B Question about bendability of objects in 4 dimensions

[ESponge2000]

TL;DR Summary

A long bus or a stick , vs a multi-car freught train with bendabikity around curves or an elastic rubber band

In 2 spatial dimensions we have the ability to see there’s 2 types of long vehicles : busses vs trains… Buses must yield when making a very sharp turn since the front to the back are aligned at all times . Freight trains can make much sharper turns the more the cars are small and the gaps between cars are “bendable” allowing curves and zigzags

Now in 1 dimension of space + time do we have the same properties as well, or do we have how it looks on space-time diagram ,
Draw a very very long bus accelerating from 0 to a very high relativistic velocity
Now draw a very very long elastic rubberband with some ability to stretch when forces are unequal ?

 

[Dale]

What you are calling "bendability" in 1+1D spacetime is simply the mass.

 

[ESponge2000]

I don’t understand the concept of bendability is just the mass. Is this about gravity bending space time around mass?

Or do we say a stick that cannot stretch is forced to stretch like the rubberband when it accelerates vertically?
Which my limited understanding is per the front of the stick perspective acceleration happens before the back accelerates even when removing the Doppler shift
Per the back of the stick perspective, the front accelerates faster than the back when removing Doppler shift , hence with removing Doppler shift on both ends of stick , for each end there’s a net 0 stick change however it’s a stretch to cancel out the length contraction of the universe , or is it ?

 

[ESponge2000]

I don’t understand the concept of bendability is just the mass. Is this about gravity bending space time around mass?

Or do we say a stick that cannot stretch is forced to stretch like the rubberband when it accelerates vertically?
Which my limited understanding is per the front of the stick perspective acceleration happens before the back accelerates even when removing the Doppler shift
Per the back of the stick perspective, the front accelerates faster than the back when removing Doppler shift , hence with removing Doppler shift on both ends of stick , for each end there’s a net 0 stick change however it’s a stretch to cancel out the length contraction of the universe , or is it ?

This is elasticity-like but maybe it’s a different kind of stretch. Or am I describing this correctly?

Is what I’m saying elementary and obvious, or is what I’m saying not making sense?

 

[Ibix]

Is this about gravity bending space time around mass?

No. If you have a 1+1 dimensional spacetime, an object that is long in the timelike direction is one that exists for a while. A curved object is an accelerating one. How hard it is to make it curve is its mass.

 

[Dale]

I don’t understand the concept of bendability is just the mass. Is this about gravity bending space time around mass?

This is not directly to do with curved spacetime, but a lot of the geometric concepts carry over.

“Bendability” is how much force it takes to bend an object. The bus is not very bendable because it takes a lot of force to make it bend. The train is bendable because the hinges let it bend without a lot of force.

In spacetime, an object’s bending is its acceleration. A particle moving in a straight line at a constant speed forms a straight line in spacetime. A particle accelerating in a straight line at a constant acceleration forms a bent line in spacetime (specifically a hyperbola for constant acceleration).

So a very massive object requires a lot of force to accelerate, meaning a lot of force to bend in spacetime. Thus a very massive object is not very “bendable” in spacetime. Conversely with an object of low mass. So mass is directly the thing that determines bendability in spacetime.

 

[ESponge2000]

This is not directly to do with curved spacetime, but a lot of the geometric concepts carry over.

“Bendability” is how much force it takes to bend an object. The bus is not very bendable because it takes a lot of force to make it bend. The train is bendable because the hinges let it bend without a lot of force.

In spacetime, an object’s bending is its acceleration. A particle moving in a straight line at a constant speed forms a straight line in spacetime. A particle accelerating in a straight line at a constant acceleration forms a bent line in spacetime (specifically a hyperbola for constant acceleration).

So a very massive object requires a lot of force to accelerate, meaning a lot of force to bend in spacetime. Thus a very massive object is not very “bendable” in spacetime. Conversely with an object of low mass. So mass is directly the thing that determines bendability in spacetime.

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?

 

[Ibix]

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?

