Re: A constant speed of light in all reference frames? Surely you can't be serious.
jem
> On Mar 4, 10:31 am, Ste <ste_ro...@hotmail.com> wrote:
>> On 4 Mar, 13:40, mpalenik <markpale...@gmail.com> wrote:
>>
>>
>>
>>> On Mar 4, 3:12 am, Ste <ste_ro...@hotmail.com> wrote:
>>>> On 3 Mar, 20:01, mpalenik <markpale...@gmail.com> wrote:
>>>>> On Mar 3, 12:52 pm, Ste <ste_ro...@hotmail.com> wrote:
>>>>>> No. In SR, clocks *appear* to run slower as you are increasing your
>>>>>> distance from the clock. The effect is entirely apparent in SR.
>>>>> You must just go through the entire thread and not pay any attention
>>>>> to what anybody says. Ever.
>>>>> 1) What you've stated above is not an effect of SR. It is an effect
>>>>> of propagation delay, which was used to calculate c from the motion of
>>>>> the moons of jupiter hundreds of years ago.
>>>> Ok.
>>>>> 2) If you were to move TOWARD the clock, it would appear to run
>>>>> faster. But SR says nothing about whether you are moving toward or
>>>>> away from an object.
>>>> <suspicious eyebrow raised> Ok.
>>>>> 3) The amount that the clock would appear to slow down is DIFFERENT
>>>>> from the amount that SR predicts the clock *actually* slows down
>>>> Really? I'm growing increasingly suspicious. In what way does SR
>>>> predict the "actual" slowdown, as opposed to the "apparent" slowdown?
>>>> And for example, if we racked up the value of 'c' to near infinity,
>>>> would SR still predict an "actual" slowdown, even though the
>>>> propagation delays would approach zero?
>>> With what you have described, I checked just to be sure, even though I
>>> was already pretty sure what the answer would be, the time you read
>>> moving away the clock would be:
>>> t2 = t - (x+vt)/c = t(1-v/c) - x
>>> and when you move toward the clock
>>> t2 = t + (x+vt)/c = t(1+v/c) + x
>>> so moving away from the clock:
>>> dt2/dt = 1-v/c
>>> and toward
>>> dt2/dt = 1-v/c
>>> Special relativity predicts that the moving clock will always slow
>>> down as
>>> dt2/dt = sqrt(1-v^2/c^2)
>>> What you *measure* is a combination of the actual slow down predicted
>>> by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation
>>> delays (which depend on the direction of motion).
>> Ok. So let us suppose that we take two clocks. Separate them by a
>> certain distance, synchronise them when they are both stationary, and
>> then accelerate them both towards each other (and just before they
>> collide, we bring them stationary again). Are you seriously saying
>> that both clocks report that the other clock has slowed down, even
>> though they have both undergone symmetrical processes? Because there
>> is obviously a contradiction there.
>
> Yes, that is correct. Both will report a slow down. And in fact,
> which ever one breaks the inertial frame to match speed with the other
> is the one that will be "wrong". This is still within the realm of
> SR, not GR.
Actually, according to SR, the two (ideal) clocks will be synchronized
when they meet, regardless which clock "breaks the inertial frame" via
an instantaneous acceleration. That final acceleration, which occurs
when the clocks are co-located, has no effect on the clock readings.
Reported clock "slow downs" will be exactly offset by clock speed-ups
during the initial accelerations*.
* For clarity, both effects are purely observational - SR presumes
(ideal) clock mechanisms are completely unaffected by a clock's motion.
mpalenik
>
> - Show quoted text -
Ste, just so you know, PD and Peter will back me up on my answer.
Ignore Jem. He (she?) has obviously never studied physics and has no
idea what SR does or doesn't say.
Jem e-mailed me a little while ago asking me to make this "correction"
for myself. As every physicist here knows, this "interpretation" of
SR has nothing whatsoever to do with reality.
That's all I'm going to say on the subject. I'm not going to get
sucked into another discussion trying to explain SR to an obvious
crank.
jem
Well, let's see who backs you up.
mpalenik
>
> - Show quoted text -
PD, Peter Webb, Inertial, and Eric Gisse all agree with me. I
garauntee that. I have no idea how many will bother responding to
you, though. Most likely, they know better by now than to bother.
jbriggs444
>
> - Show quoted text -
What's the assertion under dispute?
jem:
"
> > Actually, according to SR, the two (ideal) clocks will be synchronized
> > when they meet, regardless which clock "breaks the inertial frame" via
> > an instantaneous acceleration. That final acceleration, which occurs
> > when the clocks are co-located, has no effect on the clock readings.
"
jem is correct here.
That the clocks are synchronized follows by a simple symmetry
argument. The observation that the final acceleration does not affect
the respective clock readings is also true. (Do a Lorentz transform
when
the x offset in the direction of motion is zero and see what kind of t
offset you get.)
jem:
"
> > Reported clock "slow downs" will be exactly offset by clock speed-ups
> > during the initial accelerations*.
"
jem is correct here in the sense that the elapsed time on B's clock as
accounted for in A's frame is
exactly the same as the elapsed proper time on A's clock. Speed up
and slow down exactly offset one another. The clocks agree when they
are reunited.
You do understand about clock speed up seen when doing a Lorentz
transform from a pair of tangent inertial
frames anchored at A to in order to compare "time now" at B with "time
a moment ago" at B while A is accelerating toward B, right?
Sweeping the hyper-plane of simultaneity forward on B's timeline has
an apparent speed-up effect, right?
mpalenik:
"
> > > Yes, that is correct. Both will report a slow down. And in fact,
> > > which ever one breaks the inertial frame to match speed with the other
> > > is the one that will be "wrong". This is still within the realm of
> > > SR, not GR.
"
Your initial assertions are correct. Both will report a slow down....
During part of the trip. And a different slowdown during a different
part of the trip.
But both will _also_ report a speed up during the period of initial
acceleration.
The question of which clock is "wrong" doesn't even arise. We have a
coherent accounting for of elapsed times from both points of view.
Those accounts may not agree on whose clock was running at what rate
at what time, but they do agree on all observables. The clocks
started synchronized and they finished synchronized.
If you're trying to base this shouting match on the question of which
clock was "really" slow during the tail end of the inertial portion of
their respective flights then you're being a butthead. That's not a
question of interest. It has no observable answer.
PD, Inertial or Eric Gisse can back me up on that if they wish.
PD
I think the issue is as I described in "plain English" to Ste.
Let's take 3 clocks, one (A) which gets left behind, one (B) that
sallies out at speed v for a distance L and returns, and one (C) that
sallies out at speed v for a distance 2L and returns. At speed v, the
clocks sent forth are running slow by, say, 2%. I'm really not going
to worry about the accelerations at all because, as all have said, the
acceleration profiles are all the same. Clock B will arrive back at
home having run slow by 2% for the time of its trip and it is now 2
hours behind clock A, say. Once B arrives home and comes to rest
alongside A, its rate is no longer slower than A's by 2%, and so while
it sits there and waits for C to come home it will *continue* to be 2
hours behind A. Finally, C comes home, having run slower than A by 2%
for twice as long as B did, and so it will be 4 hours behind A. Since
B is only 2 hours behind A, clock B and clock C are no longer
synchronized.
What am I missing?
PD
jbriggs444
[discussing the scenario in which two clocks symmetrically accelerate
toward each other and this is observed from the point of view of one
of the clocks]
> Your initial assertions are correct. Both will report a slow down....
> During part of the trip. And a different slowdown during a different
> part of the trip.
I should correct myself here.
The part I wrote about "a different slowdown during a different part
of the trip" was an erroneous reference to a scenario in which a clock
is reporting the Doppler shift that is _seen_ (i.e. without accounting
for transit delays).
In such a scenario, the clock sees an interval after it has
accelerated and before its peer has accelerated and then a second
interval after the peer has accelerated. The Doppler shifts for the
intervals are, of course, different.
In the scenario at hand we are trying to discuss what is _observed_
(i.e. accounting for transit delays and adopting particular standards
of simultaneity)
What is _observed_ is only a single interval during which the two
clocks are in constant relative motion, not two such intervals. [From
an "observed" point of view, the period when the peer clock remains
motionless does not fall within the interval of the journey. Instead,
it is in the relative past]
jbriggs444
>
> - Show quoted text -
I'm sure you're not missing a thing. I hadn't gone far enough back to
see that scenario discussed.
The context I was working from is the "two clocks at rest
simultaneously accelerate toward one another".
It's clear from a naive relativity crank point of view that if the
acceleration profiles are identical that the resulting clock shifts
must be identical. If jem is a naive relativity crank then he could
indeed have a viable model for the symmetric situation and an
incorrect model for the situation you pose here.
Thanks for the pointer.
mpalenik
The scenario being discussed is only symmetric until one clock
accelerates to meet the other. In this case, the clock that breaks
symmetry will report that a shorter time has passed. In this case, it
will report a time that is lower, according to the observations of the
clock that remained at a constant velocity.
>
> jem:
> "> > Reported clock "slow downs" will be exactly offset by clock speed-ups
> > > during the initial accelerations*.
>
> "
>
> jem is correct here in the sense that the elapsed time on B's clock as
> accounted for in A's frame is
> exactly the same as the elapsed proper time on A's clock. Speed up
> and slow down exactly offset one another. The clocks agree when they
> are reunited.
It depends on the frame in which they are reunited. If we take the
frame of clock B to be the rest frame, the proper time that passed for
clock B is less than clock A when they are reunited.
>
> Your initial assertions are correct. Both will report a slow down....
> During part of the trip. And a different slowdown during a different
> part of the trip.
>
> But both will _also_ report a speed up during the period of initial
> acceleration.
>
> The question of which clock is "wrong" doesn't even arise. We have a
> coherent accounting for of elapsed times from both points of view.
> Those accounts may not agree on whose clock was running at what rate
> at what time, but they do agree on all observables. The clocks
> started synchronized and they finished synchronized.
Clearly you didn't understand the question or the explanation.
mpalenik
Jem, in the e-mail I was sent (which I haven't checked to be sure that
it matches the one he posted here), said that time dilation is only an
apparent effect and said "* For clarity, both effects are purely
observational - SR presumes (ideal) clock mechanisms are completely
unaffected by a clock's motion."
I'm in my 3rd year of PhD studies, and have studied SR, GR, and
quantum field theory. I can handle a simple relativity question like
the one discussed above. If you had read a little more carefully, you
would see nothing I said is inconsistant with SR.
mpalenik
> On Mar 5, 1:54 pm, jbriggs444 <jbriggs...@gmail.com> wrote:
> [discussing the scenario in which two clocks symmetrically accelerate
> toward each other and this is observed from the point of view of one
> of the clocks]
>
> > Your initial assertions are correct. Both will report a slow down....
> > During part of the trip. And a different slowdown during a different
> > part of the trip.
>
> I should correct myself here.
>
> The part I wrote about "a different slowdown during a different part
> of the trip" was an erroneous reference to a scenario in which a clock
> is reporting the Doppler shift that is _seen_ (i.e. without accounting
> for transit delays).
In fact, what I was trying to describe to Ste was the effects that are
specifically not due to Doppler shifting, as I was specifically making
the point that the predictions of SR for time dilation are
mathematically different than those which are due to the observed rate
of change on a ticking clock to transit delays.
>
> In such a scenario, the clock sees an interval after it has
> accelerated and before its peer has accelerated and then a second
> interval after the peer has accelerated. The Doppler shifts for the
> intervals are, of course, different.
>
> In the scenario at hand we are trying to discuss what is _observed_
> (i.e. accounting for transit delays and adopting particular standards
> of simultaneity)
No, that's not actually what we were talking about at all.
>
> What is _observed_ is only a single interval during which the two
> clocks are in constant relative motion, not two such intervals. [From
> an "observed" point of view, the period when the peer clock remains
> motionless does not fall within the interval of the journey. Instead,
> it is in the relative past]
The relevant quantity was the time that each clock displays after the
two are brought into comoving frames, which once again, depends on the
frame that they are brought into.
And yes, as the clocks accelerate, due to Doppler effects, they will
each see an apparent change in the other clock's rate that brings it's
reading into what it is supposed to be for whichever frame they are
accelerating into.
But I fear if Ste reads any of this, he's going to slip back into his
whole "relativity is due to propagation delays" thing again. Let's
tackle one problem at a time.
jbriggs444
If you had read a little more carefully you would see nothing I said
said anything you said was inconsistent with SR.
[From jem:]
> SR presumes (ideal) clock mechanisms are completely
> unaffected by a clock's motion."
