The braking of the traveler twin

Post by Luigi Fortunati
The traveling twin starts and travels for 4 years at speed v = 0.866c
and then brakes, as seen in the animation
https://www.geogebra.org/m/jzc4amud
Calculate the Earth-spaceship distance, before and after braking (in
the spaceship reference).
Calculate how many turns the Earth has made around the sun, before and
after braking (in the spaceship reference).

Suppose the two twins have just been born when the traveling twin starts
his trip. So they are each zero years old then. If the traveling twin
(he) travels for 4 years of his time, he will be 4 yours old when he
stops. (His age when he stops is an EVENT that all observers must agree
about, and so she also agrees that he is 4 years old when he stops. He
says he has traveled (0.866)(4) = 3.464 lightyears away from his twin
(her) then. She says he traveled 8 years of her time ... i.e., she says
she is 8 years old when he stops. She says he is (0.866)(8) = 6.928
lightyears away when he stops.

After he stops, she says that he remains 6.928 lightyears away from her
after that. And she says they age at the same rate after he stops. He
says that, during his essentially instantaneous stopping time, his age
essentially doesn't change during his stopping. She agrees with that.
But he says that during his essentially instantaneous stopping, SHE
essentially instantaneously gets older by 4 years ... i.e., he says that
she essentially instantaneously goes from being 4 years old to being 8
years old during the essentially instantaneous time in his life it takes
him to stop. And he also says that their distance apart essentially
instantaneously increases from 3.464 lightyears to 6.928 lightyears
while he is doing his essentially instantaneous stopping.

Luigi Fortunati

2022-03-15 12:28:40 UTC

Suppose the two twins have just been born when the traveling twin starts
his trip. So they are each zero years old then. If the traveling twin
(he) travels for 4 years of his time, he will be 4 yours old when he
stops. (His age when he stops is an EVENT that all observers must agree
about, and so she also agrees that he is 4 years old when he stops. He
says he has traveled (0.866)(4) = 3.464 lightyears away from his twin
(her) then. She says he traveled 8 years of her time ... i.e., she says
she is 8 years old when he stops. She says he is (0.866)(8) = 6.928
lightyears away when he stops.
After he stops, she says that he remains 6.928 lightyears away from her
after that. And she says they age at the same rate after he stops. He
says that, during his essentially instantaneous stopping time, his age
essentially doesn't change during his stopping. She agrees with that.
But he says that during his essentially instantaneous stopping, SHE
essentially instantaneously gets older by 4 years ... i.e., he says that
she essentially instantaneously goes from being 4 years old to being 8
years old during the essentially instantaneous time in his life it takes
him to stop. And he also says that their distance apart essentially
instantaneously increases from 3.464 lightyears to 6.928 lightyears
while he is doing his essentially instantaneous stopping.

Ok.

And how many turns around the Sun did the Earth make just before and
immediately after the stop, in the spaceship reference?

Richard Livingston

2022-03-15 13:55:06 UTC

Suppose the two twins have just been born when the traveling twin starts
his trip. So they are each zero years old then. If the traveling twin
(he) travels for 4 years of his time, he will be 4 yours old when he
stops. (His age when he stops is an EVENT that all observers must agree
about, and so she also agrees that he is 4 years old when he stops. He
says he has traveled (0.866)(4) = 3.464 lightyears away from his twin
(her) then. She says he traveled 8 years of her time ... i.e., she says
she is 8 years old when he stops. She says he is (0.866)(8) = 6.928
lightyears away when he stops.
After he stops, she says that he remains 6.928 lightyears away from her
after that. And she says they age at the same rate after he stops. He
says that, during his essentially instantaneous stopping time, his age
essentially doesn't change during his stopping. She agrees with that.
But he says that during his essentially instantaneous stopping, SHE
essentially instantaneously gets older by 4 years ... i.e., he says that
she essentially instantaneously goes from being 4 years old to being 8
years old during the essentially instantaneous time in his life it takes
him to stop. And he also says that their distance apart essentially
instantaneously increases from 3.464 lightyears to 6.928 lightyears
while he is doing his essentially instantaneous stopping.

Keep in mind that when the moving twin stops, both twins will agree
on how far apart they are. During the deceleration the moving twin
will observe the apparent expansion of the universe along the axis
of travel, and what while moving appeared to be 3.464 light years
will expand to 6.928 light years.

Something to keep in mind is that neither twin can see the other at
what each considers "now". They only see the other on their own
past light cone. So while the moving twin is stopping, he will NOT
see his twin back on earth suddenly aging. This is because what
he sees is not the instantaneous state of the earth, but an image
conveyed by light. Instead he will see her as she was about 6.928
years earlier. Thus what he sees is a much younger twin, only
about a year older than when he left earth. It is on the trip back
to earth that he sees her age rapidly as he passes the light
traveling outbound from earth.

Rich L.

Mike Fontenot

2022-03-15 18:57:00 UTC

I made a careless mistake in my previous post. Here is that post, down

Post by Mike Fontenot
Suppose the two twins have just been born when the traveling twin starts
his trip. So they are each zero years old then. If the traveling twin
(he) travels for 4 years of his time, he will be 4 yours old when he
stops. (His age when he stops is an EVENT that all observers must agree
about, and so she also agrees that he is 4 years old when he stops. He
says he has traveled (0.866)(4) = 3.464 lightyears away from his twin
(her) then. She says he traveled 8 years of her time ... i.e., she says
she is 8 years old when he stops. She says he is (0.866)(8) = 6.928
lightyears away when he stops.
After he stops, she says that he remains 6.928 lightyears away from her
after that. And she says they age at the same rate after he stops. He
says that, during his essentially instantaneous stopping time, his age
essentially doesn't change during his stopping. She agrees with that.

The above is all correct. But here is the sentence where I made the

Post by Mike Fontenot
But he says that during his essentially instantaneous stopping, SHE
essentially instantaneously gets older by 4 years ... i.e., he says that
she essentially instantaneously goes from being 4 years old to being 8
years old during the essentially instantaneous time in his life it takes
him to stop.

According to him, her age right before he stops is 2 years old, not 4
years old as I said above. Immediately after he stops, they both agree
about their respective ages. He says she is now 8 years old, so he says
her age increases by 6 years during his essentially instantaneous
stopping, not by 4 years as I stated in my previous post.

I should also have pointed out that each of the twins, during his
outbound trip, are entitled to use the famous time dilation equation for
inertial observers: That equation says that any inertial observer will
conclude that a person moving at speed "v" with respect to them is
ageing at a rate gamma times slower than they are, where

gamma = 1 / sqrt { 1 / (1 - v * v) },

where the asterisk indicates multiplication. For v = 0.866, gamma =
2.0. So on the outbound trip, each twin says the other twin is ageing
half as fast as their own rate of ageing. So right before he stops, she
says he is 4 and she is 8, but he says he is 4 and she is 2. And
immediately after he stops, they both agree that he is 4 and she is 8.
So he says she essentially instantaneously gets 6 years older during his
essentially instantaneous stopping.

There is an equation that I derived long ago that makes it easy to
calculate how her age changes (according to him) whenever he essentially
instantaneously changes his velocity. I call it the "delta_CADO" equation:

delta_CADO = -L * delta_v,

where L is their distance apart, according to her, and

delta_v = v2 - v1,

where v1 is his velocity before the change, and v2 is his velocity after
the change (with positive v being taken as the velocity when they are
moving apart).

So in this example,

L = 6.928 lightyears

v1 = 0.866 ly/y

v2 = 0.0 ly/y,

and so

delta_v = -0.866

and

delta_CADO = -6.928 * (-0.866) = 6.0 years.

We didn't need this equation to get the answer in the above particular
scenario, because his new velocity was zero, which makes things
especially easy. But when the two velocities are completely general,
it's necessary to either use the delta-CADO equation, or else do it
graphically with a Minkowski diagram, drawing lines of simultaneity.

Luigi Fortunati

2022-03-16 07:22:52 UTC

Perfect, I was going to report this discrepancy between instant aging of
4 instead of 6 years, now I absolutely agree with everything you have
written.

But you didn't answer my question about the Earth's revolutions around
the Sun.

Since I want my question to be clear, I express myself with numbers and
dates.

The traveling twin leaves on January 1, 2022 and keeps his telescope
continuously pointed at the receding solar system at speed v = 0.866c,
so that he can count how many revolutions the Earth makes around the
Sun.

Obviously, the turns seen on the telescope will take place much slower
than one for each year, both for the finite speed of the light (which,
due to the recession, takes more and more time to get to the spaceship
from the solar system) and for the time dilation

And all this greatly complicates the calculations.

This is why I ask: is it possible to calculate how many turns the
traveler twin will have * seen * make from the Earth around the Sun on
the spacecraft's telescope, after its 4-year journey, before starting to
brake?

Sylvia Else

2022-03-16 17:46:16 UTC

Are they identical, mirror, or fraternal twins. If identical, are they
conjoined? How would the experiment go for the latter case?

Sylvia.

Mike Fontenot

2022-03-16 17:47:04 UTC

Post by Richard Livingston
Keep in mind that when the moving twin stops, both twins will agree
on how far apart they are. During the deceleration the moving twin
will observe the apparent expansion of the universe along the axis
of travel, and what while moving appeared to be 3.464 light years
will expand to 6.928 light years.

That's true. But of the two effects (spatial and temporal), I think
what special relativity has to say about time is more interesting than
what it says about space (and distance).

The best example of that is the case of "negative ageing": if the
distant traveling twin (he) suddenly accelerates in the direction AWAY
from the home twin (her), he will conclude that she suddenly gets
YOUNGER during his velocity change. That result drives a lot of people
(including many physicists) crazy! Some physicists maintain that the
conclusions of the traveler during those occurrences must be ruled
inadmissible. But the negative ageing can't logically be ignored, or
disallowed, for the following reason: the traveler can do two
back-to-back instantaneous velocity reversals, which, taken together,
just cancel out. So we can't allow one of those velocity reversals, but
disallow the other.

For example, take the case where he is originally moving away from her
at speed

v = V1 = +V,

where V is some positive number. And let L be their distance apart
(according to her) at some instant. Then, at that instant, he suddenly
changes his velocity to

v = V2 = -V.

So

delta_v_1 = V2 - V1 = (-V) - V = - 2 * V.

And

delta_age_1 = -L * delta_v_1 = -L * (-2 * V) = 2 * L * V.

So he concludes that her age has instantaneously increased by (2 * L * V).

But suppose he IMMEDIATELY decides to reverse course again. He will
then conclude that her age has instantaneously DECREASED by (2 * L * V).
I.e.,

delta_v_2 = V - (-V) = V + V = 2 * V.

and

delta_age_2 = -L * delta_v_2 = -L * (2 * V) = -2 * L * V.

So he concludes that her age has instantaneously decreased by (2 * L * V).

So that gets her age (according to him) right back to where it was
before he did any accelerating ... everything is as if he had done NO
accelerating at all. But that means that we CANNOT say that
instantaneous age INCREASES are OK, but that instantaneous age DECREASES
are NOT OK. You can't allow one but disallow the other.

Post by Richard Livingston
Something to keep in mind is that neither twin can see the other at
what each considers "now". They only see the other on their own
past light cone. So while the moving twin is stopping, he will NOT
see his twin back on earth suddenly aging. This is because what
he sees is not the instantaneous state of the earth, but an image
conveyed by light. Instead he will see her as she was about 6.928
years earlier. Thus what he sees is a much younger twin, only
about a year older than when he left earth. It is on the trip back
to earth that he sees her age rapidly as he passes the light
traveling outbound from earth.

That's all true. But I am not interested in what TV images he receives
from her ... those just tell him what she looked like in the past, and
how old she was in the past. Instead, I am interested in what he
DEDUCES about her CURRENT age at any given instant in his life, using
the laws of special relativity. I.e., I'm interested in his "NOW"
instant ... what does he say her age is "RIGHT NOW", at some instant in
his life. I'm purely interested in what he says about "simultaneity at
a distance".

Richard Livingston

2022-03-17 15:14:51 UTC

On Wednesday, March 16, 2022 at 12:47:08 PM UTC-5, Mike Fontenot wrote:
...

Simultaneity at a distance is not observable. In our everyday experience
we actually experience the world on our past light cone and think of that
as "now". This a misconception. The "now" that Einstein defines with
his clock synchronization procedure is a useful concept for coordinates,
but as you are realizing, at any given event in space-time there is no
unique "now", it is highly observer dependent (i.e. depends on the
observers state of motion). I suggest that the only reality is on the
past light cone. That is something that all observers can agree on.
All observers will agree on WHAT is on the light cone and on the sequence
of events on the light cone (at least in a given direction), but may disagree
on when any event is on the light cone.

In particular, no matter what you do with direct observation, you will never
observe a clock going backwards or a person aging backwards. At most
you may CALCULATE that a clock on your "now" axis has gone backwards
due to your acceleration, but you will never be able to directly observe such
a thing. And I would argue that such a calculation is meaningless.

Proof of this is two events separated in space and occurring "simultaneously"
in some reference frame. The observer in that frame directly observing these
events from some time in the future says they occurred simultaneously at
{some time in the past}. A moving observer at that same place and time as
the first observer, will also see the two events occurring at the same time,
no matter how fast they are traveling. It is only when they calculate their
"now" times that the paradox of time appears.

This is why I think it is a mistake in quantum mechanics to talk about the
instantaneous collapse of the wave function CAUSED BY a measurement
event. There cannot be a consistent causal explanation directly linking
two events that are outside the light cone.

Rich L.

Phillip Helbig (undress to reply)

2022-03-17 22:26:56 UTC

Simultaneity at a distance is not observable.

