Time dilation questions

I board my rocket ship and start travelling. Fast. I come back to earth to discover what was days for me is months/years for earthlings. We’ve all heard that type of scenario.

It seems that if time can pass differently, that indicates some sort of frame of reference. What I mean is - Why can’t my rocket ship be considered the reference point? Earth has moved away from me and back! Why aren’t I older and earthlings younger. Next, if two objects can experience dilation, then can’t we use this to determine some sort of 0 velocity in space? Whichever object ages faster must be going slower. Keep adjusting vector and speed to maximize this effect and you would be at “rest” with the cosmos?

Your rocket ship can be considered the reference point. So long as you and the Earth are both in inertial frames, the earth will see your time going slower, and you will see the earth’s time going slower.

But, if both your rocket and the earth stay in inertial frames, you can never return to earth. In order to return to earth, you spent some time in an accelerated frame to turn around, and that breaks the symmetry. If you stayed in your inertial frame and instead the earth accelerated to catch up with you, the people on earth would be younger than you when you met up.

This is called the Twin Paradox.

Excellent tutorial on it here.

To make this a bit simpler to explain, suppose that you don’t have your own spaceship, and so have to hitch rides with aliens. One alien is heading Galactic West, and takes you to another star system, and keeps on going. Just then, another alien comes by heading Galactic East, and takes you back to Earth, and then keeps on going.

There are now three relevant frames of reference in this problem, not two: You can work in the Earth’s frame, or you can work in the westward frame, or you can work in the eastward frame (or you can work in some other frame entirely, though I’m not sure why you would). In all cases, no matter what frame you’re working in, you will get the same result: When the twins meet up again, the one who stayed on Earth will be older. But you have to make up your mind on a single frame and stick with it: You can’t work in “the traveller’s frame”, because the traveller doesn’t have a single frame: He starts off in the westward frame, and then switches to the eastward frame.

For each person in every reference frame, time passes at one second per second. Time never goes faster or slower. It’s just that when the amount of elapsed time is compared, the numbers are different.

This is totally contrary to “common sense,” true, which is why people keep trying to make it conform to whatever common sense interpretations of reality they have. It won’t work. The universe does not run by what earthlings consider common sense. It has its own rules.

Isn’t it basically GR? Time-dilation occurs because the twin the in the rocket ship is the one undergoing acceleration, which is equivalent to a gravitational force.

The twin on Earth, is more or less, not undergoing acceleration, but maintaining a constant speed. So your clock is going way slower than his when you’re accelerating there and back. Or hell, just orbit the Earth close to c for a while.

General relativity is not required to explain why the Twin Paradox is not a paradox. The easiest way to analyze it is to assume the acceleration is instantaneous, so in Chronos’ suggestion, the traveler jumps onto the alien spacecraft and instantly switches reference frames. There are three inertial frames (as he mentioned) and everything can be analyzed with just special relativity.

As Neil deGrasse Tyson put it:

“The Universe is under no obligation to make sense to you.”

I see. Interesting, thanks. Curious you can work it out in both SR and GR.

Special relativity can be seen as just a “special” case of general relativity, when spacetime is pseudo-Euclidean. Special relativity can handle acceleration (it would be a very poor theory if it could not!).

By far the easiest way to see how the twin paradox is resolved is to draw a spacetime diagram, as can be found in Senegoid’s link.

Some of these are helping.

Let’s say that two aliens fly into the Milky Way on their spaceship. They are not in a frame of reference with this spinning galaxy; instead their frame of reference would be closest to the galactic center which isn’t traveling vast distances as it rotates. They watch the arms of the galaxy spinning. (Assume they don’t get bored easy and are immortal). One alien wants to land on Earth. The other isn’t interested and instead wants to stay put. The first alien takes a space walk out of the ship and harpoons Earth as it swings by at 792,000 km/h. He rides Earth for an entire revolution around the galaxy. When he gets back one full loop he ‘stops’ and boards the ship again. 250 million years have passed for him since he was on Earth the whole time.

