Re: A constant speed of light in all reference frames? Surely you can't be serious.
mpalenik
> 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).
>
> > 3) This does not explain why atomic clocks on a jet register different
> > times AFTER being brought to rest than do their counterparts which
> > have been at rest the entire time--or why the difference in time that
> > they register is exactly consistant with the predictions of
> > relativity.
>
> Indeed, because SR doesn't deal with acceleration.
SR deals with acceleration just fine. GR is only needed to describe
gravity.
I'll explain the thought experiment I posted later but I have to teach
in less than an hour.
Ste
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.
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.
Ste
What if they both "break the inertial frame"?
mpalenik
I don't know what happened to this thread, or why google registers
several different copies of this thread, each with only a few messages
in it, but I'm going to describe the thought experiment I posted
before in greater detail.
In order for the ^ to receive the pulse of light from both emitters at
the same time and half way between the two emitters, the light must be
emitted BEFORE the ^ is half way between the two emitters.
From the stationary frame, the two emitters emit light at the same
time and the ^ moves vertically until it is half way between the two
of them, when it receives both pulses.
But from the moving frame, the ^ sees the emitters emit light when it
is not half way between them, and then it sees the emitters move until
he is exactly half way between them. However, the speed of light is
the same in all frames, and unaffected by the motion of the emitters.
So the the emitter emit while one is closer to him than the other.
The fact that they are moving does not affect the way light
propagates. If the light signal from the closer one reaches him at
the same time as the light signal from the farther one, that means the
farther one must have emitted first. The fact that he is half way
between the emitters at the end doesn't matter, because in his frame,
it is the emitters that are moving, which does not affect the speed of
light.
mpalenik
Then whichever frame they both accelerate into will be the one that
has measured the "correct" time dilation.
Ste
So in other words, the clocks will register the same time, but will
have slowed in some "absolute sense"?
mpalenik
Yes--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.
Ste
Indeed.
> From the stationary frame, the two emitters emit light at the same
> time and the ^ moves vertically until it is half way between the two
> of them, when it receives both pulses.
Ok.
> But from the moving frame, the ^ sees the emitters emit light when it
> is not half way between them, and then it sees the emitters move until
> he is exactly half way between them. However, the speed of light is
> the same in all frames, and unaffected by the motion of the emitters.
>
> So the the emitter emit while one is closer to him than the other.
> The fact that they are moving does not affect the way light
> propagates. If the light signal from the closer one reaches him at
> the same time as the light signal from the farther one, that means the
> farther one must have emitted first. The fact that he is half way
> between the emitters at the end doesn't matter, because in his frame,
> it is the emitters that are moving, which does not affect the speed of
> light.
This seems implausible. You cannot possibly (that is, physically) have
the situation you describe, in the way you describe it. The receiver,
when it receives the pulse of light, cannot possibly be in more than
one position relative to the two sources. It either receives when the
sources are equidistant, or it receives when one source is closer. It
cannot receive when it is equistant *and* when one source is closer,
for that would be an obvious contradiction.
Ste
Agreed.
So let's explore an extension of this scenario. Let's say you have two
clocks, and you accelerate both of them up to a common speed, and
after they have travelled a certain distance, you turn them around and
return them to the starting point. The only difference is that one
clock goes a certain distance, and the other clock goes twice that
distance, but they *both* have the same acceleration profile - the
only difference is that one clock spends more time travelling on
inertia.
Obviously, one clock will return to the starting point earlier than
the other. But when both have returned, are their times still in
agreement with each other, or have they changed?
mpalenik
It isn't.
> It either receives when the
> sources are equidistant, or it receives when one source is closer.
The sources are equidistant.
> It
> cannot receive when it is equistant *and* when one source is closer,
> for that would be an obvious contradiction.
