The Twin Fallacy and Schrodinger Kat
Seju Strich
Kat, who has been diagnosed with cancer, has 4 years to live. She
volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
travels at 0.866c, and arrives there ~5.0 years later. According to
classical SR, Kat's time slows to a half and ages only ~2.5 years,
arriving in Alpha Centauri, fairly bright eyed and bushy tailed, with
~1.5 years more of life.
But... also according to SR, during the space flight, inside the rocket
(Kat's inertial frame), its internal clock registers time normally, and
~5.0 years pass by for Kat before the landing at Alpha Centauri. She is
dead on arrival.
As Schrodinger would ask, is Kat alive or dead, when the rocket is
opened at Alpha Centauri? According to SR, she is alive and dead at
the same time. But there is no superposition in relativity, so
therefore, we have a contradiction, and relativity must be false, as
some 'nuts' have already known.
--
Seju Strich
papa...@gmail.com
wrote:
Well...those nuts continue to be nuts and totally wrong.
Since for an observer located on Earth a moving clock will run slow.
But, for the Kat it does not matter what the Earth observer is
calculating. His reality is given by his own clock and for sure he
will be probably dead before arriving, at 5.04 years to Alpha
Centauri.
Try again.
Miguel Rios
Bryan Olson
> Kat, who has been diagnosed with cancer, has 4 years to live. She
> volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
> travels at 0.866c, and arrives there ~5.0 years later. According to
> classical SR, Kat's time slows to a half and ages only ~2.5 years,
> arriving in Alpha Centauri, fairly bright eyed and bushy tailed, with
> ~1.5 years more of life.
>
> But... also according to SR, during the space flight, inside the rocket
> (Kat's inertial frame), its internal clock registers time normally, and
> ~5.0 years pass by for Kat before the landing at Alpha Centauri. She is
> dead on arrival.
Nope. See:
http://en.wikipedia.org/wiki/Length_contraction
In Kat's inertial frame, Alpha Centauri is 2.18 light years
away from Earth, and approaching Kat at 0.866c. That makes
it a 2.5 year trip for Kat.
> As Schrodinger would ask, is Kat alive or dead, when the rocket is
> opened at Alpha Centauri? According to SR, she is alive and dead at
> the same time. But there is no superposition in relativity, so
> therefore, we have a contradiction,
Nope. Just another big red X through someone's attempt to
do a sophomore physics problem.
> and relativity must be false, as
> some 'nuts' have already known.
Get a clue already.
--
--Bryan
Seju Strich
Bryan, read carefully... (keep your lips at rest for a moment so you can
understand this simple sophomore problem).
The 2.5 years in Kat's rest frame is EXACTLY EQUAL to a 2.5 year period
in my rest frame.
Again, in my rest frame, the distance to AC is 4 lightyears. Thus Kat
has travelled faster than light, as she has been from earth to AC in
2.5 years in EARTH TIME.
amen.
--
Seju Strich
Dirk Van de moortel
Seju.Stri...@physicsbanter.com
> Bryan, read carefully... (keep your lips at rest for a moment so you can
> understand this simple sophomore problem).
>
> The 2.5 years in Kat's rest frame is EXACTLY EQUAL to a 2.5 year period
> in my rest frame.
Forget about frames. They seem to confuse you.
Think (and talk) about clocks and how they are used to
measure the duration of some process.
For her trip, Kat measures 2.5 years on her clock.
For that same trip, you measure 5.0 years on your clock.
I don't see what is "exactly equal" about 2.5 and 5.0.
>
> Again, in my rest frame, the distance to AC is 4 lightyears. Thus Kat
> has travelled faster than light, as she has been from earth to AC in
> 2.5 years in EARTH TIME.
>
> amen.
When you stumble upon something you don't understand,
it usually is a safer trategy to ask for help, than to cry for
attention.
Dirk Vdm
Strich 9
-Forget about frames. They seem to confuse you.
Think (and talk) about clocks and how they are used to
measure the duration of some process.
For her trip, Kat measures 2.5 years on her clock.
For that same trip, you measure 5.0 years on your clock.
I don't see what is "exactly equal" about 2.5 and 5.0.
Dirk Vdm-
take it easy so your mind can work better. follow me closely. it is
deceivingly simple.
according to SR, rest clocks keep equal time.
Kat's trip lasted 2.5hours in REST time (Kat frame).
It also lasted 5 hours in REST time (Earth frame).
BOTH of these are rest times, so we can COMPARE them.
Obviously 5hrs is NOT equal to 2.5 hrs!!!
One trip, and two travel times from the same unit of measure.
If you cannot see that contradiction, well, I'll leave the commentary
to the other viewers. Think first, before you talk. You might bite
your tongue, or worse when the weather is cold, it may even freeze...
--
Strich 9
The TimeLord
<Seju.Stri...@physicsbanter.com> in
Seju.Stri...@physicsbanter.com:
> Kat, who has been diagnosed with cancer, has 4 years to live. She
> volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
> travels at 0.866c, and arrives there ~5.0 years later. According to
> classical SR, Kat's time slows to a half and ages only ~2.5 years,
> arriving in Alpha Centauri, fairly bright eyed and bushy tailed, with
> ~1.5 years more of life.
So far so good.
>
> But... also according to SR, during the space flight, inside the rocket
> (Kat's inertial frame), its internal clock registers time normally, and
> ~5.0 years pass by for Kat before the landing at Alpha Centauri. She is
> dead on arrival.
No. She sees the trip as Lorentz-contracted to half the distance (ie
~2.18 ly long trip). So at 86.6%c it still only takes ~2.5 years for her.
On Earth it will take ~5 years.
>
> As Schrodinger would ask, is Kat alive or dead, when the rocket is
Schrödinger has nothing to do with this. She will be alive at Alpha-
Centauri.
[...]
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
Dirk Van de moortel
Strich.9...@physicsbanter.com
>> -Forget about frames. They seem to confuse you.
>> Think (and talk) about clocks and how they are used to
>> measure the duration of some process.
>> For her trip, Kat measures 2.5 years on her clock.
>> For that same trip, you measure 5.0 years on your clock.
>> I don't see what is "exactly equal" about 2.5 and 5.0.
>>
>> Dirk Vdm-
> take it easy so your mind can work better. follow me closely. it is
> deceivingly simple.
>
> according to SR, rest clocks keep equal time.
Yes, clocks at rest w.r.t. each other.
>
> Kat's trip lasted 2.5hours in REST time (Kat frame).
> It also lasted 5 hours in REST time (Earth frame).
> BOTH of these are rest times, so we can COMPARE them.
> Obviously 5hrs is NOT equal to 2.5 hrs!!!
Indeed, 5 Kat-hours is not equal to 2.5 Strich-hours.
>
> One trip, and two travel times from the same unit of measure.
There are two units of measurement:
Unit1: Kat's hours on Kat's clock.
Unit2: Your hours on your clock.
One trip, and two travel times from two units of measure.
It can't be any simpler than that.
>
> If you cannot see that contradiction, well, I'll leave the commentary
> to the other viewers. Think first, before you talk. You might bite
> your tongue, or worse when the weather is cold, it may even freeze...
Since there are two units of measurement, I don't see
a contradiction.
Again, if stumble onto something you don't understand, ask first,
before you proclaim.
Dirk Vdm
PD
2a88...@physicsbanter.com> wrote:
> Kat, who has been diagnosed with cancer, has 4 years to live. She
> volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
> travels at 0.866c, and arrives there ~5.0 years later. According to
> classical SR, Kat's time slows to a half and ages only ~2.5 years,
> arriving in Alpha Centauri, fairly bright eyed and bushy tailed, with
> ~1.5 years more of life.
>
> But... also according to SR, during the space flight, inside the rocket
> (Kat's inertial frame), its internal clock registers time normally, and
> ~5.0 years pass by for Kat before the landing at Alpha Centauri. She is
> dead on arrival.
No, that's incorrect. SR does not say that. On the ship's internal
clock, 2.5 years have elapsed, while on the earth-bound clock 5 years
have elapsed. She arrives bushy tailed.
You have an incorrect notion of what "registers time normally" means.
Do you need a pointer to some reference reading material to brush up
on what SR really says?
PD
wrote:
> -Forget about frames. They seem to confuse you.
> Think (and talk) about clocks and how they are used to
> measure the duration of some process.
> For her trip, Kat measures 2.5 years on her clock.
> For that same trip, you measure 5.0 years on your clock.
> I don't see what is "exactly equal" about 2.5 and 5.0.
>
> Dirk Vdm-
>
> take it easy so your mind can work better. follow me closely. it is
> deceivingly simple.
>
> according to SR, rest clocks keep equal time.
Not in different frames. It might be helpful to read something better
than what you've got so far on what SR really says.
mlut...@wanadoo.fr
SperM.hotmail.com> wrote:
> Seju Strich <Seju.Strich.2a93...@physicsbanter.com> wrote in message
>
> Seju.Strich.2a93...@physicsbanter.com
>
> > Bryan, read carefully... (keep your lips at rest for a moment so you can
> > understand this simple sophomore problem).
>
> > The 2.5 years in Kat's rest frame is EXACTLY EQUAL to a 2.5 year period
> > in my rest frame.
>
> Forget about frames. They seem to confuse you.
> Think (and talk) about clocks and how they are used to
> measure the duration of some process.
> For her trip, Kat measures 2.5 years on her clock.
> For that same trip, you measure 5.0 years on your clock.
> I don't see what is "exactly equal" about 2.5 and 5.0.
>
According to Earthling Vdm , Kat travels during
5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
on her clock.
But Kat considers herself at rest, the Earth moving at
0.866 c wrt to her.
After 5 years of her time (Kat is still alive, she
took a new drug), she claims that Dirk Vdm measured
only 2.5 years on his clock.
According to SR, both Kat and Dirk are right.
Marcel Luttgens
Yua...@gmail.com
>
> As Schrodinger would ask, is Kat alive or dead, when the rocket is
> opened at Alpha Centauri? According to SR, she is alive and dead at
> the same time. But there is no superposition in relativity, so
> therefore, we have a contradiction, and relativity must be false, as
> some 'nuts' have already known.
>
> --
You thought she had 4 years to live because some doctor told you.
Is she alive or dead? That depends upon whether the doctor was
correct!
In any case, if she IS alive when she gets there, how will you know?
You can't see her, she's too far way!
That's the Uncertainty Principle.
From a relativistic point of view, the information as to her status
can't get to you for 4.37 years. Added on to her 5 year travel time,
that's over 9 years. You might even be dead yourself by that time, so
why worry about her!
By the way, 9 years is enough time to get a degree and a PhD , which
is what I'll be doing (if the doctors are lying).
Love,
Jenny
Dirk Van de moortel
2b4a312a-74d6-4cf5...@m3g2000hsc.googlegroups.com
> On Jun 26, 2:04 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> SperM.hotmail.com> wrote:
>> Seju Strich <Seju.Strich.2a93...@physicsbanter.com> wrote in message
>>
>> Seju.Strich.2a93...@physicsbanter.com
>>
>>> Bryan, read carefully... (keep your lips at rest for a moment so you can
>>> understand this simple sophomore problem).
>>
>>> The 2.5 years in Kat's rest frame is EXACTLY EQUAL to a 2.5 year period
>>> in my rest frame.
>>
>> Forget about frames. They seem to confuse you.
>> Think (and talk) about clocks and how they are used to
>> measure the duration of some process.
>> For her trip, Kat measures 2.5 years on her clock.
>> For that same trip, you measure 5.0 years on your clock.
>> I don't see what is "exactly equal" about 2.5 and 5.0.
>>
>
> According to Earthling Vdm , Kat travels during
> 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> on her clock.
>
> But Kat considers herself at rest, the Earth moving at
> 0.866 c wrt to her.
> After 5 years of her time (Kat is still alive, she
> took a new drug), she claims that Dirk Vdm measured
> only 2.5 years on his clock.
You are cheating - and you know it. You always cheat.
After 2.5 Kat-travel-years, she jumps to and staus in the
Earth restframe, in which 5 years had passed. And 2.5 of
her non-travel-years (and thus Earth-years) later, 7.5 Earth
years have passed.
Strich, you see what happens when a person refuses to
ask questions and decides to proclaim? They lose decades
of valuable time.
Dirk Vdm
DeadKo...@gmail.com
Mad cow disease is a terrible thing.
PD
Just as a word of caution: If a theory has been around for a century,
it's unlikely that that a problem as obvious and straightforward as
the one that occurred to you over your Cap'n Crunch this morning will
have lain undiscovered all this time. (And in fact, something like
this would have been uncovered in 20 minutes.) If you think you've
landed on something this simple, it's likely you've missed something
even more obvious.
