The physics - Genuine kinematics and dynamics of spacetime
Albertito
and integrate acceleration, a, with respect to time t
to give the change in velocity, v,
v = v_0 + a*t
Now, integrate v wrt time, too, to give the change
in space, r,
r = r_0 + v_0*t + a*t^2/2
So far so good, since a, v_0 and r_0 are all constants.
And the physical meaning of each magnitude is well-known.
But, let's now integrate r wrt time t, to give the change
in ...... spacetime, s,
s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
that magnitude s is called spacetime, because if you
divide it by space you attain a time, and if you divide
it by a time you attain a space.
"space interval is the rate of change of spacetime
with respect to time"
"time interval is the rate of change of spacetime
with respect to space"
Let's start with a dynamics of constant gravity, g,
and integrate with respect to displacement r to give
the change in potential, phi,
phi = phi_0 + g*r
Continue integrating to yield psi, and omega
psi = psi_0 + phi_0*r + g*r^2/2,
omega = omega_0 + psi_0*r + phi_0*r^2/2 + g*r^3/6
At this point, we see that omega must also be a spacetime
as define above in the kinematics, because the third derivative
of a spacetime wrt to space is a gravitational field, g, and
the third derivative of spacetime wrt time is a 'kinematical'
acceleration.
Corollary: In the above dynamics, psi, plays the role of the
source of gravitation, and omega is spacetime, therefore the
'source' of the source of gravitation is spacetime itself.
Regards
PD
> Let's start with a kinematics of constant acceleration,
> and integrate acceleration, a, with respect to time t
> to give the change in velocity, v,
>
> v = v_0 + a*t
>
> Now, integrate v wrt time, too, to give the change
> in space, r,
>
> r = r_0 + v_0*t + a*t^2/2
>
> So far so good, since a, v_0 and r_0 are all constants.
> And the physical meaning of each magnitude is well-known.
> But, let's now integrate r wrt time t, to give the change
> in ...... spacetime, s,
>
> s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>
> that magnitude s is called spacetime, because if you
> divide it by space you attain a time, and if you divide
> it by a time you attain a space.
BWAHAHAHAHAHAHA!
"Hey, kids, I have a game! Let's take a word that already exists,
invent a whole new definition for it. Then we'll use the word in a
conversation and when people look confused, we'll pretend we don't
know what they're talking about! Fun!"
Albertito
> On Oct 22, 4:52 am, Albertito <albertito1...@gmail.com> wrote:
>
>
>
> > Let's start with a kinematics of constant acceleration,
> > and integrate acceleration, a, with respect to time t
> > to give the change in velocity, v,
>
> > v = v_0 + a*t
>
> > Now, integrate v wrt time, too, to give the change
> > in space, r,
>
> > r = r_0 + v_0*t + a*t^2/2
>
> > So far so good, since a, v_0 and r_0 are all constants.
> > And the physical meaning of each magnitude is well-known.
> > But, let's now integrate r wrt time t, to give the change
> > in ...... spacetime, s,
>
> > s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>
> > that magnitude s is called spacetime, because if you
> > divide it by space you attain a time, and if you divide
> > it by a time you attain a space.
>
> BWAHAHAHAHAHAHA!
Why that idiotic and schizophrenic laugh?
Is it because the notion of spacetime I've
defined has more physical meaning than the
nonsensical Minkowskian one used by Einstein
et al.?
PD
> On Oct 22, 3:43 pm, PD <thedraperfam...@gmail.com> wrote:
>
>
>
> > On Oct 22, 4:52 am, Albertito <albertito1...@gmail.com> wrote:
>
> > > Let's start with a kinematics of constant acceleration,
> > > and integrate acceleration, a, with respect to time t
> > > to give the change in velocity, v,
>
> > > v = v_0 + a*t
>
> > > Now, integrate v wrt time, too, to give the change
> > > in space, r,
>
> > > r = r_0 + v_0*t + a*t^2/2
>
> > > So far so good, since a, v_0 and r_0 are all constants.
> > > And the physical meaning of each magnitude is well-known.
> > > But, let's now integrate r wrt time t, to give the change
> > > in ...... spacetime, s,
>
> > > s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>
> > > that magnitude s is called spacetime, because if you
> > > divide it by space you attain a time, and if you divide
> > > it by a time you attain a space.
>
> > BWAHAHAHAHAHAHA!
>
> Why that idiotic and schizophrenic laugh?
> Is it because the notion of spacetime I've
> defined has more physical meaning than the
> nonsensical Minkowskian one used by Einstein
> et al.?
Well, here's why:
1) It's funny that you think that redefining words to be your
playground is of creative scientific value.
2) It's funny that you think the Minkowski notion of spacetime is
nonsensical, when it seems to make perfect sense to me.
eric gisse
> Let's start with a kinematics of constant acceleration,
> and integrate acceleration, a, with respect to time t
> to give the change in velocity, v,
>
> v = v_0 + a*t
>
> Now, integrate v wrt time, too, to give the change
> in space, r,
>
> r = r_0 + v_0*t + a*t^2/2
>
> So far so good, since a, v_0 and r_0 are all constants.
> And the physical meaning of each magnitude is well-known.
> But, let's now integrate r wrt time t, to give the change
> in ...... spacetime, s,
>
> s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
The pattern continues. This month you seem to take standard solutions and
the integrate them a few more times, 'just to see'. When you realize that
this too is a dead end path, you'll silently drop the idea everyone told you
was stupid.
Albertito
Then, why can I prove the speed of light is
invariant in any frame of reference that uses
my notion of spacetime, and if you apply a
Galilean transform to frames in that spacetime
you attain an invariant interval of that very
spacetime?
Hint: The speed of light in this genuine spacetime
behaves as a constant acceleration. So, what we
perceive as a constant speed of light is actually
a constant accelerated motion (aka propagation) in
this genuine spacetime.
BTW, your neurons are as nonsensical as the
Minkowskian ones, or perhaps you are as gullible
as a dingleberry?
Albertito
> Albertito wrote:
> > Let's start with a kinematics of constant acceleration,
> > and integrate acceleration, a, with respect to time t
> > to give the change in velocity, v,
>
> > v = v_0 + a*t
>
> > Now, integrate v wrt time, too, to give the change
> > in space, r,
>
> > r = r_0 + v_0*t + a*t^2/2
>
> > So far so good, since a, v_0 and r_0 are all constants.
> > And the physical meaning of each magnitude is well-known.