What do you find objectionable about this statement? As I read it, it
is true. [albeit a potential setup for a crank screed]
If I adopt the frame of a nearby gamma ray decay product, you do agree
that this would not retroactively affect one of the clocks timing the
men's downhill in Vancouver, surely? Yet those clocks were most
certainly moving at .999c during those events in such a frame.
A contention that SR is _only_ an illusion, yes, that I could disagree
with. But I don't see that contention being made by jem. _At least
not right there_.
jbriggs444
> On Mar 5, 2:17 pm, jbriggs444 <jbriggs...@gmail.com> wrote:
>
> > On Mar 5, 1:54 pm, jbriggs444 <jbriggs...@gmail.com> wrote:
> > [discussing the scenario in which two clocks symmetrically accelerate
> > toward each other and this is observed from the point of view of one
> > of the clocks]
>
> > > Your initial assertions are correct. Both will report a slow down....
> > > During part of the trip. And a different slowdown during a different
> > > part of the trip.
>
> > I should correct myself here.
>
> > The part I wrote about "a different slowdown during a different part
> > of the trip" was an erroneous reference to a scenario in which a clock
> > is reporting the Doppler shift that is _seen_ (i.e. without accounting
> > for transit delays).
>
> In fact, what I was trying to describe to Ste was the effects that are
> specifically not due to Doppler shifting, as I was specifically making
> the point that the predictions of SR for time dilation are
> mathematically different than those which are due to the observed rate
> of change on a ticking clock to transit delays.
Ok good. So we're both not talking about that.
> > In such a scenario, the clock sees an interval after it has
> > accelerated and before its peer has accelerated and then a second
> > interval after the peer has accelerated. The Doppler shifts for the
> > intervals are, of course, different.
>
> > In the scenario at hand we are trying to discuss what is _observed_
> > (i.e. accounting for transit delays and adopting particular standards
> > of simultaneity)
>
> No, that's not actually what we were talking about at all.
> > What is _observed_ is only a single interval during which the two
> > clocks are in constant relative motion, not two such intervals. [From
> > an "observed" point of view, the period when the peer clock remains
> > motionless does not fall within the interval of the journey. Instead,
> > it is in the relative past]
>
> The relevant quantity was the time that each clock displays after the
> two are brought into comoving frames, which once again, depends on the
> frame that they are brought into.
In the scenario in question they are not brought into co-moving frames
(whatever that means -- the notion of things being "brought into"
frames is very questionable) They are brought _TOGETHER_.
The time displayed on each clock when they become adjacent is an
_observable_. It's not frame dependent. The numbers don't change if
you decide to adopt a different frame of reference. You take a
snapshot of the clocks side by side and you look at the numbers.
There is no ambiguity. All frames get the same answer.
Even the frame in which the two clocks are mutually at rest.
mpalenik
All I can say is read what I originally wrote again.
BTW, "together" is ambiguous. Together can mean comoving or it can
mean "they pass each other." Each of those warrants a different
answer.
mpalenik
>
> - Show quoted text -
Just to elaborate, what I said was that if one clock or the other
accellerates, the situation is no longer symmetric and will affect the
time each clock reads respectively.
And the notion of comoving frames is not questionable at all. It's
standard terminology in the literature.
jbriggs444
>
> - Show quoted text -
No it doesn't, dufus.
jbriggs444
>
> - Show quoted text -
Read it yourself, dufus
>> certain distance, synchronise them when they are both stationary, and
>> then accelerate them both towards each other (and just before they
>> collide, we bring them stationary again). Are you seriously saying
What part of "both towards each other" is asymmetric, ignoramus?
mpalenik
>
> - Show quoted text -
One clock changes speed, the other doesn't. That's what I was
describing. Ste's original scenario is symmetric. When one clock
accellerates to meet the other as I described it is not.
There's really no need for name calling.
jbriggs444
No. It doesn't affect what each clock reads.
It affects what each clock reads _at some particular event_, such as
when they come together.
> And the notion of comoving frames is not questionable at all. It's
> standard terminology in the literature.- Hide quoted text -
>
> - Show quoted text -
Read for comprehension, moron. I didn't say anything about comoving
frames being questionable.
I said that the notion of "being brought into" is questionable.
And it is. The correct terminology is "brought to rest in" a frame.
That you don't understand the distinction marks you as a failure at
physics 101.
jbriggs444
>
> - Show quoted text -
I started the name calling when you became incoherent.
I've never seen you describe the asymmetric version, It is not the
version in this reply chain.
mpalenik
>
> - Show quoted text -
It is exactly what I was talking about in the first message I wrote
that Jem responded to.
I was telling Ste what happens if you accelerate one clock into the
other's frame vs. if you accelerate them symmetrically.
If you read carefully nothing I wrote is incoherent.
mpalenik
How would you phrase it in a non-cumbersome way--to say that two
objects are accelerated so that they are comoving? They entire term
"comoving frame" isn't questionable, and I think it's reasonably clear
what "brought into comoving frames" means.
>
> And it is. The correct terminology is "brought to rest in" a frame.
> That you don't understand the distinction marks you as a failure at
I avoided the term "rest frame" because that might imply the frame we
were talking about where both clocks were moving toward each other
with equal speed. Bringing them into that frame is not the only way to
make them comoving.
BURT
>
> - Show quoted text -
Light has a constant speed throught space.
Mitch Raemsch
mpalenik
In fact, you'll notice that in another reply to Ste, I said that if
both clocks accelerate in a symmetric way, they will read the same
time when they are at rest. I have never denied this.
BURT
>
> - Show quoted text -
Matter's motion must be created by acceleration through space of which
decelerates the clock rate. Entering gravity also will slow the
overall clock rate. One time rate can be going faster than the other.
There are two times.
Mitch Raemsch
jbriggs444
>
> - Show quoted text -
Let me back off, ditch some of the hostility and think through the
asymmetric scenario and see how it plays out without using such
notions as "being brought into" a frame of reference.
We have two clocks, at rest with respect to one another, synchronized
in the frame with respect to which they are both at rest.
We accelerate one toward the other. Eventually it arrives next to the
other. We look at their respective displays
at that point and see if they match.
1. I claim that any acceleration or lack thereof at or after the time
when the clocks arrive next to one another is totally and completely
irrelevant. I claim that the frame of reference, if any, that we
adopt when we make the final comparison is totally and completely
irrelevant.
1a. I thought I saw you making an opposite claim. Maybe I was wrong
about that.
2. I agree with you that the two clocks will show different displayed
times, even though they started out synchronized.
2a. To the extent that jem's claim that the two clocks will show the
same time applies to _this scenario_, I agree with you that he is to
be ignored.
One way to analyze this scenario is to immediately adopt the frame in
which the two clocks are moving with equal and opposite velocities.
With respect to the one clock, this involves a Lorentz transformation
with a (naively computed) relative velocity of v/2 in the direction of
the other clock.
With respect to the other clock, this involves a Lorentz
transformation with a (naively computed) relative velocity of -v/2 in
the direction of the other clock.
That's asymmetric. As it should be.
Without loss of generality, let's assume that the scenario started
when both clocks read zero.
In our new reference frame, the fact that the two clocks have a non-
zero separation in the direction of travel means that the time when
the one clock reads zero (as judged in our new frame) will not be
identical to the time when the other clock reads zero. (as judged in
our new frame).
The clocks have the same speed (albeit opposite directions). It
follows that (as judged in our new frame) they both tick at the same
rate.
o They started out out of synch.
o They tick at the same rate.
It follows that they are still out of synch when they arrive, side by
side somewhere in the middle.
This isn't a very friendly approach. It's only virtue is that it
retains a symmetry argument.
BURT
>
> - Show quoted text -...
There are two aether rates measured by one clock. They come from
motion and gravity strength.
Mitch Raemsch
jbriggs444
>
> - Show quoted text -
Bull fucking shit you incompetent moron.
"
> The relevant quantity was the time that each clock displays after the
> two are brought into comoving frames, which once again, depends on the
> frame that they are brought into.
"
That was incoherent. Observables are frame-independent, shit for
brains. Frames aren't something that you get brought into, loser.
mpalenik
Yes, I agree
> I claim that the frame of reference, if any, that we
> adopt when we make the final comparison is totally and completely
> irrelevant.
I agree
>
> 1a. I thought I saw you making an opposite claim. Maybe I was wrong
> about that.
No I did not. I think I slightly misunderstood Ste's original
statement and answered a different question--but I also thought the
question I was answering was pretty clear from my reply. When he
clarified, I answered his actual question in another message.
> 2. I agree with you that the two clocks will show different displayed
> times, even though they started out synchronized.
ok
>
> 2a. To the extent that jem's claim that the two clocks will show the
> same time applies to _this scenario_, I agree with you that he is to
> be ignored.
>
In this case, I think we are in agreement.
mpalenik
> Bull fucking shit you incompetent moron.
Can you calm down for two seconds?
>
> "> The relevant quantity was the time that each clock displays after the
> > two are brought into comoving frames, which once again, depends on the
> > frame that they are brought into.
>
> "
>
> That was incoherent. Observables are frame-independent, shit for
> brains. Frames aren't something that you get brought into,...
>
I'm talking about accelerating the clocks.
mpalenik
To clarify, once again, when I say that one clock is brought into a
comoving frame with the other, I am talking about accelerating the
clock so that its speed matches that of the other clock.
BURT
>
> - Show quoted text -
For every flow speed through space there is a slow motion time aether
rate by gamma.
Mitch Raemsch
BURT
>
> - Show quoted text -
Energy flow speed determines its aether rate.
mpalenik
>
> - Show quoted text -
Oh, and also, observables aren't necesarrily frame-independent.
Length, for example, is observable and not frame-independent. The
inner product of two 4 vectors is frame-independent. Rank zero
tensors are invariant. Not all observables are rank zero tensors. E
and B fields transform as a rank 2 tensor. They are observable. They
are not frame-independent.
For the record, I know what you meant, so I'm not going to call you
"shit for brains" for saying it. It would be nice if you could
maintain at least a minimum level of courtesy.
Peter Webb
>
> Well, let's see who backs you up.
In SR, time dilation is a real property, and it is observed every day in
particle accelerators and elsewhere. In SR, two twins, blah blah, one will
be older. Do not confuse this with gravitational time dilation under GR;
this is a completely different and independent thing, is orders of magnitude
smaller that SR time dilation for most real world experiments, and was not
predicted until well after the twins paradox was discussed and resolved in
SR.
BURT
wrote:
There are two rates or times in the universe; Gravity's strength and
changing flow of energy through space.
Mitch Raemsch
Peter Webb
Length, for example, is observable and not frame-independent. The
inner product of two 4 vectors is frame-independent. Rank zero
tensors are invariant. Not all observables are rank zero tensors. E
and B fields transform as a rank 2 tensor. They are observable. They
are not frame-independent.
For the record, I know what you meant, so I'm not going to call you
"shit for brains" for saying it. It would be nice if you could
maintain at least a minimum level of courtesy.
_______________________________
Here is a thought experiment that might help:
1. Visit your local zoo, and find the monkey enclosure.
2. Observe what they are doing. Some are playing with themselves, others are
throwing shit at each other, some are screeching just for fun.
3. Now start explaining tensor calculus to them through the bars of the
enclosure.
4. Note down any change in behaviour of the monkeys.
I suspect you may teach physics at Uni or similar. That means that you are
used to dealing with reasonably smart people, with some physics and maths
knowledge, and more importantly who want to learn. You are probably great at
it.
This is clearly not the case here. The people you are arguing with don't
want to learn, they want to screech and throw shit. They have no interest
whatsoever in learning anything.
Don't think you are doing them any favours by explaining physics to them.
They don't want to learn and won't. As they cannot play the "learn physics
game" with you, your only choice is to play the "shit throwing game" with
them. Do this because you enjoy it; don't pretend it will teach them
anything. If you don't enjoy the "shit throwing game", don't play it. If you
want to play the "learn physics" game, you will have more luck at Uni than
at the zoo.
Ste
>
> > > > - Show quoted text -
>
> > > Ignore Jem. He (she?) has obviously never studied physics and has no
> > > idea what SR does or doesn't say.
>
> > > Jem e-mailed me a little while ago asking me to make this "correction"
> > > for myself. As every physicist here knows, this "interpretation" of
> > > SR has nothing whatsoever to do with reality.
>
> > > That's all I'm going to say on the subject. I'm not going to get
> > > sucked into another discussion trying to explain SR to an obvious
>
> > > - Show quoted text -
>
>
> > jem:
> > "> > Actually, according to SR, the two (ideal) clocks will be synchronized
> > > > when they meet, regardless which clock "breaks the inertial frame" via
> > > > an instantaneous acceleration. That final acceleration, which occurs
> > > > when the clocks are co-located, has no effect on the clock readings.
>
> > "
>
> > jem is correct here.