PD
Strich 9
One trip, and two travel times from the same unit of measure.
-One trip, and two travel times from two units of measure.-
Therein lies your misunderstanding. A-L-L clocks at rest in A-N-Y
inertial frame keep E-Q-U-A-L time.
How much simpler do I have to spell it out for you, Dirk.
Strich
--
Strich 9
Greg Neill
>
> -One trip, and two travel times from two units of measure.-
>
> Therein lies your misunderstanding. A-L-L clocks at rest in A-N-Y
> inertial frame keep E-Q-U-A-L time.
Define E-Q-U-A-L. It must be something other than
most people's understanding of the word, since
standard clocks in different intertial frames of
reference certainly do not keep the *same* time as
observerd from other inertial frames.
Martin Hogbin
Your ignorance of SR is boring.
Martin Hogbin
>
>
mlut...@wanadoo.fr
SperM.hotmail.com> wrote:
> mluttg...@wanadoo.fr <mluttg...@wanadoo.fr> wrote in message
>
> 2b4a312a-74d6-4cf5-ac58-942235094...@m3g2000hsc.googlegroups.com
>
>
>
>
>
> > On Jun 26, 2:04 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> > SperM.hotmail.com> wrote:
> >> Seju Strich <Seju.Strich.2a93...@physicsbanter.com> wrote in message
>
> >> Seju.Strich.2a93...@physicsbanter.com
>
> >>> Bryan, read carefully... (keep your lips at rest for a moment so you can
> >>> understand this simple sophomore problem).
>
> >>> The 2.5 years in Kat's rest frame is EXACTLY EQUAL to a 2.5 year period
> >>> in my rest frame.
>
> >> Forget about frames. They seem to confuse you.
> >> Think (and talk) about clocks and how they are used to
> >> measure the duration of some process.
> >> For her trip, Kat measures 2.5 years on her clock.
> >> For that same trip, you measure 5.0 years on your clock.
> >> I don't see what is "exactly equal" about 2.5 and 5.0.
>
> > According to Earthling Vdm , Kat travels during
> > 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> > on her clock.
>
> > But Kat considers herself at rest, the Earth moving at
> > 0.866 c wrt to her.
> > After 5 years of her time (Kat is still alive, she
> > took a new drug), she claims that Dirk Vdm measured
> > only 2.5 years on his clock.
>
> You are cheating - and you know it. You always cheat.
I am not cheating, contrary to SRists.
Marcel Luttgens
>
> After 2.5 Kat-travel-years, she jumps to and staus in the
> Earth restframe, in which 5 years had passed. And 2.5 of
> her non-travel-years (and thus Earth-years) later, 7.5 Earth
> years have passed.
>
> Strich, you see what happens when a person refuses to
> ask questions and decides to proclaim? They lose decades
> of valuable time.
>
> Dirk Vdm- Hide quoted text -
>
> - Show quoted text -
PD
wrote:
> One trip, and two travel times from the same unit of measure.
>
> -One trip, and two travel times from two units of measure.-
>
> Therein lies your misunderstanding. A-L-L clocks at rest in A-N-Y
> inertial frame keep E-Q-U-A-L time.
Experimentally not true. And SR doesn't claim that, either.
Koobee Wublee
> Just as a word of caution: If a theory has been around for a century,
> it's unlikely that that a problem as obvious and straightforward as
> the one that occurred to you over your Cap'n Crunch this morning will
> have lain undiscovered all this time.
You speak of faith. You practice faith instead of scientific method.
<shrug>
Islam, Christianity, Judaism, Buddhism, Zoroastrianism, Confuciusism,
or Taoism has been around for even longer than SR. Is it also not
likely that any inclination to question any of these religions as
obvious and straightforward as the one occurred to you reading the
morning paper while taking a dump will have lain undiscovered all this
time?
> (And in fact, something like
> this would have been uncovered in 20 minutes.)
It only takes less than that to toss SR into the trashcan after the
twin’s paradox is identified. <shrug>
> If you think you've
> landed on something this simple, it's likely you've missed something
> even more obvious.
So, the list of great religions continues:
** SR and GR, God = Spacetime, Prophet = Einstein
** Islam, God = Alah, Prophet = Mohamed
** Christianity, God = Jehova, Prophet = Jesus
** Judaism, God = Jehova, Prophet = Moses
** Buddhism, God = multiple, Prophet = Buddha
** Confucianism, God = multiple, Prophet = Confucius
** Taoism, God = multiple, Prophet = Laozi
** Zoroastrianism, God = Mazda, Prophet = Zoroaster
Now, what are you missing so obvious? Oh, yes the Orwellian
education:
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
PD
> On Jun 27, 9:34 am, PD wrote:
>
> > Just as a word of caution: If a theory has been around for a century,
> > it's unlikely that that a problem as obvious and straightforward as
> > the one that occurred to you over your Cap'n Crunch this morning will
> > have lain undiscovered all this time.
>
> You speak of faith. You practice faith instead of scientific method.
> <shrug>
On the contrary, special relativity has been subjected to scores of
experimental tests designed precisely to determine whether there is
room for SR to be wrong. This is what makes science science. To date,
there has not been one reproducible experimental result that is in
conflict with special relativity, which is why we have a high
scientific confidence in its reliability.
This of course leaves open the possibility that there is ANOTHER
theory that is correct. The criterion for determining that is that the
competitor theory has to also be consistent with the entire body of
experimental evidence that relativity is also consistent with AND the
competitor theory has to match experiment where relativity does not.
So far, there has not been a viable candidate for a competitor theory.
>
> Islam, Christianity, Judaism, Buddhism, Zoroastrianism, Confuciusism,
> or Taoism has been around for even longer than SR. Is it also not
> likely that any inclination to question any of these religions as
> obvious and straightforward as the one occurred to you reading the
> morning paper while taking a dump will have lain undiscovered all this
> time?
>
> > (And in fact, something like
> > this would have been uncovered in 20 minutes.)
>
> It only takes less than that to toss SR into the trashcan after the
> twin’s paradox is identified. <shrug>
There is nothing in the twin puzzle that poses a problem for SR. It is
used as a *teaching exercise* for SR. Apparently, it has not worked
well for you.
Dirk Van de moortel
No, perhaps you are even too stupid to cheat.
Dirk Vdm
Dirk Van de moortel
no.
>
> -One trip, and two travel times from two units of measure.-
Yes, that's the entire - rather simple - idea. Time is defined
as what we read on clocks - nothing more, nothing less.
>
> Therein lies your misunderstanding. A-L-L clocks at rest in A-N-Y
> inertial frame keep E-Q-U-A-L time.
Almost correct.
Yes, A-L-L clocks at rest in T-H-E S-A-M-E inertial
frame keep E-Q-U-A-L time.
>
> How much simpler do I have to spell it out for you, Dirk.
You're almost there.
Starting from the above definition, try to explain how Kat
measures the total Kat-time of the trip, and how you would
measure the Strich-time of it. Do you see the difference?
Dirk Vdm
Androcles
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:d97dee1c-7e26-4897...@a9g2000prl.googlegroups.com...
Now, what are you missing so obvious? Oh, yes the Orwellian
education:
** MYSTICISM IS WISDOM
You must mean your aether!
Sue...
2a88...@physicsbanter.com> wrote:
> Kat, who has been diagnosed with cancer, has 4 years to live. She
> volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
> travels at 0.866c, and arrives there ~5.0 years later.
Cat's don't really have any special abilities with
thought experiments but relativity says your character
"Kat" won't alter the doctor's prognosis my moving.
<< The key to understanding special relativity is
Einstein's relativity principle, which states that:
"All inertial frames are totally equivalent
for the performance of all physical experiments."
In other words, it is impossible to perform a physical
experiment which differentiates in any fundamental sense
between different inertial frames. By definition, Newton's
laws of motion take the same form in all inertial frames.
Einstein generalized this result in his special theory of
relativity by asserting that all laws of physics take
the same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
...and Kats wrist watch won't be altered by motion either.
<< From the scientific point of view, the important
thing is to understand the clearly defined meaning of
“proper time”, based on the concept of an “ideal clock”
corrected for all local sensible conditions, justified
by the empirical fact that all physical phenomena
are affected identically
– including their rates of temporal progression –
by their state of inertial motion
(which is not a locally sensible condition). >>
http://www.mathpages.com/rr/s4-07/4-07.htm
Sue...
[...]
>
> --
> Seju Strich
papa...@gmail.com
Very well Sue. You are quoting precisely some lectures that explain
Special Relativity and, of course, totally agree with Enstein.
Miguel Rios
Koobee Wublee
> "Koobee Wublee" wrote:
> > You speak of faith. You practice faith instead of scientific method.
> > <shrug>
> >
> > Islam, Christianity, Judaism, Buddhism, Zoroastrianism, Confuciusism,
> > or Taoism has been around for even longer than SR. Is it also not
> > likely that any inclination to question any of these religions as
> > obvious and straightforward as the one occurred to you reading the
> > morning paper while taking a dump will have lain undiscovered all this
> > time?
>
> > So, the list of great religions continues:
>
> > ** SR and GR, God = Spacetime, Prophet = Einstein
> > ** Islam, God = Alah, Prophet = Mohamed
> > ** Christianity, God = Jehova, Prophet = Jesus
> > ** Judaism, God = Jehova, Prophet = Moses
> > ** Buddhism, God = multiple, Prophet = Buddha
> > ** Confucianism, God = multiple, Prophet = Confucius
> > ** Taoism, God = multiple, Prophet = Laozi
> > ** Zoroastrianism, God = Mazda, Prophet = Zoroaster
>
> > Now, what are you missing so obvious? Oh, yes the Orwellian
> > education:
>
> > ** MYSTICISM IS WISDOM
> > ** PLAGIARISM IS CREATIVITY
> > ** CONJECTURE IS REALITY
> > ** FAITH IS THEORY
> > ** LYING IS TEACHING
> > ** BELIEVING IS LEARNING
>
> You must mean your aether!
Hardly. Embracing the emission theory of light like yourself leads to
utter stupidity. Light travels as waves. There is no way for light
to propagate under the postulate of the emission theory. Now, this is
an example of (CONJECTURE IS REALTITY). <shrug>
Denying the Aether also leads to utter stupidity. <shrug>
Koobee Wublee
> On Jun 27, 3:10 pm, Koobee Wublee wrote:
> > You speak of faith. You practice faith instead of scientific method.
> > <shrug>
>
> On the contrary, special relativity has been subjected to scores of
> experimental tests designed precisely to determine whether there is
> room for SR to be wrong.
Tests means nothing if the supporting hypothesis is stupid in the
first place. <shrug>
In the past several thousands of years, there have been plenty of
proofs that God exists. The question is which God is more true than
the others.
> This is what makes science science.
No, this is what makes science into faith.
> To date,
> there has not been one reproducible experimental result that is in
> conflict with special relativity, which is why we have a high
> scientific confidence in its reliability.
To date, there has been no instance to prove a God does not exist.
<shrug>
> This of course leaves open the possibility that there is ANOTHER
> theory that is correct.
Yes, according to Professor Roberts who becomes ever more gung-ho
about the choice of terminologies, the more politically correct choice
of wording is ‘valid’. It is actually a good way of hiding ignorance
by pointing the problem at the other direction. This might work for
the general public, but for ones who are more in tuned with the issues
involved, it does not work.
For example, if a 3-year-old kid is holding something that you want to
take away from her, you merely have to point in any direction and say
“Santa Klaus’. The kid will be diverted in her attention, and that
allows you to reach in and grab what you need to have.
> The criterion for determining that is that the
> competitor theory has to also be consistent with the entire body of
> experimental evidence that relativity is also consistent with AND the
> competitor theory has to match experiment where relativity does not.
Yes.
> So far, there has not been a viable candidate for a competitor theory.
Oh, so in your opinion, the academics of the physics communities are
utter morons?
> > Islam, Christianity, Judaism, Buddhism, Zoroastrianism, Confuciusism,
> > or Taoism has been around for even longer than SR. Is it also not
> > likely that any inclination to question any of these religions as
> > obvious and straightforward as the one occurred to you reading the
> > morning paper while taking a dump will have lain undiscovered all this
> > time?
>
> There is nothing in the twin puzzle that poses a problem for SR.
Correction. Twin’s paradox not twin puzzle. <shrug>
> It is
> used as a *teaching exercise* for SR.
Yes, I know through the Orwellian education that “LYING IS TEACHING”.
<shrug>
> Apparently, it has not worked well for you.