What you don't understand is the fact that the sources were not
equidistant when they emitted the pulse. From the frame of reference
of the moving ^, THAT is the location that is important. Even though
it is equidistant from the emitters when it RECIEVES the pulse, only
the location of the emitters when they SENT the pulse matters in the ^
frame, because in the ^ frame, it is the emitters that are moving,
which does not affect the speed the light is propagated.
mpalenik
They are not in agreement because one was traveling faster longer.
Acceleration only appears indirectly in SR. Velocity is what appears
in all the SR equations, it just happens that velocity is a function
of acceleration, which is why acceleration is important.
Also, I would avoid using the term "traveling on intertia". It just
physically isn't a very meaningful thing to say. You could simply say
"moving" are "traveling at a constant velocity," or something like
that.
PD
Agreement. Both of them will agree, but will be showing a time earlier
than a third clock that was left behind at the starting point.
mpalenik
Wait, maybe I'm confused by Ste's setup. Didn't he say that one
travels twice as far as the other? But then he also says that you
turn them both around and return them to the start after traveling a
certain distance. Have they moved different distances in his scenario
or not?
mpalenik
If they have the same acceleration profile but one travels twice as
far as the other because it is allowed to move at a constant speed
longer (as I understand his question--although he seems to contradict
this at the beginning), they will NOT be in agreement.
If he accelerates them up to a common speed and turns them both around
at the same time and stops them at the same time, so they both travel
the SAME distance, which is what he says at the beginning (but then
contradicts this later on), then the WILL be in agreement.
Ste
It does not travel faster, only longer. I assume that is what you
meant.
So what you're saying is that mere movement causes a time dilation
effect?
> Acceleration only appears indirectly in SR. Velocity is what appears
> in all the SR equations, it just happens that velocity is a function
> of acceleration, which is why acceleration is important.
>
> Also, I would avoid using the term "traveling on intertia". It just
> physically isn't a very meaningful thing to say. You could simply say
> "moving" are "traveling at a constant velocity," or something like
> that.
Indeed, it was just convenient to emphasise that no force was being
applied, whereas "constant velocity" seems to beg the question of
"relative to what"..
JT
>
> - Visa citerad text -
lol you framejumping grasshoppers have just have no idea what is
***REALLY*** going on have you. Don't forget u can always use the
fudgefactor.
JT
Ste
Oh dear. Mark contends otherwise.
mpalenik
I think he misunderstood your question. The way you worded it sounds
contradictory and you will get a different answer depending on how you
interpret it.
But it seems from your last message that I interpreted the question
correctly, in which case, my response is correct.
mpalenik
What I meant is that it is moving for a longer amount of time (i.e.
during the time when the other one is stopped, it is still moving).
>
> So what you're saying is that mere movement causes a time dilation
> effect?
Right.
>
> > Acceleration only appears indirectly in SR. Velocity is what appears
> > in all the SR equations, it just happens that velocity is a function
> > of acceleration, which is why acceleration is important.
>
> > Also, I would avoid using the term "traveling on intertia". It just
> > physically isn't a very meaningful thing to say. You could simply say
> > "moving" are "traveling at a constant velocity," or something like
> > that.
>
> Indeed, it was just convenient to emphasise that no force was being
> applied, whereas "constant velocity" seems to beg the question of
> "relative to what"..
If it's a constant velocity in one frame, it's a constant velocity in
every frame, even though that constant velocity is zero in one
particular frame. Constant velocity implies that no force is being
applied, since F = dP/dt.
Traveling on inertia kind of invokes the mideivil idea that something
is required to keep the object moving, as if the object contains some
kind of inertia that if taken away would cause it to stop.
Ste
One has travelled twice as far as the other, but both at the same
speed, with the effect that the clock that has travelled the shorter
distance returns to the start point before the other clock.
mpalenik
In that case, what I said is correct.
The two clocks have the same acceleration profile
They both accelerate up to speed v
One clock travels at speed v twice as long as the other
when returned to rest, the two clocks will display different times.
mpalenik
You figured me out! Damn it. I guess the days of the lie are over.
