Re: More Trouble for Relativity

From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 09/06/04 

Date: Mon, 06 Sep 2004 18:23:24 GMT

"sal" <pragmatist@nospam.org> wrote in message
news:pan.2004.09.06.14.26.22.803225@nospam.org...
| On Mon, 06 Sep 2004 09:21:01 +0000, Androcles wrote:
|
| >
| > "sal" <pragmatist@nospam.org> wrote in message
| > news:pan.2004.09.06.02.12.02.990793@nospam.org...
| > | On Sun, 05 Sep 2004 14:14:59 +0000, Androcles wrote:
| > |
| > |
| > | > "sal" <pragmatist@nospam.org> wrote in message
| > | > news:pan.2004.09.05.12.48.24.994073@nospam.org...
| > | > | On Sun, 05 Sep 2004 06:05:26 +0000, Androcles wrote:
| > | > |
| > | > |
| > | > | > "sal" <pragmatist@nospam.org> wrote in message
| > | > | > news:pan.2004.09.05.02.08.40.533254@nospam.org...
| > | > | > | On Sat, 04 Sep 2004 08:32:33 +0000, H wrote:
| > | > | > |
| > | > | > | > On Sat, 04 Sep 2004 01:13:54 GMT, "Androcles"
| > | > | > | > <androc1es@nospamblueyonder.co.uk> wrote:
| > | > | > | >
| > | > | > | >
| > | > | > | >>"Henri Wilson" <H@..> wrote in message
| > | > | > | >>news:gnqhj0pcik1j2g2q3qf5hlo36glnlgmhfk@4ax.com...
| > | > | > | >>| This problem deserves its own thread.
| > | > | > | >>|
| > | > | > | >>| Consider a very large, remote and perfectly spherical
planet
| > | > | > | >>| that has no atmosphere. An observer on its surface
| > synchronizes
| > | > | > | >>| two clocks then sends one in circular orbit 1 mm above the
| > | > | > | >>| ground.
| > | > | > | >>|
| > | > | > | >>| Each time the OC passes the GC, the two exchange readings.
| > | > | > | >>|
| > | > | > | >>| Thus, any change in the OC's reading or rate, wrt the
| > | > | > | >>| original frame, is detected and measured.
| > | > | > | >>|
| > | > | > | >>| I say a perfect clock will not change one iota when in
| > | > | > | >>| orbit.
| > If
| > | > the
| > | > | > | >>| OCs reading does not match that of the GC, then it has
| > | > | > | >>| physically changed in some way.
| > | > | > | >>| Relativists say it WILL change, for the usual obscure
| > | > | > | >>| reasons.
| > | > | > |
| > | > | > | :-) They seem obscure, and will continue to seem obscure,
until
| > you
| > | > | > | :spend
| > | > | > | the time to grok the geometry.
| > | > | > |
| > | > | > |
| > | > | > | >>| Note, both clocks experience the same gravitational
| > | > | > | >>| potential.
| > | > Also
| > | > | > | >>| there is only ONE FRAME INVOLVED.
| > | > | > | >>|
| > | > | > | >>| Fair enough?
| > | > | > | >>|
| > | > | > | >>| To resolve the issue, another pre-synched clock is now
| > | > | > | >>| placed
| > in
| > | > the
| > | > | > | >>| same orbit but, this time, in exactly the opposite
direction
| > to
| > | > the
| > | > | > | >>| first.
| > | > | > | >>|
| > | > | > | >>| It is clear that both OCs should will read the same as
they
| > pass
| > | > the
| > | > | > | >>| GO, whether relativity is right or wrong.
| > | > | > | >>|
| > | > | > | >>| The question is, since relativity says each clock will run
| > slow
| > | > wrt
| > | > | > | >>| the GC, the same reasoning says that each should also run
| > | > | > | >>| slow wrt the other.
| > | > | > | >>|
| > | > | > | >>Been done. They couldn't.
