# How would the twins paradox be affected by wormholes?

The famous twins paradox where one twin travels to a distant star and returns to find an aged brother can be resolved with general relativity pretty easily, but I was wondering about this:

If the twin didn't accelerate at all, but instead used a wormhole to return to earth he would technically maintain his inertial frame of reference - so what would happen when he passes by earth?

Would both see the other as young while both seeing themselves as old (from the time dilation due to special relativity)? Or would the astronaut basically see a time 'jump' as the earthbound twin catches up to the astronaut?

If both twins undergo inertial motions, parting from one another at point $\mathscr{A}$ and meet again at point $\mathscr{B}$ in spacetime, you are simply describing a situation where the spacetime simply has multiple geodesics joining the two points $\mathscr{A}$ and $\mathscr{B}$.

Either twin can age more than the other, depending on which geodesic has longer "length", i.e. proper time. They may even age the same, as can happen between certain kinds of conjugate points on a manifold. But, for a given spacetime manifold, there is always a unique answer as to who ages more: the answer is well defined because the lengths of the two geodesics are well defined by the spacetime metric.

The paths don't need to be geodesics (inertial motions) for this principle to apply. You simply work out which path is the longest to find out who ages more, in the standard twin "paradox" scenario, or yours or any other variation.

The famous twins paradox [...] can be resolved with general relativity pretty easily

It can be resolved with special relativity. Clocks (and aging people) measure proper time along curves in spacetime. You can compute the length of a curve by breaking it into small pieces and approximating each small piece by a line segment and adding up the proper time of those line segments. As the pieces get smaller you get the length of the curve.

Using general relativity would be like using differential geometry to find the circumference of a circle. Completely unnecessary if you can compute lengths of shirt line segments and approximate the circle with a bunch of short line segments.

The twin paradox is resolved by learning to find out how clocks tick correctly by doing the above. For something moving in a straight line you don't need to break it into smaller and smaller pieces because you get the same answer. So the non inertial observer needs their curve to be broken into two pieces and the stay at hone twin can use just one piece.

But the point is that each twin ages an amount equal to the proper time of the curve they travel in in spacetime (and I mean spacetime, not their curve in space). You could compute the length of all the curves in any frame you want, the stay at home frame, either of the travelling twins frames or even in a fourth frame. What you can't do is pick the twins curve and falsely pretend it is a frame and say that they have no motion in space.

And this same prohibition is what comes up for the wormhole.

If the twin didn't accelerate at all, but instead used a wormhole to return to earth he would technically maintain his inertial frame of reference - so what would happen when he passes by earth?

The twin can move inertially, but there still isn't a global inertial frame that acts like Minkowski space. So in special relativity you can't have a global rigid inertial Minkowski frame follow something if that something fires a rocket or get pushed. But in general relativity you can't necessarily do that even if it moves inertially. All you can hope for in general is that there is an inertial frame that is instantaneously comoving with you at a particular event. The frame might only cover a small infinitesimal patch of spacetime.

Now, there is also an important distinction between what you see and what you compute. When we say observe we mean compute relative to your frame. Since your frame in general relativity might not cover a far away region there isn't even an analogy between the SR twins and the GR twins.

If you track what you see, then SR has a totally different resolution that is you actually see the other twin aging slower while you see them moving away. So the moving twin goes out and sees say the other one age one year while they age three years (at a constant rate of seeing one tick of your twins clock for every three if yours you see). Then they turn around, at which point they immediately (that is important) see the other one age faster so on the three year return trip you see your stay at home twin age nine years (you see three clocks ticks of your twin arrive for every one of yours you see). So the moving twin aged 6 years and the stay at home aged 10 years. What did the stay at home twin see? They saw the signal they sent out after one year of their clock get the moving twin as the moving twin turns around. So when they see that twin getting that signal is when they see the twin arrive. Since the relative speed is the same there is s 3 to 1 ratio of what you see from them compared to what you see for yourself. So the clocks ticks the moving twin saw on their way out were spread out over nine years. The stat at home twin spent nine years watching their twin age at 1/3 the rate and at the ninth year they finally saw the moving twin turn around and then they see the moving twin's clocks ticks images arrive three times as fast as their own (so nine times faster than they saw them come in before). This means they watch their twin live out three years of life over the next year.

The stay at home twin saw their moving twin moving away for nine years of their own time and saw their moving twin live three years of life. Then over the next year the stay at home twin saw their moving twin moving towards them for one year of their own time and saw their moving twin live three years of life. This is what they actually see with the telescopes and wrist watches, no computations just counting how many times you clock ticked and how many times you see their clocks tick and when you actually see the image of the other person's clock tick.

