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From: kenseto on 10 Aug 2010 17:51
On Aug 10, 12:09 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney)
wrote:
> kenseto <kens...(a)erinet.com> writes:
> >On Aug 9, 9:02 pm, "Inertial" <relativ...(a)rest.com> wrote:
>
> >> Let set up such an experiment (as a gedanken) as above
>
> >> When B passes A1, A1 records its own time, and the time on B, and when B
> >> passes A2, A2 records its own time, and the time on B
>
> >> Then you can compare the difference between the time on B that A1 recorded,
> >> and the time on B the A2 recorded, and compare that to the difference in
> >> times on A1 and A2 themselves.
>
> >> SR says that if you do that, you will find the difference in time on B is
> >> *less* than the difference in time between A1 and A2.  That is that clock B
> >> is ticking *SLOWER* than clocks in A
> >> Now lets have a second gedanken.
>
> >> In this case we have a single clock A at rest in your frame, and two
> >> co-moving moving clocks (B1, B2 say) that are synchronised in their own
> >> frame and are moving past A (with the same speed as B in the previous
> >> experiment)
> >Why is B1 and B2 moving past A...why not A moving past B1 and B2 ???
>
> Geez, you've been kooking in here for 15 years now but still don't
> understand the basic fundamentals of relativity?  All inertial frames
> in SR have the same math, so of course an observer in A's frame can
> do measurements with B1 and B2 moving past A by assuming A is stationary.

No idiot he can't do any measurements.... he predicts.

>
> >> AS in the previous experiment, clock A takes a set of two readings, one of
> >> its own time, and the other of the time shown on the moving clock.
>
> >> So we get two sets of readings from A .. one set when B1 passes and another
> >> set when B2 passes.
> >So you are assuming that the B clocks are doing the moving while A
> >remains stationary....right?
>
> All motion is relative, so of course A will see the B clocks as moving.

Ok

> (and yes, an observer in B will see the A clocks as moving, but that
> was already discussed as the first gedanken, so it's not relevant here)

It is relevant....B will use the same SR equation to predict that A
runs slow and thus the bogus concept of nutual time dilation.

>
> You got to keep the A and B frames separate.  This is where you always
> get mixed up.

I didn't mix up anything. A and B predicts each other's clock runs
slow.

>
> >> If you compare them, you'll find, in this case, a *longer* elapsed time
> >> between B1 and B2 than is shown on A  (the opposite results to the previous
> >> experiment)
> >This is a bogus assumption based on the faulty SR math that all clocks
> >moving wrt the observer (A or B) are running slow.
>
> No, just do the logic that was just shown to you!

There is no logic....A and B uses the same equation to predict each
other's clock runs slow.

>
> >> How can this be?
>
> >> The result of the second experiment implies that either clocks B1 and B2
> >> were in sync in your frame but ticking *FASTER* -- OR -- that the clocks are
> >> *NOT* in sync in your frame.
> >This is wrong....B is the observer and B1 and B2 are in sych in the B
> >frame.
>
> But not the A frame.  And that is the point.

We are not talking about the A frame....we are talking about what the
B frame predicts.

Ken Seto


- Hide quoted text -
>
> - Show quoted text -

From: spudnik on 10 Aug 2010 22:23
you two -- socratic twins, for the sake
of our little argumentarium, herein -- are missing the so-simple part
of the so-called paradox;
there can be no "travel-away" without acceleration,
whence galilean relativity takes a backseat
to the "retarded velocity of light."

of course, if they both accelerate at the same average rate,
there will be no difference in their relative timings,
summed at the place where they meet.

thus:
wasn't the one that got stuck,
between the mainland and St. George's Island?

