Moving faster but taking longer

[gonegahgah]

The twin traveling away for 1T and then returning (disregarding acceleration/deceleration completely) is said to return to a much older Earth which is older by multiple 1T and the returning twin is said to be younger at a fraction of 1T. That is correct is it not?

 

[gonegahgah]

Let me reword that; seeing I was in a rush.

A twin (A) travels away in a spaceship at high velocity to a far star and returns. They return to find the Earth, and other twin (E), have aged much more than them.

For the twin (E) on Earth it appears the traveling twin (A) has aged very little while they have grown very old.

The A twin saw themselves traveling in normal time but to the E twin the A twin animated much slower. So A traveled a long way at a fast speed but to us they took longer to get to the star and back.

That is the story is it not?

 

[gonegahgah]

I'll assume that the above is correct and just so common place that no-one is bothering to confirm it as it has been said so many times.

Okay, some new questioning I hope.
If person A is standing at the artic circle (so that they see the sun again 24 hours later) and person B is standing at the equator (again they will see the sun again 24 hours later) they will both be traveling at different relative speeds obviously (hence the cariolis effect).
I'm guessing that they will be traveling different world lines. It's all a bit confusing at the moment and I will have to find time to work through this world lines time dilation, GR time dilation, SR time dilation and acceleration time dilation stuff. But that's not important to this question.
You blokes should be able to tell me easily enough. Will one of these characters see the other character as animating slower? If so does A see B animating slower or does B see A animating slower?

 

[Fredrik]

You got it right in #2. Regarding #3, you need to know is that there are difficulties associated with comparing clocks at different locations. That has been debated a few times in this forum. You could change the scenario to this:

A and B are both standing at a point on the equator with synchronized clocks and both decide to walk to the opposite side of the planet. (Pretend that they're doing this on a spherical rotating planet with no water or any other obstacles). A takes the shortest path that goes through the north pole and B walks along the equator. They should both reach the point on the opposite side at the same time if they've been walking with the same speed. They compare their clocks when they meet.

If we naively treat this as an exercise in SR time dilation, then we conclude that A's clock will be ahead of B's. But I think the correct (GR) answer is the opposite. B's clock will be ahead of A's, because B's world line deviates less from being a geodesic than A's. A geodesic such as a low-altitude circular orbit is a world line that maximizes the time measured by the clock, and B's higher velocity makes his world line look more like a geodesic than A's. B is closer to being in orbit.

 

[gonegahgah]

Thanks Fredrik. I will have to study the world line thing, and also the rotating space station presented idea presented by someone else when I get more time. That is a good example that you provide. It provides a means of starting together and meeting together.

I don't know myself which one animates slower of the two but the one thing most people here would agree on is that they would get back together and their clocks would be different; and that this should even be calculable.

I do have a question on one part you wrote. You said that they should reach the point on the opposite side simultaneously if they both travel the same speed. If they both set out at the same speed by their own clock and one animates slower than the other won't one arrive before the other and have to wait for the second one to arrive later?

 

[gonegahgah]

Your online but haven't answered Fredrik. Isn't it similar for the twin example where twin A stays at home and twin B travels away in space and comes back.

Twin B travels away and comes back to find the Earth and their twin A are much older while twin B is still young. Twin B appeared to animate slower to twin A as did their ship travelling. To twin A the ship and twin B take longer to return then they would have by direct Newtonian application of distance and speed.

Admittedly twin A isn't traveling to their rendevous at the same self-measured speed as twin B - like in your example - but the common element between our two Earth traversers and the twins is that one animates slower than the other and the one animating slower ages slower and stays younger but takes longer to get to the rendevous.

In your example you said that traverser B's clock was ahead of A's. This would mean that A is animating slower than B. So in your example, where A walks through the North Pole and B walks along the equator, traverser B would arrive at the rendevous first and traverser A would rock up later.

This is correct isn't it?