That is correct. No lies have worked well for me. <shrug>
Sue...
[...]
> Very well Sue. You are quoting precisely some lectures that explain
> Special Relativity and, of course, totally agree with Enstein.
Then Einstein wouldn't have any problem with this notion:
(We assume you are his spokesman today)
<<...if we want to work out the potentials at position
r and time t then we have to perform
integrals of the charge density and current
density over all space (just like in the steady-state
situation). However, when we calculate the contribution
of charges and currents at position r' to
these integrals we do not use the values at time t,
instead we use the values at some earlier time
t - | r - r' | / c
$t-\vert{\bf r} - {\bf r}'\vert/c$.
What is this earlier time? It is simply the latest time
at which a light signal emitted from position r' would be
received at position r before time t. This
is called the retarded time. Likewise, the potentials
(509) and (510) are called retarded potentials. >>
"Time-dependent Maxwell's equations"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html
Sue...
>
> Miguel Rios
Dirk Van de moortel
Androfart and Wublee - two retired engineers pulling each
other's last remaining hairs out of their sclerosing skulls.
I love it.
Dirk Vdm
PD
Koobee Wublee wrote:
> On Jun 27, 1:26 pm, PD wrote:
> > On Jun 27, 3:10 pm, Koobee Wublee wrote:
>
> > > You speak of faith. You practice faith instead of scientific method.
> > > <shrug>
> >
> > On the contrary, special relativity has been subjected to scores of
> > experimental tests designed precisely to determine whether there is
> > room for SR to be wrong.
>
> Tests means nothing if the supporting hypothesis is stupid in the
> first place. <shrug>
This is what separates you from science, then. Science does not
prejudge a hypothesis until the experimental tests are conducted. If
the experimental tests confirm the predictions which follow from the
hypotheses, no matter how odd the hypothesis appears to be, then the
hypothesis is supported. This is how science works.
I know this is not how you'd like things to work. That's a pity. It
would help if you understood a little more about how science does its
thing.
>
> In the past several thousands of years, there have been plenty of
> proofs that God exists. The question is which God is more true than
> the others.
The difference is precisely experimental test.
Not at all. You are free to come up with a competitor theory.
Dirk Van de moortel
1e97e95c-c605-4443...@i76g2000hsf.googlegroups.com
He has a competitor theory.
It is called the Who-Needs-Scientists-If-We-Have-
All -These-Retired-Engineers-Theory.
Dirk Vdm
Dirk Van de moortel
He has a competitor theory.
Dirk Van de moortel
He has a competitor theory.
Androcles
So you whine but cannot prove. <shrug>
| Light travels as waves.
Embracing the wave theory of light like yourself leads to
utter stupidity.
| There is no way for light
| to propagate under the postulate of the emission theory.
There is no way for Sagnac to work without it, STUPID.
Now, this is
| an example of (CONJECTURE IS REALTITY). <shrug>
|
| Denying the Aether also leads to utter stupidity. <shrug>
|
Denying Sagnac and MMX IS utter stupidity, it doesn't lead
anywhere, STUPID IDIOT. <shrug>
Dirk Van de moortel
X6r9k.99108$P83....@newsfe20.ams2
Tear!
3 left.
Pull!
Dirk Vdm
papa...@gmail.com
No need at all, you are doing fine quoting that page!, even if it has
nothing to do, because of your usual misunderstandings, with the topic
in dicussion.
Miguel Rios
Sue...
Please accept my appologies. I failed to notice the
topic changed to religion when Miguel Rios wrote:
<< Very well Sue. You are quoting precisely some lectures that explain
Special Relativity and, of course, totally agree with Enstein. >>
http://en.wikipedia.org/wiki/Religion
Sue...
>
> Miguel Rios
mlut...@wanadoo.fr
> On Jun 25, 2:42 pm, Seju Strich <Seju.Strich.
>
> 2a88...@physicsbanter.com> wrote:
> > Kat, who has been diagnosed with cancer, has 4 years to live. She
> > volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
> > travels at 0.866c, and arrives there ~5.0 years later.
>
> Cat's don't really have any special abilities with
> thought experiments but relativity says your character
> "Kat" won't alter the doctor's prognosis my moving.
>
> << The key to understanding special relativity is
> Einstein's relativity principle, which states that:
>
> "All inertial frames are totally equivalent
> for the performance of all physical experiments."
>
> In other words, it is impossible to perform a physical
> experiment which differentiates in any fundamental sense
> between different inertial frames. By definition, Newton's
> laws of motion take the same form in all inertial frames.
> Einstein generalized this result in his special theory of
> relativity by asserting that all laws of physics take
> the same form in all inertial frames. >>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>
> ...and Kats wrist watch won't be altered by motion either.
Sue, you rightly wrote:
"...and Kat's wrist watch won't be altered by motion either."
Indeed,
"According to Earthling Vdm , Kat travels during
5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
on her clock.
But Kat considers herself at rest, the Earth moving at
0.866 c wrt to her.
After 5 years of her time (Kat is still alive, she
took a new drug), she claims that Dirk Vdm measured
only 2.5 years on his clock.
According to SR, both Kat and Dirk are right."
What does that mean?
Let's consider the following case:
Two persons, A and B, are both 1.60 m tall when their
height is mesured in the same room.
After some jogging, the distance between A and B is
x meters, and A will claim that B measure 0.80 m, whereas
B will observe that A measures 0.80 m. Of course, both
are right, but this doesn't change the intrinsic height of
A and B, i.e. 1.60 m.
Similarly, Kat considers that Dirk Vdm measured
only 2.5 years on his clock, aginst 5 years on her clock,
and Vdm claims that Kat travelled during 5 years, but
measured 5 * sqrt(1-0.866^2) =~ 2.5 years on her clock.
SRists don't realize that such paradox is due to a mere
observational illusion. They childlishly claim that clock rates
are affected by motion, exactly like primitive people
believe that distances affect the height of the observed
persons.
Marcel Luttgens
>
> << From the scientific point of view, the important
> thing is to understand the clearly defined meaning of
> “proper time”, based on the concept of an “ideal clock”
> corrected for all local sensible conditions, justified
> by the empirical fact that all physical phenomena
> are affected identically
>
> – including their rates of temporal progression –
>
> by their state of inertial motion
> (which is not a locally sensible condition). >>http://www.mathpages.com/rr/s4-07/4-07.htm
>
> Sue...
>
> [...]
>
>
>
>
>
> > --
> > Seju Strich- Hide quoted text -
papa...@gmail.com
Well, if you think your quoting "http://farside.ph.utexas.edu/teaching/
em/lectures/node108.html" or "http://farside.ph.utexas.edu/teaching/em/
lectures/node50.html" is religion, why did you quote them in the first
place?
Do you have a personal opinion on anything?
Better learn something.....this page "http://en.wikipedia.org/wiki/
Cooking" may help you.
Miguel Rios
Dono
>
> SRists don't realize that such paradox is due to a mere
> observational illusion. They childlishly claim that clock rates
> are affected by motion,
>
> Marcel Luttgens
Marcel,
You surely forgot how GPS operates. Or are you using one of the early
satelittes, left in orbit before the time they applied the GR/Sr
corrections? :-)
Your car must be missing the correct turns by several kilometers,
doesn't it, old fart? :-)
Sue...
> On Jun 28, 3:05 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
>
> > On Jun 25, 2:42 pm, Seju Strich <Seju.Strich.
>
> > 2a88...@physicsbanter.com> wrote:
> > > Kat, who has been diagnosed with cancer, has 4 years to live. She
> > > volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
> > > travels at 0.866c, and arrives there ~5.0 years later.
>
> > Cat's don't really have any special abilities with
> > thought experiments but relativity says your character
> > "Kat" won't alter the doctor's prognosis my moving.
>
> > << The key to understanding special relativity is
> > Einstein's relativity principle, which states that:
>
> > "All inertial frames are totally equivalent
> > for the performance of all physical experiments."
>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>
> > ...and Kats wrist watch won't be altered by motion either.
> Sue, you rightly wrote:
>
> "...and Kat's wrist watch won't be altered by motion either."
>
> Indeed,
>
> "According to Earthling Vdm , Kat travels during
> 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> on her clock.
Likely Vdm has never heard of the principle of relativity.
<<...it is impossible to perform a physical
experiment which differentiates in any fundamental sense
between different inertial frames. By definition, Newton's
laws of motion take the same form in all inertial frames.
Einstein generalized this result in his special theory of
relativity by asserting that all laws of physics take
the same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
"Relativistic particle dynamics"
http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
>
> But Kat considers herself at rest, the Earth moving at
> 0.866 c wrt to her.
> After 5 years of her time (Kat is still alive, she
> took a new drug), she claims that Dirk Vdm measured
> only 2.5 years on his clock.
>
> According to SR, both Kat and Dirk are right."
Where does "SR" say that ?
<<...in reality there is not the least incompatibility
between the principle of relativity and the law of
propagation of light>>
http://www.bartleby.com/173/7.html
>
> What does that mean?
>
> Let's consider the following case:
>
> Two persons, A and B, are both 1.60 m tall when their
> height is mesured in the same room.
> After some jogging, the distance between A and B is
> x meters, and A will claim that B measure 0.80 m, whereas
> B will observe that A measures 0.80 m. Of course, both
> are right, but this doesn't change the intrinsic height of
> A and B, i.e. 1.60 m.
>
> Similarly, Kat considers that Dirk Vdm measured
> only 2.5 years on his clock, aginst 5 years on her clock,
> and Vdm claims that Kat travelled during 5 years, but
> measured 5 * sqrt(1-0.866^2) =~ 2.5 years on her clock.
>
> SRists don't realize that such paradox is due to a mere
> observational illusion. They childlishly claim that clock rates
> are affected by motion, exactly like primitive people
> believe that distances affect the height of the observed
> persons.
...Careful now. You probably don't realise that Einstein's
1905 paper was chiseled into a holy stone. If you disagree
any of its loyal worshippers you are likely to be
struck down by lighnting from opposite ends of
a railway embankment. :o)
Sue...
Dirk Van de moortel
> On Jun 28, 3:05 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>> On Jun 25, 2:42 pm, Seju Strich <Seju.Strich.
>>
>> 2a88...@physicsbanter.com> wrote:
>>> Kat, who has been diagnosed with cancer, has 4 years to live. She
>>> volunteers on a space mission to Alpha Centauri (4.37 lightyears away),
>>> travels at 0.866c, and arrives there ~5.0 years later.
>>
>> Cat's don't really have any special abilities with
>> thought experiments but relativity says your character
>> "Kat" won't alter the doctor's prognosis my moving.
>>
>> << The key to understanding special relativity is
>> Einstein's relativity principle, which states that:
>>
>> "All inertial frames are totally equivalent
>> for the performance of all physical experiments."
>>
>> In other words, it is impossible to perform a physical
>> experiment which differentiates in any fundamental sense
>> between different inertial frames. By definition, Newton's
>> laws of motion take the same form in all inertial frames.
>> Einstein generalized this result in his special theory of
>> relativity by asserting that all laws of physics take
>> the same form in all inertial frames. >>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>>
>> ...and Kats wrist watch won't be altered by motion either.
>
> Sue, you rightly wrote:
Never mind, Marcel. Sue (Dennis McCarthy) is yet *another*
retired engineer. Sheesh, what is the matter with this race?
By the way, surely you didn't expect to get a reply to your
question on spr, did you?
http://groups.google.com/group/sci.physics.research/browse_frm/thread/e0e45171c6660ef5
I mean, you don't even understand the first half of page one
of special relativity, so what do you think you are doing
poking around on the borders of general relativity?
Don't you think it's about time to close your weary eyes
now?
Dirk Vdm
papa...@gmail.com
>
> Likely Vdm has never heard of the principle of relativity.
>
> <<...it is impossible to perform a physical
> experiment which differentiates in any fundamental sense
> between different inertial frames. By definition, Newton's
> laws of motion take the same form in all inertial frames.
> Einstein generalized this result in his special theory of
> relativity by asserting that all laws of physics take
> the same form in all inertial frames. >>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
> "Relativistic particle dynamics"http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
>
You see, this guy quotes, but carefully leaves out some small part of
it. The complete quote is:
"The key to understanding special relativity is Einstein's relativity
principle, which states that:
All inertial frames are totally equivalent for the performance of
all physical experiments.
In other words, it is impossible to perform a physical experiment
which differentiates in any fundamental sense between different
inertial frames. By definition, Newton's laws of motion take the same
form in all inertial frames. Einstein generalized this result in his
special theory of relativity by asserting that all laws of physics
take the same form in all inertial frames.
