A case against relativity 2
Harold Ensle
which were never answered, so I would like to try again.
I will discuss also a few of the previous replies.
First, concerning some basic requirements of special relativity.
Mathematically, it is defined by the Lorentz Transformations:
x'=g(x-vt) (1)
t'=g(t-vx/c^2) (2)
where g=1/sqrt(1-v^2/c^2)
(x is position, t is time, c is the speed of light
and v is the relative velocity)
Note that the Lorentz Transformations are based solely on
_relative_ velocity. We can derive from these a general
expression that relates the passage of time for two observers:
t'=integral from t_1 to t_2(sqrt(1-v(t)^2/c^2))dt (3)
This can be found in most relevant textbooks such as Moller
or Jackson. Note that this equation incorporates acceleration.
It is important to understand to which acceleration it refers.
The acceleration here, since it directly correlates to the
_relative_ velocity of the Lorentz Transformations is a
_relative_ acceleration. That is, it is the change in
_relative_ velocity over time. It is a purely kinematic value.
It tells us _nothing_ about what forces are present. It tells us
only that the _relative_ velocity between the two observers
is changing.
Now I want to determine the time that has passed for both
twins in the twin paradox. First for the stay-at-home one
can use equation 3. Note that with any non-zero velocity
the integrand is always less than one, therefore the
stay-at-home will see the travelling twin's passage of time
to be less than his own by:
tt=ts/k. (4)
where tt is the travelling twin's time, ts is the stay-at-home's
time, and k is some value greater than one.
Now the travelling twin has accerated during the trip, so
since the Lorentz transformations are only valid in
an inertial frame, he can't use them.
SO the question is: What equation can the travelling twin
use?
If the travelling twin uses the Lorentz transformations, he
will derive the very same differential form as equation 3.
Of course, if he applies it, he will get:
ts=tt/k
(5)
thus tt=ts*k which is contradictory to equation (4). This is
the mathematical statement of the twin paradox.
Now since equation 3 is the time relation derived in
correspondence to the Lorentz transformations for both
observers, it is evident that the solution of the contradiction
cannot come from the Lorentz transformations. Since
these are the statements of SR, it thus follows that the
twin paradox cannot be resolved at all in the context of SR.
I am not claiming here that there is an ultimate contradiction.
I have shown here only that there is a contradiction that
does arise if SR is used by itself.
Now, in regards to an equation that will work for the
travelling twin. In the last thread I received only one
candidate (from Andersen).
t' = integral from t1 to t2(sqrt(1-v(t)^2/c^2)(1 + ax/c^2))dt (6)
where a is the acceleration of the x-t frame.
Since equation 3 is the time dilatation
derived from the Lorentz transformations, this equation is
clearly not derived, or derivable from SR alone.
Because, if it were, it would have to be identical to
equation 3.
Now, if this equation has a legitimate source, does it
solve the paradox? Well...apparently, it doesn't even do that.
Note that to solve the paradox, this equation must provide
for a round trip a solution for the travelling twin such that:
ts=tt*k (7)
So that both twins will agree on their reunion. Thus at
some point in the travelling twin's trip he must see the
stay-at-home's time accumulating faster than his own.
Imagine a trip where the travelling twin accelerates for the
first half of his outward journey and then deccelerates
for the rest of the way, and then does the same process to
return home.
First 'a' is positive and then 'a' is negative when 'x' is larger.
Thus the travelling twin will see even less accumulated time
for the stay-at-home than when he used equation 3. On the
return journey the same thing happens. So it appears that
the paradox is actually worse than before.
So, unless 'a' and 'x' have some special meanings, Andersen's
equation fails miserably.
H.Ellis Ensle
pholroyd
>
> Note that the Lorentz Transformations are based solely on
> _relative_ velocity. We can derive from these a general
> expression that relates the passage of time for two observers:
>
> t'=integral from t_1 to t_2(sqrt(1-v(t)^2/c^2))dt (3)
>
> This can be found in most relevant textbooks such as Moller
> or Jackson. Note that this equation incorporates acceleration.
moller is one of books i use to study relativity and he dont say that
equation incorporates acceleration. what he says is he assumes that
acceleration
of clock relative to inertial system does not influence clock rate. in
my book this is page 49. much later in section called clock paraodox
(page 258)
he say that now he gives complete solution.
@@@ph@@@
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Harold Ensle" <hee...@ix.netcom.com> wrote in message
news:8zkPb.19391$1e.1...@newsread2.news.pas.earthlink.net...
> In the previous thread of this name I asked some questions
> which were never answered, so I would like to try again.
> I will discuss also a few of the previous replies.
So an answer is something *you* decide is an answer, but replies don't
count?
David A. Smith
Harold Ensle
"pholroyd" <phol...@REMOVEno-spam-allowed.com> wrote in message
news:20040120215901.655$N...@news.newsreader.com...
quibbling point accepted.
It doesn't change the argument, however.
H.Ellis Ensle
Harold Ensle
"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:n...@nospam.com> wrote in
message news:81mPb.7656$bg1.7390@fed1read05...
Quibbling point accepted.
I meant answered in the sense that it resolved the problem.
H.Ellis Ensle
pholroyd
what argument? you use wrong formula and wrong idea so make no
argument. you talk about only relative accelration but that because
you use wrong formula.
formula in section starting page 258 is for acceleration of one
accelrated clock. that correct formula and correct idea. yours not
either. wrong ideas give wrong concluding.
@@@ph@@@
Bilge
>which were never answered, so I would like to try again.
>I will discuss also a few of the previous replies.
>
>First, concerning some basic requirements of special relativity.
>Mathematically, it is defined by the Lorentz Transformations:
>
>x'=g(x-vt) (1)
>t'=g(t-vx/c^2) (2)
>
>where g=1/sqrt(1-v^2/c^2)
That is not true harold. Special relativity is defined by its
postulates. The postulates lead to the lorentz transforms because those
transforms fulfill the requirement that the laws of physics are the same
in all inertial frames. Your choice of arguments here is reduced to one of
the following: (A) You do not accept that the laws of physics are the same
in all inertial frames, or (B) You do not accept that the lorentz
transforms fulfill that requirement. I just proved that the lorentz
transforms _do_ fulfill that requirement in another thread by deriving
them based upon the assumption that the laws of physics were the same in
all inertial frames, leaving you with (A) as your only objection. I'm
willing to accept (A) as an argument for a different theory having
different postulatess, but you have no evidence to support that
hypothesis.
>
>(x is position, t is time, c is the speed of light
>and v is the relative velocity)
>
>Note that the Lorentz Transformations are based solely on
>_relative_ velocity. We can derive from these a general
>expression that relates the passage of time for two observers:
>
>t'=integral from t_1 to t_2(sqrt(1-v(t)^2/c^2))dt (3)
>
>This can be found in most relevant textbooks such as Moller
>or Jackson. Note that this equation incorporates acceleration.
>It is important to understand to which acceleration it refers.
>The acceleration here, since it directly correlates to the
>_relative_ velocity of the Lorentz Transformations is a
>_relative_ acceleration.
No, it doesn't harold and it's dishonest of you say that
this wasn't answered for you. I answered it in detail,.
>That is, it is the change in
>_relative_ velocity over time. It is a purely kinematic value.
>It tells us _nothing_ about what forces are present. It tells us
>only that the _relative_ velocity between the two observers
>is changing.
>
>Now I want to determine the time that has passed for both
>twins in the twin paradox. First for the stay-at-home one
>can use equation 3.
[...]
>SO the question is: What equation can the travelling twin
>use?
I've already explained to you that you can apply that to
both twins. Both twins perform the calculation with respect
to the same inertial frame in which they started. The inertial
frame in which both twins begin, is the frame from which their
proper times are calculated. You simply insist on using equations
it incorrectly. Using equations correctly isn't new to relativity,
harold. If you apply newtonian mechanics incorrectly, you will
get non-sense for an answer, too.
[...]
Harold Ensle
"pholroyd" <phol...@REMOVEno-spam-allowed.com> wrote in message
> Harold Ensle wrote:
> > "pholroyd" <phol...@REMOVEno-spam-allowed.com> wrote in message
> > news:20040120215901.655$N...@news.newsreader.com...
> >> Harold Ensle wrote:
> >>>
> >>> Note that the Lorentz Transformations are based solely on
> >>> _relative_ velocity. We can derive from these a general
> >>> expression that relates the passage of time for two observers:
> >>>
> >>> t'=integral from t_1 to t_2(sqrt(1-v(t)^2/c^2))dt (3)
> >>>
> >>> This can be found in most relevant textbooks such as Moller
> >>> or Jackson. Note that this equation incorporates acceleration.
> >>
> >> moller is one of books i use to study relativity and he dont say
> >> that equation incorporates acceleration. what he says is he
> >> assumes that acceleration
> >> of clock relative to inertial system does not influence clock
> >> rate. in my book this is page 49. much later in section called
> >> clock paraodox (page 258)
> >> he say that now he gives complete solution.
> >
> > quibbling point accepted.
> >
> > It doesn't change the argument, however.
> >
>
> what argument? you use wrong formula
Which formula is wrong? Be specific.
>and wrong idea
Which idea is wrong? Be specific.
>so make no
> argument. you talk about only relative accelration but that because
> you use wrong formula.
Then give me the correct formula. Is that so hard?
> formula in section starting page 258 is for acceleration of one
> accelrated clock.
I do not have Moller at present. It should not be difficult
for you to write a single formula. Why all the secrecy?
[...]
H.Ellis Ensle
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:8zkPb.19391$1e.1...@newsread2.news.pas.earthlink.net...
You still don't know what an utterly silly question this is, do you?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CannotApply.html
Face it, you don't have the intelligence nor the will
http://users.pandora.be/vdmoortel/dirk/Stuff/Monopoly.html
http://groups.google.com/groups?&threadm=inuC9.8670$Ti2....@afrodite.telenet-ops.be
to grasp the very basics, so what are you trying to understand the
tough part?
There are two possibilities, either you are a dishonest troll,
or you are utterly stupid.
Don't you know that feeling that we call embarrassment?
Shame maybe?
Ding ding?
Dirk Vdm
Harold Ensle
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
news:slrnc0s7uf....@radioactivex.lebesque-al.net...
> Harold Ensle:
> >In the previous thread of this name I asked some questions
> >which were never answered, so I would like to try again.
>
> Nonsense harold.
>
> >I will discuss also a few of the previous replies.
> >
> >First, concerning some basic requirements of special relativity.
> >Mathematically, it is defined by the Lorentz Transformations:
> >
> >x'=g(x-vt) (1)
> >t'=g(t-vx/c^2) (2)
> >
> >where g=1/sqrt(1-v^2/c^2)
>
> That is not true harold. Special relativity is defined by its
> postulates.
Are you saying that these equations do not represent SR?????
You can derive these equations from two postulates, but it is also
possible start with these and come back to the two postulates
again. Now, if there are other equations needed to represent
**special** relativity, then please reveal them. That is all I ask.
If you can't do that, then it is proof that you are a liar.
I can't even believe this. I have not yet come to anything
controversial and you are already complaining. If I claimed that
relativity was correct, you would probably argue, just because
I was the one who said it.
You obviously have issues that prevent you from discussing
anything with me in a rational manner.
>The postulates lead to the lorentz transforms because those
> transforms fulfill the requirement that the laws of physics are the same
> in all inertial frames. Your choice of arguments here is reduced to one of
> the following: (A) You do not accept that the laws of physics are the same
> in all inertial frames, or (B) You do not accept that the lorentz
> transforms fulfill that requirement. I just proved that the lorentz
> transforms _do_ fulfill that requirement in another thread by deriving
> them based upon the assumption that the laws of physics were the same in
> all inertial frames, leaving you with (A) as your only objection. I'm
> willing to accept (A) as an argument for a different theory having
> different postulatess, but you have no evidence to support that
> hypothesis.
