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(Sorry, I wall o' text. For ease of skipping to the question Blue text is a remedial preface. Orange text is the explanation someone else gave for the problem in the blue text. Black text is my question about the orange explanation of the blue problem.)
I saw an explanation of the twin paradox online that tried to explain the twin paradox. I understand the theory that relative motion dilates time. Hense, a twin travels at .8c to an object 5 lightyears away. From their perspective about 15 years has passed, but when they come back to earth, everyone else has experienced closer to 25 years. The problem comes from the notion that all frames of reference are equally valid. As two objects pass each other, each will experience their time as normal and see the other person's clock running slow. Thus, if the twin that leaves could see the clock of the twin on earth the whole time, the twin that left would come back and everyone on earth will have experienced less time, thus would be younger. So, when the two are reunited... how does each see the other as nearly a decade younger than themselves?
They offer that time dilation is asymetrical due to the asymetrical nature of the trip. They explain that if each brother sends out pings every second. Since the two are traveling apart, each will experience pings from the other party as coming in slower than once per second. But, when the ship turns around, it will receive pings faster than one per second, while the earth still receives pings at a rate slower than one per second. Thus, time is subjective to the same extent that they each experience different number of pings per second.
Now, we'll ignore the G-force dyssemetry for now, since that's a whole different set of problems and this is tangled enough as it is, but... there seems to be a serious flaw in this reasoning in that they changed how they treat those pings half way through the trip.
If we assume these pings travel at a constant speed relative to a medium, then yes, the ship will receive pings more slowly as it travels away from the earth to the same exact extent that it will receive pings faster as it travels back toward the earth. The earth will receive pings at a normal rate throughout the journy. They exactly even out.
If, instead, we assume the pings travel at a speed constant to their source... we get the same exact result with a different explanation. Pings coming from the earth will put along at normal speed. The ship will receive them red-shifted as it travels away and blue-shifted to the same extent on the way back. Yes, it'll get more pings on the way back because the pings have to catch up to the shift... but that doesn't mean "more time has passed" ... it's just lag. The difference of having pings move common to their source rather than a medium is that the earth will also receive red-shifted pings as the ship travels away and blueshifted pings as the shift travels toward it. Again, evening out exactly.
There's only a discrepant number of pings if you change the rules halfway through the trip.
Further, if Asymmetry is the problem... remove that variable. Assume they're triplets. One person stays home, the other two take the same journy in opposite directions at a speed of about .8c for 5 lightyears and back. How old does each person see the other ones when they all come back? (For ease of explanation, Say Bob stays home, Al goes to left and Bert goes right).

Seems that Bob will see Al and Bert as the same (younger) age. But Al and Bert both see the other one as significantly younger than themselves (and younger than Bob sees either of them).
Edited by Didymus

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One person stays home, the other two take the same journy in opposite directions at a speed of about .8c for 5 lightyears and back. How old does each person see the other ones when they all come back? (For ease of explanation, Say Bob stays home, Al goes to left and Bert goes right).

It doesn't matter which direction Al and Bert go, what matters is the relative velocity of everyone. If Bob is in the inertial frame, i.e. non-accelerated, he will see both Al and Bert as younger. Al and Bert will see each other as the same age, and see Bob as older.

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From Bob's perspective, yes.

But when as Al travels, it's Bob's clock that runs slower. Bert's clock runs even slower as his speed relative to Al is so much closer to the speed of light. So how do Al and Bert see eachother from their frames of reference?

Note, avoid general relativity's affect on time dilation by assuming that they're all on space ships in negligable gravity and Al and Bert's ships have the technology to accelerate/decelerate very quickly without smashing the occupants. After Al and Bert get to speed, they are no longer accelerating, so still count as an inertial frame just like Bob.

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Skipping the perceptual aspects, if Bob, Al, and Bert are all reunited after these trips, or absences from one another, what age will a doctor assign to each of them from a medical examination at that point?

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After Al and Bert get to speed, they are no longer accelerating, so still count as an inertial frame just like Bob.

Yes, once they stop accelerating, they all occupy an inertial frame. But up till then, Al and Bert are NOT in an inertial frame, and that's what establishes the difference between them and Bob. Further, to reunite so that they can compare clocks in the same inertial frame, at least two of them must again undergo acceleration.

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So, to eliminate variables, assume Al and Bert are in space ships that accelerate and decelerate instantly without killing anyone. Acceleration is another topic... We're going with straight special relativity time dilation based on relative motion at different inertial frames. If it makes you feel better, assume they start accelerating on the other side of the ship and Al and Bert simply begin their journies by passing the space station simultaneously.

How will Al and Bert compare their clocks with eachothers and Bobs?