>
> > That the clocks are synchronized follows by a simple symmetry
> > argument. The observation that the final acceleration does not affect
> > the respective clock readings is also true. (Do a Lorentz transform
> > when
> > the x offset in the direction of motion is zero and see what kind of t
> > offset you get.)
>
> > "> > Reported clock "slow downs" will be exactly offset by clock speed-ups
> > > > during the initial accelerations*.
>
> > "
>
> > jem is correct here in the sense that the elapsed time on B's clock as
> > accounted for in A's frame is
> > exactly the same as the elapsed proper time on A's clock. Speed up
> > and slow down exactly offset one another. The clocks agree when they
> > are reunited.
>
> > transform from a pair of tangent inertial
> > frames anchored at A to in order to compare "time now" at B with "time
> > a moment ago" at B while A is accelerating toward B, right?
>
> > Sweeping the hyper-plane of simultaneity forward on B's timeline has
> > an apparent speed-up effect, right?
>
> > mpalenik:
> > "> > > Yes, that is correct. Both will report a slow down. And in fact,
> > > > > which ever one breaks the inertial frame to match speed with the other
> > > > > is the one that will be "wrong". This is still within the realm of
> > > > > SR, not GR.
>
> > "
>
> > During part of the trip. And a different slowdown during a different
> > part of the trip.
>
> > But both will _also_ report a speed up during the period of initial
> > acceleration.
>
> > The question of which clock is "wrong" doesn't even arise. We have a
> > coherent accounting for of elapsed times from both points of view.
> > Those accounts may not agree on whose clock was running at what rate
> > at what time, but they do agree on all observables. The clocks
> > started synchronized and they finished synchronized.
>
> > clock was "really" slow during the tail end of the inertial portion of
> > their respective flights then you're being a butthead. That's not a
> > question of interest. It has no observable answer.
>
> > PD, Inertial or Eric Gisse can back me up on that if they wish.
>
> I think the issue is as I described in "plain English" to Ste.
>
> Let's take 3 clocks, one (A) which gets left behind, one (B) that
> sallies out at speed v for a distance L and returns, and one (C) that
> sallies out at speed v for a distance 2L and returns. At speed v, the
> clocks sent forth are running slow by, say, 2%. I'm really not going
> to worry about the accelerations at all because, as all have said, the
> acceleration profiles are all the same. Clock B will arrive back at
> home having run slow by 2% for the time of its trip and it is now 2
> hours behind clock A, say. Once B arrives home and comes to rest
> alongside A, its rate is no longer slower than A's by 2%, and so while
> it sits there and waits for C to come home it will *continue* to be 2
> hours behind A. Finally, C comes home, having run slower than A by 2%
> for twice as long as B did, and so it will be 4 hours behind A. Since
> B is only 2 hours behind A, clock B and clock C are no longer
> synchronized.
>
> What am I missing?
What confuses me is that, if the clocks run slow by 2% for all the
time that they are moving, how does one reconcile this with the fact
that, if one uses the frame of one of the moving clocks, say clock B,
then it seems to be to be your argument that there is no slowdown at
all for B, and it is the other clocks, A and C, that slow down (i.e.
*disregarding* both acceleration and propagation delays).
Peter Webb
>
> Let's take 3 clocks, one (A) which gets left behind, one (B) that
> sallies out at speed v for a distance L and returns, and one (C) that
> sallies out at speed v for a distance 2L and returns. At speed v, the
> clocks sent forth are running slow by, say, 2%. I'm really not going
> to worry about the accelerations at all because, as all have said, the
> acceleration profiles are all the same. Clock B will arrive back at
> home having run slow by 2% for the time of its trip and it is now 2
> hours behind clock A, say. Once B arrives home and comes to rest
> alongside A, its rate is no longer slower than A's by 2%, and so while
> it sits there and waits for C to come home it will *continue* to be 2
> hours behind A. Finally, C comes home, having run slower than A by 2%
> for twice as long as B did, and so it will be 4 hours behind A. Since
> B is only 2 hours behind A, clock B and clock C are no longer
> synchronized.
>
> What am I missing?
What confuses me is that, if the clocks run slow by 2% for all the
time that they are moving, how does one reconcile this with the fact
that, if one uses the frame of one of the moving clocks, say clock B,
then it seems to be to be your argument that there is no slowdown at
all for B, and it is the other clocks, A and C, that slow down (i.e.
*disregarding* both acceleration and propagation delays).
________________________________
A, B and C are not equivalent. A stays in one inertial reference frame, B
and C do not. You would be well advised to read a standard explanation of
the twin's paradox. http://en.wikipedia.org/wiki/Twin_paradox is OK, and if
you look at the first few paragraphs of "Resolution of the paradox in
Special Relativity" on that page it discusses exactly that point, and
explains its significance.
http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm is
similar.
jem
> Ignore Jem. He (she?) has obviously never studied physics and has no
> idea what SR does or doesn't say.
>
> Jem e-mailed me a little while ago asking me to make this "correction"
> for myself. As every physicist here knows, this "interpretation" of
> SR has nothing whatsoever to do with reality.
>
> That's all I'm going to say on the subject. I'm not going to get
> sucked into another discussion trying to explain SR to an obvious
> crank.
>
So I skimmed the responses, Palenik, and I didn't see anybody at all
backing you up on your answer to Ste's question - did I miss something?
At any rate, I did see both your acknowledgment that what I told you
was correct*, and your TRULY pathetic attempt to weasel-out-of rather
than own-up-to a fundamental misconception about SR ("I think I
slightly misunderstood Ste's original statement and answered a
different question")**.
But tell me, Palenik, now that it's dawned on you that you were wrong,
why haven't you publicly apologized for that juvenile outburst above,
where you showed your gratitude for my having tried to spare you some
embarrassment?
*
jbriggs444: 1. I claim that any acceleration or lack thereof at or
after the time when the clocks arrive next to one another is totally
and completely irrelevant. I claim that the frame of reference, if
any, that we adopt when we make the final comparison is totally and
completely irrelevant.
Palenik: Yes, I agree
**
Ste's original statement (emphasis added):
Ok. So let us suppose that we take two clocks. Separate them by a
certain distance, synchronise them when they are both stationary, and
then *accelerate them both* towards each other (and just before they
collide, we bring them stationary again). Are you seriously saying
that both clocks report that the other clock has slowed down, even
though *they have both undergone symmetrical processes*? Because there
is obviously a contradiction there.
Palenik's "slightly misunderstood" version of the original statement:
jem
>>>> - Show quoted text -
>>> Ignore Jem. He (she?) has obviously never studied physics and has no
>>> idea what SR does or doesn't say.
>>> Jem e-mailed me a little while ago asking me to make this "correction"
>>> for myself. As every physicist here knows, this "interpretation" of
>>> SR has nothing whatsoever to do with reality.
>>> That's all I'm going to say on the subject. I'm not going to get
>>> sucked into another discussion trying to explain SR to an obvious
>>> - Show quoted text -
>> What's the assertion under dispute?
>>
>> jem:
>>>> when they meet, regardless which clock "breaks the inertial frame" via
>>>> an instantaneous acceleration. That final acceleration, which occurs
>>>> when the clocks are co-located, has no effect on the clock readings.
>> "
>>
>>
>> That the clocks are synchronized follows by a simple symmetry
>> argument. The observation that the final acceleration does not affect
>> the respective clock readings is also true. (Do a Lorentz transform
>> when
>> the x offset in the direction of motion is zero and see what kind of t
>> offset you get.)
>>
>> jem:
>>>> during the initial accelerations*.
>> "
>>
>> accounted for in A's frame is
>> exactly the same as the elapsed proper time on A's clock. Speed up
>> and slow down exactly offset one another. The clocks agree when they
>> are reunited.
>>
>> You do understand about clock speed up seen when doing a Lorentz
>> transform from a pair of tangent inertial
>> frames anchored at A to in order to compare "time now" at B with "time
>> a moment ago" at B while A is accelerating toward B, right?
>>
>> Sweeping the hyper-plane of simultaneity forward on B's timeline has
>> an apparent speed-up effect, right?
>>
>> mpalenik:
>>>>> which ever one breaks the inertial frame to match speed with the other
>>>>> is the one that will be "wrong". This is still within the realm of
>>>>> SR, not GR.
>> "
>>
>> During part of the trip. And a different slowdown during a different
>> part of the trip.
>>
>> But both will _also_ report a speed up during the period of initial
>> acceleration.
>>
>> The question of which clock is "wrong" doesn't even arise. We have a
>> coherent accounting for of elapsed times from both points of view.
>> Those accounts may not agree on whose clock was running at what rate
>> at what time, but they do agree on all observables. The clocks
>> started synchronized and they finished synchronized.
>>
>> If you're trying to base this shouting match on the question of which
>> clock was "really" slow during the tail end of the inertial portion of
>> their respective flights then you're being a butthead. That's not a
>> question of interest. It has no observable answer.
>>
>> PD, Inertial or Eric Gisse can back me up on that if they wish.
>
>
> Let's take 3 clocks, one (A) which gets left behind, one (B) that
> sallies out at speed v for a distance L and returns, and one (C) that
> sallies out at speed v for a distance 2L and returns. At speed v, the
> clocks sent forth are running slow by, say, 2%. I'm really not going
> to worry about the accelerations at all because, as all have said, the
> acceleration profiles are all the same. Clock B will arrive back at
> home having run slow by 2% for the time of its trip and it is now 2
> hours behind clock A, say. Once B arrives home and comes to rest
> alongside A, its rate is no longer slower than A's by 2%, and so while
> it sits there and waits for C to come home it will *continue* to be 2
> hours behind A. Finally, C comes home, having run slower than A by 2%
> for twice as long as B did, and so it will be 4 hours behind A. Since
> B is only 2 hours behind A, clock B and clock C are no longer
> synchronized.
>
> What am I missing?
>
The pertinent scenario.
Peter Webb
>> That's all I'm going to say on the subject. I'm not going to get
>> sucked into another discussion trying to explain SR to an obvious
>> crank.
>>
>
> So I skimmed the responses, Palenik, and I didn't see anybody at all
> backing you up on your answer to Ste's question - did I miss something?
>
I backed him up. For what its worth. This isn't some voting game or
popularity contest.
> At any rate, I did see both your acknowledgment that what I told you was
> correct*, and your TRULY pathetic attempt to weasel-out-of rather than
> own-up-to a fundamental misconception about SR ("I think I slightly
> misunderstood Ste's original statement and answered a different
> question")**.
>
He was trying to teach you something. You were rude to him, and now he
apparently couldn't be bothered.
mpalenik
The one person that responded to this part of the thread said you were
wrong. The only other response was Peter Webb telling me not to
bother with you people, so I'll take his advice.
mpalenik
>
> - Show quoted text -
Jem, I went back and carefully read your e-mail--it seems you were
right, so I suppose I owe you an apology for that, however there are
still two issues:
1) your statement: "For clarity, both effects are purely
observational - SR presumes (ideal) clock mechanisms are completely
unaffected by a clock's motion." -- I agree the physical mechanism of
the clock is unaffected, but this is a really misleading statement,
since the amount of proper time that the clock consumes is affected by
its motion. Are you trying to say that the rate at which the clock
ticks is unaffected by its motion and is only an apparent affect? If
so I still disagree with what you're saying.
2) What you posted wasn't really a correction to what I wrote. I
wrote:
a) "Yes, that is correct. Both will report a slow down. And in
fact,
which ever one breaks the inertial frame to match speed with the
other
is the one that will be "wrong". This is still within the realm of
SR, not GR."
b) "Yes [they will be in synch]--assuming they both accelerated by the
same amount (that is to
say, assuming they both broke the inertial frame in a symmetric way).
Otherwise, they will register different times. "
Part b) was my answer to Ste's actual question--what happens when the
two clocks are brought to rest (in what we initially defined as our
rest frame). It is correct.
a) was my answer to the slightly misunderstood question but it is
still correct. First of all, Ste said the two clocks are brought to
rest just before they meet, which implies to me that they are not
exactly in the same place. In my first answer I had read it as a
question of what happens when we accelerate one clock to match speed
with the other. My answer is correct in this context.
I had just written a message to Ste trying to explain that you can
separate out the effects of actual time dilation predicted with SR
from the aparent time dilation due to the finite speed of light.
Let's say we have two moving clocks, A and B, moving toward each
other, as in the original setup. If we look at things from clock B's
frame, we can figure out the "actual" time on clock A by accounting
for the distance between clocks and the finite speed of light. If
clock A accelerates to meet clock B, clock B will always measure the
"actual" time on clock B in a way that is consistant with the time
dilation predicted by SR for a moving object. It will always say that
the relative rate at which clock A is ticking is 1/sqrt(1-v^2/c^2) IF
the observer at clock B corrects for delays due to the speed of light.
