3-part relativity problem, split from functionally faster than light

[michel123456]

He's traveled 7 LY of distance measured from Earth's frame, in 1 year measured from the ship's frame.

 

How is that faster than light? How long did it take light to make the same journey? (Example answer: It took 7 years according to a clock on Earth, but the ship took over 7 years according to the same clock. Is there any clock which records the ship traveling faster than light? Or do you have to switch between clocks and frames in order to measure the ship differently than you measure light?)

 

 

He's traveled less than 1 LY in 1 year, both measured from the ship. Light is faster according to the traveler. He's traveled 7 LY in a little over 7 years, according to Earth. Light is faster. According to anyone else's measurement: Light is faster.

Now take 3 observers A, B, C

 

A is on Earth "at rest"

B travels to the left

C travels to the right (in the opposite direction of B)

Lets say all observers are all 20 years old (for the sake of simplicity) at departure time.

 

B<----------A---------->C

 

A observes B reaching the planet X after 7 years.

A observes C reaching the opposite planet Y after 7 years

 

Theory says that when B and C come back to Earth, only 2 years of their time has passed (they are 22 y. old) although for observer A 14 years have passed (he is 34 y.o.)

 

But what does B observe concerning C? If he observes A aging faster than him, he must observe that C is aging much more than A.

 

1._when B reaches planet X, observer A on Earth has aged 7 years (A is 27) and C has aged twice as much as A (C is 34 as observed from B), or what???

2._symmetrically when C reaches planet Y, observer A on Earth has aged 7 years (A is 27) and B has aged twice as much as A (B is 34 as observed from C), or what???

3. _ or point 1 & 2 above are wrong and in this case, what is B observing concerning C and reversely?

Edited March 12, 2013 by michel123456

[md65536]

But what does B observe concerning C? If he observes A aging faster than him, he must observe that C is aging much more than A.

B observes C appear to age faster than A only while approaching, but this happens for a much shorter duration than C appears to recede.

 

The answer is similar to the Doppler shift analysis of the twin paradox: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

With pictures: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html

1._when B reaches planet X, observer A on Earth has aged 7 years (A is 27) and C has aged twice as much as A (C is 34 as observed from B), or what???

2._symmetrically when C reaches planet Y, observer A on Earth has aged 7 years (A is 27) and B has aged twice as much as A (B is 34 as observed from C), or what???

3. _ or point 1 & 2 above are wrong and in this case, what is B observing concerning C and reversely?

1. No, according to B at this point, C has aged even less than A.

3. When B reaches X, it has seen C appear to age VERY slowly, because C's clock is slowed and also there is an increasing delay in incoming light as the distance increases.

 

B turns around and heads back to Earth. It observes that the outbound journey is (and appears) to take the same time on its own clocks. Due to the delay of light, C is not seen to reach its turnaround for some time. Thus B sees C appear to age very slowly for a very long time, then---I think---appear to age at a normal rate while B is returning to Earth but C still appears to be receding, and then after C appears to turn around it appears to age very quickly (faster than A) for a short time, eventually catching up to B at the same time they reach Earth.

 

In a Doppler analysis, differing amounts of time a distant clock spends appearing to tick slow vs. fast make everything work out properly.

In a Lorentz transformation or spacetime diagram analysis, changes in relative simultaneity make everything work out properly.

Edited March 12, 2013 by md65536

[michel123456]

B observes C appear to age faster than A only while approaching, but this happens for a much shorter duration than C appears to recede.

 

The answer is similar to the Doppler shift analysis of the twin paradox: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

1. No, according to B at this point, C has aged even less than A.

3. When B reaches X, it has seen C appear to age VERY slowly, because C's clock is slowed and also there is an increasing delay in incoming light as the distance increases.

 

B turns around and heads back to Earth. It observes that the outbound journey is (and appears) to take the same time on its own clocks. Due to the delay of light, C is not seen to reach its turnaround for some time. Thus B sees C appear to age very slowly for a very long time, and then age very quickly for a short time, eventually catching up to B at the same time they reach Earth.

 

In a Doppler analysis, differing amounts of time a distant clock spends appearing to tick slow vs. fast make everything work out properly.

In a Lorentz transformation or spacetime diagram analysis, changes in relative simultaneity make everything work out properly.