> > But, let's now integrate r wrt time t, to give the change
> > in ...... spacetime, s,
>
> > s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>
> The pattern continues. This month you seem to take standard solutions and
> the integrate them a few more times, 'just to see'. When you realize that
> this too is a dead end path, you'll silently drop the idea everyone told you
> was stupid.
It's not everyday that one discovers the genuine
and true notion of spacetime as opposed to the
nonsensical Minkowskian one.
Juan R.
Good you learn *some* basic math, not many but congrats!
Now you need to learn some physics too...
--
http://www.canonicalscience.org/
BLOG:
http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
Juan R.
Albertito said:
spacetime I've defined has more physical meaning than the nonsensical
Minkowskian one used by Einstein et al.
But what definition of nonsensical is using Albertito? Hint: Next is the
definition of sensical he is using,
The speed of light in this genuine spacetime behaves as a constant
acceleration. So, what we perceive as a constant speed of light is
actually a constant accelerated motion (aka propagation) in this genuine
spacetime.
;-D ;-D ;-D ;-D ;-D
Juan R.
> On Oct 22, 7:55 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>> Albertito wrote:
>> > Let's start with a kinematics of constant acceleration, and integrate
>> > acceleration, a, with respect to time t to give the change in
>> > velocity, v,
>>
>> > v = v_0 + a*t
>>
>> > Now, integrate v wrt time, too, to give the change in space, r,
>>
>> > r = r_0 + v_0*t + a*t^2/2
>>
>> > So far so good, since a, v_0 and r_0 are all constants. And the
>> > physical meaning of each magnitude is well-known. But, let's now
>> > integrate r wrt time t, to give the change in ...... spacetime, s,
>>
>> > s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>>
>> The pattern continues. This month you seem to take standard solutions
>> and the integrate them a few more times, 'just to see'. When you
>> realize that this too is a dead end path, you'll silently drop the idea
>> everyone told you was stupid.
>
> It's not everyday that one discovers the genuine and true notion of
> spacetime as opposed to the nonsensical Minkowskian one.
Fortunately, because all of us have a limited ability to laugh as now.
>
>>
>>
>> > that magnitude s is called spacetime, because if you divide it by
>> > space you attain a time, and if you divide it by a time you attain a
>> > space.
>>
>> > "space interval is the rate of change of spacetime with respect to
>> > time"
>>
>> > "time interval is the rate of change of spacetime with respect to
>> > space"
>>
>> > Let's start with a dynamics of constant gravity, g, and integrate
>> > with respect to displacement r to give the change in potential, phi,
>>
>> > phi = phi_0 + g*r
>>
>> > Continue integrating to yield psi, and omega
>>
>> > psi = psi_0 + phi_0*r + g*r^2/2,
>>
>> > omega = omega_0 + psi_0*r + phi_0*r^2/2 + g*r^3/6
>>
>> > At this point, we see that omega must also be a spacetime as define
>> > above in the kinematics, because the third derivative of a spacetime
>> > wrt to space is a gravitational field, g, and the third derivative of
>> > spacetime wrt time is a 'kinematical' acceleration.
>>
>> > Corollary: In the above dynamics, psi, plays the role of the source
>> > of gravitation, and omega is spacetime, therefore the 'source' of the
>> > source of gravitation is spacetime itself.
>>
>> > Regards
--
PD
Bully for you. But you've just proved that the "speed" of something in
your "spacetime" is invariant, and that the "interval" in your
"spacetime" is invariant. However, the "speed" in your "spacetime"
doesn't bear any relationship to that which is measured in real
spacetime, and the "interval" in your "spacetime" doesn't bear any
relationship to that which is measured in real spacetime. I can come
up with a quantity that is invariant in O(6) as well, but it doesn't
have anything to do with the invariant quantities that are measured in
our reality, even if I decide to give that invariant the label "c" or
"interval".
>
> Hint: The speed of light in this genuine spacetime
> behaves as a constant acceleration. So, what we
> perceive as a constant speed of light is actually
> a constant accelerated motion (aka propagation) in
> this genuine spacetime.
>
> BTW, your neurons are as nonsensical as the
> Minkowskian ones, or perhaps you are as gullible
> as a dingleberry?
You still haven't said what is nonsensical about Minkowski space.
eric gisse
[...]
>
> It's not everyday that one discovers the genuine
> and true notion of spacetime as opposed to the
> nonsensical Minkowskian one.
What's nonsensical about SO(3,1)?
[...]
Paul B. Andersen
Albertito
<juanREM...@canonicalscience.com> wrote:
> Albertito wrote on Thu, 22 Oct 2009 11:15:57 -0700:
>
> Albertito said:
>
> spacetime I've defined has more physical meaning than the nonsensical
> Minkowskian one used by Einstein et al.
>
> But what definition of nonsensical is using Albertito? Hint: Next is the
> definition of sensical he is using,
>
> The speed of light in this genuine spacetime behaves as a constant
> acceleration. So, what we perceive as a constant speed of light is
> actually a constant accelerated motion (aka propagation) in this genuine
> spacetime.
>
> ;-D ;-D ;-D ;-D ;-D
You didn't proved anything, but your ignorance.
BTW, most dynamical computations can be overly
simplified if you use my genuine spacetime.
Example: an action can be easily computed by means
of my genuine spacetime, as
S = F*s
where S is the action, F is a force, * is inner product,
and s is genuine spacetime vector.
;-) ;-D ;-P ;-) ;-D ;-P ;-) ;-D ;-P ;-) ;-D ;-P ;-) ;-D ;-P
>
> --http://www.canonicalscience.org/
>
> BLOG:http://www.canonicalscience.org/en/publicationzone/canonicalscienceto...
Albertito
Twin paradox (aka clock paradox)
BTW, you are very proud of your SO(3,1)
aka Lorentz group, but I can challenge
you to prove that all Einstein additions
of velocities is also a group. A group is
the most primary algebraic structure, so if
Einstein addition can't even form a group,
why the hell is it defined by means of
an algebraic binary operator?
>
> [...]
Albertito
Twin paradox. It is impossible that the stay-at-home
twin aged slower than the travelling twin, in the rest
frame of the latter, unless there were a preferred
rest frame, which btw is forbidden in SR by the very
PoR. There is nothing in SR pointing to any asymmetry
that could discriminate between a stay-at-home rest
frame and a travelling rest frame.