| > | > | > |
| > | > | > |
| > | > | > |
| >
>>http://groups.google.co.uk/groups?hl=en&lr=&ie=UTF-8&threadm=d8097fcc.0312110926.29ff02da%40posting.google.com
| > | > | > | >>Androcles.
| > | > | > | >>
| > | > | > | >>
| > | > | > | > Yes.
| > | > | > | >
| > | > | > | > I don't expect any intelligent responses from the DHR CDs,
A.
| > Even
| > | > | > | > Dinky the ultimate moron is conspicuously absent.
| > | > | > | >
| > | > | > | > They cannot answer this one.
| > | > | > |
| > | > | > | Most can't, it's true, because this problem is stinkin' hard.
| > | > | >
| > | > | > Learn the difference between "difficult" and "impossible".
| > | > | >
| > | > | >
| > | > | > | This has been discussed before in this NG, occasionally at
great
| > | > length
| > | > | > | (though not at much length in the thread A. posted a reference
| > to).
| > | > | > |
| > | > | > | I've posted a solution here before, as have others. The math
| > | > | > | gets
| > a
| > | > bit
| > | > | > | messy, but the conclusions are pretty clear and I've shown
them
| > | > | > | graphically -- in the MCRF of each ship, the other ship's
clock
| > runs
| > | > | > | _faster_ when the two ships are opposite each other; by the
time
| > | > | > | they meet the two clocks are back in sync. If anyone reading
| > | > | > | this thread
| > | > is
| > | > | > | interested, my version of this is on
| > | > | > |
| > | > | > | http://www.physicsinsights.org/revolving_twins.html
| > | > | >
| > | > | > Two clocks going in the same direction at the same speed, and
each
| > | > | > ticks faster than the other. Right. Got it. George Hammond has
| > | > | > given his official blessing to relativity, did you know that? At
| > | > | > one
| > point,
| > | > | > sal, I thought you were simply joking, but now I realize you
| > actually
| > | > | > are
| > | > serious
| > | > | > and expect others to believe your stupidity. Androcles
| > | > |
| > | > | Yes, of course I expect that, because it's the "conventional
| > stupidity".
| > | > | Mainstream physicists like Hawking and Wheeler believe it; why
| > shouldn't
| > | > | readers of this newsgroup believe it too? (That's a rhetorical
| > question
| > | > | -- you needn't answer it!)
| > | > |
| > | > | The clocks are spatially separated and both are accelerating.
| > | >
| > | > Toward each other. Proof : they meet again (and again and again).
| > |
| > | Yes, that's correct. It's the moment when they're accelerating
directly
| > | toward each other, and are at maximum separation, that each computes
the
| > | other's clock to be running fastest. Since they're on parallel paths
at
| > | that moment, it also occurs at the same time on each ship, and the
| > | ground-based observer agrees, as well, that it's at the moment when
each
| > | ship has moved 90 degrees since their last conjunction.
| >
| > No, no, no. That is the point where we say "If we place x' = x-vt",
then
| > each computes the other's clock to be running at the same speed as the
| > other fastest, but if we say "If we place x' = x+vt", then each
computes
| > the other's clock to be running at the same speed as the other slowest.
|
| No, I don't say that. I'd never say such a thing.