Now we can compare to the moving twin. The moving twin saw their stay at home twin moving away for three years of their own time and saw their moving twin live one year of life. Then the moving twin fires their rockets and since they are the one moving they immediately start getting signals differently. The other twin won't see the relative motion change until they see this happen but since the moving twin just accelerated (yes SR can handle acceleration just fine, just pick an inertial frame and compute in that frame) the accelerating twin is right there when it happens so sees it right away they see signals from their twin start coming in faster right away. And again, you can keep the old inertial frame and compute in that frame so keep using special relativity we are just having beams of light move and comparing how many intersect the world line of the twins if they are sent out so that there is an equal amount of proper time between beams).

Then over the next three years the moving twin sees their left behind twin moving towards them for three year of their own time and saw their moving twin live nine years of life. This is what they actually see with the telescopes and wrist watches, no computations just counting how many times you clock ticked and how many times you see their clocks tick and when you actually see the image of the other person's clock tick.

So the moving twin sees their twin live one year of life while seeing themself live three and sees them moving away from each other. Then the moving twin sees their twin live nine years of life while seeing themself live three and sees them moving towards each other.

So the stay at home twin sees their twin live three years of life while seeing themself live nine and sees them moving away from each other. Then the stay at home twin sees their twin live three years of life while seeing themself live one and sees them moving towards each other.

That's what they see. The asymmetry is the moving twin sees the change from moving away to moving towards each other happen sooner because they were there when it happens so didn't have to wait for it to see it.

They each see the other one age at one third the rate when they see each other moving away and they each see the other one age at three times the speed when they see each other moving towards each other.

The moving twin aged 6 years the stay at home aged 10 years. This is not mysterious. Now when you talk about what you observe what you mean is you take what you see then you compute a way the travel time of the light signals to compute when you think it left. In that case you compute that your someone moving at that speed ages at 6/10 the normal speed and ages slower regardless of whether they are moving away from you or towards you. That is called time dilation and is about what you compute not what you see (what you see is 1/3 aging when you see them moving away and you see true aging when you see them moving towards you).

Would both see the other as young while both seeing themselves as old (from the time dilation due to special relativity)? Or would the astronaut basically see a time 'jump' as the earthbound twin catches up to the astronaut?

If you try to compute (observe) then there might not even be a single frame that covers both the twins. If you look at what you see you realize that it is possible that you can look out towards the wormhole and can see an older earth. Then you watch your twin take off towards the wormhole but you also see that image of earth getting older.

You watch your twin aging slowly (and you see them age slower than time dilation predicts, such as 1/3 instead of 6/10) but you watch the wormhole image of earth age at the normal rate. For simplicity say the wormhole is six lightyears above the north pole and opens up six lightyears above the south pole. So at the north pole you look up and see the south pole from 12 years ago (you always see he past when you look at things far away). Now 0.8c is the speed that makes you see your twin age at 1/3 or 3 times when moving straight away or straight towards you (and be time dilated by 6/10 when you compute instead of seeing). So for people that compute it takes 15 years on earth before your twin arrives. But for you to see that event from this angle you have to wait for the light to travel the 12 light years, so it takes 15 years before you can walk to the south pole and shake hands but you have to wait 27 years to see them land on earth if you just sit there at the north pole and watch. You see them age at 1/3 the rate so you see them 27/3=9 years older when they arrive. What happens if you go to the south pole and watch?

So what happens if you look up from the south pole? Your twin has 12 light years to travel at 0.8c so 15 years later they will arrive. But you start out seeing the north pole from 12 years ago (you always see the past when you at far away things). So for 12 years you see nothing. Then you see your twin take off. They are moving towards you so you see them age at triple speed so over the next three of your birthdays you see them get nine years older.

Everyone agrees when you do everything properly. Note that those were both for the stay at home twin. For 12 years they see their twin going away and aging slowly. Then they see two images of their twin, one that is taking off and aging quickly and one that is still aging slowly and moving away. Then three years later their twin arrives and is nine years older. And you stop seeing the faster aging twin. But the images of the twin running away are still coming in from another 12 years since it takes time to see things from far away. So you can look at his rear bumper for 27 years and see him age at 1/3 speed or you can wait 12 years then watch his front bumper and see him age at 3x speed for 3 years. Either way he is nine years older.

