Observations about the Familiar Example of Relativity Slowing Time

I'm now reading the recent Scientific American issue about Einstein. One article repeats a version of the familiar example of someone aging more slowly while traveling at very high speed. I'll just quote their example:

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If Krikalev (a Russian cosmonaut) left Earth in 2015 and made a round-trip to Betelgeuse - a star that is about 520 light-years from Earth - at 99.995 percent the speed of light, by the time he returned to Earth he would be only 10 years older. Sadly, everyone he know would be long dead because 1,000 years would have passed on Earth; it would be the year 3015.

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There's a couple minor inaccuracies in the example that I'll note but ignore. First, according to Wikipedia Betelgeuse is 643146 light-years away. 520 light years is within the error range, so that's fine, but if Betelgeuse *is* 520 light-years away, then it would take more than 1,000 years to travel there and back. It would take at least 1,040 years at 100% the speed of light, allowing no time for acceleration or deceleration. As I said, I'll ignore these minor innacuracies.It's the acceleration at the beginning of the journey that I first want to focus on. How long it would take Krikalev to accelerate to 99.995% the speed of light at 1g? I decided to use Newtonian physics just to get a very rough idea, and also because I couldn't solve the problem using Einsteinian equations. So we start with this very simple equation: Solving for t, and since V_i is 0, we get: Plugging in 300,000,000 m/s (the speed of light, close enough to 99.995%) for V_f and 9.8 m/s² for a we get 135,848,167 seconds or about 4.3 years (I'm going to round to 4 going forward). Not bad! Anyone know the Einsteinian answer?But if we assume it takes 4 years for acceleration and deceleration, then that's 4 years at the beginning of the flight, then 8 years at Betelgeuse (because he has to decelerate for 4 years then turn around and accelerate back toward Earth for 4 years), then decelerate again as he nears Earth, for a total of 16 years.Of course, that's our time, not his time, and it doesn't take relativity into account, so I next wondered how long Krikalev thinks it takes him to accelerate to 99.995% the speed of light. Anyone know?The last thing I wondered about, and it's something I've wondered about before, is what Krikalev would observe about the passage of time on Earth, and how it would differ from what Earth observes about the passage of time of Krikalev. If you ignore the accelerations and decelerations, like the Scientific American article, then as Krikalev travels out to Betelguese he observes clocks on Earth slowing to around 1/100th of the normal rate. But as people on Earth watch Krikalev, they observe his clocks slowing to around 1/100th of the normal rate. They both see each other's clocks slowing by the same amount.And after Krikalev turns around to return to Earth the observed slowing of both their clocks will continue. When Krikalev finally returns to Earth, he will have observed Earth clocks ticking away at 1/100th the normal rate during the entire journey, and Earth will have observed Krikalev's clocks ticking away at 1/100th the normal rate during the entire journey. What will Krikalev actually observe when he exits his spaceship? Will all his friends be there to greet him? Or will they be long dead?I know the standard answer is that they'll be long dead, but in that case it isn't possible that they both observed each other's clocks slowing by the same amount. What did each actually see, and why?--PercyEdited by Percy, : 9.8 m/s => 9.8 m/s²

Maybe this is a stupid question that misses the essential point, but how would either observe what was happening to the other?

Yes, they would both see each other's clocks appear to slow down by the same amount at least during the coasting part of the trip. I think it is the same as the Newtonian answer. However Einstein and Newton would disagree on the energy required to maintain a 1g acceleration up to .999995c.For the purpose of computing the time travel, given the 1000 year journey, I think ignoring the time to accelerate is reasonable since your calcs show about 4 years of acceleration. However the acceleration and deceleration cannot be completely ignored as those acceleratons are what create the asymmetry that resolves the twin paradox.During the acceleration and deceleration, the traveling astronaut experiences a time dilation effect that is not experienced by the stationary astronaut based on the equivalence principle between gravity and acceleration. It is entirely during these periods that the difference in aging can be accounted for. Although I believe it is possible to show the difference using special relativity, using GR to equate the acceleration to a gravitational field and then using the appropriate time dilation effect for the gravitational field produces the proper aging effect and even explains why the effect is proportional to the length of the trip.If it were possible to observe the clocks during the acceleration period. The observers would agree on who was accelerating and experiencing the "gravitational effects" and who was not and would agree on the differences in clock rate although their own clocks would of course seem normal. They would agree that the accelerated clock was running slower than the constant velocity clock.I vaguely remember a discussion with cavediver during which the twin paradox is discussed. Cavediver indicated that there was a way to resolve the twin paradox using special relativity alone. I don't recall if we got into the details, but the key is the acceleration and deceleration periods. Those periods must exist if the astronaut is to return home.There is also a fairly long debate between myself and a fairly stubborn creationist about special relativity that includes a discussion of the twin paradox, but it might make your eyes bleed to read that. Good question. I don't know if this can be calculated using special relativity.Edited by NoNukes, : No reason given.

