twin clock problem - SR experts help!
dsep...@austin.rr.com
when I change the parameters to the problem, I realize that those
explanations are wrong. So I'm looking for an expert who can explain
the following SR problem.
Let there be two clocks, Clock A and Clock B, moving along the x-axis.
They have a relative velocity of 3 meters / second. At time t0, they
are at the same position in space. At this point in space and time,
both clocks are set to zero. They now continue to move away from each
other still traveling at 3 meters / second. According to Einstein's
theory, observers in Frame A (Clock A's rest frame) measure that Clock
B is running at a slower rate than identical synchronized clocks that
are at rest in Frame A. After 10**16 seconds, observers in Frame A
measure Clock B to be one second slower than Clock A. (The clocks are
about 10**8 light-seconds apart).
When Clock A reads 10**16 seconds, let Clock A undergo an
acceleration over a 0.1 second interval in the direction toward Clock
B. Likewise, when Clock B reads 10**16 seconds, let Clock B undergo an
acceleration over a 0.1 second interval towards Clock A. When the two
accelerations have ended, Clock A and Clock B are now approaching each
other with a relative velocity of 3 meters / second. Now to clarify
who's frame the 0.1 second interval was measured over, all frames
used in this problem measure 0.1 seconds as the acceleration interval
plus or minus some extremely small delta since the relative velocity
of 3 meters / second is so slow relative to c. So any variation in the
measurement of 0.1 second versus the 10**16 seconds has negligible
effect on the answer to this problem.
Observers in this new Clock A frame observe that as Clock B is
approaching Clock A, Clock B is running slower than Clock A
(Einstein's theory). When Clock A meets Clock B, both clocks
indicate the same time. This is true because of the symmetry of the
problem.
The question is, as measured by Clock A, if Clock B was running
slower than Clock A for the 10**16 seconds of inertial motion when the
clocks are moving apart, and Clock B is running slower than Clock A
for the 10**16 seconds of inertial motion when the clocks are moving
toward each other, and both clocks were set to zero when they first
meet, how do they end up showing the same time when they meet again?
Now some may think something happens to the clocks when they are
accelerated to to 3 meters / second. However, as is easily
demonstrated experimentally, when an electronic clock or most
mechanical clocks change their velocity by 3 meters / second over a
0.1 second interval, they do not gain or lose anywhere close to a
second or two relative to nearby clocks that haven't accelerated.
Observers in the original rest frame of Clock A near the position of
Clock A do not measure a substantial change in Clock A as it changes
its velocity by 3 meters / second. Likewise observers in the original
rest frame of Clock B in the vicinity of Clock A do not observe any
substantial change in Clock A as it changes its velocity by 3 meters /
second. And Observers moving with Clock A through the acceleration
likewise do not measure any substantial change in the Clock A time
relative to nearby clocks in the original Frame A and Frame B that
haven't accelerated. So experimentally, the acceleration of Clock A
(and of Clock B) has negligible effect on the time shown on Clock A
(or on Clock B).
Also we can demonstrate logically that Clock A and Clock B do not
gain or lose sufficient time during the acceleration to make any
difference in the indicated time when they meet. We do this by noting
that the amount of time they must gain or lose to read the same time
when they meet the second time is a function of their separation when
the acceleration starts and ends. Logically, the amount of time Clock
A gains or loses during an acceleration does not depend in any known
way on another Clock 10**8 light-seconds away or some other distance
away that it might encounter sometime in the future.
So if both Clocks were set to zero when they first met, and if
ClockB was running slower than Clock A for 2*(10**16) seconds, and no
measureable change in their readings is observed during the
accelerations, how does it happen that they show the same time when
they meet the second time?
Thanks,
David Seppala
we...@operamail.com
When the acceleration is over, A would have to see B to be one second
ahead of his clock (as opposed to one second behind his clock before
the acceleration). That way, their clocks will show the same time when
they meet, despite A still seeing B's clock running slower than his. I
can't tell exactly how the clock shift happens, but I imagine that A
quickly cathes up with the light from B in that 0.1 seconds (more
quickly than he would if there was no acceleration), kind of a doppler
effect but involving acceleration. So, in my view, mutual time dilation
is kind of an illusion, yet it is real, since pretty much everthing
depends on light. Disclaimer: This explanation is probably wrong :)
Nick
Only the twin traveling rapidly *through space* will
have his clock run slow.
Einstein was wrong. Physics isn't about an equal appearence
of motion and therefore isn't about an equal appearence of
the relativistic effects of motion.
The situation is one sided.
Bilge
>when I change the parameters to the problem, I realize that those
>explanations are wrong. So I'm looking for an expert who can explain
>the following SR problem.
sue jahn
<dsep...@austin.rr.com> wrote in message news:430e662f...@news-server.austin.rr.com...
A clock simulates the passage of time.
If it responds to motion, temperature or other conditions then
it is not a clock.
Sue...
>
>
JanPB
During the turnaround A measures B's clock to quickly gain a lot of
seconds. This compensates for the time dilation periods. Keep in mind
that this "measures" means "reading the A-synchronised clocks at the
relevant events along B's trajectory" - it doesn't mean B's clocks
suddenly go bananas because A turned his head.
This is fully analogous to the Euclidean situation with two cars going
from P to Q, one car along PXQ, the other along PYQ, like so:
Q
/ \
/ \
/ \
/ \
/ \
X Y
\ /
\ /
\ /
\ /
\ /
P
Both cars start at the same time and drive at the same speed. Each
driver looks *perpendicularly* out the window in the direction of the
other driver and concludes "he is driving slower than me" (that
corresponds to "his clock is slower") since every foot (say) of one
driver's path corresponds to more than a foot of the other driver's
path when viewed in the perpendicular direction. This holds for both
the outgoing and the incoming parts of their journey.
One may then ask your type of question: how is it possible then that
they meet at the same time? Well, during the turnaround the
perpendicularly stretched out finger will trace a very large and quick
*backward* sweep against the other driver's path. The amount of
distance "lost" that way compensates for the distance gained during the
straight ("inertial") portions of the journey.
Someone might say: "but this makes the notion of simultaneity into a
mere convention". And it's true - simultaneity in relativity is a
convention. It allows one to set up reference frames with particularly
nice properties plus for v << c they correspond to Newtonian frames.
Simultaneity is not really needed for SR (just like algebra of
coordinates is not needed for Euclidean geometry), it's "merely" mighty
convenient because it allows us to quantify things.
> Now some may think something happens to the clocks when they are
> accelerated to to 3 meters / second. However, as is easily
> demonstrated experimentally, when an electronic clock or most
> mechanical clocks change their velocity by 3 meters / second over a
> 0.1 second interval, they do not gain or lose anywhere close to a
> second or two relative to nearby clocks that haven't accelerated.
> Observers in the original rest frame of Clock A near the position of
> Clock A do not measure a substantial change in Clock A as it changes
> its velocity by 3 meters / second. Likewise observers in the original
> rest frame of Clock B in the vicinity of Clock A do not observe any
> substantial change in Clock A as it changes its velocity by 3 meters /
> second. And Observers moving with Clock A through the acceleration
> likewise do not measure any substantial change in the Clock A time
> relative to nearby clocks in the original Frame A and Frame B that
> haven't accelerated.
I think the Euclidean analogy above answers some of your questions.
> Also we can demonstrate logically that Clock A and Clock B do not
> gain or lose sufficient time during the acceleration to make any
> difference in the indicated time when they meet.
It's not that the clocks A and B suddenly "do" something during the
acceleration. What physically happens is that you have two
symmetrically moving clocks so they display the same elapsed times upon
reunion. The whole "time dilation" during the inertial portion of the
trip and the sudden burst of "time contraction" during the accelerated
portion of the trip is an artifact of the observation method called
"synchronised clocks" and "using instantaneously co-moving inertial
frames for periods of acceleration" - ponder that Euclidean analogy
again. Of course different elapsed times for non-symmetrically moving
cloks is a real phenomenon, it's only that the method of keeping track
of the clocks that uses the above coordinate method features those
coordinate-related artifacts.
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message
news:1125036437....@o13g2000cwo.googlegroups.com...
[snip crap]
You stupid fucking moron!
Why do you post here when you don't understand this stuff?
Anything wrong with being decent and honest?
I see you ran way from me quickly enough, fuckwit.
Androcles.
dsep...@austin.rr.com
When I draw a line perpendicular to each path and use that to
determine if the other driver is going faster or slower than me,
although I don't necessarily agree with this analogy, I observe that
when I go from P to Y, the other driver's position trails this
perpendicular line so yes you might say he's traveling slower.
However, when I pass Y and continue on to point Q, I find that the
other driver is now ahead of this perpendicular. Part of the path
he's slower and part of the path he's faster and we both end up at Q
at the same time. In the problem I posted there is no part of the
path in which clock B rusn faster than Clock A (as measured from A's
point of view)
In Einstein's theory, the clocks must physically run at different
rates just as identical physical rods in different frames cannot
occupy the same points in space at the same time. These aren't
artifacts.
David
Dirk Van de moortel
"sue jahn" <susyse...@yahoo.com.au> wrote in message news:430ea504$0$18638$1472...@news.sunsite.dk...
[snip]
> A clock simulates the passage of time.
That's only for philosophers who never held a clock in their hands.
For physicists and engineers and the rest of the world, a clock
*defines* the passage of time.
> If it responds to motion, temperature or other conditions then
> it is not a clock.
That's only for philosophers who never held a clock in their hands.
For physicists and engineers and the rest of the world, it remains a
clock if it responds to motion, temperature or other conditions, but
the more it responds, the less useful it is to define "the passage of
time" and to use it to make life easy.
Dirk Vdm
Dirk Van de moortel
> when I change the parameters to the problem, I realize that those
> explanations are wrong. So I'm looking for an expert who can explain
> the following SR problem.
Back to square one?
Again?
Again again?
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/00a9f68bca6e37b3
Dirk Vdm
Harry
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:%uAPe.11197$4b2....@news.cpqcorp.net...
>
> "sue jahn" <susyse...@yahoo.com.au> wrote in message
news:430ea504$0$18638$1472...@news.sunsite.dk...
>
> [snip]
>
> > A clock simulates the passage of time.
>
> That's only for philosophers who never held a clock in their hands.
>
> For physicists and engineers and the rest of the world, a clock
> *defines* the passage of time.
Hmm... a correctly *calibrated* clock defines the passage of time
> > If it responds to motion, temperature or other conditions then
> > it is not a clock.
>
> That's only for philosophers who never held a clock in their hands.
>
> For physicists and engineers and the rest of the world, it remains a
> clock if it responds to motion, temperature or other conditions, but
> the more it responds, the less useful it is to define "the passage of
> time" and to use it to make life easy.
Thus:
The the more it is affected, the less useful that clock becomes to indicate
"the passage of time".
Harald
Harry
I didn't read all, but I suspect that the two undergo fully symmetrical
motions. Is that right? Then, if both clocks undergo the same physical
situations, how could there be any difference in their clock readings?
Just regard both clocks from a third, inertial frame, and in which both
clocks tick equally slow.
Harald
Dirk Van de moortel
"Harry" <harald.v...@epfl.ch> wrote in message news:430edf05$1...@epflnews.epfl.ch...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:%uAPe.11197$4b2....@news.cpqcorp.net...
> >
> > "sue jahn" <susyse...@yahoo.com.au> wrote in message
> news:430ea504$0$18638$1472...@news.sunsite.dk...
> >
> > [snip]
> >
> > > A clock simulates the passage of time.
> >
> > That's only for philosophers who never held a clock in their hands.
> >
> > For physicists and engineers and the rest of the world, a clock
> > *defines* the passage of time.
>
> Hmm... a correctly *calibrated* clock defines the passage of time
No. As soon as you have a clock - any kind of clock, like
your heart beat, you can define time and the passage of time
with it, and even do experiments with it. You can find out that
the distance covered by a falling stone is more or less
proportional to the square of the falling time, which you define
as the number of heart beats that you count.
>
> > > If it responds to motion, temperature or other conditions then
> > > it is not a clock.
> >
> > That's only for philosophers who never held a clock in their hands.
> >
> > For physicists and engineers and the rest of the world, it remains a
> > clock if it responds to motion, temperature or other conditions, but
> > the more it responds, the less useful it is to define "the passage of
> > time" and to use it to make life easy.
>
> Thus:
> The the more it is affected, the less useful that clock becomes to indicate
> "the passage of time".
First of all, to *define* "the passage of time".
Dirk Vdm
Kim B
Immediately after the acceleration, A and B will observe the other
clocks as one second ahead - this is due to changed FOR. When they
meet again, they will both have noticed the other's clock slowed a
second, and clocks will match.
This is easily seen in a space time diagram
Kim
Dirk Van de moortel
"Kim B" <spam...@use.net> wrote in message news:l8stg19j2492s8d5o...@4ax.com...
During all these years, he never made one diagram
and he ignored every diagram that was ever shown
to him.
Dirk Vdm
Androcles
HAHAHAHAHAHAHA!
The first two clocks leave and come back from opposite directions,
each running slower than the other, to meet a third clock that moved
away
and was replaced by a fourth clock at the other end of a rigid rod.
Too funny, watching puppies like you chasing their own tails.
Androcles.
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:gECPe.19258$5m3....@fe1.news.blueyonder.co.uk...
[snip]
> | I didn't read all, but I suspect that the two undergo fully symmetrical
> | motions. Is that right? Then, if both clocks undergo the same physical
> | situations, how could there be any difference in their clock readings?
> | Just regard both clocks from a third, inertial frame, and in which
> | both clocks tick equally slow.
>
> HAHAHAHAHAHAHA!
>
> The first two clocks leave and come back from opposite directions,
> each running slower than the other, to meet a third clock that moved
> away
> and was replaced by a fourth clock at the other end of a rigid rod.
> Too funny, watching puppies like you chasing their own tails.
We look at each other through a gap between our fingers.
We both say that the other one is smaller, until we meet
again to find that we have the same length.
Yes, that is *very* funny indeed.
Dirk Vdm
Harry
>
> "Harry" <harald.v...@epfl.ch> wrote in message
news:430edf05$1...@epflnews.epfl.ch...
> >
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
> > in message news:%uAPe.11197$4b2....@news.cpqcorp.net...
> > >
> > > "sue jahn" <susyse...@yahoo.com.au> wrote in message
> > news:430ea504$0$18638$1472...@news.sunsite.dk...
> > >
> > > [snip]
> > >
> > > > A clock simulates the passage of time.
> > >
> > > That's only for philosophers who never held a clock in their hands.
> > >
> > > For physicists and engineers and the rest of the world, a clock
> > > *defines* the passage of time.
> >
> > Hmm... a correctly *calibrated* clock defines the passage of time
>
> No. As soon as you have a clock - any kind of clock, like
> your heart beat, you can define time and the passage of time
> with it, and even do experiments with it. You can find out that
> the distance covered by a falling stone is more or less
> proportional to the square of the falling time, which you define
> as the number of heart beats that you count.
Just try to sell that idea to a normalisation institute. ;-)
- On a serious note: when it was discovered that the solar clock is
irregular compared to atomic clocks, the atomic clock standard took over for
a more precise (regular) calibration. From then on the solar clock does not
anymore define the "passage of time", as it's not anymore the reference
standard.
Harald
Dirk Van de moortel
"Harry" <harald.v...@epfl.ch> wrote in message news:430f059e$1...@epflnews.epfl.ch...
When you are alone in the desert without an atomic clock or
wrist watch or sun or stars, you can define time and "the passage
of time" with your heart beat, and do physics and engineering.
You don't need reference standards. You need a clock,
something that does pom pom pom pom pom pom.
You count the poms and call it time.
When you have more than one kind of clock, you can
compare one against the other and compare the times you
define and read on those clocks.
If you can define regularity or "correct calibration" without
using the concept of time, then I agree with your suggestion
that "a correctly *calibrated* clock defines the passage of
time"
Dirk Vdm
Henry Haapalainen
"sue jahn" <susyse...@yahoo.com.au> kirjoitti viestissä
news:430ea504$0$18638$1472...@news.sunsite.dk...
Henry Haapalainen
Henry Haapalainen
"Androcles" <Androcles@ MyPlace.org> kirjoitti viestissä
news:nyyPe.18663$5m3....@fe1.news.blueyonder.co.uk...
Do we really need that kind of discussion? Is this a place for street fight?
Henry Haapalainen
Androcles
"Henry Haapalainen" <kir...@kolumbus.fi> wrote in message
news:denaih$dd3$1...@phys-news1.kolumbus.fi...
|
| Henry Haapalainen
Yes, of course it is. I've been very patient with the idiot, he snips
and refused to argue logically and has run from me. Now I see him
trying to defraud dseppala. He deserves a punch in the mouth, so I gave
him one. I'd lock him in jail if I could, along with this tin god,
Einstein,
but since only words are permitted, a street fight it must be.
