1. Feb 17, 2016

### FallenApple

So say that there are twins. Twin A is near a heavy planet and twin B is out far away in space.

Why does time slow for twin A compared to B? Can't they say that they both are in inertial frames because there are no forces(gravity isn't a force) acting upon them?

Therefore for each of them can consider the other younger. After all, twin A can just say that he is stationary and that its B that is under a high gravitational field and is accelerating and vice versa.

2. Feb 17, 2016

### PAllen

In special relativity (no gravity), reciprocal time dilation only applies to inertial frames. Thus your argument is already wrong even without gravity.

Further, you propose no twin scenario, though you apparently think you have by labeling observers as twins. What characterizes a twin scenario is twins that synchronize clocks at common place and moment, go their separate ways, and meet again, to compare their clocks. This removes any reciprocal relation even in SR - in general one twin will have aged less, and both will agree on it. It is easy to set up twin scenarios involving gravity, but you have not done so.

In your scenario, A is a non-inertial observer (it is undergoing proper acceleration relative to a free fall = inertial world line; it is experiencing a force from the planet surface), B is presumably an inertial observer (far enough away, the difference between a static observer and an inertial observer is inconsequential). This characterization is invariant, and mutually agreed, not relative. They will both agree A's clock runs slower (e.g. comparing them via signals each emits 1 x per second for their seconds). This non-reciprocal clock rate situation is called gravitational time dilation (I think you know this part).

3. Feb 17, 2016

### drvrm

In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more.
so humbly i suggest to please explain or state your 'twin's status' and 'problem more clearly.

in the above statement you are assuming that there are no forces/accelerations operating so the physical laws will be same for both twins- no problem- gravity is not a usual force but its effect is felt by you on earth otherwise you will fly off in space- so please clarify your statements so that one can understand you!

4. Feb 17, 2016

### stevendaryl

Staff Emeritus
If you have two twins that travel nearly the same path through spacetime, then the general rule of thumb is that the one who travels the inertial path (no accelerations) will experience the most elapsed time---he will be the oldest when they get back together. In Special Relativity, there is exactly one inertial path connecting two spacetime points. However, in General Relativity, there can be multiple inertial paths connecting two spacetime points. They do not need to be the same amount of elapsed time.

An analogy is on the surface of the Earth; if you're traveling within a small region on the Earth, there will typically be exactly one "straightest" path connecting two points on the Earth, that's the "great circle route" or "geodesic" path connecting those points. But there can be more than one geodesic path connecting two points, if you allow travel through a larger region. For instance, to go between two points on the equator, you can either travel East or West. Both directions are geodesics and are "straight" paths (or as straight as you can go, staying on the surface of the Earth), but the two paths don't have the same length, in general.

5. Feb 17, 2016

### Staff: Mentor

If the two twins do not start at the same place at the same time, separate for a while, and then reunite so that they again are at the same place at the same time again, you don't have a twin paradox problem at all. You have an (uninteresting, once you realize what's going on) discussion about simultaneity conventions and how it is basically meaningless to speak of one twin being older or younger than the other if you can't set them side by side to compare their ages.

However, it's easy to fix that defect, and doing so leads to one of the more interesting variations of the twin paradox. Let's say that the two twins start at the same place at the same time, far from the heavy planet. Twin B remains there, floating in free fall in empty space, while twin A moves towards the planet, executes a tight hairpin hyperbola sport of orbit that takes him deep into the gravity well and then sends him back out in the opposite direction so eventually he meets up with twin B again. Thus they take different paths through space yet both are in free fall and moving inertially throughout.

Twin A will be younger when they meet again. This happens because both twins travelled through spacetime, starting at the separation event and ending at the reuniting event - but they travelled on different paths, and twin A's path was shorter than twin B's path. It's no different, (except that we're doing it with time in spacetime instead of distance in space) from what happens when two drivers in two cars leave from somewhere at the same time, arrive at their common destination at the same time, and their odometers show a different number of miles covered - they took different routes of different length.

