Twin Paradox (repost for McCullough)
Harold Ensle
I had made explicit points which were never rebutted.
I give you another opportunity here. I have
confidence that you will make some reasonable statement.
But you need to be specific in relation to my
points made here. If you go off on some hand-waving
argument or flip-flop to some other example, I will
know that you are simply avoiding the arguments.
Hint: Read each of my sentences. The one that is
false, simply point it out.
You just claimed on your current thread that I
misrepresented SR and it was my own misrepresentations
that were causing the contradiction. So..........It's
easy, just show me in this post which statements
of mine our misrepresentations.
...................................................................
start of repost:////////////////////////////////
There has been some discussion of the twin paradox
on the threads "(An) Argument Against Relativity (2)" and
"Denial of the Twin Paradox". The threads have grown
rather large and I though I would start a new thread using
my responses to three important posts.
////////////////////////////////Throop//////////////////////////////////////
Wayne Throop <thr...@sheol.org> wrote in message
news:10071...@sheol.org...
> :: [..Ensle..] claimed that the non-inertial twin's viewpoint wasn't
> :: represented. I showed it was.
> : "Harold Ensle" <hen...@ix.netcom.com>
> : No you didn't.
>
> http://sheol.org/throopw/sr-twin-01.html
> shows the non-inertial twin's viewpoint quite clearly.
> It is represented by the coordinate systems drawn in green and blue.
OK I went there and examined it closely. Here is the problem:
Even though you showed the green as perpendicular (in the
second to last diagram) you scaled the time on the axis as it
was seen by red. Thus you have biased it toward red's view.
You had green meet blue at t=2, but with reciprocity O2 would
see O1's time over that separation to be t=2, BUT his own
time would be t=4.
I really don't understand why you play this shell game.
It is like you are purposely tricking yourself.
Is that what you really want?
H.Ellis Ensle
//////////////////////////////////McCullough////////////////////////////////
////////////////
Harold says...
>>I obviously understand relativity better than anyone who thinks
>>it is not self-contradictory, because it *is* self-contradictory.
>If that were really true, then you should be able to derive
>a contradiction from the theory. Let me give you the basic
>postulates about relativity, and then you show me wherein
>lies the contradiction:
> 1. Postulate: There exists a coordinate system assigning
> coordinates x,y,z,t to each event in spacetime such that
> A. Light travels along a path (x(t),y(t),z(t)) such that
> (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = c^2
OK....the speed of light is constant in the frame of reference you chose.
(Nothing more yet.)
> B. A test particles that is not acted upon by external
> forces will follow a path (x(t),y(t),z(t)) such that
> dx/dt = constant, dy/dt = constant, dz/dt = constant.
OK....this is classically called inertia, Newton's 1st law.
> C. For any spacetime path (x(t),y(t),z(t)) the proper
> time T for the path obeys the equation
> dT^2 = (1 - 1/c^2 ((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)) dt^2
OK....this is the length of the path (squared) in space-time. Note that
no physical significance has been attached to this value yet.
> 2. (Definition) A coordinate system will be called an "inertial
> coordinate system" if it obeys properties A,B, and C.
OK....since this is a definition, you can certainly do it, but you must be
alert to the fact that this is not the typical definition of inertial. The
typical inertial coordinates would only need to satisfy B.
Now since you have confiscated the term 'inertial', I will create a new
term that means a coordinate system which satisfies at least B. I will
call it '_simple_ inertial'.
> 3. (Theorem) If S is an inertial coordinate system, and S' is
> a second coordinate system related to the first by the Lorentz
> transformations, then S' is also an inertial coordinate system.
OK....this is also true. In fact I already conceded this in past posts.
I stated that the proper time is invariant under a Lorentz
transformation. The problem here is that the theorem has a limited
domain of applicability. It only deals with inertial coordinate
systems and that is only a subset of simple inertial coordinate
systems. In fact, I gave an example of a coordinate system which
satisfied B but not C. That would be the reciprocal frame of the
travelling twin. The reason the theorem doesn't cover the frame
is because one cannot get to the reciprocal twin's view using the
Lorentz transformation.
So ironically, your statements here are without contradiction, but
relativity is still wrong, because of the sin of omission.
> 4. (Relativity Postulate) The laws of physics have the same
> form when expressed in any inertial coordinate system.
>Exercise 1: Derive a contradiction from these postulates.
It is obvious. Two views that must represent a physically consistent
reality yield two different proper times.
>Exercise 2: Explain your notion that "SR demands reciprocity"
>in terms of the postulates.
It is obvious. If constant velocity is relative than each observer
must be able to work the same problem in physics as if his frame
was the rest frame. This is identical to 4. in that this would be
the simplest way for each observer to express the physical laws.
"Physical laws" includes the observed velocity of each other.
/////////////////////////////////////McCullough/////////////////////////////
/////////////////////
Harold says...
>>When the travelling twin is on the out-going part of the journey,
>>he doesn't know yet that he will be accelerating. During that
>>time he must be able to view himself at rest. Now what happens
>>when he finally accelerates? Does he have to rewrite history
>>and place himself in the space-time diagram of the stay-at-home?
>What SR tells us that the notion of which events are
>simultaneous is a function of what inertial reference
>frame you are in. Let's look at the example of a twin
>(initially 20 years old) who travels out for 5 years
>(Earth time) and back 5 years at speed .8c. Let's label
>six different events
> e1 = the travelling twin blasts off
> e2 = the stay-at-home twin's clock shows t = 1.8 years.
> e3 = the stay-at-home twin's clock shows t = 5 years.
> e4 = the travelling twin turns around
> e5 = the stay-at-home twin's clock shows t = 8.2 years.
> e6 = the twins reunite
>In the reference frame of the stay-at-home twin,
>e3 and e4 are simultaneous.
>In the outward-going frame of the travelling twin,
>e2 and e4 are simultaneous.
Only from the point of view of the stay at-home.
To the travelling twin, 5 years elapsed in order
for the stay-at-home to get that distance away.