That cheeky bastard said to me
Since you don't understand this stuff, why don't you refrain from
posting on this thread? What's your point? Anything wrong with being
decent and honest? -- Jan Bielawski
Then he said OF me:
"It was interesting to find out how far he could be pushed into
irrelevancies and nonsense. ", but his own words (and my replies) were:
Jan Phuckwit:
(Even ignoring everything else I posted, this sort of mistake would
have been caught immediately by the AdP reviewer.
Androcles:
Not relevant. It wasn't and it wasn't caught by you either.
It was caught by Androcles one hundred years later.
Anything wrong with being decent and honest?
Phuckwit:
It's WAY too simple-minded.
Androcles:
It's very simple, I'll agree.
Only the simple-minded would fall for Einstein's bullshit.
Anything wrong with being decent and honest?
Phuckwit:
It would be like Stephen Hawking dividing by zero or
something equally trivial.
Androcles:
I'm not interested in your Hawking irrelevancies.
I met him once, I was not impressed.
Anything wrong with being decent and honest?
Phuckwit:
Einstein wouldn't even bother to submit his
paper either.)
Androcles:
I can't change the facts.
Einstein intended to perpetrate a hoax, and succeeded.
Anything wrong with being decent and honest?
Then he ran away like the stinking coward he is.
So yes, its a street fight. I can lick the lying fraudulent coward with
10,000,000 neurons tied behind my back.
Androcles.
Dirk Van de moortel
"Henry Haapalainen" <kir...@kolumbus.fi> wrote in message news:dena8r$co4$1...@phys-news1.kolumbus.fi...
>
> "sue jahn" <susyse...@yahoo.com.au> kirjoitti viestissä
> news:430ea504$0$18638$1472...@news.sunsite.dk...
[snip]
> > A clock simulates the passage of time.
> > If it responds to motion, temperature or other conditions then
> > it is not a clock.
> > Sue...
> >
> I could't say that any better.
I think I could - and have :-)
Dirk Vdm
JanPB
An important clarification: when I wrote "he is driving slower then me"
I meant by this "his *speed* according to my perpendicular arm measure
is smaller than my speed": i.e. while in one second of my time my car
advances 1 unit of my distance coordinate, the other car in one second
advances *less* than 1 unit of my distance coordinate. Note that this
holds for *both* segments PY and YQ of my journey.
So each measures the other driving slower at all times except at the
turns, when each measures the other being repositioned far ahead at
high speed - that's like the period of brief time contraction.
So I'm talking about rates here, not distances (just like time
dilation/contraction in SR is a statement about coordinate clock rates,
not just plain clock readings which are, incidentally, actual physical
results, unlike clock rates which are coordinate-derived numbers).
> Part of the path
> he's slower and part of the path he's faster and we both end up at Q
> at the same time. In the problem I posted there is no part of the
> path in which clock B rusn faster than Clock A (as measured from A's
> point of view)
Right, I meant the rates.
> >It's not that the clocks A and B suddenly "do" something during the
> >acceleration. What physically happens is that you have two
> >symmetrically moving clocks so they display the same elapsed times upon
> >reunion. The whole "time dilation" during the inertial portion of the
> >trip and the sudden burst of "time contraction" during the accelerated
> >portion of the trip is an artifact of the observation method called
> >"synchronised clocks" and "using instantaneously co-moving inertial
> >frames for periods of acceleration" - ponder that Euclidean analogy
> >again.
>
> In Einstein's theory, the clocks must physically run at different
> rates just as identical physical rods in different frames cannot
> occupy the same points in space at the same time. These aren't
> artifacts.
In Einstein's theory clocks accumulate elapsed proper times that are
trajectory-dependent. That's all that's physically observable. (SR does
not provide any underlying model for *why* this is so, it only notes
that this is what follows if we presume inertial frame equivalence and
the speed of light contancy).
All statements regarding the "rates" are merely convenient
coordinate-based quantifications. Different coordinate conventions
yield different rates and ascribe different coordinate times to when
certain events occur. As far as physically observable end results go
(like total elapsed proper times, etc.), all such coordinate
conventions yield the same predictions - as they must.
So in SR clocks do not run "physically" slower or faster in any
meaningful sense of the word. Of course they would have to if the
universe was Newtonian.
--
Jan Bielawski
Cyde Weys
Can you back this up? My understanding of the twin paradox is that it
cannot be explained by special relativity, only general relativity.
But Einstein also came up with general relativity, so how exactly was
he "wrong"?
Ben Rudiak-Gould
> So if both Clocks were set to zero when they first met, and if
> ClockB was running slower than Clock A for 2*(10**16) seconds, and no
> measureable change in their readings is observed during the
> accelerations, how does it happen that they show the same time when
> they meet the second time?
Hint: look at the Doppler shift.
The last time you showed up I spent a lot of time answering your questions,
because you seemed like an intelligent person who would be able to
understand and learn. You gave every impression of doing both. But now you
seem to have forgotten it all, or never to have understood in the first
place. And Dirk Vdm says this has been going on for years and years. I can't
find the motivation to keep helping you in these circumstances. Special
relativity is just not that hard. You should move on to quantum field theory
or something.
-- Ben
Dirk Van de moortel
"Ben Rudiak-Gould" <br276d...@cam.ac.uk> wrote in message news:deo0t4$m1p$1...@gemini.csx.cam.ac.uk...
The mathematics is very easy, but interpreting the physical
meaning of the variables in the equations... that seems to
be incredibly difficult - at least on this forum.
> You should move on to quantum field theory
> or something.
They won't, because - for starters - the mathematics is a
bit harder ;-)
Dirk Vdm
Androcles
"Cyde Weys" <cy...@umd.edu> wrote in message
news:1125090274.2...@g47g2000cwa.googlegroups.com...
Hmm...
Look at this graph, x (distance) vertical and t (time) horizontal.
You'll need a fixed-width font, drawing in characters is a pain.
x
| /\B
| / \
| / \C
| /
x'/
| /
|/______+_+___t
A B C
There is a train on a track, locomotive at x' and caboose at A.
As time progresses, the train moves 'up' in distance and to the right
in time.
A ray of light start at A beside the caboose, moves along the track,
racing the train. The light wins, catching up with the locomotive at B.
There it meets a mirror and is reflected back to the caboose which is
moving forward to meet it at C.
The mirror can be on the track or on the locomotive.
All very simple, the time interval between A and B is greater than
the time interval between B and A.
Now along comes Einstein the genius, and says that to someone
riding the train, it will take the same time for light to travel from
the
caboose to the locomotove as it does from the locomotove to
the caboose. In his own words:
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
It's nothing to do with physics, it's a royal decree.
A "definition". It is because Einstein says so, no other reason.
You can define pi = 3 if you like, but most of us accept
pi = 3.1415926535897932384626433832795....
Anyway, the time for light to go from caboose to locomotive
is the same time it takes to return.
So he gets all mathematical to hoodwink you, and writes
―[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)),
the ― being the one-way time.
The equation is big enough and ugly enough that you'll ignore
the ― as "obvious".
From that he derives his cuckoo transformations, blames them
on Lorentz by giving credit to Lorentz and makes himself famous,
sits on his arse in Princeton playing the violin for the rest of his
life, doing sweet nothing.
He was the greatest huckster of all "time", if you'll pardon my
deliberate pun.
Now there's a story you can write, but mention me, huh?
I can help you with the math.
On the court docket, Science v Einstein.
Judge:
"The defendant stands before this court accused of fraud. How does the
defendant plead?"
Defense counsel:
"Not guilty, your honour."
Prosecution opens:
Ladies and Gentlemen of the jury, good morning.
Please reference: http://www.fourmilab.ch/etexts/einstein/specrel/www/
The first transformation we are given is the Galilean,
x' = x-vt
y = y
z = z
t = t
You have to agree with that, Einstein states:
"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time."
We have completed the transform from the stationary system, K, to the
moving system which I'm going to name k' because Einstein doesn't give
it a name.
"Objection! cries defense counsel.
"Yes?" asks the judge.
"Prosecuting counsel is making up names!" exclaims defense counsel,
"my client has already named the moving system 'k'."
"Why have you changed the name from k to k'?", the judge asks
prosecuting counsel.
Your honour, the name kappa (k) refers to the system of values xi, eta,
zeta, tau which are dependent upon velocity according to the accused
and should not be confused with the system of values x',y,z,t. I merely
chose a suitable name. If the court directs me to use another, I shall
abide by the court's wishes.
"Overruled, it is clear that Einstein gave no name to the system of
values x', y, z and I see no reason why prosecuting counsel should not
do so, do not waste the court's time on trivialities or I shall hold you
in contempt", says the judge.
Prosecution continues.
Applying the Galilean function g(),
For all x in K, x' in k', (x',y,z,t) = g(x,y,z,t).
It is clear, so ist klar, in agreement with experience, it being
immediately apparent and because Einstein says so,
a point at rest in system k' is independent of time.
There can be no function f such that x' = f(t), the judge's bench
will be the same length tomorrow as it is today. Hence there can
be no inverse function t = f^-1(x').
"Objection" cries the defense council.
"Shut up...err...overruled", says the judge, "I've heard this before in
another case. Don't make me hold you in contempt of Mathematics"
Defense council sits, looking sheepish.
Prosecution continues:
We have now completed the transformation from K to k' with the function
g, and can place system K on the back burner.
Now we come to the defendant's transformation. Not Lorentz's,
not Galileo's, but Einstein's own.
For all x in k', xi in kappa, (xi, eta, zeta, tau) = cuckoo(x',y,z,t)
We have a second transformation from the moving system k' to the
moving system kappa.
Einstein would have you believe that
tau = cuckoo_tau(g(x,y,z,t))
xi = cuckoo_xi(g(x,y,z,t))
is called the "Lorentz transformation".
I call it the cuckoo transformation, there is no relative motion between
k' and kappa, the time in k' has been found to be x'/(c-v) and x'/(c+v),
admitted by the defendant in his signed statement: "But the ray moves
relatively to the initial point of k, when measured in the stationary
system,
with the velocity c-v, so that x'/(c-v) = t."
"Objection!" cries the defense counsel, "it is universally known that
the velocity of light is c in all inertial frames of reference! It is a
postulate and the basis of my client's theory."
"How say you to that?", asks the judge of the prosecutor.
I can only ask the court's indulgence and request the defense counsel
produce the relevant passage in the evidence before the court, your
honour. The document referred to is "On the Electrodynamics of
Moving Bodies", the author being the defendant, and that authorship
has been stipulated to by both parties.
"The court will recess for lunch", say the judge, "I shall conduct a
computerized search for the term "inertial" in the relevant document.
============Lunch break=============
The clerk cries "Be upstanding in court!" and the judge seats himself.
"I find no reference to defense counsel's claim, the objection is
overruled".
Thank you, your honour. As I was saying:
I call it the cuckoo transformation, there is no relative motion between
k' and kappa, the time in k' has been found to be x'/(c-v) and x'/(c+v),
admitted by the defendant in his statement: "But the ray moves
relatively
to the initial point of k, when measured in the stationary system, with
the velocity c-v, so that x'/(c-v) = t".
The only purpose to the function cuckoo_tau is to satisfy Einstein's
fraudulent whim,
"we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A."
Thus it cannot be that x'/(c-v) = x'/(c+v).
As Counsel for the Physicists, I rest my case.
As Counsel for the Mathematicians, we have yet to prove that cuckoo_tau
is not a linear function.
"Objection!" screams the defense, prosecuting counsel is attempting
to prejudice the jury with the use of the word "cuckoo"!
"Sustained", says the judge, " the term 'cuckoo' will be stricken from
the record and the jury is instructed to disregard it."
Sidebar conference is requested by the prosecutor, jury is removed
from the court.
Prosecutor:
Your honour,
Einstein read "Time Machine" by H.G. Wells as a teenager, it was a
current best seller at the time. He became a clerk in the Swiss Patent
Office, saw many patent applications for cuckoo clocks, the main
industry of Switzerland in 1900-1905.
Not too many patents for chocolate or cheese were needed.
He got a hard-on... err... erection for time. The rest of us prefer
women, time is a fetish.
"I'm sustaining the objection",says the judge, "that is conjecture".
As the court pleases. (Prosecutor bows before the judge)
Jury is brought back, prosecution continues:
Here is that proof:
―[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
(given)
Juror passes a note to the judge via the bailiff. The note reads
"I don't understand this stuff, that equation is too long."
Judge: "The court should recess for 10 minutes."
==========ten minute recess =======
Prosecution continues:
Doubling both sides:
tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,t+x'/(c-v))
Taking out the t for 3:00pm on a Friday afternoon:
tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
Synchronize clocks at t = 0, tau(0,0,0,0) = 0, we remove tau(0,0,0,0)+
tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
There is no relative motion between k' and kappa, the coordinate x'
is independent of time. We do not have xi = x'-ut or x' = x'+ut or
any other function xi = fuckup(x') for Lorentz's sake, there is n...
"Objection to 'fuckup' ", says defense counsel, "it's not nice"
"Overruled".
As I was saying, there is no relative motion between k' and kappa,
the coordinate x' is independent of time. We do not have xi = x'-ut
or x' = x'+ut or any other function xi = fuckup(x') for Lorentz's
sake, there is no u, v, w or velocity between system k' and system
kappa.
The time at point zero is the same time at x', same at xi;
no translation between stationary and movng frames, this is the
moving frame only, the stationary frame K is simmering on the back
burner.
Hence:
tau(x',0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v)) =
tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v)) =
tau(a,b,c,x'/(c-v)+x'/(c+v)) = 2 * tau(a,b,c,x'/(c-v)),
the coordinate x' has no effect upon the time and is independent
of time, the time is independent of the coordinate. Were it not so,
the time at the front of the moving train, an example the defendant
uses in another document, would differ from the back of th...."
"Objection!", interjects defense counsel, "that document is not
in evidence before the court".
"I can produce it if necessary", says the prosecutor, but I'll withdraw
the statement".
"Objection sustained, the reference to a train will be stricken from
the record", orders the judge. "Continue with the mathemetics,
trains are physical and you rested on that."
My apology to the court. Let me reiterate, time is independent of
the coordinate. I have NOT stated that time was independent of velocity,
that is for the jury to decide.
Removing the superfluous coordinates:
tau(x'/(c-v)+x'/(c+v)) = 2 * tau(x'/(c-v))
Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity:
tau(a+b) = 2*tau(a)
Renaming tau as f,
f(a+b) = 2f(a) or
―f(a+b) = f(a)
Chosing a > b, we have an example
―f(1+0) = f(1)
"In the first place it is clear that the equations must be linear
on account of the properties of homogeneity which we attribute to
space and time." -- Albert Phuckwit/Huckster Einstein.
In the second place tau is not a linear function. -- Androcles.
In the third place there are no coordinates to transform.
In the fourth place you've been had!
I ask the jury to convict Einstein on the charge of fraud.
Prosecution reserves the right to cross-examine the witnesses.
I now rest my case as a mathematician also.
Prosecution counsel whispers to his learned colleague,
"I'd enter him an insanity plea if I were you, he's going down".
Counsel for the defense has the floor.
Androcles.
Daryl McCullough
General relativity is not needed to compute the effects of
acceleration on the ages of the twins, and is not even
*helpful* in computing the answer. The fact is, if there
are no huge massive objects involved, then there is no
difference between General Relativity and Special Relativity.
The reason that people sometimes say that GR is involved
is that if you use a coordinate system in which the accelerated
twin is at *rest*, then you need to use noninertial coordinates.
Understanding the mathematics of noninertial coordinates is the
first, necessary step towards understanding General Relativity.
So, the insight actually goes in the *opposite* direction:
Understanding Special Relativity in noninertial coordinates
helps you understand General Relativity, not the other way
around.
--
Daryl McCullough
Ithaca, NY
Henry Haapalainen
"Daryl McCullough" <stevend...@yahoo.com> kirjoitti viestissä
news:deo4k...@drn.newsguy.com...
There are a lot of misunderstandings in what inertial and noninertial
motions are. "Motion" caused by gravity is noninertial, because mass has no
influence in it. The deviation from free-fall motion is inertial, and that's
why these two factors are separated in FS (gravity as falling space).
Henry Haapalainen
Gravity as Falling Space http://www.wakkanet.fi/~fields/
Nick
it should see the accelerator's clock going slow if the effects are
reciprocal. It doesn't happen. Only one clock is going slow. Period.
How does the accelerated particle see the other particles
clocks?
sal
In a desert without sun or stars?
What desert might that be? Deserts I'm familiar with are all very
dry, and generally provide an excellent view of the sun during the day
and the stars at night.
I suppose you could be referring to a polar desert in white-out
conditions, but if I were wandering around alone outdoors in such a
situation I would certainly not be thinking about physics experiments.
> you can define time and "the passage of time"
> with your heart beat, and do physics and engineering.
Alone in the middle of white haze, unable to see anything, in sub-zero
temperatures ... most likely _far_ subzero ... and you're doing
physics _and_ engineering.
That's what I call dedication!
> You don't need reference standards. You need a clock,
No you don't, you need a compass! Preferably one of those cool Sagnac
gyro-compasses, since you may be too close to one magnetic pole or
another for a magnetic compass to do much for you.