Pretty soon the physicists absolute control over government, politics,
and economics will come to an end. Damn you for uncovering our
secret. Damn you all to hell!
PD
My understanding is that both clocks accelerate up to speed v, but one
goes for x kilometers and the other goes for 2x kilometers before
turning around and returning at speed v. Naturally, one arrives back
in half the time it takes the other one.
Oh, I see the issue. When the first clock comes back, it has to stop,
and therefore its rate will be different than when traveling.
Yes, I agree now, the two clocks will read different times.
PD
PD
Right. I misunderstood. He's right. I was wrong.
Ste
Indeed, but it's neither here nor there where the sources where at the
time of emission.
The question is where the sources are at the time of reception, and at
that time they must be equidistant if both pulses were emitted at the
same time. No matter what frame you look at it from, the sources are
not equidistant at the time of emission.
bert
>
> - Show quoted text -
Photon has set speed period TreBert
mpalenik
Not in the rest frame. But it is in the moving frame.
>
> The question is where the sources are at the time of reception,
Not in the ^ frame. The fact that the sources have moved does not
affect the way the light propagates. They could run over the moons of
jupiter and back, and it wouldn't affect the way the light propagates.
Lets say that you have two sets of emitters, A and B. One set is
moving and the other set is not. If they both emit
They both emit like this
A B
A B
But the B set is moving like this
A
B
A
B
The light from the A set and B set propagates exactly the same way.
It doesn't matter that the B set has moved.
Ste
But as I say, acceleration is absolute, not relative, whereas a
measure of velocity could be interpreted to be relative.
> Traveling on inertia kind of invokes the mideivil idea that something
> is required to keep the object moving, as if the object contains some
> kind of inertia that if taken away would cause it to stop.
I was treating inertia simply as "the tendency of matter to maintain
its absolute velocity through space".
Ste
>
> lol you framejumping grasshoppers have just have no idea what is
> ***REALLY*** going on have you. Don't forget u can always use the
> fudgefactor.
Lol. One has to laugh.
PD
Movement relative to an observer.
PD
Where did you get that definition? It is terrible. There is no
absolute velocity through space.
mpalenik
> On 4 Mar, 18:04, mpalenik <markpale...@gmail.com> wrote:
>
> > On Mar 4, 1:00 pm, Ste <ste_ro...@hotmail.com> wrote:
>
> > If it's a constant velocity in one frame, it's a constant velocity in
> > every frame, even though that constant velocity is zero in one
> > particular frame. Constant velocity implies that no force is being
> > applied, since F = dP/dt.
>
> But as I say, acceleration is absolute, not relative, whereas a
> measure of velocity could be interpreted to be relative.
Constant velocity always implies zero acceleration. No matter what
that velocity is, no matter what frame you look at it from, constant
velocity implies zero acceleration.
And actually, acceleration is also relative under SR (but not under
Newton), although in a different way from velocity.
>
> > Traveling on inertia kind of invokes the mideivil idea that something
> > is required to keep the object moving, as if the object contains some
> > kind of inertia that if taken away would cause it to stop.
>
> I was treating inertia simply as "the tendency of matter to maintain
> its absolute velocity through space".
Which is kind of a mideivil idea. Now days we know that objects don't
need a tendency to keep moving. There simply is no absolute motion
and thus, no reason that an object should ever change its velocity
unless a force is acting on it.
BURT
>
> - Show quoted text -
Then it has a set kinetic energy.
Mitch Raemsch
Ste
So what you're saying (and I had recognised this problem before you
said it) is that it is the "original" position of emission that
matters?
And the "original" position changes depending on the frame (i.e. in
the source frame, the source does not move, whereas in the receiver
frame, the sources are constantly moving from their "original"
positions)?
Ste
Ok. So what you're (both) saying is that time dilation (in SR) is a
simple function of speed and distance, so that the quicker you travel
the more time dilates, and the further you travel the more time
dilates? And, to boot, you're saying that it's only *relative*
distance and speed that counts (i.e. there is no absolute measure of
movement in space)?