Consider a wave-like disturbance. In general, such a disturbance
propagates at a fixed velocity with respect to the medium in which the
disturbance takes place. For instance, sound waves (at S.T.P.)
propagate at 343 meters per second with respect to air. So, in the
inertial frame in which air is stationary, sound waves appear to
propagate at 343 meters per second. Sound waves appear to propagate at
a different velocity any inertial frame which is moving with respect
to the air. However, this does not violate the relativity principle,
since if the air were stationary in the second frame then sound waves
would appear to propagate at 343 meters per second in this frame as
well. In other words, exactly the same experiment (e.g., the
determination of the speed of sound relative to stationary air)
performed in two different inertial frames of reference yields exactly
the same result, in accordance with the relativity principle.
Consider, now, a wave-like disturbance which is self-regenerating, and
does not require a medium through which to propagate. The most well-
known example of such a disturbance is a light wave. Another example
is a gravity wave. According to electromagnetic theory, the speed of
propagation of a light wave through a vacuum is
c = 1/sqrt{\epsilon_0 \mu_0}=2.99729 x 10^8 meters per second,
(1323)
where \epsilon_0 and \mu_0 are physical constants which can be
evaluated by performing two simple experiments which involve measuring
the force of attraction between two fixed changes and two fixed
parallel current carrying wires. According to the relativity
principle, these experiments must yield the same values for \epsilon_0
and \mu_0 in all inertial frames. Thus, the speed of light must be the
same in all inertial frames. In fact, any disturbance which does not
require a medium to propagate through must appear to travel at the
same velocity in all inertial frames, otherwise we could differentiate
inertial frames using the apparent propagation speed of the
disturbance, which would violate the relativity principle."
Miguel Rios
mlut...@wanadoo.fr
Dear Dono,
Don't you think that A or B have to apply a correction
to get the true height?
Marcel Luttgens
Dono
>
> Dear Dono,
>
> Don't you think that A or B have to apply a correction
> to get the true height?
>
> Marcel Luttgens
Huh?
Gerald Ray
<dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>Never mind, Marcel. Sue (Dennis McCarthy) is yet *another*
>retired engineer. Sheesh, what is the matter with this race?
Are you saying that the contributor to this group who goes by the name
"Sue" is really a man named Dennis McCarthy? That seems odd. Is this
well known in the group? Does "Sue" acknowledge this as a fact? Can
you share with us the evidence that leads you to believe this? Is it
based on stylistic similarities, or is there more objective evidence?
Sue...
No doubt those unfortunates who lack a web-browser are
eternally grateful to you for reposting the entire page on usenet.
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
dvm is likely one such unfortunate soul but I don't think
it will help him unless you can explain to him why
motion sensitive clocks are neither implied nor required
in that resolution of SR's postulates.
Sue...
> Miguel Rios
Sue...
> On Sat, 28 Jun 2008 17:46:19 +0200, "Dirk Van de moortel"
>
> >Never mind, Marcel. Sue (Dennis McCarthy) is yet *another*
> >retired engineer. Sheesh, what is the matter with this race?
>
> Are you saying that the contributor to this group who goes by the name
> "Sue" is really a man named Dennis McCarthy? That seems odd. Is this
> well known in the group? Does "Sue" acknowledge this as a fact? Can
> you share with us the evidence that leads you to believe this? Is it
> based on stylistic similarities, or is there more objective evidence?
One of Dirk's many fantasies is that a professional in the
field would actually take the time to add him to a kill-file.
If welching on a $1000 wager doesn't earn him a listing
I don't know what else Dirk can try.
Perhaps malicious redirections to alt.moron etc etc.
<<Dirk is the only person in world who would wager that Dennis
McCarthy doesn't know his own identity. It comes as little
suprise that he also writes equations to support causality
violations. >>
http://groups.google.com/group/sci.physics.relativity/search?group=sci.physics.relativity&q=mccarthy+wager+sue&qt_g=Search+this+group
Sue...
Dirk Van de moortel
48667dd9....@news.gte.net
> On Sat, 28 Jun 2008 17:46:19 +0200, "Dirk Van de moortel"
> <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>> Never mind, Marcel. Sue (Dennis McCarthy) is yet *another*
>> retired engineer. Sheesh, what is the matter with this race?
>
> Are you saying that the contributor to this group who goes by the name
> "Sue" is really a man named Dennis McCarthy?
Yep.
> That seems odd. Is this
> well known in the group? Does "Sue" acknowledge this as a fact? Can
> you share with us the evidence that leads you to believe this? Is it
> based on stylistic similarities, or is there more objective evidence?
There is objective evidence, but there is also a history. It's is up
to Dennis to guess how I found out. He never had the guts.
I have stopped holding my breath though. Too bad.
You can find out as well, provided you know how to use some
aspects of a some search engine - and provided you Really
Want To Know. That is, if you don't know already - should you
happen to be another McCartyInstance.
Dirk Vdm
Dirk Vdm
papa...@gmail.com
> On Jun 28, 12:13 pm, "papar...@gmail.com" <papar...@gmail.com> wrote:
>
> No doubt those unfortunates who lack a web-browser are
> eternally grateful to you for reposting the entire page on usenet.
>
> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>
>
>
More than they are with your incomplete and nonsensical quoting, for
sure.
Miguel Rios
Strich 9
Relativists typically give us 2 body relativity problems, and the
illogical results in one frame are reinterpreted in the other frame to
cover the error. If multibody relativity problems are analyzed, the
results become clearly and certainly contradictory.
In physics talk is cheap, calculations are costly (to the one in
error).
Shown below is a simple 6 body relativity problem. Numbers have been
rounded for clarity.
_Analyzing_Multi-Body_Special_Relativity_Problems
_
Five neutrinos are A, B, C, D, E are fired from an accelerator 10km
away to a muon at rest in an MIT lab with parameters below.
A B C D E
Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0
(E)
Gamma 10 (A) 2 (B) 1.25 (C) 1.1 (D) 1 (E)
Distance (m) 10000 (A) 10000 (B) 10000 (C) 10000 (D) 10000 (E)
Length Contraction (m) 1000 (A) 5000 (B) 8000 (C) 9090 (D) 10000 (E)
Muon Lifespan (microsec) 2 (A) 2 (B) 2 (C) 2 (D) 2 (E)
Distance M** (m) 600 (A) 520 (B) 360 (C) 240 (D) 0 (E)
Time Dilation (microsec) 20 (A) 4 (B) 2.5 (C) 2.2 (D) 2 (E)
Distance N*** (m) 6000 (A) 1040 (B) 450 (C) 264 (D) 0 (E)
*velocity relative to muon
**distance “traveled” by muon in its frame prior to
disintegration=velocity x time
***distance “traveled” by neutrino in its frame prior to
muon disintegration=velocity x dilated time
The M distances are in the muon frame and hence comparable while the N
distance are in the neutrino frames which are different and not
comparable. Since in our experiment the muon is "motionless", the M
distance represents the distance reduction between the muon and a
neutrino at the time of its death.
It is seen that the muon covers different distances depending on who is
observing it, and when there are five observers, then the muon dies
after covering 0, 240, 360, 520, AND 600 meters in its frame, which is
obviously IMPOSSIBLE.
Again, bring this to your gurus in relativity for their analyses.
Watch how many furrows develop in their foreheads :) Some may even
turn pale because they will realize that the grant money for their
research may be recalled. Meanwhile the relativity high priest may
declare me excommunicated ::)
--
Strich 9
Greg Neill
news:Strich.9...@physicsbanter.com
> It is seen that the muon covers different distances depending on who
> is observing it, and when there are five observers, then the muon dies
> after covering 0, 240, 360, 520, AND 600 meters in its frame, which is
> obviously IMPOSSIBLE.
You said the muon was at rest; the muon covers no distance in
its own frame.
There is nothing wrong with different observers measuring
different time intervals and distances for a single
spacetime event; such measurements are observer dependent.
Nor does the muon care what *other* observers measure. As
far as it is conserned it has a given lifespan in its own
frame.
What you fail to mention is that all observers will agree
on the location of the decay event in spacetime. That is,
if all observers were to transform their measurements via
the Lorentz tansform to a single frame of reference for
comparison, they would all obtain the same numbers. There
is no contradiction. Furthermore, since Relativity is
mathematically correct and complete (in the same sense
that Euclidean geometry is correct and complete), there
can be no contradiction arrising from its correct application.
Dirk Van de moortel
> Relativists typically give us 2 body relativity problems, and the
> illogical results in one frame are reinterpreted in the other frame to
> cover the error. If multibody relativity problems are analyzed, the
> results become clearly and certainly contradictory.
>
> In physics talk is cheap, calculations are costly (to the one in
> error).
>
> Shown below is a simple 6 body relativity problem. Numbers have been
> rounded for clarity.
>
> _Analyzing_Multi-Body_Special_Relativity_Problems
> _
> Five neutrinos are A, B, C, D, E are fired from an accelerator 10km
> away to a muon at rest in an MIT lab with parameters below.
>
>
> A B C D E
> Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
> Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0
> (E)
Good grief.
Surely you're not another equation challenged imbecile like
say, Androcles or Ken Seto? If you are, you better forget
it and find another hobby.
Otherwise, try expressing yourself with variables, so you can
replace these two lines with one single letter, for instance v.
Give the 10 km the name L and the 2 seconds the name D.
Put the accelerator at x = 0 and the muon at x = L.
Let the event of neutrino-production take place at t = 0,
i.o.w. at event (x,t) = (0,0).
Let the muon desintegrate at time t = T, i.o.w. at event
(x,t) = (L,T).
Let the neutrino reach the muon (or the place where the
muon was after it has desintegrated) at time R (as measured
by the accelerator and the Muon), i.o.w.at event (x,t) = (L,R)
Now describe these three events in terms of the
(x',t')-coordinates of the neutrino and tell us if you still have
a problem.
If you do this properly, you should see where you went wrong.
So, where did you go wrong?
Dirk Vdm
Dirk Van de moortel
> illogical results in one frame are reinterpreted in the other frame to
> cover the error. If multibody relativity problems are analyzed, the
> results become clearly and certainly contradictory.
>
> In physics talk is cheap, calculations are costly (to the one in
> error).
>
> Shown below is a simple 6 body relativity problem. Numbers have been
> rounded for clarity.
>
> _Analyzing_Multi-Body_Special_Relativity_Problems
> _
> Five neutrinos are A, B, C, D, E are fired from an accelerator 10km
> away to a muon at rest in an MIT lab with parameters below.
>
>
> A B C D E
> Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
> Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0
> (E)
Good grief.
Surely you're not another equation challenged imbecile like
say, Androcles or Ken Seto? If you are, you better forget
it and find another hobby.
Otherwise, try expressing yourself with variables, so you can
replace these two lines with one single letter, for instance v.
Give the 10 km the name L and the 2 seconds the name T. (not D)
Dirk Van de moortel
> illogical results in one frame are reinterpreted in the other frame to
> cover the error. If multibody relativity problems are analyzed, the
> results become clearly and certainly contradictory.
>
> In physics talk is cheap, calculations are costly (to the one in
> error).
>
> Shown below is a simple 6 body relativity problem. Numbers have been
> rounded for clarity.
>
> _Analyzing_Multi-Body_Special_Relativity_Problems
> _
> Five neutrinos are A, B, C, D, E are fired from an accelerator 10km
> away to a muon at rest in an MIT lab with parameters below.
>
>
> A B C D E
> Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
> Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0
> (E)
Good grief.
Surely you're not another equation challenged imbecile like
say, Androcles or Ken Seto? If you are, you better forget
it and find another hobby.
Otherwise, try expressing yourself with variables, so you can
replace these two lines with one single letter, for instance v.
Give the 10 km the name L and the 2 seconds the name T (not D).
Strich 9
> Relativists typically give us 2 body relativity problems, and the
> illogical results in one frame are reinterpreted in the other frame to
> cover the error. If multibody relativity problems are analyzed, the
> results become clearly and certainly contradictory.
>
> In physics talk is cheap, calculations are costly (to the one in
> error).
>
> Shown below is a simple 6 body relativity problem. Numbers have been
> rounded for clarity.
>
> _Analyzing_Multi-Body_Special_Relativity_Problems
> _
> Five neutrinos are A, B, C, D, E are fired from an accelerator 10km
> away to a muon at rest in an MIT lab with parameters below.