In his most recent book*, Carlo Rovelli summarizes it like this:
Galilean relativity expresses the idea that the concept of being in the
same place at different times is ill-defined (relative to what?), while
Special Relativity expresses the idea that the concept of happening at
the same time in different places is equally ill-defined.

Does anyone disagree?

Considering that one recovers non-relativistic physics in the limit that
the speed of light goes to infinity, is it fair to say that the finite
speed of light is the SOLE reason for differences between
non-relativistic and relativistic physics?

__
* _General Relativity: The Essentials_, by Carlo Rovelli (Cambridge
University Press), 2021. Pp. 180, 23 =D7 15.5 cm. Price =A314.99
(paperback, ISBN 978 1 00 9013697).

Stefan Ram

2022-03-18 21:27:16 UTC

Post by Richard Livingston
Simultaneity at a distance is not observable.

In an inertial system: Place a detector D midway between A
and B (using a yardstick). I assume that A, B and D are at
rest. Generate two signals at A and B. If they arrive at D
at the same time, they were sent at A and B at the same time.
This would be a kind of observation, albeit delayed.

Post by Richard Livingston
In our everyday experience
we actually experience the world on our past light cone and think of that
as "now". This a misconception.

For /everyday applications/ it does /not/ seem to be
a misconception, and when the difference between past
light cone surface and "now" starts to matter, it's
not an "everyday experience" anymore.

Post by Richard Livingston
I suggest that the only reality is on the
past light cone. That is something that all observers can agree on.

The everyday reality consists of the things I can interact
with, like an automated teller machine (ATM). To get this
type of reality, the "now", in everyday life, needs to be
extended somewhat to an extended period of time like "today".
It cannot be only infinitesimally short, like a point of time.

This is the highest degree of reality: Something is very
real when one can interact with it, i.e., observe it /and/
affect it. Causality implies:

Systems of the past only have a semi-reality:
You can sometimes observe them, but not affect them.
For example, the Boston Tea Party. Also, systems of
the past are only inferred from records, so it is never
completely sure whether they even existed at all.

Systems of the future only have a semi-reality:
You can sometimes affect them, but not observe them.
For example, the Earth of the year 2023. Also, one
cannot be completely sure whether there will be such
an Earth.

Post by Richard Livingston
This is why I think it is a mistake in quantum mechanics to talk about the
instantaneous collapse of the wave function CAUSED BY a measurement
event. There cannot be a consistent causal explanation directly linking
two events that are outside the light cone.

When we say "nothing can travel faster than light", this is
actually not quite correct. An imagined point can travel faster
than light. If I make a spot on the surface of the moon with
a laser beam, the spot can move there faster than light.

It is /energy-impulse transports/, which cannot move faster
than light (and therefore also information transports).

The collapse of an imagined function can be imagined in an
inertial frame quite as "everywhere at the same time",
as long as no energy-impulse is transported with superluminal
speed by this collapse.

Richard Livingston

2022-03-20 17:49:12 UTC

Post by Stefan Ram

Post by Richard Livingston
Simultaneity at a distance is not observable.

Delayed is the key word. You cannot observe "now" now, you
can only observe it when it becomes on your past light cone.

...

Post by Stefan Ram
This is the highest degree of reality: Something is very
real when one can interact with it, i.e., observe it /and/

Only events on the past light cone can affect you. Only
events on your future light cone can be affected by you
at this moment. Events inside your future light cone can
be affected in your future.

Post by Stefan Ram
You can sometimes observe them, but not affect them.
For example, the Boston Tea Party. Also, systems of
the past are only inferred from records, so it is never
completely sure whether they even existed at all.

Events in the past, to the extent they are recorded or
(accurately) remembered are something that everyone
can agree on. That is as close to reality as we will
ever get.
...

Post by Stefan Ram
The collapse of an imagined function can be imagined in an
inertial frame quite as "everywhere at the same time",
as long as no energy-impulse is transported with superluminal
speed by this collapse.

I agree wrt a mathematical function as it applies to any single
observer. Yet the wave function does represent something
that is real, imperfectly. I'm not saying the wave function is
real or that it is exactly something real, but that it somehow
does capture some aspect of reality. The question is exactly
what of the wave function is real and what is a mathematical
fiction, or represents various alternate possibilities. I am
coming to the conclusion it would be better to call it a
possibility function rather than a probability function.

Rich L.

Mike Fontenot

2022-03-17 22:26:10 UTC

Post by Luigi Fortunati
This is why I ask: is it possible to calculate how many turns the
traveler twin will have * seen * make from the Earth around the Sun on
the spacecraft's telescope, after its 4-year journey, before starting to
brake?

Yes. I've already shown how the traveling twin (he) can determine the
current age of the home twin (her) at any instant in his life. And, if
you know what her mom says the position of the earth in its orbit around
the sun was when her daughter (the home twin) was born, then that
establishes a one-to-one correspondence between the home twin's age and
the position of the earth in its orbit about the sun, and the number of
complete orbits it has made since she was born. So that allows the
traveling twin to determine the position of the earth in its orbit, and
the number of completed orbits since the home twin was born, at each
each instant in the life of the traveler (he), according to him. But
I'm not at all interested in that ... I just want to know what he
concludes about her current age at each instant of his life.

Tom Roberts

2022-03-19 18:27:34 UTC

Post by Richard Livingston
Simultaneity at a distance is not observable.

In an inertial system: Place a detector D midway between A and B
(using a yardstick). I assume that A, B and D are at rest. Generate
two signals at A and B. If they arrive at D at the same time, they
were sent at A and B at the same time. This would be a kind of
observation, albeit delayed.

That is not observing simultaneity at A and B, it is observing
simultaneity at D. From that observation and the setup and the inertial
frame, we can INFER that the signals from A and B were sent at the same
time IN THAT INERTIAL FRAME. Inference is not observation.

Observation implies a physical quantity is being observed. Simultaneity
at different spatial points is not any kind of physical quantity, it is
a CONVENTION based on the time coordinate of a particular inertial
frame; such coordinate-dependent quantities cannot possibly be physical
quantities.

[...]

Attempting to apply modern physical theories to our everyday experience
is hopeless, because there are too many approximations involved.

Similarly, attempting to discuss "collapse of the wave function" in
terms of simultaneity and SR is also hopeless, because such "collapse"
is not observable (one can observe the system transitioning between
states, but not any "collapse").

A major lesson of modern physics is to discuss only measurable
(observable) quantities [#]. Both "everyday lives" and "wavefunction
collapse" violate that dictum (in very different ways).

[#] Interestingly, this applies to both QM and GR (for
very different reasons).

Tom Roberts

Mike Fontenot

2022-03-20 10:44:29 UTC

Post by Tom Roberts
Observation implies a physical quantity is being observed. Simultaneity
at different spatial points is not any kind of physical quantity, it is
a CONVENTION based on the time coordinate of a particular inertial
frame

Tom, what is your "take" on the use of an array of clocks, stationary in
some inertial frame, that have been synchronized using only the
assumption that the speed of light is equal to the universal constant
"c" in that frame. Each of those clocks is attended by a human "helper
friend" (HF), who can observe his immediate local surroundings. Another
particular observer (whom I'll refer to as "he"), also stationary in
that inertial frame, wants to know (when his watch shows time tau) the
current reading on a particular distant clock which is NOT stationary
with respect to that array of clocks. It seems reasonable that he is
entitled to say that the current reading on that distant clock is what
the HF, who happens to be colocated with that distant clock at the
instant when the HF's clock reads tau, directly observes it to be.
Doesn't that convey a sense of "meaningfulness", to the observers who
are stationary in that frame? I think Einstein thought it does.

Mike Fontenot

2022-03-25 10:57:15 UTC

Tom hasn't responded yet. Maybe he will shortly. But while waiting,
I'd like to add a bit to my above argument.

I think a sizeable number of physicists DO believe that simultaneity at
a distance is a meaningless concept, and I've even noticed a trend of
trying to de-emphasize talking about simultaneity-at-a-distance in
introductory special relativity courses. But I think that is a big
mistake. It certainly wasn't Einstein's view, at least for inertial
observers. (I've seen a quote from Einstein somewhere where he said he
hardly recognizes his theory when he reads some modern descriptions of
it.)

My above question of Tom Roberts was prompted mostly by his use of the
word "Convention" in describing simultaneity at a distance for an
inertial observer. To me, the term "Convention" implies that there is
more than one possible answer to the question "How old is that distant
person (she), right now", when asked and answered by a particular
inertial observer. "Convention" implies that one can pick from among
multiple alternatives, all equally good.

But I don't believe that is true, given my above description of how an
array of synchronized clocks (all permanently stationary with respect to
the given inertial observer) can be set up, creating a common "NOW"
instant for him and for all of the HF's ("helper friends") co-located
and co-stationary with the clocks. At any given instant "tau" in the
life of the given inertial observer, it's clear that there is just a
single answer to the question "How old is that particular distant person
(she) right now (at the given time "tau" in the life of the inertial
observer): it is what the particular HF (he) who happens to be
momentarily co-located with the distant person (she), says it is, at the
instant when he is age "tau". The only way there could be any other
allowable answer is if the synchronization of the clocks isn't valid,
and that is impossible if the velocity of light in that inertial
reference frame is equal to the universal constant "c".

jtmpreno

2022-03-26 20:31:39 UTC

Tom hasn't responded yet. Maybe he will shortly. But while waiting,
I'd like to add a bit to my above argument.
I think a sizeable number of physicists DO believe that simultaneity at
a distance is a meaningless concept, and I've even noticed a trend of
trying to de-emphasize talking about simultaneity-at-a-distance in
introductory special relativity courses. But I think that is a big
mistake. It certainly wasn't Einstein's view, at least for inertial
observers. (I've seen a quote from Einstein somewhere where he said he
hardly recognizes his theory when he reads some modern descriptions of
it.)
My above question of Tom Roberts was prompted mostly by his use of the
word "Convention" in describing simultaneity at a distance for an
inertial observer. To me, the term "Convention" implies that there is
more than one possible answer to the question "How old is that distant
person (she), right now", when asked and answered by a particular
inertial observer. "Convention" implies that one can pick from among
multiple alternatives, all equally good.
But I don't believe that is true, given my above description of how an
array of synchronized clocks (all permanently stationary with respect to
the given inertial observer) can be set up, creating a common "NOW"
instant for him and for all of the HF's ("helper friends") co-located
and co-stationary with the clocks. At any given instant "tau" in the
life of the given inertial observer, it's clear that there is just a
single answer to the question "How old is that particular distant person
(she) right now (at the given time "tau" in the life of the inertial
observer): it is what the particular HF (he) who happens to be
momentarily co-located with the distant person (she), says it is, at the
instant when he is age "tau". The only way there could be any other
allowable answer is if the synchronization of the clocks isn't valid,
and that is impossible if the velocity of light in that inertial
reference frame is equal to the universal constant "c".

What if the twins (or clocks) were quantum-entangled?

Mike Fontenot

2022-03-28 09:37:26 UTC

Post by jtmpreno
What if the twins (or clocks) were quantum-entangled?

If we were talking about molecular-size objects or smaller, we'd need to
use quantum mechanics in the analysis. But we're not.

Richard Livingston

2022-03-28 09:36:55 UTC

On Friday, March 25, 2022 at 5:57:18 AM UTC-5, Mike Fontenot wrote:
...

Post by Mike Fontenot
I think a sizeable number of physicists DO believe that simultaneity at
a distance is a meaningless concept, and I've even noticed a trend of
trying to de-emphasize talking about simultaneity-at-a-distance in
introductory special relativity courses. But I think that is a big
mistake. It certainly wasn't Einstein's view, at least for inertial
observers. (I've seen a quote from Einstein somewhere where he said he
hardly recognizes his theory when he reads some modern descriptions of
it.)
My above question of Tom Roberts was prompted mostly by his use of the
word "Convention" in describing simultaneity at a distance for an
inertial observer. To me, the term "Convention" implies that there is
more than one possible answer to the question "How old is that distant
person (she), right now", when asked and answered by a particular
inertial observer. "Convention" implies that one can pick from among
multiple alternatives, all equally good.
But I don't believe that is true, given my above description of how an
array of synchronized clocks (all permanently stationary with respect to
the given inertial observer) can be set up, creating a common "NOW"
instant for him and for all of the HF's ("helper friends") co-located
and co-stationary with the clocks. At any given instant "tau" in the
life of the given inertial observer, it's clear that there is just a
single answer to the question "How old is that particular distant person
(she) right now (at the given time "tau" in the life of the inertial
observer): it is what the particular HF (he) who happens to be
momentarily co-located with the distant person (she), says it is, at the
instant when he is age "tau". The only way there could be any other
allowable answer is if the synchronization of the clocks isn't valid,
and that is impossible if the velocity of light in that inertial
reference frame is equal to the universal constant "c".

I think I mostly agree with you, but still think that the problem with
"now" at a distant location is that people take it as something
real and meaningful, and I think an argument can be made that it
is not very meaningful.

What is useful for physics is establishing a coordinate framework
that allows us to describe physics in a consistent way. For this
purpose "now" at a distant location makes sense. But our
mathematics is in a sense all knowing in that we keep track of
events at all locations and times, before they are able to interact
with other objects in the future.

For an individual, however, what is "now" at a distant location is
not something that has a consistent answer. Two different
observers at the same event can have very different ideas about
what is happening "now" someplace else. What is indisputably
real, however, is what these observers can see on their past
light cone. All observers at an event will see the same things
on their past light cones.

Rich L.

