Part 2: THE EVER GREATER MUON HOAX.
Henry Wilson
accelerated to precisely known velocities near c, then retained in a circular
orbit so that their decay rate can be monitored. From this, an estimate is made
of their average lifetimes at this high speed.
These lifetimes are much longer than those for muons at rest. Srians assure us
that the facts agree exactly with the predictions of SR.
That seems pretty straightforward. How can anyone argue against this apparently
conclusive confirmation of SR?
Well, please read on!!!!
Firstly, by way of example, you all know that the life expectancy of humans is
somewhere around 55 years (the actual figure doesn't matter). However, if you
survive the first year, that figure jumps to about 75 years. If you survive the
first forty years, it goes higher, to maybe 90 years.
Well, the same type of thing happens, on a much larger scale, to muons in this
now (in)famous experiment.
The muons are produced almost simultaneously and immediately start to decay at a
rate determined by their half life (1.56 usec).
Many decay during the acceleration period but, like the humans, those which
don't can be expected to live to a ripe old age.
By the time the muons reach their maximum velocity and the measurements begin,
only a small percentage of the original muons may remain. It is easily to show
that the expected lifetimes of these will be greatly extended, purely on
statistical grounds.
Now you are all familiar with exponential decay curves and the equation for
deriving average lifetimes of radioactive atoms. Consider what happens to the
average lifetime of these experimental muons if we cut off a significant part of
the initial stage of the decay curve?
*
Ta
* |
* |
* |
*
*
*
-----------------------------------------------------------------------------------------------------------------------------------------------------------*
THE AVERAGE LIFETIME OF THE REMAINDER IS GREATLY INCREASED!
The maths goes like this:
The average lifetime of decaying particles is e^(-0.69t/T) x (t-T/0.69) for t
between 0 and infinity. T is the half life and t is time.
Normally that would equal just T/0.69 if the decay starts at t=0.
In this case, however, t DOES NOT start at zero but at the higher value Ta, thus
giving the false impression that the remaining muons are living for much longer
times than they should. THEY ARE NOT!!!!! Their half lives are still 1.56us. The
result is as expected.
BUT THAT IS NOT THE END OF THE DECEPTION!!!!!
By manipulating the acceleration time, Ta, and the final velocity, it is easy
to make the average lifetimes appear to match EXACTLY those predicted by SR.
Ha! Bloody Ha!
You people have been exposed again!
I invite Paul Andeson, Tom Roberts, Franz Heymann, Eric Prebys and the others to
try to find a way out of this one!!!!
Dirk Van de moortel
> The SRians here often refer to cyclotron experiments in which muons are
> accelerated to precisely known velocities near c, then retained in a circular
> orbit so that their decay rate can be monitored. From this, an estimate is made
> of their average lifetimes at this high speed.
> These lifetimes are much longer than those for muons at rest. Srians assure us
> that the facts agree exactly with the predictions of SR.
>
> That seems pretty straightforward. How can anyone argue against this apparently
> conclusive confirmation of SR?
>
> Well, please read on!!!!
>
> Firstly, by way of example, you all know that the life expectancy of humans is
> somewhere around 55 years (the actual figure doesn't matter). However, if you
> survive the first year, that figure jumps to about 75 years. If you survive the
> first forty years, it goes higher, to maybe 90 years.
>
> Well, the same type of thing happens, on a much larger scale, to muons in this
> now (in)famous experiment.
> The muons are produced almost simultaneously and immediately start to decay at a
> rate determined by their half life (1.56 usec).
> Many decay during the acceleration period but, like the humans, those which
> don't can be expected to live to a ripe old age.
> By the time the muons reach their maximum velocity and the measurements begin,
> only a small percentage of the original muons may remain. It is easily to show
> that the expected lifetimes of these will be greatly extended, purely on
> statistical grounds.
HUH?
Having a go at probability and statistics now?
Like:
"Number 14 hasn't shown up this entire evening on the roulette, so I'm
putting all my money on it. Surely it must show up now!"
or
"Look, I have thrown 5 aces with these dice. Surely the probability that
I will do it again on my next throw, will be *much* smaller!"
or
"In our street 10 babies were born last week. All girls.Surely the next
baby must be a boy!"
Shall we go on or do you get the picture?
So let's [snip] the consequences of your misunderstanding of statistics...
> [snip]
>
> Ha! Bloody Ha!
> You people have been exposed again!
> I invite Paul Andeson, Tom Roberts, Franz Heymann, Eric Prebys and the others to
> try to find a way out of this one!!!!
What you are trying to do here, is pathetic.
Ask Androcles if he wants to put your devastating argument on his website.
Alternative: take a short introductory course on probability and
statistics and find your way out yourself. A good exercise.
Dirk Vdm
Martin Hogbin
--
Martin Hogbin
"Henry Wilson" <He...@the.edge> wrote in message news:3b90104f...@nsw.nnrp.telstra.net...
Fantastic! I shall keep a copy of this.
> Now you are all familiar with exponential decay curves and the equation for
> deriving average lifetimes of radioactive atoms.
Yes, and so were the people who did the muon experiments!
Martin
>
Martin Hogbin
--
Martin Hogbin
"Martin Hogbin" <mar...@hogbin.org.uk> wrote in message news:999338594.392.0....@news.demon.co.uk...
>
>
> --
> Martin Hogbin
> "Henry Wilson" <He...@the.edge> wrote in message news:3b90104f...@nsw.nnrp.telstra.net...
I accidentally did a cut-and-paste rather than copy-and-paste.
Oh well!
> Fantastic! I shall keep a copy of this.
>
> > Now you are all familiar with exponential decay curves and the equation for
> > deriving average lifetimes of radioactive atoms.
>
> Yes, and so were the people who did the muon experiments!
>
Martin Hogbin
Androcles
of posts to sci.physics.relativity
4:53 for the cut and paste, 6:17 for the copy and paste, a
difference of 1:24, thereby extending the lifespan of his post
by almost one and half minutes.:)
I wonder how long it would take the average reader
to read all of Martin's posts, reading half of them
every minute and 24 seconds, the rate at which Martin
produces them :)
Henry Wilson
You people repeatedly quote "lifetime of the muons".
The 'increased lifetime' is an expected consequence of cutting off the top end
of the exponential decay curve.
The half life of these muons is exactly the same.
Just use the equation if you want to know why.
SR is now one step closer to death!
Tom Roberts
> THE AVERAGE LIFETIME OF THE REMAINDER IS GREATLY INCREASED!
How stupid. Your long-winded argument boils down to: the muons which
survive a long time have a longer than average survival time. Yes,
that's true. But your conclusion does not follow at all -- after a muon
survives for a long time, IT IS STILL A MUON and will still decay with
its usual decay rate. Note that this is not merely a guess on my part,
or a theoretical prediction, they are _OBSERVED_ to behave that way.
Let's apply some real numbers to your argument, to see how wrong it is:
A muon at rest has a mean lifetime of 2.1970 usec, giving it a half-life
of 1.5228 usec, which I'll round to 1.5 usec. So only 1 in 1024 muons
will survive 15 usec, and only 1 in 1,048,576 muons will survive 30 usec,
and only one in 1.074*10^9 will survive 45 usec. Bailey et al _measured_
their stored muons to have a mean lifetime of 64.419 usec, which means
that _HALF_ the muons survived for 44.652 usec -- this is incredibly
different from the 1 in a billion your argument says should survive.
Henry, you really need to _LEARN_ about physics. You keep making things
up and are INCREDIBLY bad at that.
Tom Roberts tjro...@Lucent.com
Dirk Van de moortel
Here is a little lesson in simple words:
Suppose the half-life of anything (muons, wilsons, you name it...) is
T seconds. Suppose you filter a bunch that has been alive for T seconds.
This bunch has a half-life of T seconds from then on.
So suppose you filter out another (smaller) bunch after 2T seconds.
These guys have been alive for 2T seconds, and their half-life
from then on is still T. And so on.
After a while you could end up with one guy that has been alive
for let's say 10T seconds. Its half-life from then on is.... right: T seconds.
If it happens to be still alive at 11T seconds, it's half-life from then on is...
indeed: T seconds.
All these words are expressed with the exponential decay formula
and standard mathematics.
The thing is: a muon does not know how old it is: it does not say
"Hey, I have had 2 half-lifes now, I'll go to sleep" or "Sheesh, I have
had 125 half-lives, I must be immortal. They have been wrong
about my half-life, let's call Henry".
If you don't like this, you'll have to propose another decay curve
than the exponential one, but make sure it fits the observed data.
It won't be easy.
> SR is now one step closer to death!
You are one step closer to Androcles.
Perhaps you could ask him to write a program to produce the data
that fits the decay curve of your dreams. Do some team work.
Dirk Vdm
Roman Arce
> Like:
> "Number 14 hasn't shown up this entire evening on the roulette, so I'm
> putting all my money on it. Surely it must show up now!"
> or
> "Look, I have thrown 5 aces with these dice. Surely the probability that
> I will do it again on my next throw, will be *much* smaller!"
> or
> "In our street 10 babies were born last week. All girls.Surely the next
> baby must be a boy!"
People say both stupid and smart things, aetherists have said a lot of
stupid things continuosly, surely the next thing they say...
nevermind.
And to the muon moron, go publish your shit in a journal.
Androcles
"Tom Roberts" <TomRo...@avenew.com> wrote in message
news:3B9114F4...@avenew.com...
> Henry Wilson wrote:
> > THE AVERAGE LIFETIME OF THE REMAINDER IS GREATLY INCREASED!
>
> How stupid. Your long-winded argument boils down to: the muons which
> survive a long time have a longer than average survival time. Yes,
> that's true. But your conclusion does not follow at all -- after a muon
> survives for a long time, IT IS STILL A MUON and will still decay with
> its usual decay rate. Note that this is not merely a guess on my part,
> or a theoretical prediction, they are _OBSERVED_ to behave that way.
What Tom did was to lay some down on the laboratory bench, then started a
clock
and timed how many were left as time passed. Then he put them in a powerful
magnetic field, and time how long they lived. From that he can conclude that
magnetic fields make muons live longer, but he prefers to conclude that it
is there velocity that makes them live longer. The reason for his conclusion
is that has been taught relativity, and has unlearned logic (if he ever
learned any in the first place). This entitles him to call other people
stupid.
> Let's apply some real numbers to your argument, to see how wrong it is:
>
> A muon at rest
(meaning laying on the lab bench)
> has a mean lifetime of 2.1970 usec, giving it a half-life
> of 1.5228 usec, which I'll round to 1.5 usec. So only 1 in 1024 muons
> will survive 15 usec, and only 1 in 1,048,576 muons will survive 30 usec,
> and only one in 1.074*10^9 will survive 45 usec. Bailey et al _measured_
> their stored muons to have a mean lifetime of 64.419 usec, which means
> that _HALF_ the muons survived for 44.652 usec -- this is incredibly
> different from the 1 in a billion your argument says should survive.
>
> Henry, you really need to _LEARN_ about physics. You keep making things
> up and are INCREDIBLY bad at that.
And of course Tom has to _LEARN_ about logic, he keeps making stupid
assumptions as to the cause of the increased lifespan of the muons, which is
a physical cause, and not the product of some fanciful mathematical cause.
Math doesn't cause things to happen, math describes how things happen. When
Tom has _LEARNED_ the difference between physical cause and mathematical
cause, he'll be the better off for it.
Vertner Vergon
> "Henry Wilson" <He...@the.edge> wrote in message
news:3b90104f...@nsw.nnrp.telstra.net...
> > The SRians here often refer to cyclotron experiments in which muons are
> > accelerated to precisely known velocities near c, then retained in a
circular
> > orbit so that their decay rate can be monitored. From this, an estimate
is made
> > of their average lifetimes at this high speed.
> > These lifetimes are much longer than those for muons at rest. Srians
assure us
> > that the facts agree exactly with the predictions of SR.
> >
> > That seems pretty straightforward. How can anyone argue against this
apparently
> > conclusive confirmation of SR?
VERGON:
I can.
In his paper Einstein gave as an example a clock that travels in a circle
and then
comes back to its origin.
That is exactly what the H-Keating experiment was.
And that's the same situation with a particle -- or atomic clock -- in a
storage
ring.
Now it so happens that half of the time the particle is in the approach
mode, and
half the time in the recession mode. In between - in the change over mode --
it is at the
two points of the circle that are orthogonal to the observer.
This turns out to be the median frequency of the whole circle.
It is also the transverse frequency. And by coincidence the transverse
frequency, irrespective of velocity or clock rate, is the same as the
dilation
rate given by Einstein.
so it is not a confirmation -- just a coincidence.
Eric Prebys please note:- There is wiggle room after all. I never lost
faith.
*********************************************************
Androcles
"Henry Wilson" <He...@the.edge> wrote in message
> SR is now one step closer to death!
It is already dead, Henry.
It just hasn't laid down yet (in Western civilization, the Eastern bunch
doesn't revere it).
The next generation will "see the light" (excuse the deliberate pun) and let
it become a cult, just as the aetherialists have theirs...
David Evens
wrote:
>"Tom Roberts" <TomRo...@avenew.com> wrote in message
>news:3B9114F4...@avenew.com...
>> Henry Wilson wrote:
>> > THE AVERAGE LIFETIME OF THE REMAINDER IS GREATLY INCREASED!
>>
>> How stupid. Your long-winded argument boils down to: the muons which
>> survive a long time have a longer than average survival time. Yes,
>> that's true. But your conclusion does not follow at all -- after a muon
>> survives for a long time, IT IS STILL A MUON and will still decay with
>> its usual decay rate. Note that this is not merely a guess on my part,
>> or a theoretical prediction, they are _OBSERVED_ to behave that way.
>
>What Tom did was to lay some down on the laboratory bench, then started a
>clock
>and timed how many were left as time passed. Then he put them in a powerful
>magnetic field, and time how long they lived. From that he can conclude that
>magnetic fields make muons live longer, but he prefers to conclude that it
>is there velocity that makes them live longer. The reason for his conclusion
>is that has been taught relativity, and has unlearned logic (if he ever
>learned any in the first place). This entitles him to call other people
>stupid.
In reality, of course, it is observed that when you place slow-moving
muons in a powerful magnetic field, their lives are not extended by as
much as when you place fast-moving muons in a field of identicle
strength. In fact, it has been OBSERVED that field strength is
UNCONNECTED to the lifetimes of muons.
>> Let's apply some real numbers to your argument, to see how wrong it is:
>>
>> A muon at rest
>(meaning laying on the lab bench)
>> has a mean lifetime of 2.1970 usec, giving it a half-life
>> of 1.5228 usec, which I'll round to 1.5 usec. So only 1 in 1024 muons
>> will survive 15 usec, and only 1 in 1,048,576 muons will survive 30 usec,
>> and only one in 1.074*10^9 will survive 45 usec. Bailey et al _measured_
>> their stored muons to have a mean lifetime of 64.419 usec, which means
>> that _HALF_ the muons survived for 44.652 usec -- this is incredibly
>> different from the 1 in a billion your argument says should survive.
>>
>> Henry, you really need to _LEARN_ about physics. You keep making things
>> up and are INCREDIBLY bad at that.
>And of course Tom has to _LEARN_ about logic, he keeps making stupid
>assumptions as to the cause of the increased lifespan of the muons, which is
>a physical cause, and not the product of some fanciful mathematical cause.
>Math doesn't cause things to happen, math describes how things happen. When
>Tom has _LEARNED_ the difference between physical cause and mathematical
>cause, he'll be the better off for it.
You forgot to support your assumption that the muons are not living as
long as they are observed to live.
Dirk Van de moortel
:-)
> And to the muon moron, go publish your shit in a journal.
or on Androcles' website.
Dirk Vdm
Dirk Van de moortel
"Androcles" <andr...@home.com> wrote in message news:0zik7.161655$EP6.45...@news1.rdc2.pa.home.com...
>
> And of course Tom has to _LEARN_ about logic, he keeps making stupid
> assumptions as to the cause of the increased lifespan of the muons, which is
> a physical cause, and not the product of some fanciful mathematical cause.
