Re: A constant speed of light in all reference frames? Surely you can't be serious.
PD
> 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:
>
> > > > > On 4 Mar, 16:20, mpalenik <markpale...@gmail.com> wrote:
>
> > > > > > 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.
>
> > > > > What if they both "break the inertial frame"?
>
> > > > 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.
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.
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.
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.
bert
>
> - Show quoted text -
Photon has set speed period TreBert
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.
BURT
>
> - Show quoted text -
Then it has a set kinetic energy.
Mitch Raemsch
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?
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.
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.
Ste
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
> 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?
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%?
Peter Webb
"Ste" <ste_...@hotmail.com> wrote in message
news:651a713d-7ae4-4048...@v20g2000yqv.googlegroups.com...
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
>
> - 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
Ste
wrote:
> "Ste" <ste_ro...@hotmail.com> wrote in message
The question is this. We'll deal with only the outbound trip (in other
words, the clocks are on the move, but time 't' has not yet elapsed,
so there has been no further acceleration). I agree with your answer
above, as it concerns A's frame.
The question is, from the frame of B, what will the slowdown be on
clock C, *after* having accounted for the increased distances between
them (i.e. having accounted for the increased propagation delays). It
seems to me that the natural answer is to say "4%".
PD
You asked this question better in a different subthread, and so I'll
answer it there.
Peter Webb
Did you look at the diagrams on the Wikipedia page on the twins paradox as I
suggested?
This shows *exactly* what the moving and stationary clocks see as happening
at all stages of the thought experiment.
If you didn't understand it, go through it again, and identify where you
start having problems, and I will be happy to help ... unfortunately if you
want to understand the Twins Paradox fully, you will need to do some work
yourself.
Ste
> suggested?
>
> This shows *exactly* what the moving and stationary clocks see as happening
> at all stages of the thought experiment.
This isn't the twins paradox, so it would be strange to find the
answer to my question there. Also, I've read that page in the past,
and I don't recall it having relevant detail.
> If you didn't understand it, go through it again, and identify where you
> start having problems, and I will be happy to help ... unfortunately if you
> want to understand the Twins Paradox fully, you will need to do some work
> yourself.
We are not even talking about the twins paradox.
Inertial
> On 9 Mar, 05:34, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
> wrote:
>> "Ste" <ste_ro...@hotmail.com> wrote in message
>>
>> Did you look at the diagrams on the Wikipedia page on the twins paradox
>> as I
>> suggested?
>>
>> This shows *exactly* what the moving and stationary clocks see as
>> happening
>> at all stages of the thought experiment.
>
> This isn't the twins paradox,
http://en.wikipedia.org/wiki/Twin_paradox
Of course it is the twins paradox. Do you even know what the twins paradox
is ? Lets see what the web page says
"In physics, the twin paradox is a thought experiment in special relativity,
in which a twin makes a journey into space in a high-speed rocket and
returns home to find he has aged less than his identical twin who stayed on
Earth. This result appears puzzling because each twin sees the other twin as
traveling, and so, according to the theory of special relativity,
paradoxically each should find the other to have aged more slowly. How the
seeming contradiction is resolved, and how the absolute effect (one twin
really aging less) can result from a relative motion, can be explained
within the standard framework of special relativity. The effect has been
verified experimentally using precise measurements of clocks flown in
airplanes.[1][2]"
> so it would be strange to find the
> answer to my question there. Also, I've read that page in the past,
> and I don't recall it having relevant detail.
Clearly you are either lying about reading it, or you didn't understand it.
Ste
I repeat myself again, the scenario we have here is *not* the twins
paradox.
> > so it would be strange to find the
> > answer to my question there. Also, I've read that page in the past,
> > and I don't recall it having relevant detail.
>
> Clearly you are either lying about reading it, or you didn't understand it.
No, perhaps you didn't understand. As I say, this is *not* the twins
paradox, because in the twins paradox only *one* twin leaves Earth.
Inertial
Then what are you talking about now?
>> > so it would be strange to find the
>> > answer to my question there. Also, I've read that page in the past,
>> > and I don't recall it having relevant detail.
>>
>> Clearly you are either lying about reading it, or you didn't understand
>> it.
>
> No, perhaps you didn't understand. As I say, this is *not* the twins
> paradox, because in the twins paradox only *one* twin leaves Earth.
Then what are you talking about now?