Albertito
<juanREM...@canonicalscience.com> wrote:
> Albertito wrote on Thu, 22 Oct 2009 11:17:18 -0700:
>
>
>
> > On Oct 22, 7:55 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
> >> Albertito wrote:
> >> > Let's start with a kinematics of constant acceleration, and integrate
> >> > acceleration, a, with respect to time t to give the change in
> >> > velocity, v,
>
> >> > v = v_0 + a*t
>
> >> > Now, integrate v wrt time, too, to give the change in space, r,
>
> >> > r = r_0 + v_0*t + a*t^2/2
>
> >> > So far so good, since a, v_0 and r_0 are all constants. And the
> >> > physical meaning of each magnitude is well-known. But, let's now
> >> > integrate r wrt time t, to give the change in ...... spacetime, s,
>
> >> > s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>
> >> The pattern continues. This month you seem to take standard solutions
> >> and the integrate them a few more times, 'just to see'. When you
> >> realize that this too is a dead end path, you'll silently drop the idea
> >> everyone told you was stupid.
>
> > It's not everyday that one discovers the genuine and true notion of
> > spacetime as opposed to the nonsensical Minkowskian one.
>
> Fortunately, because all of us have a limited ability to laugh as now.
I agree, you have a lot of limited abilities. In fact,
laughing is your best developed ability :-)
>
>
>
>
>
> >> > that magnitude s is called spacetime, because if you divide it by
> >> > space you attain a time, and if you divide it by a time you attain a
> >> > space.
>
> >> > "space interval is the rate of change of spacetime with respect to
> >> > time"
>
> >> > "time interval is the rate of change of spacetime with respect to
> >> > space"
>
> >> > Let's start with a dynamics of constant gravity, g, and integrate
> >> > with respect to displacement r to give the change in potential, phi,
>
> >> > phi = phi_0 + g*r
>
> >> > Continue integrating to yield psi, and omega
>
> >> > psi = psi_0 + phi_0*r + g*r^2/2,
>
> >> > omega = omega_0 + psi_0*r + phi_0*r^2/2 + g*r^3/6
>
> >> > At this point, we see that omega must also be a spacetime as define
> >> > above in the kinematics, because the third derivative of a spacetime
> >> > wrt to space is a gravitational field, g, and the third derivative of
> >> > spacetime wrt time is a 'kinematical' acceleration.
>
> >> > Corollary: In the above dynamics, psi, plays the role of the source
> >> > of gravitation, and omega is spacetime, therefore the 'source' of the
> >> > source of gravitation is spacetime itself.
>
> >> > Regards
>
Albertito
wrote:
> Well done.
>
> Keep it up, the sky is the limit. :-)
You think that by pronouncing the word
"nonsensicalness" you have proved something.
Indeed, what you've proved is your own
"stupidityness" :-)
Keep your "stupidityness" up, the universe as
a whole is not even the limit.
>
> --
> Paul
>
> http://home.c2i.net/pb_andersen/
Juan R.
> On Oct 22, 7:54 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Albertito wrote on Thu, 22 Oct 2009 11:15:57 -0700:
>>
>> Albertito said:
>>
>> spacetime I've defined has more physical meaning than the nonsensical
>> Minkowskian one used by Einstein et al.
>>
>> But what definition of nonsensical is using Albertito? Hint: Next is
>> the definition of sensical he is using,
>>
>> The speed of light in this genuine spacetime behaves as a constant
>> acceleration. So, what we perceive as a constant speed of light is
>> actually a constant accelerated motion (aka propagation) in this
>> genuine spacetime.
>>
>> ;-D ;-D ;-D ;-D ;-D
>
> You didn't proved anything, but your ignorance.
Where did you read I was proving anything?
Your posts alone prove a lot of :-D
> BTW, most dynamical computations can be overly simplified if you use my
> genuine spacetime.
You have no idea of dynamical computations. Do you really believe that you
are fooling someone? Not even students!
> Example: an action can be easily computed by means of my genuine
> spacetime, as
>
> S = F*s
>
> where S is the action, F is a force, * is inner product, and s is
> genuine spacetime vector.
:_-D
Juan R.
> On Oct 22, 7:57 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Albertito wrote on Thu, 22 Oct 2009 11:17:18 -0700:
>>
>>
>>
>> > On Oct 22, 7:55 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>> >> Albertito wrote:
>> >> > Let's start with a kinematics of constant acceleration, and
>> >> > integrate acceleration, a, with respect to time t to give the
>> >> > change in velocity, v,
>>
>> >> > v = v_0 + a*t
>>
>> >> > Now, integrate v wrt time, too, to give the change in space, r,
>>
>> >> > r = r_0 + v_0*t + a*t^2/2
>>
>> >> > So far so good, since a, v_0 and r_0 are all constants. And the
>> >> > physical meaning of each magnitude is well-known. But, let's now
>> >> > integrate r wrt time t, to give the change in ...... spacetime, s,
>>
>> >> > s = s_0 + r_0*t + v_0*t^2/2 + a*t^3/6
>>
>> >> The pattern continues. This month you seem to take standard
>> >> solutions and the integrate them a few more times, 'just to see'.
>> >> When you realize that this too is a dead end path, you'll silently
>> >> drop the idea everyone told you was stupid.
>>
>> > It's not everyday that one discovers the genuine and true notion of
>> > spacetime as opposed to the nonsensical Minkowskian one.
>>
>> Fortunately, because all of us have a limited ability to laugh as now.
>
> I agree, you have a lot of limited abilities.
Agree, I never could find so many new creative ways to say nonsense as you
can!
> In fact, laughing is your
> best developed ability :-)
Thanks to your many laugdable posts of course, the merit belong to you also :-D
>
>
>>
>>
>>
>>
>> >> > that magnitude s is called spacetime, because if you divide it by
>> >> > space you attain a time, and if you divide it by a time you attain
>> >> > a space.
>>
>> >> > "space interval is the rate of change of spacetime with respect to
>> >> > time"
>>
>> >> > "time interval is the rate of change of spacetime with respect to
>> >> > space"
>>
>> >> > Let's start with a dynamics of constant gravity, g, and integrate
>> >> > with respect to displacement r to give the change in potential,
>> >> > phi,
>>
>> >> > phi = phi_0 + g*r
>>
>> >> > Continue integrating to yield psi, and omega
>>
>> >> > psi = psi_0 + phi_0*r + g*r^2/2,
>>
>> >> > omega = omega_0 + psi_0*r + phi_0*r^2/2 + g*r^3/6
>>
>> >> > At this point, we see that omega must also be a spacetime as
>> >> > define above in the kinematics, because the third derivative of a
>> >> > spacetime wrt to space is a gravitational field, g, and the third
>> >> > derivative of spacetime wrt time is a 'kinematical' acceleration.