Actually, since SR is self contradictory, it then follows that (A) is wrong.
That is: The absolute rest frame must have some physical significance.
The fact that it has not yet been detected (except possibly by Silvertooth)
means nothing, since, it could be that the correct experiment simply
has never been done.
> >
> >(x is position, t is time, c is the speed of light
> >and v is the relative velocity)
> >
> >Note that the Lorentz Transformations are based solely on
> >_relative_ velocity. We can derive from these a general
> >expression that relates the passage of time for two observers:
> >
> >t'=integral from t_1 to t_2(sqrt(1-v(t)^2/c^2))dt (3)
> >
> >This can be found in most relevant textbooks such as Moller
> >or Jackson. Note that this equation incorporates acceleration.
> >It is important to understand to which acceleration it refers.
> >The acceleration here, since it directly correlates to the
> >_relative_ velocity of the Lorentz Transformations is a
> >_relative_ acceleration.
>
> No, it doesn't harold
The acceleration in equation 3 is not relative?
This is a trivial point......again not even controversial.
Think....boy............what can you get when you take
the time derivative of a _relative_ velocity?
By arguing with all the wrong points, it just proves that
you really don't know what you're talking about.
>and it's dishonest of you say that
> this wasn't answered for you. I answered it in detail,
You did not answer the question. I asked for the equation
to be used by the travelling twin to determine his view
of the time of the stay-at-home. You did not supply it.
It is dishonest for you to say that you answered when
you really didn't.
> >That is, it is the change in
> >_relative_ velocity over time. It is a purely kinematic value.
> >It tells us _nothing_ about what forces are present. It tells us
> >only that the _relative_ velocity between the two observers
> >is changing.
> >
> >Now I want to determine the time that has passed for both
> >twins in the twin paradox. First for the stay-at-home one
> >can use equation 3.
>
> [...]
> >SO the question is: What equation can the travelling twin
> >use?
>
> I've already explained to you that you can apply that to
> both twins. Both twins perform the calculation with respect
> to the same inertial frame in which they started.
I want the equation that the twin uses in his own frame of
reference, not the stay-at-home's.
So you are claiming that the travelling twin at *all* times sees
the stay-at-home's time the exact same way that the stay-at-home
sees it? So you are claiming that there is no such thing as
mutual time dilation? Do you believe in relativity or not?
>The inertial
> frame in which both twins begin, is the frame from which their
> proper times are calculated. You simply insist on using equations
> it incorrectly. Using equations correctly isn't new to relativity,
> harold. If you apply newtonian mechanics incorrectly, you will
> get non-sense for an answer, too.
Let's go with this newtonian analogy. If I am in an accelerating frame
I can, by using virtual forces, determine the exact behavior of a
projectile in reference to my own frame.
Now I want to know the equation I can use, as the travelling
twin to do the same thing. It has to give me _mutual_ time
dilation when I am in an inertial frame and it has to give me
something else when I am accelerating. It has to be done in
relation to my own frame of reference at all times.
It can be done in Newtonian mechanics. Why can't it be done
in SR? What is the big secret?
BTW You and Andersen are apparently in disagreement
about which formula the traveling twin should use. Are
you claiming here that he is wrong also?
H.Ellis Ensle
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:_uBPb.21191$1e.1...@newsread2.news.pas.earthlink.net...
[snip]
> BTW You and Andersen are apparently in disagreement
> about which formula the traveling twin should use.
No embarrassment?
No shame?
No ding ding?
So actually, you are a dishonest troll *and* utterly stupid:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CannotApply.html
Dirk Vdm
Bill Hobba
Harold Ensle wrote:
> Are you saying that these equations do not represent SR?????
> You can derive these equations from two postulates, but it is also
> possible start with these and come back to the two postulates
> again. Now, if there are other equations needed to represent
> **special** relativity, then please reveal them. That is all I ask.
> If you can't do that, then it is proof that you are a liar.
Wrong again Harold. The equations your referring to (a variant of the
Lorentz transformations) do not imply the POR which is the basis of SR. It
is a one way implication not an iff. Either your not putting any real
thought into it or your being intellectually dishonest.
Thanks
Bill
Harold Ensle
"Bill Hobba" <bill...@yahoo.com.au> wrote in message
news:400efbd0$1...@news.iprimus.com.au...
>
> Harold Ensle wrote:
> > Are you saying that these equations do not represent SR?????
> > You can derive these equations from two postulates, but it is also
> > possible start with these and come back to the two postulates
> > again. Now, if there are other equations needed to represent
> > **special** relativity, then please reveal them. That is all I ask.
> > If you can't do that, then it is proof that you are a liar.
>
> Wrong again Harold. The equations your referring to (a variant of the
> Lorentz transformations) do not imply the POR which is the basis of SR.
It
> is a one way implication not an iff.
I believe that in a space without gravity it *is* an iff. However, it
actually
occurred to me that I might be wrong, so I finished out the paragraph
with the request for additional equations just to cover that possibility.
Of course, nobody provided any, so I suspect I was right all along.
>Either your not putting any real
> thought into it or your being intellectually dishonest.
I think you need to be more concerned with yourself.
H.Ellis Ensle
Harold Ensle
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:%EAPb.646$dB3.1...@phobos.telenet-ops.be...
[...]
Why is it silly? If it is simple, why don't you just write it out.
If it is not simple, why is it a silly question?
Surely an equation exists, so what is it?
H.Ellis Ensle
John Anderson
Bilge wrote:
That unduly restricts the answer. It doesn't matter whichframe you use to do
the calculation. The result is Lorentz
invariant.
> The inertial
> frame in which both twins begin, is the frame from which their
> proper times are calculated. You simply insist on using equations
> it incorrectly. Using equations correctly isn't new to relativity,
> harold. If you apply newtonian mechanics incorrectly, you will
> get non-sense for an answer, too.
>
Yes, that's the real point of Ensle's error. He's using equationsthat are
correct and assigning the wrong meaning to the symbols
in the equations. He wrote down valid equations and then
made incorrect claims about how the variables in those equations
are related to measurements that are made in experiments.
John Anderson
Harold Ensle
"John Anderson" <and...@attglobal.net> wrote in message
news:400F32DA...@attglobal.net...
>
> Yes, that's the real point of Ensle's error. He's using equationsthat are
> correct and assigning the wrong meaning to the symbols
> in the equations. He wrote down valid equations and then
> made incorrect claims about how the variables in those equations
> are related to measurements that are made in experiments.
Where? Which claims were incorrect?
(Do try and actually read the original post.)
H.Ellis Ensle
Bilge
>"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
>news:slrnc0s7uf....@radioactivex.lebesque-al.net...
>>
>> postulates.
>
>Are you saying that these equations do not represent SR?????
Those equations are a consequence of special relativity. Special
relativity is physics. The equations follow from the physics. If the
postulates led to different equations, different equations would apply.
Theories in physics aren't built around equations. Equations follow the
physics. A lot of your difficulty seems to be confusing the two.
>You can derive these equations from two postulates, but it is also
>possible start with these and come back to the two postulates
>again.
No, it isn't harold. Those equations are nothing more than
equations related to conic sections. Tell me harold, what
physical theory can you derive from:
x' = x cos(A) + y sin(A)
y' = y cos(A) - x sin(A)
Since I've written the lorentz transforms in hyperbolic form
a zillion times, you should note the resemblence to the lorentz
boosts.
>Now, if there are other equations needed to represent
>**special** relativity, then please reveal them. That is all I ask.
>If you can't do that, then it is proof that you are a liar.
You seem to mistake mathematics for physics, harold. Special
relativity is a theory of physics. The equations and the correct
usage of the equations is dictated by physics.
>I can't even believe this.
That's no surprise. You can't believe anything.
>I have not yet come to anything controversial and you are already
>complaining. If I claimed that relativity was correct, you would
>probably argue, just because I was the one who said it.
I would argue if you thought it was correct for the wrong reason.
>You obviously have issues that prevent you from discussing anything
>with me in a rational manner.
Right, harold. That's why you've spent the entire post so far
just ranting because I said that the postulates of relativity
are the theory, not the equations.
>>The postulates lead to the lorentz transforms because those
>> transforms fulfill the requirement that the laws of physics are the same
>> in all inertial frames. Your choice of arguments here is reduced to one of
>> the following: (A) You do not accept that the laws of physics are the same
>> in all inertial frames, or (B) You do not accept that the lorentz
>> transforms fulfill that requirement. I just proved that the lorentz
>> transforms _do_ fulfill that requirement in another thread by deriving
>> them based upon the assumption that the laws of physics were the same in
>> all inertial frames, leaving you with (A) as your only objection. I'm
>> willing to accept (A) as an argument for a different theory having
>> different postulatess, but you have no evidence to support that
>> hypothesis.
>
>Actually, since SR is self contradictory, it then follows that (A) is wrong.
I'll note that you think the geometry of the conic sections is
inconsistent.
>That is: The absolute rest frame must have some physical significance.
>The fact that it has not yet been detected (except possibly by Silvertooth)
>means nothing, since, it could be that the correct experiment simply
>has never been done.
If an absolute rest frame existed, it would have physical
significance. If you find an experiment that depends upon
that frame, let me know.
[...]
>> >The acceleration here, since it directly correlates to the
>> >_relative_ velocity of the Lorentz Transformations is a
>> >_relative_ acceleration.
>>
>> No, it doesn't harold
>
>The acceleration in equation 3 is not relative?
>This is a trivial point......again not even controversial.
I've told you precisely how to apply that equation.
I don't care what semantics games you want to play.
Either tell me what's wrong with applying as I told
you, or shut up.
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:DXEPb.22597$zj7....@newsread1.news.pas.earthlink.net...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:%EAPb.646$dB3.1...@phobos.telenet-ops.be...
> >
> > "Harold Ensle" <hee...@ix.netcom.com> wrote in message
> news:8zkPb.19391$1e.1...@newsread2.news.pas.earthlink.net...
[snip]
> > > SO the question is: What equation can the travelling twin
> > > use?
> >
> > You still don't know what an utterly silly question this is, do you?
>
> [...]
>
> Why is it silly? If it is simple, why don't you just write it out.
> If it is not simple, why is it a silly question?
> Surely an equation exists, so what is it?
You don't even want to know:
http://users.pandora.be/vdmoortel/dirk/Stuff/Monopoly.html
http://groups.google.com/groups?&threadm=inuC9.8670$Ti2....@afrodite.telenet-ops.be
Dirk Vdm
Androcles
"Harold Ensle" <hee...@ix.netcom.com> wrote in message
news:8zkPb.19391$1e.1...@newsread2.news.pas.earthlink.net...
> which were never answered, so I would like to try again.
> I will discuss also a few of the previous replies.
>
> First, concerning some basic requirements of special relativity.
> Mathematically, it is defined by the Lorentz Transformations:
>
> x'=g(x-vt) (1)
> t'=g(t-vx/c^2) (2)
>
> where g=1/sqrt(1-v^2/c^2)
[snip]
So, unless 'a' and 'x' have some special meanings, Andersen's
> equation fails miserably.
>
> H.Ellis Ensle
x certainly does.
If you examine Einstein's paper,
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/
You will find in section 3 the definition x' = x-vt.
Obviously this can be rearranged as x = x' +vt.
Further down the page you'll see that Einstein takes x' as infinitessimally
small;
Hence x = (almost nothing) + vt.
So...