Peter Webb
about simultaneity, time dilation, length contraction and c being the same
in every inertial reference frame?
Learned anything yet?
Inertial
"Ste" <ste_...@hotmail.com> wrote in message
news:b7a59ca3-e362-4a7d...@d2g2000yqa.googlegroups.com...
Of course .. because that is where the light was emitted. What happens to
its emitter AFTER that makes no difference. The light is already 'out
there'
> And the "original" position changes depending on the frame (i.e. in
> the source frame, the source does not move, whereas in the receiver
> frame, the sources are constantly moving from their "original"
> positions)?
Yeup
mpalenik
>
> - Show quoted text -
Right. The sources send out one pulse at one particular point in
time. The only thing that matters is where they were located when
they sent out that pulse. That location is the "source" of the pulse.
mpalenik
>
Well, it's really only the relative velocity that matters. Distance
doesn't enter into the equations at all. It's just that time slows
down for the moving observer. So, lets say that an observer is moving
so that his clock has slowed down by a factor of two. After 10
seconds, his clock is 5 seconds behind. After 20 seconds, his clock
is 10 seconds behind. After 30 it is 15 seconds behind, etc.
Distance is not part of the equations, although the longer you are
moving, also, the farther you will have gone with respect to a
stationary observer.
And yes, it is only the relative velocities that matter.
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.
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.
PD
The time dilation *factor* (by what factor is the clock moving more
slowly) is a simple function of relative speed. The difference in the
time *elapsed* between the two clocks is also a function of the
relative distance.
This should make perfect sense to you. If a clock is running 2%
slower, then it is running 2% slower regardless of distance. But if,
as a result of running 2% slower, it falls behind 6 minutes after
running a certain amount of time, then it will fall behind 12 minutes
after running for twice as long.
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
> On Mar 5, 3:55 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > So what you're saying (and I had recognised this problem before you
> > said it) is that it is the "original" position of emission that
> > matters?
>
> > And the "original" position changes depending on the frame (i.e. in
> > the source frame, the source does not move, whereas in the receiver
> > frame, the sources are constantly moving from their "original"
> > positions)?
>
> Right. The sources send out one pulse at one particular point in
> time. The only thing that matters is where they were located when
> they sent out that pulse. That location is the "source" of the pulse.
Ok. But one observation I would make first is that, I presume, from
the source inertial frame, both the rising and falling edges of the
wave have the same origin. However, in the receiver inertial frame,
the rising edge does not have the same origin as the falling edge - so
there is a lack of symmetry between what is being described in these
inertial frames.
Secondly, we talk of the sources being in a "particular place" when
the pulse is emitted, and yet by your own argument they are not in a
"particular place" at all - in one frame, the sources are in the same
place at all times, and in the other frame, the sources are never in
the same place for more than an instant. So is it really meaningful to
talk of the "place of origin" of the source as a well-defined, single
point in space and time?
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).
Ste
> On Mar 5, 2:55 am, Ste <ste_ro...@hotmail.com> wrote:
>
>
>
>
>
> > On 4 Mar, 18:12, PD <thedraperfam...@gmail.com> wrote:
>
> > > On Mar 4, 12:04 pm, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > On 4 Mar, 17:46, PD <thedraperfam...@gmail.com> wrote:
>
> > > > > On Mar 4, 11:17 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > > > On 4 Mar, 16:49, mpalenik <markpale...@gmail.com> wrote:
>
> > > > > > > On Mar 4, 11:45 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > > > > > On 4 Mar, 16:32, mpalenik <markpale...@gmail.com> wrote:
>
> > > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > > > > > > Then whichever frame they both accelerate into will be the one that
> > > > > > > > > has measured the "correct" time dilation.
>
> > > > > > > > So in other words, the clocks will register the same time, but will
> > > > > > > > have slowed in some "absolute sense"?