>
>
> A B C D E
> Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
> Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0
Gamma 10 (A) 2 (B) 1.25 (C) 1.1 (D) 1 (E)
Distance (m) 10000 (A) 10000 (B) 10000 (C) 10000 (D) 10000 (E)
Length Contraction (m) 1000 (A) 5000 (B) 8000 (C) 9090 (D) 10000 (E)
Muon Lifespan (microsec) 2 (A) 2 (B) 2 (C) 2 (D) 2 (E)
Distance M** (m) 600 (A) 520 (B) 360 (C) 240 (D) 0 (E)
Time Dilation (microsec) 20 (A) 4 (B) 2.5 (C) 2.2 (D) 2 (E)
Distance N*** (m) 6000 (A) 1040 (B) 450 (C) 264 (D) 0 (E)
BRAVO, BRAVO, Dirk. You are now halfway up genius level. You could
have given a simpler answer that the different M distances are correct
because they represent the relative distance between the neutrino and
the muon, and of course the faster moving neutrinos will have shorter
distances.
But don't stop there. This is not as simple as you think. If you
examine the N distances, it looks like the neutrinos, except E,
traveled faster than light. Now, before you give your final answer,
don't stop there. Analyze the data some more. Try not to use
variables to make it easier for yourself. Before you realize it, you
may be at genius level sooner than you think. Best wishes.
--
Strich 9
Dirk Van de moortel
>> Put the accelerator at x = 0 and the muon at x = L.
>> Let the event of neutrino-production take place at t = 0,
>> i.o.w. at event (x,t) = (0,0).
>> Let the muon desintegrate at time t = T, i.o.w. at event
>> (x,t) = (L,T).
>> Let the neutrino reach the muon (or the place where the
>> muon was after it has desintegrated) at time R (as measured
>> by the accelerator and the Muon), i.o.w.at event (x,t) = (L,R)
>> Now describe these three events in terms of the
>> (x',t')-coordinates of the neutrino and tell us if you still have
>> a problem.
>> If you do this properly, you should see where you went wrong.
>> So, where did you go wrong?
>>
>> Dirk Vdm
>
> A B C D E
> Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
> Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0
So are you indeed the kind of equation challenged imbecile
who can't replace this ugly mess with 1 single letter?
Dirk Vdm
mlut...@wanadoo.fr
SperM.hotmail.com> wrote:
> mluttg...@wanadoo.fr <mluttg...@wanadoo.fr> wrote in message
>
> 2b4a312a-74d6-4cf5-ac58-942235094...@m3g2000hsc.googlegroups.com
>
>
>
>
>
> > On Jun 26, 2:04 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> > SperM.hotmail.com> wrote:
> >> Seju Strich <Seju.Strich.2a93...@physicsbanter.com> wrote in message
>
> >> Seju.Strich.2a93...@physicsbanter.com
>
> >>> Bryan, read carefully... (keep your lips at rest for a moment so you can
> >>> understand this simple sophomore problem).
>
> >>> The 2.5 years in Kat's rest frame is EXACTLY EQUAL to a 2.5 year period
> >>> in my rest frame.
>
> >> Forget about frames. They seem to confuse you.
> >> Think (and talk) about clocks and how they are used to
> >> measure the duration of some process.
> >> For her trip, Kat measures 2.5 years on her clock.
> >> For that same trip, you measure 5.0 years on your clock.
> >> I don't see what is "exactly equal" about 2.5 and 5.0.
>
> > According to Earthling Vdm , Kat travels during
> > 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> > on her clock.
>
> > 0.866 c wrt to her.
> > After 5 years of her time (Kat is still alive, she
> > took a new drug), she claims that Dirk Vdm measured
> > only 2.5 years on his clock.
>
>
> After 2.5 Kat-travel-years, she jumps to and staus in the
> Earth restframe, in which 5 years had passed. And 2.5 of
> her non-travel-years (and thus Earth-years) later, 7.5 Earth
> years have passed.
Clearly, the cheater is you!
Try to solve this contradiction, without obfuscating
the subject with jumps.
"According to Earthling Vdm , Kat travels during
5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
on her clock.
But Kat considers herself at rest, the Earth moving at
0.866 c wrt to her.
After 5 years of her time (Kat is still alive, she
took a new drug), she claims that Dirk Vdm measured
only 2.5 years on his clock."
Marcel Luttgens
>
> Strich, you see what happens when a person refuses to
> ask questions and decides to proclaim? They lose decades
> of valuable time.
>
> Dirk Vdm
Dirk Van de moortel
But the jumps are essential, imbecile.
>
> "According to Earthling Vdm , Kat travels during
> 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> on her clock.
>
> But Kat considers herself at rest,
No, retard, she does not consider herself at rest. She jumps
twice.
> the Earth moving at
> 0.866 c wrt to her.
> After 5 years of her time (Kat is still alive, she
> took a new drug), she claims that Dirk Vdm measured
> only 2.5 years on his clock."
No, imbecile. After 2.5 Kat-years on her clock, she decides to
stop travelling and now stay with the alpha centauri clock which
is synchronized with the earth clock. Both read 5 earth-years at
the event when Kat stops travelling, and from then on Kat's
clock keeps pace with these clocks, so still 2.5 years later on
Kat's clock, the Earth and alpha clocks read another 2.5 years
later than the 5 years. That makes 7.5 years.
She jumped, Imbecile, first away from the earth-alpha-frame
onto a travelling clock, and then back into the earth-alpha-frame.
That's two jumps, moron.
I know, you never managed to understand this - not even
remotely.
Go ahead, blame it on me - I never managed to make a retard
like you understand what it means to look at a clock and write
the time it shows. So don't let that put you down - it is entirely
my fault. I take full blame.
Dirk Vdm
Sue...
The math is simple when Kat is far from a mass.
<<
As Einstein said:
The weakness of the principle of inertia lies in
this, that it involves an argument in a circle: a
mass moves without acceleration if it is sufficiently
far from other bodies; we know that it is sufficiently
far from other bodies only by the fact that it
moves without acceleration.>>
http://www.mathpages.com/rr/s4-07/4-07.htm
Regardless of Kat's motion, half the universe's
mass moves toward Kat; half moves away.
The net change in Kat's gravito-inertial field is zero.
The net chgage in the gravito-inertial field of Kat's
wrist watch is zero.
<< The key to understanding special relativity is
Einstein's relativity principle, which states that:
All inertial frames are totally equivalent for
the performance of all physical experiments.
In other words, it is impossible to perform a physical
experiment which differentiates in any fundamental sense
between different inertial frames. By definition,
Newton's laws of motion take the same form in all
inertial frames. Einstein generalized this result
in his special theory of relativity by asserting
that all laws of physics take the same form in
all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
Sue...
Dirk Van de moortel
"Strich 9" <Strich.9...@physicsbanter.com> wrote in message news:Strich.9...@physicsbanter.com...
>
> ok, ok, this is usually the time where most physicists become busy or
> have something better to do, when they realize that further answers to
> my questions lead to their sure contradiction. instead of waiting for
> DIRK to restudy his relativity and consult his peers and masters, i'll
> give you the final answers. not specifically to the problem above, but
> to the twin paradox type problems in general.
>
> ________________________________________________________________
>
> Case:
> A muon and neutrino are approaching each other at relative velocity
> ~c.
> The muon has a rest life of 2 microseconds. Gamma is 10.
Good grief, here we go again with the numbers....
[snip the numbers]
> Now where is relativity wrong. It is easy to find if we go back to the
> equations.
Ha...here come the equations.
Finally.
Let's prepare for the laughs.
>
> ======================================================
>
>
> Given two frames O and Q, with relative velocity V, the distances dQ in
> the moving frame are translated to dO in the stationary frame by the
> transformation equation:
>
> dO = dQ / gamma
Wrong from the start.
This equation is *only* valid for distances between two events
that are measured simultaneously in the O-frame, i.o.w. for
events with tO = 0.
>
> And the times tQ are translated into tO by the transformation equation
>
> tO = tQ x gamma
Wrong again.
This equation is *only* valid for a time difference between two
events that are measured at the same place in the Q-frame, i.o.w
for events with dQ = 0.
>
> where gamma = (1-v^2/c^2)^(1/2)
>
> now velocity vO and vQ are equal since this is the relative velocity.
Wrong again
vO = - vQ
>
> And vO = dO/tO
This is equation is *only* valid for the O-coordinates of the origin
of the Q-frame, i.o.w. for events with dQ = 0.
>
> And vQ=dQ/tQ
If you write it correctly as
vQ= - dQ/tQ
then it is *only* valid for the Q-coordinates of the origin of the
O-frame, i.o.w. for events with dO = 0.
>
> vQ = vO = dO/tO = dQ/tQ
No, for starters
vQ = - vO,
but never mind that.
If you write dO/tO and dQ/tQ in one equation, then it is valid for
events taking place on the origin of the O-frame, *and* on the
origin of the Q-frame, in other words, for only one single event,
namely the event of coincidence of both origins, in other words
in your notation, for one single event with coordinates
dO = dQ = tO = tQ = 0 .
So you are effectively writing
0 = 0 = 0 = 0
Congratulations!
>
> but dO = dQ / gamma, and tO = tQ x gamma
yes
0 = 0 and 0 = 0
Congratulations!
>
> so vO = (dQ / gamma) / ( tQ x gamma) = (dQ/tQ) / (gamma^2)
No, since tQ = 0, you are not allowed to write this.
We don't divide by zero. Sorry.
>
> but also vQ=dQ/tQ so vO = (dQ/tQ) / (gamma^2) = vQ / gamma^2
tQ = 0, remeber?
>
> and since vO = vQ then vQ = vQ / gamma ^2
yes, since 0 = 0, then 0 = 0.
Congratulations!
>
> which is a contradiction since vQ vQ / gamma^2
A contradiction?
0 = 0?
Well well...
>
>
> Try to understand it my friends. Watch relativity unravel before your
> eyes.
Try to understand the meanings of the variables. Watch what happens
with imbeciles who take their failing to understand these meanings
for errors made by the rest of the world.
Dirk Vdm
mlut...@wanadoo.fr
>
> No, perhaps you are even too stupid to cheat.
>
> Dirk Vdm- Hide quoted text -
>
> - Show quoted text -
Where did you find a reference to the "twin paradox"
in what I wrote?
Are you so SR brainwashed not to realize that
Kat's trip is limited to Alpha Centauri?
"According to Earthling Vdm , Kat travels during
5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
on her clock.
But Kat considers herself at rest, the Earth moving at
0.866 c wrt to her.
After 5 years of her time (Kat is still alive, she
took a new drug), she claims that Dirk Vdm measured
only 2.5 years on his clock."
Try to solve this contradiction, without obfuscating
the subject with the "twin paradox" jumps!
And cease to stupidly redirect your reponses, cf. your
Followup-To: alt.imbeciles.and.retards
Marcel Luttgens
Eric Gisse
It isn't a contradiction. Try to learn something, instead of being a
write-only poster.
Dirk Van de moortel
I do find meta-references to your uncurable stupidity in
everything you write.
Dirk Vdm
Strich 9
Dirk Van de moortel;1174848 Wrote:
>
>
> Try to understand the meanings of the variables. Watch what happens
> with imbeciles who take their failing to understand these meanings
> for errors made by the rest of the world.
>
> Dirk Vdm
Dirk, it is very easy to see things are wrong if you don't understand
them. Where did you study physics? A community college perhaps?
Read the example below, because it is a broader analyses of the muon
experiment. After reading it, simply substitute the -*EARTH*- for the
-*neutrino*-, and you get the details of the muon experiment.
Case:
A muon and neutrino are approaching each other at relative velocity ~c.
The muon has a rest life of 2 microseconds. Gamma is 10.
The letter in brackets signify the reference frame, the number signify
the set.
NEUTRINO at rest. Set 1
From the point of view of the neutrino, the muon is approaching at
velocity ~c.
It is time dilated to 20us [N1] in the neutrino frame.
At speed ~c, the muon covers a distance of *_7000m_* [N1] in the
neutrino frame.
From the point of view of the moving muon with respect to the
stationary neutrino, it is moving at speed ~c.
It has a life span of 2us [M1] in the muon frame.
At speed ~c, it covers 700m [M1] in the muon frame, which is equal to a
length contracted *_7000m_* [N1x] in the neutrino frame. Note [N1] =
[N1x].
MUON at rest. Set 2
From the point of view of the muon, the neutrino is approaching at
velocity ~c.
The muon has only a 2us [M2] lifespan in the muon frame, which reflects
a time dilation from 0.2us [N2] in the neutrino frame as it is moving at
~c.
At speed ~c, the neutrino covers a distance of *_700m_* [M2] in the
muon frame.
From the point of view of the moving neutrino, it is moving at speed
~c.
It is only allowed a travel time of 0.2us [N2] in the neutrino frame as
this time dilates to 2us in the muon frame.