> Math doesn't cause things to happen, math describes how things happen. When
> Tom has _LEARNED_ the difference between physical cause and mathematical
> cause, he'll be the better off for it.
Henry Wilson:
--------------
| Moving muons living longer confirm special relativity.
| Androcles has a proof that SR is crap.
| So the muon data is all faked! It's a hoax!
| oh... sorry... the muons have been artificially selected.
| That's why they live longer. They are cheating.
| So SR is still wrong.
| Oh... I made a mistake... about exponential decay....
| let's be quiet for a while. Androcles will come and help me out
Androcles:
-----------
| It's probably not faked. Let's accept the data.
| There must be a physical cause.
| It can't be relativistic because that's crap. That I proved with my
| killfile, C-program, and my personal system of sheesh-logic.
| So there must be another physical cause.
| What else is there?
| Aha! Magnetism! That's why they live longer.
| Wilson, I have nailed them down. Keep a low profile for a while.
Dirk Vdm
Dirk Van de moortel
Androcles continued:
| Oops, David Evens just said that magnetism can't possibly have
| any influence. He seems to be quite sure about it. He probably has
| some data to back it up.
| Can't just call him a liar.
| I'm running out of fresh underpants.
| I'll get together with Wilson, so we can adjust our strategies.
| I'll keep a low profile for a while too.
| For a short while anyway.
Dirk Vdm
Martin Hogbin
--
Martin Hogbin
"Androcles" <andr...@home.com> wrote in message news:i64k7.159590$EP6.44...@news1.rdc2.pa.home.com...
>
> "Martin Hogbin" <mar...@hogbin.org.uk> wrote in message
> news:999338924.542.0....@news.demon.co.uk...
> >
> > --
> > Martin Hogbin
> > "Martin Hogbin" <mar...@hogbin.org.uk> wrote in message
> news:999338594.392.0....@news.demon.co.uk...
> > >
> > >
> > > --
> > > Martin Hogbin
> > > "Henry Wilson" <He...@the.edge> wrote in message
> news:3b90104f...@nsw.nnrp.telstra.net...
> >
> > I accidentally did a cut-and-paste rather than copy-and-paste.
> >
> It looks as if Martin has greatly extended the lifespan
> of posts to sci.physics.relativity
> 4:53 for the cut and paste, 6:17 for the copy and paste, a
> difference of 1:24, thereby extending the lifespan of his post
> by almost one and half minutes.:)
I was explaining how I had accidentally deleted all of the
original post in my first response.
> I wonder how long it would take the average reader
> to read all of Martin's posts, reading half of them
> every minute and 24 seconds, the rate at which Martin
> produces them :)
So what is your opinion of Wilson's theory?
Do you support everything that agrees with SR on principle?
You do your own cause no good at all by making no comment
about an idea that is clearly ludicrous.
Martin Hogbin
Androcles
>
>
> --
> Martin Hogbin
> "Androcles" <andr...@home.com> wrote in message
news:i64k7.159590$EP6.44...@news1.rdc2.pa.home.com...
> >
> > "Martin Hogbin" <mar...@hogbin.org.uk> wrote in message
> > news:999338924.542.0....@news.demon.co.uk...
> > >
> > > --
> > > Martin Hogbin
> > > "Martin Hogbin" <mar...@hogbin.org.uk> wrote in message
> > news:999338594.392.0....@news.demon.co.uk...
> > > >
> > > >
> > > > --
> > > > Martin Hogbin
> > > > "Henry Wilson" <He...@the.edge> wrote in message
> > news:3b90104f...@nsw.nnrp.telstra.net...
> > >
> > > I accidentally did a cut-and-paste rather than copy-and-paste.
> > >
> > It looks as if Martin has greatly extended the lifespan
> > of posts to sci.physics.relativity
> > 4:53 for the cut and paste, 6:17 for the copy and paste, a
> > difference of 1:24, thereby extending the lifespan of his post
> > by almost one and half minutes.:)
>
> I was explaining how I had accidentally deleted all of the
> original post in my first response.
I know... didn't you see the ":) " at the end... its a smile. I've done it
too.
> > I wonder how long it would take the average reader
> > to read all of Martin's posts, reading half of them
> > every minute and 24 seconds, the rate at which Martin
> > produces them :)
>
> So what is your opinion of Wilson's theory?
Henry hasn't shown that the decay rate of muons is sufficient to cause a
misreading of the empirical data, IMHO. However, it may color the results a
little. On the other hand, the empirical data is that muons accelerated in
powerful magnetic fields have their lifespan extended. The relativists claim
this is caused by their velocity and a belief in time dilation (no physical
cause) and refuse to investigate other causes, their interest being to
confirm what they already "know" and is "predicted" by relativity. This is
how astrologers convince the gullible. Find a fact and claim it is a
prediction, reproduce the fact, and say "I told you so, now cross my palm
with silver".
> Do you support everything that agrees with SR on principle?
I do not deny empirical data. The clock on the GPS satellite runs faster
than a clock on the ground. Such clocks are moving at high speed through the
Earth's magnetic field, and are weightless. The muon has its its life
extended when accelerated in a powerful magnetic field. Magnetic and
electric fields are part of physics and our understanding of the nature of
light. I do not know the full mechanics of why a muon lives longer, or a
cesium clock will speed up. All I can tell you is that "time dilation" is
ludicrous, SR is certainly contradicted by a faster clock on the satellite,
and the physical causes of the phenomena are assumed, not investigated. Nor
do I claim that the magnetic field is the cause, I merely offer it as a
possible line of investigation to confirm or deny an hypothesis. Nature does
not give up her secrets easily, and dicovering them is a long, hard road.
Jumping to conclusions as the relativists have done is not the way of
science.
>
> You do your own cause no good at all by making no comment
> about an idea that is clearly ludicrous.
Comment given.
Tom Roberts
> What Tom did was to lay some down on the laboratory bench, then started a
> clock
> and timed how many were left as time passed. Then he put them in a powerful
> magnetic field, and time how long they lived. From that he can conclude that
> magnetic fields make muons live longer, but he prefers to conclude that it
> is there velocity that makes them live longer.
1) It was not I who did his, but Bailey et al.
2) they don't "lay down onthe laboratory bench", but must be moving
~0.9994 c around the storage ring.
3) The lifetime of moving muons agrees to ithin 0.1% or so with the
predictions of SR, independent of any magnetic or electric fields
present. This includes the velocity dependence of their lifetime.
4) The fact that kilometer-long muon beams exist, without any magnetic
field along most of their length, clearly refutes your fuzzy-
worded claim that the magnetic field "causes" this.
Tom Roberts tjro...@Lucent.com
Henry Wilson
Yes that's very interesting.
Henry Wilson
Tom, having gained your attention with my deliberate oversimplification of this
problem, I can now get down to the nitty gritty.
Firstly, take any group of random numbers and find its average value. Then
remove, say, the bottom half of those numbers, and calculate the average. It
will have inreased by about 50%.
In the case of an exponential type curve, if you remove all the members with
expected life times less than several 'half times', the average life of the
remainder is greatly increased.
Average lifetime cannot be calculated as = T/.69 (or whatever).
In the muon experiment, the acceleration time is therefore critical in
determining the 'average lifetime' of the remainder. You are only looking at the
'tail end' of the decay curve.
A second point. Why should the decay rate of a sample of particles, which were
produced simultaneously, follow an exponential curve, at all?
This is a vastly different situation from the natural radioactivity decay
process in which the atoms were produced at entirely different times.
In fact, I go as far as to speculate that the whole principle of exponential
decay RELIES ON random time of origin.
If you add two exponential decay curves, do you get another exponential curve?
No.
I suggest that decay rate of particles produced simultaneously, IS NOT
EXPONENTIAL.
I suggest it should be something along the lines: the probability of a particle
surviving is proportional to the time it has already survived.
That cannot be tested with natural radioactivity because the origins are
unknown.
Have you ever considered just WHY decay rates appear to be exponential?
On a slightly different note, a third point is that there is no reason to
believe that the decay curve during acceleration is anything like that when the
ring has been stabilized.
There is no reason to believe that the decay process is not massively affected
by the strong confining magnetic fields coupled with the enormous reaction to
that centripetal force.
>
>
>Tom Roberts tjro...@Lucent.com
Androcles
> Androcles wrote:
> > What Tom did was to lay some down on the laboratory bench, then started
a
> > clock
> > and timed how many were left as time passed. Then he put them in a
powerful
> > magnetic field, and time how long they lived. From that he can conclude
that
> > magnetic fields make muons live longer, but he prefers to conclude that
it
> > is there velocity that makes them live longer.
>
> 1) It was not I who did his, but Bailey et al.
experiment yourself, I was being facetious.
> 2) they don't "lay down onthe laboratory bench", but must be moving
> ~0.9994 c around the storage ring.
Where powerful magnetic fields are present...
> 3) The lifetime of moving muons agrees to ithin 0.1% or so with the
> predictions of SR, independent of any magnetic or electric fields
> present.
I'm not denying it. I'm saying that the field is the cause, not Einstein's
mathematics or any fanciful time dilation.
This includes the velocity dependence of their lifetime.
There is no velocity dependence of their lifetime. An extended lifetime is
not a function of velocity, but a function of a physical cause, such as an
electric or magnetic or gravitational field. The simple fact is you nor
anyone else has investigated this, all you do is say it works by the magic
of Einstein's SR.
> 4) The fact that kilometer-long muon beams exist, without any magnetic
> field along most of their length, clearly refutes your fuzzy-
> worded claim that the magnetic field "causes" this.
The fact that there are steering and accelerating fields *anywhere* along
the length clearly demonstrates your prejudice, bias and willingness to
approach falsehood to defend SR.
Henry Wilson
What causes exponential decay anyway?
Why do some particles live billions of times longer than others?
Why should the decay curve of particles produced simultaneously (as in the muon
ring) have similar characteristics to that of a quantity of randomly formed
radioactive matter?
PaulDanaher
> What causes exponential decay anyway?
Angels. Or demons, depending on whether you're on the side of the angels or
not. Henry, you've just demonstrated yet again that you were lying in your
teeth (if you have any) when you said you have a physics degree, acquired
25, 30 0r 40 years ago. As for your second degree in "psychology +
genetics", students have to learn some statistics for both of those subjects
(let alone a combination of them), so I think we can forget about that claim
too.
> Why do some particles live billions of times longer than others?
> Why should the decay curve of particles produced simultaneously (as in the
muon
> ring) have similar characteristics to that of a quantity of randomly
formed
> radioactive matter?
How can anybody express such far-reaching ignorance in so few words? You are
truly awesome.
Dirk Van de moortel
> On Sat, 01 Sep 2001 12:03:48 -0500, Tom Roberts <TomRo...@avenew.com> wrote:
>
> >Henry Wilson wrote:
> >> THE AVERAGE LIFETIME OF THE REMAINDER IS GREATLY INCREASED!
> >
> >How stupid. Your long-winded argument boils down to: the muons which
> >survive a long time have a longer than average survival time. Yes,
> >that's true. But your conclusion does not follow at all -- after a muon
> >survives for a long time, IT IS STILL A MUON and will still decay with
> >its usual decay rate. Note that this is not merely a guess on my part,
> >or a theoretical prediction, they are _OBSERVED_ to behave that way.
> >
> >Let's apply some real numbers to your argument, to see how wrong it is:
> >
> >A muon at rest has a mean lifetime of 2.1970 usec, giving it a half-life
> >of 1.5228 usec, which I'll round to 1.5 usec. So only 1 in 1024 muons
> >will survive 15 usec, and only 1 in 1,048,576 muons will survive 30 usec,
> >and only one in 1.074*10^9 will survive 45 usec. Bailey et al _measured_
> >their stored muons to have a mean lifetime of 64.419 usec, which means
> >that _HALF_ the muons survived for 44.652 usec -- this is incredibly
> >different from the 1 in a billion your argument says should survive.
> >
> >Henry, you really need to _LEARN_ about physics. You keep making things
> >up and are INCREDIBLY bad at that.
>
>
> problem, I can now get down to the nitty gritty.
Ha, Deliberate Oversimplification To Gain The Attention.
Tactics!
> Firstly, take any group of random numbers and find its average value. Then
> remove, say, the bottom half of those numbers, and calculate the average. It
> will have inreased by about 50%.
> In the case of an exponential type curve, if you remove all the members with
> expected life times less than several 'half times', the average life of the
> remainder is greatly increased.
> Average lifetime cannot be calculated as = T/.69 (or whatever).
> In the muon experiment, the acceleration time is therefore critical in
> determining the 'average lifetime' of the remainder. You are only looking at the
> 'tail end' of the decay curve.
You are repeatig exactly what you said before. Have you read
the objections? Or is this the same tactics as before? Why would
you that? You already have the attention.
> A second point. Why should the decay rate of a sample of particles,
> which were
> produced simultaneously, follow an exponential curve, at all?
> This is a vastly different situation from the natural radioactivity decay
> process in which the atoms were produced at entirely different times.
> In fact, I go as far as to speculate that the whole principle of exponential
> decay RELIES ON random time of origin.
> If you add two exponential decay curves, do you get another exponential
> curve?
> No.
>
> I suggest that decay rate of particles produced simultaneously, IS NOT
> EXPONENTIAL.
Of course! Why hasn't anyone else come up with that idea?
Measurements are made. Every observed curve is incredibly exponential
but Wilson has decided it can't be exponential. Why not? because it's
embarrassing. SR can't possibly be right, so it must be another kind
of curve. What would we try now?
> I suggest it should be something along the lines: the probability of a particle
> surviving is proportional to the time it has already survived.
Along these lines. Okay, propose away. See if you can come up with
something that would at least *look* like an exponential. Start with 1/x^n.
> That cannot be tested with natural radioactivity because the origins are
> unknown.
> Have you ever considered just WHY decay rates appear to be exponential?
Perhaps Quantum Physics... unless that is wrong too of course (after all,
Einstein was somehow involved in its development, so QP can't be right)
Exponential decay is observed. But it can't be observed.
So Quantum Physics must be the next Giant Hoax to tackle.
> On a slightly different note, a third point is that there is no reason to
> believe that the decay curve during acceleration is anything like that when the
> ring has been stabilized.
> There is no reason to believe that the decay process is not massively affected
> by the strong confining magnetic fields coupled with the enormous reaction to
> that centripetal force.
Only experiments. So that wouldn't be a reason of course.
Besides, Androcles just formulated the brilliant idea about magnetism
influencing lifetimes. And Androcles can't be wrong.
Harry, go play with the marbles. Don't swallow any.
See you on your next thread "THE GREAT QUANTUM HOAX."
Dirk Vdm
Tom Roberts
> Yes, Tom. I didn't think for one moment that you would actually perform an
> experiment yourself, I was being facetious.
Well, you are wrong. I have participated in numerous experiments which
are rather similar to this (in that they measured aspects of high-energy
particles). I spent 10 years performing and analyzing HEP experiments at
Fermilab, Brookhaven, and Argonne's ZGS.
> I'm saying that the field is the cause, not Einstein's
> mathematics or any fanciful time dilation.
Your claim does not hold up to experimental scrutiny. Muons and pions
in beamlines also travel much further than (proper lifetime)*(velocity)
would permit (pions are a much more stringent test of this than are
muons). The only magnetic field is the ~0.6 gauss of the earth, which
is of course present when the proper lifetime is measured for the
particles at rest.
> There is no velocity dependence of their lifetime. An extended lifetime is
> not a function of velocity, but a function of a physical cause, such as an
> electric or magnetic or gravitational field.
You are just plain wrong. As I have said so often, this best described
as a _geometrical_ effect directly comparable to perspective.
> The simple fact is you nor
> anyone else has investigated this,
Again you are wrong. This has been extensively studied experimentally.
> all you do is say it works by the magic
> of Einstein's SR.
Not true. I say that the experiments are all consistent with the
predictions of SR. And I also say that except for the ether theories
equivalent to SR, nobody has ever come up with any theory which
compares with SR in its breadth of applicbility; and none of them
compare to SR in elegance and ability to lead to new physics.