He sees only 1.8 years for the stay-at-home. He
also sees that the events e2 and e4 are
simultaneous for the stay-at-home.
People sometimes balk at this statement and reject
it immediately _because_ it_ is_ a_ contradiction_.
But that's the whole point! It IS a contradiction, but
that is what SR must predict.
>In the returning frame of the travelling twin,
>e5 and e4 are simultaneous.
ditto
>From the point of view of the travelling twin,
>on the outgoing leg, the travelling twin ages
>3 years (going from e1 to e4) while the stay-at-home
>twin ages 1.8 years (going from e1 to e2). On the
>return leg, the travelling twin ages 3 years
>(going from e4 to e6) while the stay-at-home
>twin ages 1.8 years (going from event e5 to event
>e6). The travelling twin sees the stay-at-home twin
>age less during each leg of his journey.
You have done something interesting here. You have
given the proper reciprocity to the time dilation itself,
but you preserved the simultaneity as seen from the
stay-at-home. So it is still biased toward the stay-at-home's
view. You have done a tricky manuever that
makes it look like there is reciprocity, but it is only
the pseudo reciprocity that I have mentioned before.
>However, by piecing together two different inertial
>coordinate systems, the travelling twin doesn't cover
>all the events. All the events on Earth between e2 and e5
>are not covered by *either* coordinate system.
>In the outgoing coordinate system, they happen
>after turnaround, but in the return coordinate
>system, they happen before turnaround.
So at the instant when we accelerate, the stay-at-home
suddenly gains 6.4 years, but if we didn't accelerate,
he would gain nothing. You see, even if you got out of
the reciprocity problems discussed above, you would
still have to deal with the de facto acceleration
dependence.
There is simply no hope for relativity. It cannot possibly
be correct.....not now......... and not in a thousand years.
>There is nothing remarkable about SR in this regard.
No it's not remarkable, it's a miracle!
>The same thing happens in Euclidean geometry if you
>try to piece together two Cartesian coordinate systems
>to make a single nonCartesian coordinate system. Some
>points will get left out, and some events will be
>covered twice.
whatever....
H.Ellis Ensle
///////////////////////////////////////////////////////////////////////////
Daryl McCullough
>
>Here is a repost from the end of our last discussion.
>I had made explicit points which were never rebutted.
>> 1. Postulate: There exists a coordinate system assigning
>> coordinates x,y,z,t to each event in spacetime such that
>> A. Light travels along a path (x(t),y(t),z(t)) such that
>> (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = c^2
>> B. A test particles that is not acted upon by external
>> forces will follow a path (x(t),y(t),z(t)) such that
>> dx/dt = constant, dy/dt = constant, dz/dt = constant.
>> C. For any spacetime path (x(t),y(t),z(t)) the proper
>> time T for the path obeys the equation
>> dT^2 = (1 - 1/c^2 ((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)) dt^2
>> 2. (Definition) A coordinate system will be called an "inertial
>> coordinate system" if it obeys properties A,B, and C.
>> 3. (Theorem) If S is an inertial coordinate system, and S' is
>> a second coordinate system related to the first by the Lorentz
>> transformations, then S' is also an inertial coordinate system.
>> 4. (Relativity Postulate) The laws of physics have the same
>> form when expressed in any inertial coordinate system.
>>Exercise 1: Derive a contradiction from these postulates.
>
>It is obvious. Two views that must represent a physically consistent
>reality yield two different proper times.
Harold, once again: a proof from a set of assumptions
is a sequence of statements, each annotated with its justification.
The possible justifications are (1) by assumption (and you must identify
which assumption), or (2) by logical deduction from previous assumptions.
A proof of a contradiction is a proof such that one of the statements
is the negation of an earlier statement.
Surely you realize that your line above is not a proof of a contradiction.
Where is your sequence of statements? Where is the justification of each?
What statement is the negation of what other statement?
You have failed miserably at what claimed to be able to do.
You claimed that you had a proof of a contradiction from SR,
but you don't. Why do you keep claiming that you have something
that you don't have?
--
Daryl McCullough
Ithaca, NY
Harold Ensle
Daryl McCullough <da...@cogentex.com> wrote in message
news:aqc2p...@drn.newsguy.com...
You snipped everything and then claim it failed. To which of my statements
do you refer? In fact.....you very carefully snipped out my arguments.
What is your problem?
Read the post! You are simply ignoring my arguments and then
claiming they are wrong. I know you can do better than that.
Try harder.
> You claimed that you had a proof of a contradiction from SR,
> but you don't. Why do you keep claiming that you have something
> that you don't have?
Like I said before: You don't need a physicist to tell you what's going
on, you need a psychologist to tell you why you are in such obvious
denial.
No.............honestly, I do not know why you are ignoring my arguments.
Are you afraid that they will actually make sense?
H.Ellis Ensle
Daryl McCullough
>
>
>Daryl McCullough <da...@cogentex.com> wrote
>> Harold, once again: a proof from a set of assumptions
>> is a sequence of statements, each annotated with its justification.
>> The possible justifications are (1) by assumption (and you must identify
>> which assumption), or (2) by logical deduction from previous assumptions.
>> A proof of a contradiction is a proof such that one of the statements
>> is the negation of an earlier statement.
>>
>> Surely you realize that your line above is not a proof of a contradiction.
>> Where is your sequence of statements? Where is the justification of each?
>> What statement is the negation of what other statement?
>>
>> You have failed miserably at what claimed to be able to do.
>
>You snipped everything and then claim it failed. To which of my statements
>do you refer? In fact.....you very carefully snipped out my arguments.
>What is your problem?
I didn't ask for an argument. I asked for a proof. I asked for
a numbered sequence of statements such that each statement is either
an assumption or follows by logical deduction from previous
statements. You didn't provide a proof.
Here's your entire post. There is not a proof in it.
---------Below is Harold's Proof Which Isn't A Proof-------
Here is a repost from the end of our last discussion.
I had made explicit points which were never rebutted.