(I'll use xi here as Einstein does so as to not confuse it with x' , even
though it is the same)
xi=g(x-vt) = g(vt-vt) = 0
tau =g(t-tv^2/c^2) = gt.(1-v^2/c^2) = tg^2/g = t.sqrt(1-v^2/c^2)
Since xi = 0, there has been no displacement from 0 between the frames,
hence v = 0
and tau = t.sqrt(1- 0/c^2) = t.sqrt(1) = t.1 = t
So, Harold, all your suspicions are justified, relativity is complete and
utter nonsense.
although it is quite valid in the trivial case of v = 0.
Androcles
Paul B. Andersen
"Harold Ensle" <hee...@ix.netcom.com> skrev i melding news:8zkPb.19391$1e.1...@newsread2.news.pas.earthlink.net...
Of course.
B is accelerating away from A, so A is at a lower
"pseudo force potential" (or gravitational potential, if you like).
> On the
> return journey the same thing happens.
When B is breaking, yes.
Same as above.
> So it appears that
> the paradox is actually worse than before.
Nonsense.
You have ignored what happens when B is accelerating towards A.
Then B is at a higher potential, and the greater distance will more
than compensate for the time "lost" when accereating away and breaking.
> So, unless 'a' and 'x' have some special meanings, Andersen's
> equation fails miserably.
>
> H.Ellis Ensle
So let's see what the equation really has to say about your scenario.
I have run it throught a computer program doing the integration
numerically.
Home twin A, coordinates x,t.
Travelling twin B, coordinates x',t'.
B is accelerating away from A for 1 LY i A's frame,
reversing the direction of the acceleration, until he again
is 1 LY from home in A's frame.
Reversing his acceleration again, to break.
The acceleration is c per year, which happens to be ca. 1g.
In A's inertial frame:
t t' dt'/dt v/c x
[Y] [Y] [LY]
0.0 0.0 1.00 0.00 0.00
0.1 0.1 1.00 0.10 0.00
0.2 0.2 0.98 0.20 0.02
0.3 0.3 0.96 0.29 0.04
0.4 0.4 0.93 0.37 0.08
0.5 0.5 0.89 0.45 0.12
0.6 0.6 0.86 0.51 0.17
0.7 0.7 0.82 0.57 0.22
0.8 0.7 0.78 0.62 0.28
0.9 0.8 0.74 0.67 0.35
1.0 0.9 0.71 0.71 0.41
1.1 1.0 0.67 0.74 0.49
1.2 1.0 0.64 0.77 0.56
1.3 1.1 0.61 0.79 0.64
1.4 1.1 0.58 0.81 0.72
1.5 1.2 0.55 0.83 0.80
1.6 1.2 0.53 0.85 0.89
1.7 1.3 0.51 0.86 0.97
1.8 1.4 0.52 0.86 1.06
1.9 1.4 0.54 0.84 1.14
2.0 1.5 0.56 0.83 1.23
2.1 1.5 0.59 0.81 1.31
2.2 1.6 0.62 0.78 1.39
2.3 1.6 0.65 0.76 1.47
2.4 1.7 0.68 0.73 1.54
2.5 1.8 0.72 0.69 1.61
2.6 1.9 0.76 0.65 1.68
2.7 1.9 0.79 0.61 1.74
2.8 2.0 0.83 0.55 1.80
2.9 2.1 0.87 0.49 1.85
3.0 2.2 0.91 0.42 1.90
3.1 2.3 0.94 0.34 1.94
3.2 2.4 0.97 0.26 1.97
3.3 2.5 0.99 0.16 1.99
3.4 2.6 1.00 0.06 2.00
3.5 2.7 1.00 -0.04 2.00
3.6 2.8 0.99 -0.13 1.99
3.7 2.9 0.97 -0.23 1.97
3.8 3.0 0.95 -0.32 1.95
3.9 3.1 0.92 -0.40 1.91
4.0 3.1 0.88 -0.47 1.87
4.1 3.2 0.84 -0.54 1.81
4.2 3.3 0.81 -0.59 1.76
4.3 3.4 0.77 -0.64 1.70
4.4 3.5 0.73 -0.68 1.63
4.5 3.5 0.69 -0.72 1.56
4.6 3.6 0.66 -0.75 1.49
4.7 3.7 0.63 -0.78 1.41
4.8 3.7 0.60 -0.80 1.33
4.9 3.8 0.57 -0.82 1.25
5.0 3.8 0.55 -0.84 1.17
5.1 3.9 0.52 -0.85 1.08
5.2 4.0 0.50 -0.87 1.00
5.3 4.0 0.52 -0.85 0.91
5.4 4.1 0.55 -0.84 0.83
5.5 4.1 0.57 -0.82 0.74
5.6 4.2 0.60 -0.80 0.66
5.7 4.2 0.63 -0.78 0.58
5.8 4.3 0.66 -0.75 0.51
5.9 4.4 0.70 -0.72 0.43
6.0 4.4 0.73 -0.68 0.36
6.1 4.5 0.77 -0.64 0.30
6.2 4.6 0.81 -0.59 0.24
6.3 4.7 0.85 -0.53 0.18
6.4 4.8 0.88 -0.47 0.13
6.5 4.9 0.92 -0.39 0.09
6.6 4.9 0.95 -0.31 0.05
6.7 5.0 0.97 -0.22 0.03
6.8 5.1 0.99 -0.13 0.01
6.9 5.2 1.00 -0.03 0.00
In B's accelerated frame:
t' t dt/dt' v/c x'
[Y] [Y] [LY]
0.0 0.0 1.00 0.00 0.00
0.1 0.1 0.99 -0.10 0.00
0.2 0.2 0.96 -0.20 -0.02
0.3 0.3 0.92 -0.29 -0.04
0.4 0.4 0.86 -0.38 -0.07
0.5 0.5 0.79 -0.46 -0.11
0.6 0.5 0.71 -0.54 -0.16
0.7 0.6 0.63 -0.60 -0.20
0.8 0.7 0.56 -0.66 -0.25
0.9 0.7 0.49 -0.72 -0.30
1.0 0.8 0.42 -0.76 -0.35
1.1 0.8 0.36 -0.80 -0.40
1.2 0.8 0.31 -0.83 -0.45
1.3 0.9 0.26 -0.86 -0.49
1.4 0.9 0.86 -0.84 -0.61
1.5 1.0 1.02 -0.81 -0.75
1.6 1.1 1.20 -0.78 -0.89
1.7 1.3 1.39 -0.73 -1.04
1.8 1.4 1.60 -0.68 -1.19
1.9 1.6 1.83 -0.63 -1.34
2.0 1.8 2.06 -0.56 -1.48
2.1 2.0 2.28 -0.49 -1.62
2.2 2.2 2.50 -0.41 -1.74
2.3 2.5 2.69 -0.32 -1.84
2.4 2.8 2.84 -0.23 -1.92
2.5 3.1 2.95 -0.13 -1.97
2.6 3.4 3.00 -0.03 -2.00
2.7 3.7 2.99 0.07 -1.99
2.8 4.0 2.92 0.16 -1.96
2.9 4.2 2.80 0.26 -1.90
3.0 4.5 2.63 0.35 -1.81
3.1 4.8 2.43 0.44 -1.70
3.2 5.0 2.21 0.51 -1.58
3.3 5.2 1.98 0.58 -1.44
3.4 5.4 1.75 0.64 -1.29
3.5 5.6 1.53 0.70 -1.14
3.6 5.7 1.33 0.75 -0.99
3.7 5.8 1.14 0.79 -0.85
3.8 5.9 0.97 0.82 -0.70
3.9 6.0 0.82 0.85 -0.57
4.0 6.1 0.27 0.85 -0.48
4.1 6.1 0.32 0.82 -0.43
4.2 6.1 0.38 0.79 -0.39
4.3 6.2 0.44 0.75 -0.34
4.4 6.2 0.51 0.70 -0.29
4.5 6.3 0.58 0.65 -0.24
4.6 6.3 0.66 0.58 -0.19
4.7 6.4 0.74 0.51 -0.14
4.8 6.5 0.81 0.44 -0.10
4.9 6.6 0.88 0.35 -0.06
5.0 6.7 0.93 0.26 -0.03
5.1 6.8 0.97 0.17 -0.01
5.2 6.9 1.00 0.04 0.00
The twins will obviously agree on the proper times of
their clocks when reunited.
A ages 6.9 years, B ages 5.2 years during the journey.
The equation seems to work just fine.
Paul
Harold Ensle
> Harold Ensle:
> >
> >"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
> >news:slrnc0s7uf....@radioactivex.lebesque-al.net...
[........lies snipped.....]
> >>The postulates lead to the lorentz transforms because those
> >> transforms fulfill the requirement that the laws of physics are the
same
> >> in all inertial frames. Your choice of arguments here is reduced to
one of
> >> the following: (A) You do not accept that the laws of physics are the
same
> >> in all inertial frames, or (B) You do not accept that the lorentz
> >> transforms fulfill that requirement. I just proved that the lorentz
> >> transforms _do_ fulfill that requirement in another thread by deriving
> >> them based upon the assumption that the laws of physics were the same
in
> >> all inertial frames, leaving you with (A) as your only objection. I'm
> >> willing to accept (A) as an argument for a different theory having
> >> different postulatess, but you have no evidence to support that
> >> hypothesis.
> >
> >Actually, since SR is self contradictory, it then follows that (A) is
wrong.
>
> I'll note that you think the geometry of the conic sections is
> inconsistent.
Gad...you are so stupid. I didn't say this! I said (A)!!!! can't
you even understand your own posts?
> >That is: The absolute rest frame must have some physical significance.
> >The fact that it has not yet been detected (except possibly by
Silvertooth)
> >means nothing, since, it could be that the correct experiment simply
> >has never been done.
>
> If an absolute rest frame existed, it would have physical
> significance. If you find an experiment that depends upon
> that frame, let me know.
>
> [...]
> >> >The acceleration here, since it directly correlates to the
> >> >_relative_ velocity of the Lorentz Transformations is a
> >> >_relative_ acceleration.
> >>
> >> No, it doesn't harold
> >
> >The acceleration in equation 3 is not relative?
> >This is a trivial point......again not even controversial.
>
> I've told you precisely how to apply that equation.
> I don't care what semantics games you want to play.
> Either tell me what's wrong with applying as I told
> you, or shut up.
And you snipped everything that follows because you
have no reply.
Face it, you screwed up your previous post
(and this one) royally.
H.Ellis Ensle
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:buphbm$m65$1...@dolly.uninett.no...
[snip]
> So let's see what the equation really has to say about your scenario.
> I have run it throught a computer program doing the integration
> numerically.
>
> Home twin A, coordinates x,t.
> Travelling twin B, coordinates x',t'.
>
> B is accelerating away from A for 1 LY i A's frame,
> reversing the direction of the acceleration, until he again
> is 1 LY from home in A's frame.
> Reversing his acceleration again, to break.
> The acceleration is c per year, which happens to be ca. 1g.
>
> In A's inertial frame:
>
> t t' dt'/dt v/c x
> [Y] [Y] [LY]
> 0.0 0.0 1.00 0.00 0.00
[snip]
> 6.9 5.2 1.00 -0.03 0.00
>
> In B's accelerated frame:
>
> t' t dt/dt' v/c x'
> [Y] [Y] [LY]
> 0.0 0.0 1.00 0.00 0.00
[snip]
> 5.2 6.9 1.00 0.04 0.00
>
> The twins will obviously agree on the proper times of
> their clocks when reunited.
> A ages 6.9 years, B ages 5.2 years during the journey.
>
> The equation seems to work just fine.