Tom Roberts tjro...@Lucent.com
Tom Roberts
> What causes exponential decay anyway?
Any collection of objects whose individual probability of decaying is
constant in time will have an exponential decay curve.
> Why do some particles live billions of times longer than others?
Because their individual probability of decaying in a given period of
time is billions of times smaller that that of the others.
> Why should the decay curve of particles produced simultaneously (as in the muon
> ring) have similar characteristics to that of a quantity of randomly formed
> radioactive matter?
Because that is a property of exponential decay curves. Just think
about it -- if the individual probability of a particle decaying is
constant in time, then at any time T, whatever sample of particles
you have will decay exponentially, _REGARDLESS_ of when they were
actually created. Particles which were also created but which decayed
before time T are simply not part of the sample.
This is so basic to this whole discussion. I repeat: you _REALLY_ need
to actually _LEARN_ about physics, and stop making it up -- you are
_REALLY_BAD_ at that.
Tom Roberts tjro...@Lucent.com
Tom Roberts
> I suggest that decay rate of particles produced simultaneously, IS NOT
> EXPONENTIAL.
You are wrong. Real experiments observe exponential decay to extremely
high accuracy. This implies that their individual probability of
decaying is constant in time. C.f. my other recent post in this thread.
Tom Roberts tjro...@lucent.com
Androcles
"Henry Wilson" <He...@the.edge> wrote in message
decay"?
If the former, I don't know. If the latter, then e is the base of the
natural logarithm.
exp(x) = e^x.
It's derivative is itself, its integral is itself. It's inverse function is
the natural logarithm, ln(x).
If you charge a capacitor with a fixed voltage through a resistor, the
voltage across the capacitor will climb until no more current flows to
charge the capacitor, which occurs when the potential difference across the
resistor has fallen to zero and the potential across the capacitor has
reached the impressed voltage. The charge rate is an exponential function.
If you connected two water tanks, one full and one empty, with a very thin
pipe at their base, the height of the tanks being equal, water would slowly
flow from the full tank into the empty. As the level fell in the full tank
and rose in the empty tank, the pressures would eventually balance when the
empty tank was half full and the full tank was half empty. The height of
water in the empty tank is an exponential function of time. The function, y
= A.exp(-tb) describes the behaviour. The rate of change of exp(x) is
exp(x). It is related to the Fibonacci series, which you can create by
summing the two previous numbers in the series.
1,1,2,3,5,8,13,21,34,55,89,144...
The differences form the series
1-1=0
2-1=1
3-2=1
5-3=2
8-5=3
13-8=5
and so on, which is the same series.
> Why do some particles live billions of times longer than others?
That question is really "Why do particles decay?"
I don't know.
All we know is that they do, that they appear to do so randomly, and a large
number of randomly decaying particles will follow an exponential curve as a
function of time. When there are only a few particles remaining the
randomness dominates and no prediction is possible, but for a large number
the statistical result is the exponential curve. The 'half-life' is a
convenient way of expressing this, but you'll also see that it is the
Fibonnacci series as well.
144,89,55,34,21,13,8,5,3,2, 1 = Fib(t1)
144, 72, 36, 18, 9, 4, 2, 1 = half-life(t2)
The intervals t1 and t2 being different. In (say) one hour, all 144
particles have decayed to nothing, we are just dividing the hour into 6
minutes intervals and 10 minute intervals (roughly). its just a different
way of describing the same thing. Of course a sample of 144 is too small to
use in a real situation, the randomness of thedecay would predominate, but
you get the idea. If we used a 144,000,000,000 instead, we end up with
1,000,000,000 after 1 hour, assuming a half-life of 10 minutes. The smaller
the sample, the less reliable the half-life will be. If you are down to 2
particles, when they decay is totally unpredictable.
> Why should the decay curve of particles produced simultaneously (as in the
muon
> ring) have similar characteristics to that of a quantity of randomly
formed
> radioactive matter?
Because both follow the same natural decay law, for a large sample. This
same law describes birth and death rates for human (or animal) populations,
given that people are born and die randomly. The current life span of a
human being is (last time I heard) 72 for a male and 75 for a female, but
some die in childhood and some make it to 100 years or more. Just why we die
is a question we haven't answered yet, but even if we never died from a
natural cause, eventually we would all perish from accident. Being invaded
by a disease is really an accidental death, as much as being hit by a truck,
and every death certificate gives a cause of death. There is really no "died
of old age" to consider.
I don't know why we die, or muons decay. Nor does anyone else.
Androcles
"PaulDanaher" <wa...@earthlink.net> wrote in message
news:Q9Lk7.4276$ln4.3...@newsread1.prod.itd.earthlink.net...
reply "Angels" is the most absurd thing I have ever come across. You are
only interested in boosting your own ego to put others down, and your
character stinks.
Androcles
I almost gave the same reply to Tom Roberts as I gave to Paul Danaher,
except that Tom has at least attempted to answer the question. He just has
to twist the knife at the end, and he is _REALLY_BAD_ at that.
Androcles
> Androcles wrote:
> > Yes, Tom. I didn't think for one moment that you would actually perform
an
> > experiment yourself, I was being facetious.
>
> Well, you are wrong. I have participated in numerous experiments which
> are rather similar to this (in that they measured aspects of high-energy
> particles). I spent 10 years performing and analyzing HEP experiments at
> Fermilab, Brookhaven, and Argonne's ZGS.
>
>
> > I'm saying that the field is the cause, not Einstein's
> > mathematics or any fanciful time dilation.
>
> Your claim does not hold up to experimental scrutiny.
So where do we go from here?
> Muons and pions
> in beamlines also travel much further than (proper lifetime)*(velocity)
> would permit (pions are a much more stringent test of this than are
> muons). The only magnetic field is the ~0.6 gauss of the earth, which
> is of course present when the proper lifetime is measured for the
> particles at rest.
>
>
> > There is no velocity dependence of their lifetime. An extended lifetime
is
> > not a function of velocity, but a function of a physical cause, such as
an
> > electric or magnetic or gravitational field.
>
> You are just plain wrong.
You are just plain wrong. (I can make assertions too!)
So where do we go from here?
As I have said so often, this best described
> as a _geometrical_ effect directly comparable to perspective.
Nonsense. Physical causes can be described mathematically, but mathematics
cannot BE a cause. The BEST description is the cause itself, then you can
express that geometric terms if you wish. Newton's description of gravity as
a force is the "best" description of how a body falls, it predicts
accurately, but it doesn't tell us what gravity IS.
> > The simple fact is you nor
> > anyone else has investigated this,
>
> Again you are wrong. This has been extensively studied experimentally.
No, *you* are wrong. No matter how much you experiment, in the back of your
mind is the certainty that Einstein's guess was correct; that colors your
approach, making it subjective instead of objective. The experimenter is
part of the experiment, and his views are colored by what he believes.
Copernicus thought planets moved in circles, not ellipses, and Wegener
thought continents floated. Both were right, and both were wrong. You are no
different, and neither am I.
>
> > all you do is say it works by the magic
> > of Einstein's SR.
>
> Not true. I say that the experiments are all consistent with the
> predictions of SR. And I also say that except for the ether theories
> equivalent to SR, nobody has ever come up with any theory which
> compares with SR in its breadth of applicbility; and none of them
> compare to SR in elegance and ability to lead to new physics.
The way to new physics hasn't changed, it is still objectivity. You are
advocating we retain the geocentric universe because it can accurately
predict eclipses, and the best available. Thankfully, new physics began when
Copernicus changed that way of thinking. You way of thinking is that of the
Inquisition, resistant to change.
Your relativity isn't logical, it is a religion.
John Holland
Get your mathematics right Androcles. The level of the empty tank does not
start at level A, but at level zero. That is why you called it the empty
tank. And by the way, if you take one full reservoir, which is not connected
to another one and assume the outflow to be a linear function of its storage
(which is a normal assumption in hydrology), you get a simple first degree
linear differential equation, of which the solution as we all know is the
exponential decay function.
If you want to introduce some mathematics, better stick to the mathematical
statement of the problem which is usually a differential equation of some
sort, and the solution thereoff. Growth rates of rabbit populations may
follow a Fibonacci series, exponential decay certainly does not.
Cheers,
JH
Dirk Van de moortel
>
> [snip]
>
> If you want to introduce some mathematics, better stick to the mathematical
> statement of the problem which is usually a differential equation of some
> sort, and the solution thereoff. Growth rates of rabbit populations may
> follow a Fibonacci series, exponential decay certainly does not.
Androfumble has difficulties with differential equations:
http://groups.google.com/groups?q=author:androcles+%22%40t%27/%40t%22&hl=en&newwindow=1&safe=off&rnum=1&selm=BAH37.2415%24p7.696442%
40news1.rdc2.pa.home.com
| "You see, the only way
| @t/@t' = @t'/@t
| is if
| t' = t,
| which is all the exercise was intended to show anyway.
| It's so nice to have a relativist tell me about clueless high school
| students and then prove how clued in they were after all. To tell me I'm
| wrong, then prove I'm right."
Dirk Vdm
Dirk Van de moortel
>
> "Tom Roberts" <TomRo...@avenew.com> wrote in message
> news:3B93C99C...@avenew.com...
> > Androcles wrote:
> > > Yes, Tom. I didn't think for one moment that you would actually perform
> an
> > > experiment yourself, I was being facetious.
> >
> > Well, you are wrong. I have participated in numerous experiments which
> > are rather similar to this (in that they measured aspects of high-energy
> > particles). I spent 10 years performing and analyzing HEP experiments at
> > Fermilab, Brookhaven, and Argonne's ZGS.
> >
> >
> > > I'm saying that the field is the cause, not Einstein's
> > > mathematics or any fanciful time dilation.
> >
> > Your claim does not hold up to experimental scrutiny.
>
>
> So where do we go from here?
Let's all go to:
> > Muons and pions
> > in beamlines also travel much further than (proper lifetime)*(velocity)
> > would permit (pions are a much more stringent test of this than are
> > muons). The only magnetic field is the ~0.6 gauss of the earth, which
> > is of course present when the proper lifetime is measured for the
> > particles at rest.
> >
> >
> > > There is no velocity dependence of their lifetime. An extended lifetime
> is
> > > not a function of velocity, but a function of a physical cause, such as
> an
> > > electric or magnetic or gravitational field.
> >
> > You are just plain wrong.
>
> [added two blank lines and this one to make it somewhat more readable]
>
> You are just plain wrong. (I can make assertions too!)
> So where do we go from here?
Let's all go to:
once more.
> As I have said so often, this best described
> > as a _geometrical_ effect directly comparable to perspective.
> Nonsense. Physical causes can be described mathematically, but mathematics
> cannot BE a cause. The BEST description is the cause itself, then you can
> express that geometric terms if you wish. Newton's description of gravity as
> a force is the "best" description of how a body falls, it predicts
> accurately, but it doesn't tell us what gravity IS.
>
> > > The simple fact is you nor
> > > anyone else has investigated this,
> >
> > Again you are wrong. This has been extensively studied experimentally.
>
> [added two blank lines and this one to make it somewhat more readable]
>
> No, *you* are wrong. No matter how much you experiment, in the back of your
> mind is the certainty that Einstein's guess was correct; that colors your
> approach, making it subjective instead of objective. The experimenter is
> part of the experiment, and his views are colored by what he believes.
> Copernicus thought planets moved in circles, not ellipses, and Wegener
> thought continents floated. Both were right, and both were wrong. You are no
> different, and neither am I.
Let's all go to:
once more again.
> > > all you do is say it works by the magic
> > > of Einstein's SR.
> >
> > Not true. I say that the experiments are all consistent with the
> > predictions of SR. And I also say that except for the ether theories
> > equivalent to SR, nobody has ever come up with any theory which
> > compares with SR in its breadth of applicbility; and none of them
> > compare to SR in elegance and ability to lead to new physics.
> The way to new physics hasn't changed, it is still objectivity. You are
> advocating we retain the geocentric universe because it can accurately
> predict eclipses, and the best available. Thankfully, new physics began when
> Copernicus changed that way of thinking. You way of thinking is that of the
> Inquisition, resistant to change.
> Your relativity isn't logical, it is a religion.
We have gone there enough now.
Dirk Vdm
Bill Rowe
He...@the.edge(Henry Wilson) wrote:
>What causes exponential decay anyway?
In any system in which events happen at random and the probability of an
event is independent of those events which previously happened, the
number of events per unit time will have a Poisson distribution. For
such events the time from one event to the next will have an
exponetional distribution. See any good standard statistics texts for
derviations and discussions.
In the case of exponential decay, what is being discussed is a system
where individual events (decays) occur at random independently of past
events, i.e., times to the next decay are exponentially distributed.
>Why do some particles live billions of times longer than others?
Other than saying some particles just happen to live longer than others,
I know of no way to explain this.
>Why should the decay curve of particles produced simultaneously (as in the muon
>ring) have similar characteristics to that of a quantity of randomly formed
>radioactive matter?
Both cases are examples of systems where events occur at random and are
*essentially* independent of past events. In reality, there is a small
deviation from exponential since the number of possible decays in the
system is finite and decreasing. That means the probability of the next
decay isn't completely independent of previous decays.
However, there are a large number of physical systems in which events
are truly independent of past events. The times between events in all
such systems where the events occur at random will follow an exponential
distribution.
--
-
PGPKey fingerprint: 6DA1 E71F EDFC 7601 0201 9243 E02A C9FD EF09 EAE5
Androcles
"John Holland" <hans_...@compuserve.com> wrote in message
news:9n0rpc$dj8$1...@suaar1aa.prod.compuserve.com...
>
exp(0) = 1
exp(-1) = 0.36...
exp(-2) = 0.135..
y (the height) = A-A times exp(-infinity).. yeah, you are right :)
Next you'll be telling that the derivative of e^x is not e^x....
If you can express exponential decay in terms of half-life, you can select
different units of time to express it as a Fibonacci series. Your statement
makes no sense.
John Holland
Wow, getting sarcastic ...? From what i have seen of your postings, i am
realy shivering of getting into a mathematical contest with you,
Cheers,
JH
Dirk Van de moortel
>
> Androcles <andr...@home.com> wrote in message
> news:u3Zk7.167105$EP6.48...@news1.rdc2.pa.home.com...
>
>
> > Next you'll be telling that the derivative of e^x is not e^x....
>
> Wow, getting sarcastic ...? From what i have seen of your postings, i am
> realy shivering of getting into a mathematical contest with you,
> Cheers,
> JH
Another remarkable demonstration can be found at:
http://groups.google.com/groups?sourceid=navclient&q=author%3Aandrocles+%22sqrt+has+two+answers%22
"No, no, no. sqrt has two answers."
Dirk Vdm
John Holland
Alles goed met je ? Goede vakantie gehad ? Ben zelf net 4 dagen terug uit
India, maar heb in het vliegtuig een voedsel vergiftiging opgelopen,
die me een beetje leeg doet lopen. Ben daarom even een paar dagen thuis, en
deze nieuwsgroep houdt me een beetje aangenaam bezig. Ben nog niet zo'n
doorgewinterde relativist als jijzelf, maar heb de stellige indruk dat de
wiskundige
bagage van een aantal lieden niet al te hoog aangeslagen moet worden. Ik
verbaas me nog over het geduld dat sommigen voor die permanente
hoogmoedswaanzin kunnen opbrengen, en ook over het verbale geweld dat
anderen voortdurend
menen te moeten hanteren. Aan de andere kant levert het ook genoeg humor op,
dus dat houd mekaar wel een beetje in evenwicht. Merkwaardige manier van
communicatie, maar zo gaat
het blijkbaar. Hoe dan ook, ik heb het boek van Taylor en Wheeler
("Spacetime physics") aangeschaft via Amazon.uk, en dat is aardig leesvoer,
waarmee ik weer even vooruit kan,
groeten,
Hans
Dirk Van de moortel
He ja! Ik was vergeten dat je Hans bent ;-)
Welkom terug, pech dat je ziek bent.