B, A and C are always in visual contact throughout the travel. No matter the delays caused by the distance light has to travel we have this:

B has observed A eating 14 times 365 breakfasts.

B has observed C eating only 2 times 365 breakfasts

 

Although the relative travel between B & C is much larger than between B & A.

[md65536]

B, A and C are always in visual contact throughout the travel. No matter the delays caused by the distance light has to travel we have this:

B has observed A eating 14 times 365 breakfasts.

B has observed C eating only 2 times 365 breakfasts

 

Although the relative travel between B & C is much larger than between B & A.

Agreed.

 

And yes, before C turns around, it ages less than A as measured by B, because it is moving away at a greater relative velocity. But when C turns around, there is a change in relative simultaneity. As always, calculations using length contraction and time dilation and relative simultaneity work out the same as relativistic Doppler calculations, which work out as expected that symmetrical observers will see symmetrical things.

[michel123456]

 

B, A and C are always in visual contact throughout the travel. No matter the delays caused by the distance light has to travel we have this:

B has observed A eating 14 times 365 breakfasts.

B has observed C eating only 2 times 365 breakfasts

 

Although the relative travel between B & C is much larger than between B & A.

Agreed.

 

And yes, before C turns around, it ages less than A as

measured by B, because it is moving away at a greater relative velocity.

But when C turns around, there is a change in relative simultaneity. As

always, calculations using length contraction and time dilation and

relative simultaneity work out the same as relativistic Doppler

calculations, which work out as expected that symmetrical observers will

see symmetrical things.

That makes no sense.

When B turns around what does he see?

1.B is half the trip so he sees himself aged 1 year (he is 21y.o.)

2.B sees observer A aged less than 7 years, because he traveled together with the light signal coming from A. Say he sees A has aged 4 years (as seen from B)

3.B sees observer C aged less than 1 year, for the same reason. Say he sees C has aged 0.5 year.

 

On the return trip, what does B observe?

4. B observes A aging 14-4=10 years while seeing himself aging 1 year.

5. B observes C aging 2-0.5=1.5 years while seeing himself aging 1 year.

 

But C should aged faster than A. The result should be C aging say 20 years, not 1,5 year.

There is something wrong.

Edited March 13, 2013 by michel123456

[Delta1212]

That makes no sense.

When B turns around what does he see?

1.B is half the trip so he sees himself aged 1 year (he is 21y.o.)

2.B sees observer A aged less than 7 years, because he traveled together with the light signal coming from A. Say he sees A has aged 4 years (as seen from B)

3.B sees observer C aged less than 1 year, for the same reason. Say he sees C has aged 0.5 year.

 

On the return trip, what does B observe?

4. B observes A aging 14-4=10 years while seeing himself aging 1 year.

5. B observes C aging 2-0.5=1.5 years while seeing himself aging 1 year.

 

But C should aged faster than A. The result should be C aging say 20 years, not 1,5 year.

There is something wrong.

Why would anyone see C aging 20 years?

[michel123456]

Why would anyone see C aging 20 years?

B does (should) look at C aging more than A

 

the line of sight is B---A---C

 

B observes A aging more than himself

B should observe C aging more than A

[md65536]

But C should aged faster than A. The result should be C aging say 20 years, not 1,5 year.

There is something wrong.

If you're calculating numbers, use math.

I have no idea what results in your "should" conclusions, but that's the something that is wrong.

 

From B's perspective, C is always aging slower than A. It catches up by "leaping forward" with a change in relative simultaneity when it turns around. This leap is not apparent to B, but is seen gradually as C closes the distance to B and appears to age quickly for a short period.

 

 

Math:

- Calculate velocity of B relative to A using the known Lorentz factor of 7.

- Find velocity of C relative to B using composition of velocities.

- Calc how much C appears to age according to B using relativistic Doppler equation.

- Determine when C appears to turn around, using Doppler.

- Add up the different amounts of C's aging that B sees (I think there are 3 different phases, each can use the Doppler equations).

 

B observes A aging more than himself

B should observe C aging more than A

No, A and C don't behave the same in this example.

 

IF B turns around and catches up to C without C ever turning around, then B will observe that C ages a lot more than B. But if this happens, B would spend a long time catching up to C, all the while C is appearing to age quicker. In your example where C turns around, it appears to age quickly for a very short period of time.

