B Question about bendability of objects in 4 dimensions

[ESponge2000]

We want to find the measurement of time on earth that over the longterm, relative to all n points on the surface of the earth j1 j2 j3… jn call each point j k= 1,2,3,4….n

Sigma (time elapse at jsub(k) - time elapsed at reference point) ^2 Using reference point such that the sum of squares of these variances is minimized

To be super precise , points on the equator will observe less elapsed time relative to rest of earth due to the spin rotation , by roughly 7 seconds every one million years due to relativity compared to the Poles

 

[PeterDonis]

We want to find the measurement of time on earth that over the longterm, relative to all n points on the surface of the earth j1 j2 j3… jn call each point j k= 1,2,3,4….n

Sigma (time elapse at jsub(k) - time elapsed at reference point) ^2 Using reference point such that the sum of squares of these variances is minimized

Why do you want to find this? What do you think it means?

Also, how are you defining "time elapsed"?

 

[ESponge2000]

Why do you want to find this? What do you think it means?

Also, how are you defining "time elapsed"?

The reference point will assume to be stationary and so using an atomic instrument we can get time

 

[PeterDonis]

The reference point will assume to be stationary and so using an atomic instrument we can get time

Stationary relative to what? Are you proposing putting an atomic clock at the center of the Earth?

Also, how do you define "time elapsed" at all the other points?

 

[ESponge2000]

Stationary relative to the average point on earth. On average all points of longitude on a latitude line the same time relative to each other and latitude lines they don’t because we deal with rotation speed difference

 

[Dale]

When a long object accelerates rapidly, along the time-like axis does it hold the same length for an observer riding the object as it accelerates?

Acceleration for an object of constant mass is always in a spacelike direction, never along a timelike axis.

Regarding the length, that depends on how the acceleration is physically achieved. In the usual assumption, as @Ibix mentioned, you push from one end and the length is constant if the acceleration is constant. If you were able to push from each point in a pre-programmed pattern then it is possible to have “Born rigid” acceleration which would change over time but still preserve distances on the object.

Most of the recent back and forth seems to have nothing to do with your original question regarding mass.

 

[ESponge2000]

Stationary relative to the average point on earth. On average all points of longitude on a latitude line the same time relative to each other and latitude lines they don’t because we deal with rotation speed difference

Forget the earth orbiting the sun and the sun flow with other galaxies we cannot get time exact to any of that but we can compare points on earth to each other

 

[Dale]

Can I suggest that this conversation is fairly pointless? @ESponge2000 is not engaging with the question that would actually lead him to understanding of simultaneity in relativity, which is: how do you know that two clocks are synchronised?

All the detailed specifications of clock conventions on Earth are irrelevant to this. The necessity of having precisely specified conventions rest on understanding why you can only have convention, not absolute time.

And that topic is also a big departure from the OP

 

[PeterDonis]

Stationary relative to the average point on earth. On average all points of longitude on a latitude line the same time relative to each other and latitude lines they don’t because we deal with rotation speed difference

No, this is not correct, because the distance from the center of the Earth also varies with latitude since the Earth is not a sphere, it's an oblate spheroid. The "geoid" I referred to before is the "sea level" surface on which the potential, meaning the "rate of time flow", is the same everywhere, after accounting for both the rotation speed and the distance from the Earth's center. That surface is what is used to define the standard "rate of time flow" in the ECI frame, and which is used to set the clocks we use to coordinate activities on Earth.

 

[ESponge2000]

No, this is not correct, because the distance from the center of the Earth also varies with latitude since the Earth is not a sphere, it's an oblate spheroid. The "geoid" I referred to before is the "sea level" surface on which the potential, meaning the "rate of time flow", is the same everywhere, after accounting for both the rotation speed and the distance from the Earth's center. That surface is what is used to define the standard "rate of time flow" in the ECI frame, and which is used to set the clocks we use to coordinate activities on Earth.