I said it on behalf of Einstein, your tin god who taught others to teach
you relativity his way. See, I even put what he said in quotes and
paraphrased
it with a plus replacing a minus. I'm allowed to do that because he started
with "If".
Of course if you disagree with him, that's fine. And you DID say "fastest".
|
|
| > Anyway, they are always accelerating directly toward each other, your
logo
| > shows that quite clearly,
|
| When they're colocated at the top of the orbit they're both accelerating
| in the same direction (straight "down"). Later when they pass again
| they're both accelerating in the same direction again (this time it's
| straight "up"). They're only accelerating _directly_ toward each other
| when they're 180 degrees apart in the orbit.

When is that not toward each other then? If I drive East and you drive South
with the intention of meeting at the intersection of 1st and Main, we are
moving toward each other.
(I sometimes use the manhattan metric rather than the euclidean.;-)
Anyway, they don't get any closer when they 179 degrees apart, they get
further away,
so they can't be accelerating toward each other at 180 either:-)
|
| > and since that acceleration is transverse to the
| > direction of motion it plays no part in SR whatsoever and need not be
| > considered. I can stand underneath Big Ben's clock at 3:00pm and my
| > wristwatch keeps the same time as the clock, even though we are both
| > accelerated toward the centre of the Earth and travelling transversely.
I
| > know this, because I've done it. Nor do I ever compute a wrong time. t'
=
| > t. dtau/dt = 1. Always. I NEVER compute dtau/dt = 0 <1. Only a
relativist
| > would do that, because they've never stood underneath a clock and
compared
| > it with their wristwatch.
| >
| >
| > | > | The
| > | > | acceleration, combined with the spatial separation, is the key
here.
| > | > | Since ship A is accelerating, it is constantly changing its MCRF;
in
| > | > | that constantly changing MCRF, the reading of the clock on ship B
is
| > | > | also constantly changing. But this is just the result A
| > | > | _calculates_ -- A is too far from B to actually _see_ this taking
| > | > | place. The situation on B is, of course, symmetric.
| > | >
| > | > And by symmetry you have both clocks recording the same time. By
logic
| > | > you have (tA <= tB) .AND. (tB <= tA) implies tA = tB.
| > |
| > | From the point of view of the stationary observer, you're correct --
| > | both clocks always read the same time.
| >
| > Then why would you say "each computes the other's clock to be running
| > fastest"? Don't "they" know how to do hard sums?
|
| Actually what they compute is the value that would be seen by an observer
| who happened to be right next to ship B at that moment, and who
| happened to be moving exactly the same direction and speed as ship A, and
| who had previously synchronized his clock with another observer who
| happened to be moving in the same direction at the same speed and who
| happened to be right next to ship A at that moment. In that way the times
| on the clocks on each ship can be "compared" even though they're far
apart.
|
| That is what a computation done in the "MCRF" of the ship means.
|
| It is not a very realistic computation, you see.

Yeah... I thought as much.

|
| > And even then you are
| > wrong. For the stationary observer, the clock that is approaching
appears
| > to run the faster. That's called Doppler shift. You've never heard of
it,
| > huh?
|
| Yes, I have, and it's an observed effect, not just a computed one.

Don Koks had a go at it. It is not a very realistic computation, you see.
 http://www.androc1es.pwp.blueyonder.co.uk/KoksDoppler.htm

|
|
| > | > | If the two ships were right next to each other going at the same
| > | > | speed, they'd actually _see_ each others' clocks going at the same
| > | > | rate, whether or not they were accelerating, because they'd share
an
| > | > | MCRF.
| > | >
| > | > And that is the condition you have in your graph where the red curve
| > | > labelled "Time on other clock at present time" is sinusoidal about
the
| > | > straight line labelled "Present time on ship A", at pi/2 and 3pi/2.
| > | > "Right next to" means separated by a distance delta (= chord)
| > |
| > | When I said "right next to each other" in the post to which you're
| > | responding, I meant "adjacent and separated by infinitesimal
distance".
| > | In other words, at the same place -- colocated, sharing an event, able
| > | to compare clocks just by holding them next to each other.
| >
| > That's just being silly. There is always a finite distance between the
| > clocks.
| > It is physically impossible to colocate them until you can walk through
| > walls.
| > We can miniaturize them, make them as close as we like, but we cannot
| > colocate them. The algebra doesn't change based on the diameter of the
| > orbit.
| >
| >
| > | "Right next to each other", as in directly across a circle from each
| > | other, is a different situation.
| >
| > No. You are simply varying the distance between them, it is NEVER zero.
|
| The distance can be made small, and because the variance in the actual
| readings is linear in the distance the variance will also be small.

