[Halc]

The passenger in an accelerating  rocket may read 1 g on his accelerometer because he is being pushed from behind. Having landed on Earth his instrument will read 1 g because he is being pulled from below.

No, both cases are due to being pushed from below. Were it not for this push, both cases would locally experience no proper acceleration. The two cases being locally indistinguishable is the essence of the equivalence principle, and not the below suggestion:

But acceleration does not produce gravity as [chatGTP] implies

I think it's the implication from equivalence principle.

You show signs of not having read the critique at all, since it is that assertion that was colored red every time it was made.  The equivalence principle is not in any way applicable to the mathematics of an accelerating frame, either locally or not. It makes it sound like the mathematics of special relativity was derived from general relativity and not the other way around.

Two balls are connected by a string, floating freely in outer space. They are spun until having a constant angular speed. No energy is expended afterwards, even though the balls are continuously accelerated.

Excellent counterexample.

Consider one ball moving in a straight line at constant speed. If we want to change its direction we need to exert a force perpendicular to the track, so work is done.

A force perpendicular to motion does no work, which is why the 2 balls on a string don't slow in their rotation over time.

[alancalverd]

No, both cases are due to being pushed from below.

In my universe, gravity sucks. If I let go of the accelerometer, it flies towards the ground, not away from it.

[varsigma]

No, it is not a true explanation of a theory, which is why such assertions should not be posted in the main sections of the forum. That area is reserved for questions and answers to them.

I apologise for what I'm about to post then.

Einstein, and Feynman have both said that relativity says we move in the time direction. Time isn't really a thing with extent, but it is a direction. We move at the speed of light in that direction.

John Baez, Roger Penrose, say it too.
 

[Halc]

I had to move the group of posts due to the assertive tone they took, rather than a questioning one.

Einstein, and Feynman have both said that relativity says we move in the time direction. Time isn't really a thing with extent, but it is a direction. We move at the speed of light in that direction.

John Baez, Roger Penrose, say it too.

Actual quotes (and context) would really help. Are these statements in peer reviewed publications or are they casual comments or from pop articles?
What they are trying to say is that time and space are the same thing, meaning time and length can be used interchangably. So for instance, a worldline segment one second long (the worldline of any clock over one second measured by it) is exactly one light-second in length. The statement is one to define this ratio of translating space to time. But to call it 'speed' totally contradicts the definition of speed, as Paul has correctly pointed out. There is no meaningful motion through spacetime except for the one example I posted in the other topic, and that motion is completely undetectable, given that the interpretation posits a completely undetectable thing.
In particular, with motion, when the object gets to the destination, it is no longer at the former location. With a worldline, the object is present at every event along the line. It doesn't move from one event to another, leaving the former event 'vacant' so to speak.

It is also a common shorthand to identify a frame by an observer stationary in it, but it must be made clear at some point that frame selection is an arbitrary abstract choice, not a physical property of an object or observer. If Penrose did not make that clear in his book, then he's not doing a very good job of explaining relativity.
The vast majority of human observers choose a different frame than the inertial one in which they are stationary. It is far more pragmatic to use say the accelerating rotating frame of the surface of Earth when walking past another person.

[varsigma]

"Einstein based special relativity on the idea that the speed of light is the same for all observers. You cannot do this in a Euclidean space where all the signs are plusses. But you can do it if one of the signs is different relative to the others.

That's because a space-time distance that is zero for one observer is zero for all observers. This is also the case in Euclidean space, but in Euclidean space, this just means zero in each of the directions of space. But what does a zero distance mean in space-time? Well, let's find out. For simplicity, let us look at only one dimension of space. So if the distance in space-time is zero, this means that the distance in space divided by the distance in time equals plus or minus c. And that?s the same for all observers. So this speed, c, is an invariant speed.

But, well, we are not light, so we do not travel with the speed of light through space, and we do actually cover a distance in space-time. So let us look at this equation for the space-time distance again. Now let us divide this by the time difference. Now what you have on the left side is the space-time distance per time. And under the square root you have roughly something like the squares of the velocities in each of the directions of space. Plus c².

And there you have it. Relative to yourself, you do not move through space, so these velocities are zero. You then only move into the time-like direction, and in this direction, you move with the speed of light. So, we indeed all travel through time with the speed of light."--from a Ph.D blog

[varsigma]

There might be a problem with using the word motion, or moving, it's usually a reference to space, or different positions.

In Minkowski spacetime you have a relativistic 4-velocity. It cannot be zero or, well, lots of things just don't happen.

[Halc]

"Einstein based special relativity on the idea that the speed of light is the same for all observers.

That it is the same relative to any inertial frame. It is not true relative to non-inertial frames, and there is no mention of observers in the premises.

That's because a space-time distance that is zero for one observer is zero for all observers.

It's called an interval, not a distance, and any interval (zero or otherwise) between two events or along any worldline is frame independent. Again, observers play no role in this.

So this speed, c, is an invariant speed.

You make it sound like frame invariant light speed is derived from this zero interval, rather than the frame invariant interval (of any value) being derived from the fixed light speed postulate.

You then only move into the time-like direction, and in this direction, you move with the speed of light.

