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# I have a twin paradox question 80 percent speed of light (nobody seems to answer for me)?

Ok, if you are the traveling twin traveling 4 light-years from earth then reversing back to earth ...

At 80 percent of speed of light

Earth will perceive it’s a 10-year voyage and the traveler ages only 6 years at Lorentz factor 1.67 (3 years each way , aging 60 percent slower

Now.... on the leg journey away from earth, the Line of simultaneity suggests that for the traveling twin, it’s the Earth whose clock is aging 60 percent slower

That means from the traveler’s perspective, just BEFORE the de-acceleration to turn around, while the length contraction is in play for the traveler, The traveler would perceive (not see visually but perceive) that the earth is at Time 1.8 around the time traveler reaches destination away from earth

60 percent of the 3 years aged by the traveler. Before de-accelerating

But as we know the traveler SEES the light of earth at time 1 earth year later . First as red shifted then on return home as blue shifted

If this is all accurate let’s move onto my question

The light of earth that the traveler sees when reaching outward destination must have left earth at a prior time during which the traveler was en route , as the beginning of the journey both were at Time 0 in the same reference frame

Can we calculate how far on the journey the traveler was when the light that left earth reached the eyes of the traveler later at the turnaround point ? I think so .

It was when 60 percent of the age of the traveler was equal to 1 year ,so

When the traveler’s clock was 1.67 years past the start point ..:when that line of simultaneity equated 1 year earth time for traveler’s perspective . Is this Right?

If so moving on. In what the traveler perceived to be the time from 1.67 years age to 3 years aged, traveler observes the light of earth traveled “in length contracted reference frame), 2.4 light years or 4 times 60percent.

Now here’s what I don’t follow. That means the traveler observed light travel 2.4 light years while

His clock moved 3 minus 1.67 = 1.33 years of duration

If light travels at the speed of light in all reference frames, how can traveler see the earth light of 2.4 light years ago after 1.33 years passed for his clock from the calculated time of simultaneity (60 percent time dilation) to the visual observation

Edit: my question is how do we properly reconcile that the

light beams that the traveler is seeing are traveling at 1 light year per year from the traveler’s reference frame as light is supposed to travel at the speed of light in all reference frames

Edit: what then is a line of simultaneity represent then ? Why do those red lines of simultaneity for the traveler slant diagonally ? If what you all are saying is true they should be horizontal lines

### 8 Answers

- 1 month ago
Sustained acceleration from 0 to 0.80C and from 0.80C to 0 cannot exceed 1.5 g.

Plug that in.

- PhilomelLv 72 months ago
No one Knows. It is all theoretical and conjecture. If you don't like the answers you got make up your own.

- AmyLv 72 months ago
Not digging through all of this, but remember that from the traveler's perspective, Earth is moving away from him. So the distance that light traveled from Earth-in-the-past is less than the distance Earth is at right now.

Your own line of simultaneity is horizontal in your own reference frame. A diagonal line represents which events (which are not simultaneous to you) somebody else perceives as simultaneous.

- VamanLv 72 months ago
Twin paradox is for just discussions. You can not expect to happen. It can happen to a stone and the other nonliving beings. For people like us you need certain numbers of molecular reactions per-second has to take place. This can not be reduced. If a person is traveling with c will age as faster as the person sitting in the house. One can discuss that at tea time.

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- nebLv 72 months ago
ARGGHH. I’m not going to take the time to do calculations, but let me give you some basic facts.

Spacetime paths are INVARIANT meaning the path length something travels in spacetime is exactly the same in EVERY reference frame. Light takes the null ds^2=0 path and is LOCALLY always measured at exactly ‘c’ regardless of the observer (inertial, non-inertial, in gravitational field, etc)

NON-LOCALLY - depending on path - the measured speed of light by an ACCELERATED (non-inertial) observer over a path will NOT be ‘c’. It’s a little easier to understand within the context of general relativity. Non-local measurements in a gravitational field by stationary observers (actually undergoing an equivalent acceleration to stay at rest per equivalence principle), will NOT measure the speed of light as ‘c’ in other locations in the gravitational field. The extremal case with a black hole metric is that a far away observer will measure the outward radial speed of light at the event horizon to be zero.

Lastly, you have to factor in the different definitions of simultaneity for events on earth for the earth twin and events for the inertial (and non-inertial) portions of the trip by the spacefaring twin.

So, when you properly take into account that path lengths are invariant, the time between events for different observers is relative, and the speed of light is always MEASURED only locally as ‘c’ for accelerated observers, you will find everything works out properly.

Last time I answer this line of questioning. It’s imperative you get the basics down and I don’t think you have done that. It’s very easy to make relativity very complex very quickly and lose sight of basic principles, in particular when focusing on coordinate dependent quantities.

Update: I told you what a line of simultaneity meant in a previous answer

- MorningfoxLv 72 months ago
>> ... not see visually but perceive ...

Sorry, I don't understand that part. Each person can see only the light that hits him/her. You can work out the Earth time that the light left Earth ... especially if the light is an image of a clock on Earth.

There is no sudden "leap forward" of the Earth clock. The traveler will see the light from the Earth clock, and he will see that clock ticking at a smooth rate. That rate will change during the trip, but the clock doesn't jump from one time to another, skipping the times in-between.

You can also work out what the Earth observers see, as they look at the light from the travelers clock.

- DixonLv 72 months ago
I think you need to find a quick and simple way to ask the essence of your question because frankly it is unfollowable.

But near the start I think you lost it already by saying:

"Now.... on the leg journey away from earth, the Line of simultaneity suggests that for the traveling twin, ****it’s the Earth whose clock is aging 60 percent slower***"

No. During the leg itself, ie when both are inertial, both measure the other has a slow running clock. They are both correct in their observations. Similarly for the return journey.

The only thing that ages the Earth twin more than the traveller is what happens when the traveller turns around. And from the perspective of the traveller, depending on how exactly they turn round, the Earth twin either goes into fast forward during the traveller slowing down, reversing direction and speeding up again, or, the traveller sees Earth do a leap forward in time if the traveller can somehow change direction instantly but keep the same relative speed.

IE for the simplified model when the twin travels out and back at constant velocity but with an instant direction change, then all the action happens at the turn around and the traveller misses a whole bunch of Earth events.

- billrussell42Lv 72 months ago
so what is your question ?

link has more info on so called "twin paradox"