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Introduction: The Illusion of Design [Richard Dawkins]
Natural History Magazine ^ | November 2005 | Richard Dawkins

Posted on 12/07/2005 3:31:28 AM PST by snarks_when_bored

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To: snarks_when_bored

Oh! Snarky.. your tie, well, you know..


981 posted on 12/11/2005 9:42:48 PM PST by hosepipe (CAUTION: This propaganda is laced with hyperbole..)
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To: Alamo-Girl
#980 [ Considering these views of reality, I don't really see where we have anything to explore. ]

LoL...

982 posted on 12/11/2005 9:45:21 PM PST by hosepipe (CAUTION: This propaganda is laced with hyperbole..)
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To: hosepipe
Oh! Snarky.. your tie, well, you know..

It's these danged wings, you know...can't use 'em to tie a double Windsor to save my life...

983 posted on 12/11/2005 9:58:41 PM PST by snarks_when_bored
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To: snarks_when_bored

Great photo! That duck is quite evolved.

Now, could you explain what this has to do with Al Gore?


984 posted on 12/12/2005 4:46:09 AM PST by ChessExpert (Democrats: Sore/Losermen 2000, 2004, 2008, 2012)
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To: betty boop
a point in time when two species of men existed together --

I'm sure that would be a most interesting speculation, don. But I confess my real interest in this question is the succession, or transition, of one to the other, and why that transition was successful.
We do not get that sort of information from the consideration of biological data alone. Or so it seems to me.

Hmmm, -- are you suggesting that we can surmise why our species survived without using scientific data?

Why do we need such data, when we clearly and readily see that we have "survived?"

You're the one who suggested -- "interest in this question --", not I.. Forget my comment.

(Otherwise you wouldn't be around to ask this question, nor me to reply to it.)

btw.. What would be your reaction if were proved that Neanderthal genes live on in modern humans? That hybrid vigor contributed to our 'successful transition'?

If Neanderthals were actually protohuman (which designation seems to cover a whole lot these days), that wouldn't surpise me at all. And it seems they were.
Are these "trick questions," don?

Nope, not from my view Betty.. What about them give you that idea?

985 posted on 12/12/2005 6:33:23 AM PST by don asmussen
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To: ChessExpert; hosepipe
Great photo! That duck is quite evolved.

Now, could you explain what this has to do with Al Gore?

That was a hosepipe comment...you'll have to ask him!

986 posted on 12/12/2005 10:53:18 AM PST by snarks_when_bored
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To: Alamo-Girl
I don't really see where we have anything to explore

Of course we do, not so much exploration as discovery and invention. We create patterns and laws; these things do not exist in nature until we create them. It's what we do. Man the Engineer. Most of our laws are discovered in our own machines, in our own creations, and in a few simple cases we can find an example in nature. Nature itself has very little by way of pattern and law; nature including man has all the pattern and law--work in progress.

It might be said that God created the universe, but it might equally be said the He does not manage it. That is our job, and we are having trouble managing even this speck of dust where we find ourselves. If we are otherwise alone in the universe, and so far it appears so, we are slacking off. Somebody else can discuss the nature of reality until they stumble across a clue; the rest of us are working quite hard, if ineffectively, doing things.

987 posted on 12/12/2005 11:42:08 AM PST by RightWhale (Not transferable -- Good only for this trip)
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To: Physicist; Doctor Stochastic; js1138; Right Wing Professor; RadioAstronomer
This reply is addressed to js1138 and physicist who commented, those that where pinged and all.

"Consider: the B and anti-B are entangled. Until one of them decays--thereby "picking" a flavor--the other one doesn't know how it is supposed to decay. The "second" one to decay will subsequently oscillate back and forth between being a Bo and an anti-B meson: effectively, it has two decay rates, one in its role as a Bo and one in its role as an anti-Bo)."

The 2 particles were in causal contact during the the initial decay, with each other and the original Y(4S) they came from. That's the point where the entangled state was established. It's within the light cone, the region where all the info is passed out at. The Y(4S) is a particle containing 2 quarks, b and anti-b. When it decays, each b will get a newly generated d quark eventually, as follows:

Bo = d,anti-b
anti-Bo = anti-d, b

The decay mechanism on one of the particles is stabilized by by raising the activation energy for the normally available decay paths as follows. The 2 components that make up the particle are oscillating back and forth with their own anti-particle with a high enough frequency, that before any path can be taken, it changes to the wrong particle and that path is unavailable then.