Heel goeie vakantie gehad, jammer dat het allemaal zo snel voorbij gaat.
Ik ben helemaal niet zo'n "doorwinterde" relativist hoor. D'r zitten hier
veel straffere gasten!
En grappige ook, vooral degene waarmee je het een beetje aan de stok
had. Niet te doen. Een eertseklas keikop. Intelligent en oerdom tegelijkertijd.
En zeer kwaadwillig. Eigenaardig verschijnsel ;-)
Mag ik je aanraden om eerst die G.R. from A to B te lezen? Daarna zal
St.P. waarschijnlijk helderder "klinken" en vlotter binnenlepelen.... t.t.z.
minder stroef ;-)
Amuzeer je nog en denk niet dat je deze meneer ooit zal "He ja, je
hebt gelijk!" zal doen zeggen. Hij heeft per definitie gelijk ;-)
Hou je goed, verzorg je, drink genoeg, geniet...
Dirk
Henry Wilson
>"Henry Wilson" <He...@the.edge> wrote in message
>news:3b9378d1...@nsw.nnrp.telstra.net...
>> On Mon, 03 Sep 2001 09:08:50 GMT, "Androcles" <andr...@home.com> wrote:
>>
><snipped>
>> What causes exponential decay anyway?
>
>Angels. Or demons, depending on whether you're on the side of the angels or
>not. Henry, you've just demonstrated yet again that you were lying in your
>teeth (if you have any) when you said you have a physics degree, acquired
>25, 30 0r 40 years ago. As for your second degree in "psychology +
>genetics", students have to learn some statistics for both of those subjects
>(let alone a combination of them), so I think we can forget about that claim
>too.
You silly bugger.
I am quite familiar with the equations governing radioactive decay. They are
elementary. They were obtained empirically from the observation that a constant
FRACTION of a radioactive mass will decay in a given time.
dn=-n(lambda)dt, where lambda is the radioactive constant.
Integrating that give the exponential.
UNLIKE YOU PEOPLE, I DON'T JUST ACCEPT EVERYTHING THAT IS THROWN AT ME. I AM
CAPABLE OF RECOGNISING A PHENOMENON WHOSE INVESTIGATION MIGHT LEAD TO SOMETHING
NEW.
So why is there a radioactive constant?
Have you ever bothered to ask yourself why these nucleii decay exponentially and
not, for instance, all at the same time? After all, they ARE all identical.
You will probably refer me to QM which is just another statistical theory
anyway.
I want the physics!!!
Have you ever bothered to consider that a group of atoms created simultaneously
should not decay in the same way as a group that was created randomly over a
long time?
I know much of this stuff has been investigated for natural radioactive elements
but the muon case is different. ALL the muons are produced more or less
simultaneously.
I say they DO NOT decay exponentially under these conditions.
>
>> Why do some particles live billions of times longer than others?
>> Why should the decay curve of particles produced simultaneously (as in the
>muon
>> ring) have similar characteristics to that of a quantity of randomly
>formed
>> radioactive matter?
>
>How can anybody express such far-reaching ignorance in so few words? You are
>truly awesome.
Well if you know the answers, please tell me!
Henry Wilson
>Henry Wilson wrote:
>> What causes exponential decay anyway?
>
>Any collection of objects whose individual probability of decaying is
>constant in time will have an exponential decay curve.
I know that Tom. That was discovered empirically and I am quite familiar with
the maths.
I want to know why some nucleii live much longer than others. Is this something
which is coded into them when they are created? Are they basically different.
Statistics handles the results but it doesn't tell us the reasons. Same with QM.
>
>
>> Why do some particles live billions of times longer than others?
>
>Because their individual probability of decaying in a given period of
>time is billions of times smaller that that of the others.
and why should that be, Tom? Are they not all identical?
>
>
>> Why should the decay curve of particles produced simultaneously (as in the muon
>> ring) have similar characteristics to that of a quantity of randomly formed
>> radioactive matter?
>
>Because that is a property of exponential decay curves. Just think
>about it -- if the individual probability of a particle decaying is
>constant in time, then at any time T, whatever sample of particles
>you have will decay exponentially, _REGARDLESS_ of when they were
>actually created. Particles which were also created but which decayed
>before time T are simply not part of the sample.
But if new particles are added at a constant rate, the decay curve is not
exponential. If they are added as a result of the exponential decay of a higher
atom, the curve is not exponential. If they are added at exactly the rate they
decay, then the decay curve is straight line.
If the creation of the particles followed a poisson distribution or even a
normal one, we wouldn't expect the same decay curve as the one applying to
simultaneously created particles.
In the muon ring experiment, the muons themselves are never actually counted.
Are not the biproducts of muon decay used as an indicator? If significant
quantities of these are missed, the decay rate appears longer.
Also, how do you know the muon sample is constant. If more muons are being added
regularly, the decay curve can take on any shape you like.
>
>
>This is so basic to this whole discussion. I repeat: you _REALLY_ need
>to actually _LEARN_ about physics, and stop making it up -- you are
>_REALLY_BAD_ at that.
Tom, I am looking beyond the current physics to the next level.
Like I said, WHY should nucleii decay rate be constant? There is a theory here,
waiting to be discovered.
>
>
>Tom Roberts tjro...@Lucent.com
Henry Wilson
How can the decay curve of simultaneously produced particles be the same as that
of continuously or randomly produced ones.
If you add many exponentials together you don't get an exponential.
Please correct me if I'm wrong.
>
>
>Tom Roberts tjro...@lucent.com
Henry Wilson
>In article <3b9378d1...@nsw.nnrp.telstra.net>,
> He...@the.edge(Henry Wilson) wrote:
>
>>What causes exponential decay anyway?
>
>In any system in which events happen at random and the probability of an
>event is independent of those events which previously happened, the
>number of events per unit time will have a Poisson distribution. For
>such events the time from one event to the next will have an
>exponetional distribution. See any good standard statistics texts for
>derviations and discussions.
No problem there.
I'm asking why the decay events just happen to occur randomly.
That is something we have all accepted without asking questions.
>
>In the case of exponential decay, what is being discussed is a system
>where individual events (decays) occur at random independently of past
>events, i.e., times to the next decay are exponentially distributed.
but the decay characteristics must depend on the way the particles originate.
The system you describe is one in which the decaying particles presumeably
originated at different times, maybe over a very long period.
>
>>Why do some particles live billions of times longer than others?
>
>Other than saying some particles just happen to live longer than others,
>I know of no way to explain this.
This is my question. We know the stats but we have no theory telling us why.
Look at it this way.
If I drop a million marbles on the ground and subsequently measure their
distances from the centre, I will get a nice normal distribution curve.
It is also theoretically possibly to mathematically simulate every individual
marble's path including what happens in all the collisions, knowing the starting
conditions, the elasticity, g, air drag, etc. (I wouldn't like to try to do
it).
Both methods will give the same result. One is pure maths, the other pure
physics.
I say there is a purely physical explanation for the exact time each individual
nucleus will decay.
>
>>Why should the decay curve of particles produced simultaneously (as in the muon
>>ring) have similar characteristics to that of a quantity of randomly formed
>>radioactive matter?
>
>Both cases are examples of systems where events occur at random and are
>*essentially* independent of past events. In reality, there is a small
>deviation from exponential since the number of possible decays in the
>system is finite and decreasing. That means the probability of the next
>decay isn't completely independent of previous decays.
Natural radioactive decay curves are critically dependent on the formation rate
as you are no doubt aware. If, for instance, particles are being added at the
same rate as they decay, the decay rate remains constant.
Nowhere in nature are great numbers of particles produced simultaneously as are
the muons. So how can we say that their decay is exponential?
>
>However, there are a large number of physical systems in which events
>are truly independent of past events. The times between events in all
>such systems where the events occur at random will follow an exponential
>distribution.
Because they are all in some long term equlibrium state.
Not so, the muons.
How do you know the measured sample has constant overall numbers?
Henry Wilson
>"Henry Wilson" <He...@the.edge> wrote in message
>news:3b9378d1...@nsw.nnrp.telstra.net...
>> On Mon, 03 Sep 2001 09:08:50 GMT, "Androcles" <andr...@home.com> wrote:
>>
><snipped>
>> What causes exponential decay anyway?
>
>Angels. Or demons, depending on whether you're on the side of the angels or
>not. Henry, you've just demonstrated yet again that you were lying in your
>teeth (if you have any) when you said you have a physics degree, acquired
>25, 30 0r 40 years ago. As for your second degree in "psychology +
>genetics", students have to learn some statistics for both of those subjects
>(let alone a combination of them), so I think we can forget about that claim
>too.
What are those two framed certificates doing up on my wall then?
>
>> Why do some particles live billions of times longer than others?
>> Why should the decay curve of particles produced simultaneously (as in the
>muon
>> ring) have similar characteristics to that of a quantity of randomly
>formed
>> radioactive matter?
>
>How can anybody express such far-reaching ignorance in so few words? You are
>truly awesome.
I think I'm several levels above you.
Please try to lift yourself.
>
>
Henry Wilson
wrote:
>
>Androcles <andr...@home.com> wrote in message
>> flow from the full tank into the empty. As the level fell in the full tank
>> and rose in the empty tank, the pressures would eventually balance when
>the
>> empty tank was half full and the full tank was half empty. The height of
>> water in the empty tank is an exponential function of time. The function,
>y
>> = A.exp(-tb) describes the behaviour.
>
>Get your mathematics right Androcles. The level of the empty tank does not
>start at level A, but at level zero. That is why you called it the empty
>tank. And by the way, if you take one full reservoir, which is not connected
>to another one and assume the outflow to be a linear function of its storage
>(which is a normal assumption in hydrology), you get a simple first degree
>linear differential equation, of which the solution as we all know is the
>exponential decay function.
dh/dt=-kh
dh=-khdt
Same as the decay equation, capacitor discharge and many other natural
processes.
Why should a quantity of nucleii behave like water in a tank?
I wonder if there's a connection.
PaulDanaher
> On Mon, 03 Sep 2001 13:00:32 GMT, "PaulDanaher" <wa...@earthlink.net>
wrote:
>
> >"Henry Wilson" <He...@the.edge> wrote in message
> >news:3b9378d1...@nsw.nnrp.telstra.net...
> >> On Mon, 03 Sep 2001 09:08:50 GMT, "Androcles" <andr...@home.com>
wrote:
> >>
> ><snipped>
> >> What causes exponential decay anyway?
> >
> >Angels. Or demons, depending on whether you're on the side of the angels
or
> >not. Henry, you've just demonstrated yet again that you were lying in
your
> >teeth (if you have any) when you said you have a physics degree, acquired
> >25, 30 0r 40 years ago. As for your second degree in "psychology +
> >genetics", students have to learn some statistics for both of those
subjects
> >(let alone a combination of them), so I think we can forget about that
claim
> >too.
> What are those two framed certificates doing up on my wall then?
Perhaps attesting to the exponential decay of a great mind? What do they
say, Henry?
> >
> >> Why do some particles live billions of times longer than others?
> >> Why should the decay curve of particles produced simultaneously (as in
the
> >muon
> >> ring) have similar characteristics to that of a quantity of randomly
> >formed
> >> radioactive matter?
> >
> >How can anybody express such far-reaching ignorance in so few words? You
are
> >truly awesome.
> I think I'm several levels above you.
> Please try to lift yourself.
I don't even aspire to your heights.
Paul B. Andersen
>
> A second point. Why should the decay rate of a sample of particles, which were
> produced simultaneously, follow an exponential curve, at all?
Because the probabilty for the muon to survive a certain time
from NOW is constant.
That is, if we know that the muon exists now, the probability
that it will survive another second is always the same
independent how long time it has existed up to now.
> This is a vastly different situation from the natural radioactivity decay
> process in which the atoms were produced at entirely different times.
No, it is exactly the same. The probabilty for the isotop to survive
a certain time from NOW is constant.
> In fact, I go as far as to speculate that the whole principle of exponential
> decay RELIES ON random time of origin.
> If you add two exponential decay curves, do you get another exponential curve?
> No.
>
> I suggest that decay rate of particles produced simultaneously, IS NOT
> EXPONENTIAL.
You are wrong.
Paul
PaulDanaher
The phenomenon (radioactive decay) isn't new, and its investigation did
indeed lead to something radically new - QM. The constant in this equation
simply expresses the empirical fact that the rate of decay in a sample is
directly proportional to the amount of the radioactive element present. The
underlying process is random (there *is* no causal mechanism involved in the
breakup of the nucleus).
> Have you ever bothered to ask yourself why these nucleii decay
exponentially and
> not, for instance, all at the same time? After all, they ARE all
identical.
> You will probably refer me to QM which is just another statistical theory
> anyway.
> I want the physics!!!
QM (or, ultimately, perhaps a string or brane theory) *is* the physics,
Henry. The exponential decay reflects precisely this random, statistical
nature of the universe. The world isn't determinist - guess what, Einstein
was wrong!
> Have you ever bothered to consider that a group of atoms created
simultaneously
> should not decay in the same way as a group that was created randomly over
a
> long time?
Just how do you think elements are created, by the way? What's the process
that randomly creates uranium?
> I know much of this stuff has been investigated for natural radioactive
elements
> but the muon case is different. ALL the muons are produced more or less
> simultaneously.
> I say they DO NOT decay exponentially under these conditions.
I don't think the universe has a First Amendment, but say on - the muons
aren't listening either.
> >> Why do some particles live billions of times longer than others?
> >> Why should the decay curve of particles produced simultaneously (as in
the
> >muon
> >> ring) have similar characteristics to that of a quantity of randomly
> >formed
> >> radioactive matter?
They don't, if you're going to claim that the atoms in the "quantity of
randomly formed radioactive matter" are created (randomly?) at different
times. Can you give an example of such a mixture?
> >How can anybody express such far-reaching ignorance in so few words? You
are
> >truly awesome.
> Well if you know the answers, please tell me!
I've no idea how you do it.
Androcles
[snip]
> Wow, getting sarcastic ...? From what i have seen of your postings, i am
> realy shivering of getting into a mathematical contest with you,
Try:
http://members.home.net/androcles/fumble2
Androcles
"Henry Wilson" <He...@the.edge> wrote in message
balancing out..
So you are perhaps suggesting that instead of random events, there is an
analog of pressure involved in some way with radiation, and a back-pressure
which is reducing the number of particles that will decay? Is that the way
you are thinking?
I suppose if we can discuss virtual particles and exchange particles, there
is no reason why we cannot hypothesize a virtual radiation "pressure".
So in your view, a radioactive isotope should decay almost instantly
short-circuit the capacitor or take out a single separating wall in a tank
of water)
Then we have a substance with a "resistor" or "narrow pipe" through which
the radiation must pass to escape, and it is this "virtual resistance" which
causes the exponential decay? From that we could deduce the value of this
virtual resistance as it applies to plutonium, uranium, carbon14 or
whatever? Then the virtual pressure is proportional to the quantity of
isotope remaining ( does density play a role? )
Why don't I feel comfortable with this?
Let's consider a simple decay. A neutron, decaying to a proton and a
negative charge, or vice versa. Both being part of the heavy hydrogen
nucleus, there is no reason why the charge cannot flit between two protons,
alternating, leaving one of them as the proton and the other as the neutron,
and we could not detect it if it did. We now introduce another neutron into
the group, and two charges are rotating around a triangle clockwise, or one,
counterclockwise. (The hole left by the charge is moving backwards). This
configuration is unstable, let's add a proton. Now we have helium, which is
stable. Ok, back to the unstable triple proton with two charges. A charge
escapes the nucleus, repelled by the other charge. It may be also repelled
by the electron and fall back into the triple proton, reestablishing the
previous configuration, and nothing is detected, or it may escape entirely,
leaving two protons and a neutron (light helium)..
SO... the virtual resistor is the random distribution of the electrons that
push the charge back into the nucleus.
Once past the electron, the charge is as likely to be kept out as it is of
getting back in. However, it may find a way into the nucleus of a different
atom which has lost a charge too, or escape entirely.
BUT...