You may need sunglasses, too, come to think of it, white haze or not.
And did I mention long underwear?
> something that does pom pom pom pom pom pom. You count the poms and
> call it time. When you have more than one kind of clock, you can
> compare one against the other and compare the times you define and
> read on those clocks.
You're alone in the Antarctic wilderness in the middle of a white-out
and your backpack turns out to be stuffed with nothing but clocks, of
all different kinds.
This does not reflect good planning.
> If you can define regularity or "correct calibration" without using
> the concept of time, then I agree with your suggestion that "a
> correctly *calibrated* clock defines the passage of time"
Of course it must be "correctly calibrated".
But "correctly calibrated" is, after all, a rather loose term. A
"time candle" whose stripes are equidistant from each other is
"correctly calibrated", for instance, though it's not likely to be of
much value in a blizzard in Antarctica.
Any heart you happen to have along which goes on going "pom pom
pom..." in a nice soothing regular way while you're stuck out there in
the middle of the ice fields also counts as a "correctly calibrated"
clock, under the circumstances. (You don't want to find out you've
brought along Dick Cheney's by mistake, for instance -- if the
batteries in the onboard defibrillator run down before you figure out
the way back to Ice Station Zebra you've got a big problem.)
>
> Dirk Vdm
--
Nospam becomes physicsinsights to fix the email
dsep...@austin.rr.com
<br276d...@cam.ac.uk> wrote:
>dsep...@austin.rr.com wrote:
>> So if both Clocks were set to zero when they first met, and if
>> ClockB was running slower than Clock A for 2*(10**16) seconds, and no
>> measureable change in their readings is observed during the
>> accelerations, how does it happen that they show the same time when
>> they meet the second time?
>
>Hint: look at the Doppler shift.
I don't understand what the hint in regard to Doppler shift is in
regards to this problem. What would help is if someone explains why I
just can't look at the local clocks at the turn around point. With
these clocks all I have to do to compare times is to compare the
readings on the clocks. And when I do that, no clocks have
significantly gained or lost time during the acceleration. If before
the acceleration I say one of the local clocks equals the time of a
clock 10**8 light-seconds away, and if I say that another one of the
local clocks reads one second more than the time shown on a clock
10**8 seconds away, and then I undergo an acceleation, and notice NO
change in the local clocks, why would I suddenly conclude that the
local clocks gained one second wrt to the clock 10**8 light-seconds
away. I don't grasp how nothing has changed with respect to the local
clocks, but now the local clocks are one second faster than a clock
10**8 light-seconds away that had absolutely nothing to do with my
acceleration of my local clock. That's what I don't grasp. Why did
the clocks that had nothing to do with acceleration now change their
readings wrt to each other?
Thanks,
David
dsep...@austin.rr.com
wrote:
If you would have read it all, you would see that I agree that because
of the symmetry of the problem there is no difference in their
readings when they meet the second time. But that is not the question
I'm seeking the answer to. The question is if Clock B always runs
slower than Clock A as measured by Clock A how can they start with the
same time and end up showing the same time. There is no where in
their path that I could find where the slower clock is measured to
speed up such that they have the same time when they meet. Where in
their path is such a measurement made?
David
>
>
we...@operamail.com
um.. If by "local clocks" you mean clocks local to frame A, they never
change, flow of time never changes in your frame according to you,
right? But if by "local clocks" you mean an extension of B frame with
clocks on it synchronized to B clock in B frame, you should first look
up "relativity of simultaneity". In short, clocks synchronized in B
frame will not look synchronized in A frame, even while inertial motion
not to mention while acceleration. But if you compare clock A to clock
B by using the light received from B, that's where doppler effect comes
in. Also, I would like to point out that since clock B is one second
behind clock A (according to A), they won't be accelerated at the same
time, according to either frame. And generally, there is no "real" gain
or loss of seconds, it all depends on your point of view, it is what is
measured in one particular frame and it is the reality for that frame.
Hope I didn't sound cocky..
Androcles
<dsep...@austin.rr.com> wrote in message
news:430fd354...@news-server.austin.rr.com...
| If you would have read it all, you would see that I agree that because
| of the symmetry of the problem there is no difference in their
| readings when they meet the second time.
I didn't read it all either, but I took that conclusion as your
sensible bottom line.
| But that is not the question I'm seeking the answer to.
Oh?
The question is if Clock B always runs
| slower than Clock A as measured by Clock A how can they start with the
| same time and end up showing the same time.
Ah... an IF question... I see.
The question I seek the answer to is IF bright green flying elephants
lay their eggs in black holes, how do they get them out?
| There is no where in
| their path that I could find where the slower clock is measured to
| speed up such that they have the same time when they meet. Where in
| their path is such a measurement made?
| David
Oh, that's an easy one. Put the clocks on the surface of the Earth
by your front door, do it right there. Hmm... that's a little dangerous,
clocks whizzing by at low altitude. Better to put them up higher,
so they miss trees and mountains... better yet, above atmosphere.
send one East and the other West at 17,000 mph, they'll
pass each other at 34,000 mph, go right around the Earth and
pass each other again at your front door and you can read the
time again. Being super accurate GPS clocks, which as we know
speed up with altitude, the SR effect is exactly counterbalanced
haha... cough.... by the GR effect...
giggle...
hahaha....
err... sorry, I'm being serious....
ROFLMAO!!!!
Androcles.
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2005.08.27....@nospam.org...
During a sand storm.
>
> What desert might that be?
> Deserts I'm familiar with are all very
> dry, and generally provide an excellent view of the sun during the day
> and the stars at night.
I'm in Tormentarena Desert ;-)
>
> I suppose you could be referring to a polar desert in white-out
> conditions, but if I were wandering around alone outdoors in such a
> situation I would certainly not be thinking about physics experiments.
Really?
Don't worry - no one will hold that against you.
>
>
> > you can define time and "the passage of time"
> > with your heart beat, and do physics and engineering.
>
> Alone in the middle of white haze, unable to see anything, in sub-zero
> temperatures ... most likely _far_ subzero ... and you're doing
> physics _and_ engineering.
>
> That's what I call dedication!
I guess I can't help it...
Please don't hold it against me?
>
>
> > You don't need reference standards. You need a clock,
>
> No you don't, you need a compass! Preferably one of those cool Sagnac
> gyro-compasses, since you may be too close to one magnetic pole or
> another for a magnetic compass to do much for you.
A Sagnac gyro-compass?
You are alone in your freezing desert, having a white-out,
and you are thinking of a way to invite Androcles for a party?
Ha... poor man. That must be worse than hell.
>
> You may need sunglasses, too, come to think of it, white haze or not.
>
> And did I mention long underwear?
I need *strong* underwear.
>
>
> > something that does pom pom pom pom pom pom. You count the poms and
> > call it time. When you have more than one kind of clock, you can
> > compare one against the other and compare the times you define and
> > read on those clocks.
>
> You're alone in the Antarctic wilderness in the middle of a white-out
> and your backpack turns out to be stuffed with nothing but clocks, of
> all different kinds.
Don't overdo it. I have my heart beat, and I noticed that
the wind is doing woosh woosh woosh woosh woosh.
I might be able to use that as a clock as well.
>
> This does not reflect good planning.
Indeed, sorry.
But I do have two clocks now :-)
>
>
> > If you can define regularity or "correct calibration" without using
> > the concept of time, then I agree with your suggestion that "a
> > correctly *calibrated* clock defines the passage of time"
>
> Of course it must be "correctly calibrated".
>
> But "correctly calibrated" is, after all, a rather loose term. A
> "time candle" whose stripes are equidistant from each other is
> "correctly calibrated", for instance, though it's not likely to be of
> much value in a blizzard in Antarctica.
>
> Any heart you happen to have along which goes on going "pom pom
> pom..." in a nice soothing regular way while you're stuck out there in
> the middle of the ice fields also counts as a "correctly calibrated"
> clock, under the circumstances. (You don't want to find out you've
> brought along Dick Cheney's by mistake, for instance -- if the
> batteries in the onboard defibrillator run down before you figure out
> the way back to Ice Station Zebra you've got a big problem.)
Ah, but I don't understand what you mean with
<< "pom pom pom..." in a nice soothing regular way >>
All I know, is that my heart goes pom pom pom pom pom pom...
And the wind does woosh woosh woosh woosh woosh woosh...
That's all I have - apart from all that sand finding it's way into every
conceivable orifice :-(
But it still is an interesting situation... I can still do physics!
Actually, I can do two "physicses" now and compare.
Yes!!!
;-)
Dirk Vdm
Kim B
>I don't understand what the hint in regard to Doppler shift is in
>regards to this problem. What would help is if someone explains why I
>just can't look at the local clocks at the turn around point. With
>these clocks all I have to do to compare times is to compare the
>readings on the clocks. And when I do that, no clocks have
>significantly gained or lost time during the acceleration. If before
>the acceleration I say one of the local clocks equals the time of a
>clock 10**8 light-seconds away, and if I say that another one of the
>local clocks reads one second more than the time shown on a clock
>10**8 seconds away, and then I undergo an acceleation, and notice NO
>change in the local clocks, why would I suddenly conclude that the
>local clocks gained one second wrt to the clock 10**8 light-seconds
>away. I don't grasp how nothing has changed with respect to the local
>clocks, but now the local clocks are one second faster than a clock
>10**8 light-seconds away that had absolutely nothing to do with my
>acceleration of my local clock. That's what I don't grasp. Why did
>the clocks that had nothing to do with acceleration now change their
>readings wrt to each other?
>Thanks,
>David
It is incredible simple to understand, when you have understood the
basics of time space diagrams ... and how the "line of simultaneity"
tips when FOR is changed; it is this tipping, that disturbs the
clocks, not the acceleration by itself.
Kim
Kim B
>If you would have read it all, you would see that I agree that because
>of the symmetry of the problem there is no difference in their
>readings when they meet the second time. But that is not the question
>I'm seeking the answer to. The question is if Clock B always runs
>slower than Clock A as measured by Clock A how can they start with the
>same time and end up showing the same time. There is no where in
>their path that I could find where the slower clock is measured to
>speed up such that they have the same time when they meet. Where in
>their path is such a measurement made?
>David
When the FOR changes
Kim
Bilge
>> So if both Clocks were set to zero when they first met, and if
>> ClockB was running slower than Clock A for 2*(10**16) seconds, and no
>> measureable change in their readings is observed during the
>> accelerations, how does it happen that they show the same time when
>> they meet the second time?
>
>Hint: look at the Doppler shift.
>
>The last time you showed up I spent a lot of time answering your questions,
>because you seemed like an intelligent person who would be able to
>understand and learn. You gave every impression of doing both. But now you
>seem to have forgotten it all, or never to have understood in the first
>place.
That's his modus operendi. He'll waste as much time as it takes
someone to recognize his/her time is being wasted.
>And Dirk Vdm says this has been going on for years and years.
I'll confirm that. Anyone who has taken the time to explain something to
him has wasted his/her time. His scheme is to post questions, then use the
responses to create new, more convoluted questions in which he attempts to
evade the the solutions to his previous questions. He apparently ``learns''
from the replies, but what he learns is how to bury his original miscon-
ception under a more complex facade. Answering his questions only provides
him with additional ways to string the next person along with his ruse
of feigned sincerity.
>I can't find the motivation to keep helping you in these
>circumstances.
Welcome to the club. On the other hand, there's a lot of motivation
to try and alert his next potential victim to the ploy.
>Special relativity is just not that hard. You should move on
>to quantum field theory or something.
Ironically, the reason that there are so many kooks doing the relativity
gig is precisely because relativity is rather simple. They don't stand a
chance of coming up with a kook act for a topic that would require studying
enough physics to be cured of their kookiness.
It would be a rather interesting psychological study to try and figure
out what motivates kooks to reject something they don't understand and
spend more effort trying to assemble a repertoire of cliches to argue
against it than the effort required to actually understand what they
are arguing against.
Bilge
>The mathematics is very easy, but interpreting the physical
>meaning of the variables in the equations... that seems to
>be incredibly difficult - at least on this forum.
The reason for that would be a good thesis project for a psychology
student as the difficulty seems to be self-imposed and blatantly flies in
the face of logic. In particular, the objections to relativity posted to
this newsgroup begin with misconceptions about the theory. What does the
average (sane) person do when such a misconception is pointed out?
Usually, sane people work through the same example that generated the
objection and see if that resolves the problem. What does the kook do? The
kook insists that you reject the correct premises, accept his
misconceptions and agree that the result disproves more than his
misconceptions. Only in kookland does one find people who argue in all
seriousness that relativity is illogical because their misconceptions
prove it, regardless of what the actual theory has to say.
Bilge
><br276d...@cam.ac.uk> wrote:
>
>>dsep...@austin.rr.com wrote:
>>> So if both Clocks were set to zero when they first met, and if
>>> ClockB was running slower than Clock A for 2*(10**16) seconds, and no
>>> measureable change in their readings is observed during the
>>> accelerations, how does it happen that they show the same time when
>>> they meet the second time?
>>
>>Hint: look at the Doppler shift.
>I don't understand what the hint in regard to Doppler shift is in
>regards to this problem. What would help is if someone explains why I
>just can't look at the local clocks at the turn around point.
for a future post intended to waste someone's time, but I don't
think that is the sort of help anyone is interested in providing.
Your misconceptions are only going to fixed by first trying to
understand the really simple stuff.
You might have difficulty understanding the logic here, but
most people examine their premises when they get a result which
appears inconsistent with their assumptions. For example, the
unwary might presume that you are interested in understanding
the answer to the question you posted just because you said
you wanted to understand how to analyze the scenario you posted.
Once someone discovers that you aren't interested in re-examining
your premises about relativity, he/she is re-examines his/her
premise regarding your interest in a reply. Just because your
real interest is wasting someone's time, doesn't mean no one will
realize it.
Martin Hogbin
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
news:slrndh15o3....@radioactivex.lebesque-al.net...
> >I don't understand what the hint in regard to Doppler shift is in
> >regards to this problem. What would help is if someone explains why I
> >just can't look at the local clocks at the turn around point.
>
> It might help you to develop an even more convoluted question
> for a future post intended to waste someone's time, but I don't
> think that is the sort of help anyone is interested in providing.
> Your misconceptions are only going to fixed by first trying to
> understand the really simple stuff.
He agreed some while ago that he did not understand the
simple stuff.
David is a genuine troll in that he only post to elicit
the maximum response from the group. Why, I
have no idea.
Martin Hogbin
Sue...
Dirk Van de moortel wrote:
Brilliant! The world's energy needs can be solved by serving cafienated
beverages to motorists as they fill up.
<<invariance with respect to time translation gives the well known law
of conservation of energy; >>
http://en.wikipedia.org/wiki/Noether's_theorem
Sue...
geraldk...@hotmail.com
The pre-Copernican astronomers who came up with the principles of the
equable 24 hour day and subsequently the pace of hours,minutes and
seconds already knew that using natural noon varied from one rotation
to the next hence the noon Equation of Time correction.
The early heliocentrists adapted the equable 24 hour day to the newly
discovered Copernican insight of independent and constant axial
rotation at 15 degrees per hour.
Clocks are exquisite devices which keep pace with axial rotation as
defined by the pre-Copernican and Copernican astronomers and that clock
which begins with a conceptual principle,ends as one.
Only physicists and cataloguers get the relationship between axial and
orbital motion wrong and wishful think a dimension into a clock in
order to justify a 1898 fictional novel.
"'Scientific people,' proceeded the Time Traveller, after the pause
required for the proper assimilation of this, 'know very well that
Time is only a kind of Space."
http://www.bartleby.com/1000/1.html
Even at the level of a comedy it is not worth it but then it does not
seem to bother any of you.Strange people !.
"
JanPB
> Take a particle in an accelerator. If you accelerate the particle
> it should see the accelerator's clock going slow if the effects are
> reciprocal.
In SR the effects are reciprocal only between inertial frames. Such
reciprocity follows from the Lorentz transformation which by definition
is a mapping between coordinates belonging to two *inertial* frames.
So if at least one of the objects/observers is accelerating, it is
necessary to derive the relevant transformation of coordinates
corresponding to the case at hand.
If the accelerated particle uses the curvilinear (accelerated)
coordinate system obtained from momentarily co-moving inertial frames
(a standard SR approach to accelerated motion) then there is an easy
formula giving you the reading of any stationary (lab) clock as
observed by the moving particle.
> It doesn't happen. Only one clock is going slow. Period.
>
> How does the accelerated particle see the other particles
> clocks?
Like this: let's say the particle moves in the lab along some 3D
trajectory eta(t). I mean at the lab time t the particle is at the
lab's xyz-coordinate eta(t).
This, incidentally, means that according to the lab observer at time t
the particle has velocity:
v(t) = eta'(t)
OK so far?
Now let's say that the particle is interested in reading off a lab
clock positioned at the lab xyz-coordinate r0.
When that lab clock reads time t0, then - at *the same instant
according to the lab* - the particle measures the *same* lab clock at
r0 as displaying some (possibly) other time t:
t = t0 + v(t0).(r0 - eta(t0)) (*)
...where the dot means the dot product in 3D (in the lab).