Edited by Didymus
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No matter how quickly or for how long they accelerate, they still leave the inertial frame of reference for an accelerated frame, and this establishes the time differential between the three. Al and Bert's clocks will read the same, Bob's clock will read faster.

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While this tangent is worth discussing, I feel like others may disagree with your assertion.

Are you suggesting that if two objects are traveling past eachother at nearly the speed of light, they'll each see the other's clock running more slowly than their own clocks. yet, when they meet up, their clocks will be synched again as long as their accelerations were identical?

Do you suggest that neither clock is actually ever running slower, that they just appear to be doing so from the perspective of the other person? Or do you suggest that each is actually running slower than the other, but that they speed back up when the two decelerate?

Or am I misunderstanding you?

Edited by Didymus
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Are you suggesting that if two objects are traveling past eachother at nearly the speed of light, they'll each see the other's clock running more slowly than their own clocks. yet, when they meet up, their clocks will be synched again as long as their accelerations were identical?

As long as they keep all accelerations symmetrical, yes.

Or do you suggest that each is actually running slower than the other, but that they speed back up when the two decelerate?

It's not how fast or slow the clock runs, it's the rate that time passes.

Edited by ACG52
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So if two people are moving relative to eachother, yet watching the other one's clock the whole time, each will see the other's clock moving at a slower rate and get farther and farther behind their own clock. Say they travel like this for a year and one clock is a month behind the other clock.

If, when they come to the same inertial frame, their clocks are once again synched... How do you suppose that works? Does do you believe deceleration will make the clock on the opposite ship seem to jump forward however much is needed to catch up to the normal time?

How would this change if they each accelerated to .8c for 10 seconds then decelerated vs. Staying at that pace for 10 years... Seeing the other clock fall farther behind the whole time.... And then catching up all at once?

Or, are you suggesting that when they stop, the rates will cat hbup to one another, but each pilot will still see the other clock as reading a different time according to how far behind that clock appeared to be from their perspective before they decelerated?

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Or, are you suggesting that when they stop, the rates will cat hbup to one another, but each pilot will still see the other clock as reading a different time according to how far behind that clock appeared to be from their perspective before they decelerated?

Yes. The rate is what is the same. Unless all acceleration was symmetrical, in which case the clocks will read the same.

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How would this change if they each accelerated to .8c for 10 seconds then decelerated vs. Staying at that pace for 10 years... Seeing the other clock fall farther behind the whole time.... And then catching up all at once?

The clocks don't *appear* to be ticking slow when the two are approaching, due to the diminishing delay of light. See http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler for explanation of what they see.

They both see the other's clock appearing to tick very slow while receding, and relatively fast while approaching. The twin that turns around sees each for the same duration. The inertial twin sees the receding phase appear to take longer due to the delay of the image of the other's turnaround.

The inertial twin measures the other's clock as slow the whole time. When the traveling twin turns around, there is a change in relative simultaneity for it that result in a "leap forward" of the inertial twin's clock. HOWEVER this leap is not seen, because of delay of light and a corresponding change in the time it takes for images of the other twin to arrive. The traveling twin *measures* the other clock as ticking slowly and getting behind, then leaping ahead of its own clock as it turns around, then ticking slow on the return trip. But it *sees* the other clock tick even slower on the trip out, sees nothing immediately happen during turnaround, and the other clock appears to tick faster on the way back.

I didn't quite address the details of your question, but no one would see the other's clock suddenly leap forward in any case of slow or fast or instant acceleration. The leap is a change of relative simultaneity that corresponds with a change in inertial frame, but it is not immediately visible.

Edited by md65536
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So then the net result of time dilation is a function of acceleration rather than time spent moving at a certain velocity?

If so, how do you differentiate between acceleration and deceleration?

I.e. for the sake of direction, imagine a 2D map of a planet sitting to the south of a neutron star. You accelerate (say at .5c) toward the north to get a closer look from within the star's gravity well. Once you hit your speed, you just coast the rest of the way on a trajectory that will send you just close enough to curve your path close enough to your planet (since this isn't a gravitational force slingshoting you, but a straight path through curved spacetime, there's no further acceleration).

As you get to the neutron star, you maintain your inertial frame... Just rotate your ship to keep the star in your forward window even on your way back. Now, from your inertial frame, you're just floating through space, and you see your planet wizzing by from behind you at a speed of .5c. You have to accelerate to catch up to it.

Did your rate of time normalize relative to your planet when you went through curved space to change direction without accelerating?

Or did you normalize when you accelerated to catch up to the planet after it wizzed by?

Or did time slow relative to your planet both when you left and then again once you caught back up with it? Forever giving you Flash-like super powers since everyone else is so slow?

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So then the net result of time dilation is a function of acceleration rather than time spent moving at a certain velocity?