Only if it was always accelerating and never stops accelerating. You can often idealise "stays at a constant proper acceleration for much longer than $l/c_s$, where $l$ is the length of the object and $c_s$ is the speed of sound in the object" as eternally accelerating, but any time it changes acceleration you cannot do so.

 

[ESponge2000]

I still have a problem understanding “for objects spaced apart moving at a certain same velocity, there is seemingly no way to answer “if they pause at the same time interval will they remain the same distance apart …. It looks every time ive asked this there’s a problem “at the same time” is not definable

 

[Ibix]

It looks every time ive asked this there’s a problem “at the same time” is not definable

It's definable, but there's always a free choice in the definition somewhere. There is no absolute definition of simultaneity in relativity.

 

[PeterDonis]

I still have a problem understanding “for objects spaced apart moving at a certain same velocity, there is seemingly no way to answer “if they pause at the same time interval will they remain the same distance apart …. It looks every time ive asked this there’s a problem “at the same time” is not definable

One way of approaching this is to ask, how would each object know when to pause? What physical mechanism enforces that? Does each object just look at its own clock and pause when their clock shows a certain time? Do they exchange light signals? And so on.

Taking that approach forces you to recognize that relying on "simultaneity" does not work, because "simultaneity" does not correspond to any actual physical mechanism. It's just a coordinate abstraction. You have to find some actual physical mechanism to do what you want to do.

 

[ESponge2000]

I still have a problem understanding “for objects spaced apart moving at a certain same velocity, there is seemingly no way to answer “if they pause at the same time interval will they remain the same distance apart …. It looks every time ive asked this there’s a problem “at the same time” is not definable

Can “at the same tim

One way of approaching this is to ask, how would each object know when to pause? What physical mechanism enforces that? Does each object just look at its own clock and pause when their clock shows a certain time? Do they exchange light signals? And so on.

Taking that approach forces you to recognize that relying on "simultaneity" does not work, because "simultaneity" does not correspond to any actual physical mechanism. It's just a coordinate abstraction. You have to find some actual physical mechanism to do what you want to do.

But simultaneity for time intervals for 2 to objects at rest with each other has a measurable exactitude, obviously, because distance between 2 objects at rest with each other has a measurable exactitude, so measured distance, c, time is derivative, units of distance for a light signal in a vacuum where no accelerations occur for either object, Using conventions for both objects at 0km/h, Time

 

[PeterDonis]

simultaneity for time intervals for 2 to objects at rest with each other has a measurable exactitude, obviously, because distance between 2 objects at rest with each other has a measurable exactitude

How would you obtain this "measurable exactitude"? Please explicitly describe the measurement process that gives you this.

 

[ESponge2000]

And maybe what I mean to say is “duration” but same thing . How can we say there’s no such thing as simultaneity when synchronized time is the most fundamental property of global operations? Why we know exactly when we will be late for work , or when a solar eclipse is coming, or when taxes are due, or when a new law goes into effect, or when a store closes , etc

 

[PeterDonis]

How can we say there’s no such thing as simultaneity when synchronized time is the most fundamental property of global operations?

Again, what does "synchronized time" actually mean? What actual physical mechanism enforces it? (For purposes of this question you can--and should--ignore time zones, and assume everyone on the entire planet uses UTC.)

And just to bend your brain some more: when you answer the above question for us sitting at rest on the rotating Earth, your answer will not be compatible with any inertial frame in which any of us are at rest!

 

[Ibix]

How can we say there’s no such thing as simultaneity when synchronized time is the most fundamental property of global operations?

We very carefully agree a convention for what we mean by simultaneous, actually. There is only about $\pm 0.04\mathrm{s}$ range of variation in what "now" can mean across a region the size if the globe.

Rhetoric isn't science, either. Can you come up with a process to define "simultaneous"? Hint: any answer you give will contain an assumption from you somewhere that is what actually defines "simultaneous".

 

[ESponge2000]

How would you obtain this "measurable exactitude"? Please explicitly describe the measurement process that gives you this.