If clock A undergoes instantaneous acceleration to match the speed of
clock B, the observer at B will find that his measurements of the
actual time on clock A are consistant before and after the change in
velocity.
Now, let's look at things from the frame of clock A. Clock A can
calculate the "actual" time on clock B at any given instant. Now,
clock A accelerates to meet B. The rate that it calculates clock B
ticking as it accelerates will NOT be 1/sqrt(1-v^2/c^2), where v is
the relative velocities of the two clocks. At the end, clock B will
find that the time it measured for the actual time on clock A was
"incorrect".
In the context of instantaneous acceleration again, the actual
(corrected for light speed) time on clock B measured in clock A's
frame will be slighly earlier just after the acceleration than just
before.
The two clocks will be in synch if clock A instantaneously accelerates
to match clock B when they are at exactly the same point only because
the synching is performed in a different reference frame. This means
that when the two clocks start moving, even AFTER correcting for
propagation delays of light, they will be out of synch in either of
their respective frames, and this does NOT affect which one has
measured the "correct" rate that the other clock has been ticking.
The fact that they are exactly in synch when they cross is a special
case when the differences in rates and simultanaity both happen to
line up in just the right way that they both register the same time.
If one clock changes speed with any finite distance between the clocks
at all, it will have been "wrong" about the time on the other clock.
The clock that did not change speed, however, will be "right".
mpalenik
Reading this last paragraph again, I can see it's not very clear what
I'm saying. All I'm pointing out in this last paragraph is that in
clock A's frame and clock B's frame, they are initially out of synch
just after they start moving.
The case where clock A instantaneously accelerates to match B's speed
when the two are at exactly the same location is a special case.
If there is any finite distance between clock A and B when clock A
accelerates, clock B's measurements of time on clock A (after
correcting for propagation delays) will be consistant. Clock A's
measurements of time on clock B (again, after correcting for
propagation delays) will be inconsistant.
jem
I was "trying to say" exactly what I did say. If you didn't find it
clear enough, try this: relative slow-downs/speed-ups observed in the
readings of SR's ideal clocks aren't due to changes in the tick
mechanisms of those clocks.
that the rate at which the clock
> ticks is unaffected by its motion and is only an apparent affect? If
> so I still disagree with what you're saying.
>
> 2) What you posted wasn't really a correction to what I wrote. I
> wrote:
>
Don't worry about convincing me that you weren't wrong here, worry
about convincing yourself that you weren't wrong, if you were wrong.
Admitting mistakes is character building. To see what not admitting
mistakes builds - just take a look around SPR.
Inertial
news:imOkn.22019$wr5....@newsfe22.iad...
[snip heaps for brevity]
> I was "trying to say" exactly what I did say. If you didn't find it clear
> enough, try this: relative slow-downs/speed-ups observed in the readings
> of SR's ideal clocks aren't due to changes in the tick mechanisms of those
> clocks.
It is due to the different in simultaneity in difference frames of reference
(ie how the clocks are set). The clocks all tick at their 'correct' rate.
They are MEASURED as being slower. Just as a rod is MEASURED as being
shorter by a relatively moving observer (and for the same reason .. clock
settings).
eg A one metre rod in my inertial frame of reference is the same length as a
one metre rod in your inertial frame of reference. Its only when we try to
measure each others rods lengths, that clock settings get in the way. I see
you as looking at the positions of the ends of the rods at two different
times, and so am not surprised when you get the 'wrong' answer. You see the
same when I try to measure your rods. Its the same sort of situation with
clocks.
Peter Webb
>> 1) your statement: "For clarity, both effects are purely
>> observational - SR presumes (ideal) clock mechanisms are completely
>> unaffected by a clock's motion." -- I agree the physical mechanism of
>> the clock is unaffected, but this is a really misleading statement,
>> since the amount of proper time that the clock consumes is affected by
>> its motion. Are you trying to say
>
> I was "trying to say" exactly what I did say. If you didn't find it clear
> enough, try this: relative slow-downs/speed-ups observed in the readings
> of SR's ideal clocks aren't due to changes in the tick mechanisms of those
> clocks.
>
I still don't find it clear, as it begs the question - it says what doesn't
cause the change, not what does cause the change.
The standard SR answer is much more direct - the clocks slow down due to
relativistic time dilatation from them being in different reference frames.
Is that standard position of SR also your position? Or is your somehow
different?
mpalenik
The tick *mechanism* doesn't change but the tick *rate* does change.
Is your viewpoint different?
>
> that the rate at which the clock
>
> > ticks is unaffected by its motion and is only an apparent affect? If
> > so I still disagree with what you're saying.
>
> > 2) What you posted wasn't really a correction to what I wrote. I
> > wrote:
>
> Don't worry about convincing me that you weren't wrong here, worry
> about convincing yourself that you weren't wrong, if you were wrong.
> Admitting mistakes is character building. To see what not admitting
> mistakes builds - just take a look around SPR.
>
Your answer applies to one special case, when the two clocks are
exactly at the same location and one undergoes an instantaneous change
in velocity to match the speed of the other clock. In any other
situation, if there is any acceleration when the two clocks are not at
exactly the same spacetime point, the two clocks will not be in synch
(unless they undergo equal accelerations). If one clock accelerates
to match the speed of the other, however, and ANY accceleration occurs
when they are not at the exact same location, they will not be in
synch.
Agree or disagree?
mpalenik
> mpalenik wrote:
> I was "trying to say" exactly what I did say. If you didn't find it
> clear enough, try this: relative slow-downs/speed-ups observed in the
> readings of SR's ideal clocks aren't due to changes in the tick
> mechanisms of those clocks.
>
The tick *mechanism* doesn't change but the tick *rate* does change.
Is your viewpoint different?
> that the rate at which the clock
>
> > ticks is unaffected by its motion and is only an apparent affect? If
> > so I still disagree with what you're saying.
>
> > 2) What you posted wasn't really a correction to what I wrote. I
> > wrote:
>
> Don't worry about convincing me that you weren't wrong here, worry
> about convincing yourself that you weren't wrong, if you were wrong.
> Admitting mistakes is character building. To see what not admitting
> mistakes builds - just take a look around SPR.
>
Your answer applies to one special case, when the two clocks are
exactly at the same location and one undergoes an instantaneous change
in velocity to match the speed of the other clock. In any other
situation, if there is any acceleration when the two clocks are not at
exactly the same spacetime point, the two clocks will not be in synch
(unless they undergo equal accelerations). If one clock accelerates
to match the speed of the other, however, and ANY accceleration occurs
when they are not at the exact same location, they will not be in
synch.
Do you agree or disagree?
mpalenik
> On Mar 7, 9:03 am, jem <x...@xxx.xxx> wrote:
>
> > mpalenik wrote:
> > I was "trying to say" exactly what I did say. If you didn't find it
> > clear enough, try this: relative slow-downs/speed-ups observed in the
> > readings of SR's ideal clocks aren't due to changes in the tick
> > mechanisms of those clocks.
>
> The tick *mechanism* doesn't change but the tick *rate* does change.
> Is your viewpoint different?
>
To make this question a little bit more clear, do you agree that dTau/
dt is less than 1, where Tau is the proper time for the moving clock?
Do you agree that the moving clock actually ticks less than once every
second of time in the stationary clock's frame?
Inertial
news:4b93bf73$0$28464$afc3...@news.optusnet.com.au...
SR says that the difference in clock sync (clock settings) cause the
measurement of length to be contracted and measurement of clock ticking
rates to be dilated.
BURT
> "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote in message
>
> - Show quoted text -
There are no flat atoms in physics. Flat matter is a failure in
physics.
Mitch Raemsch
Peter Webb
"Inertial" <relat...@rest.com> wrote in message
news:4b942bcf$0$27789$c3e...@news.astraweb.com...
More or less.
But I asked you about *your* position, not SR's position.
Do you agree that that the clocks slow down due to relativistic time
dilation, as predicted by SR, or not?
Peter Webb
BURT
wrote:
> "Inertial" <relativ...@rest.com> wrote in message
>
> news:4b942bcf$0$27789$c3e...@news.astraweb.com...
>
>
>
>
>
> > "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote in message
>
> - Show quoted text -
Energy and field flow through space and have time or aether flowing
over them.
Atom and light flow through the hypersphere surface with time flowing
over them.
Mitch Raemsch
Inertial
That's what it is :)
> But I asked you about *your* position, not SR's position.
My position is SR's position
> Do you agree that that the clocks slow down due to relativistic time
> dilation, as predicted by SR, or not?
They are measured as slower, just as a rod is measured as shorter. This is
due to the difference in simultaneity. They don't slow down because a
moving observer is looking at them any more than a rod shrinks because a
relatively moving observer is looking at it.
Here's a little example you might follow .. with time differences exagerated
for clarity
Here are six clocks, in tow rows (S and S'), all ticking at the correct
rate, but set with different times...
S' 10:30 11:00=A 11:30 <--v
S 11:30=C 11:00=B 10:30 -->v
Clocks B sees the A is synchronized with it.
Now .. the clocks are moving in opposite directions so after an hour we have
S' 11:30 12:00=A 12:30
S 12:30=C 12:00=B 11:30
Clock A has moved away from clock B .. but another clock (C) in S can see
the time on it. Clock C sees that clock A is half an hour slow (A shows
12:00 when C shows 12:30). So according to the clocks in S, clock A is
ticking slower. We also note that clock B now sees a *different* S' clock
next to it as being fast (it shows 12:30 when B shows 12:00)
If you look at the same scenario but from the point of view of the other row
of clocks, you get symmetric results.
This is how clock synch affects measured ticking rates for moving clocks in
SR. Even though the clocks themselves do NOT change their intrinsic ticking
rates.
Bruce Richmond
> "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote in message
>
> news:4b943853$0$11336$afc3...@news.optusnet.com.au...
>
>
>
>
>
>
>
> > "Inertial" <relativ...@rest.com> wrote in message
> >news:4b942bcf$0$27789$c3e...@news.astraweb.com...
> >> "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote in message
Looks good, but let's take it one step further. The observer with
clock A jumps to frame S" which is traveling in the same direction as
S relative to S' but at twice the velocity.
S" 1:00 12:00=A 11:00 -->2v
S' 11:30 12:00=A 12:30 <--v
S 12:30=C 12:00=B 11:30 -->v
Clocks A and B continue to tick at there same intrinsic ticking rate
and an hour later A has overtaken B.
S" 2:00 1:00=A 12:00 -->2v
S' 12:30 1:00=A 1:30 <--v
S 1:30=C 1:00=B 12:30 -->v
The above provides the same situation as the twins paradox. Clock A
left clock B and returned. So why doesn't clock A show less time
elapsed than B? (Note the clocks in S" are further out of sync than
those in S due to the higher velocity.)
Peter Webb
"Bruce Richmond" <bsr...@my-deja.com> wrote in message
news:1c6a2640-39f5-4ea4...@33g2000yqj.googlegroups.com...
_________________________________________
I can't exactly follow your experiment or its conclusions.
But, if as you say, it is the same situation as the twins paradox, then
http://en.wikipedia.org/wiki/Twin_paradox should explain it.
More generally, once you understand the "normal" twins paradox, and the
diagrams on the web page, it is very simple to change this to model
additional frames-of-reference, either by simpy adding a third object to the
diagrams or considering the various objects in pairs.
If, OTOH, you don't understand the "vanilla" twins paradox, making it more
complicated won't help.
So, as a starting point, do you understand and accept the resolution of the
"vanilla" twins paradox as explained at
http://en.wikipedia.org/wiki/Twin_paradox and many other reputable sites?
Inertial
"Bruce Richmond" <bsr...@my-deja.com> wrote in message
news:1c6a2640-39f5-4ea4...@33g2000yqj.googlegroups.com...
The three clock situation cannot be so easily drawn .. bit like trying to
drawing a three dimensional figure in 2d :) This sort of diagram only
really works for a single pair of clocks looking from a third frame in which
they move with the same speed. Things are trickier when there is frame
jumping going on :):)
Bruce Richmond
wrote:
> "Bruce Richmond" <bsr3...@my-deja.com> wrote in message
>
> More generally, once you understand the "normal" twins paradox, and the
> diagrams on the web page, it is very simple to change this to model
> additional frames-of-reference, either by simpy adding a third object to the
> diagrams or considering the various objects in pairs.
>
> If, OTOH, you don't understand the "vanilla" twins paradox, making it more
> complicated won't help.
>
> So, as a starting point, do you understand and accept the resolution of the
For starters it wasn't my experiment. As for it being the same as the
twins paradox, clock A left clock B and then returned to clock B while
clock B stayed inertial the whole time. I think we can agree that is
a perfect description of the twins paradox.