I give you another opportunity here. I have
start of repost:////////////////////////////////
////////////////////////////////Throop//////////////////////////////////////
H.Ellis Ensle
//////////////////////////////////McCullough////////////////////////////////
////////////////
Harold says...
> 1. Postulate: There exists a coordinate system assigning
> coordinates x,y,z,t to each event in spacetime such that
> A. Light travels along a path (x(t),y(t),z(t)) such that
> (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = c^2
OK....the speed of light is constant in the frame of reference you chose.
(Nothing more yet.)
> B. A test particles that is not acted upon by external
> forces will follow a path (x(t),y(t),z(t)) such that
> dx/dt = constant, dy/dt = constant, dz/dt = constant.
OK....this is classically called inertia, Newton's 1st law.
> C. For any spacetime path (x(t),y(t),z(t)) the proper
> time T for the path obeys the equation
> dT^2 = (1 - 1/c^2 ((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)) dt^2
OK....this is the length of the path (squared) in space-time. Note that
no physical significance has been attached to this value yet.
> 2. (Definition) A coordinate system will be called an "inertial
> coordinate system" if it obeys properties A,B, and C.
OK....since this is a definition, you can certainly do it, but you must be
alert to the fact that this is not the typical definition of inertial. The
typical inertial coordinates would only need to satisfy B.
Now since you have confiscated the term 'inertial', I will create a new
term that means a coordinate system which satisfies at least B. I will
call it '_simple_ inertial'.
> 3. (Theorem) If S is an inertial coordinate system, and S' is
> a second coordinate system related to the first by the Lorentz
> transformations, then S' is also an inertial coordinate system.
OK....this is also true. In fact I already conceded this in past posts.
I stated that the proper time is invariant under a Lorentz
transformation. The problem here is that the theorem has a limited
domain of applicability. It only deals with inertial coordinate
systems and that is only a subset of simple inertial coordinate
systems. In fact, I gave an example of a coordinate system which
satisfied B but not C. That would be the reciprocal frame of the
travelling twin. The reason the theorem doesn't cover the frame
is because one cannot get to the reciprocal twin's view using the
Lorentz transformation.
So ironically, your statements here are without contradiction, but
relativity is still wrong, because of the sin of omission.
> 4. (Relativity Postulate) The laws of physics have the same
> form when expressed in any inertial coordinate system.
>Exercise 1: Derive a contradiction from these postulates.
It is obvious. Two views that must represent a physically consistent
reality yield two different proper times.
>Exercise 2: Explain your notion that "SR demands reciprocity"
Harold Ensle
> Harold says...
> >
> >
> >Daryl McCullough <da...@cogentex.com> wrote
>
> >> Harold, once again: a proof from a set of assumptions
> >> is a sequence of statements, each annotated with its justification.
> >> The possible justifications are (1) by assumption (and you must
identify
> >> which assumption), or (2) by logical deduction from previous
assumptions.
> >> A proof of a contradiction is a proof such that one of the statements
> >> is the negation of an earlier statement.
> >>
> >> Surely you realize that your line above is not a proof of a
contradiction.
> >> Where is your sequence of statements? Where is the justification of
each?
> >> What statement is the negation of what other statement?
> >>
> >> You have failed miserably at what claimed to be able to do.
> >
> >You snipped everything and then claim it failed. To which of my
statements
> >do you refer? In fact.....you very carefully snipped out my arguments.
> >What is your problem?
>
> I didn't ask for an argument. I asked for a proof. I asked for
> a numbered sequence of statements such that each statement is either
> an assumption or follows by logical deduction from previous
> statements. You didn't provide a proof.
This post is where we left of last time and you are still not responding.
You are clearly in denial.
You asked for a "proof", but in order for you to see it, you are going
to have to put forth a little effort.
Well.....regardless I plan on posting something new on the subject
soon. I bet that even if I provide what you want, you will find some
excuse to ignore it. (after all, it is psychology, not physics).
H.Ellis Ensle
Daryl McCullough
>> I didn't ask for an argument. I asked for a proof. I asked for
>> a numbered sequence of statements such that each statement is either
>> an assumption or follows by logical deduction from previous
>> statements. You didn't provide a proof.
>
>This post is where we left of last time and you are still not responding.
>You are clearly in denial.
Yes, I'm denying that you posted a proof, because it is clear that
you didn't.
>You asked for a "proof", but in order for you to see it, you are going
>to have to put forth a little effort.
What effort do you expect me to do? You want *me* to construct a proof
for you? But I don't believe it is possible to construct a proof out
of what you have given me. You can prove me wrong by actually posting
a proof.
Once again, to construct a proof, you write down a numbered sequence
of statements. For each statement, you write down the justification.
The possible justifications are (1) the statement is an assumption, or
(2) the statement follows other statements by logical deduction.
Daryl McCullough
Let me try once more to show why the huge post that
you made is *not* a proof. Once again, let me cite
what I consider to be the relevant assumptions in
the twin paradox:
1. Postulate: There exists a coordinate system assigning
coordinates x,y,z,t to each event in spacetime such that
A. Light travels along a path (x(t),y(t),z(t)) such that
(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = c^2
B. A test particles that is not acted upon by external
forces will follow a path (x(t),y(t),z(t)) such that
dx/dt = constant, dy/dt = constant, dz/dt = constant.
C. For any spacetime path (x(t),y(t),z(t)) the proper
time T for the path obeys the equation
dT^2 = (1 - 1/c^2 ((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)) dt^2
2. (Definition) A coordinate system will be called an "inertial
coordinate system" if it obeys properties A,B, and C.
3. (Theorem) If S is an inertial coordinate system, and S' is
a second coordinate system related to the first by the Lorentz
transformations, then S' is also an inertial coordinate system.
Now to Harold's so-called proof:
>>>When the travelling twin is on the out-going part of the journey,
>>>he doesn't know yet that he will be accelerating. During that
>>>time he must be able to view himself at rest. Now what happens
>>>when he finally accelerates? Does he have to rewrite history
>>>and place himself in the space-time diagram of the stay-at-home?