The exactitude is I found this on Wikipedia
.
Since 2019, the metre has been defined as the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second, where the second is defined by a hyperfine transition frequency of caesium.[2]

The second [...] is defined by taking the fixed numerical value of the caesium frequency, ΔνCs, the unperturbed ground-statehyperfine transition frequency of the caesium 133atom, to be 9192631770when expressed in the unit Hz, which is equal to s−

So if the atom is stationary relative to us or is planted , then we have a consistent second , with that we have a consistent fractal such that light takes that long which equates a metre

 

[ESponge2000]

We very carefully agree a convention for what we mean by simultaneous, actually. There is only about $\pm 0.04\mathrm{s}$ range of variation in what "now" can mean across a region the size if the globe.

Rhetoric isn't science, either. Can you come up with a process to define "simultaneous"? Hint: any answer you give will contain an assumption from you somewhere that is what actually defines "simultaneous".

Simultaneous across a region the size of a globe means deriving the time at the point in the center of the earth’s core, and then applying a geocentric convection

 

[PeterDonis]

The exactitude is I found this on Wikipedia
.
Since 2019, the metre has been defined as the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second, where the second is defined by a hyperfine transition frequency of caesium.[2]

The second [...] is defined by taking the fixed numerical value of the caesium frequency, ΔνCs, the unperturbed ground-statehyperfine transition frequency of the caesium 133atom, to be 9192631770when expressed in the unit Hz, which is equal to s−

So if the atom is stationary relative to us or is planted , then we have a consistent second , with that we have a consistent fractal such that light takes that long which equates a metre

None of this has anything to do with "simultaneity" as you are using the term. In spacetime terms, all of the measurements involved happen at one event--one point in spacetime. "Simultaneity" as you are using the term involves two (or more) points in spacetime.

 

[jbriggs444]

Simultaneous across a region the size of a globe means deriving the time at the point in the center of the earth’s core, and then applying a geocentric convection

You do not need or want a time standard based on time passage at the center of the Earth. Better to use a standard at the geoid. Which is what is actually done.

https://en.wikipedia.org/wiki/International_Atomic_Time

 

[ESponge2000]

The exactitude is I found this on Wikipedia
.
Since 2019, the metre has been defined as the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second, where the second is defined by a hyperfine transition frequency of caesium.[2]

The second [...] is defined by taking the fixed numerical value of the caesium frequency, ΔνCs, the unperturbed ground-statehyperfine transition frequency of the caesium 133atom, to be 9192631770when expressed in the unit Hz, which is equal to s−

So if the atom is stationary relative to us or is planted , then we have a consistent second , with that we have a consistent fractal such that light takes that long which equates a metre

We assume it to be a truth that light moves across 2999792458/299792458 * 9192631770 caesium133 atom frequencies in the time of 9192631770 caesium133 atom frequencies … Which = 1 for 1 , which tells us nothing actually

 

[PeterDonis]

Simultaneous across a region the size of a globe means deriving the time at the point in the center of the earth’s core, and then applying a geocentric convection

That's not quite how it works.

What is true is that the frame in which "simultaneity" is defined for everyday use by us at rest on the rotating Earth is centered on the Earth's center of mass, and is a non-rotating frame (this is called the "Earth-centered inertial" or ECI frame). The clock rate in this frame (the rate at which coordinate time advances) is then adjusted to the rate of clocks at rest on the rotating Earth's geoid, i.e., the idealized "sea level" surface of constant potential.

However, although this frame is called "inertial" as a matter of nomenclature, it is not an inertial frame in the sense of relativity, because the Earth's gravitational field is not negligible within it. Further, even if we leave that out, we at rest on the rotating Earth are all moving in this frame, and different points on Earth are moving in different directions, so we all have different inertial frames that are comoving with us, and none of them are the same as the ECI frame. That means the simultaneity convention we all use to coordinate our activities on the rotating Earth is not the same as the "natural" simultaneity convention for any of us according to how inertial frames are defined in relativity. The simultaneity convention we all use is just that--a convention that we adopt.