Peter Webb
No, perhaps you didn't understand. As I say, this is *not* the twins
paradox, because in the twins paradox only *one* twin leaves Earth.
________________________
Its functionally the same. It is exactly the twins paradox, but with two
twins apparently doing exactly the same thing.
Even if you cannot see that, the explanation on the Wikipedia page of the
Twins Paradox is trivially adapted for two twins.
I assume that you do not understand the Wikipedia twins paradox page, or
else you would know the answers to your questions already. Which parts don't
you understand?
Ste
Perhaps you should review the previous postings, but as a quick
summary, this scenario essentially involves *both* twins leaving
Earth, in diametrically opposite directions. Hence when the twins
return to Earth, they are the same age as each other (although both
younger than a third sibling who remained on Earth).
Ste
wrote:
Let's just go through it step by step Peter, as we have been doing.
It's pointless spending 10 more postings arguing about how the
Wikipedia page does or does not answer the question, or how it is or
is not relevant. As I've just said in a post to Inertial, the only
analogy between my scenario and the twins paradox is that, in my
scenario, both twins leave Earth, and both return the same age as each
other - hence no paradox, and hence bearing no resemblance at all to
the twins paradox.
Peter Webb
Well, I never thought there was a paradox, and as you agree, I have to
wonder what you are going on about.
If you do still have a problem, then the easiest way is by drawing a diagram
like the ones on http://en.wikipedia.org/wiki/Twin_paradox . That is hard to
do on usenet.
The only other way of answering these questions is to use the equations of
SR, eg
If one twin travels at 0.9c for one year then his elapsed time in his frame
of reference is sqrt(1-0.9^2) maybe 0.4 years, and that will show on his
clock, and that is the time that other people see his clock as showing.
I don't know if that will answer your question; if you really want to answer
it you will need to understand the diagrams on the web page.
PD
First of all, let's establish what you think is paradoxical at all
about the description of the twins in the twin puzzle. Then let's see
whether this paradox is present in the case you mention.
Ste
As I understand it, the supposed "paradox" in the twins paradox was
that one returned younger than the other. It was, of course, not a
paradox at all, but that's besides the point.
In our scenario however, we have already agreed that both clocks
return to the origin displaying the same time, hence there is no
correspondence at all with the twins paradox.
PD
No, then you do not understand the paradox, because there is nothing
contradictory in that statement at all. It may be surprising, but it's
not contradictory, not paradoxical. Disagreement of clocks is not a
paradox.
The paradox, which is what is perceived (normally) by freshmen when
first introduced to this statement, is embodied in their immediate
classroom question: "But in the frame of the traveling twin, it is the
earth twin that is moving away and returning. Since this is symmetric
to the case of the traveling twin moving away and returning, then
shouldn't the traveling twin expect the earth twin to be younger when
they meet again?" Now perhaps the paradox is more apparent to you.
However, the puzzle is specifically designed to emphasize the danger
of oversimplifying. In fact, the two twins are NOT symmetric, because
one unambiguously experiences acceleration and the other unambiguously
experiences no acceleration. This then leads to a discussion of what
produces the asymmetry in the time.
Perhaps if you had started out by asking, "Since I don't see any
obvious paradox here at all, perhaps someone could illuminate me as to
why this is called the twin paradox?" Then at least you would have
been on square one.
Inertial
So the same twins scenarion .. Just doing it twice at the same time.
You can have as many 'twins' leaving earth in as many directions as you
like. it doesn't change that for each of them the same explanation of the
twins paradox applies.
So the posts referring to the wikipedia pages on the twins paradox is
perfectly applicable .. twice.
G. L. Bradford
"PD" <thedrap...@gmail.com> wrote in message
news:160d014f-cb0a-4f4a...@q16g2000yqq.googlegroups.com...
=========================
It's not a 4-d || 4-d scenario....to a 4-d result! It's strictly a 1-d ||
1-d scenario, so naturally the result comes out strictly 1-d.
GLB
========================
Ste
I know Paul. I know.
> Perhaps if you had started out by asking, "Since I don't see any
> obvious paradox here at all, perhaps someone could illuminate me as to
> why this is called the twin paradox?" Then at least you would have
> been on square one.
Really I just wanted to avoid going off on a long tangent about the
twins paradox. As I said, the scenario that were were addressing is
different from the twins paradox, in that we have three clocks, and
the two clocks with which we are now concerned (B and C) both return
to the origin point *synchronised* (albeit both lagging behind A),
whereas the twins' ages are not synchronised on the return of the
astronaut twin.