This still leaves a random element, because of the quantum nature of the
particles.
I think that whatever model you use to describe the exponential curve,
whether random decay or a virtual pressure and a virtual resistor, the
outcome is not going to be any different from a simple random decay.
Bill Rowe
He...@the.edge(Henry Wilson) wrote:
>If you add many exponentials together you don't get an exponential.
>Please correct me if I'm wrong.
I can interpret your comment one of two ways. In either case you are
wrong.
First, I can interpret your "exponential" to mean an exponential
function. If this is what you meant consider Exp[x]/2 and Exp[-x]/2.
Both are exponential functions but the sum cosh[x] is not normally
called an exponential function. However, I don't think this is what you
had in mind.
If I assume you meant the distribution of the sum of things each haveing
an exponential distribution, then in the simplest case where each of the
items being summed come from the same exponential distribution, the sum
will have a gamma distribution. This is something easily shown from the
moment generating functions for a gamma distribution and exponential
distribution.
Bill Rowe
He...@the.edge(Henry Wilson) wrote:
>On Tue, 04 Sep 2001 00:51:43 GMT, Bill Rowe <bjr...@earthlink.net> wrote:
>>In any system in which events happen at random and the probability of an
>>event is independent of those events which previously happened, the
>>number of events per unit time will have a Poisson distribution. For
>>such events the time from one event to the next will have an
>>exponetional distribution. See any good standard statistics texts for
>>derviations and discussions.
>No problem there.
>I'm asking why the decay events just happen to occur randomly.
I've no answer. Saying decay events happen randomly is simply another
way of saying I've no way to predict the occurance of the next event.
>That is something we have all accepted without asking questions.
No. It is something we accept since we've currently no way to predict
the occurance of the next event.
>>In the case of exponential decay, what is being discussed is a system
>>where individual events (decays) occur at random independently of past
>>events, i.e., times to the next decay are exponentially distributed.
>but the decay characteristics must depend on the way the particles originate.
>The system you describe is one in which the decaying particles presumeably
>originated at different times, maybe over a very long period.
No. An exponential distribution results any time you've got a system
where event happen at random that has no memory of past events. In fact,
this memoryless feature of such system is a hallmark characteristic of
the exponential distribution. It literally says the probability of a
given muon decaying in the next unit of time is totally independent of
the past history of that muon, i.e., it does not matter one bit how long
that muon has lived.
>>>Why do some particles live billions of times longer than others?
>>Other than saying some particles just happen to live longer than others,
>>I know of no way to explain this.
>This is my question. We know the stats but we have no theory telling us why.
>Look at it this way.
>If I drop a million marbles on the ground and subsequently measure their
>distances from the centre, I will get a nice normal distribution curve.
No you won't get a normal distribution. Take a look at Knuth for a
discussion of how to generate random normal deviates from uniform
deviates. One of the standard techniques is to generate pairs of uniform
deviates, reject those outside of the unit circle and transform the
accepted pair. In essence generating random pairs of uniform deviates is
equivalent to dropping marbles and measuring there distance from a point.
>It is also theoretically possibly to mathematically simulate every
>individual marble's path including what happens in all the collisions,
>knowing the starting conditions, the elasticity, g, air drag, etc. (I
>wouldn't like to try to do it).
>Both methods will give the same result. One is pure maths, the other pure
>physics.
Yes, both methods done right should get the same result. But I would
argue saying one method is pure physics the other pure math is a
distinction that has no basis in fact. I stongly suspect you are
suggesting the method of dropping a million marbles and computing the
solution via a distribution is the "pure math" method. I would argue
this is an empirical method and closer to "pure physics" then the second
method. The point is the distinction you are making here is in the eye
of the beholder.
>I say there is a purely physical explanation for the exact time each
>individual nucleus will decay.
Fine, There may be a better answer than assuming each decay occurs at
random. But until someone finds it, all that is left is the assumption
of randomness.
>>Both cases are examples of systems where events occur at random and are
>>*essentially* independent of past events. In reality, there is a small
>>deviation from exponential since the number of possible decays in the
>>system is finite and decreasing. That means the probability of the next
>>decay isn't completely independent of previous decays.
>Natural radioactive decay curves are critically dependent on the
>formation rate as you are no doubt aware. If, for instance, particles
>are being added at the same rate as they decay, the decay rate remains
>constant.
Correct.
>Nowhere in nature are great numbers of particles produced
>simultaneously as are the muons.
Not relevant.
> So how can we say that their decay is exponential?
See comments above.
>>However, there are a large number of physical systems in which events
>>are truly independent of past events. The times between events in all
>>such systems where the events occur at random will follow an exponential
>>distribution.
>Because they are all in some long term equlibrium state.
No. One can derive a Poisson distribution from first principles that are
clearly applicable to a large number of physical systems. Additionally,
there are even a larger number of physical systems which nearly fit the
assumptions needed to derive a Poisson distribution. Once you have a
Poisson distribution, it is quite easy to show rigourously the time
between events must be an exponential distribution. Refer to any
standard statistics texts for a discussion of this. If you need a
specific reference, email me I and I will look it up in one of the texts
I have at my office.
>Not so, the muons.
And this doesn't matter as the assumption of long term equilibrium is
not one of the assumptions needed to derive the Poisson distribution or
the exponential distribution.
John Holland
thread:
"As a consequence of what does the observer see the bullet as tilted ?"
Cheers,
JH
Henry Wilson <He...@the.edge> wrote in message
news:3b955acc...@nsw.nnrp.telstra.net...
Paul B. Andersen
>
> On Mon, 03 Sep 2001 13:28:24 -0500, Tom Roberts <TomRo...@avenew.com> wrote:
>
> >Henry Wilson wrote:
> >> Why should the decay curve of particles produced simultaneously (as in the muon
> >> ring) have similar characteristics to that of a quantity of randomly formed
> >> radioactive matter?
> >
> >Because that is a property of exponential decay curves. Just think
> >about it -- if the individual probability of a particle decaying is
> >constant in time, then at any time T, whatever sample of particles
> >you have will decay exponentially, _REGARDLESS_ of when they were
> >actually created. Particles which were also created but which decayed
> >before time T are simply not part of the sample.
> But if new particles are added at a constant rate, the decay curve is not
> exponential. If they are added as a result of the exponential decay of a higher
> atom, the curve is not exponential. If they are added at exactly the rate they
> decay, then the decay curve is straight line.
> If the creation of the particles followed a poisson distribution or even a
> normal one, we wouldn't expect the same decay curve as the one applying to
> simultaneously created particles.
Quite right, Henry. But why the "but"?
What you describe above is what we actually observe.
(You are however using the term "decay curve" in an unconventional way.)
An example:
Say we have a sample with an isotope A which decays into another isotope B
with a very long half time - say thousands of years.
Let's assume the isotope B decays into isotope C with a short half time,
say days. Let's further assume that the decays A->B and B->C emits
different radiation, like alpha and beta, so that we can discriminate
between them.
Since the decay A->B has such a long half time, it will appear to
happen according to a Poisson distribution because the population
of A will not change significantly during the observation time.
So there is a steady supply of the B isotope.
I think you will realize that the result is that the decay B->C
will happen at the same (quasi) constant rate as the decay A->B;
B will decay a short time after it is created.
The population of B will be a constant fraction of the population
of A.
That doesn't mean that the decay B->C doesn't follow an exponential
decay curve; the law n = n0*e^-lambda*t still applies.
Paul
Androcles
> I am still waiting for your answer in the "Wherever did this idea
originate"
> thread:
> "As a consequence of what does the observer see the bullet as tilted ?"
> Cheers,
> JH
John Holland
the same person ?
Cheers,
JH
Androcles <andr...@home.com> wrote in message
news:x5vl7.173483$EP6.49...@news1.rdc2.pa.home.com...
Bilge
>Firstly, take any group of random numbers and find its average value. Then
>remove, say, the bottom half of those numbers, and calculate the average.
>It will have inreased by about 50%.
>In the case of an exponential type curve, if you remove all the members with
>expected life times less than several 'half times', the average life of the
>remainder is greatly increased.
>Average lifetime cannot be calculated as = T/.69 (or whatever).
Firstly, Your ability to turn the simplest thing into an excruciaitingly
painful ordeal for no particular reason goes virtually unmatched by those
of with opposing thumbs:
dN = -N dt/T
=> dN/N = -dt/T *Now Integrate
=> N = N_0 exp(-t/T) *Now Define half-life => T' by setting N = N_0/2
=> 1/2 = exp(-t/T') *Can you deal with the algebra from here?
>In the muon experiment, the acceleration time is therefore critical in
>determining the 'average lifetime' of the remainder. You are only
>looking at the 'tail end' of the decay curve.
No, (1) Muons aren't created during any acceleration. Muons are created
by pion decay in flight and then captured with bending magnets, (2)
you're looking at a poisson statistic, dodo. It isn't _supposed_ to be
symmetric about the average value. You would know this had you occasionally
read the literature written over the last 300 years. Statistics is
not a new discipline. (3) Now, because you won't admit that you wasted
a lot of time by typing before engaging your brain, you will mindlessly
defend yet another incorrect conclusion you've drawn.
>A second point. Why should the decay rate of a sample of particles,
>which were produced simultaneously, follow an exponential curve, at all?
Gee. I dunno. Oh. Wait a second, I Know. Because every particle
(of a particular type, e.g., pions, muon, luserons), has exactly the
same probability to decay in any interval \Delta t. Duh.
>This is a vastly different situation from the natural radioactivity decay
>process in which the atoms were produced at entirely different times.
No it isn't. The number of decays is always proportional to the number
of particles that remain. If that number increases, so does the number
of decays. Read any book wriiten in the last couple hundred years about
statistics.
>In fact, I go as far as to speculate that the whole principle of
>exponential decay RELIES ON random time of origin.
No, it relies on spontaneous decay. The time of origin is irrelavent.
Don't be a moron. By a book on statistics. You'll discover that your
modem, your radio, your cd player and everything else has exactly
the same probabilistic basis.
>If you add two exponential decay curves, do you get another exponential
>curve?
>No.
Yes. y1 = exp(-x), y2 = exp(-2x)
=> y = y1 + y2 = exp(-x) + exp(-2x)
What's the problem?
>I suggest that decay rate of particles produced simultaneously, IS NOT
>EXPONENTIAL.
Suggest all you like, but I've measured the half-lives of isotopes
with 200 ms half-lives, and you get an exponential decay curve regardless
of the tims it takes to accumulate the a sample from which to count decays.
Need a reference? Need raw data (if I can still locate some)?.
>I suggest it should be something along the lines: the probability
>of a particle surviving is proportional to the time it has already
>survived.
Suggestion noted and deep sixed as not in accordance with any
observation made on this planet.
>That cannot be tested with natural radioactivity because the origins are
>unknown. Have you ever considered just WHY decay rates appear to be
>exponential?
Because decay times depend on 3 things: Available phase space,
the existence of an interaction which connects the initial and
final states (the "matrix elements"), and the strength of the
interaction (the electric charge, the weak charge or strong charge),
and every decay being a _TOTALLY RANDOM PROCESS_. See fermi's
golden rule. There exist only 3 fundamental decay rates. The rest
is circumstance. In fact, the only difference between a decay and
a scatter is the energy of the initial state.
Just for you henry, I have a short program that is so obvious
as to be almost pointless to see that it works. In each half life,
the program chooses a random number. If it's > RAND_MAX/2, it decays,
if not, it doesn't. (this defines the half-life and so isn't subject
to argument). The particles that decay are subtracted from the total,
and the procedure is iterated until there exist no more particles.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]){
int hlife = 50; /* half life */
int npart = 10000000; /* number of particles to start with */
int count = 100*hlife; /* count for 100 half-lives */
int decayed = 0, n = 0, isec = 0;
if((argc == 2) &&
(npart = strtoul(argv[1], NULL, 10)) <= 0){
fprintf(stderr, "Use a _number_ greater than 0, moron...\n");
exit(1);
}
srand(time());
fprintf(stdout, "\n Number Particles\n"
" Half Lives | Remaining\n"
"------------+------------\n");
for(isec=0; (isec<=count) && (npart > 0); isec+=hlife){
fprintf(stdout, " %-4d |%10d\n", isec/hlife, npart);
for(n=0; n<=npart; n++)
if(rand() > (RAND_MAX/2))
decayed += 1;
npart -= decayed;
decayed = 0;
}
exit(0);
}
>On a slightly different note,
Which is a substantial fraction of a semitone flat...
> a third point is that there is no reason to
>believe that the decay curve during acceleration is anything like that
>when the ring has been stabilized.
That's not how they are prooduced and for that matter, the effects
of strong fields are known, since nuclei are often studied through
mu-mesic atoms. An atom captures a muon, which because it's heavier
than an electron by a factor of 200, orbits _inside_ the nucleus.
I suggest that you try calculating the E,B fields at a distance of
less than 1 fermi from a nucleus of Z=50, moving with the velocity
of the muon orbiting.
>There is no reason to believe that the decay process is not
>massively affected by the strong confining magnetic fields
There's no reason to believe you've ever looked into the matter.
>coupled with the enormous reaction to that centripetal force.
As I noted above, you're wrong. Those forces are miniscule.
For a real force, try the afformentioned calculation.
Henry Wilson
wrote:
>Henry Wilson wrote:
>>
>> A second point. Why should the decay rate of a sample of particles, which were
>> produced simultaneously, follow an exponential curve, at all?
>
>Because the probabilty for the muon to survive a certain time
>from NOW is constant.
>That is, if we know that the muon exists now, the probability
>that it will survive another second is always the same
>independent how long time it has existed up to now.
So it seems. but why? Don't you think that is rather strange?
>
>> This is a vastly different situation from the natural radioactivity decay
>> process in which the atoms were produced at entirely different times.
>
>No, it is exactly the same. The probabilty for the isotop to survive
>a certain time from NOW is constant.
Yes, but the decay curve is very dependent on the the way new particles are
being added.
Say you have a quantity of particles which are millions of years old, then you
add a similar quantity which have just been created. Each group should folIow
its own exponential decay curve. The sum of two exponentials is not an
exponential. That doesn't comform to the equation for the overall decay
dn=-n(l)dt. There is something strange here.
I know that's not quite the same thing but how do we know that new muons aren't
being added in the ring. Added to the ones being counted, that is.
>
>> In fact, I go as far as to speculate that the whole principle of exponential
>> decay RELIES ON random time of origin.
>> If you add two exponential decay curves, do you get another exponential curve?
>> No.
>>
>> I suggest that decay rate of particles produced simultaneously, IS NOT
>> EXPONENTIAL.
>
>You are wrong.
Is the sum of two exponentials and exponential, Paul?
>
>Paul
Dirk Van de moortel
> On Wed, 05 Sep 2001 01:18:48 +0200, "Paul B. Andersen" <paul.b....@hia.no>
> wrote:
>
> >Henry Wilson wrote:
> >>
> >> A second point. Why should the decay rate of a sample of particles, which were
> >> produced simultaneously, follow an exponential curve, at all?
> >
> >Because the probabilty for the muon to survive a certain time
> >from NOW is constant.
> >That is, if we know that the muon exists now, the probability
> >that it will survive another second is always the same
> >independent how long time it has existed up to now.
> So it seems. but why? Don't you think that is rather strange?
> >
> >> This is a vastly different situation from the natural radioactivity decay
> >> process in which the atoms were produced at entirely different times.
> >
> >No, it is exactly the same. The probabilty for the isotop to survive
> >a certain time from NOW is constant.
>
>
> Yes, but the decay curve is very dependent on the the way new particles are
> being added.
> Say you have a quantity of particles which are millions of years old, then you
> add a similar quantity which have just been created. Each group should folIow
> its own exponential decay curve.
You say it should. But it just doesn't :-)
Apparently nature isn't paying much attention to what we think it
should do. It does not have to follow our intuition.
No matter what you add, the new bunch follows the same decay curve.
> The sum of two exponentials is not an
> exponential. That doesn't comform to the equation for the overall decay
> dn=-n(l)dt. There is something strange here.