Of course they work fine ;-)
Arbitrarily accelerated motion:
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
Travelling twin B, time coordinate T
t(T) = c/a sinh(aT/c)
T(t) = c/a argsinh(at/c)
Your case:
a/c = 1
4 phases: acceleration, deceleration acceleration, deceleration
4*argsinh(6.90/4) = 5.25
4*sinh(5.25/4) = 6.89
Picture of half trip:
http://users.pandora.be/vdmoortel/dirk/Stuff/AccelDecel.gif
Dirk Vdm
Dirk Van de moortel
[snip]
> Of course they work fine ;-)
> Arbitrarily accelerated motion:
> http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
> Travelling twin B, time coordinate T
> t(T) = c/a sinh(aT/c)
> T(t) = c/a argsinh(at/c)
> Your case:
> a/c = 1
> 4 phases: acceleration, deceleration acceleration, deceleration
> 4*argsinh(6.90/4) = 5.25
> 4*sinh(5.25/4) = 6.89
> Picture of half trip:
> http://users.pandora.be/vdmoortel/dirk/Stuff/AccelDecel.gif
with slightly different notation of course:
T in text ==> t' on gif picture
6.90 in text ==> 4T on gif picture
Dirk Vdm
Cesar Sirvent
Which would be perfect if you calculated the ellapsed time for the Earth
twin A from the PoV of the traveller twin and would show that it coincides
with the ellapsed time as measured by A herself.
Benno Muilwyk
In this form, x should actually read x(t), because it changes
over time. Naturally, all variables are as measured in the
accelerating frame.
> Since equation 3 is the time dilatation
> derived from the Lorentz transformations, this equation is
> clearly not derived, or derivable from SR alone.
> Because, if it were, it would have to be identical to
> equation 3.
Note that equation 3 is only applicable when v is constant
and thus a=0 between t1 and t2. When you fill in a=0 in
equation 6, you get equation 3.
> Now, if this equation has a legitimate source, does it
> solve the paradox? Well...apparently, it doesn't even do that.
> Note that to solve the paradox, this equation must provide
> for a round trip a solution for the travelling twin such that:
>
> ts=tt*k (7)
>
> So that both twins will agree on their reunion. Thus at
> some point in the travelling twin's trip he must see the
> stay-at-home's time accumulating faster than his own.
>
> Imagine a trip where the travelling twin accelerates for the
> first half of his outward journey and then deccelerates
> for the rest of the way, and then does the same process to
> return home.
>
> First 'a' is positive and then 'a' is negative when 'x' is larger.
During the first half of the outward journey, the travelling twin
feels an outward acceleration, while he sees the stay-at-home
twin move in the opposite direction. This means that a and x
have opposite signs (ax < 0) in this phase.
The next phase, the accelerating force is reversed and points
in the same direction as the stay-at-home twin, so a and x
have the same sign (ax > 0).
So the term ax in equation 6 is positive for the 'far away' part
of the journey, where the absolute value of x is the greatest.
When you apply equation 6 correctly, with the right sign of
the term ax, you will find that it does resolve the paradox.
And when you find a different result, you'd better ask where
you went wrong yourself, instead of blaming the formula or SR.
>
> H.Ellis Ensle
>
Benno Muilwijk
Dirk Van de moortel
"Cesar Sirvent" <8UMU...@SPAxMterrMAPSa.esSPAxM> wrote in message news:sqYPb.2818825$uj6.7...@telenews.teleline.es...
... which is what Paul did numerically :-)
Dirk Vdm
Bilge
>
>Gad...you are so stupid. I didn't say this! I said (A)!!!! can't
>you even understand your own posts?
wrote in (A):
(A) You do not accept that the laws of physics are the same
in all inertial frames,
rotations:
x' = x cos(A) + y sin(A)
y' = y cos(A) - y sin(A)
I have to assume you mean that this coordinate transformation also
is invalid, since you have given no other theory in which the laws
of physics would be identical in inertial frames which differ by a
rotation in the x-y plane but would differ in frames which would be
rotations in the x-t plane:
x' = x cosh(A) - t sinh(A)
t' = t cosh(A) - x sinh(A)
The point here harold, is that you don't understand the physics in
special relativity. All you know are that some equations exist for
transforming coordinates, but you don't know what they mean or how
they are derived in any way doesn't appear to be some variation on
lorentz' original derivation.
>> I've told you precisely how to apply that equation.
>> I don't care what semantics games you want to play.
>> Either tell me what's wrong with applying as I told
>> you, or shut up.
>
>And you snipped everything that follows because you have no reply.
No, I snipped it because there is no point in addressing numerous
objections based upon the same misunderstanding. Surely you can resolve
one of those points at a time. I'll address each and every one in
turn, upon resolving the ones preceeding it, if that's what you wish.
I'm only going to address one thing at a time, however, so that you
focus on it and don't digress.
>Face it, you screwed up your previous post (and this one) royally.
You always say something to that effect, so you should put it in
your signature block.
Cesar Sirvent
So the analytical solution must be quite hard? In any case, where is dt_A /
dt_B ? I am curious.
Bilge
>
>> I've already explained to you that you can apply that to
>> both twins. Both twins perform the calculation with respect
>> to the same inertial frame in which they started.
>
>That unduly restricts the answer. It doesn't matter whichframe you
>use to do the calculation. The result is Lorentz invariant.
Sure, but harold provided the restriction. He insists that the equation
has to apply to both twins equivalently _and_ that the acceleration be
considered only relative, to try and find a semantics loophole to exploit.
The simplest way to avoid the semantics arguement was to take the starting
frame to be inertial an put both twins in it and then compute the elapsed
time of each twin relative to the starting frame, which is inertial by
definition.
>> The inertial
>> frame in which both twins begin, is the frame from which their
>> proper times are calculated. You simply insist on using equations
>> it incorrectly. Using equations correctly isn't new to relativity,
>> harold. If you apply newtonian mechanics incorrectly, you will
>> get non-sense for an answer, too.
>>
>
>Yes, that's the real point of Ensle's error. He's using equationsthat are
>correct and assigning the wrong meaning to the symbols
>in the equations. He wrote down valid equations and then
>made incorrect claims about how the variables in those equations
>are related to measurements that are made in experiments.
Right. I simply provided the means to render his objections moot
rather than try to convice him of anything more general. He seems
to have difficulty if the meanings of the variables are not what
he thinks they are.
Dirk Van de moortel
"Benno Muilwyk" <be...@muilwijk-met-wyk.nl> wrote in message news:40105...@news3.prserv.net...
> > In the previous thread of this name I asked some questions
> > which were never answered, so I would like to try again.
> > I will discuss also a few of the previous replies.
> >
> > First, concerning some basic requirements of special relativity.
> > Mathematically, it is defined by the Lorentz Transformations:
> >
> > x'=g(x-vt) (1)
> > t'=g(t-vx/c^2) (2)
> >
> > where g=1/sqrt(1-v^2/c^2)
> >
> > (x is position, t is time, c is the speed of light
> > and v is the relative velocity)
> >
> > Note that the Lorentz Transformations are based solely on
> > _relative_ velocity. We can derive from these a general
> > expression that relates the passage of time for two observers:
> >
> > t'=integral from t_1 to t_2(sqrt(1-v(t)^2/c^2))dt (3)
> >
[snip]
> >
> > t' = integral from t1 to t2(sqrt(1-v(t)^2/c^2)(1 + ax/c^2))dt (6)
> > where a is the acceleration of the x-t frame.
>
> In this form, x should actually read x(t), because it changes
> over time. Naturally, all variables are as measured in the
> accelerating frame.
>
> > Since equation 3 is the time dilatation
> > derived from the Lorentz transformations, this equation is
> > clearly not derived, or derivable from SR alone.
> > Because, if it were, it would have to be identical to
> > equation 3.
>
> Note that equation 3 is only applicable when v is constant
> and thus a=0 between t1 and t2. When you fill in a=0 in
> equation 6, you get equation 3.
Hm, I do't go along with that. Eq. 3 is valid for any v, provided
t is the time coordinate used by an inertial observer and t' is the
time read on an arbitrarily accelerated clock, that has velocity
v(t) at time t according to the inertial observer. This velocity
can be any function since the clock can be arbitrarily
accelerated.
Eq. 6 is valid provided t is the time kept by an arbitrarily
accelerated observer and t' is the proper time of the
inertial observer.
Actually, keeping the same convention as in eq 3, eq. 6
should be
t = integral from t1' to t2'(sqrt(1-v'(t')^2/c^2)(1 + ax'/c^2))dt' (6')
where a=0 would again not imply v'(t') = 0.
Dirk Vdm
Dirk Van de moortel
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:u7ZPb.882$xt4.1...@phobos.telenet-ops.be...
[snip]
> Eq. 6 is valid provided t is the time kept by an arbitrarily
> accelerated observer and t' is the proper time of the
> inertial observer.
> Actually, keeping the same convention as in eq 3, eq. 6
> should be
> t = integral from t1' to t2'(sqrt(1-v'(t')^2/c^2)(1 + ax'/c^2))dt' (6')
> where a=0 would again not imply v'(t') = 0.
erm sorry.... a=0 would indeed imply v'(t') is constant because
in this special case we have that t is the time of an *inertial* observer.
If he would be allowed to be accelerated too, the a=0 would of
course not imply that v'(t') is constant.
Dirk Vdm
pholroyd
you another silly guy. i talk to you because you use book refernce
which i now reading but you do not read book? no secrecy just read
book which you refer.
too many people here change subject all time. i try to explaIn best i
can to you about accelkeration and what moller say but you not read
what moller say. maybe some one explain better to you. i still
learning relativity but even understand what moller say. why you speak
of moller for refer if you no read him?maybe bilge of pmb explain
better to you.
@@@ph@@@
Paul B. Andersen
"Paul B. Andersen" <paul.b....@hia.no> skrev i melding news:buphbm$m65$1...@dolly.uninett.no...
>
> "Harold Ensle" <hee...@ix.netcom.com> skrev i melding news:8zkPb.19391$1e.1...@newsread2.news.pas.earthlink.net...
> > Thus the travelling twin will see even less accumulated time
> > for the stay-at-home than when he used equation 3.
>
> Of course.
> B is accelerating away from A, so A is at a lower
> "pseudo force potential" (or gravitational potential, if you like).
>
> > On the
> > return journey the same thing happens.
>
> When B is breaking, yes.
> Same as above.
>
> > So it appears that
> > the paradox is actually worse than before.
>
> Nonsense.
> You have ignored what happens when B is accelerating towards A.
> Then B is at a higher potential, and the greater distance will more
> than compensate for the time "lost" when accereating away and breaking.
That should have been:
Then A is at a higher potential, ...
Paul
Gauge
You have Moller? Is it a library copy or were you able to purchase it
somewhere? I'd love to buy a copy of that text! If you did buy a copy
can you tell me where and how I can get a copy?
Thanks
Pmb
Cesar Sirvent
Can you give a step-by-step derivation of 6'? Thanks...
Bill Hobba
> > Wrong again Harold. The equations your referring to (a variant of the
> > Lorentz transformations) do not imply the POR which is the basis of SR.
> It
> > is a one way implication not an iff.
Bill Hobba replied:
>
> I believe that in a space without gravity it *is* an iff. However, it
> actually
> occurred to me that I might be wrong, so I finished out the paragraph
> with the request for additional equations just to cover that possibility.
>
That being the case then you will be able to post your derivation of the
POR from the Lorentz transformations. The POR being all the laws of physics
are the same in all inertial reference frames. Good luck - you will need it
because it is not logically possible. Indeed I would be very interested in
any set of equations that implies all the laws of physics - very interested
indeed. Think a bit harder in future.