From this you should conclude that the common explanation of non-the simpler gravity-free version of the twin paradox (one twin experienced acceleration, the other didn't, and that's what made the difference) is actually quite misleading. Even in that case, what's really going on is the stay-at-home twin took a longer path through spacetime, so more time elapsed on his path. The acceleration only comes into the picture because we needed to accelerate one or the other twins to set them on different paths through spacetime.

6. Feb 17, 2016

### FallenApple

Thanks, your explanation cleared up a lot.

Twin A took a shorter path through spacetime. Hence the shorter time. That makes sense.

But theres one thing that I'm unclear on. A can just claim it is still that it's B that moved to a "gravity well" in a hyperbolic orbit mirroring your description. In this case. It would be B that took the shorter path through spacetime.

7. Feb 18, 2016

### FallenApple

But can't you say that it's twin A that is stationary and that twin B is on a geodesic on an imaginary planet? Then from A's perspective, B is acclerating while A is stationary.

8. Feb 18, 2016

### pervect

Staff Emeritus
No, you can't. You might try reformulating your thought experiment to be compatible with special relativity. This involves replacing gravitating bodies with Einstein's elevator.

Unfortunatley, special relativity is just not able to handle gravity. GR is required. There are some useful things you can figure out within the limits of SR, for instance by using Einstein's elevator as a thought experiment to create a form of "gravity" that SR can actually handle.

I don't like to be negative, but at this point all I can say is that your guesses so far are wrong. It might be helpful to figure out how to rephrase your concerns as questions, rather than just making speculations. It's easier to point someone on to the right track than it is to get them off the wrong track and then try to redirect them.

9. Feb 18, 2016

### Staff: Mentor

Both twins can consider themselves to be at rest while the other one moves away, executes a sharp hairpin turn, and then returns. Nonetheless, the situation is not symmetrical because one path goes closer to the planet than the other, and will be the path of less elapsed time - the spacetime curvature is stronger closer to the planet and that makes the difference.

Check out the "Doppler analysis" section that FAQ that I linked to. The same technique works for analyzing this gravitational version of the twin paradox, as long as you allow for the effects of gravitational time dilation.

10. Feb 18, 2016

### FallenApple

I'm new to this. So I thought that the term relativity means that observations are relative, general or special.

Ok, got it. For me, speculations are tantamount to questions. It makes it more succinct. I do see your point though.

11. Feb 18, 2016

### pervect

Staff Emeritus
OK, lets try an explanation on Einstien's elevator. Suppose you're on such an elevator, which is accelerating constantly - it's pulled by a rope, perhaps, or perhaps it's powered by rocket engines, it doesn't matter to the problem.

Suppose you are riding in the elevator, hold something 10 feet off the floor, and then drop it. It falls down.

Now, let's look at it from an inertial frame of reference. Initially, when you're holding it "in place", there is a force on the object, and it's accelerating. When you drop it, the force is removed, the object is in free fall, it does not accelerate - and the elevator floor catches up to it, because the floor is accelerating, powered by the cable (or the rocket).

So, the object in free-fall in the elevator is the one following a geodesic - the elevator floor is not following a geodesic, because it's accelerating.

You can draw an imprefect analogy to the case with real gravity from Einstein's elevator, though it's risky to stretch it too far. It does work to say though, that the object in free fall is the one following a geodesic, an object that's not in free fall is NOT following a geodesic.

As far as time goes, in SR you can say that a geodesic maximizes proper time. In GR a bit more care is needed. If you consider short enough time intervals and a correspondingly small local neighborhood, you can say that the geodesic path maximizes proper time as it does in SR, but in general you can only say it's an extremal point - to go into more detail than that and explain the remark needs a bit of math.

12. Feb 18, 2016

### stevendaryl

Staff Emeritus
I think you missed the point of my post. "Stationary" versus "not stationary" is relative, but being on a geodesic is not. Given two paths that are almost the same (they only differ slightly), the geodesic will have the greatest elapsed time---that twin will be oldest when they get back together. If two paths are very different (as is the case with two different orbits), then you can't answer the question without actually studying the geometry of the situation. That is, you have to do a calculation.