When you add the two groups, you do it as follows:
p * exp(-a*t) + q * exp(-b*t)
which is *not* an exponential indeed: this function cannot be written
in an exponential form like r * exp(-p*t).
Instead the groups must be added as folllows:
p*exp(-a*t) + q*exp(-a*t)
which is
(p+q) * exp(-a*t)
which is an exponential.
You see, the 'old' muons don't *know* they are old.
If they survided thusfar, they have exactly the same properties
as the ones that were just created. It's bizarre, but that's the
way it works. Counter-intuitive.
> I know that's not quite the same thing but how do we know that new muons aren't
> being added in the ring. Added to the ones being counted, that is.
> >
> >> In fact, I go as far as to speculate that the whole principle of exponential
> >> decay RELIES ON random time of origin.
> >> If you add two exponential decay curves, do you get another exponential curve?
> >> No.
> >>
> >> I suggest that decay rate of particles produced simultaneously, IS NOT
> >> EXPONENTIAL.
> >
> >You are wrong.
>
> [added two blank lines and this one to make it somewhat more readable]
>
> Is the sum of two exponentials and exponential, Paul?
Yes, if the exponential have the same argument.
And they have in this case. See above.
Old muons behave as if they were just created.
Cheers,
Dirk Vdm
John Holland
is too embarrassing for him,
Cheers,
JH
Androcles <andr...@home.com> wrote in message
news:x5vl7.173483$EP6.49...@news1.rdc2.pa.home.com...
Henry Wilson
wrote:
In that situation no. But consider if A->B is about equal to B->C.
If we are watching B, its curve could be anything.
The point I was making is that, if natural systems result in exponential decays,
why should simultaneous creation do the same.
>
>Paul
Henry Wilson
wrote:
>I am still waiting for your answer in the "Wherever did this idea originate"
>thread:
>"As a consequence of what does the observer see the bullet as tilted ?"
>Cheers,
>JH
The moving observer sees the bullet's axis as being tilted wrt its apparent
diagonal path. the bullet always points vertically upward, no matter how fast
the movement.
Last year, I showed the principle of this with light. I try to draw it again.
You might have to use fixed pitch fonts to view it properly.
Consider how a moving observer views each 'wave crest' of a vertically emitted
beam. The '/' represents the individual path an infinitesimally small element of
the light beam.
^
9 / 9 . . . . . . . . . 9
8 / 9 8 . . . . . . . . 8
7 / 9 / 7 . . . . . . . 7
6 / 9 / / 6 . . . . . . 6
5 / 9 / / / 5 . . . . . 5
4 / 9 8 / / / 4 . . . . 4
3 / 9 / 7 / / / 3 . . . 3
2 / 9 / / 6 / / / 2 . . 2
1 / 9 8 7 6 5 4 3 2 1 .___<-v observer 1
S S
The figures on the left represent the alignment of numbered 'wavecrests' of an
extremely thin light ray, emitted vertically by a source 'at rest'.
The diagonal lines show the paths of each individual wave crest in the frame of
a moving observer. The point is that during the time interval between the
emission of two wave crests, the moving observer travels a short distance. Thus,
the beam as a whole, remains vertical in the moving frame (as represented by the
numbers) The same principle applies to bullets fired at fixed intervals.
You might have to think about this for a while. It isn't all that obvious.
Diagrams as used in SR and the standard MMX, show a diagonal light beam. This is
completely wrong.
Henry Wilson
>In article <3b95597d...@nsw.nnrp.telstra.net>,
> He...@the.edge(Henry Wilson) wrote:
>
>>If you add many exponentials together you don't get an exponential.
>>Please correct me if I'm wrong.
>
>I can interpret your comment one of two ways. In either case you are
>wrong.
>
>First, I can interpret your "exponential" to mean an exponential
>function. If this is what you meant consider Exp[x]/2 and Exp[-x]/2.
>Both are exponential functions but the sum cosh[x] is not normally
>called an exponential function. However, I don't think this is what you
>had in mind.
I mean, if you add e^-at to e^-a(t+x), you might not get e^-a(t+y).
The relevance is that if you take a newly created sample of decaying particles,
let them decay for a period and then mix in another similar quantity of newly
created ones, presumeably the two quantities should follows the decay curves
they would have followed if they hadn't been mixed.
Let's see.
That is: e^-at + (e^-at)(e^-x) = e^-at * (1+e^-x)
Or const x e^-at.
So that does seem to be a exponential.
But if you add new particles at exactly the rate at which the whole sample is
decaying, you get a constant decay rate or horizontal line.
Something doesn't add up here.
Henry Wilson
>> Is the sum of two exponentials an exponential, Paul?
>
>Yes, if the exponential have the same argument.
>And they have in this case. See above.
>Old muons behave as if they were just created.
Yes OK, I get that now. But see my above message about adding constantly to the
sample.
>
>Cheers,
>
>Dirk Vdm
>
>
Bill Rowe
He...@the.edge(Henry Wilson) wrote:
>I mean, if you add e^-at to e^-a(t+x), you might not get e^-a(t+y).
In fact you won't. There is no way to get something in the form of e^-at
+ e^-a(t+x) to the form e^-a(t+y)
>The relevance is that if you take a newly created sample of decaying particles,
>let them decay for a period and then mix in another similar quantity of newly
>created ones, presumeably the two quantities should follows the decay curves
>they would have followed if they hadn't been mixed.
They will
>Let's see.
>That is: e^-at + (e^-at)(e^-x) = e^-at * (1+e^-x)
>Or const x e^-at.
>So that does seem to be a exponential.
Right. That addition also has no relevance to mixed groups of particles
as you suggest above.
David Evens
>On Sat, 01 Sep 2001 08:58:47 GMT, "Dirk Van de moortel"
><dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>>"Henry Wilson" <He...@the.edge> wrote in message news:3b90104f...@nsw.nnrp.telstra.net...
>>> The SRians here often refer to cyclotron experiments in which muons are
>>> accelerated to precisely known velocities near c, then retained in a circular
>>> orbit so that their decay rate can be monitored. From this, an estimate is made
>>> of their average lifetimes at this high speed.
>>> These lifetimes are much longer than those for muons at rest. Srians assure us
>>> that the facts agree exactly with the predictions of SR.
>>>
>>> That seems pretty straightforward. How can anyone argue against this apparently
>>> conclusive confirmation of SR?
>>>
>>> Well, please read on!!!!
>>>
>>> Firstly, by way of example, you all know that the life expectancy of humans is
>>> somewhere around 55 years (the actual figure doesn't matter). However, if you
>>> survive the first year, that figure jumps to about 75 years. If you survive the
>>> first forty years, it goes higher, to maybe 90 years.
>>>
>>> Well, the same type of thing happens, on a much larger scale, to muons in this
>>> now (in)famous experiment.
>>> The muons are produced almost simultaneously and immediately start to decay at a
>>> rate determined by their half life (1.56 usec).
>>> Many decay during the acceleration period but, like the humans, those which
>>> don't can be expected to live to a ripe old age.
>>> By the time the muons reach their maximum velocity and the measurements begin,
>>> only a small percentage of the original muons may remain. It is easily to show
>>> that the expected lifetimes of these will be greatly extended, purely on
>>> statistical grounds.
>>
>>HUH?
>>Having a go at probability and statistics now?
>>Like:
>> "Number 14 hasn't shown up this entire evening on the roulette, so I'm
>> putting all my money on it. Surely it must show up now!"
>>or
>> "Look, I have thrown 5 aces with these dice. Surely the probability that
>> I will do it again on my next throw, will be *much* smaller!"
>>or
>> "In our street 10 babies were born last week. All girls.Surely the next
>> baby must be a boy!"
>>
>>Shall we go on or do you get the picture?
>>So let's [snip] the consequences of your misunderstanding of statistics...
>>
>>> [snip]
>>>
>>> Ha! Bloody Ha!
>>> You people have been exposed again!
>>> I invite Paul Andeson, Tom Roberts, Franz Heymann, Eric Prebys and the others to
>>> try to find a way out of this one!!!!
>>
>>What you are trying to do here, is pathetic.
>>Ask Androcles if he wants to put your devastating argument on his website.
>>
>>Alternative: take a short introductory course on probability and
>>statistics and find your way out yourself. A good exercise.
>>
>>Dirk Vdm
>You people repeatedly quote "lifetime of the muons".
>The 'increased lifetime' is an expected consequence of cutting off the top end
>of the exponential decay curve.
>The half life of these muons is exactly the same.
>Just use the equation if you want to know why.
So how do you pretend that the muons live longer when you state that
you now understand that they don't?
>SR is now one step closer to death!
How does your self-contradiction lead towards the end of science?
Henry Wilson
>Henry Wilson said some stupid stuff again:
>
> >Firstly, take any group of random numbers and find its average value. Then
> >remove, say, the bottom half of those numbers, and calculate the average.
> >It will have inreased by about 50%.
> >In the case of an exponential type curve, if you remove all the members with
> >expected life times less than several 'half times', the average life of the
> >remainder is greatly increased.
> >Average lifetime cannot be calculated as = T/.69 (or whatever).
>
> Firstly, Your ability to turn the simplest thing into an excruciaitingly
>painful ordeal for no particular reason goes virtually unmatched by those
>of with opposing thumbs:
>
>
> dN = -N dt/T
> => dN/N = -dt/T *Now Integrate
>
> => N = N_0 exp(-t/T) *Now Define half-life => T' by setting N = N_0/2
>
> => 1/2 = exp(-t/T') *Can you deal with the algebra from here?
Bilgey why are you forever under the delusion that you are the only person who
knows anything? Your natural aggression must have gotten you into a lot of
trouble during your life.
>
> >In the muon experiment, the acceleration time is therefore critical in
> >determining the 'average lifetime' of the remainder. You are only
> >looking at the 'tail end' of the decay curve.
>
> No, (1) Muons aren't created during any acceleration. Muons are created
>by pion decay in flight and then captured with bending magnets, (2)
>you're looking at a poisson statistic, dodo. It isn't _supposed_ to be
>symmetric about the average value. You would know this had you occasionally
>read the literature written over the last 300 years. Statistics is
>not a new discipline. (3) Now, because you won't admit that you wasted
>a lot of time by typing before engaging your brain, you will mindlessly
>defend yet another incorrect conclusion you've drawn.
Bilgey, I know quite a lot of statistics.
> >A second point. Why should the decay rate of a sample of particles,
> >which were produced simultaneously, follow an exponential curve, at all?
>
> Gee. I dunno. Oh. Wait a second, I Know. Because every particle
>(of a particular type, e.g., pions, muon, luserons), has exactly the
>same probability to decay in any interval \Delta t. Duh.
All right, they probably should and do. I wasn't seriously suggesting that they
might not. .
>
> >This is a vastly different situation from the natural radioactivity decay
> >process in which the atoms were produced at entirely different times.
>
> No it isn't. The number of decays is always proportional to the number
>of particles that remain. If that number increases, so does the number
>of decays. Read any book wriiten in the last couple hundred years about
>statistics.
Bilgey, I know all the maths.
>
> >In fact, I go as far as to speculate that the whole principle of
> >exponential decay RELIES ON random time of origin.
>
> No, it relies on spontaneous decay. The time of origin is irrelavent.
>Don't be a moron. By a book on statistics. You'll discover that your
>modem, your radio, your cd player and everything else has exactly
>the same probabilistic basis.
That is the question I am raising, Bilgey. Why?
and please just don't tell me about poisson distribution. I know that.
>
> >If you add two exponential decay curves, do you get another exponential
> >curve?
> >No.
>
> Yes. y1 = exp(-x), y2 = exp(-2x)
> => y = y1 + y2 = exp(-x) + exp(-2x)
>
> What's the problem?
>
> >I suggest that decay rate of particles produced simultaneously, IS NOT
> >EXPONENTIAL.
>
> Suggest all you like, but I've measured the half-lives of isotopes
>with 200 ms half-lives, and you get an exponential decay curve regardless
>of the tims it takes to accumulate the a sample from which to count decays.
>Need a reference? Need raw data (if I can still locate some)?.
OK we wont worry about that any more.
>
> >I suggest it should be something along the lines: the probability
> >of a particle surviving is proportional to the time it has already
> >survived.
>
> Suggestion noted and deep sixed as not in accordance with any
>observation made on this planet.
>
> >That cannot be tested with natural radioactivity because the origins are
> >unknown. Have you ever considered just WHY decay rates appear to be
> >exponential?
>
> Because decay times depend on 3 things: Available phase space,
> the existence of an interaction which connects the initial and
> final states (the "matrix elements"), and the strength of the
> interaction (the electric charge, the weak charge or strong charge),
> and every decay being a _TOTALLY RANDOM PROCESS_. See fermi's
> golden rule. There exist only 3 fundamental decay rates. The rest
> is circumstance. In fact, the only difference between a decay and
> a scatter is the energy of the initial state.
Which brings us to the whole point I wished to discuss.
We know that physical processes at the macro level can often be described
statistically. We also know they can be analysed physically to produced the same
result. For instance the probability that a '2' will appear on the throw of a
die can be calculated simply. However, we could also argue that the result of
every throw was predetermined purely on physical grounds.
So at what level do we draw the line? Do the laws of mechanics suddenly stop?
Where does causality end and true randomicity begin?
Why does one particle decay in ten seconds and the adjacent one in ten years? It
could be argued that there IS a physical reason for this. The two particles were
not the initially same.
Of course, we wouldn't do that with any confidence, because many of the factors
which cause decay are probably external to the nucleus and are genuinely randon
in nature.
But how can we be certain of that
Good for you Bilgey.
I also have written a couple of simple programs which do much the same.
I have to use basic because that's all I have..
>
>
> >On a slightly different note,
>
> Which is a substantial fraction of a semitone flat...
>
> > a third point is that there is no reason to
> >believe that the decay curve during acceleration is anything like that
> >when the ring has been stabilized.
>
> That's not how they are prooduced and for that matter, the effects
>of strong fields are known, since nuclei are often studied through
>mu-mesic atoms. An atom captures a muon, which because it's heavier
>than an electron by a factor of 200, orbits _inside_ the nucleus.
>I suggest that you try calculating the E,B fields at a distance of
>less than 1 fermi from a nucleus of Z=50, moving with the velocity
>of the muon orbiting.
You could not really say that is the same as orbiting in an acelerator.
Besides all this stuff is pretty vaguely interpreted as far as I can see.
>
> >There is no reason to believe that the decay process is not
> >massively affected by the strong confining magnetic fields
>
> There's no reason to believe you've ever looked into the matter.
I haven't
>
> >coupled with the enormous reaction to that centripetal force.
>
> As I noted above, you're wrong. Those forces are miniscule.
>For a real force, try the afformentioned calculation.
and what is the muon lifetime in that situation Bilgey. 1000years?
Scott Fluhrer
Henry Wilson <He...@the.edge> wrote in message
> On Wed, 05 Sep 2001 05:10:40 GMT, Bill Rowe <bjr...@earthlink.net> wrote:
>
> >In article <3b95597d...@nsw.nnrp.telstra.net>,
> > He...@the.edge(Henry Wilson) wrote:
> >
> >>If you add many exponentials together you don't get an exponential.
> >>Please correct me if I'm wrong.
> >
> >I can interpret your comment one of two ways. In either case you are
> >wrong.
> >
> >First, I can interpret your "exponential" to mean an exponential
> >function. If this is what you meant consider Exp[x]/2 and Exp[-x]/2.
> >Both are exponential functions but the sum cosh[x] is not normally
> >called an exponential function. However, I don't think this is what you
> >had in mind.
> I mean, if you add e^-at to e^-a(t+x), you might not get e^-a(t+y).
(or, in other words, e^(-ay) = 1 + e^-(ax)), then
e^(-at) + e^-a(t+x) =
e^(-at)( 1 + e^-(ax) ) =
e^(-at) e^(-ay) =
e^-a(t+y)
(and, assuming a != 0, such a y always exists, and is independent of t...)