I understand the "vanilla" twins paradox. It is often explained using
light clocks with a photon bouncing vertically between mirrors as the
twin moves horizontally. The moving photon follows a zig zag path
that is longer than the vertical path of the stationary twin,
resulting in fewer ticks during the trip than the stationary twin. In
that presentation the moving clock does tick slower.
Inertial started with clocks A and B moving in opposit directions at
the same speed, so their clock photons would plot symetric paths in
opposit directions, yeilding the same tick rate. Combine that with
the clock sync and you can show how the clock sync creates mutual time
dialation with the clocks tick rate unchanged by the motion of the
clock. But adding the turn around shows that the clocks are affected
by their motion. If clock A continues to tick at the same rate, as
Inertial claimed it would, then it would read the same as clock B when
they met up again. If OTOH you consider that the zig zag path of the
photon in clock A gets streched out further, resulting in a slower
tick rate, you get clock A showing less time than clock B when they
get back together.
I'm not trying to be a troll here. I agree that clock A should show
less elapsed time than clock B when they get back together. I also
agree that the primary cause for the measured slowing of moving clocks
is the syncronization of the clocks doing the measuring. But from
what I see that can't be the only reason or there would be no age
difference when the twin returns.
In the Wiki explaination it says of the turn around, "That is when he
must adjust his calculated age of the twin at rest." That calculation
doesn't change his age any more than the intrinsic length of a rod is
changed by an observer looking at it.
Bruce
Bruce Richmond
> "Bruce Richmond" <bsr3...@my-deja.com> wrote in message
>
> - Show quoted text -
The question still remains, if there is no change in the tick rate of
the clock, how can clock A have fewer ticks recorded when it is
brought back to clock B?
Bruce
Peter Webb
__________________________________
Excellent. You should have no problem.
Inertial started with clocks A and B moving in opposit directions at
the same speed, so their clock photons would plot symetric paths in
opposit directions, yeilding the same tick rate. Combine that with
the clock sync and you can show how the clock sync creates mutual time
dialation with the clocks tick rate unchanged by the motion of the
clock. But adding the turn around shows that the clocks are affected
by their motion. If clock A continues to tick at the same rate, as
Inertial claimed it would, then it would read the same as clock B when
they met up again. If OTOH you consider that the zig zag path of the
photon in clock A gets streched out further, resulting in a slower
tick rate, you get clock A showing less time than clock B when they
get back together.
___________________________________
I still don't understand the experimental design. Try explaining it without
your own editorial about zizagging photons, or your own predictions as to
what might result.
For example:
"Two clocks B and C are sent in opposite directions at 0.9c for 1 second, at
the end of the 1 second they reverse direction and head back to the origon
at 0.9c for another second, when they meet again at the origon, how would
their times compare with each other and a reference clock A that stayed at
the origin and didn't move"?
If that in fact is your question, then both B and C will show the same
elapsed time, which will be a bit over a second less elapsed time than shown
by clock A.
If that is not your question, if you are able to state it simply as per my
example I am happy to help.
I'm not trying to be a troll here. I agree that clock A should show
less elapsed time than clock B when they get back together. I also
agree that the primary cause for the measured slowing of moving clocks
is the syncronization of the clocks doing the measuring. But from
what I see that can't be the only reason or there would be no age
difference when the twin returns.
In the Wiki explaination it says of the turn around, "That is when he
must adjust his calculated age of the twin at rest." That calculation
doesn't change his age any more than the intrinsic length of a rod is
changed by an observer looking at it.
____________________________________
The turnaround is when things have to be recalculated, because he has
changed intertial frames. His age is a function of the inertial reference
frame in which it is measured. Exactly this happens every day when the age s
measured of unstable sub-atomic particles in pareticle accelerators or in
cosmic ray debris.
Bruce
Peter Webb
Bruce
_________________________________
Again, state your experimental design exactly. It should be quite easy to
say exactly what SR would predict happening. If you do not understand or
agree with what SR predicts, we can go through it step by step in the same
manner as for the vanilla twins paradox.
G. L. Bradford
==========================
Two clocks one light year apart that God certifies are in as perfect a
sync as if each of them were sitting in the other's space would each have a
year equal to 2010. But each observes the other to have the year 2009 as the
year on the clock, not 2010. That year 2009, not 2010, is any traveler's
starting block for any voyage to that destination ONE LIGHT YEAR FROM THE
TRAVELER'S DEPARTURE POINT!
You two like so many, many others totally ignore -- or are so totally
ignorant of -- the LIGHT-TIME (the LIGHT-HISTORY) universe surrounding you!
Light-time (light-history), light second by light second (second by
historical second), light year by light year (year by historical year) that,
really, would have to be accounted for in all travel. Even the slightest
fraction of a light second (the slightest fraction of a historical second)
has to be accounted for.....HAS TO BE PASSED OVER OR THROUGH! and thus
observed, second of history by second of history, by second of history!
Thus a ship departing planet A (March 2010) for planet B one light year
away (thus observed time-date for planet B being Mar 2009), taking one year
of ship's time to reach B (arrival scheduled for March 2011), has to deal
with a time period during its space travel of Mar 2009, Apr 2009, May
2009........, Jan 2010, Feb 2010, Mar 2010, Apr 2010, May 2010......, Jan
2011, Feb 2011, to its arrival at planet B one light year away and one year
later, March 2011. Ship's travel time per ship's clock, one year (Mar
2010-Mar 2011). Ship's travel time per closest observations made of the
space-time of the external environment, the light-time of the external
universe, two years (Mar 2009-Mar 2011).
Upon arrival at planet B, what does the traveler observe of planet A? He
observes Planet A one light year (or Mar 2010) in distance from planet B (or
Mar 2011) where he stands now. The same as any of planet B's homebodies who
never step foot, whatever, off planet.
Even if you are just going to walk down the street, the merest fraction of
a light second's distance, you are observing a historical universe (a
light-time universe) extending out in every direction from the departure
point, and you have to account for it each and every step of the walk. Your
starting point (vis-a-vis the universe of light) is not the time on your
watch or the time on the atomic clock you are standing next to, but the
space-time in the universe you would observe for your destination (-) from
where you stand (0). Arrival at your destination will have you at (0) from
that originally observed distant (-). You traveled the history (-) (the
light-time (-)) -- and the travel time -- all in the clocked time of your
travel from A to B.
In order to get from A to B, at all, you have to do some contracting,
followed by expanding, of the universe. Some contracting / expanding of
space and time (of space as well as time, both at once). Some contracting /
expanding of the SINGLE ENTITY of space-time (out there especially, the
single entity of light-time).
Yes travelers do leap forward into the future, but they don't do that leap
from the present. They leap forward into the future from the past. Any
competent observer should know what he is observing at any distance from
him, of course including all travelers, is history and not simultaneity. All
travelers observed to be coming on out of the distances of the universe are
observed to be fast forwarding (+) up through history (-) and cannot be
observed to be coming on present (0) to present (0).
All travelers observed to be going away are observed to be going away into
the past (-), going away into that same history (-) above, observably
becoming an observed piece (-) of that observed universe (-), and cannot be
observed to be going away present (0) to present (0).
Yet there are physicists who claim that all their observations of
travelers are observations of the present of the travelers no matter how
distant the travelers from them. And, the strangest and funniest thing of
all is that they observe, and describe, exactly the same picture for their
THERE-NOW "present" (0) of travelers that anyone else would be observing and
describing for a THERE-THEN distance (-) in space-time concerning exactly
the same travelers.
Our distant THERE-THEN universe (-), and the equally distant THERE-THEN
travelers in it (-), we observe and describe is exactly the same picture,
exactly the same universe and travelers, as those particular physicists'
no-distance-at-all (instantaneous) THERE-NOW universe (0), and the THERE-NOW
travelers in it (0), they purport to observe and describe. What differences
we observe for THERE-THEN and HERE-NOW, they claim exist for THERE-NOW and
HERE-NOW.
To go at it somewhat differently, ask them to describe the historical
space-time the historical traveler is observed to be in at 200 light seconds
from here and now, [versus] a forwardly projected picture of that traveler
200 seconds to the unobservable there-now space-time of the universe, and
they will not do it. They cannot do it. They cannot project the traveler
forward in space and time from a negative space-time picture. They cannot do
a "versus" picture because the two pictures are exactly one, and the same,
picture to them. They ascribe exactly the same negative time of the
historical picture observed of the distant traveler to the literal physical
being of the traveler who arrives home here on Earth.
GLB
============================
Inertial
Look at the Lorentz transforms to see. Its all due to clock synch.
jem
> On Mar 7, 9:03 am, jem <x...@xxx.xxx> wrote:
>> mpalenik wrote:
>> I was "trying to say" exactly what I did say. If you didn't find it
>> clear enough, try this: relative slow-downs/speed-ups observed in the
>> readings of SR's ideal clocks aren't due to changes in the tick
>> mechanisms of those clocks.
>>
>
> The tick *mechanism* doesn't change but the tick *rate* does change.
> Is your viewpoint different?
>
The tick rate of an ideal* clock doesn't change, by definition, since
SR presumes the mechanism that produces the ticks doesn't change.
If the accumulated tick counts on a pair of SR's ideal clocks differ,
it's not because those clocks' tick rates differed (which would be the
case if time were absolute), but because they experienced different
amounts of time by virtue of having traveled differently.
* SR is perfectly capable of dealing with non-ideal clocks, whose tick
mechanisms ARE affected by their motion (i.e. whose tick rates can
change), it's just that, in such cases, the model's standard equations
(e.g. the Lorentz Transformation) won't apply.
Give it a rest now, Palenik - the loose expressions "slow down" and
"speed up" used in this thread strongly imply that clock tick
mechanisms are being affected; my clarification merely pointed out
that's not the case.
>> that the rate at which the clock
>>
>>> ticks is unaffected by its motion and is only an apparent affect? If
>>> so I still disagree with what you're saying.
>>> 2) What you posted wasn't really a correction to what I wrote. I
>>> wrote:
>> Don't worry about convincing me that you weren't wrong here, worry
>> about convincing yourself that you weren't wrong, if you were wrong.
>> Admitting mistakes is character building. To see what not admitting
>> mistakes builds - just take a look around SPR.
>>
>
> Your answer applies to one special case, when the two clocks are
> exactly at the same location and one undergoes an instantaneous change
> in velocity to match the speed of the other clock. In any other
> situation, if there is any acceleration when the two clocks are not at
> exactly the same spacetime point, the two clocks will not be in synch
> (unless they undergo equal accelerations). If one clock accelerates
> to match the speed of the other, however, and ANY accceleration occurs
> when they are not at the exact same location, they will not be in
> synch.
>
> Do you agree or disagree?
And why are you asking this?
Is it because you need my confirmation that your answer would have
been correct if Ste's question had specified "only one clock
accelerates" rather than "both clocks accelerate", or is it that you
think my correction would somehow become less valid if I happened to
think the "special case" treatment applied across-the-board?
If it's the former reason, then, yes, the revised problem would reduce
to the latter half of the Twin Paradox and the clocks wouldn't be
synchronized when they meet. If it's the latter reason, then grow up.
jem
Do you agree that February has fewer than 29 days? Can you discern
the point and the relevance of these rhetorical questions?
Look, Palenik, my clarification (which I've now spent 3 posts
clarifying for someone who supposedly understands the SR model) wasn't
intended to explain SR but merely to prevent people who don't
understand SR (i.e. almost all of the SPR audience) from drawing the
wrong conclusions about what it says - which is exactly what happens
when expressions like "clock slow down" are left unclarified, because
to somebody who doesn't understand the model, "clock slow down" means
the clock is being affected, and in SR, it isn't. Capiche?
mpalenik
> mpalenik wrote:
> > On Mar 7, 9:03 am, jem <x...@xxx.xxx> wrote:
> >> mpalenik wrote:
> >> I was "trying to say" exactly what I did say. If you didn't find it
> >> clear enough, try this: relative slow-downs/speed-ups observed in the
> >> readings of SR's ideal clocks aren't due to changes in the tick
> >> mechanisms of those clocks.
>
> > The tick *mechanism* doesn't change but the tick *rate* does change.
> > Is your viewpoint different?
>
> The tick rate of an ideal* clock doesn't change, by definition, since
> SR presumes the mechanism that produces the ticks doesn't change.
>
> If the accumulated tick counts on a pair of SR's ideal clocks differ,
> it's not because those clocks' tick rates differed (which would be the
> case if time were absolute), but because they experienced different
> amounts of time by virtue of having traveled differently.