 

If you're calculating numbers, use math.

Alright I got curious and did it quick and dirty. Pasting from spreadsheet and ignoring significant digits...

(Also, I've been sloppy so I'll just say I'm using a symmetric configuration, where the velocity of O relative to P is the same as the velocity of P relative to O, and not worry about which is which. -- Sigh, so many little mistakes to fix...)

 

gamma = 7

v_BA = 0.9897433186 c, (velocity of B relative to A) by solving Lorentz factor for v

v_BC = 0.999946858 c using composition of velocities, where v_CA = v_BA

 

The observed tick rate of a moving clock is given by the reciprocal of the Doppler factor, let's call this epsilon, where

[math]\varepsilon = \sqrt{\frac{1-v_{BC}/c}{1+v_{BC}/c}}[/math] = 0.0051547761

 

On the way back, negate the velocity, which results in the reciprocal of epsilon.

 

So the 3 phases:

1: B ages 1 year while C appears to age epsilon * 1 year = 0.0051547761 years.

2: B appears to travel at the same speed as C for some time, until C is seen to turn around. I'll skip this step for laziness.

3: B sees C age one year at an accelerated rate of 1/epsilon = 193.9948452238, so this happens while B ages 0.0051547761 years.

 

From 3, we can calculate that phase 2 takes 1 - 0.0051547761 = 0.9948452239 years, during which C appears to age at the same rate as B.

 

Adding up: B ages 1 + 0.9948452239 + 0.0051547761 = 2 years.

C appears to age 0.0051547761 + 0.9948452239 + 1 = 2 years.

 

 

Note: To double check, one should be able to calculate the time it takes light to travel from C's turnaround, to where it would intercept B, and you should find that it happens at 0.9948452239 along B's return journey.

 

 

Note that when B turns around, C has only appeared to age 1.88 days, and remains .99 of a year behind B for most of the return trip. On the last 1.88 days of B's return trip, C appears to age a full year.

 

---------

 

Now just for FUN, what does B see of A?

There are only 2 phases, because A is never seen changing velocity.

 

The reciprocal of the Doppler factor comes out to epsilon = 0.0717967697

 

Phase 1: epsilon * 1 year = 0.0717967697 years

Phase 2: 1/epsilon * 1 year = 13.9282032303 years

 

for a total of 14 years seen aging.

Edited March 13, 2013 by md65536

[michel123456]

I don't get it.

Why A and C have a so radical difference?

Since there is no acceleration involved, B could be at rest and A could be traveling. The situation ought to be symmetric. Or C could be at rest. The results should show the same either way.

As seen from B, all the objects between B and A are looking aging faster than B. And all of a sudden, objects on the other side of A are looking aging slower. That makes no sense. C is a traveling object just like A. The only difference is their relative speed to B.

 

The configuration is exactly the same as if C was the origin and A & B were traveling.

[Delta1212]

Since there is no acceleration involved, B could be at rest and A could be traveling. The situation ought to be symmetric. Or C could be at rest. The results should show the same either way.

In order for B and C to turn around, they have to accelerate.

[michel123456]

Well, they have to accelerate to go out of Earth in the first place. This detail is not part of the calculations.

[md65536]

Well, they have to accelerate to go out of Earth in the first place. This detail is not part of the calculations.

It's already been explained in the original thread I think that acceleration is not a part of the calculation. You already know v from gamma, you don't have to use acceleration for anything.

 

Anyway, at Earth the distances between everyone is essentially zero. Any change in relative simultaneity at that length is zero. For example, it wouldn't make any difference if you had started B and C far apart and had them pass by Earth at time t=0 with no acceleration occurring near Earth. Similarly at the end they could pass Earth without ever stopping. As long as B and C move symmetrically and compare clocks at the same location, the clocks will mark the same time.

 

 

 

Remember no one *measures* anyone else's clock running faster than their own (only slower in SR or absence of gravity), even if they see it. Going by what they measure, accounting for delay of light, B's measurement of A's advancement in age occurs during the change in relative simultaneity that happens when B changes inertial frames.

 

Going by what they see, the asymmetry is: B sees A's clock running 13.9 times faster than its own, for a full year.

It sees C's clock running 194 times faster than its own, but only for 1.88 days.

Edited March 13, 2013 by md65536