>>What SR tells us that the notion of which events are
>>simultaneous is a function of what inertial reference
>>frame you are in. Let's look at the example of a twin
>>(initially 20 years old) who travels out for 5 years
>>(Earth time) and back 5 years at speed .8c. Let's label
>>six different events
>> e1 = the travelling twin blasts off
>> e2 = the stay-at-home twin's clock shows t = 1.8 years.
>> e3 = the stay-at-home twin's clock shows t = 5 years.
>> e4 = the travelling twin turns around
>> e5 = the stay-at-home twin's clock shows t = 8.2 years.
>> e6 = the twins reunite
>>In the reference frame of the stay-at-home twin,
>>e3 and e4 are simultaneous.
>>In the outward-going frame of the travelling twin,
>>e2 and e4 are simultaneous.
>
>Only from the point of view of the stay at-home.
>To the travelling twin, 5 years elapsed in order
>for the stay-at-home to get that distance away.
Harold, in an actual proof, every claim that you
make must be justified. What is the justification
for saying "To the travelling twin, 5 years
elapsed in order for the stay-at-home to get that
distance away." Answer: there is no justification.
Once again, in a proof, you aren't allowed to just
make a claim. You have to *justify* that claim.
You justify it by either claiming that it is an
assumption of whatever theory you are working with,
or else it is a pure mathematical fact (mathematical
facts actually should be proved, as well, but if the
fact is not too complex, we can probably see that it
is true) or else it follows from other claims by
logical deduction.
Your claim "To the travelling twin, 5 years elapsed
in order for the stay-at-home to get that distance
away" does not have a justification.
I actually know why you *think* that it follows. You
think it follows because of the following reasoning:
If in the frame of the stay-at-home twin, the
travelling twin turns around in year 5, then in the
frame of the travelling twin, the stay-at-home
twin turns around in year 5.
But that is not what relativity says. Relativity
doesn't say that everything is the same for both
twins. What it says is that the relevant
LAWS OF PHYSICS are the same for both twins.
The twins don't do the same thing: one twin turns
his rocket around. The other twin doesn't. Since
they do different things, there is no reason to
think that the same things happen to both twins.
It doesn't say that if one twin gets run
over by a bus, then the other twin will get run
over by a bus.
The fact that the travelling twin turns around
after 5 years (2.5 in ship-time) *isn't* a law
of physics. It's a choice made by the travelling
twin (or it's a choice made by whoever programmed
the rocket controls).
Okay, so you seem to believe that, to the travelling
twin, the stay-at-home twin takes 5 years to get to
his maximum distance away. Fine. Believe anything
you want. But if you claim to be proving something,
you have to say *why* you believe it. You have to
cite an assumption, or a mathematical fact to
justify your claim.
Harold Ensle
Wrong, this is based on the theory of relativity. I
am simply looking at the problem from the point of
view of the travelling twin as if he were at rest and
applying the Lorentz transformations. This is
completely valid especially as the travelling twin has
not yet accelerated. Why should the travelling twin
see anything different than the stay at home (up to
this point)? Their viewpoints cannot be distinguished
by SR. That _is_ based on the axioms of SR.
> Once again, in a proof, you aren't allowed to just
> make a claim. You have to *justify* that claim.
It was done above......trivially.
> You justify it by either claiming that it is an
> assumption of whatever theory you are working with,
> or else it is a pure mathematical fact (mathematical
> facts actually should be proved, as well, but if the
> fact is not too complex, we can probably see that it
> is true) or else it follows from other claims by
> logical deduction.
>
> Your claim "To the travelling twin, 5 years elapsed
> in order for the stay-at-home to get that distance
> away" does not have a justification.
>
> I actually know why you *think* that it follows. You
> think it follows because of the following reasoning:
>
> If in the frame of the stay-at-home twin, the
> travelling twin turns around in year 5, then in the
> frame of the travelling twin, the stay-at-home
> twin turns around in year 5.
>
> But that is not what relativity says.
The problem is that this _is_ what relativity claims.
It is simply mutual time-dilation.
>Relativity
> doesn't say that everything is the same for both
> twins.
It does at least up to the point of acceleration, the
only difference being -v instead of v. This whole
thing is blatently obvious and I do not even
understand why you are arguing it.
>What it says is that the relevant
> LAWS OF PHYSICS are the same for both twins.
>
> The twins don't do the same thing: one twin turns
> his rocket around. The other twin doesn't. Since
> they do different things, there is no reason to
> think that the same things happen to both twins.
> It doesn't say that if one twin gets run
> over by a bus, then the other twin will get run
> over by a bus.
>
> The fact that the travelling twin turns around
> after 5 years (2.5 in ship-time) *isn't* a law
> of physics. It's a choice made by the travelling
> twin (or it's a choice made by whoever programmed
> the rocket controls).
>
> Okay, so you seem to believe that, to the travelling
> twin, the stay-at-home twin takes 5 years to get to
> his maximum distance away. Fine. Believe anything
> you want. But if you claim to be proving something,
> you have to say *why* you believe it. You have to
> cite an assumption, or a mathematical fact to
> justify your claim.
I am not citing _new_ assumptions, I am simply using
the postulates of SR. In this particular case your claim
that it is not 5 years is based on the stay-at-home's view
of what is simultaneous. But if the travelling twin is at
rest (which up to this point is allowable), there is no
reason for him to see this same interval.
H.Ellis Ensle
Daryl McCullough
You can't just say "based on the theory of relativity".
Tell me which assumption of relativity you used. Tell
me how your claim logically follows from that assumption
or assumptions.
>I am simply looking at the problem from the point of
>view of the travelling twin as if he were at rest and
>applying the Lorentz transformations.
Just tell me what axioms of relativity are relevant here.
State your assumptions, and state your axioms, tell how
your conclusion logically follows.
>Why should the travelling twin
>see anything different than the stay at home (up to
>this point)?