No. Time dilation is strictly a matter of relative velocity. Acceleration just breaks the symmetry and determines who's subject to dilation.

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So then the net result of time dilation is a function of acceleration rather than time spent moving at a certain velocity?

Time dilation is due to relative velocity only, The net accumulated time difference between two observers can be the result of many different effects, depending on the observer.

For instance, in the Classic Twin Paradox, if we take the position of the Twin that does not under go any acceleration, then the end difference in ages when they reunite is entirely due to time dilation of the other twin.

Example: Twin 1 and Twin 2 pass at 0.6c with respect to each other. When they are separated by 0.6 light years, Twin 2 turns around and heads back to twin 1. This will take 1 yr by Twin 1's clock and Twin 2's clock will time dilate by a factor of 0.8. Thus in the two years it takes for twin 2 to return, he ages 1.6 yrs all due to time dilation.

However, for twin 2, thing play out differently. He still passes twin 1 at 0.6c, however, because of length contraction, the distance measured as 0.6 light year by twin 1 is contracted to 0.48 light years by his measurement. Thus he turns around when he is 0.48 light years from Twin 1. Since it only takes 0.8 years to cross 0.48 light years at 0.6c, the total round trip takes 1.6 years.

During the outbound and inbound legs of the trip, by his reckoning, twin 1's clock is time dilated by a factor of 0.8, and accumulates 1.28 yrs. Here is were the turn around becomes important. When he starts his turn around, his clock reads 0.8 years and twin 1's clock reads 0.64 years, meaning his clock is 0.16 years ahead of Twin 1's. After he finishes turning around, his clock still reads 0.64 yrs ( assuming a negligible turn around time), but Twin 1's clock will now read 1.36 years. (he won't see the clock jump ahead, but this is what time he will reckon Twins 1's clock reads when taking the distance between them, there relative speed and the speed of light into account. ) This is due to the Relativity of Simultaneity.(If you are not familiar with this, I suggest you become so, as nothing in Relativity will make sense to you until you grasp this concept.) Twin 1's clock then advances at the time dilated rate to add an additional 0.64 years and reads 2 years when they meet up.

The result is that both twins agree that Twin 1 aged 2 yrs and Twin 2 aged 1.6 years between their separation and rejoining, however, for different reasons.

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Curious about how dilation differentiates between "deceleration" and "acceleration" in the opposite direction? See the illustration a couple posts up.

Edited by Didymus
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Curious about how dilation differentiates between "deceleration" and "acceleration" in the opposite direction?

Time dilation does not differentiate between direction, just relative velocity.

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Curious about how dilation differentiates between "deceleration" and "acceleration" in the opposite direction? See the illustration a couple posts up.

"Deceleration" is just acceleration seen in a different frame. If you look at my scenario above, from the frame of Twin 1, Twin 2 decelerates to a stop and then accelerates back towards him at turn around. However, from the frame of Twin 2's "outbound" leg. Twin 2 starts at a rest and just accelerates towards twin 2. BuT in the frame of Twin 2's inbound leg, Twin 2 starts at some high velocity and then decelerates to a rest(letting Twin 1 catch up to him). But from all these frames you get the same result. Twin 1 ages 2 years and Twin 2 ages 1.6 yrs between their separation and rejoining.

As far as your illustration above, it's better if we avoid incorporating gravity and curved space-time into the mix until you are comfortable with how things work in flat space-time. Curved space-time requires invoking General Relativity vs. just sticking with Special Relativity.

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I'm aware that it will alter the calculations a bit relative to the time spent in the gravity well. But the observer on the home planet will also be experiencing Gr time dilation for the gravity well he's been sitting in the whole trip. Assume the two to be proportional (less gravity, but more time in the field) to cancel out the GR variable. The gravity well is just to provide a situation where a person is inarguably accelerating to leave a planet, then accelerating in the same direction to catch up to the planet again without the planet experiencing any acceleration itself. Just to eliminate variables.

If acceleration sets the time dilation, dilation may be asymmetrical, but every acceleration/deceleration would compound the slowing of time. There would be no function for acceleration in one direction to slow time, but equal acceleration in the opposite direction to speed it up again.

If acceleration doesn't play a role, and only relative velocity affects time... Then there is no function for asymmetric time dilation.

How can they be reconciled?

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I'm aware that it will alter the calculations a bit relative to the time spent in the gravity well. But the observer on the home planet will also be experiencing Gr time dilation for the gravity well he's been sitting in the whole trip. Assume the two to be proportional (less gravity, but more time in the field) to cancel out the GR variable. The gravity well is just to provide a situation where a person is inarguably accelerating to leave a planet, then accelerating in the same direction to catch up to the planet again without the planet experiencing any acceleration itself. Just to eliminate variables.