Mike Fontenot

2022-03-29 10:19:15 UTC

[...] At any given instant "tau" in the
life of the given inertial observer, it's clear that there is just a
single answer to the question "How old is that particular distant person
(she) right now (at the given time "tau" in the life of the inertial
observer): it is what the particular HF (he) who happens to be
momentarily co-located with the distant person (she), says it is, at the
instant when he is age "tau". The only way there could be any other
allowable answer is if the synchronization of the clocks isn't valid,
and that is impossible if the velocity of light in that inertial
reference frame is equal to the universal constant "c".

My argument above is that, IF those clocks are synchronized (according
to the given observer), then he can't help but conclude that the current
age of that distant person IS completely meaningful TO HIM. And the
only way that those clocks AREN'T synchronized according to him, is if
the velocity of light in his inertial reference frame ISN'T equal to the
universal constant "c". But the fundamental assumption of special
relativity IS that light will be measured in all inertial reference
frames to have the value "c". Therefore, FOR any given inertial
observer (he), the current age of a distant person is completely
meaningful to him.

But what about a non-inertial observer? In particular, what about a
given observer who is undergoing a constant acceleration? What does HE
say the current age of a distant person is? It turns out to be possible
for such an accelerating observer to rely on an array of clocks and
associated "helper friends" (HF's) to give him the answer. Unlike in
the inertial case, those clocks DON'T run at the same rate. But the
ratio of the rates of those clocks can be CALCULATED by the given
observer. And if he (and the HF's) are initially stationary and
unaccelerated, they can start out with synchronized clocks (and ages).
Then, if they all fire their identical rockets at the same instant, they
can each CALCULATE the current reading of each of the other clocks, at
each instant in their lives. The calculations of each of the HF's all
agree. So, at any instant in their lives during that acceleration, they
each share the same "NOW" instant with all of the other HF's. That
means that the given observer (he), at any instant "tau" in his life,
can obtain the current age "T" of some distant person (her), by asking
the HF, who happens to be momentarily co-located with her at that NOW
instant, what her age is then.

Mike Fontenot

2022-04-02 17:36:41 UTC

Post by Mike Fontenot
But what about a non-inertial observer? In particular, what about a
given observer who is undergoing a constant acceleration? What does HE
say the current age of a distant person is? It turns out to be possible
for such an accelerating observer to rely on an array of clocks and
associated "helper friends" (HF's) to give him the answer. Unlike in
the inertial case, those clocks DON'T run at the same rate. But the
ratio of the rates of those clocks can be CALCULATED by the given
observer. And if he (and the HF's) are initially stationary and
unaccelerated, they can start out with synchronized clocks (and ages).
Then, if they all fire their identical rockets at the same instant, they
can each CALCULATE the current reading of each of the other clocks, at
each instant in their lives. The calculations of each of the HF's all
agree. So, at any instant in their lives during that acceleration, they
each share the same "NOW" instant with all of the other HF's. That
means that the given observer (he), at any instant "tau" in his life,
can obtain the current age "T" of some distant person (her), by asking
the HF, who happens to be momentarily co-located with her at that NOW
instant, what her age is then.

In the above, I said that the given accelerating observer (he)
(abbreviated, the "AO"), at each instant of his life, can CALCULATE the
current reading on each of the HF's clocks. What IS that calculation?

Let t = 0 be the reading on his clock at the instant that the constant
acceleration "A" begins, and let all the HFs' clocks also read zero at
that instant. Thereafter, he and all of the HFs are accelerating at "A"
ls/s/s, and the ratio R of any given HF's clock rate to his (the
observer's (he) whose conclusions we are seeking) clock rate is

R(t) = [ 1 +- L A sech^2 (A t) ],

where L is the constant distance between him and the given HF, and
sech() is the hyperbolic secant (which is the reciprocal of cosh(), the
hyperbolic cosine). The "^2" after the sech indicates the square of the
sech. The "+-" in the above equation means that the second term is
ADDED to 1 for the HF's who are LEADING the accelerating observer, and
the second term is SUBTRACTED from 1 for the HF's who are TRAILING the
accelerating observer. For brevity, I'll just take the case where the
HF of interest is a leading HF.

The limit of R(t), as "t" goes to zero, is 1 + L A. The limit of R(t),
as "t" goes to infinity, is 1.0 So R(t) starts out at some positive
number greater than 1, and then approaches 1.0 as t goes to infinity.
So eventually, all the clocks essentially tic at the same rate, but
early in the acceleration, the ratio of the tic rates varies
significantly with time.

The current reading of the HF's clock (the "Age Change" or "AC"), when
the AO's clock reads "tau", is

AC(tau) = integral, from zero to tau, of { R(t) dt }

= tau + L tanh( A tau ).

The above result depends on the fact that

sech^2(u) = d{tanh(u)} / d{u}.

As tau goes to zero, AC goes to zero. As tau goes to infinity, AC goes
to tau + L, which goes to infinity, approaching a slope of 1.0 from above.

So there you have it. That's the calculation that defines "NOW" for the
AO and all of the HF's, and makes simultaneity at a distance a
meaningful concept for them. Simultaneity at a distance is not a choice.

Mike Fontenot

2022-04-03 17:13:48 UTC

Post by Mike Fontenot
So there you have it. That's the calculation that defines "NOW" for the
AO and all of the HF's, and makes simultaneity at a distance a
meaningful concept for them. Simultaneity at a distance is not a choice.

But what does the above say about the current age of the home twin
(she), according to the traveling twin (he), for each instant in his
life on his trip? The answer is that the above equations give the same
results as the Co-Moving-Inertial-Frames (CMIF) simultaneity method.
That is very fortuitous, because the CMIF method is relatively easy to
use. The value of the array of clocks discussed above (which establish
a "NOW" moment for the accelerating observer that extends throughout all
space) is that they GUARANTEE that the CMIF results are fully meaningful
to the traveler, and that the CMIF method is the ONLY correct
simultaneity method for him. He has no other choice.

Tom Roberts

2022-04-05 19:53:33 UTC

Post by Mike Fontenot
So there you have it. That's the calculation that defines "NOW"
for the AO and all of the HF's, and makes simultaneity at a
distance a meaningful concept for them. Simultaneity at a
distance is not a choice.

Yes, simultaneity at a distance is a choice. Your approach is
outrageously unphysical -- maneuvering the ship causes unphysical
changes in the "age" ascribed to a distant friend.

Post by Mike Fontenot
But what does the above say about the current age of the home twin
(she), according to the traveling twin (he), for each instant in his
life on his trip? The answer is that the above equations give the
same results as the Co-Moving-Inertial-Frames (CMIF) simultaneity
method. That is very fortuitous, because the CMIF method is
relatively easy to use. The value of the array of clocks discussed
above (which establish a "NOW" moment for the accelerating observer
that extends throughout all space) is that they GUARANTEE that the
CMIF results are fully meaningful to the traveler, and that the CMIF
method is the ONLY correct simultaneity method for him.

As I keep pointing out and you keep ignoring, this is outrageously
unphysical. If the traveling twin maneuvers his spaceship, the "age" he
ascribes to a distant friend can change very rapidly. No sensible person
would believe that local actions he takes can "change" his friend's age.

Post by Mike Fontenot
He has no other choice.

Not at all. The traveling twin can recognize that his current time is
completely divorced from that of his friend far away. If the age of his
friend is important to him, he would keep track of his motion relative
to the ICRF, knowing that regardless of his motion or location, the
current time of the ICRF can be used to calculate the current age of his
friend (for all practical purposes his friend on earth is at rest in
the ICRF). Any sensible astronaut would do that.

Tom Roberts

Mike Fontenot

2022-04-05 19:54:03 UTC

(This is an improved version of a posting that I submitted yesterday
[4-3-22], but which hasn't shown up yet)

In 1907, Einstein published a VERY long paper (in several volumes) on
his "relativity principle". In volume 2, section 18, page 302, titled
"Space and time in a uniformly accelerated reference frame", he
investigated how the tic rates compare for two clocks separated by the
constant distance L, with both clocks undergoing a constant acceleration
"A". He restricted the analysis to very small accelerations (and very
small resulting velocities). His result (on page 305) was that the
leading clock tics at a rate

R = 1 + L A

faster than the rear clock. Note that that result agrees with my
equation, for very small "L" and "A". But he then said:

"From the fact that the choice of the coordinate origin must not
affect the relation, one must conclude that, strictly speaking, equation
(30) should be replaced by the equation R = exp(L A). Nevertheless, we
shall maintain formula (30)."

I've never understood that one sentence argument he gave, for replacing
his linear equation with the exponential equation. But I DID assume he
was right (because he was rarely wrong), until I tried applying his
exponential equation to the case of essentially instantaneous velocity
changes that are useful in twin "paradox" scenarios in special
relativity. Specifically, I worked a series of examples where the
separation of the two clocks is always

L = 7.52 ls (lightseconds)

and where the final speed (with the initial speed being zero) is always

v = 0.866 ls/s.

That speed implies a final "rapidity" of

theta = atanh(0.866) = 1.317 ls/s.

("Rapidity" is a non-linear version of velocity. They have a one-to-one
correspondence. In special relativity, velocity can never exceed 1.0
ls/s in magnitude, but rapidity can have an infinite magnitude. An
acceleration of "A" ls/s/s lasting for "t" seconds changes the rapidity
theta by the product of "A" and "t".)

So in this case, when we are starting from zero velocity and thus zero
rapidity, at the end of the acceleration,

theta = A tau,

where tau is the duration of the acceleration (according to the rear
clock). So, if we know theta and tau, we then know the acceleration:

A = theta / tau.

I do a sequence of calculations, each starting at t = 0, with the two
clocks reading zero, and with zero acceleration for t < 0.

First, I set the duration tau of the acceleration to 1 second. The
acceleration then needs to be

A = theta / tau = 1.317 / 1.0 = 1.317 ls/s/s (that's roughly
40 g's).

So for this first case, the tic rate of the leading clock during the one
second acceleration is

R = exp( L A ) = exp(9.9034), or about 20000.

(I picked that weird value of "L" so that this value of "R" for the fist
case would be a round number, just to make the calculations easier.)

Since R is constant during the acceleration, the CURRENT reading on the
leading clock (which I'll denote as AC, for "Age Change") is

AC = tau R = 2x10sup4 = 20000

(where 10sup4 means "10 raised to the 4th power").

I then start over and work a second case, with ten times the
acceleration (13.17 ls/s/s), but with tau ten times smaller (0.1
second). That keeps the final rapidity the same as in the first case,
and the final speed is also 0.866, as before. For the second case,

AC = 1.02x10sup42.

So when we made the acceleration an order of magnitude larger, and the
duration an order of magnitude smaller, the current reading "AC" on the
leading clock got about 38 orders of magnitude larger.

Next, I start over again and work a third case, again increasing the
acceleration by a factor of 10, and the decreasing the duration by a
factor of 10, so "A" = 131.7 ls/s/s and tau = 0.01 second. Then, AC =
1.27x10sup428. So this time, when we increased "A" by a factor of 10,
and decreased tau by a factor of ten, AC got about 380 orders of
magnitude larger.

AC is not approaching a finite limit as tau goes to zero and "A" goes to
infinity. In each iteration, the change in AC compared to the previous
change gets MUCH larger. Clearly, the clock reading is NOT converging
to a finite limit. It is going to infinity as tau goes to zero.

We can see this, even without doing the above detailed calculations. Since

AC = tau exp(L A),

the tau factor goes to zero LINEARLY as tau goes to zero, but exp(L A)
goes to infinity EXPONENTIALLY as tau goes to zero, so their product is
obviously not going to be finite as tau goes to zero.

So, for the idealization of an essentially instantaneous velocity
change, the change of the reading on the leading clock is INFINITE
during the infinitesimal change of the rear clock. That means that,
when the traveling twin instantaneously changes his speed from zero to
0.866 (toward the home twin), the exponential version of the R equation
says that the home twin's age becomes infinite. But we know that's not
true, because the home twin is entitled to use the time dilation
equation for a perpetually-inertial observer, and that equation tells
her that for a speed of 0.866 ls/s, the traveler's age is always
increasing half as fast as her age is increasing. So when they are
reunited, she is twice as old he is, NOT infinitely older than he is, as
the exponential form of the gravitational time dilation equation claims.
The time dilation equation for a perpetually-inertial observer is the
gold standard in special relativity. Therefore the exponential form of
the gravitational time dilation equation is incorrect.

The correct gravitational time dilation equation turns out to
approximately agree with what Einstein used in his "small acceleration"
analysis, for very small accelerations, but differs substantially for
larger accelerations. And the correct gravitational time dilation
equation agrees with the ages of the twins when they are reunited. It
also exactly agrees with the CMIF simultaneity method for the traveler's
conclusions about the sudden increase in the home twin's age when the
traveler suddenly changes his velocity. The CMIF method provides a
practical way to compute the change in the home twin's age when the
traveler instantaneously changes his velocity. But it is the new
gravitational time dilation equation, and its array of clocks with a
common "NOW" moment, that guarantees that the CMIF result is fully
meaningful to the traveling twin, and that the CMIF method is the ONLY
correct simultaneity method for the traveling twin.

Tom Roberts

2022-04-02 06:22:53 UTC

Post by Mike Fontenot
My above question of Tom Roberts was prompted mostly by his use of
the word "Convention" in describing simultaneity at a distance for
an inertial observer. To me, the term "Convention" implies that
there is more than one possible answer to the question "How old is
that distant person (she), right now", when asked and answered by a
particular inertial observer. "Convention" implies that one can pick
from among multiple alternatives, all equally good.