(This formula follows from intersecting the particle's simultaneity
space with the lab clock's worldline.)
It is easy to see that this formula reduces to the usual time dilation
factor (as resulting from a Lorentz transformation) if the particle
happens to move at a constant velocity with respect to the lab
(exercise).
In the other interesting case - that of a circular track of some radius
R in the XY-plane around a lab clock sitting at the origin - we get the
following. In this case eta(t) is:
eta(t) = (R cos wt, R sin wt, 0)
...where w is the particle's angular speed.
Thus:
v(t) = (-Rw sin wt, Rw cos wt, 0)
The lab clock the particle is observing sits at the origin:
r0 = (0, 0, 0)
Using the general formula (*) in this special case tells us that when
the lab clock at the origin reads t0, its reading according to the
particle *at the same lab-instant* is:
t = t0 + (-Rw sin wt0, Rw cos wt0, 0).(-R cos wt0, -R sin wt0, 0) =
= t0 + R^2w sin(wt0) cos(wt0) - R^2w cos(wt0) sin(wt0) =
= t0,
or:
t = t0
Careful! This doesn't mean the rates are the same! Remember that the
particle uses a different elapsed time measure - let's call it tau.
According to the lab observer:
-----------------------------
Since the particle moves at a constant *speed* (equal to Rw), the lab
observer can apply the usual Lorentz-type time dilation formula and
concludes that at his time t the particle clock reads time tau =
t/gamma. Hence, TIME DILATION.
According to the particle:
-------------------------
Since the particle observer is noninertial, the usual Lorentz-type time
dilation formula doesn't apply but the formula (*) does. And we've just
used it to conclude that at the lab time t0 the particle (whose own
clock shows tau = t0/gamma at this lab-instant) measures that the lab
clock at the origin reads t0 as well. So the observer sitting on the
particle concludes that the lab clock moves *fast* by the factor of
gamma:
tau = t0/gamma
solve for t0 to get the lab clock time:
t0 = tau*gamma
Hence, TIME CONTRACTION.
If you did the exercise (the one with the particle moving at a constant
velocity), you'd obtain the opposite result, namely time dilation:
t0 = tau/gamma
...as you'd expect by the usual inertial mutual time dilation thing.
(Aside: it's also interesting to use the (*) to check how the particle
measures lab clocks *not* at the origin and/or for non-circular
particle trajectories.)
--
Jan Bielawski
Henry Haapalainen
offered a reward of 1000 US dollars, and one "easy way" is to prove that GPS
clocks speed up with altitude.
Henry Haapalainen
http://www.wakkanet.fi/~fields/ FS (gravity as falling space)
"Androcles" <Androcles@ MyPlace.org> kirjoitti viestissä
news:nASPe.24080$5m3....@fe1.news.blueyonder.co.uk...
Androcles
| The same unverified claims over and over again (below by "Androcles").
I
| offered a reward of 1000 US dollars, and one "easy way" is to prove
that GPS
| clocks speed up with altitude.
Hey phuckwit! I'm offering a thousand dollars reward to anyone that
can prove bright green flying elephants lay their eggs in black holes.
Only an idiot would try to prove something that doesn't happen.
Fuck off, moron.
*plonk*
Androcles
bz
> Ah, but I don't understand what you mean with
> << "pom pom pom..." in a nice soothing regular way >>
> All I know, is that my heart goes pom pom pom pom pom pom...
> And the wind does woosh woosh woosh woosh woosh woosh...
> That's all I have - apart from all that sand finding it's way into every
> conceivable orifice :-(
> But it still is an interesting situation... I can still do physics!
> Actually, I can do two "physicses" now and compare.
> Yes!!!
> ;-)
>
sounds like whish-ful thinking to me :)
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
we...@operamail.com
Martin Hogbin wrote:
> He agreed some while ago that he did not understand the
> simple stuff.
>
> David is a genuine troll in that he only post to elicit
> the maximum response from the group. Why, I
> have no idea.
>
> Martin Hogbin
Hey everyone, assume good faith please. Troll would be a person who
pretends he doesn't understand things while in fact he knows it all. I
believe this person does not understand the given explanations (I don't
understand much myself). Anyway, when one doesn't understand the
explanations, he can't argue anymore and he goes silent for a while,
then just comes up with a different example. That's the case here if
you ask me. I think the solution is to explain things in plain words,
with analogies and graphs (preferably links to real images, these ascii
graphs don't do much good in my opinion).
we...@operamail.com
try to explain this the way I understand it, may not be entirely
correct, but it's simple, all based on the constant speed of light for
all observers.
In your example, the acceleration seems to cause a shift in the clocks.
According to A, his own clock is still the same, but the other clock
seems to have gained a few seconds during the acceleration. You argue
that this doesn't happen for such an acceleration in normal life. Yes,
but the reason is not the acceleration, it is the change of frame of
reference. Please assume that there was another observer, C, coming
from the other direction. While C was passing by A, his clock was
showing the same time as A at that moment. Such that, after A turned
around, A and C have the same speed and their clocks are showing the
same time. They are essentially the same frame of reference now. Only,
there was no acceleration for C. Yet, C is thinking that clock B is one
second ahead of his clock (despite running slower), which is also what
A agrees on. So the question is, why was C thinking that? Problem is in
how you compare clocks. You can't assume that an observer can instantly
observe or know everything in the universe. For any distant object, you
have to rely on the light received from it. Of course you can take into
account the time it took light to reach you. But that's the problem.
The speed of light being constant for all observers makes a difference
in the calculations for different observers. In your example, observer
B thinks that he is stationary and A is moving away from him, so he
thinks light takes a little bit longer to reach him. But A thinks he
himself is the stationary one, and that the speed of light doesn't
depend on its source and is constant. Therefore the time it takes for
light from B to A is calculated differently by B and A. Coming back to
C, the situation is the reverse. So, that's the source of disagreement
and different measurements. As I wrote before, these different
measurements and calculations are kind of an illusion but still real,
because nothing is more real than what you can measure... Hope this
helps...
Nick
will have a relativstic effect.
It is one sided.
Einstein was wrong to introduce appearences
into the laws of physics.
JanPB
> Jan? Only the one moving rapidly *through space*
> will have a relativstic effect.
Both the accelerated observer and the lab observer "have" relativistic
effects: the lab observes the particle as time-dilated, the particle
observes the lab as time-contracted.
> It is one sided.
What do you mean one-sided? Only one observer here is accelerated, it's
a physically different situation. Hence the lack of symmetry between
their observations.
> Einstein was wrong to introduce appearences
> into the laws of physics.
He didn't introduce any appearances. All he said was:
1. Different objects accumulate different elapsed times (this is a
physical - not coordinate - statement).
2. And if you want to quantify things by measurements of distance and
time then set up your coordinates like so: (here comes the sync
procedure, etc.)
3. There is no simultaneity. So if you want to introduce it for
quantification purposes then here is how you do it - just keep in mind
it's only a coordinate convention.
--
Jan Bielawski
Martin Hogbin
<we...@operamail.com> wrote in message news:1125193897....@g44g2000cwa.googlegroups.com...
>
> Martin Hogbin wrote:
>
> > He agreed some while ago that he did not understand the
> > simple stuff.
> >
> > David is a genuine troll in that he only post to elicit
> > the maximum response from the group. Why, I
> > have no idea.
> >
>
Have a look at any of my posts here and you will see that
I always treat posters as genuine to start with. Sadly
this rarely turns out to be justified.
>Troll would be a person who
> pretends he doesn't understand things while in fact he knows it all.
That is certainly true of some trolls but the definition of
a troll is someone more interested in stirring up a reaction
on the newsgroup rather than the subject of the group
itself.
Wekipedia says, ' It is widely thought to be a diminutive
of the phrase "trolling for suckers" '. That description is
very apt in this case.
> I
> believe this person does not understand the given explanations...
That may be true. It is also possible that he is a very able physicist
who enjoys testing the knowledge of others on the subject, although
I consider this unlikely. (If that is the case, David, why not come
clean and post your questions as puzzles, with your own answers.)
>(I don't
> understand much myself). Anyway, when one doesn't understand the
> explanations, he can't argue anymore and he goes silent for a while,
> then just comes up with a different example. That's the case here if
> you ask me. I think the solution is to explain things in plain words,
> with analogies and graphs (preferably links to real images, these ascii
> graphs don't do much good in my opinion).
David has been trolling here for many years (over ten I believe)
I have spent a considerable time trying to explain things to him
and I have seen many other physicists do the same. We have all
given up utterly frustrated.
His questions nearly always include one or more of the
following:
Non-inertial frames
Rotation
Accelerating rigid objects
Unfamiliar distances or timescales.
I and many others have told him that he has no chance
whatever of understanding the answers to his questions
until he has understood SR of inertial objects in inertial
frames. He has admitted in conversations with me, and
no doubt others, that he does not understand the very
basics of the subject.
He has had books, such as 'Spacetime Physics'
recommended to him many times but as far as I
know has never studied it.
You will see from my posts here that I very rarely
accuse others of trolling. I never call the persistent
aether crackpots trolls because they genuinely want
to discuss the subject.
Have a look back over the history of this group and you
will see hundreds of similar questions from this person.
Perhaps then you will understand why I call him a troll.
Martin Hogbin
Androcles
"JanPB" <fil...@gmail.com> wrote in message
news:1125205380.2...@o13g2000cwo.googlegroups.com...
According to Bielawski, relativity works because
An error in Relativity "would be like Stephen Hawking dividing by zero
or
something equally trivial." -- Bielawski.
It's WAY too simple-minded.-- Bielawski.
"would have been caught immediately by the AdP reviewer." -- Bielawski.
It is boring to find out how far Bielawski can be pushed into
irrelevancies and nonsense.
Androcles.
Bilge
>Martin Hogbin wrote:
>
>> He agreed some while ago that he did not understand the
>> simple stuff.
>>
>> David is a genuine troll in that he only post to elicit
>> the maximum response from the group. Why, I
>> have no idea.
>>
>> Martin Hogbin
>
>Hey everyone, assume good faith please.
He ran out a long time ago and he isn't making much of an
effort to earn any new credits.
>Troll would be a person who
>pretends he doesn't understand things while in fact he knows it all. I
>believe this person does not understand the given explanations (I don't
>understand much myself). Anyway, when one doesn't understand the
>explanations, he can't argue anymore and he goes silent for a while,
>then just comes up with a different example. That's the case here if
>you ask me. I think the solution is to explain things in plain words,
>with analogies and graphs (preferably links to real images, these ascii
>graphs don't do much good in my opinion).
I'll tell you what. I have a rather standard way to determine
how interested someone really is (it's also a good and standard
way to teach, in general). You try it and see how far you get.
The plot is to make him answer his own question, by breaking
the question down into smaller chunks that he can go work on.
If he gets stuck, he can tell you how far he got and why he's
stuck so you can give him a hint. Lead him through with hints
until _he_ answers the question.
Anyone who really wants the answer to a question, will put in
some effort to get the answer. Anyone who wants someone else to
spend time answering his/her question but does not want to
expend any effort him/herself, must think his/her time is worth
more than my time, in which case, his/her good faith credit
has been spent. So, try it and see. I used to make the mistake
of thinking that someone who asked a question must be intersted
in the reply. After a while, I got tired of typing lengthy
replies for no reason other than to give some moron additional
opportunities to digress with some silly argumentm like ``no
it isn't.'' As soon as someone has to actually think, the
trolls drop out real fast.
JanPB
>
> According to Bielawski, relativity works because
Did I say that?
> An error in Relativity "would be like Stephen Hawking dividing by zero
> or something equally trivial." -- Bielawski.
What I was referring to was your claim that there was a trivial
mathematical error in SR.
> It's WAY too simple-minded.-- Bielawski.
>
> "would have been caught immediately by the AdP reviewer." -- Bielawski.
Again, I meant the type of mistake you are claiming exists in SR.
--
Jan Bielawski
Androcles
| Androcles wrote:
| >
| > According to Bielawski, relativity works because
|
| Did I say that?
In as many words, yes.
You can't derive the cuckoo transfromations, as I PROVED,
so your irrelevancies are all you have left.
|
| > An error in Relativity "would be like Stephen Hawking dividing by
zero
| > or something equally trivial." -- Bielawski.
|
| What I was referring to was your claim that there was a trivial
| mathematical error in SR.
|
| > It's WAY too simple-minded.-- Bielawski.
| >
| > "would have been caught immediately by the AdP reviewer." --
Bielawski.
|
| Again, I meant the type of mistake you are claiming exists in SR.
The time at (x-vt, 0,0) a coordinate which is independent of time,
differs from the time at (0,0,0), another coordinate in the k-frame
which is independent of time.
As I PROVED, the function tau is not linear as claimed.
Too bad, Bielawski, all you have left are your way too simple-minded
silly assertions.
Wanna see it again?
Sending the light from x' to 0 and back again, simultaneously
with sending it from 0 to x' and back,
x /x'
| /\/
| / /\
|x' / / \0'
|\ / /
| \/ /
| /\/
|/_________t
0 A B
tau(A) = tau(B)
You are screwed and so is your religion, Bielawski.
Androcles.
Jon Bell
Martin Hogbin <ngr...@hogbin.org> wrote:
>
>David has been trolling here for many years (over ten I believe)
The earliest thread started by him in sci.physics.*, that I can find in
Google Groups, is this one from almost exactly nine years ago:
http://groups.google.com/group/sci.physics/browse_frm/thread/268607421a057283
--
Jon Bell <jtb...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1125245752....@g49g2000cwa.googlegroups.com...
> | Androcles wrote:
> | >
> | > According to Bielawski, relativity works because
> |
> | Did I say that?
>
> In as many words, yes.
> You can't derive the cuckoo transfromations, as I PROVED,
> so your irrelevancies are all you have left.
You haven't proved anything. The proof you presented is based e.g. on
misinterpreting what "taking infinitesimal x'" means.
> | > An error in Relativity "would be like Stephen Hawking dividing by
> zero
> | > or something equally trivial." -- Bielawski.
> |
> | What I was referring to was your claim that there was a trivial
> | mathematical error in SR.
> |
> | > It's WAY too simple-minded.-- Bielawski.
> | >
> | > "would have been caught immediately by the AdP reviewer." --
> Bielawski.
> |
> | Again, I meant the type of mistake you are claiming exists in SR.
>
> The time at (x-vt, 0,0) a coordinate which is independent of time,
> differs from the time at (0,0,0), another coordinate in the k-frame
> which is independent of time.
> As I PROVED, the function tau is not linear as claimed.
Your proof is based on a mathematical error on your part. Specifically,
you let x' tend to zero on one side of the equation but not on the
other. When the result is 0=1 you complain loudly.
> Too bad, Bielawski, all you have left are your way too simple-minded
> silly assertions.
>
> Wanna see it again?
>
> Sending the light from x' to 0 and back again, simultaneously
> with sending it from 0 to x' and back,
>
>
> x /x'
> | /\/
> | / /\
> |x' / / \0'
> |\ / /
> | \/ /
> | /\/
> |/_________t
> 0 A B
>
> tau(A) = tau(B)
What's A and B? The way you drew it it's not clear what these labels
denote.
--
Jan Bielawski
Androcles
Bollocks, you have no idea of mathematical proof.
| Specifically,
| you let x' tend to zero on one side of the equation but not on the
| other. When the result is 0=1 you complain loudly.
The fuck I do, but since you claimed to be able to derive
the cuckoo transformations without calculus, go ahead and do
it.
You are to use Einstein's light postulate, Galileo's relativity
exemplified by Einstein, "the reciprocal electrodynamic action
of a magnet and a conductor. The observable phenomenon here
depends only on the relative motion of the conductor and the magnet",
Einstein's definition of time "we establish by definition that the
"time"
required by light to travel from A to B equals the "time" it requires
to travel from B to A", and NO assumptions.
You will not succeed, I shall rip your "proof" to shreds.
|
| > Too bad, Bielawski, all you have left are your way too simple-minded
| > silly assertions.
| >
| > Wanna see it again?
| >
| > Sending the light from x' to 0 and back again, simultaneously
| > with sending it from 0 to x' and back,
| >
| >
| > x /x'
| > | /\/
| > | / /\
| > |x' / / \0'
| > |\ / /
| > | \/ /
| > | /\/
| > |/_________t
| > 0 A B
| >
| > tau(A) = tau(B)
|
| What's A and B? The way you drew it it's not clear what these labels
| denote.
| --
| Jan Bielawski
|
A and B are times on the x-axis, of course.
Way too simple-minded for you to understand, but the light
reflects at A and returns to x'. It also reflects at B and returns to
zero.
1/2tau(A+B) = tau(A)
1/2tau(B+A) = tau(B)
tau(A) = tau(B), that's a 2-1 mapping unless you wish to challenge
the associativity of A+B = B+A, you are crazy enough to try it.