>
> > > > > > > Yes--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.
>
> > > > > > Agreed.
>
> > > > > > So let's explore an extension of this scenario. Let's say you have two
> > > > > > clocks, and you accelerate both of them up to a common speed, and
> > > > > > after they have travelled a certain distance, you turn them around and
> > > > > > return them to the starting point. The only difference is that one
> > > > > > clock goes a certain distance, and the other clock goes twice that
> > > > > > distance, but they *both* have the same acceleration profile - the
> > > > > > only difference is that one clock spends more time travelling on
> > > > > > inertia.
>
> > > > > > Obviously, one clock will return to the starting point earlier than
> > > > > > the other. But when both have returned, are their times still in
> > > > > > agreement with each other, or have they changed?
>
> > > > > Agreement. Both of them will agree, but will be showing a time earlier
> > > > > than a third clock that was left behind at the starting point.
>
> > > > Oh dear. Mark contends otherwise.
>
> > > Right. I misunderstood. He's right. I was wrong.
>
> > Ok. So what you're (both) saying is that time dilation (in SR) is a
> > simple function of speed and distance, so that the quicker you travel
> > the more time dilates, and the further you travel the more time
> > dilates? And, to boot, you're saying that it's only *relative*
> > distance and speed that counts (i.e. there is no absolute measure of
> > movement in space)?
>
> The time dilation *factor* (by what factor is the clock moving more
> slowly) is a simple function of relative speed. The difference in the
> time *elapsed* between the two clocks is also a function of the
> relative distance.
>
> This should make perfect sense to you. If a clock is running 2%
> slower, then it is running 2% slower regardless of distance. But if,
> as a result of running 2% slower, it falls behind 6 minutes after
> running a certain amount of time, then it will fall behind 12 minutes
> after running for twice as long.
Agreed.
The question now is, if we agree that both clocks suffer time dilation
in this way, then when they return to the start point, how do they
each reconcile the fact that (after accounting for the effects of
acceleration) it ought to be the other clock which is slow, when in
fact one clock (the one that went furthest from the start point) will
be slower than the other? And a third clock, left at the start point,
will be running ahead of both?
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.
Peter Webb
> slower, then it is running 2% slower regardless of distance. But if,
> as a result of running 2% slower, it falls behind 6 minutes after
> running a certain amount of time, then it will fall behind 12 minutes
> after running for twice as long.
Agreed.
The question now is, if we agree that both clocks suffer time dilation
in this way, then when they return to the start point, how do they
each reconcile the fact that (after accounting for the effects of
acceleration) it ought to be the other clock which is slow, when in
fact one clock (the one that went furthest from the start point) will
be slower than the other? And a third clock, left at the start point,
will be running ahead of both?
_________________________
They know that the operations were not symmetric. Only one clock remained in
the same inertial reference frame throughout. The other two clocks spent
different amounts of time in at least 3 different inertial reference frames.
Everybody can see this is true, and so nobody expects that the clocks will
remain synchronised.
If you really want to understand the twin paradox, read
http://en.wikipedia.org/wiki/Twin_paradox and feel free to ask any questions
you may have. When you read and understand this, then you will understand
what is going on. The question with 3 clocks is not materially different to
that for two clocks, and it would be trivial to change their diagrams to
also include the third clock.
This web page obviously took far more time to put together than you can
expect that I or anybody else will provide in newsgroup message, and you are
not going to understand the twin paradox until you sit down and go through
an explanation similar to this. I note it involves no maths beyond the
equations of SR itself (simple algebra).
So, why don't you?
Peter Webb
"Ste" <ste_...@hotmail.com> wrote in message
news:e75f3463-ac9f-4b6c...@z4g2000yqa.googlegroups.com...
On 5 Mar, 14:46, mpalenik <markpale...@gmail.com> wrote:
> On Mar 5, 3:55 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > So what you're saying (and I had recognised this problem before you
> > said it) is that it is the "original" position of emission that
> > matters?