The b quark can't oscillate alone, or a net charge would be seen magnets and the "delay clock" could be seen running. The net result is that it is neither Bo, nor anti-Bo. This oscillation is called a CP eigenstate, f. f looks like:

d anti-b <=> anti-d, b

I said one of the particles would begin as an f, but there's no reason, that I know of to constrain it that way. I'm not too familiar with the Std Model. Both particles could begin in that state, with one more likely to have a short life as such. It will become the tag. It seems the flavor symmetry takes some time to develop and One of the particles gets a head start in the region of the original decay. No particle in particular becomes a tag as far as I know.

"you can't say that it was pre-informed about the decay fate of the "first" meson because that doesn't help: such a scheme will necessarily obey Bell's inequality

I looked, but was unable to find, or access any description of the technique and data. Page 4 of the second ref. contains statements I can't construct a picture of. They also name the states differently. In that paper, it seems to me that they're referring to some initial decay, before particle decay begins. Others gave me the impression that they used the first decay to tag both particles ID and then extract curves. I think the 2nd particle's decay curve contains the ttag/decay. ?

Suppose someone pokes their finger in the detector and pokes at random particles giving them enough activation energy to decay. The probability is 1/2, that anyone poked will be an f. The f must decay to the correct particle even though it never saw the tag. The quantum correlations for the finger experiment must match those for the tag run.

Regardless of the details of where ttag/decay ends up in the second particles decay curve. Flavor is a symmetry and if ttag/decay defines to for the second particles decay, then all I see is a phase change(shape) in the fields, Higgs and EM. Phase velocities do exceed c. In this case, the CP eigenstate is maintained by the field between the particles. The finger experiment has the same effect as ttag.

"what about non-EPR correlated nuclear decays?"

A decay here is simply a high energy state of a system falling to a lower energy state with the release of equivalent energy. The system has an inherent stability and an activation energy for any decay path available. Nuclear and particle decays are normally considered with the concept of half life and the equation,

N = Noe-t/t1/2.

Nuclear decays around room temperature are relatively insensitive to temperature, until T times Boltzmann's constant approaches the activation E. In short, in terms of energy, decay(non tunneling) is generally proportional to,

e(-Eact(or delta E)/kT)
, or is the sum of terms containing this factor.

In addition to activation energy driven processes, there is tunneling. One of the first verifications for QM was Gamow, Condon, and Gurney's calc of predicting the number of alpha particles emitted per second in alpha decay. They considered the problem as tunneling through a barrier. The calculated the transmission coefficient and took the number of trials per second, N, as:

N = 1/2 v/R, where v is the velocity of the particle and R is the radius of the nucleus.

The eq for the number of particles/sec, n, is,

n=v/2R*e-2*integral from R to infinity of U,

where U = (2m/hbar2)((zZe2/r-E)dr

E is the energy of the particle. z and Z are the charges of the alpha particle and nucleus respectively. e is the electron charge and m is the mass of the alpha particle.

The following were calculated for n.

n(U238) = 5*10-18 events/sec
n(Po212) = 2*10-6 events/sec

The reference values are.

n(U238) =7.05 *10-18 events/sec
n(Po212) = 3.04*10-7 events/sec

Here's a web calc within a factor of 2. Here's a paper that shows more detail. Stability against decay in isotopes depends on binding energy/nucleon(protons and neutrons). The binding energy represents the height of the potential well. Iron has the highest binding energy.

"Occam's Razor leads us to conclude that all subatomic decays have a random, causeless element, somewhere."

The driving force, or cause, is simply the fact that the system has a path to a lower energy configuration. Even if that path is over an intermediate higher energy hill, or by tunneling. The path over the hill simply requires an activation E. Tunneling simply requires a successful attempt. Even with tunneling, there are underlying fluctuations, that effect "wall height" and particle E. Random just refers to the distribution of events that amount to, "a path has been taken."

I pulled some results from some CLEO B factory analysis here.

The average lifetimes for the 2 particles are, Bo = 1.55ps and anti-Bo = 1.49ps.