That's like saying that a school bus is parked in my driveway (our reference frame), front bumper against my garage door. 20 ns later the rear bumper is 10 meters away at the street?  Did the bus move at faster than c? No. It's just a different part of an extended object and the thing isn't moving at all. Likewise, you are a worldline, an extended object that is present at both events delimiting the measurement. You don't move from one event to the other, it's just different parts of you that are present at each event, just like the bumpers being different parts of one bus. It isn't motion.

All that said, yes, different definitions are used, and it is commonplace to express motion through spacetime as taking place at c, expressed as a four-velocity, not a normal velocity.

[varsigma]

All that said, yes, different definitions are used, and it is commonplace to express motion through spacetime as taking place at c, expressed as a four-velocity, not a normal velocity.

I'm trying to figure out a response to the phrase "normal velocity"; if you mean the Euclidean part of the spacetime metric, I guess. But in the case of non-Euclidean solutions, I don't know what to say about it.

If you accept that "normal" is restricted to a locally defined region of the universe, the only thing moving at c is light, and the universal expansion is effectively zero; I would say that then, the universe is indeed "normal". But I know enough about the equations that I can say, that is a trivial solution. I was hoping we could lean towards what an exact reason for the different ages of twins is. Part of the reason is the exact speed of light, of course. Another part is the relativistic 4-velocity any object with mass "inherits" from the ah, causal structure.

[paul cotter]

"I was hoping we could lean towards what an exact reason for the different ages of the twins is". Am I missing something? I was of the opinion that this question was comprehensively answered at an early stage in this query. Surely 1/√1-vsq/csq is all one needs to answer this without digressing into ever more complicated reasoning.

[Halc]

I was hoping we could lean towards what an exact reason for the different ages of twins is.

Am I missing something? I was of the opinion that this question was comprehensively answered at an early stage in this query.

That was indeed done in the very first reply, the important part being this:

There are many ways to explain the difference, so there is no one correct way.

That means that anything claiming or asking for 'the correct way' is already presuming something wrong.

The rest of the post lists only one of the ways to explain it, and also provides counterexamples to every one of the 'explanations' mentioned in the videos, both of which were shown to be mistaken in subsequent posts. Takeaway is to use a peer reviewed text to get your information, not pop videos, even those put out by supposedly knowledgeable celebrities.  An arm-chair amateur like me should not so easily be able to find the mistakes in them.

If you want another explanation, it can be something as simple as "That's the nature of Minkowskian geometry". A straight (timelike) path between two spacetime points (events) is longer in length than one that isn't straight. This is unintuitive since in Euclidean geometry, the opposite is true.

[varsigma]

I think this thread has followed the same kind of pattern I've seen on other forums, any discussion of 4-dimensional spacetime is full of disagreements about whether this or that is a good explanation or even a correct one.

I have to bear in mind I suppose, that this pattern emerged as soon as Einstein published his general theory. The idea that a spacetime existed in which the notions of fixed time and space were only relative, weren't met with enthusiasm.

[A-wal]

Lots of wrong information in this topic. Yes of course it's caused by acceleration. There's two frames, from the inertial frame's perspective the accelerating clock's slow down is purely a function of relative velocity, T = t(√((c^2−v^2)/c^2)), we want the accelerating frame's perspective of the inertial frame's clocks to show how the acceleration creates the time difference.

There must be a simple formula to calculate the time dilation by the acceleration.  This seems to be a way to know who is correct.

Of course there is, T = (2(√(c^2/(c^2−v^2))))((vx)/c^2) where x is space applies a non‐reciprocal time dilation in the form of a time jump on the inertial frame from the perspective of the accelerating frame that's proportional to distance. It simplifies to this because we're using constant relative velocities and instantaneous accelerations.

This overcooks the time difference so that despite the Earth twin's clocks slowing during the inertial phases of the journey from the perspective of the accelerating twin both twins will agree on the time difference once the traveling twin returns because the time jump is proportional to distance, this is a simultaneity shift.


A simpler way to look at it, if the turnaround point is at rest relative to Earth then the distance is shortened by the square root of ((c^2−v^2)/c^2) for an observer that is in motion relative to that frame, so the time that it takes to travel that shortened distance is reduced by the same amount, so the shortened time for the traveling twin is simply T = t(√((c^2−v^2)/c^2)).

[A-wal]

This should be in this topic.

There must be a simple formula to calculate the time dilation by the acceleration.  This seems to be a way to know who is correct.

Of course there is, T = (2(√(c^2/(c^2−v^2))))((vx)/c^2)

In the topic where this comment originated, I gave several examples of constant acceleration with negligible dilation, and examples of different accelerations that gave identical dilation.

Meanwhile, the formula posted makes no mention at all of acceleration, only speed, making it a function of speed, not of acceleration at all.

If you had a smooth acceleration profile or an acceleration profile that kept accelerating in the opposite direction to slow its velocity relative to some other reference frame the second profile would have a great deal more acceleration than the first smoother acceleration profile but the simultaneity shift would be identical. So in that way it's not acceleration directly that causes the twin's age different, it's the change in velocity relative to another frame scaled by distance.