I said it was never zero. You can make it as small as you like, but not
zero.

|
| When they're across a macroscopic chord of the circle OTOH the distances
| are not "small" and can't be ignored.

 I can make the diameter of the orbit smaller than your choice of 'small',
and the chord is smaller than the diameter except at the point where the
satellites
are relatively at rest with no time dilation between them. Unless you are
claiming
there is?
|
|
| > Each sees the other's clock lagging, but I will agree that each sees the
| > other's clock going at the same rate, since they always do anyway.
| > dtau/dt = 1, always. If you want to model the situation mathematically,
| > don't make up impossibilities, I won't accept that. The wheels of a
| > train are clocks, no matter how wide the track. They all keep the same
| > time. Place a solid axle between them, rigidily attached to each wheel.
| > They keep the same time even on a curved track, one or both will slip.
| > |
| > | > and the
| > | > difference is simply tA-tB = chord/c. The other clock (B) always
| > | > lags behind the reference clock (A). It never appears to be ahead.
| > | > Since tA-tB is always positive, half of your red curve is upside
| > | > down.
| > |
| > | Not at all.
| >
| > Yes at all. I've given you a proof. You've given an assertion.
|
| And a pile of equations on the rest of the page.

You can't derive the Lorentz transforms, so you have only assertion.
Assertion includes Lorentzian garbage. See the Seven Sins of Relativity. |
|
| > | But note that the red curve -- labeled 'Time on "other" clock at
| > | "present time"' -- is what the astronauts on A _compute_ regarding B.
It
| > | is absolutely not what they'd see in a telescope.
| >
| > So A got his sums wrong. Too bad, give A an F. Besides, who needs a
| > telescope? The same math applies not matter how large or small the
| > situation is.
| >
| >
| > | The "porthole view"
| > | -- what they see in a telescope -- certainly always lags behind.
| >
| > Well then! A got his sums wrong. His mathematical model did not
correctly
| > apply to the physical situation. This clearly demonstrates the
difference
| > between real physics and theoretical physics.
|
| But it doesn't, not by itself. If the Doppler shift which is measured
| actually matches Einstein's Doppler then the computations will all work
| out.
An "IF", I see.
If the Doppler shift which is measured actually doesn't match Einstein's
Doppler then the computations will NOT all work out.

 If the Doppler shift matches Newton's Doppler (and if that makes any
| sense!) then it won't work out. A disproof would require an experiment;
| you can't do it just using paper.

Why not? The whole of relativity is a paper exercise and has nothing to do
with physics.
Koks' Doppler gives a disproof of relativity, and closely resembles the
Newtonian verdict. You chose to ignore the aetherialistic "light is constant
in empty space"
verdict, but then, you deliberately choose to ignore any argument that
conflicts with your religion, wishing to remain willfully ignorant.

|
|
| > | You are
| > | correct about that, and that's shown in the graph as the blue line,
| > | which is labeled 'Time on "Other" Clock as Seen in Telescope'. It is
| > | purely concave upward.
| > |
| > | Note that, since the astronauts on A can see that the clocks on B are
| > | running "slow" when they pass,
| >
| > No they can't. The clocks show the same time as they pass. The only
| > thing that makes one see the other running slow or fast is Doppler
| > shift, but that is exactly 1 as they pass.
|
| You are correct, when we're discussing a single clock passing a single
| clock. But what I actually proposed on the page was something little more
| complex.