1 CP Violation in B Meson Decays

2 Time-Dependent CP Asymmetries in b --> s qbar q Transitions and sin2phi1 in B0 --> J/psi K0 Decays with 386 Million BBbar Pairs
Authors: The Belle Collaboration: K. Abe, et al

988 posted on 12/12/2005 6:20:54 PM PST by spunkets
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To: spunkets
I'm afraid I didn't understand much of what you were trying to say, and some of the statements sounded wrong. Can you summarize your point, and how it differs from mine?
989 posted on 12/12/2005 9:27:23 PM PST by Physicist
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To: Physicist
My reply was essentially to js1138, who I had a discussion with regarding cause. I don't have any disagreement. Most of what I wrote regarding your reply was a restatement of how I understood it. I also posed some questions, since I was unclear on the decay details and not very familiar with the STD Model. I did leave a dangling error in there, but that was to js1138.

If anything sounded wrong, go ahead and comment.

As far as the action at a distance trick, I would appreciate your comment on this.

all I see is a phase change(shape) in the fields, Higgs and EM. Phase velocities do exceed c. In this case, the CP eigenstate is maintained by the field between the particles. The finger experiment has the same effect as ttag.

990 posted on 12/12/2005 9:52:22 PM PST by spunkets
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To: caffe
And even though their ideas break the Laws of Thermodynamics, gravity, general relativity, quantum mechanics and the Big Bang, they insist they are being scientific. This is , to unscientific devotion to naturalism.

Once again, everybody, for the 10150 time, there is absolutely NOTHING about evolution that breaks the Laws of Thermodynamics. Every time someone claims so, somewhere a physicist or engineer dies laughing. Please stop it. Learn what thermodynamics actually is and how it is actually applied before postulating such absurd assertions.

BTW, the Big Bang is a prediction of general relativity and has nothing to do with evolution, either.

How about, one, just one, testable prediction that can be made with a supernatural assumption?

991 posted on 12/13/2005 9:04:56 AM PST by Quark2005 (No time to play. One post per day.)
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To: js1138; spunkets; Physicist
Quantum theory posits there is no distinction between atoms. Could be wrong, but QM has not been wrong yet.

My 2¢ (if anyone still cares):

Not exactly true. QM theory posits there is no distinction between particles with the same quantum numbers. For example, two free electrons in a positive spin state are truly indistinguishable.

Nuclei are a bit more complex. There can be several spin states (and thus energy levels & quantum numbers) of the protons & neutrons, for example (although still a limited number of states). This can (and will) affect decay rates; but you would have to isolate and identify which atoms had which spin states (by separation with a magnetic field, for example) to get this information. Problem is, such an apparatus would have the effect of changing some of the quantum numbers (some of the spins would "realign" with the magnetic field). There would thus be some "smearing" of the original identity of the nuclei caused by the mere fact that you are trying to isolate them - an effect that is totally unavoidable. This blurry zone is statistical in nature; one can only be sure within a statistical certainty of which original particles they are looking at. With lone neutrons (the simplest decaying "nucleus"), you have no certainty at all; with plutonium nuclei, where there are many more quantum numbers to work with, you can (in theory) separate them with less alteration to the initial identities.

If you consider large molecules, there are so many possible quantum states that you start to approach the limit of a classical (i.e. normal) system; they are almost certain to retain their "original" identity upon observation. Nuclei sort of ride that "middle zone" of being truly indistinguishable and distinguishable; they can be properly said to have a limited number of distinguishable states which can be observed on an individual basis with a limited certainty.

(Hope I got all this right; I'm a much newer player to this game than Physicist et.al. QM can really give you a headache.)

992 posted on 12/13/2005 9:36:25 PM PST by Quark2005 (No time to play. One post per day.)
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To: Quark2005

So returning to the original question, will two nuclei having the same spin state decay at the same time? Is there any property of an atom that predicts decay time?


993 posted on 12/13/2005 10:05:05 PM PST by js1138 (Great is the power of steady misrepresentation.)
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To: Quark2005
That's all fine. The way I think about atoms is that, while indentical particles will mix their wavefunctions, you can in practice "mark" one of a pair of otherwise identical atoms by adding or subtracting a neutron or electron. (Easier to visualize than a spin state.) Throw the two of them into a box together and their wavefunctions almost won't mix. I say "almost" because there's a finite probability the spare neutron or electron can "tunnel" from one atom to the other.
994 posted on 12/13/2005 10:25:06 PM PST by Physicist
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To: Quark2005
"For example, two free electrons in a positive spin state are truly indistinguishable."