Thanks
Bill
Dirk Van de moortel
"Cesar Sirvent" <8UMU...@SPAxMterrMAPSa.esSPAxM> wrote in message news:FqfQb.2849869$uj6.7...@telenews.teleline.es...
You could start from equations 6.17 of &6.6 in MTW Gravitation
(page 173), but let's, for example, start from equations 3.3.9 On
David Waite's site:
http://www.geocities.com/zcphysicsms/chap3.htm
with c = 1
{ t = (1/a+x')sinh(at') [1]
{ x = (1/a+x')cosh(at') - 1/a [2]
(t,x) are coordinates in inertial frame, and (t',x') in accelerated frame,
a is proper acceleration.
differentiate [1] and [2]
dt = dx' sinh(at') + (1/a+x') a cosh(at') dt' [3]
dx = dx' cosh(at') + (1/a+x') a sinh(at') dt' [4]
Since we are interested in time intervals on the worldline
of x we have dx = 0, so [4] becomes
dx' = - (1/a+x') a tanh(at') dt'
Inserting this into [3] gives
dt = - (1+ax') sinh(at') tanh(at') dt' + (1+ax') cosh(at') dt'
= ( -sinh^2(at')/cosh(at') + cosh(at') ) (1+ax') dt'
= 1/cosh(at') (1+ax') dt'
= 1/gamma (1+ax') dt' (see eq. 3.3.7)
= sqrt(1-v'(t')^2) (1+ax') dt'
and thus
t = int{ t1' to t2', sqrt(1-v'(t')^2) (1+ax') dt' }
Dirk Vdm
Cesar Sirvent
Ok. Thanks.
Tomorrow I will present my idea that this t in your last formula can be
presented directly in two different ways, with no need to your previous
calculations. I think.
Good night :-)
Tom
"Cesar Sirvent" <8UMU...@SPAxMterrMAPSa.esSPAxM> wrote in message
news:FqfQb.2849869$uj6.7...@telenews.teleline.es...
>
at
http://www.geocities.com/slithytove5/mypage.html
Unfortunately, I am very inexperienced with constructing web pages. The
best I am able to do is provide a link to a Microsoft Word document. So,
you will need Word in order to view it.
The derivation is for the rate of one accelerating clock as 'observed' from
another accelerating clock and, so, it does not presume that either clock is
at rest in an inertial frame. I thought out this derivation a few months
ago as a result of a discussion in another thread in sci.physics.relativity.
I make no claim of priority, originality, elegance, or conciseness for the
derivation. (Especially conciseness). Critcisms of the derivation are, of
course, welcome.
Tom
(remove 'z' to email)
Harold Ensle
OK I want to get this straight. When I did the problem, positive
acceleration (a) was always _away_ from the stay-at-home and
distance (x) was always the distance to the stay-at-home.
Now I did not transform either of these values as I assume they
are what is observed by the travelling twin. So when the travelling
twin accelerates, his view of the resulting velocity and displacement
are just derivatives (due to invariance).
Is this what you did?
H.Ellis Ensle
Harold Ensle
Ahhh......that is what I did wrong. I called the initial acceleration
of the travelling twin as positive.
Thank you. I will redo the problem as you describe here.
H.Ellis Ensle
Cesar Sirvent
Well, the one before the last equation can be writen directly.
Only suffices to realize that the primed observer undergoes a constant
proper acceleration a.
She is passing from one inertial observer to another one constantly.
This "differential passing" goes associated with a time correction which is
x'dv'.
As dv' = adt' we get ax'dt'.
This term must be added to any calculation of differential proper time in
the inertial system, but multiplied by 1/gamma.
Thus we get
dt = sqrt(1-v'(t')^2) dt' + sqrt(1 - v'(t')^2) * a * x' dt'
and integrating
t = int{t1' to t2', [ sqrt(1-v'(t')^2) + ax'*sqrt(1-v'(t')^2)] dt' }
which is what we wanted. No need of starting with [1] and [2].
For a discussion on the technic of the "set of comoving inertial systems",
you can see C. Moller.
Notice that my derivation is almost words and concepts. But it has universal
validity, i.e., you can use it to calculate dt/dt' in any kind of complex
movement.
Dirk Van de moortel
"Tom" <tsnyd...@insightbb.com> wrote in message news:GHiQb.105504$Rc4.735767@attbi_s54...
[snip]
> For anyone interested, I have provided a derivation of (6) that is available
> at
>
> http://www.geocities.com/slithytove5/mypage.html
>
> Unfortunately, I am very inexperienced with constructing web pages. The
> best I am able to do is provide a link to a Microsoft Word document. So,
> you will need Word in order to view it.
>
> The derivation is for the rate of one accelerating clock as 'observed' from
> another accelerating clock and, so, it does not presume that either clock is
> at rest in an inertial frame. I thought out this derivation a few months
> ago as a result of a discussion in another thread in sci.physics.relativity.
> I make no claim of priority, originality, elegance, or conciseness for the
> derivation. (Especially conciseness). Critcisms of the derivation are, of
> course, welcome.
>
> Tom
>
*Very* nice and well done - i.m.o. as elegant and concise as
it can possibly get!
Note that your equation (16) but with Bert coinciding with the
inertial Earth frame directly gives the desired result since
x_Bert(t) = v_Bert(t) = 0 for all t.
By the way, I'm sure someone is willing to convert the MS-word
file to a pdf. If you don't find anyone to do it, I'll do it on Monday
at the office...
Cheers,
Dirk Vdm
Tom
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:datQb.3128$YA5.2...@phobos.telenet-ops.be...
Thanks for your comments, Dirk. I didn't think about converting the file to
pdf (!) I believe we have the capabiltiy of doing this at my work. So I'll
try on Monday. If not, I might take you up on your offer.
Tom S.
Dirk Van de moortel
"Tom" <tsnyd...@insightbb.com> wrote in message news:JPuQb.110573$sv6.548930@attbi_s52...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:datQb.3128$YA5.2...@phobos.telenet-ops.be...
> >
[snip]
> > By the way, I'm sure someone is willing to convert the MS-word
> > file to a pdf. If you don't find anyone to do it, I'll do it on Monday
> > at the office...
> >
> > Cheers,
> > Dirk Vdm
> >
>
> Thanks for your comments, Dirk. I didn't think about converting the file to
> pdf (!) I believe we have the capabiltiy of doing this at my work. So I'll
> try on Monday. If not, I might take you up on your offer.
No problem.
By the way, did you know that MS-Word can save as html?
I'll send you a zipped version and some instructions...
Dirk Vdm
Tom S.
>
> For anyone interested, I have provided a derivation of (6) that is
available
> at
>
> http://www.geocities.com/slithytove5/mypage.html
>
> Unfortunately, I am very inexperienced with constructing web pages. The
> best I am able to do is provide a link to a Microsoft Word document. So,
> you will need Word in order to view it.
>
> The derivation is for the rate of one accelerating clock
Dirk Vdm kindly converted my Word doc to html. It is now available at
http://www.geocities.com/slithytove5/mypage.html
Tom S.
(Remove the 'z' to email)
Benno Muilwyk
> > feels an outward acceleration, while he sees the stay-at-home
> > twin move in the opposite direction. This means that a and x
> > have opposite signs (ax < 0) in this phase.
> > The next phase, the accelerating force is reversed and points
> > in the same direction as the stay-at-home twin, so a and x
> > have the same sign (ax > 0).
> > So the term ax in equation 6 is positive for the 'far away' part
> > of the journey, where the absolute value of x is the greatest.
>
> Ahhh......that is what I did wrong. I called the initial acceleration
> of the travelling twin as positive.
There is nothing wrong with defining the initial acceleration of the
travelling twing as positive. Then, from the stay-at-home twin's
perspective, the travelling twin moves away with (increasing)
positive x coordinate. But from the travelling twin's perspective,
the stay-at-home twin is moving away in the opposite direction,
with negative x coordinate.
Note that a is the proper acceleration of the travelling twin's frame
of refererence and _not_ the apparent acceleration of the stay-at-
home twin from the travelling twin's perspective!
Benno Muilwijk
Harold Ensle
OK I redid the problem and was able to get the correct time
for the stay-at-home from the travelling twin's perspective.
However, how do I deal with the travelling twin's view of his
own time?
H.Ellis Ensle
Benno Muilwyk
"Harold Ensle" <hee...@ix.netcom.com> schreef in bericht
news:hnHQb.26605$zj7....@newsread1.news.pas.earthlink.net...
dt = (c/a) d(phi) where phi = arctanh(v/c)
For constant a:
The travelling twin's own time is linear to arctanh(v/c).
The time he needs to accelerate to velocity v is:
t = (c/a) arctanh(v/c)
v/c = tanh(at/c)
Benno Muilwijk
pholroyd
i get old book from library. book very old pages are brown. i not know
where to buy old book like this. used bookstres might be. if i had
real own copy i would give to you. i try learn from book and then
others.
@@@ph@@@
Paul B. Andersen
"Harold Ensle" <hee...@ix.netcom.com> skrev i melding news:AQoQb.25597$zj7....@newsread1.news.pas.earthlink.net...
Typo. Should have been:
Then A is at a higher potential, and the greater distance will more
than compensate for the time "lost" when accereating away and breaking.
>
> OK I want to get this straight. When I did the problem, positive
> acceleration (a) was always _away_ from the stay-at-home and
> distance (x) was always the distance to the stay-at-home.
In the accelerated frame, the accelerated twin is at origo.
So if the "positiv direction" is to the right, "the inertial twin" will
"fall" to the left, e.g. x will be negative
A B -> a
-----|------> x
So the acceleration and distance will have opposite sign
when accelerating away and breaking, equal sign (both negative)
when accelerating to return.
Same sign - A is at a lower potential than B,
opposite sign - A is at a higher poptential than B.
>
> Now I did not transform either of these values as I assume they
> are what is observed by the travelling twin. So when the travelling
> twin accelerates, his view of the resulting velocity and displacement
> are just derivatives (due to invariance).
>
> Is this what you did?
Yes, but it is not quite as simple as that.
The co-ordinate acceleration of the inertial A in B's frame is
NOT equal to B's proper acceleration.
Paul
Harold Ensle
OK. I used your formula and was able to get the matching amount
of time for the stay-at-home on reunion. I then wanted to see
the travelling twins view of his own time so that at reunion he
sees the correct time passing for himself. How do I apply your
equation for this case?
Since he is always at the origin of his own frame I am not sure what
x or v to use...even if he does have a non-zero 'a'.
H.Ellis Ensle
greywolf42
news:e7203033.0401...@posting.google.com...
> You have Moller? Is it a library copy or were you able to purchase it
> somewhere? I'd love to buy a copy of that text! If you did buy a copy
> can you tell me where and how I can get a copy?
>
> Thanks
>
> Pmb
Try Amazon.com. They often broker the sale of old or out-of-print books.
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}
Benno Muilwyk
Thanks for correcting me. I was too hasty when I wrote that!
Only the observer must have a constant velocity for eq. 3 to be
applicable (or better: his FOR must be inertial).
In eq. 6, a is the acceleration of the observer's frame, which can
be independent of v(t). Since a and x can also vary with time,
eq. 6 can be applied for arbitrary accelerations of both observer
and observed object.
Taking a=0 means that the observer's frame is inertial and filling in
in eq. 6 indeed gives eq. 3 for inertial frames.
BTW, In the special case of the twins, the stay-at-home twin is
assumed to be inertial, so v(t) does depend on a. When a is
constant during outward acceleration, the relation is:
v(t) = -c tanh(at/c)
Benno
Benno Muilwyk
> > for the stay-at-home from the travelling twin's perspective.
> > However, how do I deal with the travelling twin's view of his
> > own time?
