Does this scenario disprove relativity?


by erics
Tags: disprove, relativity, scenario
erics
erics is offline
#1
Apr5-11, 09:16 PM
P: 7
Suppose you have two ships departing from Earth at the same time, Ship #1 and Ship #2. Both ships depart Earth in the year 2000, both in the same direction from earth, and both Ships return to Earth in the year 2020, so 20 Earth Years for BOTH ships.....

Ship #2 does something somewhat different from Ship #1, you'll see


From Earth's Reference Frame:
Ship #1 Cruises at a Velocity of 86.6025% of C, which is a Lorentz Factor of 2.... for half the 20-year duration.... then the ship returns to Earth. Based on Twin Paradox, the ship's clock ticks half of 20 years.... so the ship ages 10 years.

Ship #2 Cruises at the same velocity as Ship #1, but only for the first 6 Earth Years of time.

Between Years 7 thru 10:
After 6 years time, Ship #2 gains additional velocity and passes ahead of Ship #1 at a Velocity of 98.7433% of C with respect to Earth (Lorentz Factor of exactly 7).

Then at some point in time, Ship #2 de-accelerates to 0 km/h with respect to Earth and waits patiently for Ship #1 to catch up to Ship #2. Ship #1 catches up to Ship #2 after 10 Earth Years Time.

Note Ship #1's Reference Frame Between
But based on U - V / (1 + U*-V), the Ship #2 was moving ahead of Ship #1 at 86.6025 of C, Lorentz Factor of 2.... and then would have appeared to head back towards ship #1 also at 86.6025% of C.


After 10 Earth Years:
Both Ship #1 and Ship #2 return to Earth side by side at 86.6025% of C with respect to Earth, arriving home to Earth in the year 2020....


Now looking at Years 7 thru 10, however much Ship #1 was aging, Ship #2 would be aging half as much, because with respect to Ship #1's frame of reference, Ship #2 accelerated to Lorentz Factor 2 speed.... then moved back towards ship #1 at Lorentz Factor 2 speed (while ship #2 was moving 0 with respect to Earth it was moving backwards with respect to Ship #1).......
Based on the same logic as the Twin Paradox.... this would result in a situation where Ship #2 aged one year less than Ship #1, because during the 4 Earth Years Ship #2 parted from Ship #1....... Ship #1 would age 2 years and Ship #1 would have aged half of that... 1 years less aging than Ship #2..... or is this wrong?


During years 7 thru 10, Ship #1 did not exert any force as it kept its contant cruising velocity. The same would hold true for Earth.... eventhough they were traveling at different velocities.

So here is what gets confusing to me.

From Ship #1's Reference Frame:

It would appear that between Earth years 7-10, Ship #2 spent 50% of the time moving farther away from Ship #1, and 50% of the time moving back towards Ship #1.... for with respect to the ship.... the relative speed of Ship #2 was 86.6025% of C


From the Earth's Reference Frame:

I would appear that between Earth years 7-10, Ship #2 spent much more time moving away from Ship #1 compared to the time it took Ship #1 to catch back up to Ship #2.

Ship #2 would have appeared to move 98.7433% of C away from Earth while Ship #1 was cruising at 86.6025% of C away from Earth..... a velocity difference of only 12.1408% of C....! Then when Ship #2 stopped moving and waited for Ship #1 to catch up, this would have appeared to be a movement of 86.6025% of C for BOTH Earth and the ship...


so what's confusing to me is given the Earth AND Ship #1 are not accelerating/deacclerating since only Ship #2 is exerting force and shifting in speeds....

Why does it make sense relativity wise that Earth and Ship #1 would disagree on the PERCENTAGE of duration it took Ship #2 to move farther away from Ship #1... VS. the percentage of duration the ship #1 was moving closer to Ship #2....?

Can someone explain this discrepancy or correct me if my assumptions are incorrect?

Thanks!
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DaleSpam
DaleSpam is offline
#2
Apr5-11, 09:24 PM
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P: 16,466
This scenario does not disprove relativity. It only proves that you can make a scenario which is so absurdly complicated that it is confusing for a novice to describe or analyze.