>>
>> >> > Corollary: In the above dynamics, psi, plays the role of the
>> >> > source of gravitation, and omega is spacetime, therefore the
>> >> > 'source' of the source of gravitation is spacetime itself.
>>
>> >> > Regards
>>
>> --http://www.canonicalscience.org/
>>
>> BLOG:http://www.canonicalscience.org/en/publicationzone/canonicalscienceto...
--
PD
Ah. It seems you have a minor misconception about a few things, and
that is confusing you.
The traveling twin does not have "a" rest frame. Do you see why?
Secondly, you may have misread the puzzle in the first place. The stay-
at-home twin did not age more slowly.
> There is nothing in SR pointing to any asymmetry
> that could discriminate between a stay-at-home rest
> frame and a travelling rest frame.
There is no single traveling rest frame.
And of course there's an asymmetry -- *physics* points to it. The
traveling twin experiences something the stay-at-home twin does not.
Do you know what that is?
Once we clear up a couple of your misconceptions, then tell me again
what is nonsensical about Minkowski spacetime.
PD
> On Oct 22, 11:01 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>
> > Albertito wrote:
>
> > [...]
>
> > > It's not everyday that one discovers the genuine
> > > and true notion of spacetime as opposed to the
> > > nonsensical Minkowskian one.
>
> > What's nonsensical about SO(3,1)?
>
> Twin paradox (aka clock paradox)
>
> BTW, you are very proud of your SO(3,1)
> aka Lorentz group, but I can challenge
> you to prove that all Einstein additions
> of velocities is also a group.
I believe they are. They satisfy closure, associativity, invertibility
and identity.
Almost by inspection. Do you find a problem with any of them?
> A group is
> the most primary algebraic structure, so if
> Einstein addition can't even form a group,
> why the hell is it defined by means of
> an algebraic binary operator?
I don't think it is. We sure don't use "+" to convey the operation.
These all look like minor confusions on your part. Once they're
cleared up, maybe you'll feel better about relativity.
>
>
>
> > [...]
>
>
Albertito
Again coming with the stinky "you do not understand
relativity" argument?
> The traveling twin does not have "a" rest frame.
That's wrong. Now, who is having a misconception here?
The travelling twin can define his rest frame as far as
he is not accelerating, but moving inertially. And do not
come again telling me that "acceleration plays a role in
the scenario", because that's not true. Your gullibility
allowed you to stomach the twin paradox without any
objection. That's bad, very badddddddddddddddddd!
> Do you see why?
> Secondly, you may have misread the puzzle in the first place. The stay-
> at-home twin did not age more slowly.
Wrong. The stay-at-home twin aged more slowly in the rest
frame of the travelling twin. I have showed that the travelling
twin has a rest frame most of the time as far as he doesn't
accelerate.
>
> > There is nothing in SR pointing to any asymmetry
> > that could discriminate between a stay-at-home rest
> > frame and a travelling rest frame.
>
> There is no single traveling rest frame.
So, are you saying that once a body has accelerated and
stopped accelerating, it can never be at rest in any
proper frame? How stupid can you get to be?
> And of course there's an asymmetry -- *physics* points to it. The
> traveling twin experiences something the stay-at-home twin does not.
Sure, there is an asymmetry: relativists have the left
cerebral hemisphere less developed than the right one.
> Do you know what that is?
Yes, I do. Maybe, that I do not understand relativity
but you do, because you are smart and I don't? The
stinky "you do not understand relativity" argument?
>
> Once we clear up a couple of your misconceptions, then tell me again
> what is nonsensical about Minkowski spacetime.
I'm bored now. T W I N P A R A D O X X X X X X X!
Albertito
> On Oct 23, 4:04 am, Albertito <albertito1...@gmail.com> wrote:
>
>
>
> > On Oct 22, 11:01 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>
> > > Albertito wrote:
>
> > > [...]
>
> > > > It's not everyday that one discovers the genuine
> > > > and true notion of spacetime as opposed to the
> > > > nonsensical Minkowskian one.
>
> > > What's nonsensical about SO(3,1)?
>
> > Twin paradox (aka clock paradox)
>
> > BTW, you are very proud of your SO(3,1)
> > aka Lorentz group, but I can challenge
> > you to prove that all Einstein additions
> > of velocities is also a group.
>
> I believe they are. They satisfy closure, associativity, invertibility
> and identity.
That's not a matter of belief, but of mathematical proof.
If I weren't too lazy now, I could perform the proof showing
Einstein addition of velocities is not a group, but then you
would come with the "so what? we can live without that" argument.
> Almost by inspection. Do you find a problem with any of them?
Your inspection falied. Try again.
>
> > A group is
> > the most primary algebraic structure, so if
> > Einstein addition can't even form a group,
> > why the hell is it defined by means of
> > an algebraic binary operator?
>
> I don't think it is. We sure don't use "+" to convey the operation.
I said algebraic, not arithmetic. An operator involving
several arithmetic operator is also a algebraic operator.
Maybe, you can't distinguish between arithmetic and
algebraic?
>
> These all look like minor confusions on your part.
No, there are no confusions, it is all very easy to be
understood correctly.
> Once they're
> cleared up, maybe you'll feel better about relativity.
I'm bored now. T W I N P A R A D O X X X X X X X!
>
>
>
> > > [...]
Inertial
news:06b597b7-773d-4166...@j19g2000vbp.googlegroups.com...
Only because you don't
>> The traveling twin does not have "a" rest frame.
>
> That's wrong.
Seee .. you don't understand
> Now, who is having a misconception here?
You are
> The travelling twin can define his rest frame as far as
> he is not accelerating, but moving inertially.
If he only travels inertially he cannot get back to his brother
> And do not
> come again telling me that "acceleration plays a role in
> the scenario", because that's not true.
Changing frames does. Acceleration is another name for changing inertial
frames
> Your gullibility
> allowed you to stomach the twin paradox without any
> objection. That's bad, very badddddddddddddddddd!
You clearly don't understand the twins 'paradox'
>> Do you see why?
>> Secondly, you may have misread the puzzle in the first place. The stay-
>> at-home twin did not age more slowly.
>
> Wrong.
No .. you are wrong
> The stay-at-home twin aged more slowly in the rest
> frame of the travelling twin.
Nope .. not by the time the twins reunite
> I have showed that the travelling
> twin has a rest frame most of the time as far as he doesn't
> accelerate.