Right, that's what it means: there are an infinite number of coordinate
systems from which to choose, and in GR (and modern physics in general),
all are equally good for describing what happens. (Note that none of
them actually affects what happens, they only describe it differently.)

Post by Mike Fontenot
But I don't believe that is true, given my above description of how
an array of synchronized clocks (all permanently stationary with
respect to the given inertial observer) can be set up, [...]

As I said before, you simply applied the convention of using Einstein
synchronization in an inertial frame. There is no compulsion to do so,
it just so happens that it is (usually) the easiest and simplest system
of coordinates to use in the case where the observer is at rest in the
inertial frame.

IOW the frame does not determine the coordinates, the human observer
does so. We humans generally select the coordinates in which
calculations are simplest, but that is all it is. Don't confuse an
ordinary human desire for simplicity (laziness) with anything more profound.

Post by Mike Fontenot
I think I mostly agree with you, but still think that the problem
with "now" at a distant location is that people take it as something
real and meaningful, and I think an argument can be made that it is
not very meaningful.

Hmmm. It's not merely an "argument", it's an inescapable aspect of
modern physics: "now at a distant location" does not contribute to any
current model of the world. In GR such spacelike intervals are outside
the past lightcone and so cannot affect what happens at the event in
question (i.e. now at the speaker's location); in QFT, fields evaluated
at such a spacelike interval always commute (and thus do not contribute
to any amplitudes).

Post by Mike Fontenot
What is useful for physics is establishing a coordinate framework
that allows us to describe physics in a consistent way.

Yes.

Post by Mike Fontenot
For this purpose "now" at a distant location makes sense.

For describing what happened, AFTER THE FACT, based on reports received
from distant 'helper friends', sure. But for modeling the world, it is
not useful at all (see my previous paragraph).

Post by Mike Fontenot
IF those clocks are synchronized (according to the given observer),
then he can't help but conclude that the current age of that distant
person IS completely meaningful TO HIM.

Given the CONVENTION used, sure. But such clock synchronization is only
a convention. For an inertial observer, using the coordinates of their
rest frame is the simplest way to interpret "now" at distant locations,
but they are well advised to remember it is merely a convention to do it
that way, one based on simplicity TO HUMANS, not any physical basis.
Don't confuse an ordinary human desire for simplicity (laziness) with
anything more profound.

Post by Mike Fontenot
But what about a non-inertial observer?

That is what you have spent a lot of time writing nonsense about. To an
accelerating observer, "now" has no definite meaning, and attempting to
give it one is just foolish, and no sensible person would ever believe
it. They can calculate whatever they like, but no sensible person would
think that just by turning their spaceship around their distant friend
gets younger -- they would KNOW that this is purely an artifact of the
math used in the calculation and their current acceleration, and that
has nothing whatsoever to do with their friend. Any astronaut in a
spacecraft capable of traveling at an appreciable fraction of c would
already know that their time is completely divorced from that of their
friends back home.

Tom Roberts

Tom Roberts

2022-04-02 06:22:52 UTC

Post by Tom Roberts
Observation implies a physical quantity is being observed.
Simultaneity at different spatial points is not any kind of
physical quantity, it is a CONVENTION based on the time coordinate
of a particular inertial frame

Tom, what is your "take" on the use of an array of clocks,
stationary in some inertial frame, that have been synchronized using
only the assumption that the speed of light is equal to the universal
constant "c" in that frame. Each of those clocks is attended by a
human "helper friend" (HF), who can observe his immediate local
surroundings. Another particular observer (whom I'll refer to as
"he"), also stationary in that inertial frame, wants to know (when
his watch shows time tau) the current reading on a particular
distant clock which is NOT stationary with respect to that array of
clocks. It seems reasonable that he is entitled to say that the
current reading on that distant clock is what the HF, who happens to
be colocated with that distant clock at the instant when the HF's
clock reads tau, directly observes it to be.

Sure, one could do that. Though there seems to be little motivation to
do so.

The particular observer NEVER knows "the current reading" of that
distant clock (for any meaning of "current"), he can only learn what it
was at some time in the past (e.g. at time tau), after the HF transmits
their result to him. Note that in your scenario the particular observer
must transmit the value of tau to all HFs well in advance;
alternatively, each HF could record values whenever the distant clock
passes by, and they all send those records to the particular observer,
who can then pick and choose among them, after the fact.

Post by Mike Fontenot
Doesn't that convey a sense of "meaningfulness", to the observers who
are stationary in that frame? I think Einstein thought it does.

You simply implemented the CONVENTION of Einstein synchronization. If
you synchronized the clocks differently you would get a different
result. It says nothing at all about other frames.

Tom Roberts

Mike Fontenot

2022-04-02 17:51:23 UTC

Post by Tom Roberts

Post by Mike Fontenot
Doesn't that convey a sense of "meaningfulness", to the observers who
are stationary in that frame? I think Einstein thought it does.

You simply implemented the CONVENTION of Einstein synchronization. If
you synchronized the clocks differently you would get a different
result.

If you synchronized those clocks differently, you would be ignoring the
fundamental assumption that special relativity is based on: that the
speed of light in any inertial frame is always equal to the universal
constant "c".

Tom Roberts

2022-04-03 22:15:18 UTC

Post by Tom Roberts
You simply implemented the CONVENTION of Einstein synchronization.
If you synchronized the clocks differently you would get a
different result.

If you synchronized those clocks differently, you would be ignoring
the fundamental assumption that special relativity is based on: that
the speed of light in any inertial frame is always equal to the
universal constant "c".

As you say, that assumption of SR applies in any INERTIAL FRAME. But if
one synchronized those clocks differently, they would not yield the time
coordinate of an inertial frame, so the assumptions of SR simply would
not apply, and can be ignored.

[Don't ask me why anyone would do that, I'm merely pointing
out that it is possible. The usual Einstein synchronization
is used because it simplifies the math. As I said before:
Don't confuse an ordinary human desire for simplicity
(laziness) with anything more profound.]

Tom Roberts

Tom Roberts

2022-04-03 22:15:19 UTC

Post by Tom Roberts
Observation implies a physical quantity is being observed.
Simultaneity at different spatial points is not any kind of
physical quantity, it is a CONVENTION based on the time coordinate
of a particular inertial frame

Tom, what is your "take" on the use of an array of clocks,
stationary in some inertial frame, that have been synchronized using
only the assumption that the speed of light is equal to the universal
constant "c" in that frame. Each of those clocks is attended by a
human "helper friend" (HF), who can observe his immediate local
surroundings. Another particular observer (whom I'll refer to as
"he"), also stationary in that inertial frame, wants to know (when
his watch shows time tau) the current reading on a particular
distant clock which is NOT stationary with respect to that array of
clocks. It seems reasonable that he is entitled to say that the
current reading on that distant clock is what the HF, who happens to
be colocated with that distant clock at the instant when the HF's
clock reads tau, directly observes it to be.

Post by Mike Fontenot
Doesn't that convey a sense of "meaningfulness", to the observers who
are stationary in that frame? I think Einstein thought it does.

You simply implemented the CONVENTION of Einstein synchronization. If
you synchronized the clocks differently you would get a different
result. It says nothing at all about other frames.

Tom Roberts

Stefan Ram

2022-03-20 17:50:48 UTC

Post by Tom Roberts
A major lesson of modern physics is to discuss only measurable
(observable) quantities

In quantum theory, the comprehensive framework theory of
physics, the state of a system plays the central role.
This state is not an observable, but discussed extensively!

In almost all areas of physics, also outside of quantum
physics, for example in GR, quantities are modeled by real
numbers, which are also not observable.

Richard Livingston

2022-03-20 18:30:33 UTC

On Saturday, March 19, 2022 at 1:27:38 PM UTC-5, Tom Roberts wrote:
...

Post by Tom Roberts
A major lesson of modern physics is to discuss only measurable
(observable) quantities [#]. Both "everyday lives" and "wavefunction
collapse" violate that dictum (in very different ways).
[#] Interestingly, this applies to both QM and GR (for
very different reasons).
Tom Roberts

Tom, I would be very interested in your expanding a bit on how this all applies to GR. -Rich L.

[[Mod. note -- I too would be interested in what Tom says.

My take would be that in GR, there is no preferred coordinate system
and all physical quantities (= those that are measurable, at least in
a gedanken sense) should be independent of the coordinate system in use.

Notably:
* the coordinate "time" of an event, or the difference between the
coordinate times of two events) is merely a coordinate; it has no
inherent physical meaning and can be changed arbitrarily by changing
our coordinate system
* *proper* time along some (timelike) worldline is measurable (it's
what an (ideal) clock moving along that worldline would measure),
can be said to have an inherent physical meaning (as the observable
result of that measurement), and *doesn't* change when we change
coordinates
* similarly, the coordinate position of an object, or the (coordinate)
distance between two objects, is also a coordinate, has no inherent
physical meaning, and can be changed arbitrarily by changing our
coordinate system
* the *proper* distance along a given path is measurable (at least in
a gedanken sense: one can imagine laying down a sequence of standard
rulers end-to-end along the path), and doesn't change when we change
coordinates;
* coordinate singularities (and the set of events where they occur)
have no inherent physical meaning, and a change in coordinates can
change the set of events where there is a coordinate singularity;
only singularities in observable quantities like curvature invariants,
proper times/distances, etc, are physically meaningful
-- jt]]

Tom Roberts

2022-04-02 06:22:52 UTC

Post by Tom Roberts
A major lesson of modern physics is to discuss only measurable
(observable) quantities [#]. Both "everyday lives" and
"wavefunction collapse" violate that dictum (in very different
ways).
[#] Interestingly, this applies to both QM and GR (for very
different reasons).

Tom, I would be very interested in your expanding a bit on how this
all applies to GR. -Rich L.

Note that coordinates are arbitrary human constructs, which we use to
simplify, codify, and quantify our observations and descriptions. Nature
clearly uses no coordinates, so the choice of coordinates used to
describe some natural phenomenon cannot possibly affect that phenomenon.
That is the importance of coordinate independence in GR (and in all of
theoretical physics) -- for a quantity to correspond to some physical
phenomenon, it must be independent of coordinates (aka invariant).

For instance, each and every measurement is a definite value for
whatever physical phenomenon is being measured, and is inherently
invariant. So observer A can construct a locally inertial frame and use
it to measure the kinetic energy of a baseball, and all other observers
will agree that is the value A measures in that frame, even though they
themselves use other frames to measure a different value.

[This has been called a "coordinate-dependent invariant
quantity -- the value depends on which coordinates are
used, but the result is invariant because it is
inextricably bound to the coordinates used.]

Take careful note of the wording: kinetic energy is NOT invariant,
but the kinetic energy of a designated object relative to a specified
inertial frame is indeed invariant. So observers using other frames can
make measurements of the designated object, transform them to the
specified frame, and agree on the value obtained in the specified frame.

Post by Richard Livingston
Mod. note -- I too would be interested in what Tom says.
My take would be that in GR, there is no preferred coordinate system
and all physical quantities (= those that are measurable, at least in
a gedanken sense) should be independent of the coordinate system in use.

Yes, "should be" => "are".

Also beware of "preferred coordinate system", because those words are
ambiguous -- physicists use that phrase in the sense of a coordinate
system that appears explicitly in the equations of the dynamics; but in
many/most cases there is a particular choice of coordinates relative to
which the calculations are simplified, and we invariably prefer to use
them. The invariance of physical quantities ensures we can do so.

Post by Richard Livingston
* the coordinate "time" of an event, or the difference between the
coordinate times of two events) is merely a coordinate; it has no
inherent physical meaning and can be changed arbitrarily by changing
our coordinate system
* *proper* time along some (timelike) worldline is measurable (it's
what an (ideal) clock moving along that worldline would measure),
can be said to have an inherent physical meaning (as the observable
result of that measurement), and *doesn't* change when we change
coordinates
* similarly, the coordinate position of an object, or the (coordinate)
distance between two objects, is also a coordinate, has no inherent
physical meaning, and can be changed arbitrarily by changing our
coordinate system
* the *proper* distance along a given path is measurable (at least in
a gedanken sense: one can imagine laying down a sequence of standard
rulers end-to-end along the path), and doesn't change when we change
coordinates;
* coordinate singularities (and the set of events where they occur)
have no inherent physical meaning, and a change in coordinates can
change the set of events where there is a coordinate singularity;
only singularities in observable quantities like curvature invariants,
proper times/distances, etc, are physically meaningful

Yes to all. The Moderator and I agree, except for details in wording.

Tom Roberts

Tom Roberts

2022-04-03 22:15:19 UTC

Post by Tom Roberts
A major lesson of modern physics is to discuss only measurable
(observable) quantities [#]. Both "everyday lives" and
"wavefunction collapse" violate that dictum (in very different
ways).
[#] Interestingly, this applies to both QM and GR (for very
different reasons).

Tom, I would be very interested in your expanding a bit on how this
all applies to GR. -Rich L.

Yes to all. The Moderator and I agree, except for details in wording.

Tom Roberts

Richard Livingston

2022-04-06 18:22:40 UTC

On Tuesday, April 5, 2022 at 2:54:06 PM UTC-5, Mike Fontenot wrote:
... {long derivation related to Ehrenfest Paradox and Rendler Coordinates}

Post by Mike Fontenot
So, for the idealization of an essentially instantaneous velocity
change, the change of the reading on the leading clock is INFINITE
during the infinitesimal change of the rear clock. That means that,
when the traveling twin instantaneously changes his speed from zero to
0.866 (toward the home twin), the exponential version of the R equation
says that the home twin's age becomes infinite. But we know that's not
true, because the home twin is entitled to use the time dilation
equation for a perpetually-inertial observer, and that equation tells
her that for a speed of 0.866 ls/s, the traveler's age is always
increasing half as fast as her age is increasing. So when they are
reunited, she is twice as old he is, NOT infinitely older than he is, as
the exponential form of the gravitational time dilation equation claims.
The time dilation equation for a perpetually-inertial observer is the
gold standard in special relativity. Therefore the exponential form of
the gravitational time dilation equation is incorrect.