The easiest way to approach this problem is to draw a spacetime diagram in the Earth frame and clearly label the t and x coordinates of each acceleration/deceleration event. Then simply use the spacetime interval formula to determine the aging along each segment.
erics
erics is offline
#3
Apr5-11, 09:34 PM
P: 7
My guess is that perhaps nothing can be said for time scales of vehicles moving at different velocities to be proportional to each other on a percent scale of total time elapsed..... only that the end results at the end of 20 years accumulate to the proportions that they are......

and so it cannot be determined in absolute time what percentage of clock ticks occurred before the U-turn of Ship #1, the De-acceleration of Ship #2.... none of the above.... if this isn;t true what I'm saying here..... then I can't think of another answer


On another note, is it written down anywhere that this is true?

Given Obect #1 Is traveling with Respect to the Earth at Lorentz Factor X

And Object #2 Is Traveling with Respect to Object #1 at Lorentz Factor X

Then Object #2 is Traveling with Respect to the Earth at Lorentz Factor 2*(x-squared) -1

I discovered this, I wonder if it is written anywhere or can I get credit for this equality.

You can test it out and you'll see I'm correct.

DaleSpam
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#4
Apr5-11, 09:50 PM
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P: 16,466

Does this scenario disprove relativity?


There is no absolute time, so obviously you cannot determine anything in absolute time.

Just go through the exercise I suggested to solve your scenario completely.
JesseM
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#5
Apr5-11, 10:19 PM
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P: 8,470
Quote Quote by erics View Post
Suppose you have two ships departing from Earth at the same time, Ship #1 and Ship #2. Both ships depart Earth in the year 2000, both in the same direction from earth, and both Ships return to Earth in the year 2020, so 20 Earth Years for BOTH ships.....

Ship #2 does something somewhat different from Ship #1, you'll see


From Earth's Reference Frame:
Ship #1 Cruises at a Velocity of 86.6025% of C, which is a Lorentz Factor of 2.... for half the 20-year duration.... then the ship returns to Earth. Based on Twin Paradox, the ship's clock ticks half of 20 years.... so the ship ages 10 years.
OK, so in the Earth frame, ship #1 must travel a total distance of 0.866025 * 10 = 8.66025 light-years before turning around.
Quote Quote by erics View Post
Ship #2 Cruises at the same velocity as Ship #1, but only for the first 6 Earth Years of time.

Between Years 7 thru 10:
After 6 years time, Ship #2 gains additional velocity and passes ahead of Ship #1 at a Velocity of 98.7433% of C with respect to Earth (Lorentz Factor of exactly 7).

Then at some point in time, Ship #2 de-accelerates to 0 km/h with respect to Earth and waits patiently for Ship #1 to catch up to Ship #2. Ship #1 catches up to Ship #2 after 10 Earth Years Time.
From what I said earlier, if you want ship #1 to catch up with ship #2 at the same time that ship #1 turns around to return to Earth (after 10 years in the Earth frame), that means ship #2 must come to rest relative to Earth when it reaches a distance of 8.66025 light-years from Earth. After 6 years ship #2 will be at a distance of 0.866025 * 6 = 5.19615 light years from Earth, so it still has a distance of 8.66025 - 5.19615 = 3.4641 light years to go, so if it traverses that distance at 0.987433c then the time needed will be 3.4641/0.987433 = 3.5082 years, so ship #1 stops after 6 + 3.5082 = 9.5082 years in the Earth frame.
Quote Quote by erics View Post
But based on U - V / (1 + U*-V), the Ship #2 was moving ahead of Ship #1 at 86.6025 of C, Lorentz Factor of 2.... and then would have appeared to head back towards ship #1 also at 86.6025% of C.
No, your math is wrong here. In ship #1's frame, ship #2 is traveling at (0.987433c - 0.866025c)/(1 - 0.987433*0.866025) = 0.838115c.
Quote Quote by erics View Post
Now looking at Years 7 thru 10, however much Ship #1 was aging, Ship #2 would be aging half as much, because with respect to Ship #1's frame of reference, Ship #2 accelerated to Lorentz Factor 2 speed
First of all, as noted above ship #2 is actually initially moving at 0.838115c in ship #1's frame (before ship #2 comes to rest relative to Earth), for a Lorentz factor of 1.8332. Second, since ship #2 does come to rest relative to Earth at the turning point, ship #2 changes velocities in the frame of ship #1, so that ship #2 is traveling at 0.838115c for some time and then changes to moving at 0.866025c towards ship #1, until eventually they meet.