>
> > And the "original" position changes depending on the frame (i.e. in
> > the source frame, the source does not move, whereas in the receiver
> > frame, the sources are constantly moving from their "original"
> > positions)?
>
> Right. The sources send out one pulse at one particular point in
> time. The only thing that matters is where they were located when
> they sent out that pulse. That location is the "source" of the pulse.
Ok. But one observation I would make first is that, I presume, from
the source inertial frame, both the rising and falling edges of the
wave have the same origin. However, in the receiver inertial frame,
the rising edge does not have the same origin as the falling edge - so
there is a lack of symmetry between what is being described in these
inertial frames.
_____________________________________
No, they can still be symmetric, and the choice of origin is arbitrary. The
rising and falling edges of the photon's E field (which I assume you are
talking about) is a complete red herring in the context of this thread; this
is what gives rise to the Relativistic Doppler Effect, but you seem to be
discussing time dilation generally, and this is not tied to using light -
you would get the same answers using neutrinos for signalling, and as far as
we know they don't "wave" at all.
Secondly, we talk of the sources being in a "particular place" when
the pulse is emitted, and yet by your own argument they are not in a
"particular place" at all - in one frame, the sources are in the same
place at all times, and in the other frame, the sources are never in
the same place for more than an instant. So is it really meaningful to
talk of the "place of origin" of the source as a well-defined, single
point in space and time?
____________________________________
Well yes, it is, but in a somewhat more abstract sense than in Newton.
Individuals in their own co-ordinate systems in 3D space will measure these
differently, but they are the same point in spacetime viewed from different
angles.
Much earlier on, somebody replied to a question about the ladder/barn
paradox by pointing out that an 80 foot high ladder can fit under a 10 foot
high door by rotating it. As you live in a world where speeds near c are
uncommon, its as if every ladder you have ever seen is standing upright and
you had no idea they could be tilted.
You are now asking a question which use of this analogy will help answer. If
your question "is it really meaningful to talk about a single well defined
origin for an event" was translated as "is it really meaningful to talk
about the *height* of a ladder", the answer is no, because by tilting it you
can have whatever height you want. OTOH, if it is translated as "is it
really meaningful to talk about the *length* of a ladder", the answer is yes
because this is invariant; an 80 foot ladder lying flat on the ground is
still 80 feet long, even if it has almost zero height.
HTH. Really, I do hope this helps; if you take the time to frame half decent
questions civilly I will take the time to give you half decent answers,
civilly.
jem
> On Mar 5, 9:58 am, jem <x...@xxx.xxx> wrote:
>> 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.
>
> 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.
Ste
wrote:
> > This should make perfect sense to you. If a clock is running 2%
> > slower, then it is running 2% slower regardless of distance. But if,
> > as a result of running 2% slower, it falls behind 6 minutes after
> > running a certain amount of time, then it will fall behind 12 minutes
> > after running for twice as long.
>
> Agreed.
>
> The question now is, if we agree that both clocks suffer time dilation
> in this way, then when they return to the start point, how do they
> each reconcile the fact that (after accounting for the effects of
> acceleration) it ought to be the other clock which is slow, when in
> fact one clock (the one that went furthest from the start point) will
> be slower than the other? And a third clock, left at the start point,
> will be running ahead of both?
>
> _________________________
> They know that the operations were not symmetric. Only one clock remained in
> the same inertial reference frame throughout. The other two clocks spent
> different amounts of time in at least 3 different inertial reference frames.
> Everybody can see this is true, and so nobody expects that the clocks will
> remain synchronised.
Yes, but the important question here is whether they agree *after* the
effects of acceleration are taken into account. I mean, if we said
that each travelling clock slows by 2% when moving away from the start
point at a certain speed, then by rights both travelling clocks should
slow equally. Yes?