So let me say again. The twins paradox would be applicable if we were
talking about A and B, or A and C. In the event, we are talking about
what B and C observe of each other from their own reference frames.
There is, therefore, no correspondence with the twins paradox, because
unlike the twins, B and C return synchronised with each other.
PD
You can imagine my surprise, since what you said explicitly above was
that the paradox was that the twins aged differently.
>
> > Perhaps if you had started out by asking, "Since I don't see any
> > obvious paradox here at all, perhaps someone could illuminate me as to
> > why this is called the twin paradox?" Then at least you would have
> > been on square one.
>
> Really I just wanted to avoid going off on a long tangent about the
> twins paradox. As I said, the scenario that were were addressing is
> different from the twins paradox, in that we have three clocks, and
> the two clocks with which we are now concerned (B and C) both return
> to the origin point *synchronised* (albeit both lagging behind A),
> whereas the twins' ages are not synchronised on the return of the
> astronaut twin.
Well, yes, it is a different result from the application of the same
principle.
If I calculate the angle that I can tip a TV tray before the coffee
cup on it starts to slide, I find that I'm using the same principle
(equilibrium of forces) that I would use to determine the tension in
picture-hanger wire when mounting a photo on the wall.
Different situation. Very same principle.
Fine example of losing the forest for the trees. As you've done here.
ben6993
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
I have written below a twin plank paradox of length. I know the
explanation must really be very straightforward, but I don't see it
yet. This is not a time paradox, but it looks to be a similar issue.
A plank has four units of length (----) in its own frame and a garage
has two units of length (--) in its own frame.
There are two similar planks and garages in relative motion such that
plank P and garage G are moving fast towards the other plank p and
garage
g. The relative speed is such that lengths are approximately halved
in relativistic contraction.
<--- direction of motion
p ---- G' -
g -- P' --
Where ' sign indicates motion in the other twins' frame.
Here, p does not fit within G', but P' fits approx. within g.
Looked at from the other frame, below, a similar result occurs: p'
fits approx. within G, but P does not fit within g'.
---->
p' -- G --
g' - P ----
As the same two events are looked at twice, i.e. from two frameworks,
there are four outcomes in total. The shed doors are destroyed in two
outcomes and the planks fit into the sheds in the other two outcomes.
So it looks like one of the two sheds is damaged and the other is
safe. But that cannot be true as the motion is relative not absolute,
and the planks and garages are twins, and so the two outcomes should
be identical.
Inertial
"ben6993" <ben...@hotmail.com> wrote in message
news:716c1760-a5db-4ce8...@o30g2000yqb.googlegroups.com...
They are identical if you do identical things . If you do the 'close both
garage doors simultaneously in the garage's rest frame of reference when the
pole is fully within' for both garages, both are safe.
If you close the door at different times in their frame, then they may not
be safe (depending on when you do)
spudnik
to promulgate the ideal of travel "in" time, when neither is possible;
Heinlein wrote a realistic novel with twins on this, except that
they got FTL, in the middle of it.
thus:
Einstein was very far from perfect, as any examination
of his biographers will show; nor was his theorizing. insofar as
The
department of Einsteinmania/The Musical Department is such
a large section of "Science" in the bookstores & libraries,
it is certainly true what Tom said; unfortunately,
Nein Ein Stein is stuck in Minkowski's little box of "times-pace,"
a serious piece of junk from an otherwise great geometer.
> After viewing your post, as the second bona fide "kook" (although I
> think "idiot" is a better description) to post on this topic, it is
> interesting that both you and the other moron, Tom Potter, are
thus:
he actually seemed to state that
the paper may have been mistranslated, as well.
> 1. E's paper was not written clearly.
> a. That's because he revised it at the last pre-publication moment.
thus:
get with the program; Einstein wasn't perfect, and
Minkowski, like anyone else, put his pants on ...
one lightcone at a a time!
thus:
that should be, "Fossilized Fuel (TM)" and *sic*;
it is simply a trade-name of oilolgical mythology
(sediments piling-up in the ocean create enough pressure
to create oil ... even though this happens, continuously,
since time immemorial).
thus:
read _The Big Bang Never Happened" by Eric Lerner,
student of the late Johannes Alfven (I have not).
thus:
what a bunch of silliness. the only real question is,
how much energy was in "de planes,"
compared to the rather small amount that is required
for a "controlled demo?" that is to say,
were the planes not adequate bombs?