In that case shouldn’t the other alien think MORE time has passed? The alien landing on Earth actually had to accelerate twice, and traveled a very long distance. Time dilation should mean it’s been longer at the alien vessel.

Meanwhile, humans evolve into immortal robot like creatures. They send out a probe that is accelerating away from Earth in the opposite direction of the spin of the galaxy. It also travels at 792,000 km/s. They reach this time capsule in 225 million years after they have rotated one loop. In this case the capsule has done the accelerating and time dilation means less time has passed for the capsule.

Which is correct? It can’t be both because the capsule and the ship seem to be interchangeable.

Ignoring the complications gravity and general relativity:

1. As you say the Alien who stays in the “galactic centred inertial frame” (GCIF) experiences the most time.

2. In this case the probe experiences the most time. In order to rotate around the galaxy the Earthbound observer is accelerating, and though the probe also experiences acceleration, from the point of view of an observer in the GCIF, the speed of the probe is at times less than the speed of the Earthbound observer and is never more than the Earthbound observer. In this situation this is enough to show that the probe experiences the most time.

In a problem where all of the acceleration is due to gravity, you can’t do that and get any sort of meaningful answer.

It’s to illustrate a point about special relativity which does not deal with gravity. If we are allowed to set the situation up so that the “kinematic time dilation” dominates over the “gravitational time dilation” then the answers are qualitatively the same.

So particles that travel at the speed of light (e.g. neutrinos, photons) experience NO time flow-they are frozen in time, local time doesn’t exist for them/

Yes. From the photon’s perspective it is an event, not a trip. For the photon, emission and absorption happen at the same point in space/time. (The entire universe is Lorentz-contracted to a 2-D plane in the direction of the photon’s travel.)

Neutrinos, however, don’t quite travel at c. They have very low mass, and so travel very quickly, but it’s still meaningful to speak of a neutrino’s frame of reference.

Which I understand leads to problems if the universe is closed (such as a hypersphere). In that case it would be possible for two observers in relative motion to (eventually, having circumnavigated the universe!) come together again without accelerating.

A closed universe does, in fact, have a preferred reference frame, at least globally.

In fact this is known as the cosmological twin paradox.

Take for example the Einstein static Universe, which has the geometry and topology of hypersphere and space is static (neither expanding or contracting). It is perfectly possible for one observer to stay at home on his planet (which we assume doesn’t move relative to the fixed background of stars/galaxies), but for another observer to circumnavigate the Universe, without accelerating and starting and finishing at the planet. In this case the observer who circumnavigates the Universe experiences less time than the observer who stays at home. So as Chronos says there is a preferred reference frame decided by the fixed background of stars/galaxies.

In fact even if space is completely empty and has a flat geometry and you can still set up a similar situation. Take flat Minkowski space from special relativity described by an arbitrary set of (Minkwoski) inertial coordinates (x,y,z,t), if you make the identification (a,y,z,t) = (0,y,z,t) for some value of x, a, greater than 0, then you can “roll up” space into a sort of “hypercylinder” , much like you can roll up a flat newspaper into a tube without creasing it. In this situation an observer who remains in the inertial frame which was used to create the “hypercylinder” experiences more time than an observer who travels around it.

The reason is that non-accelerating observers in spacetime have timelike geodesic wordlines. In many spacetimes(like normal “non-rolled up” Minkwoski space), timelike geodesics can only intersect once, but others (like the above two examples) they can intersect more than once. The definition of a timelike geodesic is that it locally maximizes the proper time along it, but that doesn’t mean it globally maximizes the proper time, which is why one inertial observer can experience more time than another inertial observer, even when they start and finish at the same time and place.

In Riemannian space (as opposed to Lorentizan spacetime) a geodesic locally minimizes the distance along it, so an analogy would be two travellers who want to go from New York to Washington DC. One traveller sets out in a South-West direction and travels 200 miles, the other one sets out in a North-East direction, but has to travel over 24,000 miles. They both take geodesic (locally distance minimizing) paths on the spherical surface of the Earth, which start and end at the same point, but one is significantly shorter than the other.