> The relevance is that if you take a newly created sample of decaying
particles,
> let them decay for a period and then mix in another similar quantity of
newly
> created ones, presumeably the two quantities should follows the decay
curves
> they would have followed if they hadn't been mixed.
>
> Let's see.
>
> That is: e^-at + (e^-at)(e^-x) = e^-at * (1+e^-x)
> Or const x e^-at.
> So that does seem to be a exponential.
Sure looks exponential to me. It might help if you were a bit more familiar
with exponentials -- a constant times an exponential is another exponential,
as I showed above...
>
> But if you add new particles at exactly the rate at which the whole sample
is
> decaying, you get a constant decay rate or horizontal line.
> Something doesn't add up here.
That is, of course, true no matter what statistics particular particular
uses to decide when to decay. For example, if people get born at exactly
the rate at which they die, you get a constant population, even if the life
expectancy of any individual person is not particularly exponential. Using
this statement to argue that a particular life expectancy is/isn't
exponential is silly, as it is true whether or not the life expectancy is
exponential...
Bilge
>Bilgey why are you forever under the delusion that you are the only
>person who
>knows anything? Your natural aggression must have gotten you into a lot of
>trouble during your life.
I go by what I observe. I don't read minds. People that don't listen
and try to figure things out prior to reponding are especially annoying.
[...]
>Bilgey, I know quite a lot of statistics.
Then apply them to the task at hand. Spontenaity means never
having to say your stable.
>>Bilge: No, it relies on spontaneous decay. The time of origin is
>> irrelavent. Don't be a moron. By a book on statistics. You'll discover
>> that your modem, your radio, your cd player and everything else has
>> exactly the same probabilistic basis.
>That is the question I am raising, Bilgey. Why?
If there was a "why", then it wouldn't be probabilistic, would it?
That's also not the right question. The right question is "why are
any prticles stable?". If you digest any of the rest of this, perhaps
you'll find out.
[...]
>> Because decay times depend on 3 things: Available phase space,
>> the existence of an interaction which connects the initial and
>> final states (the "matrix elements"), and the strength of the
>> interaction (the electric charge, the weak charge or strong charge),
>> and every decay being a _TOTALLY RANDOM PROCESS_. See fermi's
>> golden rule. There exist only 3 fundamental decay rates. The rest
>> is circumstance. In fact, the only difference between a decay and
>> a scatter is the energy of the initial state.
>Which brings us to the whole point I wished to discuss.
>We know that physical processes at the macro level can often be described
>statistically.
Not the same thing. Thaat's a matter of convenience. If you want to
do inconvenient statistics and attempt to use quantum mechanics from
the ground up, you could. You are confusing pragmatism with fundamental
physics.
> We also know they can be analysed physically to produced
>the same result. For instance the probability that a '2' will appear on
>the throw of a die can be calculated simply.
Las Vegas counts on this.
> However, we could also argue
>that the result of every throw was predetermined purely on physical grounds.
Las Vegas considers people that believe this a bonus. They
always lose. You could try and argue that, but it would be a falacious
argument. I'll change my mind if you go to vegas and turn $10.00
into $10,000,000.00 at the craps table.
Consider something that you don't have a well-formed opinion about or
haven't gotten too old to view as anything but the brick-and-mortar
world you've cemented in. Like your modem. Your modem transfers data
how well? Pretty close to 100% or the time, you get the correct data.
Why? It's a purely probabilistic process. The data going from one place
to another is purely a random walk. W
>
>So at what level do we draw the line? Do the laws of mechanics suddenly
>stop? Where does causality end and true randomicity begin?
Apparently, it doesn't. People that believe in bohmian mechanics
believe quantum mechanics is determinstic. Personally, I think that's
totally misguided and totally overlooks the physics. Nature is
probabilistic. All of it.
>Why does one particle decay in ten seconds and the adjacent one in ten
>years? It could be argued that there IS a physical reason for this.
It often is argued. Mostly by people that have no idea what the
physical reason could possibly be or why a cause should conspire with
nature to only appear random.
> The
>two particles were not the initially same. Of course, we wouldn't do that
Particles of the same type are indistinguishable. Literally. Otherwise,
you wouldn't have atoms.
>with any confidence, because many of the factors which cause decay are
>probably external to the nucleus and are genuinely randon in nature.
There is exactly one reason particles decay. Because they can. If a
lower energy state can be reached by decay, they decay. If that were not
true, atoms would have no reason to form, either. A p and e become a
hydrogen because it's worth 13.6 eV to do so. A deuteron forms from a
p and n because it's worth 2.2 MeV to do so. Tritium decays to 3He
because it's worth 18 keV to do so.
>But how can we be certain of that
You can't. However, since decay rates can be predicted from first
principles (which I've mentioned in the post you didn't read prior to
writing your response), there's a good chance it's true. A lot better
chance than any crackpot scheme put forth for anything, that's for sure.
Need references? There are 3 decay rates: strong decay, weak decay
and e-m decay. The rest is circumstance. When you can calculate ANYTHING
in some hair-brained scheme as well as the decay rates of super-allowed
fermi-transitions in mirror nuclei, then get back to me.
>> As I noted above, you're wrong. Those forces are miniscule.
>>For a real force, try the afformentioned calculation.
>and what is the muon lifetime in that situation Bilgey. 1000years?
You tell me henry.
Androcles
"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message
news:slrn9pjlo...@radioactivex.lebesque-al.net...
> Henry Wilson said some stuff about
>
> >Bilgey why are you forever under the delusion that you are the only
> >person who
> >knows anything? Your natural aggression must have gotten you into a lot
of
> >trouble during your life.
>
> I go by what I observe.
I don't read minds.
(or use one)
>People that don't listen
> and try to figure things out prior to reponding are especially annoying.
(never heard of "look before you leap")
Bilge
Re: Part 2: THE EVER GREATER MUON HOAX. to usenet:
>
>"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message
>news:slrn9pjlo...@radioactivex.lebesque-al.net...
>> Henry Wilson said some stuff about
>>
>> >Bilgey why are you forever under the delusion that you are the only
>> >person who
>> >knows anything? Your natural aggression must have gotten you into a lot
>of
>> >trouble during your life.
>>
>> I go by what I observe.
>(Believes what he sees)
And I believe that I see you're an idiot. Toodles.
island
<snip>
Well, lookie what the cat dragged in... ;)))
WB man!
Dirk Van de moortel
>
> "Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message
> news:slrn9pjlo...@radioactivex.lebesque-al.net...
> >
> >
> I can't refute anything, so I'll just say something utterly stupid.
>
> Androcles
On this one I agree with Androcles.
Dirk Vdm
Bill Rowe
"Scott Fluhrer" <sflu...@ix.netcom.com> wrote:
>Henry Wilson <He...@the.edge> wrote in message
>news:3b981f0d...@nsw.nnrp.telstra.net...
>> I mean, if you add e^-at to e^-a(t+x), you might not get e^-a(t+y).
>Huh? If we define y = -1/a log(1 + e^-(ax)),
>(or, in other words, e^(-ay) = 1 + e^-(ax)), then
>e^(-at) + e^-a(t+x) =
>e^(-at)( 1 + e^-(ax) ) =
>e^(-at) e^(-ay) =
>e^-a(t+y)
>(and, assuming a != 0, such a y always exists, and is independent of t...)
But y is not independent of x the way you've done thinngs. On what basis
do you assume a relationship between x and y?
Baically, what you've shown is by appropriately defining relationships
you can get reduce e^(-at) + e^-a(t+x) to an obviously exponential
expression. The same trick can be done with just about any sum of
exponentials. That doesn't make the sum exponential since you really
aren't free to define relationships between variables as you like.
Henry Wilson
wrote:
I was showing how the APPARENT half life could be affected by addition to the
population.
Bilge
>>That is, of course, true no matter what statistics particular particular
>>uses to decide when to decay. For example, if people get born at exactly
>>the rate at which they die, you get a constant population, even if the
>>life expectancy of any individual person is not particularly exponential.
>>Using this statement to argue that a particular life expectancy is/isn't
>>exponential is silly, as it is true whether or not the life expectancy is
>>exponential...
>I was showing how the APPARENT half life could be affected by addition
>to the population.
Well, thank you. Now we know that failing to insure that an experiment
reflects the assumptions upon which it is based and/or poor experimental
technique leads to erroneous conclusions. Now that you've informed any of
the readers that have not yet taken their first science class in elementary
school of the caveats involved in experimental work, do you have anything
that pertains to any problems of which scientists in grown-up land might
not have been aware? Don't worry about the subtlty of such problems as you
might envision. We can handle it. On the other hand, if this is as subtle
as it gets for you, then you might find this has broader appeal in the k12
heirarchy.
Paul B. Andersen
>
> On Wed, 05 Sep 2001 01:18:48 +0200, "Paul B. Andersen" <paul.b....@hia.no>
> wrote:
>
> >Henry Wilson wrote:
> >>
> >> A second point. Why should the decay rate of a sample of particles, which were
> >> produced simultaneously, follow an exponential curve, at all?
> >
> >Because the probabilty for the muon to survive a certain time
> >from NOW is constant.
> >That is, if we know that the muon exists now, the probability
> >that it will survive another second is always the same
> >independent how long time it has existed up to now.
> So it seems. but why? Don't you think that is rather strange?
Not really. The particles have no memory, they don't know how old
they are. That IS what I said above. They have a _constant_ probability
of decaying the "next time increment" regardless of their age.
> >> This is a vastly different situation from the natural radioactivity decay
> >> process in which the atoms were produced at entirely different times.
> >
> >No, it is exactly the same. The probabilty for the isotop to survive
> >a certain time from NOW is constant.
> Yes, but the decay curve is very dependent on the the way new particles are
> being added.
You are using the phrase "decay curve" in an unconventional manner.
Of course the population of a particle type depend on how it is
created, and it may not change exponentially with time.
I have addressed this point in another posting.
> Say you have a quantity of particles which are millions of years old, then you
> add a similar quantity which have just been created. Each group should folIow
> its own exponential decay curve.
No. The particles have no memory and there is no difference between
the "old" particles and the newly created ones. They ALL have the same
probability of decaying during "the next time increment".
So they will follow the decay curve N = No*e^-labda*t
where No is the sum of the old and the new particles.
> The sum of two exponentials is not an
> exponential. That doesn't comform to the equation for the overall decay
> dn=-n(l)dt. There is something strange here.
That is probably because you think the particles have memory.
It is because they have NOT that the decay curve is exponential.
Entities with memory (like light bulbs and people) do not die
according to an exponential function. Your probability of
dying the next year is much higher if you are 80 (and still living)
than if you are 20.
The decay curve will typically be (1 - integral of a Gauss function).
(Possibly with some deviations at yang age, etc.)
> I know that's not quite the same thing but how do we know that new muons aren't
> being added in the ring. Added to the ones being counted, that is.
Muons don't come out of nothing.
> >> In fact, I go as far as to speculate that the whole principle of exponential
> >> decay RELIES ON random time of origin.
> >> If you add two exponential decay curves, do you get another exponential curve?
> >> No.
> >>
> >> I suggest that decay rate of particles produced simultaneously, IS NOT
> >> EXPONENTIAL.
> >
> >You are wrong.
> Is the sum of two exponentials and exponential, Paul?
Yes, when the mean lifetime is the same.
N = N_old*e^-lambda*t + N_new*e^-lambda*t = (N_old+N_new)*e^-lambda*t
See above.
Paul
Paul B. Andersen
Not quite anything, but not necessarily a single exponential function.
That IS what I said above.
So?
> The point I was making is that, if natural systems result in exponential decays,
> why should simultaneous creation do the same.
The point you made was that the population of a particle type (or isotope)
doesn't necessarily change with time as an exponential function
if new particles are added continuously.
You are right about that.
So?
Paul
Scott Fluhrer
Bill Rowe <bjr...@earthlink.net> wrote in message
news:bjrowe-967E52....@nnrp04.earthlink.net...
> In article <9nc3i9$a95$1...@slb6.atl.mindspring.net>,
> "Scott Fluhrer" <sflu...@ix.netcom.com> wrote:
>
> >Henry Wilson <He...@the.edge> wrote in message
> >news:3b981f0d...@nsw.nnrp.telstra.net...
>
> >> I mean, if you add e^-at to e^-a(t+x), you might not get e^-a(t+y).
>
> >Huh? If we define y = -1/a log(1 + e^-(ax)),
> >(or, in other words, e^(-ay) = 1 + e^-(ax)), then
>
> >e^(-at) + e^-a(t+x) =
> >e^(-at)( 1 + e^-(ax) ) =
> >e^(-at) e^(-ay) =
> >e^-a(t+y)
>
> >(and, assuming a != 0, such a y always exists, and is independent of
t...)
>
> But y is not independent of x the way you've done thinngs. On what basis
> do you assume a relationship between x and y?
You are create a fixed amount ("1") of some material with a halflife at time
0, and hence, at time t, the amount of material will be e^-at. Then, at
time -x, you create that same fixed amount of that same material, and so the
amount of material will be e^-a(t+x). Hence, e^-at + e^-a(t+x) is the total
amount within both piles.
We see that x is fixed within one particular run of the expirement (because,
during the run, when we created the second pile will not change). Hence, we
can legitimitely let y depend on it, as it is just a constant of the
expirement, just as "a" is.
Paul B. Andersen
Isn't it?
It is trivially simple and obvious to me.
> Diagrams as used in SR and the standard MMX, show a diagonal light beam. This is
> completely wrong.
Duh.
Why the hell do you keep repeating this utter nonsense when you have
been told so many times that it is just that?
Please read the following carefully, Henry.
I have told you this before - several times.
Either you don't care to read my postings, or you must be very slow.
So please - read it this time!
What you explained above is the transverse arm of the MMX.
The transverse arm is just that, transverse to the moving observer.
And the light beam "as a whole" is following that arm. Every photon
(or wave crest or light element or whatever you prefer to call it)
is moving along that transverse arm. So of course the light beam
"as a whole" is transverse to the moving observer.
NOBODY HAS EVER FAILED TO UNDERSTAND THAT!
You must be incredible naive to believe that this is a new
discovery not realized by anyone before!
The "diagonal light beam" you are referring to is NOT the light beam
"as a whole", but the path of an individual photon (or light element),
e.g. THE VERY SAME DIAGRAM YOU MADE YOURSELF ABOVE!
How the hell can you have missed such an obvious matter?
It is of course THIS path that is relevant when it comes to
calculating the time for the light to transverse the arm.
How the hell can you have missed such an obvious matter?
But to do that calculation YOU MUST USE A SPECIFIC THEORY.
How the hell can you have missed such an obvious matter?
In you "ballistic light theory + Galilean relativity",
the speed of the photons in the "arm frame" is c, and the
time to traverse the arm is L/c regardless of the orientation
of the arm, zero fringe shifts are predicted.
The moving observer is irrelevant, there is not really
any reason to calculate the speed of the photons in
his frame, but if you do, it will blatantly obvious be:
sqrt(c^2+v^2)
/|
sqrt(c^2+v^2)/ |
/ |c
/___|
v
However, according to "Michelson type ether theory",
the photons are moving at the speed c IN THE ETHER.
And the transverse arm is moving at the speed v IN THE ETHER.
That means that the speed of the photons in the arm frame
will be sqrt(c^2-v^2), and the time to traverse the transverse arm
= L/sqrt(c^2-v^2) = L/(c*sqrt(1-v^2/c^2))
/|
c/ |
/ | sqrt(c^2-v^2)
/___|
v
How the hell can you have missed such an obvious matter?
AND PLEASE DON'T REPEAT YOUR IDIOTIC CLAIM THAT THE PHOTONS
ARE NOT MOVING DIAGONALLY AT C IN THE ETHER FRAME!
And the "round trip speed" of the photons along the parallel
arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
and a fringe shifts is predicted.
How the hell can you have missed such an obvious matter?
According to SR, the speed of the photons are c in ANY frame,
including the "arm frame". So when measured in the arm frame,
the time to traverse the arm is L/c regardless of the orientation
of the arm, zero fringe shifts are predicted.