It’s not a perfect sphere but it’s a discrepancy that isn’t really worth adjusting for because we have too many other minor ones with even bigger effects so why bother. It’s a difference of a few kilometers over a day …

As for the 30th parallel being the weighted average spin someone said what’s the math for that ? It’s that a circle that runs verticle around the earth dividing East and Western hemispheres follows that to cross sides of the earth sticking to the same latitude , It’s the x distance across is cosine the latitude

Parallels are proportional to those thru earth out the same parallel lines , So the circumpolar distance at any latitude is the cosine of the latitude times the circumference of the earth at the equator.

The half-point for surface area is the riemann sums of all the circumpolars from 0 to 90 in each side of the equator … Integral of cos(x) is sin(x) and sin(30-deg) - sin(0-deg) = 0.5, not accounting for imperfect sphere but it’s a difference of very small adjustments a few kilometers of altitude due to topography and also due to shape , and 6 inches for moon pull on earth gravity but this is too small to worry about as a bigger issue is every time we have a big earthquake the earth rotation speed also slows indefinitely

 

[Dale]

What does any of this simultaneity and earth rotation stuff have to do with the OP?

 

[ESponge2000]

What does any of this simultaneity and earth rotation stuff have to do with the OP?

Trying to define simultaneity or “at the same time” is a component of understanding how much bend occurs when accelerating, but since we can’t define simultaneity we move to that sub-question to then move back to my original one. But I agree this tangent took us way off topic,

 

[PeterDonis]

It’s not a perfect sphere but it’s a discrepancy that isn’t really worth adjusting for

Um, yes, it most definitely is, since, as I have already said more than once now, it affects what surface is the actual equipotential surface, i.e., on what surface the "rate of time flow" on the rotating Earth is actually the same.

 

[PeterDonis]

Trying to define simultaneity or “at the same time” is a component of understanding how much bend occurs when accelerating

No, it isn't. Proper acceleration is an invariant computed along a single worldline and does not depend on any choice of simultaneity.

 

[ESponge2000]

No, it isn't. Proper acceleration is an invariant computed along a single worldline and does not depend on any choice of simultaneity.

Maybe relativity is more individualistic it’s better to focus on one object at a time and its parameters and not what time it matches another object in another frame but shift one variable at a time . It’ll make my scenarios simpler

 

[ESponge2000]

Um, yes, it most definitely is, since, as I have already said more than once now, it affects what surface is the actual equipotential surface, i.e., on what surface the "rate of time flow" on the rotating Earth is actually the same.

Well something close to sea level. But how is it not negligible ?

If I ride in an airplane for a million years I am relatively out of sync with all stationary points on earth by approximately 7 seconds due to relativity.
If I underestimate the rotation of earth by the difference in sea level distance to earth core , it’s something of a few miles discrepancy , like the highest point from the center of earth is in Colombia , but highest elevation is Everest, it’s not a big difference . The impact of this discrepancy on spin is a few miles per day, which is a spacetime relativity impact down to below nanoseconds in a year,
When an unexpected earthquake hit 5 years ago it was announced the impact of the quake was a permanent slowing of the earth’s spin by more impact than this discrepancy which leaves me to my point, I think our timekeeping is quite incredibly precise for geocentric purposes . If we have room for precision it’s not these little adjustments we are to concern with but to understand time discrepancies across the solar system

 

[Dale]

None of this is relevant to the initial question.

Do you understand how acceleration is how an object is bent in 1+1D space-time?

 

[ESponge2000]

None of this is relevant to the initial question.

Do you understand how acceleration is how an object is bent in 1+1D space-time?

Sort of but the part about constant acceleration preserves the length I don’t completely grasp that

 

[ESponge2000]

Sort of but the part about constant acceleration preserves the length I don’t completely grasp that

So then when you actually get to 0.8c you stop accelerating at a constant rate you then unbend ?

 

[PeterDonis]

something close to sea level. But how is it not negligible ?

You do realize that "sea level" is about 13 miles higher (i.e., further from the center of the Earth) at the Earth's equator than it is at the poles, right? And yet clocks still run at the same rate everywhere at "sea level"?

 

[PeterDonis]

a permanent slowing of the earth’s spin by more impact than this discrepancy which leaves me to my point, I think our timekeeping is quite incredibly precise for geocentric purposes

Um, you do realize that the reason we know about the change in the rate of the Earth's spin on such short tlime scales is that we have clocks that are much more accurate than the difference, right?