> > (I know of only one such instance .-)
>
> The iceberg in the Antarctic caused a lot of worry as the resupplying
> fuel tankers couldn't get into McMurdo.
> We get some ice off Macquarie Island about half-way between New
> Zealand and Antarctica.  at 54°37'53"S, 158°52'15"E

thus: there were also four investigations into Whitewatergate,
til some b***** from the Special Counsel's office sent a complaint
to -- I kid you, not -- the Resolution Trust Corp.,
which is a whole another story.... ah; "L. Jean Lewis."
> It ain't over till it's over.

dear editor:
Re the new mercury standard from the EPA, it seems to be assumed that
the mercury is from emmissions -- searched a recent story about
possible increase in fish e.g. -- but here's another suspect.
I've seen a special facility at new apartments for dispozing of
flourescent lights, but I have also seen the profligate tossing of the
new, small, screw-in CFLs -- most of which are probably made in
southwest Asia -- and it is hard to imagine that this stuff is not
leached from the landfills, eventually if not sooner. Alas, the long
bulbs have been around for decades.

thus: any trigon or tetragon will tile the plane. as
for Weber, I think it is in this article:
http://www.21stcenturysciencetech.com/articles/spring01/Electrodynamics.html
and the second root of two is just an additional factor.
From: Koobee Wublee on 11 Aug 2010 12:39
On Aug 11, 4:38 am, "Paul B. Andersen" wrote:
> On 10.08.2010 19:09, Koobee Wublee wrote:

> > For the record, Koobee Wublee was explaining how the Lorentz transform
> > actually predicts both blue and red shift at the same time. See the
> > post below.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/c4d248370bd...
>
> > Suddenly, professor Andersen, out of the blue, changed the discussion
> > with this post without addressing Koobee Wublee’s previous post.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/077983367e3...
>
> > Koobee Wublee, an ever so humble, good Samaritan, answered professor
> > Andersen’s questions fair and square. The good hearted Koobee Wublee
> > saw the same mistakes and garbage that professor Andersen had
> > defecated all over these newsgroups in the past few years and decided
> > not to press any further charges. See the post below.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8d...
>
> > Seeing himself checkmated once again, professor Andersen decided to
> > behave childish and defiant with the following useless post.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/29d192787e6...
>
> > The good hearted and ever so humble Koobee Wublee tried once again to
> > reach out to save professor Andersen from drowning in his cesspool of
> > fermented diarrhea of Einstein the nitwit, the plagiarist, and the
> > liar with the following post. In doing so, Koobee Wublee even showed
> > professor Andersen what a mathematical model of Doppler effect would
> > look like and guided with hints on how to arrive with that.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/22d990f50ed...
>
> > The asinine professor Andersen then refused to discuss any further by
> > repeating the same bullshit from the very beginning. Professor
> > Andersen is indeed a very small man.
>
> > <shrug>
>
> Sore again, Kooblee? :-)
> Poor looser, eh?

It looks like the little man who calls himself professor Andersen from
Trondheim is performing his gig again.

http://www.mazepath.com/uncleal/sunshine.jpg

Ahahahaha...
From: Michael Moroney on 11 Aug 2010 22:30
kenseto <kenseto(a)erinet.com> writes:

>On Aug 10, 12:09 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney)
>wrote:
>> kenseto <kens...(a)erinet.com> writes:

>> >Why is B1 and B2 moving past A...why not A moving past B1 and B2 ???
>>
>> Geez, you've been kooking in here for 15 years now but still don't
>> understand the basic fundamentals of relativity? All inertial frames
>> in SR have the same math, so of course an observer in A's frame can
>> do measurements with B1 and B2 moving past A by assuming A is stationary.

>No idiot he can't do any measurements.... he predicts.

What do you mean "he can't do any measurements"? The gedanken is very
simple, and it explicitly describes the measurements. (Of course he can
predict as well)

>> (and yes, an observer in B will see the A clocks as moving, but that
>> was already discussed as the first gedanken, so it's not relevant here)

>It is relevant....B will use the same SR equation to predict that A
>runs slow and thus the bogus concept of nutual time dilation.

But it's the first half of the gedanken. B observer sees A clocks moving
is first part. A observer seeing B clocks moving is second part.