 

[JesseM]

I do have a question on one part you wrote. You said that they should reach the point on the opposite side simultaneously if they both travel the same speed. If they both set out at the same speed by their own clock and one animates slower than the other won't one arrive before the other and have to wait for the second one to arrive later?

If they're both traveling at the same speed in the coordinate system we're using (say, the rest frame of the planet's center), then they will cover the same distance in the same time in this coordinate system, since speed is defined as (change in position coordinate)/(change in time coordinate). In other frames, like the instantaneous rest frame of one of the travelers (if they're not moving inertially they will have different inertial rest frames at different moments), the distance they travel and the time it takes may be different, although if one frame predicts they arrive at the destination at the same moment, then all frames will make the same prediction.

Your online but haven't answered Fredrik. Isn't it similar for the twin example where twin A stays at home and twin B travels away in space and comes back.

Twin B travels away and comes back to find the Earth and their twin A are much older while twin B is still young. Twin B appeared to animate slower to twin A as did their ship travelling. To twin A the ship and twin B take longer to return then they would have by direct Newtonian application of distance and speed.

No. Whatever twin B's speed in A's frame, it will be equal to the distance B travels divided by the time it takes for B to cover this path (with both distance and time defined in A's frame). Again, that's just how speed is defined.

 

[gonegahgah]

Cool. Well Jesse let's say that they are traveling by their own clocks; and not by milestones on the Earth. They jointly set a common speed that they will both stick too. Traverser A travels 1L for every 1T of their own clock. Traverser B travels 1L for every 1T of their own clock. (Whatever it is that they set 1L as and 1T as).

So A thinks he himself is traveling at 1L/1T by his clock (and speedo) and B thinks that she herself is traveling at 1L/1T by her clock (and speedo).

As we have discussed, one of them sees the other one's clock animating slower because they are in a different world lines - as per Fredrik - and in accordance with the twin paradox. Let's say that he is walking along the equator and she is walking via the North Pole. According to Fredrik, the male traveller would see the female traveller, who is going via the Pole, as animating slower. So the male traveller would see himself traveling faster than her. The male equator follower would arrive at the rendevous sooner than the female pole passer. Is that not correct?

You know what I'm saying for the second part Jesse. If traveller twin B travels at what they think is near the speed of light; home twin A will actually see them as traveling much slower than this. If a star is 50 light years away and B travels near c to the star and back it will take a lot longer than 100 years for B to get back. It will seem to A that B is not traveling near the speed of light.

To B it will seem that they took a fraction of 100 years to travel the trip but to A it will seem that B took a multiple of 100 years to travel the trip. Correct?

 

[JesseM]

Cool. Well Jesse let's say that they are traveling by their own clocks; and not by milestones on the Earth. They jointly set a common speed that they will both stick too. Traverser A travels 1L for every 1T of their own clock. Traverser B travels 1L for every 1T of their own clock. (Whatever it is that they set 1L as and 1T as).

In their own rest frames they are not changing position at all because they are at rest, by definition. If you have some set of external landmarks they're both moving with respect to, then each observer can define their speed with respect to the landmarks, in terms of the time between passing a pair of landmarks and the distance between those landmarks in the observer's rest frame.

So A thinks he himself is traveling at 1L/1T by his clock (and speedo) and B thinks that she herself is traveling at 1L/1T by her clock (and speedo).

As we have discussed, one of them sees the other one's clock animating slower because they are in a different world lines - as per Fredrik - and in accordance with the twin paradox. Let's say that he is walking along the equator and she is walking via the North Pole. According to Fredrik, the male traveller would see the female traveller, who is going via the Pole, as animating slower. So the male traveller would see himself traveling faster than her. The male equator follower would arrive at the rendevous sooner than the female pole passer. Is that not correct?

I don't think the question is sufficiently well-defined. If they are traveling along the curved surface of the Earth then they do not have a single inertial rest frame throughout the journey, and there isn't really a "standard" way to construct a frame for a non-inertial observer. Also, are we assuming the Earth is rotating? Are we assuming they are moving on the surface of a massless sphere so this is a pure SR scenario and the only type of time dilation is velocity-dependent, or are we assuming the Earth is massive and causing spacetime curvature so it's a GR situation involving gravitational time dilation?