thus:
the "official" report (NIST) does have interesting stuff in it
-- I linked to it via the link to Wiki --
for example, "Figure 9.3. Minimum heating of reinforced heavy columns
to initiate global collapse," which shows "temperature range
for a 50% redicution in steel strength," as opposed
to the typical desideratum of "melting" that is promoted.
thus:
please, don't bother
with the pro-hominemania of your supposed status
as a practicing physicist and/or trained netdoggy!
proabably most of the interpretation of the EPR "paradox"
results,
a la Alain Aspect et al, is due to the ideal of a photon,
in assinging all of the energy of the wave-front
as a "mass" (electron-voltage, say) of a particle, whence
the wave-energy was somehow collected
by the photoeletrical device. here are two ways to get over this: a)
just consider the practice of audio quantization, the phonon; b)
show how the photoelectrical device is actually tuned
to absorb a particular frequency of light.
so, is the "phonon" just one cycle of the period of the sound,
and
like-wise, is the photon just one cycle of the frequency?
--Light: A History!
http://wlym.com
--Weber's electron, Moon's nucleus!
http://www.21stcenturysciencetech.com/
--The Ides of March Are Coming:
Pro-Impeachment Democrat
Wins Nomination in Texas!
http://larouchepub.com/pr_lar/2010/lar_pac/100303kesha_victory.html
BURT
>
> - Show quoted text -
The clocks change when accelerating and decelerating in space. Time
decelerates and accelerates when there is a change in speed in space.
Mitch Raemsch
mpc755
Due to the change in the pressure associated with the aether.
BURT
>
> - Show quoted text -
In my model Aether flows over flowing energy and field without any
pressure. And time flows nowhere else that is empty but only over
particle and field.
Mitch Raemsch
mpc755
In AD, the pressure associated with the aether displaced by massive
objects is gravity. The faster a clock moves with respect to the
aether the greater the pressure associated with the aether on the
clock the slower the clock ticks.
An atomic clock 'ticks' based upon the aether pressure it exists in.
The speed of a GPS satellite with respect to the aether causes it to
displace more aether and for that aether to exert more pressure on the
clock in the GPS satellite than the aether pressure associated with a
clock at rest with respect to the Earth. This causes the GPS satellite
clock to "result in a delay of about 7 ìs/day". The aether pressure
associated with the aether displaced by the Earth exerts less pressure
on the GPS satellite than a similar clock at rest on the Earth
"causing the GPS clocks to appear faster by about 45 ìs/day". The
aether pressure associated with the speed at which the GPS satellite
moves in the aether and the aether pressure associated with the aether
displaced by the Earth causes "clocks on the GPS satellites tick
approximately 38 ìs/day faster than clocks on the ground."
(quoted text from http://en.wikipedia.org/wiki/Effects_of_relativity_on_GPS).
BURT
>
> - Show quoted text -
MPC is your aether pressure accross the extension of particle's and
light's fields equal in strength of gravity? Do they experience "even
gravity" over their whole forms while falling in a "spread out"
different tidal gravity strengths?
This must necessarily be true otherwise how would things know which
speed to fall at as the whole forms they are? If they are in more than
one strength of gravity afterall? There is only one strength that
counts for matter and light and is the strength of gravity at their
centers.
What about the aether gives the pressure? Can you define the different
strengths of gravity with one aether pressure spreading out into the
distance from mather? How does aether speed up without its time nature
going faster instead of slower?
Mitch Raemsch; The beginning of center based physics in gravity theory
BURT
>
> - Show quoted text -
Aether is flowing through space like energy flowing because it flows
over energy and its field and nowhere else.
Mitch Raemsch
BURT
Aether flows through space because it is coupled to energy that is
flowing through space.
Mitch Raemsch
mpc755
>
> > - Show quoted text -
>
> MPC is your aether pressure accross the extension of particle's and
> light's fields equal in strength of gravity? Do they experience "even
> gravity" over their whole forms while falling in a "spread out"
> different tidal gravity strengths?
>
We need to distinguish between momentum, and acceleration and gravity.
A body moving with constant momentum with respect to the aether has
the aether pressure applied equally to each and every nuclei which is
the matter which is the body.
When a body is accelerating or the body is under the influence of
gravity the pressure associated with the aether is not applied equally
to each and every part of the nuclei which is the matter which is the
body.