If acceleration sets the time dilation, dilation may be asymmetrical, but every acceleration/deceleration would compound the slowing of time. There would be no function for acceleration in one direction to slow time, but equal acceleration in the opposite direction to speed it up again.

If acceleration doesn't play a role, and only relative velocity affects time... Then there is no function for asymmetric time dilation.

How can they be reconciled?

The dilation is from the relative speed and the amount of time of the trip.

The acceleration is necessary to change the frame of reference, but need not contribute at all to the dilation. In many formulations of the problem it's assumed instantaneous so as to not confuse the issue further.

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The gravity well is just to provide a situation where a person is inarguably accelerating to leave a planet, then accelerating in the same direction to catch up to the planet again without the planet experiencing any acceleration itself. Just to eliminate variables.

In this situation all you're doing is eliminating "proper acceleration", so that the traveling twin doesn't feel any effects of acceleration. It's been used to argue that the effects of acceleration (other than the change of velocity) theoretically do not contribute to time dilation in SR.

The traveling twin still changes direction. Its spacetime path between the two points where the twins are reunited is longer spatially and shorter temporally.

Acceleration is required to get two twins to travel between two points along different paths. Once the velocity of the two along their paths is fully accounted for (whether using rockets or gravity wells or whatever), there is no further effect of acceleration that contributes to the time dilation in SR. This is theoretical (there's no theorized effect of acceleration) and agrees with experiments so far.

However I think once you add in gravitational time dilation (GR) it gets more complicated. Edit: I think in either case (SR or GR) acceleration on its own does not contribute to time dilation, but of course it is related to velocity and gravity, so it can be easy to confuse oneself that "acceleration is doing something to time" when really it's velocity and gravitation that are doing it. I'm not sure if I'm saying this correctly.

Edited by md65536
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However I think once you add in gravitational time dilation (GR) it gets more complicated.

Not by much, you only need to know how to apply the Schwazschild solution. The answer is a one line simple expression.

Edited by xyzt
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#15 janus

He still passes twin 1 at 0.6c, however, because of length contraction, the distance measured as 0.6 light year by twin 1 is contracted to 0.48 light years by his measurement. he turns around when he is 0.48 light years from Twin 1.

Assume distance markers in the T1 frame. They agree to reverse at .6 ly. The expression t=x/v works in either frame. T2 calculates t=.6 ly/.6c=1y. Unable to detect his time dilation, he arrives at .6 ly when his clock reads .8 y, and concludes the universe has contracted by a factor of .8. By symmetry, his trip takes 1.6 y.

The contracted T1 frame is his interpretation of his own time dilation, and the markers are now contracted distances. If T2 wants to reverse at .48 ly, his clock would read .64 y.

Here is a geometric method for aging.

Time dilation and aging for a pair of clocks.doc

Edited by phyti
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So if youre floating in deep space when suddenly a clock flies past. You see whether it's ticking slow from your perspective... Then catch up to it and look at it again.

Are you suggesting that you can definitively detect which one accelerated more during their trip by noting which clock jumps forward in time when the inertial frames meet back up?

If both clocks experienced equal acceleration, it's been suggested that both clocks would jump forward and synch. It seems that one could compare the ratio of how each object had accelerated during it's trip by noting how much each clock adjusts when they come together.

Al goes left at .8c and Bert goes right at .7c, but turns around early so they still meet back at the same time. They still had a relative velocity of nearly the speed of light, so the whole trip, each would see the other's clock run slower. Right before they decelerate, they compare how far behind the other clock is. Even if they don't see the clockove forward during deceleration... Once they stop, they can then compare which clock jumped forward more than the other.

Is that a fair interpretation of your suggestion? If not, where do you disagree?

New post: md65- that's why I suggested a straight line through curved space to limit the "acceleration" to a change in relative velocity. If it's not a matter of G-forces, the. What function differentiates the two inertial frames? When the ship turns around the star, from its perspective, it's going straight and hasn't accelerated at all.... Instead his home planet rapidly changed from flying away at a certain speed, decelerated, then accelerated flying toward the ship at that speed.

If SR is a matter of relative velocity, it can't be asymmetrical.

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According to the internet, the twin paradox can be resolved because "only one twin underwent acceleration." But, from the frame of reference of someone traveling away from Earth near the speed of light, they would have observed Earth moving away from themselves at nearly the speed of light instead. To both frames of reference, both observed the other undergoing near-light acceleration, so why wouldn't it be symmetric? Let's say there wasn't turn-around acceleration involved and Earth and the ship are in a hyper-doughnut where by traveling in a straight line away from Earth, one would end up back at Earth eventually, after let's say, 10 light years of distance.

Even if there was turn-around acceleration, couldn't the near-light traveler just say they saw Earth traveling in a semi-circle from their view instead?

Edited by SamBridge

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