Its what happens in the time when the travelling twin does NOT have the same
rest frame that the twins 'paradox' happens. Its like saying the if I push
you off a cliff you won't be hurt because most of the time you are in free
fall .. but its the landing that does the damage.
>> > There is nothing in SR pointing to any asymmetry
>> > that could discriminate between a stay-at-home rest
>> > frame and a travelling rest frame.
>>
>> There is no single traveling rest frame.
>
> So, are you saying that once a body has accelerated and
> stopped accelerating, it can never be at rest in any
> proper frame?
Why would you come to such a stupid conclusions as that .. really, you're an
idiot. Did you see anywhere where it was claimed that there is no rest
frame for the traveler at all? only that there is no SINGLE rest frame. If
the traveler only has a single rest frame, then there is no paradox, because
the traveler cannot return to his twin.
> How stupid can you get to be?
You said a pretty good standard for high levels of stupidity that few of us
here could ever reach
>> And of course there's an asymmetry -- *physics* points to it. The
>> traveling twin experiences something the stay-at-home twin does not.
>
> Sure, there is an asymmetry: relativists have the left
> cerebral hemisphere less developed than the right one.
You have neither, it seems
>> Do you know what that is?
>
> Yes, I do. Maybe, that I do not understand relativity
The only intelligent thing you've said
> but you do, because you are smart and I don't?
BAHAHAHA
> The
> stinky "you do not understand relativity" argument?
Its a fact that you could remedy if you bothered to learn and understand
PD
It's not an argument. It's an observation. If a student says 34+17=63,
pointing out that he's wrong is not an argument for arithmetic, it's
simply pointing out an error.
>
> > The traveling twin does not have "a" rest frame.
>
> That's wrong. Now, who is having a misconception here?
> The travelling twin can define his rest frame as far as
> he is not accelerating, but moving inertially.
Which he doesn't do for the entirety of the trip.
> And do not
> come again telling me that "acceleration plays a role in
> the scenario", because that's not true.
Of course it's true. Acceleration is the key to the difference in the
two clock readings of the twins when they return.
It's the very fact that the traveling twin's worldline is not straight
that there's a difference at all.
> Your gullibility
> allowed you to stomach the twin paradox without any
> objection. That's bad, very badddddddddddddddddd!
If there were a valid objection, sure, but you're objection is just a
mistake in understanding what's going on.
>
> > Do you see why?
> > Secondly, you may have misread the puzzle in the first place. The stay-
> > at-home twin did not age more slowly.
>
> Wrong. The stay-at-home twin aged more slowly in the rest
> frame of the travelling twin. I have showed that the travelling
> twin has a rest frame most of the time as far as he doesn't
> accelerate.
So? The steeper the acceleration, the more severe the effect it has.
>
>
>
> > > There is nothing in SR pointing to any asymmetry
> > > that could discriminate between a stay-at-home rest
> > > frame and a travelling rest frame.
>
> > There is no single traveling rest frame.
>
> So, are you saying that once a body has accelerated and
> stopped accelerating, it can never be at rest in any
> proper frame? How stupid can you get to be?
No, I didn't say that. But it's not in the same frame it was before.
It's in a new frame. That's at least two frames, you see. That's
important.
>
> > And of course there's an asymmetry -- *physics* points to it. The
> > traveling twin experiences something the stay-at-home twin does not.
>
> Sure, there is an asymmetry: relativists have the left
> cerebral hemisphere less developed than the right one.
>
> > Do you know what that is?
>
> Yes, I do. Maybe, that I do not understand relativity
> but you do, because you are smart and I don't?
Smart doesn't have anything to do with it. It's just that you haven't
learned it.
You're smart. Learn it.
PD
> On Oct 23, 1:02 pm, PD <thedraperfam...@gmail.com> wrote:
>
>
>
> > On Oct 23, 4:04 am, Albertito <albertito1...@gmail.com> wrote:
>
> > > On Oct 22, 11:01 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>
> > > > Albertito wrote:
>
> > > > [...]
>
> > > > > It's not everyday that one discovers the genuine
> > > > > and true notion of spacetime as opposed to the
> > > > > nonsensical Minkowskian one.
>
> > > > What's nonsensical about SO(3,1)?
>
> > > Twin paradox (aka clock paradox)
>
> > > BTW, you are very proud of your SO(3,1)
> > > aka Lorentz group, but I can challenge
> > > you to prove that all Einstein additions
> > > of velocities is also a group.
>
> > I believe they are. They satisfy closure, associativity, invertibility
> > and identity.
>
> That's not a matter of belief, but of mathematical proof.
> If I weren't too lazy now, I could perform the proof showing
> Einstein addition of velocities is not a group, but then you
> would come with the "so what? we can live without that" argument.
If you can prove that closure or associativity or invertibility or
identity are not satisfied, please show that.
>
> > Almost by inspection. Do you find a problem with any of them?
>
> Your inspection falied. Try again.
Do you find a problem with any of them?
>
>
>
> > > A group is
> > > the most primary algebraic structure, so if
> > > Einstein addition can't even form a group,
> > > why the hell is it defined by means of
> > > an algebraic binary operator?
>
> > I don't think it is. We sure don't use "+" to convey the operation.
>
> I said algebraic, not arithmetic. An operator involving
> several arithmetic operator is also a algebraic operator.
> Maybe, you can't distinguish between arithmetic and
> algebraic?
Thanks for the clarification. It certainly is a group.
Albertito
No it isn't.
It could help if you bother to accomplish a rigorous proof.
It seems you do not know relativity or your misconceptions
are pretty severe. See the reason why Einstein addition of
velicities can't form a group. Learn the reason why gyrovectors
were invented
http://en.wikipedia.org/wiki/Gyrovector_space#Gyrovectors
Now, don't come again with the "so what? we can live without
that" argument.
>
>
>
Albertito
> "Albertito" <albertito1...@gmail.com> wrote in message
Dear ignorant bonehead,
Just a question and a clarification.
Question: Who invited you here to yield your stinky dungs?
Clarification: All your stupid 'reasoning' (so speaking)
can be easily refuted by saying that, under SR assumptions,
time dilation is not a function of acceleration but of speed
(see Lorentz factor for details). PERIOD.
Now, fuck off.
PD
You're quite right. I was mistaken about associativity because I was
thinking of collinear examples.
> It could help if you bother to accomplish a rigorous proof.