> >
> > H.Ellis Ensle
> >
>
> dt = (c/a) d(phi) where phi = arctanh(v/c)
>
> For constant a:
> The travelling twin's own time is linear to arctanh(v/c).
> The time he needs to accelerate to velocity v is:
>
> t = (c/a) arctanh(v/c)
>
> v/c = tanh(at/c)
Where a is the proper acceleration of the traving twin's
frame, t is the time coordinate in the travelling twin's frame
and v is the velocity of the travelling twin wrt an inertial
object which was comoving at t=0.
This comoving object could be the stay-at-home twin,
in which case t=0 at the moment the travellling twin starts
accelerating away.
The comoving object could also be an object which is
released from the hull of the travelling twin's spacecraft
at the moment he start decelerating, in which case t=0
for that moment.
Benno Muilwijk
Paul B. Andersen
"Harold Ensle" <hee...@ix.netcom.com> skrev i melding news:O9bRb.28208$zj7....@newsread1.news.pas.earthlink.net...
The v is the speed between the instant inertial rest frames of the two observers.
The x is the distance in the instant inertial rest frame of the accelerated twin.
Paul
Harold Ensle
You see the problem here don't you. If the travelling twin's view of his own
time is based on "x" then there is automatically an infinite number of
contradictions since the travelling twin could be sitting in a potential
well
of any observer, each having a different "x".
H.Ellis Ensle
>
> Paul
>
>
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:it1Sb.1064$uM2...@newsread1.news.pas.earthlink.net...
[snip]
> You see the problem here don't you. If the travelling twin's view of his own
> time is based on "x" then there is automatically an infinite number of
> contradictions since the travelling twin could be sitting in a potential
> well
> of any observer, each having a different "x".
But then you can always try to find out whether he
can "apply the transformation to himself", can't you?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CannotApply.html
Dirk Vdm
Paul B. Andersen
"Harold Ensle" <hee...@ix.netcom.com> skrev i melding news:it1Sb.1064$uM2...@newsread1.news.pas.earthlink.net...
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:bv7rta$cvg$1...@dolly.uninett.no...
> >
> > "Harold Ensle" <hee...@ix.netcom.com> skrev i melding
> news:O9bRb.28208$zj7....@newsread1.news.pas.earthlink.net...
> > > of time for the stay-at-home on reunion. I then wanted to see
> > > the travelling twins view of his own time so that at reunion he
> > > sees the correct time passing for himself. How do I apply your
> > > equation for this case?
> > >
> > > Since he is always at the origin of his own frame I am not sure what
> > > x or v to use...even if he does have a non-zero 'a'.
> >
> > The v is the speed between the instant inertial rest frames of the two
> observers.
> > The x is the distance in the instant inertial rest frame of the
> accelerated twin.
>
> You see the problem here don't you. If the travelling twin's view of his own
> time is based on "x" then there is automatically an infinite number of
> contradictions since the travelling twin could be sitting in a potential
> well of any observer, each having a different "x".
I do indeed see your problem here.
You have the most naive misconception of them all.
Do you, in your potential well, have a lot of different views
of your own time because there are a lot of different clocks
at different x's and with different v's? :-)
Do you actually believe GR say you should have?
But maybe you have a lot of different views of the rate
of the other clocks at those different x's and at different
gravitational potential and with different v's?
One for each clock maybe?
Because THAT's what GR say you should have.
Paul
Harold Ensle
Exactly!! This is related indeed. You cannot apply the transformation
(which
is based on _relative_ velocity) to yourself. Why? Because you are always at
rest
in your own frame. (Of course you can apply the equations with v=0 but it
doesn't transform anything).
Gad..it is so easy. Why don't you understand it?
H.Ellis Ensle
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:iteSb.1845$uM2...@newsread1.news.pas.earthlink.net...
After all these years you still haven't got the slightest idea
what you are talking about, and you still don't even know
what coordinates are. Gad... you must be stupid. Aren't
you embarrassed by that?
Maybe stamp collecting is something for you.
Dirk Vdm
Harold Ensle
I hope not, since that would then be a contradiction.
> Do you actually believe GR say you should have?
I don't know what GR says. But to resolve the twin
paradox, the acceleration term that you provided
contains "x". Now as long as you are comparing
times _between_ two observers, their separation "x"
seems appropriate to include. However, when the
travelling twin has a view of his own time that is
somehow being transformed, then it obviously can't
depend on "x", since there is no relevant "x" that
can be applied.
> But maybe you have a lot of different views of the rate
> of the other clocks
_between_ two observers, but I am talking about
the travelling twin's view of his _own_ time.
>at those different x's and at different
> gravitational potential and with different v's?
> One for each clock maybe?
> Because THAT's what GR say you should have.
Am I explaining it badly, or are you just dense?
H.Ellis Ensle
Harold Ensle
You are using a different equation here. I would like
to know how this equation:
t'=integral from t0 to t1( sqrt(1-v^2/c^2)(1+ax/c^2))dt
can be employed by the travelling twin to determine the
true passage of time for himself (on his return home).
So all I need really is the correct physical definitions
for v,a,x in regards to this particular application of the
formula.
H.Ellis Ensle
Benno Muilwyk
OK. That's rather trivial:
v is the velocity of the observed object (himself) = 0;
x is the distance of the observed object (himself) = 0;
a is the acceleration of his own frame (felt by himself).
When you fill in v=0 and x=0 for himself you get:
t' = integral from t0 to t1 (dt) = t1 - t0
for any a (even when a is not constant).
Benno Muilwijk
Benno Muilwyk
Anybody's view of his _own_ time is simply what his
own clock reads. One can express another's time
_in_terms_of_his_own_time by using the transformation
equations, but when anybody wants to apply them
to himself, i.e. when he is observing himself, he must
always fill in v=0 and x=0 and then the transformation
equations will (and should) always return identity:
dt' = dt
t' = t
x' = x
Benno Muilwijk
Bilge
>
>"Harold Ensle" <hee...@ix.netcom.com>:
>> Exactly!! This is related indeed. You cannot apply the transformation
>> (which is based on _relative_ velocity) to yourself. Why? Because you
>> are always at rest in your own frame. (Of course you can apply the
>> equations with v=0 but it doesn't transform anything).
>>
>> Gad..it is so easy. Why don't you understand it?
>
>After all these years you still haven't got the slightest idea
>what you are talking about, and you still don't even know
>what coordinates are. Gad... you must be stupid. Aren't
>you embarrassed by that?
>Maybe stamp collecting is something for you.
Or perhaps providing source material for a new series of
twilight zone episodes.
Harold Ensle
OK...thank you, this is what I was getting at with Andersen.
In relation to himself I did not see how v or x (and thus a)
could contribute any transformation.
The problem is that the stay at home sees the travelling
twin age less on return, thus the travelling twin must
also see his own age as less (only when he actually
completes the trip).
H.Ellis Ensle
Paul B. Andersen
"Harold Ensle" <hee...@ix.netcom.com> skrev i melding news:cIeSb.1865$uM2...@newsread1.news.pas.earthlink.net...
No, you are explaining your misconception very clearly,
so I have no problem with seeing what it is.
As I said, it is just about the most naive misconceptions of all.
I note that you missed my attempt to explain the nature
of your misconception.
See the response of Benno Muilwy, he spells it
out a little more clearly, so I won't bother to repeat it.
Paul
Benno Muilwyk
> twin age less on return, thus the travelling twin must
> also see his own age as less (only when he actually
> completes the trip).
>
> H.Ellis Ensle
Er... Do you still see it as a problem?
Benno Muilwijk
Harold Ensle
Yes......but tell me, how does one determine t1 and t2
in your above solution?
H.Ellis Ensle
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:v0GSb.3550$uM2....@newsread1.news.pas.earthlink.net...
Troll alert:
Januari 2004 on
http://groups.google.com/groups?&as_umsgid=TcGSb.3560$uM2....@newsread1.news.pas.earthlink.net
| "Give it up. There is no solution to the twin paradox.
| It is definitely not in SR (though some fools would still argue it).
| And if one brings in GR, there is simply no sort of justification.
| Everything that has been done is not a true solution. It is
| all a kind of mathematical propaganda which simply hides
| the problem.
|
| How does all this subterfuge benefit anyone?
|
| It doesn't."
|
| H.Ellis Ensle
Dirk Vdm
Benno Muilwyk
> > > twin age less on return, thus the travelling twin must
> > > also see his own age as less (only when he actually
> > > completes the trip).
> > >
>
> Yes......but tell me, how does one determine t1 and t2
> in your above solution?
>
> H.Ellis Ensle
The same as anybody else.
1. It can be predicted when you know distance, velocity and
acceleration.
2. It can be measured by simply looking at one's own clock:
for example t1 when one leaves and t2 when one returns.
The two should match within the limits of accuracy.
Sigh... Dirk is right. This feels like explaining the obious
to a child. Your questions are now at a level too silly for
someone who understands integrals.
Benno
Harold Ensle
> > > > The problem is that the stay at home sees the travelling
> > > > twin age less on return, thus the travelling twin must
> > > > also see his own age as less (only when he actually
> > > > completes the trip).
> > > >
> > > Er... Do you still see it as a problem?
> >
> > Yes......but tell me, how does one determine t1 and t2
> > in your above solution?
> >
> > H.Ellis Ensle
>
> The same as anybody else.
>
> 1. It can be predicted when you know distance, velocity and
> acceleration.
> 2. It can be measured by simply looking at one's own clock:
> for example t1 when one leaves and t2 when one returns.
For this specific problem. Obviously we can make t1=0.
But for the returning twin what is t2?
You think it is trivial, but I actually had a good reason to ask.
If I use the stay-at-home's view of t2, then we are right back
where we started...i.e. a stay-at-home biased solution
which, of course will not conflict with the stay-at-home's view.
If I use any other value and the twin paradox is still a contradiction.
In fact, given this solution, your equation would make the travelling
twin's view of his own time equivalent to the stay-at-home's view
of the travelling twin's time for the entire trip.
What this then means is that Andersen's acceleration term only
solves part of the paradox.
> The two should match within the limits of accuracy.
>
> Sigh... Dirk is right. This feels like explaining the obious
> to a child. Your questions are now at a level too silly for
> someone who understands integrals.
Perhaps _you_ do not then. I cannot derive t2 from the
integral. For a definite integral, it has to be supplied.
Now I am asking you once more for a single sentence
answer: For the twin paradox scenario, what is t2?
H.Ellis Ensle
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:iiWSb.5019$uM2....@newsread1.news.pas.earthlink.net...
[snip]
> > Sigh... Dirk is right. This feels like explaining the obious
> > to a child. Your questions are now at a level too silly for
> > someone who understands integrals.
>
> Perhaps _you_ do not then. I cannot derive t2 from the
> integral. For a definite integral, it has to be supplied.
> Now I am asking you once more for a single sentence
> answer: For the twin paradox scenario, what is t2?
Here, this was something I had picked for your friend Luttgens:
http://users.pandora.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
It could have picked it for you.
Dirk Vdm
Benno Muilwyk
> > > > > twin age less on return, thus the travelling twin must
> > > > > also see his own age as less (only when he actually
> > > > > completes the trip).
> > > > >
> > > > Er... Do you still see it as a problem?
> > >
> > > Yes......but tell me, how does one determine t1 and t2
> > > in your above solution?
> > >
You just gave the answer yourself! The limits of an integral
have to be supplied! That means they are input and cannot
be derived from the integral by itself! That is why I gave
you another equation.
This particular integral can only be used to find how much
time will pass for an observed object (e.g. a twin brother),
_given_ the acceleration a of the frame of reference and
the velocity v and distance x of the observed object at any
time between the bounds of the integral (t1 and t2).