You know what I'm saying for the second part Jesse. If traveller twin B travels at what they think is near the speed of light; home twin A will actually see them as traveling much slower than this.

"Near the speed of light" relative to what? If you have some markers set up which are at rest in A's frame, and the markers are 0.8 light year apart in A's frame, then if B is traveling at 0.8c relative to the markers, that means A will measure B to take 1 year to travel between a pair of markers. Now, it's true that in A's frame, B's clock appears slowed down, so in that 1 year B's clock only advances forward by 0.6 years. But although it's true that in B's frame only 0.6 years go by between passing a pair of markers, it's also true that in B's frame the distance between the markers is shrunk to 0.6*0.8 light years = 0.48 light years due to the length contraction effect. So B sees himself take 0.6 years to pass between two markers which are 0.48 light years apart, and therefore concludes he is moving at 0.48 ly/0.6 y = 0.8c relative to the markers, the same speed that A measured B to be moving relative to the markers.

If a star is 50 light years away and B travels near c to the star and back it will take a lot longer than 100 years for B to get back. It will seem to A that B is not traveling near the speed of light.

To B it will seem that they took a fraction of 100 years to travel the trip but to A it will seem that B took a multiple of 100 years to travel the trip. Correct?

When you take both time dilation and length contraction into account, you find that whatever speed B is traveling in the rest frame of A and the star, B measures A and the star to be moving at this speed relative to himself in his own rest frame. So if B measures A and the star to be moving at 0.99c relative to himself, then A will also measure B to be moving at 0.99c relative to himself. Of course the time between B passing A and B reaching the star will be much smaller in B's frame, but then again the distance between A and the star will be much smaller in B's frame too.

 

[gonegahgah]

Can you please just answer the questions Jesse? They are not difficult questions!

 

[JesseM]

Can you please just answer the questions Jesse? They are not difficult questions!

What questions, specifically? Just the two paragraphs where you asked "correct?" If so, I told you that your first scenario involving people moving along the Earth was not sufficiently well-defined and asked you for some clarifications. As for your second "correct?", I thought I made it clear that my answer was no, you were not correct, because if A measured the relative speed between A & B to be close to the speed of light, B's measurement of the relative speed between A & B would be exactly the same. So, your statement "If a star is 50 light years away and B travels near c to the star and back ... It will seem to A that B is not traveling near the speed of light" is incorrect, they will both measure the relative speed between A and B to be exactly the same. (On the other hand, your statement 'To B it will seem that they took a fraction of 100 years to travel the trip but to A it will seem that B took a multiple of 100 years to travel the trip' was correct, but this does not imply they disagree about the relative speed, the difference in time is just because in B's frame the distance from A to the star is a lot smaller than 100 light-years due to length contraction).

 

[gonegahgah]

Let's start simple.

Is Fredrick correct about his world lines; or should I back track immediately here so that we can get somewhere?
Is Fredrick correct that the pole passer will animate slower than the equator follower?

 

[JesseM]

Let's start simple.

Is Fredrick correct about his world lines; or should I back track immediately here so that we can get somewhere?
Is Fredrick correct that the pole passer will animate slower than the equator follower?

Fredrick was considering a GR scenario rather than an SR scenario, but I don't think his argument about the equatorial path being "closer to a geodesic" is convincing. All orbits are geodesics, but there are valid orbits that go through the poles and valid orbits that go through the equator, and they both require a particular orbital speed (as measured in a center-of mass coordinate system, not measured in terms of speed relative to the rotating surface of the Earth), so Fredrick's argument might be right if the speed of the equatorial path was closer to the speed for a circular orbit at that radius (although even then the argument is pretty handwavey), but we could equally well have a scenario where the polar path was closer to the orbital speed.