>
> * SR is perfectly capable of dealing with non-ideal clocks, whose tick
> mechanisms ARE affected by their motion (i.e. whose tick rates can
> change), it's just that, in such cases, the model's standard equations
> (e.g. the Lorentz Transformation) won't apply.
>
> Give it a rest now, Palenik - the loose expressions "slow down" and
> "speed up" used in this thread strongly imply that clock tick
> mechanisms are being affected; my clarification merely pointed out
> that's not the case.
>
>
Nobody was implying that the mechanism was affected by the velocity.
I wasn't implying that, and I didn't see anyone else implying that.
Can you show
>
>
>
> >> that the rate at which the clock
>
> >>> ticks is unaffected by its motion and is only an apparent affect? If
> >>> so I still disagree with what you're saying.
> >>> 2) What you posted wasn't really a correction to what I wrote. I
> >>> wrote:
> >> Don't worry about convincing me that you weren't wrong here, worry
> >> about convincing yourself that you weren't wrong, if you were wrong.
> >> Admitting mistakes is character building. To see what not admitting
> >> mistakes builds - just take a look around SPR.
>
> > Your answer applies to one special case, when the two clocks are
> > exactly at the same location and one undergoes an instantaneous change
> > in velocity to match the speed of the other clock. In any other
> > situation, if there is any acceleration when the two clocks are not at
> > exactly the same spacetime point, the two clocks will not be in synch
> > (unless they undergo equal accelerations). If one clock accelerates
> > to match the speed of the other, however, and ANY accceleration occurs
> > when they are not at the exact same location, they will not be in
> > synch.
>
> > Do you agree or disagree?
>
> And why are you asking this?
>
> Is it because you need my confirmation that your answer would have
> been correct if Ste's question had specified "only one clock
> accelerates" rather than "both clocks accelerate"
I made two posts. One dealt with what happens if only one clock
accelerates. The other dealt with what happens when two clocks
accelerate. You replied to the one that dealt with one clock
accelerating. In fact, as far as I can tell, your response deals with
one clock accelerating as well, since you say, "the two (ideal) clocks
will be synchronized when they meet, regardless which clock "breaks
the inertial frame" via an instantaneous acceleration." Again, you,
too, are referring to one clock accelerating.
, or is it that you
> think my correction would somehow become less valid if I happened to
> think the "special case" treatment applied across-the-board?
Given that we were BOTH talking about one clock accelerating, can you
show any point where anybody said anything about instantaneous change
in velocity or the clocks being colocated when the acceleration occurs
before your message?
I've already admitted that I was too hasty for saying you were wrong
and apologized for it, because what you posted is correct--however, my
post wasn't discussing what happens under instantaneous acceleration
when the two clocks are at the same location, and there is no reason
to believe it was.
>
> If it's the former reason, then, yes, the revised problem would reduce
> to the latter half of the Twin Paradox and the clocks wouldn't be
> synchronized when they meet. If it's the latter reason, then grow up.
>
I don't know why this is such a big deal to you. I already apologized
for saying you were wrong. I didn't read your message carefully--but
again, an instantaneous change in velocity when the two clocks were at
the exact same location was not what I was talking about.
mpalenik
>
Ok, but since I never wrote that, and I didn't think anyone else had
implied that either, I thought you were trying to correct something I
had already written in the thread. The statement "both effects are
purely observational" is also not very clear, since Ste had been
arguing that time dilation effects are purely observational in the
sense that they are only due to regular non-relativistic doppler
shifting. When I first saw that statement, I thought it was meant to
be an agreement with him, and I wanted to be absolutely sure what you
were saying.
mpalenik
> mpalenik wrote:
> > On Mar 7, 9:03 am, jem <x...@xxx.xxx> wrote:
> >> mpalenik wrote:
> >> I was "trying to say" exactly what I did say. If you didn't find it
> >> clear enough, try this: relative slow-downs/speed-ups observed in the
> >> readings of SR's ideal clocks aren't due to changes in the tick
> >> mechanisms of those clocks.
>
> > The tick *mechanism* doesn't change but the tick *rate* does change.
> > Is your viewpoint different?
>
> The tick rate of an ideal* clock doesn't change, by definition, since
> SR presumes the mechanism that produces the ticks doesn't change.
>
> If the accumulated tick counts on a pair of SR's ideal clocks differ,
> it's not because those clocks' tick rates differed (which would be the
> case if time were absolute), but because they experienced different
> amounts of time by virtue of having traveled differently.
Nobody was implying that the mechanism was affected by the velocity. I
wasn't implying that, and I didn't see anyone else implying that.
>
> * SR is perfectly capable of dealing with non-ideal clocks, whose tick
> mechanisms ARE affected by their motion (i.e. whose tick rates can
> change), it's just that, in such cases, the model's standard equations
> (e.g. the Lorentz Transformation) won't apply.
>
> Give it a rest now, Palenik - the loose expressions "slow down" and
> "speed up" used in this thread strongly imply that clock tick
> mechanisms are being affected; my clarification merely pointed out
> that's not the case.
Actually, we had specifically told Ste that this wasn't the case a
couple of times, already (i.e. the clock mechanisms are not
affected). This was (and may still be) his viewpoint, but several
posters have been working to correct this, and I don't think anyone
else involved in this thread thinks that is the case.
I made two posts. One dealt with what happens if only one clock
accelerates. The other dealt with what happens when two clocks
accelerate. You replied to the one that dealt with one clock
accelerating. In fact, as far as I can tell, your response deals with
one clock accelerating as well, since you say, "the two (ideal) clocks
will be synchronized when they meet, regardless which clock "breaks
the inertial frame" via an instantaneous acceleration." Again, you,
too, are referring to one clock accelerating.
Given that we were BOTH talking about one clock accelerating, can you
show any point where anybody said anything about and instantaneous
change in velocity and the clocks being colocated when the
acceleration occurs before your message?
I've already admitted that I was too hasty for saying you were wrong
and apologized for it, because what you posted is correct--however, my
post wasn't discussing what happens under instantaneous change in
velocity when the two clocks are at the same location, and there is no
reason to believe it was.
>
> If it's the former reason, then, yes, the revised problem would reduce
> to the latter half of the Twin Paradox and the clocks wouldn't be
> synchronized when they meet. If it's the latter reason, then grow up.- Hide quoted text -
>
I don't know why this is such a big deal to you. I already apologized
for saying you were wrong. I didn't read your message carefully--but
again, an instantaneous change in velocity when the two clocks were at
the exact same location was not what I was talking about.
Anyway, I'm done responding to you. This isn't a productive use of
time.
mpalenik
>
Ok, but I never wrote that, and I didn't think anyone else (except
maybe Ste) had implied that either. I thought you were trying to
correct something I had already written in the thread. The statement
"both effects are purely observational" is also not very clear in the
context of this thread, since Ste had been arguing that time dilation
mpalenik
>
Ok, but I never wrote that, and I didn't think anyone else (except
maybe Ste) had implied that either. I thought you were trying to
correct something I had already written in the thread. The statement
"both effects are purely observational" is also not very clear in the
context of this thread, since Ste had been arguing that time dilation
effects are "purely observational" in the sense that they are only due
to regular non-relativistic doppler shifting. When I first saw that
statement, I thought it was meant to be an agreement with him, and I
wanted to be absolutely sure what you were saying.
I'm sure that I'm most likely to get a nasty response back from you
about how I need to grow up and how absolutely important it is that I
admit that I was wrong for some reason--because admitting that you
were right isn't enough--but I have way too many things to do to
quibble over the language in a few stupid usenet posts. I've
explained what I wrote and either you will understand at this point,
or you'll continue to try to have an argument with me, and I've
apologized for telling you what you said was wrong, even if it wasn't
pertinent to what I had written (we all do things like that--just as
my first response to Ste was not wrong, even if it wasn't pertinent to
what he had written).
At this point, I don't think anything further can be accomplished by
this discussion. Say whatever you want. I'm done.
PD
> On 5 Mar, 19:12, PD <thedraperfam...@gmail.com> wrote:
>
>
>
> > On Mar 5, 12:54 pm, jbriggs444 <jbriggs...@gmail.com> wrote:
>
>
> > > > On Mar 5, 9:58 am, jem <x...@xxx.xxx> wrote:
>
> > > > > mpalenik wrote:
> > > > > when they meet, regardless which clock "breaks the inertial frame" via
> > > > > when the clocks are co-located, has no effect on the clock readings.
> > > > > Reported clock "slow downs" will be exactly offset by clock speed-ups
> > > > > during the initial accelerations*.
>
> > > > > (ideal) clock mechanisms are completely unaffected by a clock's motion.- Hide quoted text -
>
> > > > > - Show quoted text -
>
> > > > Ignore Jem. He (she?) has obviously never studied physics and has no
> > > > idea what SR does or doesn't say.
>
> > > > Jem e-mailed me a little while ago asking me to make this "correction"
> > > > for myself. As every physicist here knows, this "interpretation" of
> > > > SR has nothing whatsoever to do with reality.
>
> > > > That's all I'm going to say on the subject. I'm not going to get
> > > > sucked into another discussion trying to explain SR to an obvious
>
> > > > - Show quoted text -
>
>
> > > jem:
> > > > > when they meet, regardless which clock "breaks the inertial frame" via
> time that they are moving, how does one reconcile this with the fact
> that, if one uses the frame of one of the moving clocks, say clock B,
> then it seems to be to be your argument that there is no slowdown at
> all for B, and it is the other clocks, A and C, that slow down (i.e.
> *disregarding* both acceleration and propagation delays).
Be careful. The acceleration profiles are common between B and C, but
they are not common to A. So while there is no difference between B
and C due to the acceleration, you CANNOT say that the acceleration
has no effect whatsoever. In fact, it is the indisputable fact that B
and C accelerate and A does NOT accelerate that makes the situation
nonsymmetric for A. This is what makes the worldline for A straight,
and the worldline for B and C kinked.
Two places where I will try to intercept misconceptions.
1. The first temptation is to say, well, if the kink is what's
responsible for the time dilation, then all the dilating must happen
during the acceleration. That is not the case. Note the time dilation
is different for B and C, even though they have the same kink (the
same acceleration profile). The fact that there IS a kink is what
makes the elapsed time less on B and C than it is on A (where there is
no kink), but how much less depends on the steepness and length of the
straight parts of the worldline on either side of the kink.
2. The second temptation is to say, well, in B's frame it does not
accelerate, and A does accelerate. This is not correct. B *does*
accelerate and this is unambiguously detected (in much the same way
that you can fall asleep on a plane making 550 mph ground speed, but
will wake up when the plane throttles down to start the descent). In
the same way, the clock A *knows* it does not accelerate. Acceleration
cannot be made to disappear by choice of inertial reference frame.
Now, it IS fair to say that while the clock B is on its outward
journey at constant speed, it can look back at clock A and discern
that clock A is running slow relative to B. And the same goes on the
inbound journey. And so it's a fair question to say, how does it
happen then that by the time B lands back at A, it is BEHIND clock A?
And here is where things do get interesting in terms of what B sees in
terms of the clock reading on A in the turnaround.
You've seen, I suppose, this discussion of the twin puzzle on the web
Physics FAQ: http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/TwinParadox/twin_paradox.html
Bruce Richmond
>
> - Show quoted text -
You are going to give yourself a head cold with all that arm waving.
You are saying the tick rate doesn't change, yet SR says that the
returning twin's clock will show less elapsed time. You don't see any
conflict there?
Dono.
>
>
> You are going to give yourself a head cold with all that arm waving.
> You are saying the tick rate doesn't change, yet SR says that the
> returning twin's clock will show less elapsed time. You don't see any
> conflict there?
There is no conflict, the difference in elapsed time is due to the
differences in the length of the space paths. No clock tick rate
difference necessary. This is SR 101.
Peter Webb
You are going to give yourself a head cold with all that arm waving.
You are saying the tick rate doesn't change, yet SR says that the
returning twin's clock will show less elapsed time. You don't see any
conflict there?
___________________________________
You are playing word games here, and you know it. His meaning is clear to me
and I believe to you as well.
Just because English lacks words which refer specifically to relativistic
time dilation and the transformations which occur when you change reference
frames doesn't mean these aren't real; they can be easily, tersely and
unambiguously described mathematically as Einstein did in 1905.
Any "conflict" is due to the imprecision of English, not a logical conflict
in expected outcomes.
Inertial
Do you REALLY need to to give you the SR explanation of the twins paradox?
> You are saying the tick rate doesn't change,
That's what SR says. But a moving observer will measure the tick rate as
slower. Just like the intrinsic length of a rod doesn't change because a
moving observer measures it.
> yet SR says that the
> returning twin's clock will show less elapsed time.
It will
> You don't see any
> conflict there?