> This must necessarily be true otherwise how would things know which
> speed to fall at as the whole forms they are? If they are in more than
> one strength of gravity afterall? There is only one strength that
> counts for matter and light and is the strength of gravity at their
> centers.
>
The reason all objects fall at 32ft/sec^2 in a vacuum near the surface
of the Earth is the pressure associated with the aether displaced by
the Earth is greater 'on top' of each and every nuclei which is the
matter which is the body pushing the body towards the Earth. As the
body is being pushed by the aether pressure towards the Earth more
aether is continually 'on top' of the body pushing the body towards
the Earth and so on. This is why objects accelerate as the 'fall'
towards the Earth.
> What about the aether gives the pressure? Can you define the different
> strengths of gravity with one aether pressure spreading out into the
> distance from mather? How does aether speed up without its time nature
> going faster instead of slower?
>
The aether is not at rest when displaced. We know this because light
reaches us from where Jupiter was in its orbit (i.e. Jupiter does not
leave a void in its wake). Aether pressure abides by the inverse
square law.
I stay away from describing the aether as 'speeding up'. I prefer to
discuss the state of the aether as its state of displacement and its
state of at rest with respect to the matter which displaces the
aether.
ben6993
> "ben6993" <ben6...@hotmail.com> wrote in message
>
> - Show quoted text -
Thanks very much for your help, Inertial.
I have introduced time into my calculations now and have relative
speed is approx. 0.866c; contraction is to half the length; and, time
slows to double for the moving plank and garage.
The time dilation for the moving garage enables the two planks to fit
through the garage doors in the appropriate, equivalent time
intervals.
I still need to work more on fitting the stationary plank within the
moving garage, but am resolving that by taking a break from physics
for a one week contract job! Physics paradoxes are very tiring!
Androcles
============================================
Let L = 1 metre.
Contracted length = 1 metre * 1/sqrt( 1 - 0.866^2/1^2)
= 1 / sqrt( 1 - 0.75)
= 1 / sqrt(0.25)
= 1 / 0.5
= 2.0 metres.
1 metre "contracts" to 2 metres.
Your man Einstein didn't know much arithmetic, did he?
And you think double is a half, right?
What a shithead!
ben6993
>
> - Show quoted text -
http://en.wikipedia.org/wiki/Length_contraction
L' = L*√(1-v2/c2)
where
L is the proper length (the length of the object in its rest frame),
L' is the length observed by an observer in relative motion with
respect to the object,
- - - - - -
so
L =1 metre
L'=0.5 metre (whereas you claimed 2 metres, unless I am
misunderstanding your meaning?)
where v=0.866c
- - - -- -
You appear to be a mathematician judging from some of your other posts
but for some things I prefer also to use an easy-to-remember sentence
such as: "The proper length is always longer than a measure of the
length from another frame." Then if using the formula disagrees with
that, I know that I have probably made a silly numerical error in the
maths.
Abusive language in messages contracts a writer's proper IQ and EQ to
much smaller observed levels.
Androcles
============================================
That's LET, not SR.
http://en.wikipedia.org/wiki/Lorentz_ether_theory
In SR,
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
where beta = 1/sqrt(1-v^2/c^2)
See the last sentence of
§ 3. Theory of the Transformation of Co-ordinates and Times from a
Stationary System to another System in Uniform Motion of Translation
Relatively to the Former
in http://www.fourmilab.ch/etexts/einstein/specrel/www/
L is the proper length (the length of the object in its rest frame),
L' is the length observed by an observer in relative motion with
respect to the object, and it is called
"Einstein expansion", not "Lorentz contraction".
Never mind checking what is actually written, listen to bluster and bullshit
instead.
It takes a real shithead not to know the difference between multiplication
and division,
but do carry on learning with inert-brained Inertial... he's a real shithead
and you can
be too.
ben6993
<snip>
> In SR,
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
> where beta = 1/sqrt(1-v^2/c^2)
>
> See the last sentence of
> § 3. Theory of the Transformation of Co-ordinates and Times from a
> Stationary System to another System in Uniform Motion of Translation
> Relatively to the Former
Understanding a translation of Einstein's paper is quite a lot beyond
me, at the moment, as you no doubt knew from my previous posts.
However, the formula you are pointing at does not obviously give L'=2
metres (whereas I claimed L' = 0.5 metres using the standard Lorentz
contraction of length).
It boils down to
L' = 2(1- 0.866ct) as far as I can make out,
and this would only go on to produce 2 metres if t=0, and I cannot
assume that t=0.