If they're in a free state, they can be distinguished. Consider Milliken's experiment and spacial separation. In systems, distinguishability depends on spin, as you were describing. A beam energy of electrons(Fermions) with net spin, is higher than a random spin beam. A polarized photon(Boson) beam has the same energy as an unpolarized one. That's the idea in your post.

On these threads it's common to find the belief that random processes have no cause, or that random is a cause. In this subthread, the former was the topic. Random refers to the process itself, is evidenced by the outcome and at any point in the process. If all processes in Nature weren't random, the probability of a real photon "hv" being absorbed by an harmonic oscillator would be zilch, because the linewith was zero and every trial would fail. ie. the tolerance is too tight.

995 posted on 12/14/2005 12:11:41 AM PST by spunkets
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To: Physicist; Quark2005

Continuing this off-topic discussion, I was wondering if I could ask a question which has been on my mind lately. It is my understanding that the wavelength of a particle is inversely related to its momentum, and that therefore, in an accelerator, where both the velocity and the mass of the particles are continualy increasing, the wave aspect of the particle becomes less noticeable. But is there any way in which the wave nature of matter has been or potentially can be exploited in HEP experiments?


996 posted on 12/14/2005 9:41:58 AM PST by RightWingAtheist ("Why thank you Mr.Obama, I'm proud to be a Darwinist!")
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To: RightWingAtheist
But is there any way in which the wave nature of matter has been or potentially can be exploited in HEP experiments?

The answer is yes and no, depending on the context of your question. In the broad setting of the lab, particles in HEP behave very specifically like particles, as their wavelength is so small. Remember, though, that high-energy particles are used to probe tiny distances in collisions, where the wave properties of matter are inextricable from the particle properties. (Actually, relativistic QM, or quantum field theory, is used to understand the properties of these collisions. QFT calculations are very esoteric; the quantum mechanical "structure" of particles is inherent in it and the "wave" and "particle" aspects can't really be explicitly extricated.)

In low-energy physics (a whole other equally important, though less sensationalistic) subfield, approximating particles in scattering experiments as waves is done using partial wave analysis all the time.

997 posted on 12/15/2005 8:20:31 AM PST by Quark2005 (No time to play. One post per day.)
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To: js1138
So returning to the original question, will two nuclei having the same spin state decay at the same time?

No, they have a fixed half-life, but the decay time is governed by probability.

Is there any property of an atom that predicts decay time?

Exact decay time, no. You're right on the $$$, there. There are properties that predict half-lives (probable decay times), but not actual decay times.

998 posted on 12/15/2005 8:24:09 AM PST by Quark2005 (No time to play. One post per day.)
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To: spunkets
If they're in a free state, they can be distinguished.

True. I implicitly assumed free localized electrons with similar momenta when I made my statement (something I shouldn't have done).

Keep in mind though, that from a QM point of view, even unlocalized electrons are considered identical until a measurement is made, at which time they are placed into a position and/or momentum eigenstate (one of an infinite possible allowed number when in free space) - this imparted quantum "number" then gives a unique identity to the particle. You are right in this - free particles do not have discrete position/momenta spectra.

On these threads it's common to find the belief that random processes have no cause, or that random is a cause.

QM (Bell's inequality, etc.) only shows that random processes have no localized, physically measurable cause. It says nothing about globally hidden variables. (Then again, this starts getting us into the realm of the metaphysical and unmeasurable.)

999 posted on 12/15/2005 8:32:57 AM PST by Quark2005 (No time to play. One post per day.)
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To: Quark2005
" QM (Bell's inequality, etc.) only shows that random processes have no localized, physically measurable cause. It says nothing about globally hidden variables. (Then again, this starts getting us into the realm of the metaphysical and unmeasurable.)"

All processes are random. With such phenomenon as EPR, all such system states began in causal contact. The spacelike "comm" between the 2 particles is simply due to phase shifts in an existing field. No energy exchange is necessary, because that was done when the field was established. The field and it's properties are the cause. I see local as defined by the boundary of the system, in these cases of spacelike separation of bound particles. That's, because the state isn't a function of distance, only E.

1,000 posted on 12/15/2005 11:31:24 AM PST by spunkets
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