No .. do you?
jem
Come on now, I just finished telling you that the implication is in
the expressions "slow down" and "speed up".
Yes, well, when you send multiple responses to the same post it makes
it look like the earlier ones were sent in error.
One dealt with what happens if only one clock
> accelerates. The other dealt with what happens when two clocks
> accelerate. You replied to the one that dealt with one clock
> accelerating. In fact, as far as I can tell, your response deals with
> one clock accelerating as well, since you say, "the two (ideal) clocks
> will be synchronized when they meet, regardless which clock "breaks
> the inertial frame" via an instantaneous acceleration." Again, you,
> too, are referring to one clock accelerating.
>
> , or is it that you
>> think my correction would somehow become less valid if I happened to
>> think the "special case" treatment applied across-the-board?
>
> Given that we were BOTH talking about one clock accelerating, can you
> show any point where anybody said anything about instantaneous change
> in velocity or the clocks being colocated when the acceleration occurs
> before your message?
>
Please! Just read the original post - those conditions were OBVIOUSLY
intended.
> I've already admitted that I was too hasty for saying you were wrong
> and apologized for it, because what you posted is correct--however, my
> post wasn't discussing what happens under instantaneous acceleration
> when the two clocks are at the same location, and there is no reason
> to believe it was.
>
>> If it's the former reason, then, yes, the revised problem would reduce
>> to the latter half of the Twin Paradox and the clocks wouldn't be
>> synchronized when they meet. If it's the latter reason, then grow up.
>>
>
> I don't know why this is such a big deal to you.
Huh? I was ready to drop this several posts ago - you're the one
who's been pushing it.
I already apologized
> for saying you were wrong. I didn't read your message carefully--but
> again, an instantaneous change in velocity when the two clocks were at
> the exact same location was not what I was talking about.
But it is what the original problem was talking about. (Even so, I
hope you don't think there'd be a significant difference in the
outcome if the transition was just /almost/ instantaneous).
BTW, I wasn't looking for an apology for saying I was wrong. Whether
I was right or wrong, your little tirade was uncalled for, given that
I had just done you the favor of pointing out what I deemed was an
error, in a manner aimed at sparing you embarrassment.
jem
Are they working you too hard at PU? You wrote: "Both [clocks] will
report a /slow down/." And for consistency with your terminology, my
response indicated that your reported slow down would be offset by a
clock /speed up/. And those two expressions were the explicit objects
of my clarification.
and I didn't think anyone else had
> implied that either, I thought you were trying to correct something I
> had already written in the thread.
I wasn't trying to correct anything, I was trying to clarify
something, which is why I prefaced the attempt with "For clarity,".
The statement "both effects are
> purely observational" is also not very clear, since Ste had been
> arguing that time dilation effects are purely observational in the
> sense that they are only due to regular non-relativistic doppler
> shifting. When I first saw that statement, I thought it was meant to
> be an agreement with him, and I wanted to be absolutely sure what you
> were saying.
OK. And, hopefully, now you are.
Ste
> On Mar 6, 5:32 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > What confuses me is that, if the clocks run slow by 2% for all the
> > time that they are moving, how does one reconcile this with the fact
> > that, if one uses the frame of one of the moving clocks, say clock B,
> > then it seems to be to be your argument that there is no slowdown at
> > all for B, and it is the other clocks, A and C, that slow down (i.e.
> > *disregarding* both acceleration and propagation delays).
>
> Be careful. The acceleration profiles are common between B and C, but
> they are not common to A. So while there is no difference between B
> and C due to the acceleration, you CANNOT say that the acceleration
> has no effect whatsoever. In fact, it is the indisputable fact that B
> and C accelerate and A does NOT accelerate that makes the situation
> nonsymmetric for A. This is what makes the worldline for A straight,
> and the worldline for B and C kinked.
Yes, but we're supposed to have isolated the effect of acceleration,
and disregarded it. And in any event, the more important question is
the discrepancy between B and C.
> Two places where I will try to intercept misconceptions.
> 1. The first temptation is to say, well, if the kink is what's
> responsible for the time dilation, then all the dilating must happen
> during the acceleration. That is not the case. Note the time dilation
> is different for B and C, even though they have the same kink (the
> same acceleration profile). The fact that there IS a kink is what
> makes the elapsed time less on B and C than it is on A (where there is
> no kink), but how much less depends on the steepness and length of the
> straight parts of the worldline on either side of the kink.
As I say, I've stipulated that we are measuring on the outbound
journey, before any of the clocks have turned back. So we've had one
episode of acceleration and now B and C are travelling at the same
speed away from the origin point, but in opposite directions. What
amount of time dilation does C suffer relative to B? Nil? 2%? 4%?
> 2. The second temptation is to say, well, in B's frame it does not
> accelerate, and A does accelerate. This is not correct.
I know.
> Now, it IS fair to say that while the clock B is on its outward
> journey at constant speed, it can look back at clock A and discern
> that clock A is running slow relative to B. And the same goes on the
> inbound journey. And so it's a fair question to say, how does it
> happen then that by the time B lands back at A, it is BEHIND clock A?
> And here is where things do get interesting in terms of what B sees in
> terms of the clock reading on A in the turnaround.
No, the question is between B and C, not B and A. I'm using B and C
specifically in order to allow us to disregard any effects of
acceleration.
And we've already stipulated that, if B and C have the same
acceleration profiles and travel the same distance before returning to
A, then they will be in agreement with each other, but both will have
slowed relative to A. The question is, while they are still on the
outbound journey, what do B and C report about each other? Do they
both agree with each other still? Or not?
jem
Only one of us has produced a nasty response in this thread, my
friend, and it wasn't me.
> about how I need to grow up
that was a conditional comment - did you meet the condition?
and how absolutely important it is that I
> admit that I was wrong
I explicitly told you several posts ago, that I wasn't interested in
whether or not you were wrong. I merely suggested that you not lie to
yourself about it.
for some reason--because admitting that you
> were right isn't enough--but I have way too many things to do to
> quibble over the language in a few stupid usenet posts.
But that's exactly what you have been doing!
I've
> explained what I wrote and either you will understand at this point,
> or you'll continue to try to have an argument with me, and I've
> apologized for telling you what you said was wrong, even if it wasn't
> pertinent to what I had written (we all do things like that--just as
> my first response to Ste was not wrong, even if it wasn't pertinent to
> what he had written).
>
> At this point, I don't think anything further can be accomplished by
> this discussion. Say whatever you want. I'm done.
>
OK
Peter Webb
"Ste" <ste_...@hotmail.com> wrote in message
news:28440e7b-bffc-4d93...@z11g2000yqz.googlegroups.com...
I know.
___________________________________
What do you mean, "agree with each other"?
You can only ask what are the results of observation. You can ask questions
about what time clocks read, but not whether they with agree with each
other, as this depends upon the frame of reference in which you observe
them, and its not even defined anyway.
If you are describing a symmetrical arrangement then both will see the same
thing; relativity works the same way in a mirror image.
To the extent that I understand your question, when they observe each others
clocks when they are separating they will appear to run slowly, and then
speed up when they come back together, to the exact amount such that they
are back in sync.
They will calculate that time is running slowly for the other clock when
they are separating, because even accounting for the speed of light they
will see time run more slowly for the other clock, and conversely when
approaching.
JT
wrote:
> -
>
> You are going to give yourself a head cold with all that arm waving.
> You are saying the tick rate doesn't change, yet SR says that the
> returning twin's clock will show less elapsed time. You don't see any
> conflict there?
>
> ___________________________________
>
> You are playing word games here, and you know it. His meaning is clear to me
> and I believe to you as well.
>
> Just because English lacks words which refer specifically to relativistic
> time dilation and the transformations which occur when you change reference
> frames doesn't mean these aren't real; they can be easily, tersely and
> unambiguously described mathematically as Einstein did in 1905.
No there is no language conflict, the conflict is the claim that two
systems with variant units of different magnitude the d/t=v=c and d'/
t'=v'=c represent same velocities and that both are c, when in reality
c is variant.
PD
> On 8 Mar, 22:55, PD <thedraperfam...@gmail.com> wrote:
>
>
>
> > On Mar 6, 5:32 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > What confuses me is that, if the clocks run slow by 2% for all the
> > > time that they are moving, how does one reconcile this with the fact
> > > that, if one uses the frame of one of the moving clocks, say clock B,
> > > then it seems to be to be your argument that there is no slowdown at
> > > all for B, and it is the other clocks, A and C, that slow down (i.e.
> > > *disregarding* both acceleration and propagation delays).
>
> > Be careful. The acceleration profiles are common between B and C, but
> > they are not common to A. So while there is no difference between B
> > and C due to the acceleration, you CANNOT say that the acceleration
> > has no effect whatsoever. In fact, it is the indisputable fact that B
> > and C accelerate and A does NOT accelerate that makes the situation
> > nonsymmetric for A. This is what makes the worldline for A straight,
> > and the worldline for B and C kinked.
>
> Yes, but we're supposed to have isolated the effect of acceleration,
> and disregarded it.
No, we did not. We said that it cannot account for the DIFFERENCE
between B and C, but this does not discount or remove acceleration
from further consideration, particularly with regard to how clock A's
rate is seen by B.
> And in any event, the more important question is
> the discrepancy between B and C.
>
> > Two places where I will try to intercept misconceptions.
> > 1. The first temptation is to say, well, if the kink is what's
> > responsible for the time dilation, then all the dilating must happen
> > during the acceleration. That is not the case. Note the time dilation
> > is different for B and C, even though they have the same kink (the
> > same acceleration profile). The fact that there IS a kink is what
> > makes the elapsed time less on B and C than it is on A (where there is
> > no kink), but how much less depends on the steepness and length of the
> > straight parts of the worldline on either side of the kink.
>
> As I say, I've stipulated that we are measuring on the outbound
> journey, before any of the clocks have turned back. So we've had one
> episode of acceleration and now B and C are travelling at the same
> speed away from the origin point, but in opposite directions. What
> amount of time dilation does C suffer relative to B? Nil? 2%? 4%?
It's not that simple, and to get a number, you need to use the math.
Sorry, but that's the facts. The 2% is the Lorentz dilation factor,
and that is given by a function 1/sqrt(1-v^2/c^2), and you can see
that doubling the speed will not double the Lorentz dilation factor.
Furthermore, the speed of C relative to B is not twice the velocity of
B with respect to A, because the relation for combining velocities is
(v1+v2)/(1+v1*v2/c^2).
To find out what the Lorentz dilation factor is for C relative to B,
then you simply need to put in the numbers and crank.
Sue...
wrote:
> "Ste" <ste_ro...@hotmail.com> wrote in message
What makes your *stipulations* any more physical than
those of the theory's author?
<<After further consideration you cast a somewhat
disdainful glance at me—and rightly so—and you declare:
“I maintain my previous definition nevertheless, because
in reality it assumes absolutely nothing about light.
There is only one demand to be made of the definition of simultaneity,
namely, that in every real case it must
supply us with an empirical decision as to whether or
not the conception that has to be defined is fulfilled.
That my definition satisfies this demand is indisputable.
That light requires the same time to traverse the path
A —> M as for the path B —> M is in reality neither a
supposition nor a hypothesis about the physical nature
of light, but a stipulation which I can make of my own
freewill in order to arrive at a definition of simultaneity.” >>
http://www.bartleby.com/173/8.html
<<...one of Einstein's two main reasons for abandoning
special relativity as a suitable framework for physics
was the fact that, no less than Newtonian mechanics,
special relativity is based on the unjustified and epistemologically
problematical assumption of a preferred
class of reference frames, precisely the issue raised
by the twins paradox. Today the "special theory" exists
only, aside from its historical importance, as a convenient
set of widely applicable formulas for important limiting
cases of the general theory, but the epistemological
foundation of those formulas must be sought in the
context of the general theory. >>
http://www.mathpages.com/rr/s4-07/4-07.htm
Classical Electromagnetism:
An intermediate level course
Richard Fitzpatrick
Associate Professor of Physics
The University of Texas at Austin
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
Sue...
> So we've had one
[...]
PD
No, there is no conflict. When you say that the tick rate does not
change, this is a LOCAL statement. What it means is that a process
measured locally with this clock will still have the same duration.
For example, if the half-life of a muon that is slow in frame A and is
measured with a clock at rest in A is 2.2 us, then if you make the
same measurement of the half-life of a muon that is slow in frame B
and is measured with a clock at rest in B, clock B will still show the
half-life to be 2.2 us. In this sense, we say that the clock tick rate
has not changed, because measurements of local phenomena are
unchanged.
However, this does NOT mean that the tick rate of clock B will agree
with the clock rate of all other clocks, nor that it will read the
durations of nonlocal processes to be the same.