He doesn't. In the reference frame of the travelling twin we
have:
When the travelling twin has aged 3 years, the stay-at-home
twin has only aged 1.8 years. At that time, the stay-at-home
twin is 2.4 light-years away.
In the reference frame of the stay-at-home twin we have
When the stay-at-home twin has aged 3 years, the travelling
twin has only aged 1.8 years. At that time, the travelling
twin is 2.4 light-years away.
It's exactly symmetrical.
Since the travelling twin turns around at 3 years (according
to his own clocks) the situations stop being symmetrical after
that.
>> Once again, in a proof, you aren't allowed to just
>> make a claim. You have to *justify* that claim.
>
>It was done above......trivially.
No. Cite the axiom, and show that plugging in the particular
values lead to your conclusion.
>> I actually know why you *think* that it follows. You
>> think it follows because of the following reasoning:
>>
>> If in the frame of the stay-at-home twin, the
>> travelling twin turns around in year 5, then in the
>> frame of the travelling twin, the stay-at-home
>> twin turns around in year 5.
>>
>> But that is not what relativity says.
>
>The problem is that this _is_ what relativity claims.
>It is simply mutual time-dilation.
You can't just say "mutual time dilation". Cite what assumptions
you are using. At the top of the page, I've listed the
relevant assumptions.
>>Relativity
>> doesn't say that everything is the same for both
>>twins.
>
>It does at least up to the point of acceleration,
Right. That happens at 3 years, ship time. So the
first 3 years for the stay-at-home twin are exactly
symmetrical with the first 3 years for the travelling
twin. After 3 years, different things happen.
>> Okay, so you seem to believe that, to the travelling
>> twin, the stay-at-home twin takes 5 years to get to
>> his maximum distance away. Fine. Believe anything
>> you want. But if you claim to be proving something,
>> you have to say *why* you believe it. You have to
>> cite an assumption, or a mathematical fact to
>> justify your claim.
>
>I am not citing _new_ assumptions, I am simply using
>the postulates of SR.
Say which one. Name the assumption. I've listed the
relevant assumptions at the top of the page. Tell me
which one is relevant to your claim.
The travelling twin travels without accelerating for 3
years (ship time). The stay-at-home twin travels without
acceleration for 10 years. If you want to use relativity
to compare them, you can only compare the first three
years for the travelling twin to the first 3 years for
the stay-at-home twin. After that point, things are
not symmetrical.
Harold Ensle
I am not even sure what more you want. Apply the
Lorentz transformations from the frame of the
moving twin. Is this or is this not what relativity
claims that you can do?
If relativity claims otherwise, please specify.
I am pasting this from your scenario above:
///////////////////////////////////////////////////////////////////////////
> >> >>In the outward-going frame of the travelling twin,
> >> >>e2 and e4 are simultaneous.
///////////////////////////////////////////////////////////////////////////
This is the point of disagreement. You claim that this is
the point of view of the travelling twin, but it is not. It
is the stay-at-home's view of the travelling twin's view.
The travelling twin doesn't know that he is moving (up
to this point), so there is no way he can manipulate
any simultaneity in any way, he can only see a mirror
image of the stay-at-home's motion away. Therefore,
we can only assume that he must see 5 years for
himself as the stay-at-home did for himself. ANY
correction to this value must be based on the
travelling twin's velocity relative to the stay-at-home,
but from his viewpoint he is at rest. So just as the
stay at home does not transform his own time,
neither can the travelling twin (in his own view and
up to the point of acceleration).
[............]
H.Ellis Ensle
Daryl McCullough
>> You can't just say "based on the theory of relativity".
>> Tell me which assumption of relativity you used. Tell
>> me how your claim logically follows from that assumption
>> or assumptions.
>
>I am not even sure what more you want.
All right, let's try again:
Let's look at the example of a twin
(initially 20 years old) who travels out for 5 years
(Earth time) and back 5 years at speed .8c. Let's label
six different events
e1 = the travelling twin blasts off
e2 = the stay-at-home twin's clock shows t = 1.8 years.
e3 = the stay-at-home twin's clock shows t = 5 years.
e4 = the travelling twin turns around
e5 = the stay-at-home twin's clock shows t = 8.2 years.
e6 = the twins reunite
In the frame of the stay-at-home twin, these events have
the following coordinates:
e1 = (x=0,t=0)
e2 = (x=0,t=1.8)
e3 = (x=0,t=5)
e4 = (x=4,t=5)
e5 = (x=0,t=8.2)
e6 = (x=0,t=10)
I claim that in the outgoing reference frame of the
travelling twin, e2 and e4 are simultaneous. My proof
is this:
1. Definition: e2 and e4 are simultaneous in the
reference frame of the outgoing twin if the time
coordinate of e2 in that frame is equal to the
time coordinate of e4 in that frame.
2. By the Lorentz transformations, the time coordinate
in the outgoing twin's frame is given by
t' = gamma( t - v x/c^2)
where x and t are the coordinates in the stay-at-home frame.
3. For our example, gamma = 1.66..., v = .8. For
event e2, t = 1.8, x = 0.
4. Plugging in these values gives
t' = 1.666 * (1.8 - 0) = 3 years for event e2.
5. In our example, gamma = 1.66..., v = .8. For event e4,
t = 5, x = 4.
6. Plugging in these values gives
t' = 1.66 * (5 - .8 * 4)
= 3 years for event e4
7. By 4 and 6, the time coordinates of e2 and e4 in the
travelling twin's outward-going frame are both 3 years.
8. Therefore, e2 and e4 are simultaneous in the
travelling twin's outward-going frame.
All right. That's my derivation of the claim that
e2 and e4 are simultaneous in the outward-going
frame of the travelling twin.
You made the claim that
>To the travelling twin, 5 years elapsed in order
>for the stay-at-home to get that distance away.
I want you to show what formula you used to derive
that result, 5 years.
>I am pasting this from your scenario above:
>///////////////////////////////////////////////////////////////////////////
>> >> >>In the outward-going frame of the travelling twin,
>> >> >>e2 and e4 are simultaneous.