Androcles.
dsep...@austin.rr.com
What I don't grasp is the following. No one in the local vicinity of
Clock A sees Clock A's time change significantly during the turn
around. All local observers in the frames used in this problem agree
to this. And this is an experiment we can do. So now, how do we
determine that Clock B has gained several seconds during the turn
around. The way I understand how Einstein and company do it, is that
they ignore the initial reference frame, and start using the inertial
reference frame Clock A is in after Clock A's acceleration. The
problem, is that no one ever measures Clock B to gain time. The local
observers in the vicinity of Clock B do not measure Clock B to gain
time just as no one in the vicinity of Clock A observe it to gain or
lose significant time. It just doesn't happen and we can
experimentally demonstrate this. So I am left with the notion
posters here believe that we should ignore what local observers
measure, and what we can experimentally measure, and instead we should
conclude that Clock B must have gained time, because that is the only
way Clock A and Clock B can show the same time when they meet the
second time if we apply Einstein's equations.
I'm looking for someone who can explain why local observers cannot
measure any significant change in Clock A's time or Clock B's time
during the turn arounds, yet it must occur if Clock B runs slower than
Clock A and they start and end with the same readings. According to
Einstein, there are many observers that can measure that Clock B is
running slower than Clock A. Everyone believes this, that is that we
can actually make the measurement. But there is not one observer nor
any experiment we can do that shows that Clock B somewhere catches up
in time with Clock A.
That makes no sense to me.
David (the Troll)
>
dsep...@austin.rr.com
In the problem I posted, the clocks have to gain a second or two
during the 0.1 second acceleration, when their velocity changes by 3
meters / second. If you do this experiment, you will find that no
local observer can measure any significant change in the clock rate
during this acceleration. It just doesn't happen. If you think it
does, how are you measuring this 1 or 2 second speed up?
David
we...@operamail.com
Thanks for responding. I was starting to think you abandoned the
thread, but you didn't, so I still think you are sincere. Please keep
in mind that I am no expert, I was where you were a few years ago but
then I figured out a few things.
dsep...@austin.rr.com wrote:
> What I don't grasp is the following. No one in the local vicinity of
> Clock A sees Clock A's time change significantly during the turn
> around. All local observers in the frames used in this problem agree
> to this. And this is an experiment we can do.
Yes, observer A never notices any change of rate in his own clock. Even
if there was any change in the flow of time (I'm not saying there is),
one couldn't notice it because one's brain and sensations would be
affected the same way and it would feel the same. In short, there is
never any change for local clocks according to local observers, so
forget about that.
> So now, how do we
> determine that Clock B has gained several seconds during the turn
> around. The way I understand how Einstein and company do it, is that
> they ignore the initial reference frame, and start using the inertial
> reference frame Clock A is in after Clock A's acceleration. The
> problem, is that no one ever measures Clock B to gain time.
Before the acceleration, observer A thinks that clock B is one second
behind clock A. After the acceleration, observer A thinks that clock B
is one second ahead of clock A. So it follows that, according to
observer A, clock B must have gained two seconds during the
acceleration.
I tried to explain before why observer A thinks so. Suppose that
observer A is using a powerful telescope to see clock B. But what he
acutally sees is history, he has to take into account the time it took
light to get from B to A. He knows their relative speed and the time
passed since the beginning (according to his own clock), so he
calculates their distance, divides by the constant speed of light, adds
this to the actual time he sees, compares the found value with his own
clock, and arrives the conclusions above.
Now you may think something must be wrong there. The only strange thing
is the constancy of the speed of light for all observers (I hope you
fully understand what this means and how counterintuitive it is). Once
you accept that, all the weird relativity effects follow.
My response ends here, we can continue if we clarify the points above.
David Marsden
> The fuck I do, but since you claimed to be able to derive the cuckoo
Many detailed step-by step explanations of Einstein's 1905 derivation
are available on the web. Calculus is not really involved, because (as
Einstein explicitly says) the partial derivatives are all constants.
It's just a linear transformation. Since you are so fixated on that
particular derivation, it's surprising that you've never been curious
enough to actually study any of the explanations of it.
JanPB
>
> | Specifically,
> | you let x' tend to zero on one side of the equation but not on the
> | other. When the result is 0=1 you complain loudly.
>
> The fuck I do, but since you claimed to be able to derive
> the cuckoo transformations without calculus, go ahead and do
> it.
Sure, no problem but I need to get some sleep first.
> You are to use Einstein's light postulate, Galileo's relativity
> exemplified by Einstein, "the reciprocal electrodynamic action
> of a magnet and a conductor. The observable phenomenon here
> depends only on the relative motion of the conductor and the magnet",
> Einstein's definition of time "we establish by definition that the
> "time"
> required by light to travel from A to B equals the "time" it requires
> to travel from B to A", and NO assumptions.
> You will not succeed, I shall rip your "proof" to shreds.
You'll not be able to even touch it, let alone rip to shreads. I'll
derive the Lorentz transform from the assumption of linearity of the
transform, from the "tau" equation, and the usual postulates.
--
Jan Bielawski
RP
JanPB wrote:
You'll lose the argument, just ask Randy Poe. He's much worse than I
ever was. I never doubted the math, just it's interpretation. Andro OTOH
is only interested in being heard. BTW, your first mistake that Andro
will tear to shreds is the fact that you can't derive something from
assumptions without arriving at yet further assumptions. That's just
simple logic. He'll be right on that point. Better try another approach.
Richard Perry
Androcles
Circularity is not permitted.
You are to use Einstein's light postulate, Galileo's relativity,
Einstein's definition of time.
Androcles.
|
| --
| Jan Bielawski
|
Androcles
"David Marsden" <dmar...@anti-spam.com> wrote in message
news:431294a0....@news.gte.net...
| On Mon, 29 Aug 2005 "Androcles" wrote:
| > The fuck I do, but since you claimed to be able to derive the cuckoo
| > transformations without calculus, go ahead and do it. You will not
| > succeed, I shall rip your "proof" to shreds.
|
| Many detailed step-by step explanations of Einstein's 1905 derivation
| are available on the web.
I don't see you providing any URLs.
| Calculus is not really involved, because (as
| Einstein explicitly says) the partial derivatives are all constants.
I don't see you citing where he says that.
| It's just a linear transformation.
It is not a linear transformation, you are bullshitting.
| Since you are so fixated on that
| particular derivation, it's surprising that you've never been curious
| enough to actually study any of the explanations of it.
Since you are so full of bullshit, it is not surprising that you've
never
been curious enough to actually study the problem.
Androcles.
JanPB
>
> You'll lose the argument,
Oh, I know very well that Androcles won't understand a word I'm saying.
One always "loses" arguments on Usenet in that funny sense, I've done
this before with many others as I'm always curious about limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
In this case the guy would rather assume that 100 years and generations
of physicists "missed" an idiotic high-school-level "error" than admit
to himself that he has no talent for that sort of thing and doesn't
understand squat. He would invent ridiculous Hollywoodesque plots
involving Einstein and his supposed "intent" to "deceive" scientists -
anything, *anything*, except the plain truth that Androcles has no case
and is wasting his life away on chimeras.
> BTW, your first mistake that Andro
> will tear to shreds is the fact that you can't derive something from
> assumptions without arriving at yet further assumptions. That's just
> simple logic. He'll be right on that point. Better try another approach.
The assumptions are all standard physics + Einstein's postulates. So
unless Androcles insists on something exotic like the universe being
non-isotropic or observers depending on their entire motion histories
and the like, then he has no case since it's all just a mathematical
derivation at that point. It would be interesting to find out what sort
of random assumption he brings up this time. I've seen one of them
already - the idea that SR "ignores" the time it takes for light to
reflect off a mirror!
--
Jan Bielawski
JanPB
>
> | You'll not be able to even touch it, let alone rip to shreads. I'll
> | derive the Lorentz transform from the assumption of linearity of the
> | transform, from the "tau" equation, and the usual postulates.
>
> Circularity is not permitted.
Where is the circularity according to you? Assumptions:
1. Transform is a linear function on space and time,
2. The "tau" equation,
3. The definition of time and the sync convention,
4. The speed of light does not depend on the motion of the source and
is always the same numerically,
5. Equivalence of inertial frames,
6. Other assumptions which Einstein left unsaid in his paper but were
presumed obvious, for example:
- the universe is isotropic,
- clock readings to not depend on their motion histories,
- reflecting off mirrors is instantaneous, etc.
--
Jan Bielawski
Androcles
| Androcles wrote:
| >
| > | You'll not be able to even touch it, let alone rip to shreads.
I'll
| > | derive the Lorentz transform from the assumption of linearity of
the
| > | transform, from the "tau" equation, and the usual postulates.
| >
| > Circularity is not permitted.
|
| Where is the circularity according to you?
I told you NO assumptions.
PoR, light postulate, definition of time.
Assumptions:
|
| 1. Transform is a linear function on space and time,
That's a requirement, not an assumption.
| 2. The "tau" equation,
Up to you, I want a 1-1 transformation.
| 3. The definition of time and the sync convention,
Definition of time is given, time for light A to B equals
time for light B to A.
You'll have to be more explicit about "sync convention",
I can't accept vague definitions.
| 4. The speed of light does not depend on the motion of the source
I said you are to use the light postulate.
and
| is always the same numerically,
Where did you get that bullshit from? Einstein derived c =
(c+v)/(1+v/c), no circularity allowed. That's definitely out.
Fault found, [snip], try again.
Androcles
Androcles
| RP wrote:
| >
| > You'll lose the argument,
|
| Oh, I know very well that Androcles won't understand a word I'm
saying.
Bullshit, I know you will attempt circularity.
| One always "loses" arguments on Usenet in that funny sense, I've done
| this before with many others as I'm always curious about limits of
| human capacity for withstanding plain contradictions and the means
| people use to protect themselves from certain inconvenient truths
| regarding themselves.
LOL! You are talking about yourself, phuckwit.
| In this case the guy would rather assume that 100 years and
generations
| of physicists "missed" an idiotic high-school-level "error" than admit
| to himself that he has no talent for that sort of thing and doesn't
| understand squat.
You are so ridiculous its incredible. I'm always bored by the limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
| He would invent ridiculous Hollywoodesque plots
| involving Einstein and his supposed "intent" to "deceive" scientists -
| anything, *anything*, except the plain truth that Androcles has no
case
| and is wasting his life away on chimeras.
I'm always bored by the limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
|
| > BTW, your first mistake that Andro
| > will tear to shreds is the fact that you can't derive something from
| > assumptions without arriving at yet further assumptions. That's just
| > simple logic. He'll be right on that point. Better try another
approach.
See, RP knows.
You on the other hand are too stupid to know that, which is why
you can't win. I'm always bored by the limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
| The assumptions are all standard physics
What is "standard" physics? Vector addition of velocities?
| + Einstein's postulates.
He's only got one. Having trouble counting to one, are you?
The PoR is standard physics and Galileo's. Einstein's ONE
postulate is "light is always propagated in empty space with
a definite velocity c which is independent of the state of motion
of the emitting body. " No numerical value permitted in the proof
though. You can use a numerical value for explanation if need be.
I'm generously permitting you to use his definition of time,
although I do not accept it.
So
| unless Androcles insists on something exotic like the universe being
| non-isotropic or observers depending on their entire motion histories
| and the like, then he has no case
Quit bragging and start proving.
| since it's all just a mathematical
| derivation at that point. It would be interesting to find out what
sort
| of random assumption he brings up this time.
It'll be boring listening to you bragging and bringing up such total
irrelevancies as your argument, such as "would rather *assume*
that 100 years and generations of physicists "missed" an idiotic
high-school-level "error" than admit to himself that he has no talent
for that sort of thing and doesn't understand squat.
I dont assume, phuckwit, but you do, and that's why you have no
talent for this sort of thing and don't understand squat.
| I've seen one of them
| already - the idea that SR "ignores" the time it takes for light to
| reflect off a mirror!
| --
| Jan Bielawski
Are you claiming the mirror doesn't move, or that a photon
has no momentum? Which?
That's why you have no talent for this sort of thing and don't
understand squat.
Androcles.
Martin Hogbin
<we...@operamail.com> wrote in message news:1125285570.7...@z14g2000cwz.googlegroups.com...
> David,
> Thanks for responding...
Been there, done that, got the T-shirt. Treat him politely,
explain everything carefully, and he will soon begin to
understand.
I wish there was some way I could have a bet with you
that you will eventually give up in frustration. Remember,
troll is short for 'trolling for suckers'.
Martin Hogbin
Bilge
>"JanPB" <fil...@gmail.com> wrote in message
>news:1125245752....@g49g2000cwa.googlegroups.com...
>| Androcles wrote:
>| >
>| > According to Bielawski, relativity works because
>|
>| Did I say that?
>
>In as many words, yes.
what people write rather than what goes through your rather limited,
``in as many words,'' translator. I realize it's probably convenient
to reduce everything to a preconception in which you have invested
a great deal of time and effort to maintain, but you're only arguing
with your own inability to understand english, not relativity.
Harald EPFL
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:wfEPe.11208$qc2....@news.cpqcorp.net...
>
> "Harry" <harald.v...@epfl.ch> wrote in message
news:430f059e$1...@epflnews.epfl.ch...
> >
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
> > in message news:6ZBPe.11200$8d2....@news.cpqcorp.net...
> > >
> > > "Harry" <harald.v...@epfl.ch> wrote in message
> > news:430edf05$1...@epflnews.epfl.ch...
> > > >
> > > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> > wrote
> > > > in message news:%uAPe.11197$4b2....@news.cpqcorp.net...
> > > > >
> > > > > "sue jahn" <susyse...@yahoo.com.au> wrote in message
> > > > news:430ea504$0$18638$1472...@news.sunsite.dk...
> > > > >
> > > > > [snip]
> > > > >
> > > > > > A clock simulates the passage of time.
> > > > >
> > > > > That's only for philosophers who never held a clock in their
hands.
> > > > >
> > > > > For physicists and engineers and the rest of the world, a clock
> > > > > *defines* the passage of time.
> > > >
> > > > Hmm... a correctly *calibrated* clock defines the passage of time
> > >
> > > No. As soon as you have a clock - any kind of clock, like
> > > your heart beat, you can define time and the passage of time
> > > with it, and even do experiments with it. You can find out that
> > > the distance covered by a falling stone is more or less
> > > proportional to the square of the falling time, which you define
> > > as the number of heart beats that you count.
> >
> > Just try to sell that idea to a normalisation institute. ;-)
> > - On a serious note: when it was discovered that the solar clock is
> > irregular compared to atomic clocks, the atomic clock standard took over
for
> > a more precise (regular) calibration. From then on the solar clock does
not
> > anymore define the "passage of time", as it's not anymore the reference
> > standard.
>
> When you are alone in the desert without an atomic clock or
> wrist watch or sun or stars, you can define time and "the passage
> of time" with your heart beat, and do physics and engineering.
> You don't need reference standards. You need a clock,
> something that does pom pom pom pom pom pom.
> You count the poms and call it time.
> When you have more than one kind of clock, you can
> compare one against the other and compare the times you
> define and read on those clocks.
Sure. And when you discover that they don't indicate the same, you will have
some choosing to do, exactly as I desribed above...
> If you can define regularity or "correct calibration" without
> using the concept of time, then I agree with your suggestion
> that "a correctly *calibrated* clock defines the passage of
> time"
Hey Dirk, "temperature" is *not* what any alcohol thermometer indicates;
evenso "time" is not affected by your emotions while your heart beat is
affected by that! I already imagine your time go quicker when you see a
pretty girl - pom pom pom ;-)
We can use such instruments to do experiments with limited precision.
The regularity of any measurement standard is verified by for example
comparing it with other devices and processes, just as it was discovered
that an alcohol thermometer is less good than a mercury thermometer.
By comparing an abundance of instruments one can determine which ones are
less regular under what circumstances and in what intervals.
Harald
sal
> On 25 Aug 2005 23:07:17 -0700, "JanPB" <fil...@gmail.com> wrote:
>
>>dsep...@austin.rr.com wrote:
>>> Explanations of problems similar to this one have been posted, but
>>> when I change the parameters to the problem, I realize that those
>>> explanations are wrong. So I'm looking for an expert who can
>>> explain the following SR problem.
>>>
>>>
>>> Let there be two clocks, Clock A and Clock B, moving along the
>>> x-axis. They have a relative velocity of 3 meters / second. At
>>> time t0, they are at the same position in space. At this point in
>>> space and time, both clocks are set to zero. They now continue to
>>> move away from each other still traveling at 3 meters / second.
>>> According to Einstein's theory, observers in Frame A (Clock A's
>>> rest frame) measure that Clock B is running at a slower rate than
>>> identical synchronized clocks that are at rest in Frame A. After
>>> 10**16 seconds, observers in Frame A measure Clock B to be one
>>> second slower than Clock A. (The clocks are about 10**8
>>> light-seconds apart).