At speed ~c, the neutrino covers a distance of 70m [N2] in the neutrino
frame (c x 0.2us), which is equal to a length contracted *_700m_* [M2x]
in the muon frame. Note [M2] = [M2x].
The computations of EACH SET are self-consistent. This is because the
second part of each set is merely a *-restatement-* of the first part.
Strict application of the theory of relativity with one stationary
observer and one moving observee means both sets are applied. This
creates inconsistent results, shown above where
*_N1=7000m_is_NOT_equal_to_N2=70m_*. The difference is a *-factor of
gamma, squared-*. Why this is so is explained below.
--------------------------------------------------------------------------
Now obviously you don't understand Minkowski space-time. You don't
even understand the definition of a distance or time interval, which
are *ZERO* only for two bodies if they are IDENTICAL. I can see your
penchant for zeros, it must be your score in life. Seriously, these
are not partial differentials either. These are relabeled below to
avoid brain meltdown for people who read too fast but understand too
slowly, Dirk.
Given two frames F1 (stationary) and F2 (moving), with relative
velocity V, the distance intervals in L2 are translated to L1 in the
stationary frame by the transformation equation:
L1 = L2 / gamma
And the times T2 are translated into T1 by the transformation equation
T1 = T2 x gamma
where gamma = (1-v^2/c^2)^(1/2)
V1 = D1/T1
V2 = D2/T2
now velocity V1 and V2 are equal since this is the relative velocity
(ok, the sign is different, BIG ACHIEVEMENT there Dirk, but remember,
this is 1-dimensional, to make it easier for you, so it doesn't really
matter).
V2 = V1 = D1/T1 = D2/T2
D1 = D2 / gamma, T1 = T2 x gamma
V1 = (D2 / gamma) / ( T2 x gamma) = (D2/T2) / (gamma^2)
but also D2/T2 = V2, so V1 = V2 / (gamma^2)
and since V1 = V2 then *V2 = V2 / gamma ^2*
which is a contradiction unless gamma is 1 (the two bodies are
*STATIONARY* with respect to each other).
and there is the contradiction in the special theory of relativity.
this is the mathematical explanation for the "twin fallacy". it is not
only a *-linear fallacy-*, it is a *-fallacy in the second degree-*, and
that is why it is very hard to grasp for people who can only think
linearly.
and Dirk, don't giggle when you cannot understand something. you just
look more stupid. and that is me being kind
--
Strich 9
Bryan Olson
> Try to solve this contradiction, without obfuscating
> the subject with jumps.
>
> "According to Earthling Vdm , Kat travels during
> 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> on her clock.
According to SR, in the approximately-inertial rest frame
of Earth and Alpha Centauri, the events of Kat leaving
Earth and Kat arriving at Alpha Centauri are about 5 year
apart in time. In Kat's traveling frame, the two events
are about 2.5 years apart.
> But Kat considers herself at rest, the Earth moving at
> 0.866 c wrt to her.
> After 5 years of her time (Kat is still alive, she
> took a new drug), she claims that Dirk Vdm measured
> only 2.5 years on his clock."
If Kat continues in her away-from-Earth frame, after she
clocks five years she'll be twice as far from Earth as is
Alpha Centauri. In the frame of the Earth, Kat reaching
that point happens 10 years after Kat departed Earth.
Now the tricky bit: in Kat's travel frame, what does the
Earth clock read? In Kat's travel frame, the events of
Kat reaching the twice-Alpha-Centauri-distance point and
the Earth clock showing departure-time plus 2.5 years
are (approximately) simultaneous. Those events are clearly
no where near simultaneous in the Earth's frame.
Understanding this much did not come easy for me, and at
one time I had thought SR to be contradictory. There's hope
that many who are currently of anti-relativity cranks will
come around, though the odds are not great.
--
--Bryan
Sue...
[...]
>
> Understanding this much did not come easy for me, and at
> one time I had thought SR to be contradictory. There's hope
> that many who are currently of anti-relativity cranks will
> come around, though the odds are not great.
Thus far, your mis-spent youth has not made a convincing
argument for any of the issues you have offered it in
support of. Are you sure you didn't take a wrong
turn on the way to a psychology or biology group where
they not only discuss developmental problems but
also things that affect the rate that hair grows
and genetic similarities in twins.
Sue...
>
> --
> --Bryan
Bryan Olson
> The math is simple when Kat is far from a mass.
Glad you think so, Sue. Assuming SR effects dominate, can you
now do problem 1 question e, on this quiz given at MIT?
http://web.mit.edu/8.01/www/Spring05/exams/qs12-s05.pdf
What I'm citing is the answer sheet with the correct answer
marked; this isn't a challenge. The important part is the
concepts.
Sue, previously when the twins paradox came up, you pointed me
to some MIT course material, but what you cited said nothing
about the phenomenon at issue. When I found what MIT actually
teaches, it turned out your version would rate a big red X.
> << The key to understanding special relativity is
> Einstein's relativity principle, which states that:
>
> All inertial frames are totally equivalent for
> the performance of all physical experiments.
>
> In other words, it is impossible to perform a physical
> experiment which differentiates in any fundamental sense
> between different inertial frames. By definition,
> Newton's laws of motion take the same form in all
> inertial frames. Einstein generalized this result
> in his special theory of relativity by asserting
> that all laws of physics take the same form in
> all inertial frames. >>
> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
Good stuff. Sadly, Sue draws conclusions opposite to
what it implies.
--
--Bryan
Bryan Olson
> Bryan Olson wrote:
> [...]
>> Understanding this much did not come easy for me, and at
>> one time I had thought SR to be contradictory. There's hope
>> that many who are currently of anti-relativity cranks will
>> come around, though the odds are not great.
>
> Thus far, your mis-spent youth has not made a convincing
> argument for any of the issues you have offered it in
> support of.
Indeed, my efforts played no part in convincing serious
scientists that these seemingly strange theories are in
fact how the universe works. I was, at best, late to the
game.
Nevertheless, that case is made. Question do no remain
open simply because some people refuse to learn the
answers.
> Are you sure you didn't take a wrong
> turn on the way to a psychology or biology group where
> they not only discuss developmental problems but
> also things that affect the rate that hair grows
> and genetic similarities in twins.
Having been wrong before, I'm less that certain about most
everything. Still, I find it reassuring, Sue, when the very
sources you choose to cite support my understanding and
refute yours. Do I need to cite those again?
--
--Bryan
Dirk Van de moortel
>>
>>
>> Try to understand the meanings of the variables. Watch what happens
>> with imbeciles who take their failing to understand these meanings
>> for errors made by the rest of the world.
>>
>> Dirk Vdm
>
> Dirk, it is very easy to see things are wrong if you don't understand
> them. Where did you study physics? A community college perhaps?
>
> Read the example below,
[snip the silly numbers again -
first make sure you can handle the equations]
> --------------------------------------------------------------------------
>
>
> Now obviously you don't understand Minkowski space-time. You don't
> even understand the definition of a distance or time interval, which
> are *ZERO* only for two bodies if they are IDENTICAL. I can see your
> penchant for zeros, it must be your score in life. Seriously, these
> are not partial differentials either. These are relabeled below to
> avoid brain meltdown for people who read too fast but understand too
> slowly, Dirk.
I was taking the (0,0) events in both frames as the events to
calculate the distances and time intervals, bozo.
I'l be more explicit if you like.
>
>
> Given two frames F1 (stationary) and F2 (moving), with relative
> velocity V, the distance intervals in L2 are translated to L1 in the
> stationary frame by the transformation equation:
>
> L1 = L2 / gamma
No, they are translated by the transformation equations
{ L2 = gamma ( L1 - v T1 )
{ T2 = gamma ( T1 - v L1/c^2 )
{ L1 = gamma ( L2 + v T2 )
{ T1 = gamma ( T2 + v L2/c^2 )
{ gamma = 1 / sqrt(1-v^2/c^2)
The first equation
L2 = gamma ( L1 - v T1 )
which is valid for *all* events, reduces to your special equation
L1 = L2 / gamma
if and only if
T1 = 0,
i.o.w. *only* for events that are simultaneous in the F1-frame.
You clearly don't understand that equation.
>
> And the times T2 are translated into T1 by the transformation equation
>
> T1 = T2 x gamma
The fourth equation
T1 = gamma ( T2 + v L2/c^2 )
which is valid for *all* events, reduces to your special equation
T1 = T2 gamma
if and only if
L2 = 0,
i.o.w. *only* for events that happen at the same place in the F2-frame.
You clearly don't understand that equation either
So, bozo, if you use both equations
L1 = L2 / gamma
T1 = T2 gamma
together to arrive at any conclusion, the result is valid if and only if
T1 = 0
L2 = 0,
i.o.w, only for *only* for events that are simultaneous in the F1-frame
*and* happening at the same place in the F2-frame, which as you can
see from your own two eqations, if
L1 = L2 = T1 = T2 = 0,
i.o.w. the combination is invalid for any two *distinct* events
>
> where gamma = (1-v^2/c^2)^(1/2)
Actually, where
gamma = 1 / (1-v^2/c^2)^(1/2)
>
> V1 = D1/T1
>
> V2 = D2/T2
You know what? We'll let this pass this pass this time.
How about that?
>
> now velocity V1 and V2 are equal since this is the relative velocity
> (ok, the sign is different, BIG ACHIEVEMENT there Dirk, but remember,
> this is 1-dimensional, to make it easier for you, so it doesn't really
> matter).
>
> V2 = V1 = D1/T1 = D2/T2
And you know what? We'll let this pass as well. Be my guest.
>
> D1 = D2 / gamma, T1 = T2 x gamma
Sure, but *only* if T1 = 0 for the first equation, and if D2 = 0 for
the second equation.
Let me guess... now you are now going to use your equations
together, so what you write is only valid if
D1 = D2 = T1 = T2 = 0
Let's see...
>
> V1 = (D2 / gamma) / ( T2 x gamma) = (D2/T2) / (gamma^2)
>
> but also D2/T2 = V2, so V1 = V2 / (gamma^2)
Voila, here you go.
That amounts to nonsense like
V1 = (0/gamma ) / (0 gamma) = (0/0) / gamma^2
but also 0/0 = V2, so V1 = V2 / gamma^2
Congratulations!
>
> and since V1 = V2 then *V2 = V2 / gamma ^2*
Sometimes they say in a sloppy way that "0/0 can be anything",
remember? You have proven that anything is equal to everything!
You see what happens with people who have turned juggling
equations while miserably failing to understand the meanings
of the variables into a fine art?
I'll snip here to save you the embarrasment. How about that?
But wait... just let be guess for a minute...
You are a retired engineer, right?
Dirk Vdm
mlut...@wanadoo.fr
On Jun 30, 10:23 pm, Eric Gisse <jowr...@gmail.com> wrote:
On Jun 30, 12:03 pm, mluttg...@wanadoo.fr wrote:
<snip>
> > Try to solve this contradiction, without obfuscating
> > the subject with the "twin paradox" jumps!
> It isn't a contradiction. Try to learn something, instead of being a
> write-only poster.
Gisse,
I was trying again and again to get a straightforward
reponse from Vdm, but he prefers to sidestep the question.
At least, *you* have an opinion, i.e. the fact that
Kat and the Earthling,both using SR, find symmetric
values doesn't represents a contradiction in SR.
How do you justify your position?
N.B.:
I don't expect a meaningful response from you, or from
any other pseudo-SR expert, whose main specialty is
'redirection' :
Newsgroups: sci.physics.relativity
*Followup-To: alt.morons*
From: Eric Gisse <jowr...@gmail.com>
Date: Mon, 30 Jun 2008 13:23:32 -0700 (PDT)
Local: Mon, Jun 30 2008 10:23 pm
Subject: Re: Finally, Special Relativity Is Proven False
Marcel Luttgens
Dirk Van de moortel
[snip Luttgens clumsy quoting mess]
> On Jun 30, 10:23 pm, Eric Gisse <jowr...@gmail.com> wrote:
> On Jun 30, 12:03 pm, mluttg...@wanadoo.fr wrote:
>
> <snip>
>
>>> Try to solve this contradiction, without obfuscating
>>> the subject with the "twin paradox" jumps!
>
>> It isn't a contradiction. Try to learn something, instead of being a
>> write-only poster.
>
> Gisse,
>
> I was trying again and again to get a straightforward
> reponse from Vdm, but he prefers to sidestep the question.
Luttgens, you are too dumb to judge answers given to you.
>
> At least, *you* have an opinion, i.e. the fact that
> Kat and the Earthling,both using SR, find symmetric
> values doesn't represents a contradiction in SR.