How about from the moving twin's view. What does he see? He sees his twin on the north pole age slowly and when he arrives at the south pole he is still seeing his twin from 12 years ago because that is the light that just there. He aged 9 years and saw only 3 of his twin's birthdays so far. So he thinks the distance traveled was (9 years)( 0.8 c) = 0.72 lightyears. That's time dilation. Is that consistent? He arrive 15 years later, and 9 years older and he sees his twin from 12 years ago if he looks behind him, so a twin that is 3 years older. So to be clear if the moving twin looks behind him he sees a twin that ages at 1/ 3 speed so sees a three years old twin's image as he arrives nine years later and can talk to his twin that is 15 years older than when he left (you are always 12 years older than the images you get from looking at earth through the wormhole).

OK but what if the moving twin looks forward at his destination? He starts out seeing the earth from 12 years ago. He travels for 9 years and sees it age at triple speed (he is moving towards it) so over that nine subjective years worth of travel he sees 27 years of activity on the earth he sees in front of him. It started out 12 years behind so it is 15 years older when he arrives. Everything is consistent.

The moral is if you want to focus on what people see it is consistent but you have to say when, where and in which direction they look.

he would technically maintain his inertial frame of reference

No. In general relativity there don't have to be global inertial frames, that is literally what the word general means (special relativity allows acceleration just fine by computing in a global inertial frame that observers the acceleration).

so what would happen when he passes by earth?

I've answered that, he is 9 years older, his twin is 15 years older. And they sometimes see multiple images of each other one aging at 1/3 speed and one aging at 3/1 speed.

Would both see the other as young while both seeing themselves as old (from the time dilation due to special relativity)?

If they look where they see the twin at first they see them get younger as they move away. If they look the other way they have to wait before they see the launch and then they see them age at triple speed. And it all works out.

Stay at home sees slow twin travelling away over the next 27 years. Stay at home can wait 12 years then start seeing fast twin travelling toward them for the next three years. At which point wormhole twin lands. And wormhole twin is 9 years older when they land.

Wormhole twin sees slowing aging twin age three years by the time they land, but they know that image is from 12 years ago so it is 15 years later. Looking the other way the wormhole twin starts out seeing the earth from 12 years ago and sees it age at triple speed for nine years so arrives nine years older to an earth that is 15 years older (27 years older than the original image it saw looking at the originally far away earth).

The asymmetry comes from the fact that the wormhole connects times that look the same to the earth. From the example were the moving twin fires rockets the asymmetry came from moving twin seeing the change earlier.

You can get similar math if the universe just repeats (no wormhole needed just cut together a flat spacetime) or just have a series of identical earths spaces out every 12 lightyears with identical people and rockets and such all taking off at the same time (same time to the stay at home twin, as computed/observed not as seen).

Or would the astronaut basically see a time 'jump' as the earthbound twin catches up to the astronaut?

You never see a jump. If you switch coordinates then the value of a coordinate might be very different in the two coordinate systems.

I think Duffield's answer can be summarized better this way: a putative "worm hole" is our view of a bending of 3-space thru 4 spatial dimensions such that the Euclidean distance from Earth to \$DISTANT_PLANET is extremely small. In our 3-D universe, we can't move along this putative 4th dimension, so we're forced to go the "long way" thru space which is, undetectably to us, warped in 4-dimensions.

the analog would be living on the corner of a 2-dimensional sheet of paper. Travel to the far corner takes a long time, but if the paper is warped in 3-space and you can find a "wormhole" to jump from one corner, thru 3-space, to the far corner, then the distance is very small

Thus, a wormhole obviates the need for high-speed, long-distance travel, and the difference between the time clocks for the twins is negligable.

How would the twins paradox be affected by wormholes?

Not at all. The so-called paradox is that each twin claims time is "passing more slowly" for the other twin. People say Whoa! Paradox! to this, but it isn't a paradox at all. When you and I are separated by distance, I say you look smaller than me and you say I look smaller than you. But we don't say Whoa! Paradox! It's similar when we're separated by relative motion. My time looks slower than yours and your time looks slower than mine. But there's absolutely nothing weird about it at all.

Take a look at the Simple inference of time dilation due to relative velocity. Imagine you're the light reflecting back and forth in one parallel-mirror light clock, and I'm the light reflecting back and forth in another, and that we have motion relative to each other. I see me like this || and you like this /\ . You see you like this || and me like this /\ . Our times are the number of reflections. So I claim your time is less than mine and you claim mine is less than yours. That's it. It's that simple.

The famous twins paradox where one twin travels to a distant star and returns to find an aged brother can be resolved with general relativity

Actually Lemon, it's resolved via Special Relativity.

If the twin didn't accelerate at all, but instead used a wormhole to return to earth he would technically maintain his inertial frame of reference - so what would happen when he passes by earth?