Your calculations are all related to the Ehrenfest Paradox and Rindler
Coordinates. What myself and others have been trying to get you
to understand is that the acceleration of the observer does not actually
change anything about the distant twin. By your own calculations
the age of the distant twin can be anything +/- c*d depending on the
relative state of motion and accelerations. Furthermore, with sufficiently
high acceleration the observer cannot "see" the distant twin at all.

Please note that an inertial observer at the same location as your
traveling twin, watching everything as the traveling twin accelerates,
does not see any change in the rest of the universe. All these
coordinate transformations are entirely observer dependent and
do not represent anything real for the physics. They only affect the
APPEARANCE of the world to that observer. Any attempt to place
greater significance to these appearances is misguided.

Post by Mike Fontenot
The correct gravitational time dilation equation turns out to
approximately agree with what Einstein used in his "small acceleration"
analysis, for very small accelerations, but differs substantially for
larger accelerations. And the correct gravitational time dilation
equation agrees with the ages of the twins when they are reunited. It
also exactly agrees with the CMIF simultaneity method for the traveler's
conclusions about the sudden increase in the home twin's age when the
traveler suddenly changes his velocity. The CMIF method provides a
practical way to compute the change in the home twin's age when the
traveler instantaneously changes his velocity. But it is the new
gravitational time dilation equation, and its array of clocks with a
common "NOW" moment, that guarantees that the CMIF result is fully
meaningful to the traveling twin, and that the CMIF method is the ONLY
correct simultaneity method for the traveling twin.

Again, you should study the Ehrenfest Paradox and Rindler Coordinates.
You still don't seem to appreciate the true significance (or lack thereof)
of the calculated "now".

Rich L.

Mike Fontenot

2022-04-07 04:04:05 UTC

Post by Richard Livingston
Your calculations are all related to the Ehrenfest Paradox and Rindler
Coordinates.

I looked up Ehrenfest Paradox on Wiki, and it says it's about a rotating
disk. The work I've been doing has nothing to do with a rotating disk.

Richard Livingston

2022-04-07 18:02:15 UTC

Post by Richard Livingston
Your calculations are all related to the Ehrenfest Paradox and Rindler
Coordinates.

I looked up Ehrenfest Paradox on Wiki, and it says it's about a rotating
disk. The work I've been doing has nothing to do with a rotating disk.

Mike,

Sorry about that. I should have referenced the Bell Spaceship paradox.
That is a closely related effect for a linearly accelerating reference
frame that is closer to what you are analyzing.

Rich L.

Mike Fontenot

2022-04-08 07:05:52 UTC

What myself and others have been trying to get you (Mike Fontenot)
to understand is that the acceleration of the observer does not actually
change anything about the distant twin.

It's not clear what that statement even MEANS. Obviously, the distant
twin (she) doesn't suddenly feel like she's getting younger. At each
instant in her life, her brain is in a state that is different from all
of her other brain states. Nothing can change those states. But the
accelerating observer DOES conclude that she instantaneously gets
younger when he instantaneously changes his velocity in the direction
away from her. And so, FOR HIM, she ACTUALLY gets younger. All
perpetually-inertial observers disagree with him about her getting
younger when he instantaneously changes his velocity, but they also all
disagree among themselves about what her current age is when the
accelerating observer changes his velocity. And FOR EACH OF THEM, she
ACTUALLY has the current age they compute. That's just the way special
relativity IS ... different observers disagree, they all think they are
right, and none of them is wrong!

What is really new, though, in my latest results, is the fact that the
accelerating observer can assemble an array of clocks (and attending
"helper friends" (HF's)), which give him a "NOW" that extends throughout
all space (analogous to what Einstein did for inertial observers). And
THAT guarantees that the accelerating observer's conclusions about the
home twin's age are fully MEANINGFUL to him. His conclusions agree with
the CMIF simultaneity method, which means that the CMIF simultaneity
method is the only correct simultaneity method.

[[Mod. note -- I think you're mistaken in a couple of places:

1. An accelerating obserer ("he") does not (or to be pedantic, should
not, if he is doing physics correctly) conclude that the distant twin
("she") instantaneously gets younger when he instantaneously changes
his velocity in the direction away from her. Rather he concludes
that her age coordinate in inertial reference frame #2 (after his
velocity change) < her age coordinate in inertial reference frame #1
(before his velocity change). But these are two DIFFERENT inertial
reference frames, with DIFFERENT time coordinates. Attributing
physical meaning to a comparison between DIFFERENT inertial frame's
time coordinates is no more valid than (say) attributing physical
meaning to the difference between 2022 (the current year on Earth
in the Gregorian calendar) and 4720 (the current year on Earth in
the Chinese calendar). If I install new calendar software on my
computer, I don't suddenly get 4720-2022 years younger or older for
any sensible meaning of "younger" or "older". :)

2. What does it mean to say a time coordinate is "physically meaningful"?
I would argue that it means that you can write the laws of physics
in a sensible form in terms of that time coordinate. So, what would
(say) Newton's 2nd law look like using the CMIF time coordinate of
an accelerating observer? Ick, not nice at all. Or how about Maxwell's
equations? Or even something very simple like the radioactive decay
law
N_atoms(t) = N_atoms(0) * exp(-lambda*t)
for a fixed lambda. Again, not nice at all if "t" on the left-hand
side and "t" on the right-hand-side are the time coordinate of different
inertial frames.

The fact that these and other laws of physics don't have a sensible
form when written in a mixture of different time coordinates (such
as CMIF times for accelerating observers) is, I would argue, prima
facie evidence that such a mixture of time coordinates is *not*
physically meaningful.

3. You write that "the CMIF simultaneity method is the only correct
simultaneity method". But this begs the question of how to define
"correct". There are other ways of doing distant clock synchronization
which differ from Einstein synchronization (e.g., slow (adiabatic)
clock transport, which gives a different synchronization result
for each choice of the inertial reference frame in which the clock
transport is "slow").

[That is, suppose we are at (fixed) position A in some inertial
reference frame F0, and set a (gedanken) ideal clock M to match
our A clock. Then we transport M at velocity v << c to some
other (fixed) position B a distance d away in this same inertial
reference frame F0. This takes a time d/v. Since M's Lorentz
time Lorentz time dialation factor is quadratic in v (for v << c),
the accumulated time dialation effect on effect on M's clock
by the time M arrives at B is linear in v, and hence can be
made arbitrarily small by choosing v small enough (and waiting
long enough for M to arrive at B). Then when M (eventually)
arrives at B, we set B's clock to M's reading.

This defines the "slow clock transport" clock synchronization
scheme.

The interesting -- and slightly counterintuitive -- thing is
that if we observe this entire process from some other inertial
reference frame F1 which is moving (along the A-B direction)
with respect to our original inertial reference frame F0, and
use F1's definition of "slow motion", then it turns out that
we'll get a *different* clock synchronization.]

Can you point to a law of physics which specifically picks out
Einstein synchronization as "correct" and other synchronizations
as "incorrect"? If not, what basis do we have for saying that
one of these is "correct".

-- jt]]

Mike Fontenot

2022-04-09 02:03:52 UTC

Post by Mike Fontenot
What is really new, though, in my latest results, is the fact that the
accelerating observer can assemble an array of clocks (and attending
"helper friends" (HF's)), which give him a "NOW" that extends throughout
all space (analogous to what Einstein did for inertial observers). And
THAT guarantees that the accelerating observer's conclusions about the
home twin's age are fully MEANINGFUL to him. His conclusions agree with
the CMIF simultaneity method, which means that the CMIF simultaneity
method is the only correct simultaneity method.

I WOULD like to hear your "take" on my arguments here:

First, consider a perpetually-inertial observer (PIO). Einstein showed
us how that PIO (he) can construct an array of synchronized clocks that
are stationary wrt him, extending throughout all of space. The clocks
have been synchronized by using light signals. The fundamental (and
really only) assumption that defines special relativity is that, in ANY
inertial reference frame, the velocity of light is always equal to the
universal constant "c". We can also imagine that, co-located with each
clock is a "helper friend" (HF), whose age is always the same as the
PIO's age.

So, if the PIO wants to know "How old is that distant person (she)
"right now" (say, when the PIO is age T1), he just needs to know which
HF is momentarily co-located with her when the HF's age is T1. He can
eventually determine that, from messages sent him by all the HF's. He
has previously told all HF's to report to him all encounters with all
people, telling him what the encountered person's age was, who it was,
and what the observing HF's age was then. The PIO reviews all those
responses, and eventually will find one that tells him that, when that
HF was T1 years old, he was momentarily co-located with the particular
distant person the PIO is interested in, and her age was T2.

Now, here is the important question: Given the above, should the PIO
regard that age of the distant person (that he has eventually
determined) to be MEANINGFUL? Many people tell me the answer is NO.
But I claim that, if the PIO says that, he will effectively be saying
that he doesn't believe that the speed of light is equal to "c" in his
inertial frame. And if he doesn't believe that, he doesn't believe in
special relativity.

All of the above applies equally well to the array of clocks and
helper-friends I've described for someone (the "AO") who is initially
unaccelerated, but who then undergoes a constant non-zero acceleration
for some length of time. That AO can also be mutually stationary with
respect to an array of clocks that establish a "NOW" moment for him,
extending throughout all space. The previous arguments all apply to the
AO as well. The only difference is that, if the AO doesn't regard the
answer he has gotten for the distant person's current age to be
MEANINGFUL, that doesn't imply that he doesn't believe that the speed of
light is "c" (he already knows that the speed of light in his frame is
NOT "c"). If the AO doesn't regard the answer he has gotten for the
distant person's current age to be MEANINGFUL, that implies that he
doesn't believe that the equations he has used to calculate the current
reading on each of the HFs' clocks are correct. If he DOES believe
those equations are correct, then he MUST conclude that the distant
person's current age he has determined IS meaningful. I believe those
equations are correct. Others may believe they are not correct. I
think they ARE potentially testable.

[[Mod. note --
What do you mean by the word "meaningful"?

If the AO accelerates, he will assign a different CMIF-time to a given
(fixed) event (e.g., the explosion of the first hydrogen bomb on Earth),
and correspondingly assign the label "now @ Earth worldline" to a
different event along the Earth's worldline.

But does anything in the universe (other than AO's motion and the
observations AO makes) change when the AO accelerates? If not, then
what is the basis for declaring changes in AO-CMIF-time "meaningful"?

It might be useful to conceptualize the AO and his CMIF definition of
"now @ Earth worldline" as a "time viewer" than can observe the Earth
at any point (event) on the Earth's past worldline. Accelerating the AO
(changing the AO's velocity with respect to some inertial reference frame)
then corresponds to turning the control knob on this "time viewer" back
and forth (and hence moving the AO's "now @ Earth worldline" observation
point forwards and backwards in time along the Earth's past worldline).
Do you consider this change in observation point to be "meaningful"
(beyond its obvious change in what AO himself observes)?

This change in observation point is certainly not unique -- another
accelerating observer AO' will in general ascribe a different observation
point. And, this observation point can move superluminally both forwards
and backwards in time. And, no observation on Earth (apart from asking
AO to report what he is observing) changes when AO moves his observation
point.

To me, this all (very strongly) suggests that the motion of this
observation point (i.e., AO's CMIF-time definition of "now" at the Earth's
position) doesn't deserve to be called "physically meaningful".
-- jt]]

Julio Di Egidio

2022-04-10 18:47:48 UTC

(IMO) You are perfectly right: relativity as it is usually presented
and interpreted is simply inconsistent and arbitrary nonsense unless
one does fix the notion of *proper time* and what that even means.
Indeed yes, if I and you synchronize our clocks, and as long as the
clocks keep working, forever and ever I and you will be reading the
same exact time at at the same exact moment, aka we age the same
just like clocks tick the same (amd I think this is already some
postulate, and if not it should be). The fact that we on the other
hand move in space-time entails we are not anymore on the same plane
of simultaneity, it does not and cannot change the synchronization
of our clocks any more than it does in Galilean physics, just we
here drift in space-time instead of just space. Proper time is
just not coordinate time which has rather to do with coordinate
systems. And if we drop that postulate I am saying, physics indeed
becomes "disconnected" and plain arbitrary...

Please look at this diagram to begin with: isn't there already, in it's elementarity indeed, the unescapable answer to all above questions?