You are beginning to sound like Seto.
Way to go!
>
> How do you justify your position?
>
> N.B.:
>
> I don't expect a meaningful response from you, or from
> any other pseudo-SR expert, whose main specialty is
> 'redirection' :
If you don't expect a meaningful response,what the hell
are you doing then?
Dirk Vdm
mlut...@wanadoo.fr
> mluttg...@wanadoo.fr wrote:
> > Try to solve this contradiction, without obfuscating
> > the subject with jumps.
>
> > "According to Earthling Vdm , Kat travels during
> > 5 years, but will measure 5 * sqrt(1-0.866^2) =~ 2.5 years
> > on her clock.
>
> According to SR, in the approximately-inertial rest frame
> of Earth and Alpha Centauri, the events of Kat leaving
> Earth and Kat arriving at Alpha Centauri are about 5 year
> apart in time. In Kat's traveling frame, the two events
> are about 2.5 years apart.
>
> > But Kat considers herself at rest, the Earth moving at
> > 0.866 c wrt to her.
> > After 5 years of her time (Kat is still alive, she
> > took a new drug), she claims that Dirk Vdm measured
> > only 2.5 years on his clock."
>
> If Kat continues in her away-from-Earth frame, after she
> clocks five years she'll be twice as far from Earth as is
> Alpha Centauri. In the frame of the Earth, Kat reaching
> that point happens 10 years after Kat departed Earth.
You said "If Kat continues in her away-from-Earth frame,.."
But this is different from what I wrote:
"After 5 years of her time (Kat is still alive, she
took a new drug), she claims that Dirk Vdm measured
only 2.5 years on his clock."
The contradiction is obvious:
1) If B moves at v = 0.866 c relative to A, A claims that
to a time interval tA = 10 seconds measured on his/her
clock corresponds a time interval
tB = tA * sqrt(1-v^2/c^2) =~ 5 seconds measured on B's
clock.
Thus, tA = 10 s, tB = 5 s.
If B considers that A is moving at -0.866 c wrt to him/her,
B claims that to a time interval tB = 10 seconds measured
on his/her clock corresponds a time interval
tA = tB * sqrt(1-v^2/c^2) = 5 seconds measured on A's
clock.
Then, tA = 5 s, tB = 10 s.
Marcel Luttgens
jem
> Sue... wrote:
>> The math is simple when Kat is far from a mass.
>
> Glad you think so, Sue. Assuming SR effects dominate, can you
> now do problem 1 question e, on this quiz given at MIT?
>
> http://web.mit.edu/8.01/www/Spring05/exams/qs12-s05.pdf
>
> What I'm citing is the answer sheet with the correct answer
> marked; this isn't a challenge. The important part is the
> concepts.
IMO, choice (iii) is a better answer than (iv). Acceleration isn't
the differentiator. E.g. let one twin accelerate in a certain manner
to a fixed speed, some time later let the other twin accelerate in the
same manner, then let the two twins simultaneously accelerate
identically, but in opposite directions in order to reunite. They
will have accelerated identically, but, according to SR, they won't be
the same age. (It can also be arranged for them to accelerate
differently and rejoin at the same age).
The rejection of choice (iii) (i.e. the ages differ because the twins
/moved/ differently), on the grounds that the laws of Physics would
then not be identical in every Inertial reference frame, is misguided.
The twins moved differently, not relative to each other, but
relative to the reference frame in which their ages were ultimately
compared (any frame in this case, since they're co-located when the
comparison is done).
Dirk Van de moortel
Yes, retard,
tB = tA / gamma,
only valid for 2 events with the same xB, i.o.w. on the clock
caried by B.
>
> If B considers that A is moving at -0.866 c wrt to him/her,
> B claims that to a time interval tB = 10 seconds measured
> on his/her clock corresponds a time interval
> tA = tB * sqrt(1-v^2/c^2) = 5 seconds measured on A's
> clock.
>
> Then, tA = 5 s, tB = 10 s.
Yes, imbecile,
tA = tB / gamma,
only valid for 2 events with the same xA, i.o.w. on the clock
caried by A.
How many zillions of times do you reckon that has been explained
to you?
Please face it, you don't have the brain power to understand
the meanings of the variables.
Get another hobby. IIRC some 5 or 6 years ago I recommendend
collecting poodle droppings. Have you tried it, bozo?
Dirk Vdm
PD
> Sue... wrote:
> > The math is simple when Kat is far from a mass.
>
> Glad you think so, Sue. Assuming SR effects dominate, can you
> now do problem 1 question e, on this quiz given at MIT?
>
> http://web.mit.edu/8.01/www/Spring05/exams/qs12-s05.pdf
>
> What I'm citing is the answer sheet with the correct answer
> marked; this isn't a challenge. The important part is the
> concepts.
>
The test suffers from something common -- it's all too easy to provide
a set of answers, none of which are right but one of which is less
displeasing than the others.
The simplest accounting for the difference, in my opinion, is that the
straightest line through spacetime has the longest proper duration.
The one with the nonstraight path therefore naturally sees less
elapsed duration. This is opposite the usual Euclidean rule about the
shortest distance between two points, and for a very simple reason:
that little minus sign in the metric that changes the duration from a
sum to a difference.
PD
PD
>
> You said "If Kat continues in her away-from-Earth frame,.."
> But this is different from what I wrote:
> "After 5 years of her time (Kat is still alive, she
> took a new drug), she claims that Dirk Vdm measured
> only 2.5 years on his clock."
>
> The contradiction is obvious:
>
> 1) If B moves at v = 0.866 c relative to A, A claims that
> to a time interval tA = 10 seconds measured on his/her
> clock corresponds a time interval
> tB = tA * sqrt(1-v^2/c^2) =~ 5 seconds measured on B's
> clock.
> Thus, tA = 10 s, tB = 5 s.
>
> If B considers that A is moving at -0.866 c wrt to him/her,
> B claims that to a time interval tB = 10 seconds measured
> on his/her clock corresponds a time interval
> tA = tB * sqrt(1-v^2/c^2) = 5 seconds measured on A's
> clock.
>
> Then, tA = 5 s, tB = 10 s.
>
Sorry, what contradiction?
Perhaps you should first denote whether you're talking about the same
pair of events.
Dirk Van de moortel
b5c642c9-dad2-4946...@l64g2000hse.googlegroups.com
As I explained.
They can be the same pair of events if they both take place
on the clock caried by A and on the clock caried by B.
Since these clocks coincide in only 1 event, we have
tA = tB = 0, in which case his equations reduce to
0 = 0 * sqrt(1-v^2/c^2) = 0 seconds measured on A's clock,
and
0 = 0 * sqrt(1-v^2/c^2) = 0 seconds measured on B's clock
which is not really a contradition in my book.
But Luttgens has a strange book.
Dirk Vdm
Sue...
Simpler still is limiting to a domain of applicability.
You will find Einstein's famous formula E = m c^2 here:
"Relativistic particle dynamics"
http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
Without it, you have no principle of inertia on which to
base calculations. A light particle moving as a bullet
under the influence of inertia is a false basis because
no such particle exist.
http://nobelprize.org/physics/articles/ekspong/index.html
Sue...
>
> PD
PD
Wow. Do your bushel baskets of chaff explode spontaneously like that
often?
PD
Sue...
>
> > Sue... wrote:
> > > The math is simple when Kat is far from a mass.
>
> > Glad you think so, Sue. Assuming SR effects dominate, can you
> > now do problem 1 question e, on this quiz given at MIT?
>
> > http://web.mit.edu/8.01/www/Spring05/exams/qs12-s05.pdf
>
> > What I'm citing is the answer sheet with the correct answer
> > marked; this isn't a challenge. The important part is the
> > concepts.
>
> The test suffers from something common -- it's all too easy to provide
> a set of answers, none of which are right but one of which is less
> displeasing than the others.
>
> The simplest accounting for the difference, in my opinion, is that the
> straightest line through spacetime has the longest proper duration.
And what changes spatial to temporal displacements in that semantic
shell game?
<< if you know about complex numbers you will notice that the
space part enters as if it were imaginary
R2 = (ct)2 + (ix)2 + (iy)2 + (iz)2 = (ct)2 + (ir)2
where i^2 = -1 as usual. This turns out to be the
essence of the fabric (or metric) of spacetime geometry
- that space enters in with the imaginary factor i
relative to time.>>
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
<<the four-dimensional space-time continuum of
the theory of relativity, in its most essential
formal properties, shows a pronounced relationship
to the three-dimensional continuum of Euclidean
geometrical space. 1 In order to give due prominence
to this relationship, however, we must replace
the usual time co-ordinate t by an imaginary magnitude
sqrt(-1)
ct proportional to it.
http://www.bartleby.com/173/17.html
> The one with the nonstraight path therefore naturally sees less
> elapsed duration.
Space-time intervals don't have a "duration".
They have both spatial and temopral components
and a rigid formalism for keeping them separate.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Sue...
mlut...@wanadoo.fr
The idiotic Vdm has just ceased jumping!
Marcel Luttgens
>
> How many zillions of times do you reckon that has been explained
> to you?
> Please face it, you don't have the brain power to understand
> the meanings of the variables.
> Get another hobby. IIRC some 5 or 6 years ago I recommendend
> collecting poodle droppings. Have you tried it, bozo?
>
mlut...@wanadoo.fr
The little scenario is clear enough:
1) B moves at v = 0.866 c relative to A
2) B *considers* that A is moving at -0.866 c
Marcel Luttgens
Dono
> Get another hobby. IIRC some 5 or 6 years ago I recommendend
> collecting poodle droppings. Have you tried it, bozo?
>
He's probably thinking that they are truffles and...
Strich 9
>
> Dirk Van de moortel;1174848 Wrote: -
>
>
> Try to understand the meanings of the variables. Watch what happens
> with imbeciles who take their failing to understand these meanings
> for errors made by the rest of the world.
>
>
> Dirk, it is very easy to see things are wrong if you don't understand
> them. Where did you study physics? A community college perhaps?
>
>
>
> [snip the silly numbers again -
> first make sure you can handle the equations]
>
> -
>
> --------------------------------------------------------------------------
>
>
> Now obviously you don't understand Minkowski space-time. You don't
> even understand the definition of a distance or time interval, which
> are *ZERO* only for two bodies if they are IDENTICAL. I can see your
> penchant for zeros, it must be your score in life. Seriously, these
> are not partial differentials either. These are relabeled below to
> avoid brain meltdown for people who read too fast but understand too
>
> I was taking the (0,0) events in both frames as the events to
> calculate the distances and time intervals, bozo.
> I'l be more explicit if you like.
>
>
> Given two frames F1 (stationary) and F2 (moving), with relative
> velocity V, the distance intervals in L2 are translated to L1 in the
> stationary frame by the transformation equation:
>
>
> No, they are translated by the transformation equations
> { L2 = gamma ( L1 - v T1 )
> { T2 = gamma ( T1 - v L1/c^2 )
> { L1 = gamma ( L2 + v T2 )
> { T1 = gamma ( T2 + v L2/c^2 )
> { gamma = 1 / sqrt(1-v^2/c^2)
>
> The first equation
> L2 = gamma ( L1 - v T1 )
> which is valid for *all* events, reduces to your special equation
> L1 = L2 / gamma
> if and only if
> T1 = 0,
> i.o.w. *only* for events that are simultaneous in the F1-frame.
> You clearly don't understand that equation.
>
> And the times T2 are translated into T1 by the transformation equation
>
>
> The fourth equation
> T1 = gamma ( T2 + v L2/c^2 )
> which is valid for *all* events, reduces to your special equation
> T1 = T2 gamma
> if and only if
> L2 = 0,
> i.o.w. *only* for events that happen at the same place in the F2-frame.
> You clearly don't understand that equation either
>
> So, bozo, if you use both equations
> L1 = L2 / gamma
> T1 = T2 gamma
> together to arrive at any conclusion, the result is valid if and only
> if
> T1 = 0
> L2 = 0,
> i.o.w, only for *only* for events that are simultaneous in the F1-frame
> *and* happening at the same place in the F2-frame, which as you can
> see from your own two eqations, if
> L1 = L2 = T1 = T2 = 0,
> i.o.w. the combination is invalid for any two *distinct* events
>
> where gamma = (1-v^2/c^2)^(1/2)-
>
> Actually, where
> gamma = 1 / (1-v^2/c^2)^(1/2)
>
> V1 = D1/T1
>
> V2 = D2/T2-
>
> You know what? We'll let this pass this pass this time.
> How about that?
