
#19
Apr1211, 10:56 AM

P: 96

So are you saying the traveler witnesses a massive warping of time on the colony during his decceleration? A warping that makes him go from being older than a colonist born in the same (Earth/colony year) as him to being younger than that colonist? Because that strikes me as really odd. 



#20
Apr1211, 12:10 PM

P: 250

The traveler will CONCLUDE that his hometwin ages suddenly during his deceleration, but the AMOUNT of that ageswing depends on his distance (according to her) from his home twin during his deceleration. If his deceleration starts when he is fairly close to the colony, he won't conclude that the people in the colony age much during his deceleration. In the simple, idealized case where his velocity changes are instantaneous, he will conclude that the colonists don't age at all during his deceleration, because his separation from them is zero then. Here is a posting that specifies how separation enters into the disagreements about how ages correspond: http://www.physicsforums.com/showpos...06&postcount=7 . (That posting refers to the standard twin "paradox" scenario, but you can probably see how to apply it to the case where the traveler remains at the turnaround location). Mike Fontenot 



#21
Apr1211, 12:28 PM

P: 848

I haven't really changed anything, just chose a different reference frame (I should have had the earth flying off to the left instead of putting the stayathome twin in a red rocketbut you get the point). The blue twin is still taking an overall shorter path through spacetime (his short return path) and therefore ages less. 



#22
Apr1211, 12:58 PM

P: 52

There are two parts to the paradox, assuming only these two frames of reference. The first part is from the perspective of "relative velocity", meaning each twin will see the other as moving. They will both see the other's clock as running slow, or fast depending on the direction of the relative motion. The second part has to do with how the velocity of an object affects length contraction and time dilation. This second part can only be separated from the first by "knowing" which observer is in motion and which is at rest. There are two ways to determine which is in motion. The first requires that the traveling twin return to the earth at some point so that the two clocks can be compared. The second is to determine which twin experienced acceleration. The twin on earth has a hypothetical velocity = 0, while the traveling twin's velocity > 0. The paradox is that both twins observe the other as moving, length contracted and time dilated and their self as unaffected. Only by introducing some method of knowing which is in motion can it be determined which is length contracted and time dilated. Adding a third frame of reference, the colony, complicates this thought experiment. While you can assume that their is no relative difference in velocity between the earth and the colony and that they can have their clocks synchronized, each represents a different perspective of the ship. While the ship is in motion the observers on the earth and on the colony will not agree, as to their observation of the ship. One will see the ship as time running slow, while the other sees time running fast. The observer on the ship will see the same thing as each observer when viewing that observer. When the ship stops at the colony, the clock on the ship can be compared to the colony clock. At that point the ship's clock will be found to have run slow. If you could communicate those results back to the earth, the earth would also see the ships clock as having run slow compared to the colony clock and it's own, when taking the light time difference between the earth and the colony into consideration. In both of these thought experiments, acceleration and deceleration only help to determine which observer was in real motion within the hypothetical. Comparing clocks and knowing which is accelerates are just two ways to determine which observe is or was in motion in the hypothetical. 



#23
Apr1211, 01:19 PM

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#24
Apr1211, 01:19 PM

P: 96

I can see there are many ways to resolve the paradox, I'm almost there myself, I just need the answer to my question in post #18 (my attempt to make both frames of references as equal as possible).




#25
Apr1211, 02:16 PM

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#26
Apr1211, 02:31 PM

P: 96

Can you tell me what went wrong with my calculation in post #18? 



#27
Apr1211, 02:35 PM

P: 52

Below is a link to a calculator for length contraction, time dilation and relativistic mass. Here is a quote from the explanation of the first calculator (for length contraction of a moving object), Measurements of distance is between two points and is not time dependent. Any individual measurement of distance is between two points. If there is relative motion between two observers two consecutive measurements of the distance, separated by a given time, can provide the relative velocity. Time will be experienced by all observers as constant and uniform. Only by comparing once synchronized clocks will a difference be apparent. Two observers will "see" each other equally time dilated and length contracted. 



#28
Apr1211, 02:49 PM

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#29
Apr1211, 02:53 PM

P: 96

For example: Imagine the speed of the ship is 0.9 c and the distance as seen from Earth is 10 lightyears, then someone on Earth (or at the end point) would expect the ship to make the journey in 11.111 years. For v=0.9 gamma is 2.294, so to someone aboard the ship the journey would take only 4.84 years. This means that he would be going faster than light, unless he sees the distance contracted to 4.36 lightyears, than he would conclude he travels at 0.9, like he should. Distance contraction going hand in hand with time dilation actually makes sense to me, even at an intuitive level (as far as that's possible with relativity), because it ensures the speed of light will be the same to everyone (which is one of Einstein's postulates). So I'm pretty sure that part of my post #18 is correct, but there has to be a flaw somewhere else. 