Is there any particular reason you feel the need to consider a GR scenario here? It really would be much simpler to imagine the Earth as a massless rotating sphere so the problem would be a pure SR one, you'd still have velocity-based time dilation in this scenario.

 

[Fredrik]

Fredrick was considering a GR scenario rather than an SR scenario, but I don't think his argument about the equatorial path being "closer to a geodesic" is convincing.

For the record, I'm not convinced either, but I still believe it's correct. (Edit: Not anymore. See my next post below). If I'm wrong I'd like to find out how.

All orbits are geodesics, but there are valid orbits that go through the poles and valid orbits that go through the equator, and they both require a particular orbital speed (as measured in a center-of mass coordinate system, not measured in terms of speed relative to the rotating surface of the Earth), so Fredrick's argument might be right if the speed of the equatorial path was closer to the speed for a circular orbit at that radius (although even then the argument is pretty handwavey), but we could equally well have a scenario where the polar path was closer to the orbital speed.

I'm assuming that the rotational axis of this planet is such that the velocity (in the inertial frame where the center of the planet is at rest) of someone standing on the surface is zero at the poles and has its maximum at the equator. I'm also assuming that this "speed of the equator" is much higher than these people's speed relative to the surface.

I'm also assuming that the mass and rotational velocity of this planet is of the same order of magnitude as the mass and rotational velocity of the Earth. I believe that the speed required to stay in a very low (almost touching the surface of the planet) circular orbit is essentially the same regardless of whether the orbit is around the equator, or goes through the poles. (GR frame dragging probably causes a small correction to that, but it has to be small. Otherwise Newton's theory of gravity wouldn't work so well in the solar system).

With those assumptions, it seems pretty clear to me that the person moving along the equator will be closer to being in a low altitude circular orbit (which is a geodesic) than the one moving on the path through the north pole. The only possible counterargument I can think of is that maybe there's some other geodesic that's closer to the other guy's world line. (E.g. a circular orbit just below the surface).

Maybe I shouldn't have talked about people who walk around the planet, because if I'm right about time dilation, it will be natural for them to walk with different speeds in the rotating frame in which the center of the planet is at rest and all fixed points on the surface are at constant spatial coordinates. Let's just assume that the travelers' coordinate speeds are the same in that frame. This guarantees that they will reach the point on the opposite side at the same time.

Your online but haven't answered Fredrik.

You do know that I'm not getting paid for this, right?

Is there any particular reason you feel the need to consider a GR scenario here?

I'm curious about this too. Gonegahgah, if you don't know what a world line is, you will probably find it more enlightening to study the basics instead of getting other people to tell you the answer to specific problems with specific complications.

 

[Ich]

For the record, I'm not convinced either, but I still believe it's correct. If I'm wrong I'd like to find out how.

In this scenario, it's fairly easy to split the GR effects into a "pure GR" part - due to gravitational potential, and a "SR part" due to velocity.
If you do that, and if you assume a spherical planet, all "pure GR" terms cancel for movement along the surface.
Then this is the Haefele-Keating-scenario, where A' clock reads more or less than B's clock, depending on the direction of the equatorial movement.
Geodesics are only local maxima, not global ones. There are sometimes accelerated paths with greater proper time between two events.

 

[Fredrik]

In this scenario, it's fairly easy to split the GR effects into a "pure GR" part - due to gravitational potential, and a "SR part" due to velocity.
If you do that, and if you assume a spherical planet, all "pure GR" terms cancel for movement along the surface.

OK, my argument was flawed, but at least yours isn't entirely correct either.

Look e.g. at the Wikipedia page for the Hafele-Keating experiment. (Link). They say that the predicted gravitational time dilation is 144±14 in the eastward direction and 179±18 in the westward direction. If altitude is the only thing that affects the GR prediction (if "all pure GR terms cancel for movement along the surface"), then these predictions should be the same. I tried to explain this difference using my "closer to being in orbit" argument, but I'm getting the opposite result.