> It seems you do not know relativity or your misconceptions
> are pretty severe. See the reason why Einstein addition of
> velicities can't form a group. Learn the reason why gyrovectors
> were inventedhttp://en.wikipedia.org/wiki/Gyrovector_space#Gyrovectors
>
> Now, don't come again with the "so what? we can live without
> that" argument.
Well, let's see if you're right. Now why do you think that being a
group is essential again?
Albertito
Murder instinct.
PD
excuse me?
eric gisse
> On Oct 22, 11:01 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>> Albertito wrote:
>>
>> [...]
>>
>>
>>
>> > It's not everyday that one discovers the genuine
>> > and true notion of spacetime as opposed to the
>> > nonsensical Minkowskian one.
>>
>> What's nonsensical about SO(3,1)?
>
> Twin paradox (aka clock paradox)
Not a paradox.
>
> BTW, you are very proud of your SO(3,1)
> aka Lorentz group, but I can challenge
> you to prove that all Einstein additions
> of velocities is also a group.
Open up a textbook on group theory and DO IT YOURSELF. The axioms of group
theory are explicit.
> A group is
> the most primary algebraic structure, so if
> Einstein addition can't even form a group,
No, a group is not the most primary algebraic structure.
> why the hell is it defined by means of
> an algebraic binary operator?
It isn't.
>
>>
>> [...]
PD
You did. When you post on an unmoderated, public newsgroup, you are
invited comments from anyone.
>
> Clarification: All your stupid 'reasoning' (so speaking)
> can be easily refuted by saying that, under SR assumptions,
> time dilation is not a function of acceleration but of speed
> (see Lorentz factor for details).
That's not quite right. The proper time measured by any system that
travels between two spacetime events depends on the shape of the
worldline between the two events. This very much depends on
acceleration.
It's really pretty simple. You can rant and foam that when others
point out that you don't know much about relativity, they are trying
to make you feel stupid. Or you can take the time to learn relativity
so that this doesn't happen so much.
PD
> PERIOD.
>
> Now, fuck off.
Albertito
Wrong. Under SR assumptions, time dilation
depends upon the length of the spacetime
interval, it never depends upon its shape.
You say that acceleration plays a role in
that scenario, I say it doesn't.
Remove the contributions to time dilation caused
by accelerations and just integrate only the
inertial parts. Then you still attain time
dilation, but under SR it is impossible to say
to what twin it belongs, because if you remove
those acceleration parts you are breaking the
hypothetical asymmetry.
Let's put a bit of maths into it. Call a1
initial acceleration of travelling twin,
a2 his turn back acceleration, and a3 his
acceleration close to home for being at rest.
The travel between the end of a1 and the
beginning of a2 is inertial, and the travel
between the end of a2 and the beginning of a3
is inertial as well. The most meaningful
contributions to time dilation are those
coming from the inertial motions.
Now, think. What happens if the stay-at-home
twin also accelerates with the same a1, a2 and a3,
in such a way that the inertial intermediate
motions spend minimal proper times as compared
to those intermediate inertial motions of the
travelling twin? Both twins accelerated with
the same accelerations, both return to home to
be at rest, but under SR assumptions, one of the
twins aged faster. Who and why? Since both twins
accelerated with the same accelerations, there
can't be asymmetry. Why SR predict such an asymmetry?
Answer: because SR is bullshit.
>
> It's really pretty simple. You can rant and foam that when others
> point out that you don't know much about relativity, they are trying
> to make you feel stupid. Or you can take the time to learn relativity
> so that this doesn't happen so much.
Did Herbert Dingle understand relativity? He devoted his
life to study relativy, and become a respectable expert
in relativity. And it was close to the end od his
professional career that he dared to critisize relativity.
Then, all enraged relativists attacked him, saying that
he never understood relativity. So, how come that
relativity is so difficult to understand that even
a guy as Herbet Dinglewas unable to understand it along
his whole life? I'm not interested in any theory that
will never be well-understood, with the farse that those
that claim to understand it are the clever guys, and
those that say to not undertand it are the stupid ones.
Who what to belong to the stupides' clan?
Inertial
>> > On Oct 23, 2:27 pm, "Inertial" <relativ...@rest.com> wrote:
>> > > "Albertito" <albertito1...@gmail.com> wrote in message
>> > > > Again coming with the stinky "you do not understand
>> > > > relativity" argument?
>> > > Only because you don't
>> > > >> The traveling twin does not have "a" rest frame.
>> > > > That's wrong.
>> > > Seee .. you don't understand
Do you understand why it is wrong? The travelling twin cannot have a single
rest frame
>> > > > Now, who is having a misconception here?
>> > > You are
>> > > > The travelling twin can define his rest frame as far as
>> > > > he is not accelerating, but moving inertially.
>> > > If he only travels inertially he cannot get back to his brother
Do you see why?
>> > > > And do not
>> > > > come again telling me that "acceleration plays a role in
>> > > > the scenario", because that's not true.
>> > > Changing frames does. Acceleration is another name for changing
>> > > inertial
>> > > frames
>> > > > Your gullibility
>> > > > allowed you to stomach the twin paradox without any
>> > > > objection. That's bad, very badddddddddddddddddd!
>> > > You clearly don't understand the twins 'paradox'
>> > > >> Do you see why?
>> > > >> Secondly, you may have misread the puzzle in the first place. The
>> > > >> stay-
>> > > >> at-home twin did not age more slowly.
>> > > > Wrong.
>> > > No .. you are wrong
Do you see that yet?
>> > > > The stay-at-home twin aged more slowly in the rest
>> > > > frame of the travelling twin.
>> > > Nope .. not by the time the twins reunite
>> > > > I have showed that the travelling
>> > > > twin has a rest frame most of the time as far as he doesn't
>> > > > accelerate.
>> > > Its what happens in the time when the travelling twin does NOT have
>> > > the same
>> > > rest frame that the twins 'paradox' happens. Its like saying the if
>> > > I push
>> > > you off a cliff you won't be hurt because most of the time you are in
>> > > free
>> > > fall .. but its the landing that does the damage.
Do you understand that?
>> > > >> > There is nothing in SR pointing to any asymmetry
>> > > >> > that could discriminate between a stay-at-home rest
>> > > >> > frame and a travelling rest frame.
>> > > >> There is no single traveling rest frame.
>> > > > So, are you saying that once a body has accelerated and
>> > > > stopped accelerating, it can never be at rest in any
>> > > > proper frame?