>///////////////////////////////////////////////////////////////////////////
>
>This is the point of disagreement. You claim that this is
>the point of view of the travelling twin, but it is not.
It follows from these assumptions:
1. In the stay-at-home frame, e2 has coordinates (x=0,t=1.8),
and e4 has coordinates (x=4, t=5). (All distances in light-years,
and all times in years).
2. To find out the coordinates of e2 and e4 in the outward-going
frame of the travelling twin, use the Lorentz transformations.
3. The result is that in the outward-going
frame of the travelling twin, both events have time
coordinate equal to 3 years.
4. By definition of "simultaneous", they are simultaneous.
>It is the stay-at-home's view of the travelling twin's view.
Who says? What does that follow from? The Lorentz transformations
don't mention anything about one observer's view of another
observer's view.
In a derivation, you aren't allowed to make claims without
justifying them. Tell me what axiom you are using.
Harold Ensle
But how will you determine these coordinates?
> 2. By the Lorentz transformations, the time coordinate
> in the outgoing twin's frame is given by
>
> t' = gamma( t - v x/c^2)
>
> where x and t are the coordinates in the stay-at-home frame.
As soon as you use the Lorentz transformations, you have based
your result on the view of the stay-at-home. Like I already stated
and you snipped. The travelling twin cannot transform his own
coordinates as he is at rest in his own frame.
> 3. For our example, gamma = 1.66..., v = .8. For
> event e2, t = 1.8, x = 0.
>
> 4. Plugging in these values gives
> t' = 1.666 * (1.8 - 0) = 3 years for event e2.
>
> 5. In our example, gamma = 1.66..., v = .8. For event e4,
> t = 5, x = 4.
>
> 6. Plugging in these values gives
> t' = 1.66 * (5 - .8 * 4)
> = 3 years for event e4
>
> 7. By 4 and 6, the time coordinates of e2 and e4 in the
> travelling twin's outward-going frame are both 3 years.
>
> 8. Therefore, e2 and e4 are simultaneous in the
> travelling twin's outward-going frame.
>
> All right. That's my derivation of the claim that
> e2 and e4 are simultaneous in the outward-going
> frame of the travelling twin.
As soon as you used the Lorentz transformations to represent
the travelling twin's view of himself, you made an error. Relative
to himself, the travelling twin is at rest (before acceleration), so
he _cannot_ apply a transformation to himself.
H.Ellis Ensle
Daryl McCullough
>> In the frame of the stay-at-home twin, these events have
>> the following coordinates:
>>
>> e1 = (x=0,t=0)
>> e2 = (x=0,t=1.8)
>> e3 = (x=0,t=5)
>> e4 = (x=4,t=5)
>> e5 = (x=0,t=8.2)
>> e6 = (x=0,t=10)
>>
>> I claim that in the outgoing reference frame of the
>> travelling twin, e2 and e4 are simultaneous. My proof
>> is this:
>>
>> 1. Definition: e2 and e4 are simultaneous in the
>> reference frame of the outgoing twin if the time
>> coordinate of e2 in that frame is equal to the
>> time coordinate of e4 in that frame.
>
>But how will you determine these coordinates?
>
>> 2. By the Lorentz transformations, the time coordinate
>> in the outgoing twin's frame is given by
>>
>> t' = gamma( t - v x/c^2)
>>
>> where x and t are the coordinates in the stay-at-home frame.
>
>As soon as you use the Lorentz transformations, you have based
>your result on the view of the stay-at-home.
Yes, I'm basing it on the assumption that the Lorentz
transformations correctly relate the two coordinate systems:
If the coordinates of an event in one frame are (x,t),
then the coordinates of the event in the other frame
is (gamma(x - vt), gamma(t - vx/c^2))
>Like I already stated and you snipped. The travelling twin
>cannot transform his own coordinates as he is at rest in his own frame.
I don't know what you are talking about. It isn't the travelling
twin who is applying the Lorentz transformations. It is Daryl
McCullough who is using the Lorentz transformations to predict
what the travelling twin will measure.
>> 3. For our example, gamma = 1.66..., v = .8. For
>> event e2, t = 1.8, x = 0.
>>
>> 4. Plugging in these values gives
>> t' = 1.666 * (1.8 - 0) = 3 years for event e2.
>>
>> 5. In our example, gamma = 1.66..., v = .8. For event e4,
>> t = 5, x = 4.
>>
>> 6. Plugging in these values gives
>> t' = 1.66 * (5 - .8 * 4)
>> = 3 years for event e4
>>
>> 7. By 4 and 6, the time coordinates of e2 and e4 in the
>> travelling twin's outward-going frame are both 3 years.
>>
>> 8. Therefore, e2 and e4 are simultaneous in the
>> travelling twin's outward-going frame.
>>
>> All right. That's my derivation of the claim that
>> e2 and e4 are simultaneous in the outward-going
>> frame of the travelling twin.
>
>As soon as you used the Lorentz transformations to represent
>the travelling twin's view of himself, you made an error.
I don't know what you are talking about. I (that is, Daryl McCullough)
used the Lorentz transformations to figure out what the coordinates
of e2 and e4 are in the coordinate system of the travelling twin.
I didn't say anything about "the travelling twin's view of himself".
Do you agree that you haven't come anywhere close to giving what
I would consider to be a "proof"? You haven't labelled your statements,
you haven't justified your statements by appealing to axioms or previously
justified statements. You've just been making claims without giving
justifications.
Dirk Van de moortel
"Harold Ensle" <hen...@ix.netcom.com> wrote in message news:aqoqf3$hs7$1...@slb6.atl.mindspring.net...
>
> Daryl McCullough <da...@cogentex.com> wrote in message
> news:aqoeu...@drn.newsguy.com...
> > Harold says...
> >
>
> As soon as you use the Lorentz transformations, you have based
> your result on the view of the stay-at-home. Like I already stated
> and you snipped. The travelling twin cannot transform his own
> coordinates as he is at rest in his own frame.