Perhaps you want to change the thought-experiment so that ship #2 initially accelerates to (0.866025c + 0.866025c)/(1 + 0.866025*0.866025) = 0.989743c relative to the Earth, then comes to rest at the turning point and waits for ship #1 to catch up, in which case in ship #1's frame ship #2 initially moves away at 0.866025c and then turns around and moves back towards ship #1 at 0.866025c until they meet at the turning point?
erics
erics is offline
#6
Apr5-11, 10:34 PM
P: 7
whoops I meant ship 2 Velocity with respect to earth would be 0.9897433186.... typo... I left out a nine.... it should then be exactly the same as the .8606 C

So knowing this, it would appear to Ship #1 that an equal amount of ship #1 clock tickings were during the getting surpassed by ship #2 as the time catching up to the ship....

Using intuition, no graph, I think what is happening here is that Ship #1's time relationship with Earth is not changing between Year 0-6 and 7-10.....

The disagreement over the time of U-turn of Ship #2 purely has to do with Earth's relationship to Ship #2 in space-time differing from Ship #1's relationship to Ship #2 in space-time.... explained by a disagreement over how much Ship #2 is accelerating/de-accelerating........... from the perception of Earth compared to Ship #1's perception

However, all 3 clocks are out of sync with each other until they meet each other so I can't use Earth's clock to measure aging of the ship's clocks except note that the total times accumulated at the end are what they are.... and trusting the formulas.
JesseM
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#7
Apr5-11, 10:51 PM
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Quote Quote by erics View Post
whoops I meant ship 2 Velocity with respect to earth would be 0.9897433186.... typo... I left out a nine.... it should then be exactly the same as the .8606 C

So knowing this, it would appear to Ship #1 that an equal amount of ship #1 clock tickings were during the getting surpassed by ship #2 as the time catching up to the ship....
Yes, that's correct. And since ship #1 only ages by 2 years between t=6 and t=10 in the Earth frame, this must mean that ship #2 only ages 1 year between t=6 and t=10 in the Earth frame. To see that this makes sense in the Earth frame, start with an altered version of my earlier comment:
From what I said earlier, if you want ship #1 to catch up with ship #2 at the same time that ship #1 turns around to return to Earth (after 10 years in the Earth frame), that means ship #2 must come to rest relative to Earth when it reaches a distance of 8.66025 light-years from Earth. After 6 years ship #2 will be at a distance of 0.866025 * 6 = 5.19615 light years from Earth, so it still has a distance of 8.66025 - 5.19615 = 3.4641 light years to go, so if it traverses that distance at 0.989743c then the time needed will be 3.4641/0.989743 = 3.5 years, so ship #1 stops after 6 + 3.5 = 9.5 years in the Earth frame.
So in the Earth frame, ship #2 is traveling at 0.989743c for 3.5 years, and then it's at rest for the next 0.5 years until ship #1 catches up with it. at 0.989743c the Lorentz factor is 7, so ship #2 only ages 3.5/7 = 0.5 years during the 3.5 years in the Earth frame for it to reach the turnaround point, then it's at rest for the next 0.5 years so it ages an additional 0.5 years, meaning the Earth frame also predicts that ship #2's total aging between leaving ship #1 and reuniting with ship #1 should be 1 year.

So is your question just about the fact that the first leg of ship #2's trip lasts 3.5 years and the second leg only lasts 0.5 years in the Earth frame, while in ship #1's frame both legs last for an equal time of 1 year? If that is your concern, I suggest you look into the relativity of simultaneity (also see this page for more info), which says that different frames disagree about which pairs of distant events happened "at the same time" (simultaneously). Specifically, in ship #1's frame the event of ship #2 coming to rest relative to the Earth happens simultaneously with ship #1's clock showing a time of 4 years and Earth's clock showing a time of 2 years, but in the Earth's frame the event of ship #2 coming to rest relative to the Earth happens simultaneously with ship #1's clock showing a time of 9.5/2 = 4.75 years and Earth's clock showing a time of 9.5 years.


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