 

[ESponge2000]

That's not quite how it works.

What is true is that the frame in which "simultaneity" is defined for everyday use by us at rest on the rotating Earth is centered on the Earth's center of mass, and is a non-rotating frame (this is called the "Earth-centered inertial" or ECI frame). The clock rate in this frame (the rate at which coordinate time advances) is then adjusted to the rate of clocks at rest on the rotating Earth's geoid, i.e., the idealized "sea level" surface of constant potential.

However, although this frame is called "inertial" as a matter of nomenclature, it is not an inertial frame in the sense of relativity, because the Earth's gravitational field is not negligible within it. Further, even if we leave that out, we at rest on the rotating Earth are all moving in this frame, and different points on Earth are moving in different directions, so we all have different inertial frames that are comoving with us, and none of them are the same as the ECI frame. That means the simultaneity convention we all use to coordinate our activities on the rotating Earth is not the same as the "natural" simultaneity convention for any of us according to how inertial frames are defined in relativity. The simultaneity convention we all use is just that--a convention that we adopt.

I would think it’s an average. And off by so small a discrepancy that we don’t note the difference. Isnt it exact to trillionths of a second or even more precise than even that ? And UTC using that marking the rest is not relevant to this topic but we have arbitrary leap seconds on upkeep and then

(analemma (equation of time) adjusting for axial tilt and eliptical orbit and timezones by Govt laws including DST override the rule of every 15 longitude multiple being a center of 12pm = solar noon on the average day, 4 times a year and one of which is on Christmas Day.)

 

[Ibix]

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.

 

[Nugatory]

So if the atom is stationary relative to us or is planted , then we have a consistent second , with that we have a consistent fractal such that light takes that long which equates a metre

This measurement is done (conceptually) with one cesium atom in one place so the question of simultaneity never comes up - we watch the atom, we count its cycles, that tells us that one second has passed at the location of that atom. It tells us absolutely nothing about how much time has passed anywhere else, or even whether we started counting cycles at the same times that the atom cycled (in fact, you know we didn’t, because of the light travel time between the atom and our cycle counter). To answer any of these questions we need some convention for defining what we mean by “at the same time” anywhere else - or as @Ibix said above “how do you know that two clocks are synchronised?”

 

[PeterDonis]

I would think it’s an average.

No, it's not.

And off by so small a discrepancy that we don’t note the difference.

The issue of simultaneity has nothing to do with the accuracy of the measurements.

analemma (equation of time)

Has nothing to do with the simultaneity convention we use to coordinate activities on Earth. The equation of time describes the movement of the Sun in the sky, but we don't use the movement of the Sun in the sky to set our clocks.

including DST override the rule of every 15 longitude multiple being a center of 12pm = solar noon on the average day

This involves time zones, which are also irrelevant to the simultaneity question (that's why I said to ignore them in my previous post).

 

[ESponge2000]

ignore the timezone stuff . Do we have the ability to take for a fixed point on earth nanoseconds between 2 times on a point on the 30th parallel (this will be the surface-area weighted average rotation spin, for both hemispheres , 0 the highest , the poles no rotation spin … This is how to do it ….

 

[PeterDonis]

Do we have the ability to take for a fixed point on earth nanoseconds between 2 times on a point on the 30th parallel (this will be the surface-area weighted average rotation spin, for both hemispheres , 0 the highest , the poles no rotation spin … This is how to do it ….

I have no idea what you mean by this.

 

[Vanadium 50]

When a long object accelerates rapidly, along the time-like axis

What does that even mean?

Do we have the ability to take for a fixed point on earth nanoseconds between 2 times on a point on the 30th parallel (this will be the surface-area weighted average

What does this mean?

 

[renormalize]

Do we have the ability to take for a fixed point on earth nanoseconds between 2 times on a point on the 30th parallel (this will be the surface-area weighted average rotation spin, for both hemispheres , 0 the highest , the poles no rotation spin … This is how to do it ….

Then show the math for this.

 

[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"?