I think I see what's wrong with my argument. The world lines we're considering aren't close to being geodesics at all, and the "closest" geodesics are clearly not orbits around the planet, but free-fall paths through the planet (the geodesics that these world lines are tangent to). So if it's possible to determine which of these world lines have the highest proper time by comparing them to geodesics, I should have been comparing them to a different set of geodesics. And maybe it isn't possible to do it this way at all.

 

[gonegahgah]

At least we are getting somewhere hopefully.

I appreciate your help Fredrik (I do wait a little impatiently at times) as I do yours Jesse. I just need you to stay closer to the question if possible Jesse; which you did in your last post, thanks. Also, if somebody does steer me slightly wrong then I do need people to pop in and say that this is happening.

Actually, with respect to people's latitude, if the spinning world were to stay spherical and consistently dense then someone at the equator ought to feel and weigh lighter than someone at the poles.
The reason has to do with momentum. Anyone at the poles is feeling the full gravity under themselves but anyone at the equator is hanging onto the outside of a spinning wheel which wants to throw them (or allow them to continue in a straight line) - except for the gravity; This would decrease their weight.
As Frerik would say I think, if you were to orbit around the Earth at ground level (clearing all those mountains out of the way), you would feel weightless.
If the equator somehow spun at orbital speed (and the Earth remained in shape) then you would feel weightless at the equator.

The world itself bulges at the equator demonstrating that it feels the overcoming effect of less weight due to rotation as well.

This is just an aside observation, and nothing to do with world lines or GR or SR but you can still tell me Jesse, is that correct that we should feel lighter at the equator?

That's all I have time to ask right now because I have to go pick up dad from the hospital with his broken leg in a cast.

 

[JesseM]

This is just an aside observation, and nothing to do with world lines or GR or SR but you can still tell me Jesse, is that correct that we should feel lighter at the equator?

Yes, by a very slight amount--any acceleration will cause you to feel a G-force in the opposite direction that you're accelerating, and moving in a circle means constantly accelerating towards the center so you feel a G-force in the opposite direction (sometimes called the centrifugal force). On the equator of a planet, this would slightly counteract the G-force towards the center you feel due to gravity.

 

[Ich]

OK, my argument was flawed, but at least yours isn't entirely correct either.

Now I still believe it's correct. I gave the derivation here, but it's obvious from the Schwarzschildmetric if you set r=const. and scale the time coordinate appropriately that you can calculate transverse movement just like in Minkowski spacetime.

 

[gonegahgah]

Thanks Jesse. It is just one of those interesting observations.

Back to the original line of query.
I will step right back; basically back to the beginning.
I'll do it step by step with corrections so that I can progress.

Let's say you have an observer and an object in gravitational fields.
Do you yourself do the following:
Do a two part process to work out the difference between the observer's and the object's clocks? That is one equation for the different gravitational fields they are in (GR portion), and another equation for their different relative speeds (SR portion)?
Then add these together to get the total difference in tick rate of the object's clock to our observer's clock?

 

[JesseM]

Let's say you have an observer and an object in gravitational fields.
Do you yourself do the following:
Do a two part process to work out the difference between the observer's and the object's clocks? That is one equation for the different gravitational fields they are in (GR portion), and another equation for their different relative speeds (SR portion)?
Then add these together to get the total difference in tick rate of the object's clock to our observer's clock?

I'm not that knowledgeable about the details of GR calculations; I would guess that in general you can't always split things this way for arbitrary paths in arbitrary spacetimes, but Ich says that in the specific case of two objects which remain at the same radius from a spherical source of gravity (a Schwarzschild spacetime) you can separate the gravitational time dilation from the velocity-based time dilation in this way, I can't verify that this is true myself, but I don't have any reason to doubt it.

 

[Fredrik]

Now I still believe it's correct. I gave the derivation here, but it's obvious from the Schwarzschildmetric if you set r=const. and scale the time coordinate appropriately that you can calculate transverse movement just like in Minkowski spacetime.