<https://jp-diegidio.github.io/STUDY.Physics.SpecialRelativity/InertialFrames/App/index.html>

Julio

[[Mod. note -- When you write

I and you synchronize our clocks, and as long as the
clocks keep working, forever and ever I and you will be reading the
same exact time at at the same exact moment

that's only true if we follow the same worldline, i.e., if our positions
are the same at all times. If our positions differ then in general we'll
see different clock readings when we get back together again -- this was
experimentally tested by (among others) the Hafele-Keating experiment
(1972)
https://paulba.no/paper/Hafele_Keating.pdf
-- jt]]

Julio Di Egidio

2022-04-12 22:55:35 UTC

Post by Julio Di Egidio

(IMO) You are perfectly right: relativity as it is usually presented
and interpreted is simply inconsistent and arbitrary nonsense unless
one does fix the notion of *proper time* and what that even means.
Indeed yes, if I and you synchronize our clocks, and as long as the
clocks keep working, forever and ever I and you will be reading the
same exact time at at the same exact moment, aka we age the same
just like clocks tick the same (amd I think this is already some
postulate, and if not it should be). The fact that we on the other
hand move in space-time entails we are not anymore on the same plane
of simultaneity, it does not and cannot change the synchronization
of our clocks any more than it does in Galilean physics, just we
here drift in space-time instead of just space. Proper time is
just not coordinate time which has rather to do with coordinate
systems. And if we drop that postulate I am saying, physics indeed
becomes "disconnected" and plain arbitrary...
Please look at this diagram to begin with: isn't there already, in it's elementarity indeed, the unescapable answer to all above questions?
<https://jp-diegidio.github.io/STUDY.Physics.SpecialRelativity/InertialFrames/App/index.html>

No comments on that? That is the most concrete and immediate alleged
proof of my point: easy to read, impossible to equivocate upon,
easy to debunk if that's the case: and should you just call that
"arbitrary" (how proper time becomes the very relation between
different frames/observers), then you should tell me what's your
picture of it and how it even works. And please notice that I clam
that that is SR, not something else: i.e. I am not proposing an
alternative theory here, just reading the existing one.

Post by Julio Di Egidio
[[Mod. note -- When you write

Post by Julio Di Egidio
I and you synchronize our clocks, and as long as the
clocks keep working, forever and ever I and you will be reading the
same exact time at at the same exact moment

that's only true if we follow the same worldline, i.e., if our positions
are the same at all times.

No, that is universally true (not even limited to inertial frames):
*locally* I and you experience the one and only universal time, aka
*proper* time (as do our clocks and everything else).

Post by Julio Di Egidio
If our positions differ then in general we'll
see different clock readings when we get back together again

"Drifting" happens in coordinate space-time: with an analogy, If I
take a trip away from you and later come back, the reading of our
(spacial) mileage would be different, yet that doesn't mean space
for me is not the same as space for you... exactly the same in SR,
just here it's the space-time mileage: to reiterate, an altogether
different notion than proper time/distance.

I risk to botch the terminology, but the problem there is essentially
mixing simultaneity (which is a bunch of operational definitions
to compute from *within* a frame) with synchronicity (isochronous
lines and so on), which is simply something else, namely *elapsed
proper time/distance*.

-- this was

Post by Julio Di Egidio
experimentally tested by (among others) the Hafele-Keating experiment
(1972)
https://paulba.no/paper/Hafele_Keating.pdf

AFAICT, I am not contradicting the theory, even less any experimental
result.

BTW, thanks for the reply, appreciated.

Julio

Mike Fontenot

2022-04-10 18:52:23 UTC

Post by Mike Fontenot
What do you mean by the word "meaningful"?

If I were ever able to take an actual long-term, high-speed space
voyage, I'm sure I would often wonder what my wife back home was doing
"right now". I definitely wouldn't believe that she had ceased to exist
just because we were separated.

If she exists "right now", then she must be doing something specific
right now. And that "doing something right now" is associated with a
certain unique state of her brain right now. And each unique state of
her brain corresponds to a unique specific time in her life. So I would
regard her current age during each instant in my life on my trip as
being completely meaningful to me.

And if my maneuvering on my trip resulted in my calculating that she was
exactly the same age at two widely separated instants in my life, I
would just say "WOW, isn't that interesting!". I would believe it. It
would be completely meaningful to me.

[[Mod. note -- So basically you're saying that what you observe is
meaningful to you, regardless of whether anything else in the universe
is affected.
-- jt]]

Mike Fontenot

2022-04-11 08:00:11 UTC

Post by Mike Fontenot
So basically you're saying that what you observe is
meaningful to you, regardless of whether anything else in the universe
is affected.

If I am an accelerating observer, and if I OBSERVE a TV image of the
distant person, that tells me what that distant person looked like a
long time ago. That's not meaningful to me, because I don't know how to
determine how much she aged while the message was in transit.

But if I'm mutually stationary wrt the array of clocks that I have
previously described, which provides a "NOW" for me extending throughout
space, that DOES give me a meaningful answer to the question of how old
she currently is. And by "meaningful", I mean that I REALLY believe
that she is currently that age. The only way I can be wrong about her
current age is if my equation for the rate ratio of the two clocks is
wrong. I'm confident that it is correct. I think it IS experimentally
testable.

I don't understand what you meant in the above, when you said
"regardless of whether anything else in the universe
is affected". Although I realize that the way I accelerate affects what
I conclude about how her age changes, I don't contend that that has ANY
effect on what other observers (including she herself) conclude about
her age changes. Different people disagree about simultaneity at a
distance. That's just the way special relativity is.

Michael Leon Fontenot

Mike Fontenot

2022-04-12 07:21:19 UTC

If I am an accelerating observer, and if I OBSERVE a TV > image of

the distant person, that tells me what that distant > person looked like
along time ago. That's not meaningful > to me, because I don't know how
to determine how much > she aged while the message was in transit.
I need to correct that last sentence. Observing the TV image of the
distant person WOULD be meaningful to me. ALL observations are
meaningful, almost by definition. But what I was thinking when I made
that statement was that, that observation wouldn't help me determine her
current age "right now", because I don't know how to determine how much
she aged while the message was in transit.

But if I'm mutually stationary wrt the array of clocks that I > have

previously described, which provides a "NOW" for >me extending
throughout space, that DOES give me a >meaningful answer to the question
of how old

she currently is. And by "meaningful", I mean that I >REALLY believe

that she is currently that age. The only >way I can be wrong about her
current age is if my equation >for the rate ratio of the two clocks is
wrong. I'm confident >that it is correct. I think it IS experimentally
testable.

The ability to construct an array of clocks (mutually stationary with
the accelerating observer) establishes a "NOW" moment for him, and
answers his question about her current age in a way that is fully
meaningful. But that hinges on my equations for the rate ratio R(t) and
the age change AC(t) being correct.

The rate ratio equation is

R(t) = [ 1 +- L A sech^2 (A t) ],

where L is the constant distance between him and the given HF, and
sech() is the hyperbolic secant (which is the reciprocal of cosh(), the
hyperbolic cosine). The "^2" after the sech indicates the square of the
sech. The "+-" in the above equation means that the second term is
ADDED to 1 for the HF's who are LEADING the accelerating observer, and
the second term is SUBTRACTED from 1 for the HF's who are TRAILING the
accelerating observer.

The current reading of the HF's clock (the "Age Change" or "AC"), when
the AO's clock reads "tau", is

AC(tau) = integral, from zero to tau, of { R(t) dt }

= tau + L tanh( A tau ).

If anyone can spot an error in my derivation of those two equations,
please let me know about it. The derivations are fairly lengthy, but
are shown completely in the paper that I put on the viXra on-line
repository:

https://vixra.org/abs/2201.0015

Michael Leon Fontenot
***@comcast.net

Mike Fontenot

2022-04-11 19:18:50 UTC

Post by Mike Fontenot
If I am an accelerating observer, and if I OBSERVE a TV image of the
distant person, that tells me what that distant person looked like a
long time ago. That's not meaningful to me, because I don't know how to
determine how much she aged while the message was in transit.

I need to correct that last sentence. Observing the TV image of the
distant person WOULD be meaningful to me. ALL observations are
meaningful, almost by definition. But what I was thinking when I made
that statement was that, that observation wouldn't help me determine her
current age "right now", because I don't know how to determine how much
she aged while the message was in transit.

Post by Mike Fontenot
But if I'm mutually stationary wrt the array of clocks that I have
previously described, which provides a "NOW" for me extending throughout
space, that DOES give me a meaningful answer to the question of how old
she currently is. And by "meaningful", I mean that I REALLY believe
that she is currently that age. The only way I can be wrong about her
current age is if my equation for the rate ratio of the two clocks is
wrong. I'm confident that it is correct. I think it IS experimentally
testable.

The ability to construct an array of clocks (mutually stationary with
the accelerating observer) establishes a "NOW" moment for him, and
answers his question about her current age in a way that is fully
meaningful. But that hinges on my equations for the rate ratio R(t) and
the ace change AC(t) being correct.

The rate ratio equation is

R(t) = [ 1 +- L A sech^2 (A t) ],

where L is the constant distance between him and the given HF, and
sech() is the hyperbolic secant (which is the reciprocal of cosh(), the
hyperbolic cosine). The "^2" after the sech indicates the square of the
sech. The "+-" in the above equation means that the second term is
ADDED to 1 for the HF's who are LEADING the accelerating observer, and
the second term is SUBTRACTED from 1 for the HF's who are TRAILING the
accelerating observer.

The limit of R(t), as "t" goes to zero, is 1 + L A. The limit of R(t),
as "t" goes to infinity, is 1.0 So R(t) starts out at some positive
number greater than 1, and then approaches 1.0 as t goes to infinity. So
eventually, all the clocks essentially tic at the same rate, but early
in the acceleration, the ratio of the tic rates varies significantly
with time.

The current reading of the HF's clock (the "Age Change" or "AC"), when
the AO's clock reads "tau", is

AC(tau) = integral, from zero to tau, of { R(t) dt }

= tau + L tanh( A tau ).

Tom Roberts

2022-04-11 19:19:47 UTC

Post by Mike Fontenot
if I'm mutually stationary wrt the array of clocks that I have
previously described, which provides a "NOW" for me extending
throughout space, [...]

If you are an astronaut in a spaceship that can travel at an appreciable
fraction of c, and which you maneuver, then such an array of clocks is
impossible -- each such clock must vary its acceleration in concert with
yours, so the spaceships carrying those clocks must be clairvoyant,
because they are separated from you by spacelike intervals. Moreover,
such clocks located sufficiently far away from you will require an
unphysical acceleration and/or velocity to remain on station; you
cannot cover the universe with them, only a region "close" to you (how
close depends on the details of your maneuvering).

[I suppose you could plan out your trip in exquisite
detail, and the spaceships carrying the other clocks
could pre-compute their accelerations to match. But
anyone who has ever driven a car knows how poorly
such plans are followed.]

If you carefully keep track of your acceleration, velocity, and position
relative to some coordinates, then you can calculate what "now" means to
you at a distant location, as you maneuver. As I have pointed out many
times before, that in general gives nonsensical and unphysical results.

If you "believe" that your wife back home can grow younger due to your
maneuvering, then you will believe anything, which is useless, and is
certainly not physics.

Tom Roberts

Mike Fontenot

2022-04-12 07:21:19 UTC

Post by Tom Roberts
If you are an astronaut in a spaceship that can travel at an appreciable
fraction of c, and which you maneuver, then such an array of clocks is
impossible -- each such clock must vary its acceleration in concert with
yours [...]

No, that's not correct. According to the accelerating observer (the
AO), whose conclusions we seek, he and each of his "helper friends"
(HF's) undergo EXACTLY the same (constant) acceleration, as recorded on
their accelerometers. This is clear by looking at the equivalent
scenario in the case of a constant gravitational field with no
accelerations (via the equivalence principle) ... all of those people
are motionless, unaccelerated, and mutually stationary. And according
to them, the distance between each of them is also constant. In the
acceleration scenario (in the infinite flat spacetime of special
relativity), perpetually-inertial observers, who are initially
stationary with respect to the AO and the HF's, WILL conclude that the
accelerations of the AO and the various HF's, and their distances apart,
DO vary with time. But it is not their conclusions that I am interested in.

Tom Roberts

2022-04-12 18:28:54 UTC

Post by Tom Roberts
If you are an astronaut in a spaceship that can travel at an
appreciable fraction of c, and which you maneuver, then such an
array of clocks is impossible -- each such clock must vary its
acceleration in concert with yours [...]

No, that's not correct.

Yes, it is correct. Finally you say what you are thinking behind the
scenes, and I can now explain that you are thinking of an inappropriate
analogy.

Post by Mike Fontenot
According to the accelerating observer (the AO), whose conclusions
we seek, he and each of his "helper friends" (HF's) undergo EXACTLY
the same (constant) acceleration, as recorded on their
accelerometers.

This is wrong. You require the helper friends to remain at the same
proper distance from the accelerating observer -- i.e. in each
successive instantaneously co-moving inertial frame (ICIF) those
distances are constant. That is Born rigid motion, and it is well
known that HFs lower than the AO must have larger proper accelerations
than the AO, and HFs above the AO must have smaller proper
accelerations; there is often a limit below which no HF can possibly
keep up, so in general the HFs cannot cover the manifold.

[Note that accelerometers display their proper acceleration.]

Moreover, whenever the observer maneuvers (changes his proper
acceleration), all of the HFs must SIMULTANEOUSLY (in the current ICIF)
make corresponding changes to their proper accelerations -- that
requires either clairvoyance or detailed pre-planning, because they are
separated by spacelike intervals from the AO. As such detailed
pre-planning cannot hold for a spacefaring astronaut (who will maneuver
his spacecraft based on current observations), I said that this array
of HFs is impossible.

Post by Mike Fontenot
This is clear by looking at the equivalent scenario in the case of a
constant gravitational field with no accelerations (via the
equivalence principle) ... all of those people are motionless,
unaccelerated, and mutually stationary. And according to them, the
distance between each of them is also constant.

(I presume you mean a static gravitational field, all
those people are located at various altitudes with each
4-velocity parallel to the timelike Killing vector, and
the field is uniform in that their proper accelerations
are all equal. I also presume that "all of those people"
are the AO and all the HFs. Note this is not "equivalent"
to the above scenario, as I discuss next.)