>>>
>>> When Clock A reads 10**16 seconds, let Clock A undergo an
>>> acceleration over a 0.1 second interval in the direction toward
>>> Clock B. Likewise, when Clock B reads 10**16 seconds, let Clock B
>>> undergo an acceleration over a 0.1 second interval towards Clock
>>> A. When the two accelerations have ended, Clock A and Clock B are
>>> now approaching each other with a relative velocity of 3 meters /
>>> second. Now to clarify who's frame the 0.1 second interval was
>>> measured over, all frames used in this problem measure 0.1 seconds
>>> as the acceleration interval plus or minus some extremely small
>>> delta since the relative velocity of 3 meters / second is so slow
>>> relative to c. So any variation in the measurement of 0.1 second
>>> versus the 10**16 seconds has negligible effect on the answer to
>>> this problem.
>>>
>>> Observers in this new Clock A frame observe that as Clock B is
>>> approaching Clock A, Clock B is running slower than Clock A
>>> (Einstein's theory). When Clock A meets Clock B, both clocks
>>> indicate the same time. This is true because of the symmetry of
>>> the problem.
>>>
>>> The question is, as measured by Clock A, if Clock B was running
>>> slower than Clock A for the 10**16 seconds of inertial motion when
>>> the clocks are moving apart, and Clock B is running slower than
>>> Clock A for the 10**16 seconds of inertial motion when the clocks
>>> are moving toward each other, and both clocks were set to zero
>>> when they first meet, how do they end up showing the same time
>>> when they meet again?
>>
>>During the turnaround A measures B's clock to quickly gain a lot of
>>seconds. This compensates for the time dilation periods. Keep in
>>mind that this "measures" means "reading the A-synchronised clocks
>>at the relevant events along B's trajectory" - it doesn't mean B's
>>clocks suddenly go bananas because A turned his head.
>>
>>This is fully analogous to the Euclidean situation with two cars
>>going from P to Q, one car along PXQ, the other along PYQ, like so:
>>
>> Q
>> / \
>> / \
>> / \
>> / \
>> / \
>> X Y
>> \ /
>> \ /
>> \ /
>> \ /
>> \ /
>> P
>>
>>Both cars start at the same time and drive at the same speed. Each
>>driver looks *perpendicularly* out the window in the direction of
>>the other driver and concludes "he is driving slower than me" (that
>>corresponds to "his clock is slower") since every foot (say) of one
>>driver's path corresponds to more than a foot of the other driver's
>>path when viewed in the perpendicular direction. This holds for both
>>the outgoing and the incoming parts of their journey.
> When I draw a line perpendicular to each path and use that to
> determine if the other driver...
Jan answered your question with the first line of his response. Go
back and reread it from the start, and ignore the cars -- they were
just an attempt to help you understand the problem geometrically, and
(with sincere apologies to Jan! It was a good post!) I think it's
obvious that they didn't help. They just distracted you.
> is going faster or slower than me, although I don't necessarily
> agree with this analogy,
Then you don't understand it.
The question isn't "do you agree with the explanation" -- you're the
one who doesn't understand the subject matter here and whether you
"agree" with the explanation or not is meaningless. The question is
"can you work to understand it and, in the process, learn something".
You don't show any signs of working to understand why the analogy
might be helpful. Instead you seem to be working to show that it's
incorrect. That helps nobody.
> I observe that when I go from P to Y, the other driver's position
> trails this perpendicular line so yes you might say he's traveling
> slower. However, when I pass Y and continue on to point Q, I find
> that the other driver is now ahead of this perpendicular. Part of
> the path he's slower and part of the path he's faster and we both
> end up at Q at the same time. In the problem I posted there is no
> part of the path in which clock B rusn faster than Clock A (as
> measured from A's point of view)
Wrong. You haven't gotten it yet.
Here's the first sentence of Jan's post again:
>>During the turnaround A measures B's clock to quickly gain a lot of
>>seconds.
Read that, read it again, understand it, and then you will understand
the problem. Forget the cars, they won't help you unless you can
understand that sentence.
Everything else is irrelevant until you understand that sentence.
[ .. ]
>>> Now some may think
Wrong phrasing. Some may understand the problem. You don't. Your
comments on what people who understand the problem "think" about it
are irrelevant and meaningless.
>>> something happens to the clocks when they are accelerated to to 3
>>> meters / second. However, as is easily demonstrated
>>> experimentally, when an electronic clock or most mechanical clocks
>>> change their velocity by 3 meters / second over a 0.1 second
>>> interval, they do not gain or lose
You don't understand the solution.
Here it is again:
>>During the turnaround A measures B's clock to quickly gain a lot of
>>seconds.
Several things should be pointed out:
B's acceleration does _NOT_ affect A's measurement of B's clock rate.
(Read that sentence again; you probably didn't get it the first time
through.)
It is A's acceleration which affects A's "measurement" of B's clock
rate. (It will be obvious in another paragraph why I put
"measurement" in quotes.)
BUT:
A's acceleration affects A's "measurement" of B's clock rate ONLY
BECAUSE THEY ARE SEPARATED IN SPACE. A clock adjacent to "A" would
show no such effect.
AND, _most_ important of all:
"A" cannot directly measure B's clock rate himself! The only way for
A to obtain a direct "measurement" of B's clock rate while A is
accelerating would be to employ an army of assistants, each traveling
at a (different) constant rate, each of whom has a clock which is
synchronized with A's at the instant when A is moving at the same rate
as that assistant, and who are all located next to B. At each
instant, the assistant who is traveling at the same rate and in the
same direction as A at that particular instant reads B's clock (ONE
clock reading -- this is not a _rate_!) and after it's all over they
compare notes and figure out what the _rate_ of B's clock must have
been at each instant in order to obtain those time measurements.
And you must always keep in mind that a SINGLE reading from a clock
does not provide a measurement of the rate! You must make at least
TWO clock readings, separated in time, in order to obtain the rate.
If A attempts to use direct observations of B's clock, using light
signals to read it, then A must do some confusing calculations
involving Doppler shift and lag in the readings due to the distance
between A and B as "measured" by A in order to _deduce_ what a
momentarily comoving assistant would have measured.
If you don't understand this read it again. If you still don't
understand it then you are probably failing to understand the
relativity of simultaneity and the basic Lorentz transforms, which are
both needed to understand this problem.
>>> anywhere close to a second or two relative to nearby clocks that
>>> haven't accelerated. Observers in the original rest frame of Clock
>>> A near the position of Clock A do not measure a substantial change
>>> in Clock A as it changes its velocity by 3 meters / second.
>>> Likewise observers in the original rest frame of Clock B in the
>>> vicinity of Clock A do not observe any substantial change in Clock
>>> A as it changes its velocity by 3 meters / second. And Observers
>>> moving with Clock A through the acceleration likewise do not
>>> measure any substantial change in the Clock A time relative to
>>> nearby clocks in the original Frame A and Frame B that haven't
>>> accelerated.
>>
>>I think the Euclidean analogy above answers some of your questions.
>>
>>> Also we can demonstrate logically that Clock A and Clock B do not
>>> gain or lose sufficient time during the acceleration to make any
>>> difference in the indicated time when they meet.
>>
>>It's not that the clocks A and B suddenly "do" something during the
>>acceleration. What physically happens is that you have two
>>symmetrically moving clocks so they display the same elapsed times
>>upon reunion. The whole "time dilation" during the inertial portion
>>of the trip and the sudden burst of "time contraction" during the
>>accelerated portion of the trip is an artifact of the observation
>>method called "synchronised clocks" and "using instantaneously
>>co-moving inertial frames for periods of acceleration" - ponder that
>>Euclidean analogy again.
> In Einstein's theory, the clocks must physically run at different
> rates
This statement makes it clear that you don't understand "Einstein's
theory", as you so quaintly call it. Jan has very kindly given you
the answer to your question. If you don't understand what he said,
don't jump to the conclusion that his answer is incorrect.
If you think the moving clock must "physically" run at a "different
rate" from the stationary clock then you haven't even got the
beginning of a glimmer of a clue regarding the theory.
Given any two non-accelerating clocks in uniform relative motion, from
a third reference frame which is moving at the average of the
velocities of the two clocks, it will always be apparent that the two
clocks are running at the same rate.
--
Nospam becomes physicsinsights to fix the email
sal
On Sun, 28 Aug 2005 20:19:30 -0700, wespe wrote:
> David,
^^^^^^^^
WAIT there is an issue here! Here's where things totally jump the
tracks, and most likely this is the curve that David can't go around
without winding up in the weeds.
In the telescope, _nothing_ strange seems to happen -- clock A
advances by, say, 1 second, and clock B advances by 1+epsilon seconds,
and where's the big deal?
It's here:
The DISTANCE from A to B, as calculated by A, CHANGED during the
acceleration. And what's worse, "simultaneity" changed, too, so we
can't even use simple "contraction" to figure out the change in
distance.
Before the acceleration, A was adjusting the time he
saw in the telescope by
pre-accel-distance/C
and afterward he needs to adjust it by
post-accel-distance/C
And during the acceleration, the computed distance is changing
_rapidly_ as a function of A's proper time, because A is changing
reference frames _rapidly_ as a function of his own proper time while
he's accelerating.
In consequence, the time A _computes_ for B changes rapidly during
acceleration and is quite different after acceleration has been
completed.
Example: A is next to Earth, and stationary. One light-hour away a
satellite is also stationary with respect to Earth. A is watching the
satellite in his telescope, and sees that the satellite's clock reads
1:00. A's clock reads 2:00. A says, "1 hour ago, this light left the
satellite, and its clock read 1:00. At that time, my clock read 1:00,
too. We're in sync."
A accelerates, suddenly, to 0.867C, directly toward the satellite. A
does this very very quickly, so he's still (momentarily) next to Earth
afterward. But now he sees distances in the Earth frame contracted by
1/gamma = 1/2, and the satellite (still stationary WRT Earth) is now
just 1/2 light hour away ... and what's more, the times of all events
in the Earth frame no longer map so simply into times in A's frame.
A _still_ sees 1:00 in his telescope when he looks at the satellite
(remember, he's just started moving, and he's still next to Earth,
though not for long).
But how far away was the satellite when that light was emitted? Make
2:00 at Earth the origin. The satellite's Earth coordinates when it
emitted the light now being observed in the telescope were, then,
(t=-1, x=1) (1 hour before 2:00, 1 light-hour away).
In A's frame, the X coordinate of that event was
x' = gamma*(x - vt) = 2(1 + 0.867) = 3.734
And the time coordinate was
t' = gamma*(t - vx) = 2(-1 - 0.867) = -3.734
The light was emitted much earlier, and came from much farther away,
than A had thought before he accelerated!
When A considers the satellite's time using this newly computed
distance, he observes, "3.734 hours ago, the satellite's clock read
1:00; at that time, my clock read about 10:15. Wow, the satellite's
clock now seems to be 2:45 _ahead_ of my clock! How did that happen??
Boy it must have been running fast while I was accelerating!"
An hour and 9 minutes later, Earth time, A arrives at the satellite.
A's clock reads just past 2:30 -- the trip took about a half hour in
A's frame. In A's frame, it's now been about 4 and a quarter hours
since the satellite's clock read 1:00. But the satellite's going at
0.867C, relative to A, so A figures its clock must be ticking at
half-speed; A figures the satellite will have seen a little over 2
hours elapse. So A expects the satellite's clock to read a little
past 3:00.
And that's just what it does read.
> divides by the constant speed of light, adds this to the actual time
> he sees, compares the found value with his own clock, and arrives
> the conclusions above.
>
> Now you may think something must be wrong there. The only strange
> thing is the constancy of the speed of light for all observers
Again, the _distance_ from A to B, as computed by A (and as measured
by an imaginary army of assistants, all comoving with A, all with
synchronized clocks) _changes_ when A changes speed. That's pretty
strange, IMHO, and it's at the heart of this problem.
> (I hope you fully understand what this means and how
> counterintuitive it is). Once you accept that, all the weird
> relativity effects follow.
>
> My response ends here, we can continue if we clarify the points
> above.
--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org
Bilge
>"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
>in message news:wfEPe.11208$qc2....@news.cpqcorp.net...
>> When you are alone in the desert without an atomic clock or
>> wrist watch or sun or stars, you can define time and "the passage
>> of time" with your heart beat, and do physics and engineering.
>> You don't need reference standards. You need a clock,
>> something that does pom pom pom pom pom pom.
>> You count the poms and call it time.
>> When you have more than one kind of clock, you can
>> compare one against the other and compare the times you
>> define and read on those clocks.
>
>Sure. And when you discover that they don't indicate the same, you will have
>some choosing to do, exactly as I desribed above...
I'm afraid that I don't see why anyone would choose to use an
``incorrectly'' calibrated clock for a thought experiment, or why
you can't just specify a ``thought calibration'' which illustrates
whatever preconception about time keeping you want to impose. I think
the basic idea behind the idea that ``time is what a clock measures,''
is that it doesn't specify a particular kind of clock that would
depend upon a detailed understanding of how the clock is supposed
to work. Having a lot of things called clocks is not a useful way
to go about it, unless you have some reason to believe the clocks
actually measure the same thing. That requires a physical model
for each clock.
>> If you can define regularity or "correct calibration" without
>> using the concept of time, then I agree with your suggestion
>> that "a correctly *calibrated* clock defines the passage of
>> time"
>
>Hey Dirk, "temperature" is *not* what any alcohol thermometer indicates;
Sure it is, except perhaps in the world of management in which managers
and marketeers like to believe whichever thermometer they can buy to reads
the best temperature for the advertising they envisioned. An alchohol
thermometer is an alcohol thermometer. If it has no numbers on it, its not
difficult to put it in some boiling water and scribble a number on it,
then put it in some water where the triple point is observed and then
divide up the interval into equally spaced divisions, which can be checked
to see if the divisions coincide with what you think temperature is
supposed to mean. How ``good'' it is, amounts to how well the thermometer
reads what your scribble marks say it should read under the same
conditions you used to decide where to scribble.
A ``calibrated thermometer'' is the same thermometer, except that
someone else does the scribbling and a disclai^H^H^H^H^H^H^Hcertificate
comes with it that tells you the extent to which the manufacturer will
will sympathize with you if you trust it for something important. In
any real experiment, if a calibration matters, the experimentor does
it himself, since he's the one whose reputation for doing careful
work is on the line.
>evenso "time" is not affected by your emotions while your heart beat is
>affected by that!
That doesn't matter unless you are completely incompetent as an
experimentalist. Experimentalists have to live with imperfect
and even crummy instruments. Clever experimentalists try to construct
experiments that don't rely to much on knowing what an instrument
reads relative to some NIST calibration standard. Instead they try
to design experiments that mainly depend on knowing that the
instruments read the same thing consistently, regardless of what
absolute number that might be. Ideally, one tries not to count on
that too much if possible.
For example, if you want to divide a bag of sand into some
number of equal piles, then you'll do better using a balance
that doesn't tell much of anything at all, but which can
be nulled out, than you will do using a calbrated scale to weigh
out each pile. (I mean, your average cocaine dealer is not the
best experimentalist in the world, but the measuring instrument
of choice is ye-olde triple beam balance, because it works
reliably under all sorts of conditions, many of which I imagine
are much less ideal than most regular labs.)
>I already imagine your time go quicker when you see a
>pretty girl - pom pom pom ;-)
>We can use such instruments to do experiments with limited precision.
Precision is a statistical quantity. It depends on how well
your instrument reproduces a number that you _think_ should
be the same and how much that uncertainty matters for the
length of the interval you are measuring. If you calibrate
your heartrate using a thousand trials for the same measurement
of something that lasts about 100 heartbeats, then you ought to
get excellent precision for a measurement that takes 1 million
heartbeats. If you madethree such measurements that were
different by a thouand heartbeats, you'd still do better than
a tenth of a percent.
>The regularity of any measurement standard is verified by for example
>comparing it with other devices and processes, just as it was discovered
>that an alcohol thermometer is less good than a mercury thermometer.
Discovering which is better depends upon how well each thermometer
yields the same result when measuring the same thing. Seeing if
two thermometers give the same result for some arbitrary measurement,
can only tell you that one of them is wrong, not which one is wrong.
Which is wrong requires specifying a definition for ``right.''
>By comparing an abundance of instruments one can determine which ones are
>less regular under what circumstances and in what intervals.
No, you can't. All you can determine is that the instruments don't all
give the same result. You can't even tell if any of them are ``regular''
much less which ones are more ``regular'' than others. Before you
can say anything about that, you have to define ``regular.'' The only
way to do that is to choose some natural phenomenon as a definition,
based on some model that you have some reason to believe applies
the way you think it does.
The reason that atomic clocks are believed to be so precise is
because their precision depends entirely upon the statistics of
a random process, which happens to be precisely how quantum
mechanics describes atomic transitions. If any particular atomic
transition occurs probabilistically (i.e., has no cause), then
such a clock is as theoretically perfect as any clock could
ever be, because it doesn't depend upon any mechanism. It only
depends upon statistics.
Sue...
Atomic clocks don't rely on statistics... idiots who don't
understand how they work do.
<<
where
m is the mass of an electron
mp is the mass of a proton
a is the fine structure constant (1/137.036)
c is the speed of light.