#30
Apr1211, 03:23 PM

P: 101

Gulli, first you need to give up thinking of the twinparadox age difference as resulting from a turnaround acceleration. It’s possible to reform the problem, leaving out the “acceleration” phase, and still get an age difference when the twins meet after initially being the same age. You need to accept that this is how our universe’s space and time are structured.
You likely accept without question the fact that R2 = X2 + Y2 leaves R unchanged under a coordinateframe rotation of (X, Y). [The 2's are superscripts.] You need also to accept that S2 = (cT)2 – X2 leaves S unchanged under a boost (i.e., a coordinateframe velocity change in the X direction). This universal behavior of time (you may be aware from your study of SR) is a result of there being a universal upper limit, c, to relative velocity. That’s really it in a nutshell. The rest is just dressing to help you accept it. My own favorite description of why time behaves so differently from our expectation is shown in each twin’s plot of the other twin’s clock during the entire trip, from departure to return. Each twin is receiving a televised image of the other twin’s clock. The imageclock time is plotted (on the yaxis) against the receiver’sclock time (on the xaxis). Do this for both twins (on the same plotpaper) and you will see why the clock times at return CANNOT be the same. 



#31
Apr1211, 03:32 PM

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One handy formula to know when thinking about the relativity of simultaneity is this: if two clocks are synchronized and a distance D apart in their own rest frame, then that means that in the frame of an observer who sees the clocks moving at speed v (parallel to the axis between them), then at a single moment in the observer's frame the clock at the rear will show a time that's ahead of the clock at the front by an amount vD/c^2. So in this example, since the Earth clock and the colony clock are synchronized and a distance of 20 lightyears apart in their own frame, then for the spaceman who sees them moving at 0.5c, at any given moment in his frame the two clocks must be outofsync by (0.5c)(20 lightyears)/c^2 = 10 years, with the rear colony clock ahead of the front Earth clock by that amount. So, at the same moment the Earth clock reads 600,000 AD, the colony clock must read 600,010 AD, according to the definition of simultaneity in the spaceman's rest frame. 



#32
Apr1211, 03:39 PM

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#33
Apr1211, 03:58 PM

P: 101

JesseM:
Start with the twins separated, with a closing velocity, and with their clocks synchronized by twinA to the same reading. When they meet twinB’s clock will read less than twinA’s. If twinB does the synchronization, then twinA’s clock will read less than twinB’s. 



#34
Apr1211, 04:05 PM

P: 96

@JesseM
So there is an additional boost of 10 years that fixes everything. Am I correct in saying that the invariance (the fact that a similar boost doesn't manifest when we consider the colony stationary, and well, the whole reason the colony and the spaceman experience different time intervals) stems from the fact that we are working with a starting point and ending point which share the same frame of reference, namely that of one of the observers (the one on the colony)? The 10 year boost then is something the spaceman notices because of the relativity of simultaneity: he's closer to the starting point, Earth, than he is to the colony when he sets out, so he sees a time difference between Earth and the colony, even though Earth and the colony share the same frame of reference and someone located halfway between them would say Earth and the colony have the same time, to which the people of Earth and the colony would agree. Am I right? 



#35
Apr1211, 04:08 PM

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#36
Apr1211, 04:11 PM

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Anyway, if you want a totally symmetrical situation, imagine that behind the spaceman is a second spaceman #2 traveling at the same velocity relative to the Earth/station, and whose distance from spaceman #1 is 20 lightyears in the spacemens' rest frame and 17.32 lightyears in the Earth/station frame. In that case, assuming the clocks of the spaceman are synchronized in their own rest frame, in the Earth/station frame the clock of spaceman #2 will read 600,010 AD at the same moment the clock of spaceman #1 reads 600,000 AD (also the moment he passes Earth). And to complete the symmetry, when spaceman #2 passes the Earth, spaceman #2's clock reads 600,040 AD while Earth's clock reads 600,034.64 AD, just like how when spaceman #1 passes the station, the station's clock reads 600,040 AD while spaceman #1's clock reads 600,034.64 AD. 


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