Two of the four transformations seem to cross-refer the measurements
of time and length, and you need to consider and work these
transformations together, I think.
A few lines later on it is written that "the X dimension appears
shortened in the ratio 1 : sqrt(1-v^2/c^2), i.e. the greater the value
of v, the greater the shortening." He can't have changed his mind
within a couple of lines on the same page.
A relativity website at Stamford is also content to assert that: "Each
observer will see the meter stick of the other as shorter than their
own, by the same factor gamma (- defined above). This is called length
contraction".
http://www2.slac.stanford.edu/vvc/theory/relativity.html
I am away now for a week.
Androcles
On Mar 13, 8:25 pm, "Androcles" <Headmas...@Hogwarts.physics_v> wrote:
<snip>
======================================================
<unsnip>
Ben6993 wrote:
http://en.wikipedia.org/wiki/Length_contraction
L' = L*?(1-v2/c2)
where
L is the proper length (the length of the object in its rest frame),
L' is the length observed by an observer in relative motion with
respect to the object,
============================================
Androcles wrote:
That's LET, not SR.
http://en.wikipedia.org/wiki/Lorentz_ether_theory
> In SR,
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
> where beta = 1/sqrt(1-v^2/c^2)
>
> See the last sentence of
> § 3. Theory of the Transformation of Co-ordinates and Times from a
> Stationary System to another System in Uniform Motion of Translation
> Relatively to the Former
Ben6993 wrote:
Understanding a translation of Einstein's paper is quite a lot beyond
me, at the moment, as you no doubt knew from my previous posts.
================================================
Androcles:
Einstein's bullshit is deliberately designed to be beyond the novice,
as I've known for many years. However, he cannot translate equations
and he lived in the USA for many years where he spoke and wrote
English, he had ample opportunity to correct any mistranslation.
So let's leave out any bullshit about translation, shall we?
The simple FACT is Einstein wrote 1/sqrt(1-v^2/c^2) and
you've used sqrt(1-v^2/c^2).
The simple FACT is inert Inertial is a shithead like Einstein and
you are heading that way.
......................................<SNIP>..................................
Androcles
On Mar 13, 8:25 pm, "Androcles" <Headmas...@Hogwarts.physics_v> wrote:
<snip>
======================================================
<unsnip>
Ben6993 wrote:
http://en.wikipedia.org/wiki/Length_contraction
L' = L*?(1-v2/c2)
where
L is the proper length (the length of the object in its rest frame),
L' is the length observed by an observer in relative motion with
respect to the object,
============================================
Androcles wrote:
That's LET, not SR.
http://en.wikipedia.org/wiki/Lorentz_ether_theory
> In SR,
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
> where beta = 1/sqrt(1-v^2/c^2)
>
> See the last sentence of
> § 3. Theory of the Transformation of Co-ordinates and Times from a
> Stationary System to another System in Uniform Motion of Translation
> Relatively to the Former
Ben6993 wrote:
Understanding a translation of Einstein's paper is quite a lot beyond
me, at the moment, as you no doubt knew from my previous posts.
BURT
If you must use the state of rest of aether and then you use state
of aether in motion then for both of those to be true the aether
necessarily must speed up.
I challenge your assumption that the Aether is everything. This is not
true. Energy and space also play a role in physics. Aether is only
part of whole formsaof matter and light in physics.
Togetherness is Unification not the sameness of everything.
Mitch Raemsch
>
>
> > Mitch Raemsch; The beginning of center based physics in gravity theory- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
Ste
> On Mar 12, 12:00 pm, Ste <ste_ro...@hotmail.com> wrote:
>
> > > first introduced to this statement, is embodied in their immediate
> > > classroom question: "But in the frame of the traveling twin, it is the
> > > earth twin that is moving away and returning. Since this is symmetric
> > > to the case of the traveling twin moving away and returning, then
> > > shouldn't the traveling twin expect the earth twin to be younger when
>
> > > However, the puzzle is specifically designed to emphasize the danger
> > > one unambiguously experiences acceleration and the other unambiguously
> > > experiences no acceleration. This then leads to a discussion of what
> > > produces the asymmetry in the time.
>
> > I know Paul. I know.
>
> You can imagine my surprise, since what you said explicitly above was
> that theparadoxwas that thetwinsaged differently.
They do, although for some reason I anticipate further conflict on
this point.