This is wrong at several levels.
1) The HFs are not "motionless" or "mutually stationary" in any of the
AO's ICIFs; this is not Born rigid motion.
2) the HFs are not "unaccelerated" -- each has a nonzero proper
acceleration (even though they have zero coordinate acceleration [#]).
3) the equivalence principle applies only in a region small enough that
the curvature of spacetime is negligibly small (compared to measurement
accuracy); it CLEARLY does not apply here.
4) the distances between the HFs and the AO are not "constant" in any
ICIF -- you are confusing constant coordinate difference [#] with
constant distance in an ICIF -- the AO and HFs have constant coordinate
values (and differences) [#], but not constant distances in any of the
ICIFs (as required by your scenario above).

[#] Implicitly using coordinates aligned with the static
gravitational field. These are not the coordinates of any
of the ICIFs.

It seems you have implicitly been thinking of this inappropriate analogy
all along. Relativity is more complicated than that (even in flat
spacetime).

Post by Mike Fontenot
In the acceleration scenario (in the infinite flat spacetime of
special relativity), perpetually-inertial observers, who are
initially stationary with respect to the AO and the HF's, WILL
conclude that the accelerations of the AO and the various HF's, and
their distances apart, DO vary with time.

But the only way to construct that collection of HFs is for the AO to
adhere to a detailed, prearranged plan of accelerations, so each HF can
pre-compute their own detailed plan of accelerations. The AO cannot
maneuver to, say, land on a discovered planet or orbit a discovered star.

Post by Mike Fontenot
But it is not their conclusions that I am interested in.

But as I keep saying, the conclusions you are interested in are
nonsensical and unphysical. And the AO + HFs scenario you have in mind
is either impossible, or requires unacceptably rigid adherence to a
pre-computed plan (and even then can be limited in scope).

Tom Roberts

Mike Fontenot

2022-04-14 13:32:50 UTC

Tom, you've misunderstood what I'm doing.

Start with the gravitational scenario, with no acceleration and no
motion at all. Imagine that there is a high-rise building, with many
floors. A clock and the "AO" (whose "viewpoint" we are seeking) is
located on the first floor. A clock and an attending HF is on each of
the higher floors. The distance between the AO and each of the HF's is
constant. They are all motionless and unaccelerated.

There is initially no gravitational field. And initially all of the
clocks are synchronized and ticking at the same rate. So initially, the
rate ratio R, for each HF's clock, is just equal to 1.0.

But at some instant (say, t = 0), there suddenly appears a constant and
uniform gravitational field, of strength "g", directed downwards, and
acting over the entire length of the building. Each person suddenly
feels exactly the same force per unit mass, trying to pull them
downwards against the floor. (But they don't move, because they were
already tethered in that position). They could be constantly standing
on a bathroom scale, displaying their weight.

The gravitational time dilation equation says that, according to the AO,
each HF's clock suddenly starts ticking faster than the AO's clock, by
the rate ratio

R(t) = [ 1 + L g sech^2 (g t) ],

where "L" is the distance between the AO and that particular HF. And,
according to the AO, the change in age (AC) of each HF (relative to his
age when the field suddenly appeared), is

AC(tau) = integral, from zero to tau, of { R(t) dt }

= tau + L tanh( g tau ).

So that's the outcome of the gravitational scenario.

We now use the equivalence principle (EP) to convert the above
gravitational scenario to the EQUIVALENT scenario with the constant and
uniform GRAVITATIONAL FIELD acting on the AO and each HF replaced by a
constant and uniform ACCELERATION acting on the AO and each HF.
Everything else stays exactly the same, except the gravitational field
is replaced by an acceleration.

Just as in the gravitational scenario, each person suddenly feels
exactly the same force per unit mass, trying to pull them against the
floor. (But they don't move, because they were already tethered in that
position). They could be constantly standing on a bathroom scale,
displaying their weight. That serves as an accelerometer.

The equivalence principle says that EVERYTHING stays the same as in the
gravitational scenario, except that the parameter "g" just gets replaced
by the parameter "A" in the equations, with each having the same
numerical value. So we still have the same equation for R and for AC as
we had above, with "g" replaced by "A", with equal numerical values. The
fact that "L" and "g" don't vary with time in the gravitational scenario
means that "L" and "A" don't vary with time in the acceleration scenario
either.

Mike Fontenot

Mike Fontenot

2022-04-15 04:25:30 UTC

Tom, you've badly misunderstood what I'm doing.

Start with the gravitational scenario, with no acceleration and no
motion at all. Imagine that there is a high-rise building, with many
floors. A clock and the "AO" (whose "viewpoint" we are seeking) is
located on the first floor. A clock and an attending HF is on each of
the higher floors. The distance between the AO and each of the HF's is
constant. They are all motionless and unaccelerated.

There is initially no gravitational field. And initially all of the
clocks are synchronized and ticking at the same rate. So initially, the
rate ratio R, for each HF's clock, is just equal to 1.0.

But at some instant (say, t = 0), there suddenly appears a constant and
uniform gravitational field, of strength "g", directed downwards, and
acting over the entire length of the building. Each person suddenly
feels exactly the same force per unit mass, trying to pull them
downwards against the floor. (But they don't move, because they were
already tethered in that position). They could be constantly standing
on a bathroom scale, displaying their weight.

The gravitational time dilation equation says that, according to the AO,
each HF's clock suddenly starts ticking faster than the AO's clock, by
the rate ratio

R(t) = [ 1 + L g sech^2 (g t) ],

where "L" is the distance between the AO and that particular HF. And,
according to the AO, the change in age (AC) of each HF (relative to his
age when the field suddenly appeared), is

AC(tau) = integral, from zero to tau, of { R(t) dt }

= tau + L tanh( g tau ).

So that's the outcome of the gravitational scenario.

We now use the equivalence principle (EP) to convert the above
gravitational scenario to the EQUIVALENT scenario with the constant and
uniform GRAVITATIONAL FIELD acting on the AO and each HF replaced by a
constant and uniform ACCELERATION acting on the AO and each HF.
Everything else stays exactly the same, except the gravitational field
is replaced by an acceleration.

Just as in the gravitational scenario, each person suddenly feels
exactly the same force per unit mass, trying to pull them against the
floor. (But they don't move, because they were already tethered in that
position). They could be constantly standing on a bathroom scale,
displaying their weight. That serves as an accelerometer.

The equivalence principle says that EVERYTHING stays the same as in the
gravitational scenario, except that the parameter "g" just gets replaced
by the parameter "A" in the equations, with each having the same
numerical value. So we still have the same equation for R and for AC as
we had above, with "g" replaced by "A", with equal numerical values.
The fact that "L" and "g" don't vary with time in the gravitational
scenario means that "L" and "A" don't vary with time in the acceleration
scenario either.

Richard Livingston

2022-04-15 18:11:54 UTC

Post by Mike Fontenot
...
But at some instant (say, t = 0), there suddenly appears a constant and
uniform gravitational field, of strength "g", directed downwards, and
acting over the entire length of the building. Each person suddenly
feels exactly the same force per unit mass, trying to pull them
downwards against the floor. (But they don't move, because they were
already tethered in that position). They could be constantly standing
on a bathroom scale, displaying their weight.

If you do this you will find that the distance to each helper observer no
longer remains constant. If you force the distances to remain constant
then the helper observers cannot have the same acceleration. Please
study the "Spaceship Paradox" and Rindler coordinates. You are
making an assumption that is not correct.

Rich L.

[[Mod. note -- Bell's "spaceship paradox" is indeed highly relevant
here. The Wikipedia article
https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
is a nice introduction.

In this context, when trying to understand the meaning and implications
of the phrase "there suddenly appears", it's important to ask "suddenly
in which inertial reference frame?", and to think about whether things
would still happen "suddenly" in other inertial reference frames.
-- jt]]

Mike Fontenot

2022-04-15 04:25:45 UTC

Mike Fontenot

2022-04-15 23:06:21 UTC

You (Richard Livingston and moderator JT) are both confusing the
conclusions of the people who are undergoing acceleration with the
conclusions of people who are perpetually-inertial and who are observing
the people who are accelerating. The perpetually-inertial people DO
conclude that the distance between the accelerating people decreases
when they accelerate, but the people who are accelerating don't agree.

Mike Fontenot

Tom Roberts

2022-04-16 06:57:11 UTC

Post by Mike Fontenot
You (Richard Livingston and moderator JT) are both confusing the
conclusions of the people who are undergoing acceleration with the
conclusions of people who are perpetually-inertial and who are
observing the people who are accelerating. The perpetually-inertial
people DO conclude that the distance between the accelerating people
decreases when they accelerate, but the people who are accelerating
don't agree.

This is wrong in two quite different ways:
1. You presume that the principle of equivalence (POE) applies,
and base your conclusion on it: you are discussing Case 2
(below), and incorrectly believe the POE permits you to apply
the constancy of Case 1 (also below); it doesn't.
2. You are claiming a coordinate dependence for measurements that
are invariant [@] -- all observers ("people") will agree on
the results of a measurement (or series of measurements),
regardless of whether the observers are inertial or accelerated.

Bottom line: the POE applies ONLY in regions of spacetime that are small
enough that any curvature is negligible (compared to measurement
accuracies). For the case you have in mind, with helper friends
("people") near a distant friend while the accelerated observer ("AO")
roams the universe, those people span an enormous spatial region, over a
very long time. (Parentheticals relate your earlier and later notations.)

Case 1: all people are at rest in a uniform and static gravitational
field [#]. They will indeed measure their pairwise distances to all be
constant over time [@]. As the field is uniform, they all have identical
proper accelerations.

[#] I.e. their 4-velocities are parallel to the timelike
Killing vector.

[@] Measuring pairwise distances in the AO's instantaneously
co-moving inertial frame (ICIF) at a given time.

Case 2: all people are in in flat spacetime, initially at rest in some
inertial frame, each with an identical proper acceleration. They will
NOT measure their pairwise distances to be constant [@], because this is
not Born rigid motion (for which lower people must have larger proper
accelerations than higher people).

[In Case 2, if one measured those distances in the initial
inertial frame, they would be constant. But the changing
inertial frames of successive measurements, and the
structure of Lorentz transforms between frames, makes them
be not constant in successive ICIFs of the AO. See Bell's
spaceship paradox.]

This has nothing to do with different measurement methods, or the use of
inertial or non-inertial coordinates, or any supposed coordinate
dependence of invariant measurements; it is due the different ways the
two collections of people evolve in the two cases.

Tom Roberts

Mike Fontenot

2022-04-17 08:03:00 UTC

Post by Tom Roberts
Bottom line: the POE applies ONLY in regions of spacetime that are small
enough that any curvature is negligible (compared to measurement
accuracies). For the case you have in mind, with helper friends
("people") near a distant friend while the accelerated observer ("AO")
roams the universe, those people span an enormous spatial region, over a
very long time.

The theory of special relativity places no limits AT ALL on the extent
of spacetime. THAT is the domain of my scenario with acceleration.

Tom Roberts

2022-04-18 15:46:23 UTC

Post by Tom Roberts
Bottom line: the POE applies ONLY in regions of spacetime that are
small enough that any curvature is negligible (compared to
measurement accuracies). For the case you have in mind, with
helper friends ("people") near a distant friend while the
accelerated observer ("AO") roams the universe, those people span
an enormous spatial region, over a very long time.

The theory of special relativity places no limits AT ALL on the
extent of spacetime. THAT is the domain of my scenario with
acceleration.

Sure. But you attempt to argue by "equivalence" to a situation in which
a uniform gravitational field is present -- that is NOT Special
Relativity, and that "equivalence" is NOT valid, as I explained earlier.

Specifically: in flat spacetime for helper friends with equal proper
accelerations, you claim that the pairwise distances [#] between helper
friends are constant, because they are constant in the uniform-gravity
spacetime you think is "equivalent". Your argument is invalid because
the physical situations aren't actually equivalent -- the region
involved is too large for the Principle of Equivalence to apply; for
your case in flat spacetime those pairwise distances [#] are not
constant, as one can explicitly calculate.

[#] Pairwise distances are measured in the instantaneously
co-moving inertial frame of the accelerating observer (AO).

[This is getting tedious and overly repetitive. Unless you come up with
something new, I won't continue.]

Tom Roberts

Tom Roberts

2022-04-16 06:57:20 UTC

Post by Mike Fontenot
Tom, you've misunderstood what I'm doing.

No, I don't think I have. But you have misunderstood basic relativity,
and have misunderstood when the equivalence principle applies, and when
it doesn't. See my recent post about this.

Post by Mike Fontenot
[... completely new scenario involving clocks at different floors of
a high-rise building]

1. In the gravitational scenario, if you have good enough
measurement accuracy to distinguish the elapsed proper times
of the clocks, then you cannot apply the equivalence principle
(EP), because the curvature of spacetime is not negligible.
2. Because of #1, your two scenarios do not correspond as you
claim -- they are NOT "equivalent" because the EP does not
apply.
3. GR does NOT say "each HF's clock suddenly starts ticking faster
than the AO's clock", because clocks always tick at their usual
(intrinsic) rate [#]. IOW: a clock's proper tick rate is
independent of its instantaneously co-moving inertial frame. --
this is a direct consequence of Einstein's first postulate
of SR, as it applies in GR.
4. But you are not actually comparing clock tick rates, you are
comparing their elapsed proper times. In GR a clock's elapsed
proper time is computed by integrating the metric over its path
through spacetime. The equation you use is the difference between
two such integrals, applied to your specific physical situation.
That difference is not due to different clock tick rates, but
rather is due to the difference in the metric at their locations
-- examine the derivation and you'll see it assumes equal proper
(intrinsic) tick rates but different values of the metric.

[#] But signals from a distant clock can tick at a
different rate from that of a local, identical clock.
This is due to the way such signals are measured, which
is the basis of all types of redshift measurements.