The issue is not the Earth's spin per se. It is the difference in "rate of time flow" of clocks at different places on Earth. We have clocks that can detect that difference even over differences in height as small as about one millimeter.

 

[PeterDonis]

So then when you actually get to 0.8c you stop accelerating at a constant rate you then unbend ?

Your worldline does. In other words, your "path through spacetime" is bent while you are accelerating (i.e., feeling a force) and straight when you are not.

 

[ESponge2000]

You do realize that "sea level" is about 13 miles higher (i.e., further from the center of the Earth) at the Earth's equator than it is at the poles, right? And yet clocks still run at the same rate everywhere at "sea level"?

Well an atomic clock can be set somewhere in London and a formula to advance clocks 0.04 hundredths of a second or some surface distance formula can most entirely do the trick… and if the clock is ported in London then it’s perpetually rescyncronized to correct the departures from London time, which is actually how smartphones keep time I think … when online it resyncs to an atomic control clock and when offline it uses analog methods to keep time, gearing off the last resync. I think laptop clocks work this way as well. This isn’t I think , I’m pretty sure this is the way electronics have been working for a decade now

 

[ESponge2000]

Your worldline does. In other words, your "path through spacetime" is bent while you are accelerating (i.e., feeling a force) and straight when you are not.

Well that is obvious but how does it look during the constant acceleration , x axis slopes upward or is it concave down?

 

[PeterDonis]

a formula to advance clocks 0.04 hundredths of a second or some surface distance formula

I'm not sure what you mean here.

if the clock is ported in London then it’s perpetually rescyncronized to correct the departures from London time, which is actually how smartphones keep time I think ... when online it resyncs to an atomic control clock and when offline it uses analog methods to keep time, gearing off the last resync

Smartphones don't sync directly with atomic clocks, they usually get their time from whichever cell tower they are connected to. Some might also check NTP servers, which is how most laptop and desktop computers get their time.

However, these sync operations have nothing to do with any relativistic effects. Smartphone clocks, and indeed any ordinary clocks, aren't accurate enough to detect those. The sync operations are actually because ordinary clocks, whether they're in smartphones, computers, bedside alarm clocks, kitchen appliances, or whatever, simply don't keep time very well at all. Their "tick rates" are so variable that they have to be reset at periodic intervals.

Actual atomic clocks that are accurate enough to detect relativistic effects are at fixed locations on Earth (unless they're being used for a special experiment like the Hafele-Keating experiment), and keeping them synchronized is just a matter of periodically having them exchange signals of some sort to check their timekeeping. The difference in their "tick rates" due to altitude relative to the geoid is already known to great accuracy and can be adjusted for locally at each clock.

 

[PeterDonis]

how does it look during the constant acceleration , x axis slopes upward or is it concave down?

Your acceleration doesn't change the $x$ axis of your coordinates.

Your worldline is concave in the direction you are accelerating. For example, if we just consider the $t$ and $x$ axes, the worldline of an object with constant proper acceleration in the positive $x$ direction is a hyperbola that is concave to the right (the positive $x$ direction). The asymptotes of the hyperbola are the 45 degree "light cone" lines through the origin.

 

[Dale]

Sort of but the part about constant acceleration preserves the length I don’t completely grasp that

Lengths will be distorted when the internal stresses change. If the acceleration is constant then the internal stresses are also constant.

 

[DrGreg]

Your worldline is concave in the direction you are accelerating. For example, if we just consider the $t$ and $x$ axes, the worldline of an object with constant proper acceleration in the positive $x$ direction is a hyperbola that is concave to the right (the positive $x$ direction). The asymptotes of the hyperbola are the 45 degree "light cone" lines through the origin.

Here is a spacetime diagram of a rod accelerating along its own length.

The green area is the rod. This is drawn under the assumption of constant proper acceleration and a Born rigid rod.

On the diagram below I have superimposed some snapshots of the rod at various times, using the definition of simultaneity of a single inertial frame.

And on the diagram below I have superimposed some snapshots of the rod at various times, using the definition of simultaneity of an accelerating frame in which the rod is at rest.