This is a much smaller perturbation than the
fine structure or Lamb shift. >>
http://en.wikipedia.org/wiki/Hyperfine_transition
Sue...
dsep...@austin.rr.com
If all during this experiment light pulses are sent from Clock A to
Clock B at 0.01 second intervals as measured by Clock A, and Clock B
encodes the time shown on the Clock B clock when the pulse is received
and sends the pulse back to Clock A, I don't see how the encoded time
of sequentional received pulses can differ by several seconds. Can
you explain that to me?
Thanks,
David
dsep...@austin.rr.com
sal
My response was actually directed at Wespe, who's sincerely interested
in learning more about relativity, and who, I figured, might not be
familiar with all that goes on during acceleration. (Or he might
already know it all "cold" -- whatever...)
You, David, I didn't expect to give a close reading to the post.
You could have surprised me, but you didn't.
Just to reiterate, you said,
> I don't see how the encoded time of sequentional received pulses can
> differ by several seconds.
Perhaps you overlooked this statement in my post:
>>In the telescope, _nothing_ strange seems to happen -- clock A
>>advances by, say, 1 second, and clock B advances by 1+epsilon
>>seconds, and where's the big deal?
"Encoding" the time is not a problem, indeed it's typically _assumed_
in such problems. Just look at a clock with a telescope -- there's
your "encoded time".
Now, go back and read my entire post, including the detailed example
of an observer next to Earth who accelerates toward a satellite a
light-hour away. As I said, A is watching an image of a B's clock.
And the time on the image A sees does not jump -- it just continues
advancing at the same rate it's been advancing at, which is dtau/dt
for B as viewed from A's rest frame.
You obviously didn't read my post before replying. Go back
and read it, including the following description of what happens when
A accelerates from a standing start, while viewing a clock 1 light
hour away. If you then post something about it which shows you
actually read it and made an attempt at understanding why it's true
I'll try to help you with it.
However, this key phrase in your OP lead me to believe you won't
bother to read carefully and sincerely try to understand:
>> but when I change the parameters to the problem, I realize that
>> those explanations are wrong.
No, _you're_ wrong. You don't understand the explanations, and when
you "vary the parameters" you don't see how to apply them. Concluding
the explanations are "wrong" because you don't understand them is
fallacious.
So, if your further posts indicate you're not reading the responses or
just not bothering to try to understand them, I won't waste any more
time replying to you.
dsep...@austin.rr.com
Your reply seemed to be a dfferent problem to me. Here's why. In
your problem, the relative velocity between Clock A and Clock B (the
telescope and satellite) changed during the acceleration. Initially,
in your explanation they were stationary (had zero relative velocity)
and after the acceleration they had a relative velocity V.
In the problem I posted, before the accelerations the relative
velocity was 3 meters / second, and after the accelerations the
velocity was 3 meters / second but in the opposite direction. Since
length contraction is independent of the direction of V, I didn't see
how the distance between the objects suddenly changed. Can you
clarify that for me. Before the acceleration, observers on Clock A
say that Clock B is 10**16 meters away. I don't see what changed
during the acceleration of Clock A and Clock B to make this distance
any different.
Thanks,
David
we...@operamail.com
dsep...@austin.rr.com wrote:
............
> Your reply seemed to be a dfferent problem to me. Here's why. In
> your problem, the relative velocity between Clock A and Clock B (the
> telescope and satellite) changed during the acceleration. Initially,
> in your explanation they were stationary (had zero relative velocity)
> and after the acceleration they had a relative velocity V.
> In the problem I posted, before the accelerations the relative
> velocity was 3 meters / second, and after the accelerations the
> velocity was 3 meters / second but in the opposite direction. Since
> length contraction is independent of the direction of V, I didn't see
> how the distance between the objects suddenly changed. Can you
> clarify that for me. Before the acceleration, observers on Clock A
> say that Clock B is 10**16 meters away. I don't see what changed
> during the acceleration of Clock A and Clock B to make this distance
> any different.
> Thanks,
> David
> >
I have just read the recent posts.. My explanation was flawed as Sal
pointed out. David, you should really examine Sal's example throughly.
It is a different example but demonstrates how time and space
coordinates of events are calculated, and how the skewed results are
obtained. Regarding to your question above, the numbers won't go away
if you change the sign of v. Likewise making the parameters small
numbers just yields to small results, but they're still there. This is
a quick post, I may write later if something comes up. Take care.
Kim B
Draw the spacetime diagram
Kim
Harald EPFL
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
news:slrndh6rfa....@radioactivex.lebesque-al.net...
My main objection to that simplistic statement about time is that it lacks
precision, it's ambiguous.
Apart of that, we basically agree on how to deal with it.
> >> If you can define regularity or "correct calibration" without
> >> using the concept of time, then I agree with your suggestion
> >> that "a correctly *calibrated* clock defines the passage of
> >> time"
> >
> >Hey Dirk, "temperature" is *not* what any alcohol thermometer indicates;
>
> Sure it is, except perhaps in the world of management in which managers
> and marketeers like to believe whichever thermometer they can buy to reads
> the best temperature for the advertising they envisioned. An alchohol
> thermometer is an alcohol thermometer. If it has no numbers on it, its not
> difficult to put it in some boiling water and scribble a number on it,
> then put it in some water where the triple point is observed and then
> divide up the interval into equally spaced divisions, which can be checked
> to see if the divisions coincide with what you think temperature is
> supposed to mean.
Here you just calibrated it, which was part of my argument.
< How ``good'' it is, amounts to how well the thermometer
> reads what your scribble marks say it should read under the same
> conditions you used to decide where to scribble.
Not if I remember correctly that mercury thermomenters are better than
alcohol thermometers because alcohol's expansion is slightly non-linear. The
check will then give unsatisfying results.
> A ``calibrated thermometer'' is the same thermometer, except that
> someone else does the scribbling and a disclai^H^H^H^H^H^H^Hcertificate
> comes with it that tells you the extent to which the manufacturer will
> will sympathize with you if you trust it for something important. In
> any real experiment, if a calibration matters, the experimentor does
> it himself, since he's the one whose reputation for doing careful
> work is on the line.
>
> >evenso "time" is not affected by your emotions while your heart beat is
> >affected by that!
[inserted back in place:]
> >I already imagine your time go quicker when you see a
> >pretty girl - pom pom pom ;-)
> That doesn't matter unless you are completely incompetent as an
> experimentalist.
I commented on time being *defined* as "the number of heart beats that you
count".
A competent experimentalist adds conditions to that overly naive point of
view.
> Experimentalists have to live with imperfect
> and even crummy instruments. Clever experimentalists try to construct
> experiments that don't rely to much on knowing what an instrument
> reads relative to some NIST calibration standard. Instead they try
> to design experiments that mainly depend on knowing that the
> instruments read the same thing consistently, regardless of what
> absolute number that might be. Ideally, one tries not to count on
> that too much if possible.
>
> For example, if you want to divide a bag of sand into some
> number of equal piles, then you'll do better using a balance
> that doesn't tell much of anything at all, but which can
> be nulled out, than you will do using a calbrated scale to weigh
> out each pile.
Very right.
< (I mean, your average cocaine dealer is not the
> best experimentalist in the world, but the measuring instrument
> of choice is ye-olde triple beam balance, because it works
> reliably under all sorts of conditions, many of which I imagine
> are much less ideal than most regular labs.)
;-))
> >We can use such instruments to do experiments with limited precision.
>
> Precision is a statistical quantity. It depends on how well
> your instrument reproduces a number that you _think_ should
> be the same and how much that uncertainty matters for the
> length of the interval you are measuring. If you calibrate
> your heartrate using a thousand trials for the same measurement
> of something that lasts about 100 heartbeats, then you ought to
> get excellent precision for a measurement that takes 1 million
> heartbeats. If you madethree such measurements that were
> different by a thouand heartbeats, you'd still do better than
> a tenth of a percent.
>
> >The regularity of any measurement standard is verified by for example
> >comparing it with other devices and processes, just as it was discovered
> >that an alcohol thermometer is less good than a mercury thermometer.
>
> Discovering which is better depends upon how well each thermometer
> yields the same result when measuring the same thing. Seeing if
> two thermometers give the same result for some arbitrary measurement,
> can only tell you that one of them is wrong, not which one is wrong.
> Which is wrong requires specifying a definition for ``right.''
Exactly - in this case likely "the same thing" was probably equal input of
heat energy.
> >By comparing an abundance of instruments one can determine which ones
are
> >less regular under what circumstances and in what intervals.
>
> No, you can't. All you can determine is that the instruments don't all
> give the same result.
Easy: measure Dirk's heart as well as several other hearts, his heart may go
pom-pom-pomperde-pom when a girl to his taste passes by, while another heart
will still go pom-pom-pom. Now compare that with a set of quartz watches and
you will know which instruments are more suited for the task.
> You can't even tell if any of them are ``regular''
> much less which ones are more ``regular'' than others.
Nonsense, see above.
> Before you
> can say anything about that, you have to define ``regular.'' The only
> way to do that is to choose some natural phenomenon as a definition,
> based on some model that you have some reason to believe applies
> the way you think it does.
>
> The reason that atomic clocks are believed to be so precise is
> because their precision depends entirely upon the statistics of
> a random process, which happens to be precisely how quantum
> mechanics describes atomic transitions. If any particular atomic
> transition occurs probabilistically (i.e., has no cause), then
> such a clock is as theoretically perfect as any clock could
> ever be, because it doesn't depend upon any mechanism. It only
> depends upon statistics.
You put the horse behind the cart, their precision is determined by
experiment and not by idealistic theory.
For example:
"Rubidium clocks are prized for their low cost, small size (commercial
standards are as small as 400 cm3), and short term stability. They are used
in many commercial, portable and aerospace applications. Hydrogen masers
(often manufactured in Russia) have superior short term stability to other
standards, but lower long term accuracy. "
http://www.nationmaster.com/encyclopedia/Atomic-clock
Harald
Kim B
You'll have to understand the difference between "observe" and "see".
Just before turnaround, they "observe" each others clock one second
slow -- meaning "the others clock shows t-1 AT THE SAME TIME as my
clock shows t".
Just after turnaorund, "AT THE SAME TIME" has changed because FOR has
changed. Now "the others clock shows t+1 AT THE SAME TIME as my clock
shows t" solely because of changed FOR. This is what you OBSERVE.
What A SEES is the lightpulses from B's clock, at he CANNOT SEE this
"jump" because of changed FOR. Draw the timespace diagram and the
clockpulses, then it is obvious what is SEEN.
Kim
Dirk Van de moortel
"Harald EPFL" <harald.v...@epfl.ch> wrote in message news:4314...@epflnews.epfl.ch...
I have something that produces a countable quantity.
I call it a clock. The count of the 'poms' I call time. I can now
attach a time to events and time intervals to event pairs.
That is perfectly unambiguous.
> Apart of that, we basically agree on how to deal with it.
I don't think so.
> I commented on time being *defined* as "the number of heart beats that you
> count".
> A competent experimentalist adds conditions to that overly naive point of
> view.
A competent experimentalist (and specially an engineer)
would do exactly the same when he only had his heart
beat at his disposal.
[snip]
> >
> > No, you can't. All you can determine is that the instruments don't all
> > give the same result.
>
> Easy: measure Dirk's heart as well as several other hearts, his heart may go
> pom-pom-pomperde-pom when a girl to his taste passes by, while another heart
> will still go pom-pom-pom. Now compare that with a set of quartz watches and
> you will know which instruments are more suited for the task.
But you don't have a quartz watch at your disposal.
You only have your heart beat. And you can do real
physics with it.
Whether you think it's bad physics, does not matter.
It can still be useful, and in fact the only useful option
you have.
Comparing heart time with quartz time comes after
you have defined time.
With your philosophical definition, there will always be
someone like yourself who will mutter: "yes, but that
atomic clock is not well *calibrated* yada yada... I can
imagine a much better calibrated clock yada yada.... in
the future there will probably be clocks that.... yada
yada yada..."
You are trying to define time with the notion of time,
disguised as regularity.
Dirk Vdm
sal
Duh. The different example I used was different. It was slightly
simpler.
But the principle is identical, and understanding it in one case means
you understand it in the other.
> Here's why. In your problem, the relative velocity between Clock A
> and Clock B (the telescope and satellite) changed during the
> acceleration. Initially, in your explanation they were stationary
> (had zero relative velocity) and after the acceleration they had a
> relative velocity V.
>
> In the problem I posted, before the accelerations the relative
> velocity was 3 meters / second, and after the accelerations the
> velocity was 3 meters / second but in the opposite direction. Since
> length contraction is independent of the direction of V, I didn't
> see how the distance between the objects suddenly changed.
You didn't actually read the post, did you?
You're using simple length contraction to try to understand it. That
won't work.
> Can you clarify that for me. Before the acceleration, observers on
> Clock A say that Clock B is 10**16 meters away. I don't see what
> changed during the acceleration
Because you didn't read the post.
> of Clock A and Clock B to make this
> distance any different. Thanks, David
OK, it's lecture time. You almost surely won't read this, either, but
maybe somebody else will find it of some slight interest.
There are three things in SR which you use to understand any problem
like this.
1) Events
2) Inertial frames
3) Lorentz transforms
The _only_ physical items which actually appear in the SR model are
"events". An "event" is something which has specific, unambiguous
coordinates in every frame of reference. In every other field of
mathematics it's called a "point". (Clocks are physical objects which
generate events. Clocks don't appear in the SR model; only the events
they generate actually appear in the model.)
The _only_ coordinate systems which are easy to use are "inertial
coordinates" or "inertial frames". An inertial frame is not
accelerating. You must be able to find the coordinates of each event
in some inertial frame, or you are noplace. You must know the
relationship of each inertial frame in the problem to each other
inertial frame, or you are noplace.
You transform the coordinates of events between inertial frames using
Lorentz transforms.
That's it. Keep that in mind, and you won't be confused.
* * *
There is ONE inertial frame specified throughout both your problem and
my example in which we know the coordinates of all the events: The
frame which is "stationary" relative to Earth. So do the problems in
that frame.
Look at the simple example I gave. You are looking through a
telescope at a clock. There are no clocks, no telescopes, no images
in the SR universe -- only events. How can we express this as a set
of events? It's simple, if you stay in our "universal" inertial
frame in which Earth is stationary:
The light left the clock on the satellite at a particular moment
... that is our first EVENT. Coordinates were (1:00, 1 lh), IIRC.
The light arrived at the eye of the observer in the spaceship
... that is our second EVENT. Coordinates were (2:00, 0), IIRC.
We know the time and location of each of those events in Earth's
inertial frame, so by using the Lorentz transforms, we can find the
time and location of those events in any other inertial frame.
DISTANCE is only well-defined if you specify a particular inertial
frame, and it's ONLY SPECIFIED BETWEEN EVENTS.
So, the spatial distance between those two events in Earth's inertial
frame is 1 LH. The distance in time between them is 1 hour.
To find the distance in any other frame, such as the frame which is
moving at 0.867C toward the satellite, just transform the coordinates
of the events into the new frame, and subtract. I did that, below,
and obtained a spatial distance of 3.734 lh and a time distance of
3.734 hours between the two events.
You can do exactly the same thing with your "two spaceship" problem.
Start with a single inertial frame in which you know the behavior of
each ship, find the coordinates of all the events in that frame, and
then transform them to every other frame you care about. After you
have the coordinates of the events in whatever frame you care about,
you can find their differences to compute physical distances and time
differences _in_ _that_ _frame_.
I will await your attempt at doing that before replying to any further
posts of yours.
Naively applying length contraction and time dilation without
understanding how they relate to coordinate differences between events
in particular frames can only lead to confusion.
--
Harry
> Comparing heart time with quartz time comes after
> you have defined time.
OK: time was originally counted (and thus defined) in "days". That unit is a
single night-day cycle. But as we know, the definition has been slightly
changed without a change of the concept of "time".
> With your philosophical definition, there will always be
> someone like yourself who will mutter: "yes, but that
> atomic clock is not well *calibrated* yada yada... I can
> imagine a much better calibrated clock yada yada.... in
> the future there will probably be clocks that.... yada
> yada yada..."
As you should know, that's what did happen in the 20th century when the
atomic clock standard was adopted - but not because of calibration but
because of precision, thus necessitating a modification of calibration
standard. It's not philosophical but technical.
> You are trying to define time with the notion of time,
> disguised as regularity.
Regularity of interval is one of the requirements for any serious
measurement tool, independent of the notion of it; another requirement is to
have a verifiable standard by which measurements can be compared.
Harald
sal
How do you define and measure regularity of interval for a clock? I
am curious.
For a standard of physical length it's obvious: It should be invariant
under longitudinal translation. You can check a measuring stick for
such invariance by making marks on a separate object and sliding it
through its own length. For a ruler with multiple marks which are
supposedly regularly spaced, this can provide you with an immediate
check on whether the marks are actually regularly spaced. (In fact
exactly this sort of test, with some slight variations and a little
cleverness, reveals that many cheap slide rules were rather
inaccurately laid out.)
