Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 years ago

# Help me!!! Relativity: Twin Paradox version 2.0!!!?

Bob and Jane are sitting in a high speed spacecraft traveling at a speed of 0.49C towards a stationary observer John who is standing on Earth. At time t0(t=0), they reach the point where john was standing and Bob immediately jumps off the spacecraft whilst Jane continues flying in the same direction as before without making a stop (John dodged the spacecraft in order to avoid being killed). Four years after t0, Jane reaches Planet-X which is located 1.96 light years away from the Earth.

Now when Bob initially jumped off the spacecraft at time t0, he instantaneously decelerated to speed zero and stayed on Earth with John for a period of 2 years. But 2 years after t0, Bob proceeded to travel towards Planet-X at the speed of 0.98C(twice Jane's speed). Bob reaches Planet-X at the exact same time as Jane (4 years after initial t0). At the exact moment when Bob and Jane reach Planet-X, Bob instantaneously decelerates to Jane's velocity whilst the Earth and Planet-X instantaneously accelerate to Jane's velocity. However, Jane continues traveling at 0.49C velocity without ever accelerating or decelerating, hence this would make it seem like Jane was the stationary observer all along.

Update:

Now when they reach Planet-X(4 years after initial t0), Bob should have aged only 2.4 years instead of 4 years because of time dilation. This is because he aged 2 years while staying on Earth with John and then during the trip to Planet-X he traveled at 0.98C and hence he only aged 0.4 years during that journey (gamma = 5.025). This result of 2.4 years is calculated from John's perspective on the basis that John was the stationary observer.

Update 2:

BUT since we accelerate everybody to Jane's frame at the end, it would make it seem like Jane was the stationary observer all along. So how do I calculate the result from Jane's perspective? If we pretend that Jane was the stationary observer, it would seem like Bob initially travels away from her at 0.49C speed and then towards her at 0.94C speed. So How do I calculate the number of years Bob has aged from Jane's perspective as the stationary observer?

Relevance
• Nick
Lv 6
6 years ago

Okay, let me get out my pen, paper and calculator and I'll get back.

From Jane's perspective Bob ages 2 years, he does this in 2*γ(0.49) = 2.29 years according to Jane. In this time he has travelled 2.29*0.49 = 1.12 ly according to Jane (who, of course, sees herself as stationary) He then travels the 1.12 ly at a velocity (0.98-0.49)/(1-0.98*0.49) = 0.94 ly/y according to Jane.

This distance he travels in a time 1.12/0.94=1.19 years according to Jane but the he must age 1.19/γ(0.94)=0.4 years which we can add to his 2 to make 2.4 years according to Jane.

It's probably easier to just do it according to John 2/γ(0.98) = 0.4 and add that on to his 2 years to make 2.4 years.

The disagreement occurs when we notice Bob observes Jane to have aged less than him whilst she observes him to have aged less than her. This is resolved by the acceleration at the end, Bob will see Jane age rapidly to her age according to her.

• Anonymous
6 years ago

"Help me!!! Relativity: Twin Paradox version 2.0!!!?"

When you do not understand relativity, why in heaven's name would you complexify a problem to the point that few could bother to read your entire problem statement?

What are you trying to get to? You understand that you are "frame jumping" by having Jane your "stationary" person, but using "Earth rest" distance measurements.

Acceleration is a boondoggle. I can reproduce the "twin paradox" with no accelerations, just three objects, two moving (one +v and one -v), and the ability to synchronize clocks on those objects.

So have Jane stationary, and have Planet-X, Earth, and Bob move away "negative" initially, see if that doesn't evaporate your quandary.

[EDIT: "No you can't"

A stationary.

B passes A at speed v, and synchronizes its clock to A's.

B travels some distance (or duration, you choose).

C passes B at speed -v, and synchronizes its clock to B's as it passes.

When C passes A, A compares clocks, and sees time dilation.

No acceleration required between initial synchronization, and final comparison.

Acceleration is not the differentiator between frames, "velocity history" (so the integral) is. Acceleration is just an easy indicator, but is not all that meaningful in and of itself.

]

[EDIT: "Okay well now that you took the time to read the problem, do you know how to work out Jane's aging from Bob's perspective? waiting....crickets chirping...."

Yes, and not interested in combing through your rat's nest. Recommended you treat Jane as stationary, and all others moving, but you have ignored so far. That is all the "help" you will get from me. You have made a problem that is too inane to bother with, at least for me.

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