> > > Perhaps if you had started out by asking, "Since I don't see any
> > > obviousparadoxhere at all, perhaps someone could illuminate me as to
> > > why this is called the twinparadox?" Then at least you would have
> > > been on square one.
>
> > Really I just wanted to avoid going off on a long tangent about the
> >twinsparadox. As I said, the scenario that were were addressing is
> > different from thetwinsparadox, in that we have three clocks, and
> > the two clocks with which we are now concerned (B and C) both return
> > to the origin point *synchronised* (albeit both lagging behind A),
> > whereas thetwins' ages are not synchronised on the return of the
> > astronaut twin.
>
> Well, yes, it is a different result from the application of the same
> principle.
>
> If I calculate the angle that I can tip a TV tray before the coffee
> cup on it starts to slide, I find that I'm using the same principle
> (equilibrium of forces) that I would use to determine the tension in
> picture-hanger wire when mounting a photo on the wall.
>
> Different situation. Very same principle.
>
> Fine example of losing the forest for the trees. As you've done here.
Of course we are talking about the "same principle", because we are of
course still talking about SR. The point is that I want to confine the
discussion to a specific scenario, and while the principles involved
may be the same, the scenario is not.
Ste
As far as I know, there is never an "acceleration of time" under any
circumstances.
mpc755
Matter moves relative to the aether. The aether does not necessarily
speed up in and of itself.
> I challenge your assumption that the Aether is everything. This is not
> true. Energy and space also play a role in physics. Aether is only
> part of whole formsaof matter and light in physics.
>
Aether is not everything. Aether and matter are different states of
mather.
> Togetherness is Unification not the sameness of everything.
>
AD is the most correct unified theory to date.
Ste
Just to qualify this dangerous statement immediately, I mean there is
never an "acceleration of time" as a direct result of a change of
velocity.
mpc755
Time is a concept. Time does not accelerate or change with velocity or
exist at all beyond our conceptualization of it. And just to be clear,
all physical life has the concept of time or there would be no
physical life. And by physical life I mean anything that is born and
dies.
The rate at which a clock ticks is dependent upon the aether pressure
in which it exists.
BURT
My theory can stand the tests of the future. I challenge that yours
cannot.
Togetherness replaces Unification.
Mitch Raemsch
mpc755
>
> > AD is the most correct unified theory to date.
>
> My theory can stand the tests of the future. I challenge that yours
> cannot.
>
> Togetherness replaces Unification.
>
If you own a battery operated clock and it begins to tick slower has
time changed or do you replace the batteries?
BURT
Please. Every clock is not broken that measures its own time slower
flow rate.
You cannot get rid of the two aether rates behind slowing time for
energy.
Mitch Raemsch
mpc755
It is a very simple question you seem to be refusing to answer. If a
battery operated clock begins to tick slower has time changed?
If not, why not and why is this any different to a clock in a GPS
satellite ticking faster than a comparable clock which remains on the
surface of the Earth?
The answer is there is no difference. The battery operated clock is
physically ticking slower because of the physical state of the
battery. The atomic clock in the GPS satellite physically ticks faster
because of the decrease in the aether pressure in which it exists.
BURT
>
> - Show quoted text -
Go shake an atomic clock mpc.
Mitch Raemsch
BURT
>
> - Show quoted text -
Slow down in space and leave gravity behind and time increases in flow
over you. There is a fastest time if time slows down.
Mitch Raemsch
Inertial
news:5VPmn.242488$8t3.1...@newsfe29.ams2...
Androcles lies about what the length contractino formula is. In doing so
just shows everyone what a moron he is.
Inertial
Wrong .. LET and Sr have the same math
> In SR,
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
> where beta = 1/sqrt(1-v^2/c^2)
You can't read simple math
> See the last sentence of
> § 3. Theory of the Transformation of Co-ordinates and Times from a
> Stationary System to another System in Uniform Motion of Translation
> Relatively to the Former
> in http://www.fourmilab.ch/etexts/einstein/specrel/www/
>
>
> L is the proper length (the length of the object in its rest frame),
> L' is the length observed by an observer in relative motion with
> respect to the object, and it is called
> "Einstein expansion", not "Lorentz contraction".
>
> Never mind checking what is actually written, listen to bluster and
> bullshit instead.
> It takes a real shithead not to know the difference between multiplication
> and division,
> but do carry on learning with inert-brained Inertial... he's a real
> shithead and you can
> be too.
Indeed .. you ARE a real shithead.