Bottom line: as before in your earlier scenarios, you have misunderstood
basic relativity, and have misapplied the equivalence principle.

Tom Roberts

Mike Fontenot

2022-04-18 21:17:57 UTC

Post by Tom Roberts
Your argument is invalid because
the physical situations aren't actually equivalent -- the region
involved is too large for the Principle of Equivalence to apply

The equivalence principle has no restrictions on the size of the region
in which it is valid.

[[Mod. note -- The EP applies if and only if we neglect tidal effects,
i.e., if tidal effects are "small". But the size of tidal effects grows
with the distance, so saying that tidal effects should be "small" (so
that we can apply the EP) is essentially saying that the size of our
region should be "small". So, effectively, the EP only applies in "small"
regions, where the precise definition of "small" depends on the curvature
of spacetime and your accuracy threshold for how small tidal effects need
to be before it's ok to neglect them.

For example, suppose we're considering some experiment near the Earth's
surface, where we're measuring accelerations on the order of 1g. We might
reasonably say that for this experiment, we're willing to apply the EP
(i.e., neglect tidal effects) if the tidal accelerations are below 1e-6 g
in magnitude, i.e., if the tidal accelerations are < 1 part-per-million
of the Earth's Newtonian "little g". To satisfy that condition requires
that our experimental region be (roughly) < 3 meters in diameter. That
is, the tidal change in the Earth's Newtonian "little g" over a vertical
distance of 3 meters is about 1 part-per-million. (In the horizontal
direction the change is probably a bit smaller.) If our experimental
region is larger than 3 meters in diameter, the tidal change in "little g"
from one part of our apparatus to another part may be > 1 part-per-million,
so we can't safely apply the EP across at our 1 part-per-million accuracy
threshold.
-- jt]]

Mike Fontenot

2022-04-19 07:19:49 UTC

Post by Tom Roberts
Your argument is invalid because
the physical situations aren't actually equivalent -- the region
involved is too large for the Principle of Equivalence to apply

Mike Fontenot replied (on 4/18/22 3:17 PM):

The equivalence principle has no restrictions on the size of the region
in which it is valid.

Post by Mike Fontenot
[[Mod. note -- The EP applies if and only if we neglect tidal effects,
i.e., if tidal effects are "small". But the size of tidal effects grows
with the distance, so saying that tidal effects should be "small" (so
that we can apply the EP) is essentially saying that the size of our
region should be "small". So, effectively, the EP only applies in "small"
regions, where the precise definition of "small" depends on the curvature
of spacetime and your accuracy threshold for how small tidal effects need
to be before it's ok to neglect them.

There are NO "tidal effects" in my gravitational scenario (nor in my
equivalent special relativity scenario, of course).

Tidal effects arise when the gravitational field lines are NOT parallel,
which occurs when the source of the gravitational field is a spherical
distribution of mass, which acts as if all of the mass is concentrated
at a point at the center of the sphere, giving an inverse-square
variation of "g" with height).

In my gravitational scenario, the gravitational lines are all EXACTLY
PARALLEL, and perpendicular to the FLAT surface (of infinite extent) of
the mass that the high-rise building is sitting on. That's why the
field strength "g" in my scenario is (by design) independent of height
... it does NOT vary as the inverse-square of height.

Tom Roberts

2022-04-20 17:57:24 UTC

Post by Mike Fontenot
The equivalence principle has no restrictions on the size of the
region in which it is valid.

Yes, it does, as has been explained several times in this thread.

Post by Mike Fontenot
There are NO "tidal effects" in my gravitational scenario (nor in my
equivalent special relativity scenario, of course).

But a calculation shows that in SR, helper friends (HFs) with proper
accelerations equal to that of the accelerated observer (AO), do NOT
remain at constant proper distance from the AO, contrary to your claim.
Look up Bell's spaceship paradox for an example calculation:
https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

Such constant proper distances requires the HFs and AO all execute Born
rigid motion [#]. As is well known, that requires that "lower" HFs have
larger proper accelerations, and "higher" HFs have smaller proper
accelerations than the AO.

[#] Born rigid motion is defined as all parts of the object
having pairwise constant proper distances. Here the "object"
is the collection of all HFs and the AO.

Bottom line: in the not-really-equivalent scenario you gave, with a
constant and uniform gravitational field, the HFs and AO all have equal
proper accelerations, and their pairwise proper distances do remain
constant. The Principle of Equivalence does NOT relate your two
scenarios, and your argument is invalid.

Tom Roberts

Mike Fontenot

2022-04-21 06:21:50 UTC

Post by Tom Roberts
But a calculation shows that in SR, helper friends (HFs) with proper
accelerations equal to that of the accelerated observer (AO), do NOT
remain at constant proper distance from the AO, contrary to your claim.
https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

I looked at that wiki page, and here's what they say:

"A delicate thread hangs between two spaceships. They start accelerating
simultaneously and equally as measured in the inertial frame S, thus
having the same velocity at all times as viewed from S."

So in the above scenario, according to the inertial frame "S", the two
spaceships accelerate equally (and thus remain separated by a fixed
distance). But in that scenario, the people in the spaceships would NOT
agree that they were accelerating at the same rate, and they would NOT
agree that their separation was constant. So that is a DIFFERENT
scenario than the one I have described.

In my scenario, the people who are accelerating conclude that their
accelerations are exactly equal, and that their separations are exactly
equal. The inertial observers there disagree.

Tom Roberts

2022-04-22 20:45:25 UTC

Post by Tom Roberts
But a calculation shows that in SR, helper friends (HFs) with
proper accelerations equal to that of the accelerated observer
(AO), do NOT remain at constant proper distance from the AO,
contrary to your claim. Look up Bell's spaceship paradox for an
https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

I looked at that wiki page, [...] So that is a DIFFERENT scenario
than the one I have described.

Yes it is, but it is an example of how to perform the calculation. You
MUST perform the calculation, as you keep making incorrect claims.

In my scenario, the people who are accelerating conclude that their
accelerations are exactly equal, and that their separations are
exactly equal.

[You said earlier that their proper accelerations
are exactly equal, and instead of "separations"
you said "proper distances". Wishy-washy words
like you use here are insufficient -- precision
in thought and word is essential.]

Your claim here is wrong. With equal proper accelerations, this
collection of people is not executing Born rigid motion, and their
(pairwise) proper distances change over time. Or the other way: with
constant (pairwise) proper distances they do execute Born rigid motion,
so their proper accelerations cannot be equal.

You clearly have an incorrect mental image of this, which you doggedly
cling to. The only way to disabuse yourself of that error is to actually
perform the calculation. Simplify it: consider just the AO accelerating
in the +x direction, and a single HF at a large negative value of x with
the same proper acceleration; keep their proper accelerations constant
and compute the proper distance between them (i.e. their distance apart
measured in successive ICIFs of the AO). Be sure to use proper
accelerations, and not accelerations relative to some inertial frame.

[Alternative: look up "Born rigid motion" and understand it
-- see what it is and how it differs from your scenario.]

Tom Roberts

Mike Fontenot

2022-04-25 07:41:35 UTC

"A delicate thread hangs between two spaceships. They start accelerating
simultaneously and equally as measured in the inertial frame S, thus
having the same velocity at all times as viewed from S."
So in the above scenario, according to the inertial frame "S", the two
spaceships accelerate equally (and thus remain separated by a fixed
distance). But in that scenario, the people in the spaceships would NOT
agree that they were accelerating at the same rate, and they would NOT
agree that their separation was constant. So that is a DIFFERENT
scenario than the one I have described.
In my scenario, the people who are accelerating conclude that their
accelerations are exactly equal, and that their separations are exactly
equal. The inertial observers there disagree.

In the above paragraph, instead of saying

"and that their separations are exactly equal",

I SHOULD have said

"and that their separations are constant."

In Tom Roberts' scenario (which is the "Bell Paradox" scenario), the
perpetually-inertial observers correctly conclude that, according to
their inertial frame, the accelerating people are all accelerating at
the same constant rate (and therefore that their separations are all
equal and constant). The accelerating people do NOT agree with them.

The perpetually-inertial observers are certainly entitled to set up that
scenario, but that is a scenario that I have NO interest in at all.

The scenario that I am interested in, is the scenario where it is the
accelerated people who correctly conclude that their accelerations are
all equal, and that their separations are all constant. In that
scenario, perpetually-inertial observers do NOT agree with them.

The REASON that I am solely interested in the immediately above scenario
is because what I want to know IS the conclusion of the accelerating
people, about how their ageing rates compare. THAT is what allows me to
define a meaningful "NOW" instant for one of them. And that is what
allows me to show that the CMIF simultaneity method is the ONLY correct
simultaneity method for an accelerating observer.

And once I know the relative ageing rates of the accelerating people in
the above acceleration scenario, the equivalence principle immediately
tells me the ageing rates of the stationary people in the equivalent
gravitational scenario. Everything remains the same, except that the
equal and constant accelerations of the people (directed along the
straight line connecting them) are replaced by an equal and constant
gravitational field acting on the stationary people (where the
gravitational field is directed along the straight line connecting the
people).

Michael Leon Fontenot

Tom Roberts

2022-05-02 08:36:20 UTC

Post by Tom Roberts
But a calculation shows that in SR, helper friends (HFs) with
proper accelerations equal to that of the accelerated observer
(AO), do NOT remain at constant proper distance from the AO,
contrary to your claim.

In Tom Roberts' scenario (which is the "Bell Paradox" scenario), the
perpetually-inertial observers correctly conclude that, according to
their inertial frame, the accelerating people are all accelerating at
the same constant rate (and therefore that their separations are all
equal and constant).

You need to read more carefully. In the paragraph you quoted above, I
specifically said "proper distance" -- and earlier in that post I
clarified that this means the distance measured in the AO's
instantaneously co-moving inertial frame (ICIF). That is NOT "according
to their inertial frame", it is according to the AO's ICIF (which
obviously changes over time as the AO is accelerating).

Post by Mike Fontenot
The accelerating people do NOT agree with them.

You claim the accelerating people measure proper distance, so they MUST
agree with ANYBODY who also measures (or calculates) proper distance.
After all, the (pairwise) proper distances are each unique and invariant
(at any given point along the AO's worldline).

Post by Mike Fontenot
The perpetually-inertial observers are certainly entitled to set up
that scenario, but that is a scenario that I have NO interest in at
all.

And it is NOT what I was discussing. You misread what I wrote.

Post by Mike Fontenot
The scenario that I am interested in, is the scenario where it is
the accelerated people who correctly conclude that their
accelerations are all equal, and that their separations are all
constant.

You may be "interested" in such a scenario, but it is inconsistent with
relativity -- equal (proper) accelerations do NOT yield Born rigid
motion (in which all pairwise proper distances remain constant).

There are two choices:
A) all have equal proper accelerations, in which case their
pairwise proper distances are changing.
or:
B) All pairwise proper distances are constant, in which case
"lower" people have larger proper accelerations than
"higher" people (here "lower" people are behind "higher"
people along the direction of acceleration).

This is directly related to the Bell spaceship paradox. The fact that
you don't realize that is a major part of your misunderstanding.

Post by Mike Fontenot
In that scenario, perpetually-inertial observers do NOT agree with
them.

When the perpetually-inertial people make measurements and compute the
proper distances, they MUST agree with the accelerated people (who do
the same). Because such (pairwise) proper distances are each unique and
invariant (at any given point along the AO's worldline).

Post by Mike Fontenot
[... further discussion based on the above error]

I give up. You keep repeating the same errors without reading what I
write, so there's no point in continuing. Goodbye. (I'm surprised the
moderators have let it go this long.)

Tom Roberts

Mike Fontenot

2022-05-04 11:11:51 UTC

[...]

In a SINGLE SCENARIO:

(1): YOU contend that the accelerating person (the AO) concludes that
the separation between himself and his helper friend ("HF") is CONSTANT.

(2): YOU contend that the perpetually-inertial observer ALSO concludes
that the separation between the AO and the HF is CONSTANT.

The special theory of relativity does NOT allow item (1) and item (2) to
BOTH be true, in a SINGLE SCENARIO. In special relativity, a
perpetually-inertial observer can NEVER agree with an accelerating
observer for longer than a single instant.

Michael Leon Fontenot

Mike Fontenot

2022-05-08 07:41:36 UTC

(1): YOU (Tom Roberts) contend that the accelerating person (the AO) concludes that
the separation between himself and his helper friend ("HF") is CONSTANT.
(2): YOU (Tom Roberts) contend that the perpetually-inertial observer ALSO concludes
that the separation between the AO and the HF is CONSTANT.
The special theory of relativity does NOT allow item (1) and item (2) to
BOTH be true, in a SINGLE SCENARIO. In special relativity, a
perpetually-inertial observer can NEVER agree with an accelerating
observer for longer than a single instant.
Michael Leon Fontenot

The simplest way, to see that my statements above are true, is to use
the well-known (and sacrosanct) length-contraction equation (LCE) of
special relativity:

According to a perpetually-inertial observer (PIO), a meter-stick
stationary in any other frame that is moving wrt the PIO will be
SHORTENED by the FACTOR

gamma = sqrt( 1 / [ 1 - v * v ],

where "v" is the current velocity of the other frame wrt the PIO's
frame, and the asterisk denotes multiplication.

At any instant in his life, the accelerating observer (AO) says that his
yardstick is three feet long. But the PIO says the yardstick's length
is 3 feet DIVIDED by gamma. Since gamma is greater than 1.0 when v is
nonzero, the PIO says that the moving yardstick is shorter than three
feet long.

Michael Leon Fontenot

Mike Fontenot

2022-07-12 02:40:42 UTC