Yes, I don't doubt that you're completely correct in the case of a non-rotating planet, but the only reason I can think of why the westward result is different from the eastward result in the Hafele-Keating GR calculation is that the effects of rotation are large enough to be significant. And the hypothetical situation we were talking about involves rotation. So the Schwarzschild metric isn't the right metric to use.

 

[Ich]

Yes, I don't doubt that you're completely correct in the case of a non-rotating planet, but the only reason I can think of why the westward result is different from the eastward result in the Hafele-Keating GR calculation is that the effects of rotation are large enough to be significant.

No, they took different planes and different routes, and were flying at different altitudes. That's all.

 

[JesseM]

No, they took different planes and different routes, and were flying at different altitudes. That's all.

Isn't the spacetime around a rotating spherical body different than the spacetime around a nonrotating one, though? At least in the case of rotating black holes you can't use the Schwarzschild solution.

 

[ZikZak]

Yes, I don't doubt that you're completely correct in the case of a non-rotating planet, but the only reason I can think of why the westward result is different from the eastward result in the Hafele-Keating GR calculation is that the effects of rotation are large enough to be significant. And the hypothetical situation we were talking about involves rotation. So the Schwarzschild metric isn't the right metric to use.

No, the result of the Hafele-Keating experiment is not due to frame-dragging. In the frame of the distant inertial observer, it is due to the differing speeds of the eastward and westward airplanes. In the rotating frame of the Earth airport, it is due to the Sagnac Effect.

I got interested in this problem this morning for some reason and did the full calculation. I hope it's right ;)

Eddie walks along the equator, and Peter walks over the pole to get to the other side of the Earth. Let's do this all in the inertial frame of a distant observer. So let the radius of the Earth be R and its rate of rotation $\omega$. Take the metric to be Schwartzshild and weak-field:
$$d\tau^2 = (1+2\Phi) dt^2 - (1-2\Phi)(dr^2 + r^2 d\theta^2 + r^2\sin^2\theta d\phi^2)$$

For Eddie, $r=R$, $\theta = \pi/2$, and $d\phi= (\omega \pm v/R) d\tau$, where the + sign is for motion with the rotation and the - sign is for motion against it. Plugging this stuff into the metric yields:
$$d\tau_E^2= (1+2\Phi) dt^2 - (1-2\Phi)R^2(\omega \pm v/R)^2d\tau_E^2$$
$$dt = d\tau_E\sqrt{\frac{1+(1-2\Phi)R^2(\omega \pm v/R)^2}{1+2\Phi}}$$

For Peter, $r=R$, $d\theta = (v/R)d\tau$, and $d\phi = \omega d\tau$. Plugging this into the metric:
$$d\tau_P^2 = (1+2\Phi) dt^2 - (1-2\Phi)(v^2 d\tau_P^2 + R^2\omega^2\sin^2\theta d\tau_P^2)$$
$$dt = d\tau_P\sqrt{\frac{1+(1-2\Phi)(v^2 + R^2\omega^2\sin^2\theta)}{{1+2\Phi}}$$

Setting dt=dt,
$$d\tau_E\sqrt{1+(1-2\Phi)R^2(\omega \pm v/R)^2} = d\tau_P\sqrt{1+(1-2\Phi)(v^2 + R^2\omega^2\sin^2\theta)}$$
$$d\tau_E\sqrt{1+(1-2\Phi)(R^2\omega^2 \pm 2\omega v R +v^2)} = d\tau_P\sqrt{1+(1-2\Phi)(v^2 + R^2\omega^2\sin^2\theta)}$$

When Eddie walks with the rotation of the earth, the sign is (+) and the left radical is always larger than the right radical, and therefore $d\tau_E < d\tau_P$, and Eddie takes less proper time to walk around the Earth than Peter does, because their gravitational potentials are equal and Eddie's total speed including rotation is always greater than Peter's. When Eddie walks against the rotation of the Earth, the sign is (-) and it's going to depend on how fast they are walking. This is all assuming that I did this right.