>
[snip]
>
> > All right. That's my derivation of the claim that
> > e2 and e4 are simultaneous in the outward-going
> > frame of the travelling twin.
>
> As soon as you used the Lorentz transformations to represent
> the travelling twin's view of himself, you made an error. Relative
> to himself, the travelling twin is at rest (before acceleration), so
> he _cannot_ apply a transformation to himself.
>
> H.Ellis Ensle
A stunning beauty!
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#CannotApply
Title: "He _cannot_ apply a transformation to himself"
Dirk Vdm
Harold Ensle
Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in
message news:OVWz9.15951$Nd....@afrodite.telenet-ops.be...
Explain what is wrong with this statement. (in context!)
H.Ellis Ensle
Harold Ensle
Whose velocity are you using in the Lorentz transformations?
The travelling twin thinks he is at rest so he sees the stay-at-home
moving away, he then can do a transformation based on that
velocity for the time and displacement of the stay-at-home. The
relationship he obtains is based on his time. He cannot use
the transformations on himself to get his own altered time in
relation to his own time. After all, his time is merely his time.
So if you used a Lorentz transformation to determine the
synchronization of the turn around in the travelling twins view,
you had to base it on the velocity as viewed by the stay-at-home.
That then biases the result to the frame of the stay-at-home.
>
>
> >> 3. For our example, gamma = 1.66..., v = .8. For
> >> event e2, t = 1.8, x = 0.
> >>
> >> 4. Plugging in these values gives
> >> t' = 1.666 * (1.8 - 0) = 3 years for event e2.
> >>
> >> 5. In our example, gamma = 1.66..., v = .8. For event e4,
> >> t = 5, x = 4.
> >>
> >> 6. Plugging in these values gives
> >> t' = 1.66 * (5 - .8 * 4)
> >> = 3 years for event e4
> >>
> >> 7. By 4 and 6, the time coordinates of e2 and e4 in the
> >> travelling twin's outward-going frame are both 3 years.
> >>
> >> 8. Therefore, e2 and e4 are simultaneous in the
> >> travelling twin's outward-going frame.
> >>
> >> All right. That's my derivation of the claim that
> >> e2 and e4 are simultaneous in the outward-going
> >> frame of the travelling twin.
> >
> >As soon as you used the Lorentz transformations to represent
> >the travelling twin's view of himself, you made an error.
>
> I don't know what you are talking about. I (that is, Daryl McCullough)
> used the Lorentz transformations to figure out what the coordinates
> of e2 and e4 are in the coordinate system of the travelling twin.
> I didn't say anything about "the travelling twin's view of himself".
Daryl McCullough used the Lorentz transformations from the rest
frame of the stay-at-home. BUT "the travelling twin's view of himself
(and the stay-at-home)" is exactly what we need to see, because
that is where the paradox is.
> Do you agree that you haven't come anywhere close to giving what
> I would consider to be a "proof"?
No I did not give a proof, but I clearly (?) demonstrated that your proof
was not a proof.
>You haven't labelled your statements,
> you haven't justified your statements by appealing to axioms or previously
> justified statements. You've just been making claims without giving
> justifications.
That is not fair. I did justify my statements, but you have to try and
understand what I am saying.
H.Ellis Ensle
Daryl McCullough
>Whose velocity are you using in the Lorentz transformations?
You seem to be missing some essential points about the Lorentz
transformations. If you are given the coordinates of an event
in frame S, then you can derive the coordinates of the same
event in frame S' according to the formula:
x' = gamma (x - vt)
t' = gamma (t - vx/c^2)
where v is the velocity of an observer at rest in S' as measured
in frame S.
>The travelling twin thinks he is at rest so he sees the stay-at-home
>moving away, he then can do a transformation based on that
>velocity for the time and displacement of the stay-at-home.
If the travelling twin wants to know the coordinates in the
frame of the stay-at-home twin, he uses the velocity of the
*stay-at-home* twin (as measured in the travelling twin's frame).
>The relationship he obtains is based on his time. He cannot use
>the transformations on himself to get his own altered time in
>relation to his own time. After all, his time is merely his time.
>So if you used a Lorentz transformation to determine the
>synchronization of the turn around in the travelling twins view,
>you had to base it on the velocity as viewed by the stay-at-home.
No. If you are given the coordinates in the travelling twin's
frame, then you use the velocity of the stay-at-home twin,
as measured in the travelling twin's frame.
>That then biases the result to the frame of the stay-at-home.
I give up. I don't know what you are talking about.
You keep making claims without backing them up. That's
why your thinking is so confused. You need to try to set down in
as clear a way as possible what you are assuming, written in
unambiguous mathematical notation. Anything you claim should
be backed up by mathematical and/or logical derivations from
those assumptions.
You've never bothered to get straight what the assumptions
behind relativity are, and you keep coming up with confusion
because of that. You need to start at the beginning again,
and try to learn the physics correctly this time.
>> Do you agree that you haven't come anywhere close to giving what
>> I would consider to be a "proof"?
>
>No I did not give a proof
That's my point. If you actually tried to formulate your
objections as a rigorous proof, you would realize how incorrect
your reasoning is. That's the point of using rigorous mathematics,
it shows mistakes that occur from sloppy reasoning.
Dirk Van de moortel
"Harold Ensle" <hen...@ix.netcom.com> wrote in message news:aqq49t$97l$1...@slb0.atl.mindspring.net...
By definition, events are independent of coordinates or
observers or frames.
Suppose that something happens in spacetime, for instance
a little explosion.
One observer makes measurements of this event in his
frame, and labels the measurements t, x, y and z.
The observer attaches the numbers (t,x,y,z) to the event to
uniquely identify it for future reference.
Another observer makes measurements in his frame of the
*same* event and labels his measurements t', x', y' and z'.
This other observer attaches the numbers (t',x',y',z') to the
same event to uniquely identify it for future reference.
The Lorentz transformation gives a relation between the
numbers (t,x,y,z) and the numbers (t',x',y',z').