Making the issue more complex than it is assinine. I really don't care
if your complete the ring, joining nose to tail and placing a miriad clocks

 Here, I'll quote:
|
| [from webpage:]
| > Suppose there's a clock in the nose and the tail of each ship, and
| > suppose that the "nose clocks" are in sync when they pass each other.
| > Then when the astronaut in the tail of ship A looks at ship B's nose
| > clock, he'll see that it's slow -- it hasn't ticked as often as it
| > should have while the ships were passing. Yet, the next time that clock
| > passes the nose of ship A, it'll be back in sync.
|
| Of course this doesn't quite match my drawing, which shows each ship
| nose-inward, but so it goes. The point is that if each ship carries two
| clocks, spatially separated along the line of motion, then by observing
| the other ship's similarly arranged pair of clocks, the astronauts on each
| ship can determine that the other ship's clocks each appear to be running
| slow.

I really don't care if your complete the ring, joining nose to tail and
placing
a miriad clocks throughout the ship. They are all relatively at rest and
have
the same time. The other ship has the same situation but revolves the
opposite
way. One clock on one ship agrees with the time of one clock on the
other ship implies all clocks agree. No ifs, ands or buts.

|
| > but at the exact moment when the clocks
| > | pass each other they are in sync, they _must_ conclude that B's clocks
| > | were _ahead_ just before the ships met.

Phooey. The excrement of the male bovine.

| >
| > Doppler. That has nothing to do with any stupid gamma.
| >
| > | Therefore, at least part
| > | of the "Other clock at present time" curve _must_ lie above the
straight
| > | line which shows the "present time". In other words, it's not possible
| > | that half the red curve is upside down, for that would make the time
| > | dilation of B's clock which is observed by A impossible, or at any
rate
| > | inexplicable.
| >
| > Connect two wheels by a differential of the type your car uses. Mount
| > them
| > on a single circular rail, is you might for a carousel. Any time
dilation
| > between the clocks (wheels) would result in a rotation of the engine
drive
| > shaft. You are claiming the drive shaft will oscillate back and forth.
It
| > will not.
|
| I'm sorry, the analogies can be carried too far. It's worth trying to
| understand the graphs using English, but the graphs are not of the
| results of an analogy, they're plots of the equations which were derived
| earlier in the page.

In other words, garbage in, garbage out.
|
|
| > | But again, the state of affairs where the other clocks are ahead is
| > | only deduced, never observed
| >
| > So you have your math wrong. There is no time dilation.
|
| Well, in that case, you should be able to find the error, right?

I already have. See the Seven Deadly Sins of Relativity.
You used the non-existent Lorentz Transform.

|
| It's got to be somewhere in the page -- all the math is there. Too much
| math, actually, or so I've been told by another relativist ("too much
| math, and you missed the point of the physics" -- something like that).
|
Too many assumptions and you have no math at all.

| > -- the time it takes the signals to get from B to
| > | A effectively negates the "lead" B's clock has over A's clock.
| >
| > Doppler shift cancels gamma?
| > I don't think so. Doppler shift is real enough, and it IS observed. B's
| > clock always appears to lag A from A's pov, and A appears to lag B from
| > B's pov. There is never a situation where B's clock is faster than A's,
| > or A's clock is faster than B's. NEVER.
|
|
| In terms of observations, I agree, as I said before.

Well then... theoretical physics bears no relation to observation.
|
|
| > | > Then perhaps you'll gain some INSIGHT into the relationship
| > | > between gamma and the eccentricity of an ellipse, but I have a
serious
| > | > doubt you will, being stubborn and ineducable as you are.
| > |
| > | And short on time -- don't forget that one.
| >
| > You seem to have plenty of time playing the silly relativity game and
none
| > for real physics.
|
| No, I don't have much time for either. I'm spending a few minutes a day
| looking over s.p.r. these days, and posting replies that I don't have to
| do any "hard sums" for. Then it's back to the salt mines.
|
| Labor day? Oh, yeah, that's a day when we labor harder, right?

We have a bank holiday here when all the banks go on strike,
so this nation of shopkeepers is deprived of making a profit.

Androcles.

| --
| I can be contacted through http://www.physicsinsights.org
|

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