And yet, the assertion seems to be that each clock will consider
itself correct, while holding that the other clock has slowed by,
what, 4%?
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.
Peter Webb
As I understand your thought experiment, no.
In SR, time dilation is a function of relative speed and the time for which
they are moving at the speed. It is not a function of accleration.
A doesn't move. B moves at speed v for time t, and its clock will read x
behind A. C moves at speed v for time 2t, and its clock will read 2x behind
A.
> And yet, the assertion seems to be that each clock will consider
> itself correct, while holding that the other clock has slowed by,
> what, 4%?
No, not as I understand your question, anyway.
BTW the clocks do not *consider* themselves correct; unless they are broken
they *are* correct. There is no absolute time to use as a reference.
BURT
wrote:
> "Ste" <ste_ro...@hotmail.com> wrote in message
>
> - Show quoted text -
If clocks slow down they must start slowing from some fastest tick.
This would corespond to zero gravity and zero motion for the clock.
The two universal rates must go fastest at the beginning of time.
Mitch Raemsch
BURT
> On Mar 4, 1:10 pm, mpalenik <markpale...@gmail.com> wrote:
>
>
>
>
>
> > On Mar 4, 1:03 pm, JT <jonas.thornv...@hotmail.com> wrote:
>
> > > On 4 mar, 18:49, mpalenik <markpale...@gmail.com> wrote:
>
> > > > On Mar 4, 12:46 pm, PD <thedraperfam...@gmail.com> wrote:
>
> > > > > On Mar 4, 11:17 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > > > On 4 Mar, 16:49, mpalenik <markpale...@gmail.com> wrote:
>
> > > > > > > On Mar 4, 11:45 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > > > > > On 4 Mar, 16:32, mpalenik <markpale...@gmail.com> wrote:
>
> > > > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...@hotmail.com> wrote:
>
> > > > > > > > > Then whichever frame they both accelerate into will be the one that
> > > > > > > > > has measured the "correct" time dilation.
>
> > > > > > > > So in other words, the clocks will register the same time, but will
> > > > > > > > have slowed in some "absolute sense"?
>
> > > > > > > say, assuming they both broke the inertial frame in a symmetric way).
> > > > > > > Otherwise, they will register different times.
>
>
> > > > > > So let's explore an extension of this scenario. Let's say you have two
> > > > > > clocks, and you accelerate both of them up to a common speed, and
> > > > > > after they have travelled a certain distance, you turn them around and
> > > > > > return them to the starting point. The only difference is that one
> > > > > > clock goes a certain distance, and the other clock goes twice that
> > > > > > distance, but they *both* have the same acceleration profile - the
> > > > > > only difference is that one clock spends more time travelling on
> > > > > > inertia.
>
> > > > > > Obviously, one clock will return to the starting point earlier than
> > > > > > the other. But when both have returned, are their times still in
> > > > > > agreement with each other, or have they changed?
>
> > > > > Agreement. Both of them will agree, but will be showing a time earlier
> > > > > than a third clock that was left behind at the starting point.
>
> > > > travels twice as far as the other? But then he also says that you
> > > > turn them both around and return them to the start after traveling a
> > > > certain distance. Have they moved different distances in his scenario
> > > > or not?- Dölj citerad text -
>
> > > > - Visa citerad text -
>
> > > lol you framejumping grasshoppers have just have no idea what is
> > > ***REALLY*** going on have you. Don't forget u can always use the
> > > fudgefactor.
>
> > > JT
>
> > You figured me out! Damn it. I guess the days of the lie are over.
> > Pretty soon the physicists absolute control over government, politics,
> > and economics will come to an end. Damn you for uncovering our
> > secret. Damn you all to hell!- Hide quoted text -
>
> > - Show quoted text -
>
>
> - Show quoted text -
If light has a set speed of C it has kinetic energy of C always. But
this is wrong. Its energy comes from its wave frequency and not its
constant motion.
Mitch Raemsch
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
> 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?