Not literally observe, but imagine that they agree to transmit clock signals between themselves at some agreed upon interval, say one pulse per second. Upon receipt we should be able to determine the speed of the clock of the transmitter.

I mean to look into this again but rather than wait I'll note something that I think I remember.One not intuitive thing that I might remember correctly is that the unaccelerated path through space-time is the longest. I am struggling to remember how to show this and can't so I'll have to look it up.That is the reason why the earth bound twin is older at the most basic level.Now were are the physicists that we need?

The two situations are not symmetrical. The outbound twin experiences two different coordinate systems, the stay-at-home only one.Think of rocket A passing the Earth and the pilot observes the stay-at-home's clock and tracks it on board. That's one coordinate system. When she gets to 643 light-years out rocket B passes her inbound and transfers her current estimate of the stay-at-home's clock to its pilot. Coordinate system 2. When B passes Earth the pilot sees gobs of difference between his clock and the stay-at-home's.There's a pretty good animation at The twin paradox: Is the symmetry of time dilation paradoxical?

Of course B sees that. The problem is that the earth bound clock sees the same descrepancy (running slow) for both A and B clocks as A and B see with the earthbound clock. I don't understand how your explanation resolves the paradox. How does introducing rocket B help with anything? The direction of travel does not affect the fact that each observe thinks the other clock is running slow by the same amount.

Look at the animation in the link.

What I read is that the earth bound and rocket ship rider see symmetrical slowing of each others clock on the outbound part of the journey and see slowing of each others clock on the inbound journey but on the inbound one it is not a symmetrical slowing.I don't get why though.

I just wanted to register my appreciation for the posts to this thread. I haven't responded because I need to set aside a little time to digest the information.--Percy

The slowing of each clock as reported by the other observer is symmetrical on the inbound trip. That's why there appears to be a paradox. Edited by NoNukes, : No reason given.

I think this observation, made in your previous post is correct. It is the key to resolving the paradox while using special relativity alone.On the other hand, I think some of the stuff on the site you linked to is either not correct or misleading. In particular there is a bunch of stuff about the frequency of receiving incoming clock ticks while moving away or moving towards that is just not relevant. The truth is that relativity is not about those kinds of frequency shifting effects just as it is not about the delay in making observations due to light travel time. We should assume that the receivers know about those things and account for them. The strange effects from SR are those which remain after we've accounted for light travel time.Any explanation that would work the same way with sound waves and the speed of sound as it does with light waves and the speed of light is just wrong.

Distance through space time:
s² = (ct)² - x²
Where t is the time dimenstion and x is the (for now 1 dim) space dimension.One rule is that all observers must measure the same spacetime distance.This applies for special relativity only and that is important to note.Thus the longest distance is when the x component is zero. So if we measure from the point of view relative to the twin on earth who stays still in space with a zero x component we see that the twin on earth has taken the longest distance through space time to get to the same place on earth when the traveling twin comes home.If we try to calculate from the point of view of the traveling twin we can not use this formula because the traveling twin spends part of their time in a non-inertial reference frame.

This is actually the formula for the proper time (Assuming one dimensional situation at constant velocity. Or rather the longest proper time. with A 'c' multiplier. Yes you can! If you use the formula for both parts of the journey you can show that the time experienced by the accelerating twin is shorter than the time experienced by the stay at home twin. The proper time is the time experienced a person in frame. The proper time will be longer for the twin who does not travel at all. (Hint: Remember that for this problem, both the starting planet and the destination planet are in the same reference frame)Edited by NoNukes, : No reason given.Edited by NoNukes, : No reason given.Edited by NoNukes, : No reason given.