Do you see the distinction?
PD
Sue...
Indeed! The maths!
<< Einstein's relativity principle states that:
All inertial frames are totally equivalent
for the performance of all physical experiments.
In other words, it is impossible to perform a physical
experiment which differentiates in any fundamental sense
between different inertial frames. By definition, Newton's
laws of motion take the same form in all inertial frames.
Einstein generalized[1] this result in his special theory of
relativity by asserting that all laws of physics take the
same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
[1]<< the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal
properties, shows a pronounced relationship to the
three-dimensional continuum of Euclidean geometrical space.
In order to give due prominence to this relationship,
however, we must replace the usual time co-ordinate t by
an imaginary magnitude
sqrt(-1)
ct proportional to it. Under these conditions, the
natural laws satisfying the demands of the (special)
theory of relativity assume mathematical forms, in which
the time co-ordinate plays exactly the same rôle as
the three space co-ordinates. >>
http://www.bartleby.com/173/17.html
<< where epsilon_0 and mu_0 are physical constants which
can be evaluated by performing two simple experiments
which involve measuring the force of attraction between
two fixed charges and two fixed parallel current carrying
wires. According to the relativity principle, these experiments
must yield the same values for epsilon_0 and mu_0 in all
inertial frames. Thus, the speed of light must be the
same in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
Sue...
[...]
bert
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
It would not be a photon if it did not go at c It would not be a
photon if it bounced. TreBert
Ste
So anyway, the point is that C will be time dilated relative to B (and
the same for B relative to C). Yes?
Right, so if both are dilated relative to the other, then exactly how
do they equalise with each other again when they return to their
original location at A? By definition, all dilation effects measured
between B and C on the outbound trip, must be totally reversed on the
inbound trip (although we accept both will be dilated relative to A).
Yes?
PD
Yes.
>
> Right, so if both are dilated relative to the other, then exactly how
> do they equalise with each other again when they return to their
> original location at A?
Well, first of all, they don't, in the scenario you suggested, as I've
noted.
> By definition, all dilation effects measured
> between B and C on the outbound trip, must be totally reversed on the
> inbound trip (although we accept both will be dilated relative to A).
> Yes?
Yes. I'll take the symmetric case here, where they both travel outward
and back the same distance.
The details here are interesting, and again, it depends a little on
what's actually being measured. This is also covered and explained
nicely in the link that I gave you, so I do recommend you read that.
You have to first keep in mind that the B and C clocks are not in the
same place, so it takes some care to explain what you mean when you
say that B sees that clock C is running slow. How exactly is the
information exchanged and the propagation time corrected for? See the
links. In a nutshell, and under one such scenario, what B sees is the
following:
1. On the outward trip, C's clock is running slow, so that it is
behind B's clock by the time the outward journey ends.
2. On the turnaround, C's clock leaps forward so that it is *ahead* of
B's clock by the time the turnaround is complete.
3. On the inbound trip, C's clock is running slow, so that it ends up
showing the same time as B's clock by the time the inward journey
ends.
It's important for you to understand that the leap-ahead of C's clock
is not due to any physical process affecting clock C, because it is
the turnaround of *B* that is change that invokes it. Again,
understanding why this happens is in the details of how the
communication of the reading of the clock is performed (and
propagation accounted for). I'll not go into more detail here, as I've
already provided a link to a more thorough discussion.
PD
mpalenik
As a matter of fact, yes--my adviser doesn't have money to pay his
students as RA's (in fact, a lot of professors don't and the
department is getting upset about that and just sent out a letter
about some changes to their financial support policy), and so I'm also
a TA--but for some reason, they gave me a split assignment this
semester, which means that I'm teaching two different classes--twice
as many exams to proctor, twice as many weekly meetings to go to--not
to mention that the original assignment they gave me interfered with
my course schedule, I had to switch a recitation and help center hours
for another lab--which wouldn't be bad, except we give quizzes in lab,
so that means more grading.
Also, my adviser made me re-register for his class, since he said all
of his students have taken it more than once--so I have 15 hours this
semester, even though the maximum is supposed to be 12 and most people
have 9. He also loves having meetings, so by the time I can actually
sit down to do any of my work for more than a 30 minute window, it's
5:00 at night.
Most of the time when I post messages here, it's while I'm waiting for
a calculation to finish running, while I'm in the middle of some other
work, and I don't always have time to read carefully, and I try to get
my responses in quickly so it doesn't interfere with what I'm doing.
Sue...
==================
> It's important for you to understand that the leap-ahead of C's clock
> is not due to any physical process affecting clock C, because it is
> the turnaround of *B* that is change that invokes it. Again,
> understanding why this happens is in the details of how the
> communication of the reading of the clock is performed (and
> propagation accounted for). I'll not go into more detail here, as I've
> already provided a link to a more thorough discussion.
Indeed... one must allow for to loss to predatory organisms that
might
inhabit the communication path.
<< University of Hohenheim researchers are hoping to receive help
from nature in their fight against ticks. Worms, fungi and wasps
that feed on ticks will now be used to reduce the number of these
dangerous disease carriers.>>
http://www.bio-pro.de/magazin/wissenschaft/archiv_2007/index.html?lang=en&artikelid=/artikel/02099/index.html
Sue...
>
> PD
Ste
> On Mar 9, 11:14 am, Ste <ste_ro...@hotmail.com> wrote:
>
> Yes. I'll take the symmetric case here, where they both travel outward
> and back the same distance.
Yes, we are back to talking about the symmetric case, although on
reflection I might not have made that clear.
> The details here are interesting, and again, it depends a little on
> what's actually being measured. This is also covered and explained
> nicely in the link that I gave you, so I do recommend you read that.
> You have to first keep in mind that the B and C clocks are not in the
> same place, so it takes some care to explain what you mean when you
> say that B sees that clock C is running slow. How exactly is the
> information exchanged and the propagation time corrected for? See the
> links. In a nutshell, and under one such scenario, what B sees is the
> following:
> 1. On the outward trip, C's clock is running slow, so that it is
> behind B's clock by the time the outward journey ends.
> 2. On the turnaround, C's clock leaps forward so that it is *ahead* of
> B's clock by the time the turnaround is complete.
> 3. On the inbound trip, C's clock is running slow, so that it ends up
> showing the same time as B's clock by the time the inward journey
> ends.
When we talk of the "turnaround", can we be a bit more specific about
what is happening? Also, bear in mind that *both* turnaround at the
same time.
I also find it implausible that C could leap ahead of B. The more
plausible explanation, surely, is that B slows down dramatically
relative to C. So for example, if both clocks stopped moving (relative
to A) while at their farthest distance from each other, then for a
short time the other clock would appear to leap ahead (actually a
slowing of the reference clock), until the effects of each clock
stopping had actually propagated to the other clock, at which point
they would snap back into synchronisation again (propagation delays
disregarded). Yes?
PD
As I said, the specifics are in the link that I provided and I don't
want to replicate them here.
>
> I also find it implausible that C could leap ahead of B.
I'm sorry, but do not confuse "don't understand how" with
"implausible". If I tell you that the pressure at the bottom of a tank
of water doesn't depend on whether the cross-sectional area at the
bottom of the tank is bigger than the cross-sectional area at the top
of the tank, or the cross-sectional area at the top of the tank is
bigger than the cross-sectional area at the bottom of the tank, you
may not understand how this is possible, but that does not make it
implausible.
> The more
> plausible explanation, surely, is that B slows down dramatically
> relative to C.
No. Again, you are making a generalization from a comic-book
representation of special relativity that changes in readings in
clocks is due to (and only due to) the relative speed of the clock.
This is not the case, not even in SR.
Inertial
"Sue..." <suzyse...@yahoo.com.au> wrote in message
news:856f758f-9a80-4184...@b7g2000yqd.googlegroups.com...
> the three space co-ordinates. >>
> http://www.bartleby.com/173/17.html
>
> << where epsilon_0 and mu_0 are physical constants which
> can be evaluated by performing two simple experiments
> which involve measuring the force of attraction between
> two fixed charges and two fixed parallel current carrying
> wires. According to the relativity principle, these experiments
> must yield the same values for epsilon_0 and mu_0 in all
> inertial frames. Thus, the speed of light must be the
> same in all inertial frames. >>
> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>
> Sue...
Oh gawd.. its back .. the sue bot.
Peter Webb
"JT" <jonas.t...@hotmail.com> wrote in message
news:085eef06-a66f-444a...@t23g2000yqt.googlegroups.com...
______________________________
Shit, if you have experimental data that says that the speed of light is not
a constant in an intertial reference frame, but as you say is "variant" then
SR is wrong.
Of course, if have no such experimental data, then you are simply inventing
non-existent physical laws.
Peter Webb
When we talk of the "turnaround", can we be a bit more specific about
what is happening? Also, bear in mind that *both* turnaround at the
same time.
I also find it implausible that C could leap ahead of B. The more
plausible explanation, surely, is that B slows down dramatically
relative to C. So for example, if both clocks stopped moving (relative
to A) while at their farthest distance from each other, then for a
short time the other clock would appear to leap ahead (actually a
slowing of the reference clock), until the effects of each clock
stopping had actually propagated to the other clock, at which point
they would snap back into synchronisation again (propagation delays
disregarded). Yes?
____________________________________
No.
As I keep saying to you - but you fail to understand (which is itself a
worry) - is that statements like "the clocks are synchronisation" are not
really meaningful, as whether they show the same time is a function of the
inertial reference frame in which the clocks are observed; it is not a
function of the underlying dynamics, it is an artifact of the co-ordinate
system you choose.
That is why you can only ask questions about what is observed to happen for
a specific observer.
In answer to your question about "snapping back into synchronisation", apart
from the fact that it is meaningless, even if we reformulated this as a
question about observable quantities it is still wrong. When B and C become
relatively stationary, there is no sudden dramatic change to the clock to
the time they read on each others clocks; no jumps or discontinuities or
"snaps".
For more detail, consult, read and try to understand any standard
explanation of the twins paradox.
Uncle Al
[snip]
> No there is no language conflict, the conflict is the claim that two
> systems with variant units of different magnitude the d/t=v=c and d'/
> t'=v'=c represent same velocities and that both are c, when in reality
> c is variant.
[snip crap]
idiot
http://arXiv.org/abs/0706.2031
Physics Today 57(7) 40 (2004)
http://physicstoday.org/vol-57/iss-7/p40.shtml
Phys. Rev. D8, pg 3321 (1973)
Phys. Rev. D9 pg 2489 (1974)
<http://cfa-www.harvard.edu/Walsworth/pdf/PT_Romalis0704.pdf>
No aether
<http://relativity.livingreviews.org/Articles/lrr-2005-5/index.html>
Phys. Rev. D 81 022003 (2010)
http://arxiv.org/abs/0801.0287
http://arxiv.org/abs/0905.1929
No Lorentz violation
idiot
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
Bruce Richmond
That is not what we were discussing. I agree that the clock continues
to tick at a rate of one second per proper second in the rest frame of
the clock. The question was whether the slowed tick rate measured in
the frame of the stay at home twin is real, or an illusion due to the
clock sync proceedure, as length contraction is. If it is an illusion
the accumulated time on the two clocks should be the same when they
are brought back together. The way Inertial described it, it came
across as an illusion caused purely by clock sync.
I hate discussing what is "real". In a sense length contraction is
real because the pole will fit into the barn. But with length
contraction there is no accumulated length to inspect when the pole is
brought to rest. I can see that the measured slowing of the clock is
due to the rotation of the time coordinates when it changed frames,
but that doesn't make it any less real.
> For example, if the half-life of a muon that is slow in frame A and is
> measured with a clock at rest in A is 2.2 us, then if you make the
> same measurement of the half-life of a muon that is slow in frame B
> and is measured with a clock at rest in B, clock B will still show the
> half-life to be 2.2 us. In this sense, we say that the clock tick rate
> has not changed, because measurements of local phenomena are
> unchanged.
Yes, but again you are measuring locally, more or less in the rest
frame of the muon. We know that if we measure the half life of a fast
moving muon it is longer, as if its personal clock was ticking slower.
> However, this does NOT mean that the tick rate of clock B will agree
> with the clock rate of all other clocks, nor that it will read the
> durations of nonlocal processes to be the same.
>
> Do you see the distinction?
If we set up a light clock with a vertical bouncing beam, and have it
send a flash back to the stay at home twin every time the beam hits
the top, we can tell just how fast the clock is ticking. The signal
will be doppler shifted, but the total count is what it is. None of
the flashes drifted off and avoided detection. I am not claiming this
slowing is due to motion wrt an ether. It is due to the finite speed
of light and how we define/construct the coordinate systems.
Bruce