It is a prescription that can be used by everyone (who
knows a little bit of algebra) to calculate the numbers
(t',x',y',z') when the numbers (t,x,y,z) are known,
or vice versa.
It can also be used
- by both observers to calculate and verify each other's
numbers, after they have been exchanging information
about which numbers they attach to certain events.
- by Daryl and by you and by me to calculate and
verify and explain about how things are supposed
to behave in special relativity.
I hope that this provides enough context...
Dirk Vdm
Harold Ensle
> Harold says...
>
> >Whose velocity are you using in the Lorentz transformations?
>
> You seem to be missing some essential points about the Lorentz
> transformations.
No, I am not.
>If you are given the coordinates of an event
> in frame S, then you can derive the coordinates of the same
> event in frame S' according to the formula:
>
> x' = gamma (x - vt)
> t' = gamma (t - vx/c^2)
>
> where v is the velocity of an observer at rest in S' as measured
> in frame S.
Yes, the inverse transforms......I know.
> >The travelling twin thinks he is at rest so he sees the stay-at-home
> >moving away, he then can do a transformation based on that
> >velocity for the time and displacement of the stay-at-home.
>
> If the travelling twin wants to know the coordinates in the
> frame of the stay-at-home twin, he uses the velocity of the
> *stay-at-home* twin (as measured in the travelling twin's frame).
Yes...as I wrote above.
> >The relationship he obtains is based on his time. He cannot use
> >the transformations on himself to get his own altered time in
> >relation to his own time. After all, his time is merely his time.
> >So if you used a Lorentz transformation to determine the
> >synchronization of the turn around in the travelling twins view,
> >you had to base it on the velocity as viewed by the stay-at-home.
>
> No. If you are given the coordinates in the travelling twin's
> frame, then you use the velocity of the stay-at-home twin,
> as measured in the travelling twin's frame.
I thought you might state this, and it is, of course, true. But
think carefully here. The velocity of the stay-at-home cannot
effect the synchronicity of the travelling twin, because the
travelling twin is at rest. What you need to do is _pretend_
that you are each of the twins, one at a time, and imagine
what would be observed if each one innocently applied
the Lorentz transformations to the problem.....without
second-guessing the contradiction and adjusting the
simultaneity to favor one of them. And when I mean
the Lorentz transformations, I mean the inverse transformations
for the travelling twin as you mentioned above.
> >That then biases the result to the frame of the stay-at-home.
>
> I give up. I don't know what you are talking about.
[.........insults snipped.......]
H.Ellis Ensle
Harold Ensle
Hardly...as it does not answer my question. How was
the statement incorrect. When you use the Lorentz
transformations, it tells you the relationship between the
moving frame and your own "rest" frame. To use them on
yourself would be nonsense, since you are at rest in your
own frame (not completely, but you would simply get
t=t' and x=x'). What I should have included was: You cannot
use the transformations on yourself and get any meaningful
information, which was quite precisely what McCullough
claimed. He obtained a different measurement of the
travelling twin's _self-view_ of the outgoing interval from
the stay-at-home's _self-view_ of the interval. But in order
to get a difference, one of them had to be meaningfully
(not an identity) transformed.
Now you did not understand it (and probably still do not) and
simply put it in your collection of "errors", which I suppose
you maintain as a way to cope with an inferiority complex.
However, seeing how poorly you misunderstood the argument,
I suspect that the site is more likely a show case for all the
things you do not understand.
H.Ellis Ensle
Daryl McCullough
>> I give up. I don't know what you are talking about.
>
>[.........insults snipped.......]
It's not an insult to say that you don't understand relativity,
or how to make a rigorous scientific argument, it's just being
realistic.
Dirk Van de moortel
"Harold Ensle" <hen...@ix.netcom.com> wrote in message news:aqrkvq$4mb$1...@slb4.atl.mindspring.net...
The Lorentz transformation tells the relationship between
coordinates of an event according to one frame and
coordinates of the same event according to another frame.
> To use them on
> yourself would be nonsense, since you are at rest in your
> own frame (not completely, but you would simply get
> t=t' and x=x').
Not really.
Since I am at rest im my own frame, all the x-values of the
events taking place on myself, are 0, while they have ever
increasing t-values that I can read off my clock.
When you fly away from me with speed v, and when you
label the events that take place on me with the coordinates
x' and t', then you can plug x=0 in the reverse transformation
x = gamma(x'+vt')
and find that all the events taking place on me, satisfy the
relation
x' = -vt'
which precisely expresses the fact that we are flying away
from each other with speed v.
This was a little example of how it works.
This is how we talk about events and coordinates and frames.
These are the rules of the game.
I can give more examples if you want.
In fact, if you like, I can precisely show how this system can
be used to explain what happens with the twins. Only if you
are interested though... It is up to you.
Dirk Vdm
Nicholas Steele
>
> This was a little example of how it works.
> This is how we talk about events and coordinates and frames.
> These are the rules of the game.
> I can give more examples if you want.
> In fact, if you like, I can precisely show how this system can
> be used to explain what happens with the twins. Only if you
> are interested though... It is up to you.
>
> Dirk Vdm
I know this was addressed to someone else but I'd be interested in seeing
the solution. Thanks in advance.
Nicholas Steele
Dirk Van de moortel
"Nicholas Steele" <a@b.c> wrote in message news:PGdA9.782606$v53.29...@news3.calgary.shaw.ca...
I will post it on a new thread:
"Twins, events and transformations".
Give me 20 minutes :-)
Dirk Vdm
Harold Ensle
> Harold says...
>
> >> I give up. I don't know what you are talking about.
> >
> >[.........insults snipped.......]
>
> It's not an insult to say that you don't understand relativity,
> or how to make a rigorous scientific argument, it's just being
> realistic.
Interesting point, but since it is you who does not understand
(noting the lack of any reply to my argument), then it can
only be assumed that it is you who doesn't understand
relativity or how to understand a reasonable argument.
So please be realistic and don't blame me for your own failings.
H.Ellis Ensle