>> > > Why would you come to such a stupid conclusions as that .. really,
>> > > you're an
>> > > idiot. Did you see anywhere where it was claimed that there is no
>> > > rest
>> > > frame for the traveler at all? only that there is no SINGLE rest
>> > > frame. If
>> > > the traveler only has a single rest frame, then there is no paradox,
>> > > because
>> > > the traveler cannot return to his twin.
Do you see the distinction between the traveller not having just one single
rest frame and not having any at all?
>> > > > How stupid can you get to be?
>> > > You said a pretty good standard for high levels of stupidity that few
>> > > of us
>> > > here could ever reach
>> > > >> And of course there's an asymmetry -- *physics* points to it. The
>> > > >> traveling twin experiences something the stay-at-home twin does
>> > > >> not.
>> > > > Sure, there is an asymmetry: relativists have the left
>> > > > cerebral hemisphere less developed than the right one.
>> > > You have neither, it seems
>> > > >> Do you know what that is?
>> > > > Yes, I do. Maybe, that I do not understand relativity
>> > > The only intelligent thing you've said
>> > > > but you do, because you are smart and I don't?
>> > > BAHAHAHA
>> > > > The
>> > > > stinky "you do not understand relativity" argument?
>> > > Its a fact that you could remedy if you bothered to learn and
>> > > understand
>> > Dear ignorant bonehead
You talking to yourself?
>> > Just a question and a clarification.
Sure
>> > Question: Who invited you here to yield your stinky dungs?
None required .. You posted in a public newsgroup.
>> > Clarification: All your stupid 'reasoning' (so speaking)
>> > can be easily refuted by saying that, under SR assumptions,
>> > time dilation is not a function of acceleration but of speed
>> > (see Lorentz factor for details).
Indeed it is .. and acceleration is a change in speed, and hence a change in
how time dilates.
You've not refuted anything as yet.
Do you actually have a point to make, by the way, other than showing that
your grasp of the so-called twins paradox is rather flimsy?
>> > PERIOD.
>> > Now, fuck off.
After you
PD
A spacetime interval doesn't have a shape. A worldline does.
And yes, the proper time measured by any system that travels between
two spacetime events depends on that worldline shape.
See Penrose, "The Road to Reality".
> You say that acceleration plays a role in
> that scenario, I say it doesn't.
Well, you say a lot of foolish things.
>
> Remove the contributions to time dilation caused
> by accelerations and just integrate only the
> inertial parts.
But that doesn't yield anything that is measured on the traveling
twin's clock.
> Then you still attain time
> dilation, but under SR it is impossible to say
> to what twin it belongs, because if you remove
> those acceleration parts you are breaking the
> hypothetical asymmetry.
>
> Let's put a bit of maths into it. Call a1
> initial acceleration of travelling twin,
> a2 his turn back acceleration, and a3 his
> acceleration close to home for being at rest.
> The travel between the end of a1 and the
> beginning of a2 is inertial, and the travel
> between the end of a2 and the beginning of a3
> is inertial as well. The most meaningful
> contributions to time dilation are those
> coming from the inertial motions.
Nonsense. "Most meaningful contributions." Pah.
>
> Now, think. What happens if the stay-at-home
> twin also accelerates with the same a1, a2 and a3,
But he doesn't. And he knows it.
> in such a way that the inertial intermediate
> motions spend minimal proper times as compared
> to those intermediate inertial motions of the
> travelling twin? Both twins accelerated with
> the same accelerations, both return to home to
> be at rest, but under SR assumptions, one of the
> twins aged faster.
Not if they both accelerate. Geez.
> Who and why? Since both twins
> accelerated with the same accelerations, there
> can't be asymmetry. Why SR predict such an asymmetry?
> Answer: because SR is bullshit.
>
>
>
> > It's really pretty simple. You can rant and foam that when others
> > point out that you don't know much about relativity, they are trying
> > to make you feel stupid. Or you can take the time to learn relativity
> > so that this doesn't happen so much.
>
> Did Herbert Dingle understand relativity? He devoted his
> life to study relativy, and become a respectable expert
> in relativity. And it was close to the end od his
> professional career that he dared to critisize relativity.
And his methodology was weak at that point.
> Then, all enraged relativists attacked him, saying that
> he never understood relativity. So, how come that
> relativity is so difficult to understand that even
> a guy as Herbet Dinglewas unable to understand it along
> his whole life?
It's NOT difficult to understand. Some people go off the deep end.
Relativity is really pretty simple. Especially if you choose to take
action to learn it.
> I'm not interested in any theory that
> will never be well-understood,
It IS well-understood. And it's remarkable that you sit here and
attempt to critique it while AT THE SAME TIME acknowledging that you
don't understand it. This is a little like criticizing brain surgery
if you don't know the first thing about brain surgery.
> with the farse that those
> that claim to understand it are the clever guys, and
> those that say to not undertand it are the stupid ones.
I didn't say you were stupid. I explicitly TOLD you that you were
smart. Which is why you should try to actually learn what relativity
says.
Albertito
You are brutally wrong! The traveling twin can detect when
his probe is accelerating, so he can stop his local clock
in those intervals of acceleration and resume it when inertial
motion is on. Your misconceptions about relativity are more
severe than expected!
>
> > Then you still attain time
> > dilation, but under SR it is impossible to say
> > to what twin it belongs, because if you remove
> > those acceleration parts you are breaking the
> > hypothetical asymmetry.
>
> > Let's put a bit of maths into it. Call a1
> > initial acceleration of
>
> ...
>
> read more »
PD
Oh, come, come. You are suggesting that the way to show that
relativity is wrong is by having the traveling twin start and stop his
clock during the trip?? In other words, it's OK to CHEAT?
John Park
Surely deleting a couple of arbitrarily small time intervals makes no
difference to the analysis of the twins "paradox"?
--John Park
PD
Oh, but it DOES. The kink in the worldline is what makes it not
straight, and that is what makes the traveling twin's proper time end
up being smaller than the earth twin's, whose wordline is straight.
(In more conventional language, this is where the traveling twin
switches between inertial reference frames.)
The steeper the kink, the more pronounced the acceleration and
therefore the more dramatic the effect. Likewise, the shorter the
amount of time spent in navigating that kink, then the more dramatic
the effect. This is what gives rise to the "snap forward" of the Earth
clock as seen by the traveling twin's clock, as discussed in some
presentations of the puzzle.
Basically what happens is this, presented in the plainest language I
can muster (but without any derivation of how this happens): Suppose
the traveling twin can keep an eye on the Earth clock throughout the
trip. On the outbound, inertial trip, the Earth clock falls steadily
more and more behind the spaceship clock, because the Earth clock is
moving inertially relative to the spaceship and so is running slow.