But how do you do that for a clock?
And for that matter, how do you verify that a measuring stick of some
sort is invariant under perpendicular translation, and in particular,
how do you verify that it's invariant under translation through time?
> another requirement is to have a verifiable standard by which
> measurements can be compared.
>
> Harald
--
we...@operamail.com
Let's call the frames before and after acceleration A and C. They are
inertial and have the same speed in opposite directions according to
earth frame. Suppose they are spaceships and they are trying to measure
the speed of light coming from B. Suppose the light enters the
spaceship's front end and exits from the rear end, and there are clocks
at these ends and they are looking at the differences the clocks
record. Now, as seen from earth, both spaceships are length contracted
and time dilated by the same factor (I think these effects cancel each
other out for this measurement). Since the spaceships are moving, the
approaching spaceship would measure the speed higher than c, and the
other lower than c, but no, they both measure c. The real reason is
relative simultaneity. As seen from earth, the clocks on these
spaceships are not in sync. This is because of how they synchronize the
two clocks: flash a light at the midpoint of the spaceship and reset
the clocks when the light hit the walls at the rear and front. This
procedure makes sense for the spaceships, they consider themselves
stationary and the light is sent from the midpoint so they should
arrive the walls what they consider at the same time. Therefore, when A
synchorizes his clocks, they will have an offset (according to earth);
the clock closer to earth will be a little bit ahead of the other
clock. Similarly for C's clocks, the clock closer to earth will be a
little behind the other. Now scale this offset up to B's location,
compare this with B's clock, and you should get the one second
ahead/behind values. Only I am not sure whether the synchronization
automatically changes during acceleration or they have to be
resynchronized. I think this explanation is right but feel free to
correct me, anyone.
Dirk Van de moortel
"Harry" <harald.v...@epfl.ch> wrote in message news:431483ab$1...@epflnews.epfl.ch...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:MDWQe.181771$xD1.10...@phobos.telenet-ops.be...
> >
> > "Harald EPFL" <harald.v...@epfl.ch> wrote in message
> news:4314...@epflnews.epfl.ch...
> SNIP
>
> > Comparing heart time with quartz time comes after
> > you have defined time.
>
> OK: time was originally counted (and thus defined) in "days". That unit is a
> single night-day cycle. But as we know, the definition has been slightly
> changed without a change of the concept of "time".
Comparing heart time with days time comes after
you have defined time.
Dirk Vdm
sal
> OK, I think I have found a simpler explanation, if you bear with
> me. Let's call the frames before and after acceleration A and
> C. They are inertial and have the same speed in opposite directions
> according to earth frame. Suppose they are spaceships and they are
> trying to measure the speed of light coming from B. Suppose the
> light enters the spaceship's front end and exits from the rear end,
> and there are clocks at these ends and they are looking at the
> differences the clocks record. Now, as seen from earth, both
> spaceships are length contracted and time dilated by the same factor
> (I think these effects cancel each other out for this measurement).
Ouch -- be careful. The effects are numerical reciprocals but saying
they "cancel out" is touchy.
Yes, the ship is gamma times longer in its rest frame. And yes, a
shipboard clock ticks 1/gamma times as fast. But naively, one might
conclude that the astronauts see a flash take 1/gamma times as much
time to go gamma times as far, which implies they'd measure the speed
of light as being gamma/(1/gamma) = gamma^2 times faster than an
Earth-based observer!
> Since the spaceships are moving, the approaching spaceship would
> measure the speed higher than c, and the other lower than c, but no,
> they both measure c. The real reason is relative simultaneity.
Right. At least, that's as "real" as one can say any cause of this
might be.
> As seen from earth, the clocks on these spaceships are not in
> sync. This is because of how they synchronize the two clocks: flash
> a light at the midpoint of the spaceship and reset the clocks when
> the light hit the walls at the rear and front. This procedure makes
> sense for the spaceships, they consider themselves stationary and
> the light is sent from the midpoint so they should arrive the walls
> what they consider at the same time.
Actually, the method chosen to sync the clocks doesn't matter. You
could throw rocks rather than light flashes (if the rock velocity was
uniform), or you could just move the clocks in opposite directions out
from the center. As long as the clocks are moved at the same speed
all will be well and the same results will be observed.
> Therefore, when A synchorizes his clocks, they will have an offset
> (according to earth); the clock closer to earth will be a little bit
> ahead of the other clock. Similarly for C's clocks, the clock closer
> to earth will be a little behind the other. Now scale this offset up
> to B's location, compare this with B's clock, and you should get the
> one second ahead/behind values. Only I am not sure whether the
> synchronization automatically changes during acceleration or they
> have to be resynchronized.
They must be resynchronized. The clocks in the back of the ship lose
time.
Strange stuff happens during acceleration of an extended object. For
example, if the ship is to hang together, the back end must accelerate
more strongly than the front end.
Uniform acceleration, as might happen while the engines are firing, is
a lot like a uniform gravitational field. In a gravity well, clocks
that are "deeper" in the well run slower; clocks "higher up" run
faster. Similarly, during acceleration clocks in the nose of the ship
will run faster than clocks in the back of the ship. In fact, this is
the same effect as that which causes the clocks on Earth to seem to
run faster while a distant astronaut is turning his ship around to
head back home.
I've got a page which takes a brief look at this, with a little math and a
graph, at
http://www.physicsinsights.org/revolving_astronaut.html
(Note, though, that the comments in the "Porthole View" section are
slightly wrong.)
> I think this explanation is right but feel free to
> correct me, anyone.
--
we...@operamail.com
> On Tue, 30 Aug 2005 10:21:22 -0700, wespe wrote:
>
> > OK, I think I have found a simpler explanation, if you bear with
> > me. Let's call the frames before and after acceleration A and
> > C. They are inertial and have the same speed in opposite directions
> > according to earth frame. Suppose they are spaceships and they are
> > trying to measure the speed of light coming from B. Suppose the
> > light enters the spaceship's front end and exits from the rear end,
> > and there are clocks at these ends and they are looking at the
> > differences the clocks record. Now, as seen from earth, both
> > spaceships are length contracted and time dilated by the same factor
> > (I think these effects cancel each other out for this measurement).
>
> Ouch -- be careful. The effects are numerical reciprocals but saying
> they "cancel out" is touchy.
>
> Yes, the ship is gamma times longer in its rest frame. And yes, a
> shipboard clock ticks 1/gamma times as fast. But naively, one might
> conclude that the astronauts see a flash take 1/gamma times as much
> time to go gamma times as far, which implies they'd measure the speed
> of light as being gamma/(1/gamma) = gamma^2 times faster than an
> Earth-based observer!
>
>
yes.. another mistake of mine. Then I assume relative simultaneity
would compansate for this too.
> > Since the spaceships are moving, the approaching spaceship would
> > measure the speed higher than c, and the other lower than c, but no,
> > they both measure c. The real reason is relative simultaneity.
>
> Right. At least, that's as "real" as one can say any cause of this
> might be.
>
> > As seen from earth, the clocks on these spaceships are not in
> > sync. This is because of how they synchronize the two clocks: flash
> > a light at the midpoint of the spaceship and reset the clocks when
> > the light hit the walls at the rear and front. This procedure makes
> > sense for the spaceships, they consider themselves stationary and
> > the light is sent from the midpoint so they should arrive the walls
> > what they consider at the same time.
>
> Actually, the method chosen to sync the clocks doesn't matter. You
> could throw rocks rather than light flashes (if the rock velocity was
> uniform), or you could just move the clocks in opposite directions out
> from the center. As long as the clocks are moved at the same speed
> all will be well and the same results will be observed.
>
I totally agree, any method would do, because that's really what the
astronauts experience. If a spaceship is moving with a speed close to
c, light emitted from midpoint could take a thousand years to reach the
front as seen from earth, but the astronauts don't experience such a
delay.
> > Therefore, when A synchorizes his clocks, they will have an offset
> > (according to earth); the clock closer to earth will be a little bit
> > ahead of the other clock. Similarly for C's clocks, the clock closer
> > to earth will be a little behind the other. Now scale this offset up
> > to B's location, compare this with B's clock, and you should get the
> > one second ahead/behind values. Only I am not sure whether the
> > synchronization automatically changes during acceleration or they
> > have to be resynchronized.
>
> They must be resynchronized. The clocks in the back of the ship lose
> time.
>
Thanks for this info, I wasn't sure.
> Strange stuff happens during acceleration of an extended object. For
> example, if the ship is to hang together, the back end must accelerate
> more strongly than the front end.
>
> Uniform acceleration, as might happen while the engines are firing, is
> a lot like a uniform gravitational field. In a gravity well, clocks
> that are "deeper" in the well run slower; clocks "higher up" run
> faster. Similarly, during acceleration clocks in the nose of the ship
> will run faster than clocks in the back of the ship. In fact, this is
> the same effect as that which causes the clocks on Earth to seem to
> run faster while a distant astronaut is turning his ship around to
> head back home.
>
> I've got a page which takes a brief look at this, with a little math and a
> graph, at
>
> http://www.physicsinsights.org/revolving_astronaut.html
>
Thanks, very interesting.
> (Note, though, that the comments in the "Porthole View" section are
> slightly wrong.)
>
>
> > I think this explanation is right but feel free to
> > correct me, anyone.
>
> --
> Nospam becomes physicsinsights to fix the email
Note to David: what do you have to say about all this?
Bilge
>Atomic clocks don't rely on statistics... idiots who don't
>understand how they work do.
Buy a clue.
Bilge
>Not if I remember correctly that mercury thermomenters are better than
>alcohol thermometers because alcohol's expansion is slightly non-linear. The
>check will then give unsatisfying results.
grasp the concept of a calibration as anything but a comparison to
a certified instrument.
[...]
>> That doesn't matter unless you are completely incompetent as an
>> experimentalist.
>
>I commented on time being *defined* as "the number of heart beats that you
>count".
Right. Whats the problem?
>A competent experimentalist adds conditions to that overly naive point of
>view.
Such as? (Try to avoid adding ``conditions'' that beg the question.)
>> For example, if you want to divide a bag of sand into some
>> number of equal piles, then you'll do better using a balance
>> that doesn't tell much of anything at all, but which can
>> be nulled out, than you will do using a calbrated scale to weigh
>> out each pile.
>
>Very right.
And all of that without ever weighing anything...
[...]
>> Discovering which is better depends upon how well each thermometer
>> yields the same result when measuring the same thing. Seeing if
>> two thermometers give the same result for some arbitrary measurement,
>> can only tell you that one of them is wrong, not which one is wrong.
>> Which is wrong requires specifying a definition for ``right.''
>
>Exactly - in this case likely "the same thing" was probably equal input of
>heat energy.
That begs the question. How do you measure ``equal input of heat energy,''
without a thermometer?
[...]
>> No, you can't. All you can determine is that the instruments don't all
>> give the same result.
>
>Easy: measure Dirk's heart as well as several other hearts, his heart may go
>pom-pom-pomperde-pom when a girl to his taste passes by, while another heart
>will still go pom-pom-pom.
Which tells you only that the heartrates are different, just as I said.
>Now compare that with a set of quartz watches and
>you will know which instruments are more suited for the task.
That begs the question. What reason do you have for assuming anything
about quartz watches?
>> You can't even tell if any of them are ``regular''
>> much less which ones are more ``regular'' than others.
>
>Nonsense, see above.
You need to re-examine your preconceptions. You take a lot of things
for granted simply because you grew up being told this stuff was true.
[...]
>> mechanics describes atomic transitions. If any particular atomic
>> transition occurs probabilistically (i.e., has no cause), then
>> such a clock is as theoretically perfect as any clock could
>> ever be, because it doesn't depend upon any mechanism. It only
>> depends upon statistics.
>
>You put the horse behind the cart, their precision is determined by
>experiment and not by idealistic theory.
Dont be an idiot. Without a theory about what the experiments
mean, the results don't tell you anything. Do you really think
archimedes is remembered because he was the only person to put
two different materials into some water, or do think perhaps
he's remembered because he came up with a theory about what
the experiment meant? You remind of a mechanical engineer I knew
whose answer to every question was to see what the ASME manuals
had to say about it. Needless to say, he was rather useless for
advice on engineering anything, since the ASME manuals don't cover
problems that haven't come up before.
>For example:
>"Rubidium clocks are prized for their low cost, small size (commercial
>standards are as small as 400 cm3), and short term stability. They are used
>in many commercial, portable and aerospace applications. Hydrogen masers
>(often manufactured in Russia) have superior short term stability to other
>standards, but lower long term accuracy. "
Gee harry, how do they know that what the clocks read represents
what they think it does? The stone tablets that god sent along with
the clocks? You'd really be in a bind if you got stuck on island.
Without NIST certified rulers, grade 8 bolts and graded lumber,
you would have no way to determine if it was possible to build a
boat that doesn't sink.
Bilge
>> My main objection to that simplistic statement about time is that it lacks
>> precision, it's ambiguous.
>
>I have something that produces a countable quantity.
>I call it a clock. The count of the 'poms' I call time. I can now
>attach a time to events and time intervals to event pairs.
>That is perfectly unambiguous.
It's also way over harry's head. Recall also that harry advocates
the idea of an ether which precludes him from even claiming any device
represents the ``real time." I wonder why he assumes a clock that
could give an accurate calibration of another clock when the time
on different clocks depends upon how the clocks are oriented in
his ether? He doesn't seem to get the point even without that added
difficulty.
[...]
>> A competent experimentalist adds conditions to that overly naive point of
>> view.
>
>A competent experimentalist (and specially an engineer)
>would do exactly the same when he only had his heart
>beat at his disposal.
One would hope so, anyway.
>> Easy: measure Dirk's heart as well as several other hearts, his heart may
>> go pom-pom-pomperde-pom when a girl to his taste passes by, while another
>> heart will still go pom-pom-pom. Now compare that with a set of quartz
>> watches and you will know which instruments are more suited for the task.
>
>But you don't have a quartz watch at your disposal.
Impossible. Quartz watches have always existed. How else could
people like kepler have known what day of the year it was when
measuring the position of planets?
>You only have your heart beat. And you can do real
>physics with it.
You mean without NIST certification and an ASME manual to explain
how to slide the beads across the dental floss for each heartbeat to
insure a conforming measurement?? Preposterous.
Harry
It's quite easy: instead of depending on measuring you depend on constant
input conditions, such as when using a heater - the same trick box as your
sand bags...
> >> No, you can't. All you can determine is that the instruments don't
all
> >> give the same result.
> >
> >Easy: measure Dirk's heart as well as several other hearts, his heart
may go
> >pom-pom-pomperde-pom when a girl to his taste passes by, while another
heart
> >will still go pom-pom-pom.
>
> Which tells you only that the heartrates are different, just as I said.
That's enough to have a first precision estimate... then add a bunch of
quartz clocks.
> >Now compare that with a set of quartz watches and
> >you will know which instruments are more suited for the task.
>
> That begs the question. What reason do you have for assuming anything
> about quartz watches?
We need to assume nothing about them, just call them boxes x1 to x10. Do you
really not understand how we know that quartz clocks are better?!
> >> You can't even tell if any of them are ``regular''
> >> much less which ones are more ``regular'' than others.
> >
> >Nonsense, see above.
>
> You need to re-examine your preconceptions. You take a lot of things
> for granted simply because you grew up being told this stuff was true.
Fine to me if you believe that heart rate is as regular as quartz clocks,
it's not my problem...
> >> mechanics describes atomic transitions. If any particular atomic
> >> transition occurs probabilistically (i.e., has no cause), then
> >> such a clock is as theoretically perfect as any clock could
> >> ever be, because it doesn't depend upon any mechanism. It only
> >> depends upon statistics.
> >
> >You put the horse behind the cart, their precision is determined by
> >experiment and not by idealistic theory.
>
> Dont be an idiot.
I'm not, but you pretend to be one...
> Without a theory about what the experiments
> mean, the results don't tell you anything. Do you really think
> archimedes is remembered because he was the only person to put
> two different materials into some water, or do think perhaps
> he's remembered because he came up with a theory about what
> the experiment meant?
No - this is about the tools that he used for his experiment...
> You remind of a mechanical engineer I knew
> whose answer to every question was to see what the ASME manuals
> had to say about it. Needless to say, he was rather useless for
> advice on engineering anything, since the ASME manuals don't cover
> problems that haven't come up before.
>
> >For example:
> >"Rubidium clocks are prized for their low cost, small size (commercial
> >standards are as small as 400 cm3), and short term stability. They are
used
> >in many commercial, portable and aerospace applications. Hydrogen masers
> >(often manufactured in Russia) have superior short term stability to
other
> >standards, but lower long term accuracy. "
>
> Gee harry, how do they know that what the clocks read represents
> what they think it does?
Exactly - it seems that you haven't got a clue.
> The stone tablets that god sent along with
> the clocks? You'd really be in a bind if you got stuck on island.
> Without NIST certified rulers, grade 8 bolts and graded lumber,
> you would have no way to determine if it was possible to build a
> boat that doesn't sink.
Nonsense, I can set up my own standards.
Harald