Obviously, I haven't included frame-dragging, which would make this a little more complicated (but still doable) but shouldn't change the results much.

 

[Ich]

Isn't the spacetime around a rotating spherical body different than the spacetime around a nonrotating one, though? At least in the case of rotating black holes you can't use the Schwarzschild solution.

Of course, but GPB shows vividly just how small the difference is. Here on Earth you can safely neglect it.

 

[Ich]

in a hurry: ZikZak, the (1-2\Phi) belong to dr² only, not the transversal coordinates. Neglect dr and scale t to get Minkowski.

 

[ZikZak]

in a hurry: ZikZak, the (1-2\Phi) belong to dr² only, not the transversal coordinates. Neglect dr and scale t to get Minkowski.

Oh, crud. That's what I get for doing GR at 6 in the morning... Oh well. Next time :)

 

[gonegahgah]

Cool thanks, Jesse. That is more specifically what I was meaning when I was asking. I meant that they remain at a constant radius as you infered.

Gotta rush, but I was just wondering something today.
We talk about the equivalence between gravity and acceleration.

I can see that if you are standing in a box and it is accelerating and also if you are standing in the same box on the Earth's surface you couldn't tell the difference (assuming smooth acceleration). Effectively you are squashed in both scenarios making you shorter. And if you are falling towards the Earth in freefall under gravitation you will feel weightless. If anything you will extremely minutely be stretched in freefall due to a extremely minutely greater gravitational pull at your feet than at your head. (I assume that is correct but feel free to correct this).

So that gives us:
In gravitation field feeling weight (on surface) | In gravitational field weightless (freefall)
------------------------------------------------------------------------------------
Accelerating in deepest space feeling weight |

Again I've left a fourth quadrant free. Obviously when we are accelerating we will feel weight, and on the surface of the Earth we will feel weight, and freefalling on Earth we will feel weightless. But how do you feel weightless where acceleration is involved? Is there no equivalence for acceleration and gravity when it comes to weightlessness?

 

[JesseM]

I can see that if you are standing in a box and it is accelerating and also if you are standing in the same box on the Earth's surface you couldn't tell the difference (assuming smooth acceleration). Effectively you are squashed in both scenarios making you shorter. And if you are falling towards the Earth in freefall under gravitation you will feel weightless. If anything you will extremely minutely be stretched in freefall due to a extremely minutely greater gravitational pull at your feet than at your head. (I assume that is correct but feel free to correct this).

The slight difference in gravity between your head and feet that leads to stretching is one type of tidal force. In the limit as the size of the region of spacetime where you do your experiments goes to zero (i.e. in the limit as the volume of your box becomes arbitrarily small and the time in which you do experiments inside it also becomes arbitrarily small), the tidal forces approach zero too, and the laws of physics observed in this freefalling box approach perfect equivalence with the laws of physics seen by an inertial observer in special relativity.

So that gives us:
In gravitation field feeling weight (on surface) | In gravitational field weightless (freefall)
------------------------------------------------------------------------------------
Accelerating in deepest space feeling weight |

Again I've left a fourth quadrant free. Obviously when we are accelerating we will feel weight, and on the surface of the Earth we will feel weight, and freefalling on Earth we will feel weightless. But how do you feel weightless where acceleration is involved? Is there no equivalence for acceleration and gravity when it comes to weightlessness?

The "equivalence" in the equivalence principle is not between acceleration and non-acceleration, it's between the laws of physics as seen in the flat spacetime of SR and the local laws of physics seen by an observer in a small region of a larger curved GR spacetime (small enough that all tidal forces are negligible in that region). So, an observer who's accelerating in SR (deep space) measures the same thing as an observer at a constant height in a gravitational field; an observer who's moving inertially in SR measures the same thing as an observer in freefall in a gravitational field (both feel weightless, for example). In fact, in the way physicists define their terms, the observer in freefall is not said to be accelerating, instead this observer is moving in a "locally inertial" manner, whereas the observer at rest on the surface of the Earth is said to be accelerating.