All these (a(t), v(t), x(t), t1 and t2) have to be known
before you can give a numerical solution. Otherwise,
the best you can do, is express the answer in terms of the
unknown variables. You should know that!
But even when when t1 and t2 are unknown, your problem
can still be solved. We have seen that the solution for the
travelling twin is trivial: t2 - t1.
You stated earlier that you have solved the integral for the
stay-at-home twin. Then you must have found a solution
such that t' > t2 - t1. What more do you want?
If you want to know how to predict t2, you will need
other equations, with other input, such as maximum
distance and maximum velocity.
Benno
Harold Ensle
OK, I looked again and found this:
dt = (c/a) d(phi) where phi = arctanh(v/c)
This looks like what you used to determine t2. Is this the case?
Where did this equation come from?
Actually, I would guess that it comes from using relativistic
addition of velocity. BUT you realize of course, that this
refers back to the rest frame of the stay-at-home.
In the travelling twin's view, at every instance, he thinks he is
at rest, just having deccelerated from a previous velocity. Thus
he would simply think that v=da/dt.
Of course, that would result in a contradiction, but
we can't have that, can we?
H.Ellis Ensle
Benno Muilwyk
>
> dt = (c/a) d(phi) where phi = arctanh(v/c)
>
> This looks like what you used to determine t2. Is this the case?
Yes, it is.
> Where did this equation come from?
Actually, I found it myself, long before I became an internet
user and found it confirmed in the faq of this newsgroup.
> Actually, I would guess that it comes from using relativistic
> addition of velocity. BUT you realize of course, that this
> refers back to the rest frame of the stay-at-home.
No, I found it by asking myself: "how is everything experienced
and measured from the travellling twin's perspective?" which is
just about the same as what you are asking, isn't it?
In fact, when I found it, I could verify its correctness by
deriving the equation for relativistic addition of velocity as well
as the Lorentz Transformation from it.
> In the travelling twin's view, at every instance, he thinks he is
> at rest, just having deccelerated from a previous velocity. Thus
> he would simply think that v=da/dt.
No. He would use a = -dv/dt. And locally (within the travelling
twin's spacecraft) that is correct; he can measure his own
acceleration in the following way:
- he can release (drop) an object with initial v=0
- he can measure the velocity v=dv of the object after dt time
- with dv << c he can calculate the "downward" acceleration of
the dropped object as dv/dt
- he feels an acceleration "upward", so a = -dv/dt
> Of course, that would result in a contradiction, but
> we can't have that, can we?
Which contradiction?
> H.Ellis Ensle
Benno Muilwijk
Harold Ensle
> > OK, I looked again and found this:
> >
> > dt = (c/a) d(phi) where phi = arctanh(v/c)
> >
> > This looks like what you used to determine t2. Is this the case?
>
> Yes, it is.
>
> > Where did this equation come from?
>
> Actually, I found it myself, long before I became an internet
> user and found it confirmed in the faq of this newsgroup.
>
> > Actually, I would guess that it comes from using relativistic
> > addition of velocity. BUT you realize of course, that this
> > refers back to the rest frame of the stay-at-home.
>
> No, I found it by asking myself: "how is everything experienced
> and measured from the travellling twin's perspective?" which is
> just about the same as what you are asking, isn't it?
> In fact, when I found it, I could verify its correctness by
> deriving the equation for relativistic addition of velocity as well
> as the Lorentz Transformation from it.
>
> > In the travelling twin's view, at every instance, he thinks he is
> > at rest, just having deccelerated from a previous velocity. Thus
> > he would simply think that v=da/dt.
>
> No. He would use a = -dv/dt.
Actually mine was a typo. I meant this.....so OK so far.
>And locally (within the travelling
> twin's spacecraft) that is correct; he can measure his own
> acceleration in the following way:
> - he can release (drop) an object with initial v=0
> - he can measure the velocity v=dv of the object after dt time
> - with dv << c he can calculate the "downward" acceleration of
> the dropped object as dv/dt
> - he feels an acceleration "upward", so a = -dv/dt
OK. This is fine.
[.....]
So how did you come from this to the above formula?
H.Ellis Ensle
Benno Muilwyk
The perspective from the travelling twin can best be compared
with somebody in a merry-go-round. He is at rest wrt his own
frame of reference (the merry-go-round), but sees the world
turn around him.
Suppose he is sitting at distance R from the center of the
merry-go-round, with a transparant floor, so he can see the
ground rotate underneath him.
The amount of ground that passes underneath him is
y=phi*R where phi is the angle of rotation.
Next suppose there is a straight line on the ground that passes
the center of the merry-go-round at (closest) distance R,
so once every round the person in the merry-go-round
passes right over that line. Let's say phi=0 at that point.
At any point he would measure his distance x to the line
along the radius of the merry-go-round, which is
perpendicular to his (forward) y direction. This distance is
x = -R(1 - 1/cos(phi)) = -R(1 - 1/cos(y/R))
so for phi=0, x=0.
In comparison with the travelling twin, the straight line is the
world line of the stay-at-home twin, and the amount of
ground passed underneath the person in the merry-go-round
maps to the elapsed time of the travelling twin:
y = ict where i = sqrt(-1)
An important difference between the merry-go-round is
that the merry-go-round (or the world around it) rotates
in the x-y (space-space) plane, so phi can be expressed as
a real number. However, the travelling twin is rotating in
the x-t (space-time) plane, so phi is an imaginary number:
phi = y/R = ict/R where R = c^2/a, which gives
phi = iat/c
So far I have assumed that R is constant, but when it is not,
we get:
dy = R*d(phi)
ic*dt = (c^2/a)*d(phi)
i dt = (c/a) d(phi)
In the equation at the top I presented phi as a real number
by leaving out i.
> H.Ellis Ensle
Benno Muilwijk
Harold Ensle
> > > > OK, I looked again and found this:
> > > >
> > > > dt = (c/a) d(phi) where phi = arctanh(v/c)
> > > >
> > > > This looks like what you used to determine t2. Is this the case?
> > >
> > > Yes, it is.
> > >
> > > > Where did this equation come from?
> > >
> > > Actually, I found it myself, long before I became an internet
> > > user and found it confirmed in the faq of this newsgroup.
> > >
> > > > Actually, I would guess that it comes from using relativistic
> > > > addition of velocity. BUT you realize of course, that this
> > > > refers back to the rest frame of the stay-at-home.
> > >
> > > No, I found it by asking myself: "how is everything experienced
> > > and measured from the travellling twin's perspective?" which is
> > > just about the same as what you are asking, isn't it?
> > > In fact, when I found it, I could verify its correctness by
> > > deriving the equation for relativistic addition of velocity as well
> > > as the Lorentz Transformation from it.
If your method was more basic, then it would have been used instead
of the postulates to derive the Lorentz transformations. What I suspect
here is that your method is consistent with these equations and thus
derivable from them just as I suggested above. So your answer is "yes
( your guess was correct)".
Yes, this is very nice. However, I think that you have focused
so much on the mechanics of the problem, you did not
consider how it relates to this thread.
Since you used the geometry of Minkowski space, which
itself is merely the geometric representation of the Lorentz
transformations, you thus did indeed ultimately use the
Lorentz transformations in determing the travelling twin's
view of his own time. And the velocity used is, of course,
the velocity relative to the stay-at-home.
It is interesting that once an origin is selected for the space,
there is only one observer who can determine his time
independent of any velocity. And I can see no way
this can be avoided if everything is to tally up.
The question is, then where is the mutual time dilatation
required by SR?
H.Ellis Ensle
Benno Muilwyk
>
> "Benno Muilwyk" <be...@muilwijk-met-wyk.nl> wrote in message
> news:401ee...@news3.prserv.net...
> > > > > OK, I looked again and found this:
> > > > >
> > > > > dt = (c/a) d(phi) where phi = arctanh(v/c)
> > > > >
> > > > > This looks like what you used to determine t2. Is this the case?
> > > >
> > > > Yes, it is.
> > > >
> > > > > Where did this equation come from?
> > > >
> > > > Actually, I found it myself, long before I became an internet
> > > > user and found it confirmed in the faq of this newsgroup.
> > > >
> > > > > Actually, I would guess that it comes from using relativistic
> > > > > addition of velocity. BUT you realize of course, that this
> > > > > refers back to the rest frame of the stay-at-home.
> > > >
> > > > No, I found it by asking myself: "how is everything experienced
> > > > and measured from the travellling twin's perspective?" which is
> > > > just about the same as what you are asking, isn't it?
> > > > In fact, when I found it, I could verify its correctness by
> > > > deriving the equation for relativistic addition of velocity as well
> > > > as the Lorentz Transformation from it.
>
> If your method was more basic, then it would have been used instead
> of the postulates to derive the Lorentz transformations. What I suspect
> here is that your method is consistent with these equations and thus
> derivable from them just as I suggested above. So your answer is "yes
> ( your guess was correct)".
Yes, they are consistent, but that does not imply that I found it by using
relativistic addition of velocity.
> > > > > In the travelling twin's view, at every instance, he thinks he is
I started using flat space geometry and derived flat space
equations from the merry-go-round's perspective. Only
in the end I made the transition from flat space to Minkowski
space-time. And I did not use velocity anywhere in above
derivations.
> It is interesting that once an origin is selected for the space,
> there is only one observer who can determine his time
> independent of any velocity. And I can see no way
> this can be avoided if everything is to tally up.
Either observer can determine his time by simply looking at
his own clock. I stated that before. This is surely
independent of any velocity.
And neither can predict his time (between departure and
arrival of the travelling twin) independent of any velocity.
> The question is, then where is the mutual time dilatation
> required by SR?
Mutual time dilation, which can be derived from the
Lorentz Transformation and its inverse, only occurs
when both observers are inertial. From an accelerating
observer's perspective, the standard Lorentz Transformation
(or its inverse) cannot be applied, so time dilation becomes
one way only.
> H.Ellis Ensle
Benno Muilwijk
Harold Ensle
Right...and I was premature. Though phi does depend on v,
right?
> > It is interesting that once an origin is selected for the space,
> > there is only one observer who can determine his time
> > independent of any velocity. And I can see no way
> > this can be avoided if everything is to tally up.
>
> Either observer can determine his time by simply looking at
> his own clock. I stated that before. This is surely
> independent of any velocity.
> And neither can predict his time (between departure and
> arrival of the travelling twin) independent of any velocity.
Yes..and my argument wasn't particularly good here.
> > The question is, then where is the mutual time dilatation
> > required by SR?
>
> Mutual time dilation, which can be derived from the
> Lorentz Transformation and its inverse, only occurs
> when both observers are inertial. From an accelerating
> observer's perspective, the standard Lorentz Transformation
> (or its inverse) cannot be applied, so time dilation becomes
> one way only.
This last sentence is the issue. _When_ does it become one-way?
And with acceleration can it be partly two-way and partly one-way,
after all, velocity is only a _singular_ special case of acceleration.
(i.e. a=0, not a=0.0000000001).
H.Ellis Ensle
Benno Muilwyk
Well, there is a relation between them, but I prefer to say that
the velocity is a result of the angle of rotation, because the angle
is more basic. Angles can even be added straight away:
When A's world line is rotated over an angle phi1 wrt B's world
line, and B's world line is rotated over an angle phi2 wrt C's world
line, then the rotation of A's world line wrt C's world line is simply
phi1 + phi2.
The relativistic addition of velocities can easily be derived from that.
> > > It is interesting that once an origin is selected for the space,
> > > there is only one observer who can determine his time
> > > independent of any velocity. And I can see no way
> > > this can be avoided if everything is to tally up.
