for them would pass at the same rate as the observer on Earth? Let's not worry about stars moving apart or the universe expanding for now. Their twin would be 5 years older but they would now both age at the same rate.

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ghwellsjr

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Yes, they would now both age at the same rate and all frames will agree, but it is only in their common rest frame that we can say the traveling twin would be 5 years younger than the earth twin. Other frames would not agree. It is only when the traveling twin makes a quick return trip while the earth twin ages another 5 years that all frames will agree on their age difference.

Does this mean that when our traveller disembarked at their destination, time for them would pass at the same rate as the observer on Earth? Let's not worry about stars moving apart or the universe expanding for now. Their twin would be 5 years older but they would now both age at the same rate.

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PAllen

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It mathematically proven that entanglement cannot be used to send messages.

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PAllen

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Also, though relativity per se sets no limit on rocket speed, there are very strong reasons to doubt it will ever be achieved. For example, if you are traveling at a mere 90% of lightspeed (which is not enough for dramatic time dilation), hitting 1 gram particle will release energy greater than the Nagasaki atom bomb. It is really hart to see how effective shielding could exist for near light speed (where a speck of dust could be more dangerous than all the H-bombs in the world).

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Plus if the experiments prove that actions on one of a tangled pair affects the other then they must have observed the effect on the other. In other words they have passed on information. It's just a matter of encoding it surely?

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PAllen

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Speed of light limit and its consequences are experimentally established (to a high degree), not mathematical facts. OPERA result, if confirmed, would be new evidence requiring theory modification. Your question was asked in light of current theory and knowledge.

Plus if the experiments prove that actions on one of a tangled pair affects the other then they must have observed the effect on the other. In other words they have passed on information. It's just a matter of encoding it surely?

The way quantum entanglement works is that no prior agreement plus local observations can distinguish an entangled particle from an ordinary particle - the results will be random. Only on receiving (by some other means - thus light speed limited) the results of the other party's measurements do you detect correlation that can only be explained by entanglement.

If you want to play the game of 'what if the world is different' this is not the proper forum for that.

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Could you show the calculations or a reference?if you are traveling at a mere 90% of lightspeed (which is not enough for dramatic time dilation), hitting 1 gram particle will release energy greater than the Nagasaki atom bomb.

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PAllen

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The calculation is simple. Using the relativistic KE formula, at around .86c the KE = mc^2 (that is, KE = rest energy). Then compare m c^2 in ergs (for example) to the Nagasaki bomb in ergs (available, for example, from wikipedia). I did it once when I was kid, and once in the last year to ensure I hadn't made a mistake. I remember this fact as as useful order of magnitude to keep in mind.Could you show the calculations or a reference?

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I am sure it is, so why not work out this example, I am sure people will appreciate it. :)The calculation is simple.

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PAllen

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This is really, really silly. I thoroughly outlined the computation. However, to humor a silly request:I am sure it is, so why not work out this example, I am sure people will appreciate it. :)

1) For v=.9c, gamma = 1/sqrt(1 - v^2/c^2) = 2.294

2) This implies KE = 1.294 m c^2. For energy in ergs, use grams, and cm/sec. Thus for 1 gram, KE will then be: 1.165 * 10^21 ergs.

3) The Nagasaki bomb was about 21 kilotons or about 88 terajoules (http://en.wikipedia.org/wiki/Nagasaki_bomb#Nagasaki). A joule is 10^7 erg. Thus 8.8 * 10^20 erg.

4) 1 gm at .9c has KE of 1.32 Nagasaki bombs.

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PAllen

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