Of course it makes perfect sense A and B measure the same thing if you
consider this: Clocks A1, A2 B1 B2 are identical, and observers A and B
are also identical. For anything else you can think of, let anything in
the two frames be identical. Therefore the two frames are identical so of
course they will both predict and measure the same thing. If not, there
would be a difference between them but I just said make everything in the
two frames be identical.

>> You got to keep the A and B frames separate. This is where you always
>> get mixed up.

>I didn't mix up anything. A and B predicts each other's clock runs
>slow.

And measure them as running slow.

>There is no logic....A and B uses the same equation to predict each
>other's clock runs slow.

And measure them as running slow.

>> >This is wrong....B is the observer and B1 and B2 are in sych in the B
>> >frame.
>>
>> But not the A frame. And that is the point.

>We are not talking about the A frame....we are talking about what the
>B frame predicts.

B observer sees B clocks as synchronized - but not the A clocks.
A observer sees A clocks as synchronized - but not the B clocks.

From: kenseto on 12 Aug 2010 09:06
On Aug 11, 10:30 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney)
wrote:
> kenseto <kens...(a)erinet.com> writes:
> >On Aug 10, 12:09 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney)
> >wrote:
> >> kenseto <kens...(a)erinet.com> writes:
> >> >Why is B1 and B2 moving past A...why not A moving past B1 and B2 ???
>
> >> Geez, you've been kooking in here for 15 years now but still don't
> >> understand the basic fundamentals of relativity?  All inertial frames
> >> in SR have the same math, so of course an observer in A's frame can
> >> do measurements with B1 and B2 moving past A by assuming A is stationary.
> >No idiot he can't do any measurements.... he predicts.
>
> What do you mean "he can't do any measurements"?  The gedanken is very
> simple, and it explicitly describes the measurements.  (Of course he can
> predict as well)

In real life no such measurement is possible....only predictions and
you runts of the SRians keep on perpetuating the myth that predictions
are measurements to give your theory more credence.

>
> >> (and yes, an observer in B will see the A clocks as moving, but that
> >> was already discussed as the first gedanken, so it's not relevant here)
> >It is relevant....B will use the same SR equation to predict that A
> >runs slow and thus the bogus concept of nutual time dilation.
>
> But it's the first half of the gedanken.  B observer sees A clocks moving
> is first part.  A observer seeing B clocks moving is second part.

Here's the flaws in you logic:
1. B sees A clock is moving and thus A clock is running slow.
2. A sees B clock is moving and thus B clock is running slow.
These conclusions include the bogus assumption that every observer's
clock is running faster than every observed clock.

>
> Of course it makes perfect sense A and B measure the same thing if you
> consider this:  Clocks A1, A2 B1 B2 are identical, and observers A and B
> are also identical.  For anything else you can think of, let anything in
> the two frames be identical. Therefore the two frames are identical so of
> course they will both predict and measure the same thing.

This makes no sense at all. It is based on the bogus assumption that
every observer assumes that his clock is running at a faster rate than
every observed clock.
No observer can claim that all clocks moving wrt him are running slow.
Therefore he must include all possibilities: that an observed clock
can run slow by a factor of 1/gamma or run fast by a factor of gamma
compared to the observer's clock.

Ken Seto

Ken Seto


 If not, there
> would be a difference between them but I just said make everything in the
> two frames be identical.
>
> >> You got to keep the A and B frames separate.  This is where you always
> >> get mixed up.
> >I didn't mix up anything. A and B predicts each other's clock runs
> >slow.
>
> And measure them as running slow.
>
> >There is no logic....A and B uses the same equation to predict each
> >other's clock runs slow.
>
> And measure them as running slow.
>
> >> >This is wrong....B is the observer and B1 and B2 are in sych in the B
> >> >frame.
>
> >> But not the A frame.  And that is the point.
> >We are not talking about the A frame....we are talking about what the
> >B frame predicts.
>
> B observer sees B clocks as synchronized - but not the A clocks.
> A observer sees A clocks as synchronized - but not the B clocks.

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