>
> now velocity V1 and V2 are equal since this is the relative velocity
> (ok, the sign is different, BIG ACHIEVEMENT there Dirk, but remember,
> this is 1-dimensional, to make it easier for you, so it doesn't really
> matter).
>
>
> And you know what? We'll let this pass as well. Be my guest.
>
> D1 = D2 / gamma, T1 = T2 x gamma-
>
> Sure, but *only* if T1 = 0 for the first equation, and if D2 = 0 for
> the second equation.
>
> Let me guess... now you are now going to use your equations
> together, so what you write is only valid if
> D1 = D2 = T1 = T2 = 0
> Let's see...
>
> V1 = (D2 / gamma) / ( T2 x gamma) = (D2/T2) / (gamma^2)
>
>
> Voila, here you go.
> That amounts to nonsense like
> V1 = (0/gamma ) / (0 gamma) = (0/0) / gamma^2
> but also 0/0 = V2, so V1 = V2 / gamma^2
>
> Congratulations!
>
> and since V1 = V2 then *V2 = V2 / gamma ^2*-
>
> Sometimes they say in a sloppy way that "0/0 can be anything",
> remember? You have proven that anything is equal to everything!
>
> You see what happens with people who have turned juggling
> equations while miserably failing to understand the meanings
> of the variables into a fine art?
> I'll snip here to save you the embarrasment. How about that?
>
> But wait... just let be guess for a minute...
> You are a retired engineer, right?
>
> Dirk Vdm
Excellent Dirk. You have shown basic knowledge of the *-theory-* of
special relativity. Now apply the theory to reality, in this case the
actual muon experiment. Show us your calculations if you have to.
Once you have done that, pick out the erroneous line in the reasoning
below. It is now a single multiple choice final exam for you. Be
careful, you don't want to lose that "upper hand". Don't clam up on me
now.
_______________________________________________________________
PROP 1: Muons formed in the upper atmosphere have a rest life of
~2.26us.
PROP 2: Traveling at speed ~c, they cover 700m as measured in the muon
frame prior to disintegration after ~2.26us,
PROP 3: Which corresponds to 7000m in the Earth frame due to Lorentz
contraction, enabling them to reach the lower atmosphere.
PROP 4: In 2.26us in the earth frame, any particle with speed ~c, (such
as a photon, or a muon in this case) can only travel 700m as measured in
the Earth frame, and not reach the lower atmosphere.
PROP 5: Since the 2.26us in the muon REST frame is EXACTLY EQUAL
(barring the trivial effects of a gravitational field) to the 2.26us in
the earth REST frame, then we have a contradiction, since 700m is not
equal to 7000m.
Conclusion: Special Relativity is erroneous.
--
Strich 9
Dirk Van de moortel
And, Marcel, when you're really hungry, like Dono suggested,
you can even pretend they are truffles, and EAT them.
How about that?
Dirk Vdm
Dirk Van de moortel
[snip]
> Excellent Dirk. You have shown basic knowledge of the *-theory-* of
> special relativity. Now apply the theory to reality, in this case the
> actual muon experiment.
My newsreader has a problem with the headers of yours in this thread.
See new thread
"Finally once more, Special Relativity is Proven False by Seju Strich"
Dirk Vdm
Strich 9
Well, well, Dirk. How much time do you need for this one
multiple-choice question? What happened to all the bravado? Don't go
weak on me now. Finish the job and prove that I am wrong. Or am I?
--
Strich 9
Eric Gisse
wrote:
Consider that several years of the same idiots making the same
mistakes dulls you to the same mistakes by a marginally different set
of idiots.
Bryan Olson
Marcel, you had asked for no frame jumps. Kat therefore
continues in the same inertial frame for those 5 years.
> she claims that Dirk Vdm measured
> only 2.5 years on his clock."
You need to state these things carefully. In Kat's inertial
frame, her clock reaching 5 years is (approximately)
simultaneous with a clock on Earth reaching 2.5 years
(assuming both are set to zero as Kat departs Earth).
> The contradiction is obvious:
>
> 1) If B moves at v = 0.866 c relative to A, A claims that
> to a time interval tA = 10 seconds measured on his/her
> clock corresponds a time interval
> tB = tA * sqrt(1-v^2/c^2) =~ 5 seconds measured on B's
> clock.
> Thus, tA = 10 s, tB = 5 s.
>
> If B considers that A is moving at -0.866 c wrt to him/her,
> B claims that to a time interval tB = 10 seconds measured
> on his/her clock corresponds a time interval
> tA = tB * sqrt(1-v^2/c^2) = 5 seconds measured on A's
> clock.
>
> Then, tA = 5 s, tB = 10 s.
The problem is that you used the same variable names for
times in different frames. Different observers using
different coordinate systems make different observations;
that's not a contradiction.
Your earlier attempt argued what would be a contradiction:
the theory predicting both that Kat would be alive upon
reaching Alpha Centauri and being dead upon reaching Alpha
Centauri. That would be real contradiction because the same
observation comes out two different ways, but of course it
turned out that had the analysis wrong.
We've recently seen a couple relativity-deniers state their
own theories that turned out to be contradictory. Ken Seto
and Robert Winn both tried the train-and-embankment problem,
but the way they thought physics works, a single observer
would have to see the lightning events as both simultaneous
and not simultaneous.
--
--Bryan
PD
> On Jul 1, 9:48 am, PD <TheDraperFam...@gmail.com> wrote:
>
>
>
> > On Jul 1, 2:22 am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
>
> > > Sue... wrote:
> > > > The math is simple when Kat is far from a mass.
>
> > > Glad you think so, Sue. Assuming SR effects dominate, can you
> > > now do problem 1 question e, on this quiz given at MIT?
>
> > > http://web.mit.edu/8.01/www/Spring05/exams/qs12-s05.pdf
>
> > > What I'm citing is the answer sheet with the correct answer
> > > marked; this isn't a challenge. The important part is the
> > > concepts.
>
> > The test suffers from something common -- it's all too easy to provide
> > a set of answers, none of which are right but one of which is less
> > displeasing than the others.
>
> > The simplest accounting for the difference, in my opinion, is that the
> > straightest line through spacetime has the longest proper duration.
>
> And what changes spatial to temporal displacements in that semantic
> shell game?
The duration is related to spacetime interval. It is a combination of
spatial and temporal displacements.
>
> << if you know about complex numbers you will notice that the
> space part enters as if it were imaginary
>
> R2 = (ct)2 + (ix)2 + (iy)2 + (iz)2 = (ct)2 + (ir)2
>
> where i^2 = -1 as usual. This turns out to be the
> essence of the fabric (or metric) of spacetime geometry
> - that space enters in with the imaginary factor i
> relative to time.>>http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
>
> <<the four-dimensional space-time continuum of
> the theory of relativity, in its most essential
> formal properties, shows a pronounced relationship
> to the three-dimensional continuum of Euclidean
> geometrical space. 1 In order to give due prominence
> to this relationship, however, we must replace
> the usual time co-ordinate t by an imaginary magnitude
>
> sqrt(-1)
>
> ct proportional to it.http://www.bartleby.com/173/17.html
>
> > The one with the nonstraight path therefore naturally sees less
> > elapsed duration.
>
> Space-time intervals don't have a "duration".
Sure they do. Every timelike spacetime interval has a reference frame
in which the spatial displacement is zero, and in this frame the
interval coincides with the duration, which is given a name -- do you
recall the name?
> They have both spatial and temopral components
> and a rigid formalism for keeping them separate.
Actually, no, a rigid formalism for keeping them muddled up together.
Sue...
> On Jul 1, 9:51 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
>
> > On Jul 1, 9:48 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > > On Jul 1, 2:22 am, Bryan Olson <fakeaddr...@nowhere.org> wrote:
>
> > > > Sue... wrote:
> > > > > The math is simple when Kat is far from a mass.
>
> > > > Glad you think so, Sue. Assuming SR effects dominate, can you
> > > > now do problem 1 question e, on this quiz given at MIT?
>
> > > > http://web.mit.edu/8.01/www/Spring05/exams/qs12-s05.pdf
>
> > > > What I'm citing is the answer sheet with the correct answer
> > > > marked; this isn't a challenge. The important part is the
> > > > concepts.
>
> > > The test suffers from something common -- it's all too easy to provide
> > > a set of answers, none of which are right but one of which is less
> > > displeasing than the others.
>
> > > The simplest accounting for the difference, in my opinion, is that the
> > > straightest line through spacetime has the longest proper duration.
>
> > And what changes spatial to temporal displacements in that semantic
> > shell game?
>
<<
> The duration is related to spacetime interval. It is a combination of
> spatial and temporal displacements.
>>
Gasoline is related to a fire apparatus but
you can't use it to extinguish a fire.
I am related to a surgeon but you don't
want to get near my kitchen knife.
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
You are not in every case taking values only for that inertial frame?
Read the formalism for "Proper [space] time" so you will know
when the spatial component can be ignored and how to remove
it in four-space.
Hint: "Kinetic energy"
Warning: Mathematics rather than semantics must be used.
http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
>
> > They have both spatial and temopral components
> > and a rigid formalism for keeping them separate.
>
> Actually, no, a rigid formalism for keeping them muddled up together.
Unless you are into parlor tricks
(and you seem to be
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
)
there is no reason you want them "muddled up together".
They are not on the same footing.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Sue...
Dirk Van de moortel
> Well, well, Dirk. How much time do you need for this one
> multiple-choice question? What happened to all the bravado? Don't go
> weak on me now. Finish the job and prove that I am wrong. Or am I?
I have given you the tools to find out for yourself. Or have I?
Dirk Vdm
Strich 9
Well, it looks like Dirk dropped out of class. Is there any other
physicist out there who is willing to help Dirk? Will you all drop
out? I would presume the silence reflects agreement with the
conclusion of this thread!
_____________________________________________________________
Which of the following is the wrong proposition?
PROP 1: Muons formed in the upper atmosphere have a rest life of
~2.26us.
PROP 2: Traveling at speed ~c, they cover 700m as measured in the muon
frame prior to disintegration after ~2.26us,
PROP 3: Which corresponds to 7000m in the Earth frame due to Lorentz
contraction, enabling them to reach the lower atmosphere.
PROP 4: In 2.26us in the earth frame, any particle with speed ~c, (such
as a photon, or a muon in this case) can only travel 700m as measured in
the Earth frame, and not reach the lower atmosphere.
PROP 5: Since the 2.26us in the muon REST frame is EXACTLY EQUAL
(barring the trivial effects of a gravitational field) to the 2.26us in
the earth REST frame, then we have a contradiction, since 700m is not
equal to 7000m.
Conclusion: If all the above is true, then -Special Relativity is
*ERRONEOUS*-.
--
Strich 9
Dirk Van de moortel
Ha, another setup with the frames swapped now.
Perhaps you have realised your mistake in the previous setup?
I doubt that, since there's really no difference with this setup,
but one never knows. Perhaps you're just trolling a bit.
But, good grief, you are *still* using silly numbers.
Give the velocity the name v, the distance of 7000 km the name L,
and the 2.26 microseconds the name T'.
Let the muon be created at x = 0 and t = 0 i.o.w. at event (x,t) = (0,0).
Let the muon desintegrate at (muon-)time t' = T', i.o.w. at event
(x',t') = (0,T').
Imagine another particle flying along with the muon with an infinite
lifetime to make sure that it can reach the ground (at x = L) at
(Earth-)time t = L/v, i.o.w. in event (x,t) = (L,L/v).
.
Now use the simple equations I gave you to describe all the events
in all the coordinates and restate your questions in terms of those
coordinates.
If you do it properly, you'll be able to answer the question you
should have asked, as opposed to making that rather silly declaration.
Don't be afraid to show your work and ask for help if you get stuck.
Dirk Vdm
Dirk Van de moortel
2OKak.226437$M63....@newsfe13.ams2
[correction]
Don't be afraid to show your work and ask for help when you get stuck.
Dirk Vdm
PD
>
> there is no reason you want them "muddled up together".
>
> They are not on the same footing.
So you say. Then again, vertical and horizontal are not on the same
footing, either, but it makes more than good sense to define a
distance between the tip of my left big toe and the tip of my right
index finger, which muddles up together the vertical and horizontal
distances involved. It turns out this distance in a rest frame is
independent of the choice of horizontal and vertical axes.
Likewise, there is good sense in defining an interval between one
event (a location in space and time) and another that muddles together
the spatial and temporal extents involved. It turns out this interval
in an inertial frame is independent of the choice of spatial and
temporal axes.
PD
wrote:
Whoops. There seems to be a problem in PROP 5, already noted earlier
but repeated again here.