> >
> > Either observer can determine his time by simply looking at
> > his own clock. I stated that before. This is surely
> > independent of any velocity.
> > And neither can predict his time (between departure and
> > arrival of the travelling twin) independent of any velocity.
>
> Yes..and my argument wasn't particularly good here.
>
> > > The question is, then where is the mutual time dilatation
> > > required by SR?
> >
> > Mutual time dilation, which can be derived from the
> > Lorentz Transformation and its inverse, only occurs
> > when both observers are inertial. From an accelerating
> > observer's perspective, the standard Lorentz Transformation
> > (or its inverse) cannot be applied, so time dilation becomes
> > one way only.
>
> This last sentence is the issue. _When_ does it become one-way?
Good question. Let's see...
The mutual time dilation is equal when a=0 (straight world lines).
For non-zero acceleration, the world line is curved.
When a is positive along the x-axis, the center of rotation is at -c^2/a.
In "backward" direction (closer to the center of rotation) time runs
slower, so time dilation is in fact stronger. At the center of rotation
time is standing still, so time dilation reaches infinity.
In "forward" direction time runs faster (on co-accelerating clocks
at rest wrt the accelerating observer), so the effect of acceleration
works opposite to the effect of relative velocity.
That means there must be some relation between a, v and x where
the two effects compensate each other.
This is also true in GR: higher clocks run faster and moving clocks
run slower. Since the velocity of a satellite depends on the altitude
(and g), there is a certain altitude where the two effects compensate
each other.
> And with acceleration can it be partly two-way and partly one-way,
> after all, velocity is only a _singular_ special case of acceleration.
> (i.e. a=0, not a=0.0000000001).
I'm not sure what you mean. If my explanation above does not
answer the question, please clarify.
> H.Ellis Ensle
Benno Muilwijk
Harold Ensle
My point still has not been addressed. Does the relativistic addition
of velocities relate back to the stay-at-home or not? If it does,
then you are surreptitiously preferring the stay-at-home's view
of time. This avoids any contradiction, but it denies SR reciprocity.
> > after all, *** velocity is only a _singular_ special case of
acceleration.
> > (i.e. a=0, not a=0.0000000001).
(***Actually I meant _constant_ velocity)
> I'm not sure what you mean. If my explanation above does not
> answer the question, please clarify.
Your mechanics are fine, but you are simply not seeing how you
are preferring the stay-at-home in the process. I don't know what
else to say.
I appreciate you taking the time to explain why you believe that
there is no contradiction in relativity.
H.Ellis Ensle
Benno Muilwyk
> of velocities relate back to the stay-at-home or not? If it does,
> then you are surreptitiously preferring the stay-at-home's view
> of time. This avoids any contradiction, but it denies SR reciprocity.
The relativistic addition of velocities works for all inertial frames.
That includes not only the stay-at-home twin's frame, but also all
instant co-moving frames of the travelling twin. Each instant
co-moving frame is rotated d(phi) wrt the previous instant co-moving
frame. So the total rotation is the integral from t1 to t2 of d(phi)
where d(phi) = (a/c) dt.
This integral relates back to the travelling twin's view of time.
> > > after all, *** velocity is only a _singular_ special case of
> acceleration.
> > > (i.e. a=0, not a=0.0000000001).
>
> (***Actually I meant _constant_ velocity)
>
> > I'm not sure what you mean. If my explanation above does not
> > answer the question, please clarify.
>
> Your mechanics are fine, but you are simply not seeing how you
> are preferring the stay-at-home in the process. I don't know what
> else to say.
The least you could do is tell me in which step(s) you think this is
the case and why.
Dirk Van de moortel
"Benno Muilwyk" <be...@muilwijk-met-wyk.nl> wrote in message news:40280...@news3.prserv.net...
>
> "Harold Ensle" <hee...@ix.netcom.com> schreef in bericht
> news:MZfVb.18709$uM2....@newsread1.news.pas.earthlink.net...
> >
[snip]
> > Your mechanics are fine, but you are simply not seeing how you
> > are preferring the stay-at-home in the process. I don't know what
> > else to say.
>
> The least you could do is tell me in which step(s) you think this is
> the case and why.
Benno, hij is verschrikkelijk met je voeten aan het spelen :-)
Dirk Vdm
Harold Ensle
All you are saying here is that it commutes. But this is a strict property
of the Lorentz transformations that works as well for the travelling
twin as he observes the relative acceleration of the stay-at-home.
Thus one could apply your rotation to the stay-at-home's
trip (as seen by the travelling twin). So the fact that the transformations
commute does not resolve the paradox.
Now I would like to use your integral in the actual twin paradox
scenario. All I need is the way to determine t2.
H.Ellis Ensle
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:pSaWb.22755$uM2....@newsread1.news.pas.earthlink.net...
[snip]
> Now I would like to use your integral in the actual twin paradox
> scenario. All I need is the way to determine t2.
All you need is a brain, moron.
Dirk Vdm
Benno Muilwyk
> scenario. All I need is the way to determine t2.
The integral of equation 6 you mean? I already answered that.
It's the same as asking to determine t_2 for the stay-at-home
twin using equation 3. How would you do that?
You can't do one without the other. You have to apply eq. 6
from both perspectives to get the round trip proof.
Start by dividing the trip in parts in which a is constant.
Define durations of those parts as measured by the travellling
twin. Let's say his total time of the whole trip is Tt.
Apply eq.6 for each part of the trip and calculate the total
time Ts (in terms of Tt and a) as measured by the stay-at-home
twin. Then apply eq. 6 for each part from the stay-at-home
twin's perspective (note that a=0 in this case) to calculate Tt
(in terms of Ts and a).
Finally substitute Ts for what you found earlier (in terms of Tt
and a) and verify that you get Tt = Tt.
Of course you can also start by assuming an initial Ts for the
stay-at-home twin, calculate Tt from there, then calculate Ts
from the travelling twin's perspective and verify that you get
Ts = Ts.
What more proof do you need?
> H.Ellis Ensle
Benno
Harold Ensle
Why did you not respond to any of my comments?
You asked me for the reasons and when I give them you simply
ignore them. You are obviously in denial.
> > Now I would like to use your integral in the actual twin paradox
> > scenario. All I need is the way to determine t2.
>
> The integral of equation 6 you mean? I already answered that.
> It's the same as asking to determine t_2 for the stay-at-home
> twin using equation 3. How would you do that?
I only asked the question so that you would see that its determination
was dependent on the stay-at-home's view.
But again you wrap yourself up in the mechanics of the problem,
still not recognizing the source. (as I see below)
> You can't do one without the other. You have to apply eq. 6
> from both perspectives to get the round trip proof.
>
> Start by dividing the trip in parts in which a is constant.
> Define durations of those parts as measured by the travellling
> twin. Let's say his total time of the whole trip is Tt.
> Apply eq.6 for each part of the trip and calculate the total
> time Ts (in terms of Tt and a) as measured by the stay-at-home
> twin. Then apply eq. 6 for each part from the stay-at-home
> twin's perspective (note that a=0 in this case) to calculate Tt
> (in terms of Ts and a).
> Finally substitute Ts for what you found earlier (in terms of Tt
> and a) and verify that you get Tt = Tt.
>
> Of course you can also start by assuming an initial Ts for the
> stay-at-home twin, calculate Tt from there, then calculate Ts
> from the travelling twin's perspective and verify that you get
> Ts = Ts.
>
> What more proof do you need?
I will write up a review of the current progress of this discussion
on a new thread.
BTW instead of the two requirements you give above, there are
_four_ calculations that need to be made.
H.Ellis Ensle
Harold Ensle
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:4WaWb.6031$zC2.5...@phobos.telenet-ops.be...
Brilliant response. Note that I take some time with people who might have
the capacity to understand. But I gave up on you long ago as you have
proven yourself to be incompetent.
H.Ellis Ensle
Benno Muilwyk
> > That includes not only the stay-at-home twin's frame, but also all
> > instant co-moving frames of the travelling twin. Each instant
> > co-moving frame is rotated d(phi) wrt the previous instant co-moving
> > frame. So the total rotation is the integral from t1 to t2 of d(phi)
> > where d(phi) = (a/c) dt.
> > This integral relates back to the travelling twin's view of time.
>
> of the Lorentz transformations that works as well for the travelling
> twin as he observes the relative acceleration of the stay-at-home.
> Thus one could apply your rotation to the stay-at-home's
> trip (as seen by the travelling twin). So the fact that the
transformations
> commute does not resolve the paradox.
Like I said in my response two days ago, equation 6 works both ways
indeed. And this commutation confirms the reciprocity you are looking
for!
So again I ask: what more proof do you need?
> H.Ellis Ensle
Benno
Benno Muilwyk
> on a new thread.
OK.
> BTW instead of the two requirements you give above, there are
> _four_ calculations that need to be made.
Only four?
BTW, you said you resolved the integral of equation 6 before.
Did you do that for all parts of the trip?
(You can answer in the new thread).
> H.Ellis Ensle
Benno
Benno Muilwyk
>
> Dirk Vdm
Ik kan wel tegen een beetje kietelen, zolang het me niet verveelt ;-)
Benno
Dirk Van de moortel
"Harold Ensle" <hee...@ix.netcom.com> wrote in message news:H8EWb.999$WW3...@newsread2.news.pas.earthlink.net...
I concede. Just like everyone else on this group I am incompetent
as far as teaching you is concerned. Maybe you should first apply
the Lorentz Transformation to yourself and then try a little game of
Monopoly - with an open mind, that is
http://users.pandora.be/vdmoortel/dirk/Stuff/Monopoly.html
Dirk Vdm
Harold Ensle
Like I said the limits to the integral for the twin's view of himself is
determined based on the stay-at-home's view. Therefore it is not
truely the travelling twin's view.
> So again I ask: what more proof do you need?
More than you have shown so far.
H.Ellis Ensle
Benno Muilwyk
How would you determine the limits to the integral for the stay-at-home
twin's view of himself???
> > So again I ask: what more proof do you need?
>
> More than you have shown so far.
That's a bit vague, isn't it?
> H.Ellis Ensle
Benno
Harold Ensle
This is not what I wrote. I wrote that the limits are set by the
stay-at-home's
view of the travelling twin's time. This is instead of the travelling twin's
view of
the stay-at-home's time. In other words, even if you use equation (6) with
the corrective acceleration term, the travelling twin's time as viewed by
himself
(to establish the limits of the integral) is determined by the
stay-at-home's
view of the time. But to properly set the limits, the travelling twin would
use
his own timing of the sequence of events from his observation of the
stay-at-home's apparent motion.
[....]
H.Ellis Ensle
Benno Muilwyk
I know, that's why I asked you the counter-question: how does the
stay-at-home twin do it? But you have answered, so now I can
answer your question in (more or less) your own words:
to properly set the limits, each twin would use his own timing of
the sequence of events from his observation of the other's apparent
motion.
But I already stated this more than two weeks ago using the following
words: anybody's view of his _own_ time is simply what his
own clock reads.
> [....]
>
> H.Ellis Ensle
Benno
Harold Ensle
You stated it many times, but _think again_, because when you actually
do the problem, you do not do this at all.
If you do not realize this, then it is something that you need to
work out. I cannot be of much help in this case.
Concerning the twin paradox, I have vacillated in the past about the
source of the problem. I knew that it was true...and rendered SR
as impossible. BUT I wasn't sure where the problem began. I
have been assuming that the problem only occurs when there is
acceleration. This is because you can solve the Lorentz transformations
to get the inverse transformations with -v instead of v. But on further
consideration it seems more intrinsic. It seems that the contradiction
has already occurred before the twin even returns (since they would
agree on the distance of there separation when the twin is at his
destination).
H.Ellis Ensle