When the traveling twin begins his acceleration to turn around, the
Earth clock suddenly advances forward (and more rapidly, the more
rapid the acceleration) so that by the time the turnaround is
complete, the Earth clock is *well* ahead of the spaceship clock. Then
on the inbound trip, the Earth clock again appears to run slower than
the spaceship clock, but not so much that it compensates for the leap
forward. In the end, the Earth clock remains ahead of the spaceship
clock.
The novice to SR will look at only the two inertial legs and not have
any idea of what happens when the ship turns around (accelerates,
moves from one inertial reference frame to another), but *assume*
(incorrectly) that the effect can be made arbitrarily small, when in
fact the opposite is true.
You can also ask the question what happens if you don't make a sharp
turnaround at all, but are constantly doing the turnaround -- as in a
big circle. Well, in that case the "snap forward" is spread over the
whole trip so that the Earth clock is still ahead of the traveling
clock by the time it comes around again. (The traveling twin's
worldline is still not straight, where the Earth twin's worldline is,
and that's the fundamental reason for the result.) This effect is what
contributes to the shift in the clocks in GPS satellites, you see.
(This is what eludes Ken Seto constantly.)
John Park
> On Oct 27, 11:41=A0am, af...@FreeNet.Carleton.CA (John Park) wrote:
>> PD (thedraperfam...@gmail.com) writes:
>> >> On Oct 27, 12:52 pm, PD <thedraperfam...@gmail.com> wrote:
>>
>> >> > On Oct 27, 5:54 am, Albertito <albertito1...@gmail.com> wrote:
>>
>> >> > > On Oct 26, 7:38 pm, PD <thedraperfam...@gmail.com> wrote:
>>
>> >> > > > > Clarification: All your stupid 'reasoning' (so speaking)
>> >> > > > > can be easily refuted by saying that, under SR assumptions,
>> >> > > > > time dilation is not a function of acceleration but of speed
>> >> > > > > (see Lorentz factor for details).
>>
for a couple of instants (which is what I meant by deleting them).
--John Park
Inertial
news:hc8cs3$qr0$1...@theodyn.ncf.ca...
[snip]
> But as far as I can see, none of this is changed if you stop the clocks
> for a couple of instants (which is what I meant by deleting them).
What do you think stopping a clock would accomplish. Do people stop aging
because you stop a clock? All it will mean is that the time shown on the
clock will be out by the amount of time for which you stop the clock.
PD
This is reminiscent of a neighbor's strategy of sleeping through her
birthday so that when she woke up, she'd still be 39.
Albertito
[snip]
> > But as far as I can see, none of this is changed if you stop the clocks
> > for a couple of instants (which is what I meant by deleting them).
>
> > --John Park
>
> This is reminiscent of a neighbor's strategy of sleeping through her
> birthday so that when she woke up, she'd still be 39.
You lost the thread and now everybody can
see you're wandering and saying stupidies.
Once again,
"Remove the contributions to time dilation
caused by accelerations and just integrate
only the inertial parts. Then, you still
attain time dilation, but under SR it is
impossible to say which twin it belongs to,
because if you remove those acceleration
parts you are breaking the hypothetical
asymmetry."
Once again. If you remove the contributions to time
dilation caused by accelerations you still have time
dilation. But, now the twin paradox show its strongest
nonsense!
"After you have removed contributions to
time dilation caused by accelerations, you
are left with the immense ignorance about
which twin (the traveller or the at-home one)
the remainder dilated time belongs to."
The twin paradox not only is a paradox, but the evidence
that SR is a model of nothing, just B U L L S H I T !
PD
Let's see. So if you strip out the part that resolves the paradox,
then the paradox remains. Hmmmm....
John Park
> On Oct 27, 10:13=A0pm, af...@FreeNet.Carleton.CA (John Park) wrote:
>> PD (thedraperfam...@gmail.com) writes:
>> >> PD (thedraperfam...@gmail.com) writes:
>> >> >> On Oct 27, 12:52 pm, PD <thedraperfam...@gmail.com> wrote:
>>
>> >> >> > On Oct 27, 5:54 am, Albertito <albertito1...@gmail.com> wrote:
>>
>> >> >> > > On Oct 26, 7:38 pm, PD <thedraperfam...@gmail.com> wrote:
>>
>> >> >> > > > > Clarification: All your stupid 'reasoning' (so speaking)
>> >> >> > > > > can be easily refuted by saying that, under SR assumptions,
> d
>> >> >> > > > > (see Lorentz factor for details).
>>
> m t=3D
>> > hat
>> >> >> > > > travels between two spacetime events depends on the shape of =
> the
>> >> >> > > > worldline between the two events. This very much depends on
>> >> >> > > > acceleration.
>>
>> >> >> > > Wrong. Under SR assumptions, time dilation
>> >> >> > > depends upon the length of the spacetime
>> >> >> > > interval, it never depends upon its shape.
>>
>> >> >> > A spacetime interval doesn't have a shape. A worldline does.
> een
>> >> >> > two spacetime events depends on that worldline shape.
>> >> >> > See Penrose, "The Road to Reality".
>>
>> >> >> > > You say that acceleration plays a role in
>> >> >> > > that scenario, I say it doesn't.
>>
>> >> >> > Well, you say a lot of foolish things.
>>
>> >> >> > > Remove the contributions to time dilation caused
>> >> >> > > by accelerations and just integrate only the
>> >> >> > > inertial parts.
>>
>> >> >> > But that doesn't yield anything that is measured on the traveling
>> >> >> > twin's clock.
>>
>> >> >> You are brutally wrong! The traveling twin can detect when
>> >> >> his probe is accelerating, so he can stop his local clock
>> >> >> in those intervals of acceleration and resume it when inertial
>> >> >> motion is on. Your misconceptions about relativity are more
>> >> >> severe than expected!
>>
>> >> > Oh, come, come. You are suggesting that the way to show that
> is
>> >> > clock during the trip?? In other words, it's OK to CHEAT?
>>
>> >> Surely deleting a couple of arbitrarily small time intervals makes no
>> >> difference to the analysis of the twins "paradox"?
>>
>> But as far as I can see, none of this is changed if you stop the clocks
>> for a couple of instants (which is what I meant by deleting them).
>>
>
> birthday so that when she woke up, she'd still be 39.
So, since it makes no difference, why did you call it cheating? I'd call
it irrelevant.
--John Park