The moving observer is irrelevant, there is not really
any reason to calculate the speed of the photons in
his frame, but if you do, you must use the Lorentz transform!
When doing that, the speed of the photons is still c, but
the path lengths are longer; the round trip time to traverse BOTH
arms will as measured in the observers frame be: 2*L/(c*sqrt(1-v^2/c^2)).
Now I think it is about time you understand this, Henry.
It shouldn't be hard at all.
So please stop repeating your nonsense.
Paul
Henry Wilson
Bilgey, we are refering to lab muon decay, right.
How do we know that what is being measured as an indicator of muon decay is
coming from a constant (or predictable) population of muons?
Henry Wilson
Bilge
>Bilgey, we are refering to lab muon decay, right.
>How do we know that what is being measured as an indicator
>of muon decay is coming from a constant (or predictable)
>population of muons?
Go think about it for a while. That way you'll answer some of the
arguments you'd otherwise pose, yourself. 90% of your objections to
everything would evaporate if you struggled awhile with your question
first and had some idea of what it was you don't understand. Then you
can ask about _specific_ items to be clarified from some _real_
experiments. Start with the particle properties data for which you will
find the list of references used to determine the published value. Then
your argument won't be hypothetical and we won't be discussing hypo-
thetical experiments based upon speculation about imaginary problems and
I'll at least get the impression you're serious. It's simple to get a
known muon flux, if that's what you want, but that's not what you want
here. (Hint: It doesn't matter whether you use N particles and count
the fraction of decays in fixed time intervals or make N measurements
of a fixed number of particles [like 1] and measure the time for each
decay., i.e., you can fill time bins with particles or particle bins
with time). Make sure you understand this if you plan to object so that
I don't think it's a troll.
Bilge
>Bilgey, we are refering to lab muon decay, right.
>How do we know that what is being measured as an indicator
>of muon decay is coming from a constant (or predictable)
>population of muons?
Henry Wilson
wrote:
>Henry Wilson wrote:
NO!!!!!!!!!!!NO!!!!!!!!!!!!
How many times do I have to repeat myself.
No complete photon moves along a particular, INFINITELY THIN diagonal.
Only an infinitesimal point does!!
A line has NO WIDTH, Paul!!
AND WHY THE HELL SHOULD AN INFINITESIMAL POINT OF ANYTHING MOVES AT VELOCITY C?
Ib this case it obvioucly moves at sqrt(c^2+v^2)
Why should the moving observer measure the travel time of the cross beam as
being anything but L/c, whatever his velocity?
>e.g. THE VERY SAME DIAGRAM YOU MADE YOURSELF ABOVE!
>How the hell can you have missed such an obvious matter?
YOU ARE THE ONE WHO IS MISSING THE POINT!!
>
>It is of course THIS path that is relevant when it comes to
>calculating the time for the light to transverse the arm.
>How the hell can you have missed such an obvious matter?
>
>But to do that calculation YOU MUST USE A SPECIFIC THEORY.
>How the hell can you have missed such an obvious matter?
I haven't missed anything. I can plainly see the stupidity of your argument.
>
>In you "ballistic light theory + Galilean relativity",
>the speed of the photons in the "arm frame" is c, and the
>time to traverse the arm is L/c regardless of the orientation
>of the arm, zero fringe shifts are predicted.
>The moving observer is irrelevant, there is not really
>any reason to calculate the speed of the photons in
>his frame, but if you do, it will blatantly obvious be:
>sqrt(c^2+v^2)
AND THAT IS WHAT IT IS IN REALITY!!!
>
> /|
>sqrt(c^2+v^2)/ |
> / |c
> /___|
> v
>
>However, according to "Michelson type ether theory",
>the photons are moving at the speed c IN THE ETHER.
>And the transverse arm is moving at the speed v IN THE ETHER.
>That means that the speed of the photons in the arm frame
>will be sqrt(c^2-v^2), and the time to traverse the transverse arm
>= L/sqrt(c^2-v^2) = L/(c*sqrt(1-v^2/c^2))
>
> /|
> c/ |
> / | sqrt(c^2-v^2)
> /___|
> v
>How the hell can you have missed such an obvious matter?
I know THAT!!
But you admit there is no aether. So why bother to mention it.
Each photon always moves at c in the direction of its axis, anyway.
>
>AND PLEASE DON'T REPEAT YOUR IDIOTIC CLAIM THAT THE PHOTONS
>ARE NOT MOVING DIAGONALLY AT C IN THE ETHER FRAME!
The bloody photons move vertically at c, in all frames.
>
>And the "round trip speed" of the photons along the parallel
>arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
>and a fringe shifts is predicted.
>How the hell can you have missed such an obvious matter?
>
The fringe shift is WRONGLY predicted.
>According to SR, the speed of the photons are c in ANY frame,
>including the "arm frame".
Bugger SR. Do you think I'm interested in hearing about a postulate of SR?
>So when measured in the arm frame,
>the time to traverse the arm is L/c regardless of the orientation
>of the arm, zero fringe shifts are predicted.
>The moving observer is irrelevant, there is not really
>any reason to calculate the speed of the photons in
>his frame, but if you do, you must use the Lorentz transform!
>When doing that, the speed of the photons is still c, but
>the path lengths are longer; the round trip time to traverse BOTH
>arms will as measured in the observers frame be: 2*L/(c*sqrt(1-v^2/c^2)).
I know all that Paul! and it is plain BULL!
>
>Now I think it is about time you understand this, Henry.
>It shouldn't be hard at all.
>So please stop repeating your nonsense.
The reason I repeat my good sense is because I DO understand!
>
>Paul
Paul B. Andersen
So call it "an ifinitesimal small light element", then.
The point is that whatever you call it, its path is "diagonal"
in the observers frame, just as you illustrated above.
You repeat yourself to state the very same thing as I said,
how the bloody hell can you miss that?
What is the point with this mindless protesting?
> AND WHY THE HELL SHOULD AN INFINITESIMAL POINT OF ANYTHING MOVES AT VELOCITY C?
> Ib this case it obvioucly moves at sqrt(c^2+v^2)
> Why should the moving observer measure the travel time of the cross beam as
> being anything but L/c, whatever his velocity?
Why indeed?
I address this point further down, which you must now by now.
Why the hell don't you edit your responses to remove the questions
which are answered in the very same posting you are responding to?
> >e.g. THE VERY SAME DIAGRAM YOU MADE YOURSELF ABOVE!
> >How the hell can you have missed such an obvious matter?
> YOU ARE THE ONE WHO IS MISSING THE POINT!!
And what the hell would the missed point be?
That every infinitesimal light element moves diagonally
in the moving observer's frame, just as you say above?
> >It is of course THIS path that is relevant when it comes to
> >calculating the time for the light to transverse the arm.
> >How the hell can you have missed such an obvious matter?
HERE IS THE ANSWER TO YOUR QUESTION ABOVE:
> >But to do that calculation YOU MUST USE A SPECIFIC THEORY.
> >How the hell can you have missed such an obvious matter?
> I haven't missed anything. I can plainly see the stupidity of your argument.
The stupidity in which argument?
That to calculate the predictions of a specific theory,
you must use that specific theory to calculate the predictions?
You HAVE missed this obvious point, Henry.
You prove that over and over in this posting!
> >In you "ballistic light theory + Galilean relativity",
> >the speed of the photons in the "arm frame" is c, and the
> >time to traverse the arm is L/c regardless of the orientation
> >of the arm, zero fringe shifts are predicted.
> >The moving observer is irrelevant, there is not really
> >any reason to calculate the speed of the photons in
> >his frame, but if you do, it will blatantly obvious be:
> >sqrt(c^2+v^2)
> AND THAT IS WHAT IT IS IN REALITY!!!
I have noticed that this is your opinion, Henry.
What is a FACT, is that THIS THEORY predicts no fringe
shifts in the MMX.
> >
> > /|
> >sqrt(c^2+v^2)/ |
> > / |c
> > /___|
> > v
> >
> >However, according to "Michelson type ether theory",
> >the photons are moving at the speed c IN THE ETHER.
> >And the transverse arm is moving at the speed v IN THE ETHER.
> >That means that the speed of the photons in the arm frame
> >will be sqrt(c^2-v^2), and the time to traverse the transverse arm
> >= L/sqrt(c^2-v^2) = L/(c*sqrt(1-v^2/c^2))
> >
> > /|
> > c/ |
> > / | sqrt(c^2-v^2)
> > /___|
> > v
> >How the hell can you have missed such an obvious matter?
> I know THAT!!
Fine. So we agree then.
According to Michelson's ether theory, the photons (infinitesimal
light elements) move diagonally at c in the ether, and
the velocity component along the transverse arm is sqrt(c^2-v^2).
> But you admit there is no aether. So why bother to mention it.
Because you have repeated over and over again that the MMX
does NOT falsify Michelsons ether theory because Michelson
misinterpreted the MMX, and that this theory predicts zero
fringe shift.
> Each photon always moves at c in the direction of its axis, anyway.
According to "ballistic light theory".
But the existence of other theories is utterly irreleavnt to
what a specific theory predicts.
> >AND PLEASE DON'T REPEAT YOUR IDIOTIC CLAIM THAT THE PHOTONS
> >ARE NOT MOVING DIAGONALLY AT C IN THE ETHER FRAME!
It is obvious from the context that this is
ACCORDING TO MICHELSON'S ETHER THEORY!
> The bloody photons move vertically at c, in all frames.
So you DO repeat your MORONIC claim that according to
Michelson's ether theory, the photons are not moving
diagonally at c in the ether.
You are very inconsistent, Henry.
Right above, you said: "I know THAT" to what you now deny.
I suspect you write without thinking.
> >And the "round trip speed" of the photons along the parallel
> >arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
> >and a fringe shifts is predicted.
> >How the hell can you have missed such an obvious matter?
> The fringe shift is WRONGLY predicted.
How so?
We agree that according to Michelson's ether theory:
the round trip speed of light in the interferometer frame is:
along the transverse arm: sqrt(c^2-v^2)
along the parallel arm: (c^2-v^2)
You are claiming that this doesn't imply fringe shifts
when the interferometer is rotated.
Show it!
> >According to SR, the speed of the photons are c in ANY frame,
> >including the "arm frame".
> Bugger SR. Do you think I'm interested in hearing about a postulate of SR?
> >So when measured in the arm frame,
> >the time to traverse the arm is L/c regardless of the orientation
> >of the arm, zero fringe shifts are predicted.
> >The moving observer is irrelevant, there is not really
> >any reason to calculate the speed of the photons in
> >his frame, but if you do, you must use the Lorentz transform!
> >When doing that, the speed of the photons is still c, but
> >the path lengths are longer; the round trip time to traverse BOTH
> >arms will as measured in the observers frame be: 2*L/(c*sqrt(1-v^2/c^2)).
> I know all that Paul! and it is plain BULL!
What does this statement mean, Henry?
Does it mean that you know that SR predicts no fringe shifts?
Or does it mean that it is bullshit that SR predicts no fringe shift?
Make up yout mind.
Does SR predict fringe shifts, or doesn't it?
And please remeber that you opinion about the corectness of the theory
is utterly irrelevant to what it predicts.
> >Now I think it is about time you understand this, Henry.
> >It shouldn't be hard at all.
> >So please stop repeating your nonsense.
> The reason I repeat my good sense is because I DO understand!
When you repeat your "wave crest diagram" above, you are repeating
what never has been disputed, and what nobody ever have failed to
understand. So what is the bloody point with that?
When you repeat that "the photons are moving vertically at c",
you repeat what is correct according to one specific theory only,
and again - nobody has ever failed to understand that
the "ballistic light theory" states so.
So what is the bloody point with that?
When you repeat that the predictions of SR and Michelson's ether theory
are not what they are because you think the theories are wrong,
then you are repeating a huge logical nonsense.
So what is the bloody point with that?
Paul
Henry Wilson
wrote:
>Henry Wilson wrote:
No we don't bloody agree! When I said "I know that" I meant I know all about
your argument.
>According to Michelson's ether theory, the photons (infinitesimal
>light elements) move diagonally at c in the ether, and
>the velocity component along the transverse arm is sqrt(c^2-v^2).
NO Paul! (tearing hair out).
A photon is not an 'infinitesimal point'. A photon has length, cross section and
structure. Otherwise it couldn't contain the information about its source,
needed to give the correct Doppler shift when it hits something. The double slit
experiment wouldn't work, either.
Each infinitesimal point moves along a unique infinitesimally thin diagonal at
sqrt(c^2+v^2). Just like each minute part of the 'ants' walking up the pole.
.
>
>> But you admit there is no aether. So why bother to mention it.
>
>Because you have repeated over and over again that the MMX
>does NOT falsify Michelsons ether theory because Michelson
>misinterpreted the MMX, and that this theory predicts zero
>fringe shift.
It is irrelevant anyway because there is no aether.
>
>> Each photon always moves at c in the direction of its axis, anyway.
>
>According to "ballistic light theory".
>But the existence of other theories is utterly irreleavnt to
>what a specific theory predicts.
>
>> >AND PLEASE DON'T REPEAT YOUR IDIOTIC CLAIM THAT THE PHOTONS
>> >ARE NOT MOVING DIAGONALLY AT C IN THE ETHER FRAME!
>
>It is obvious from the context that this is
>ACCORDING TO MICHELSON'S ETHER THEORY!
SR IS an aether theory.
What other justification do you have for claiming that the light beam moves
diagonally at c. That is just what sound would do in air.
>
>> The bloody photons move vertically at c, in all frames.
>
>So you DO repeat your MORONIC claim that according to
>Michelson's ether theory, the photons are not moving
>diagonally at c in the ether.
According to any theory, they are not. Maxwell for instance.
The whole thing is a huge blunder. Just draw the path of each 'wavecrest' in the
MMX cross beam and you will see why.
Surely there is someone else in the world capable of carrying out this simple
exercise in graphing.
Plot the body of an ant walking up the pole in the moving observer's frame. IT
NEVER ROTATES DIAGONALLY. IT ALWAYS REMAINS VERTICAL!!!
>
>You are very inconsistent, Henry.
>Right above, you said: "I know THAT" to what you now deny.
I have attended to that.
>
>I suspect you write without thinking.
No,that's David Evens.
>
>> >And the "round trip speed" of the photons along the parallel
>> >arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
>> >and a fringe shifts is predicted.
>> >How the hell can you have missed such an obvious matter?
>
>> The fringe shift is WRONGLY predicted.
>
>How so?
>We agree that according to Michelson's ether theory:
>the round trip speed of light in the interferometer frame is:
>along the transverse arm: sqrt(c^2-v^2)
It varies. The individual rays of the cross beam which meet to create fringes do
not have exactly the same travel time. They come from different parts of its
cross section and enter the eyepiece in such a away as to result in no fringe
change.
But like I said, there is no aether anyway so why worry about it.
>along the parallel arm: (c^2-v^2)
Not if light speed is source dependent over short distances.
>
>You are claiming that this doesn't imply fringe shifts
>when the interferometer is rotated.
>Show it!
I'm working on it - along with a few others.
>
>> >According to SR, the speed of the photons are c in ANY frame,
>> >including the "arm frame".
>> Bugger SR. Do you think I'm interested in hearing about a postulate of SR?
>> >So when measured in the arm frame,
>> >the time to traverse the arm is L/c regardless of the orientation
>> >of the arm, zero fringe shifts are predicted.
>> >The moving observer is irrelevant, there is not really
>> >any reason to calculate the speed of the photons in
>> >his frame, but if you do, you must use the Lorentz transform!
>> >When doing that, the speed of the photons is still c, but
>> >the path lengths are longer; the round trip time to traverse BOTH
>> >arms will as measured in the observers frame be: 2*L/(c*sqrt(1-v^2/c^2)).
>
>> I know all that Paul! and it is plain BULL!
>
>What does this statement mean, Henry?
>Does it mean that you know that SR predicts no fringe shifts?
>Or does it mean that it is bullshit that SR predicts no fringe shift?
Of course SR MUST predict NO fringe shifts. The lorentz transforms achieve that
because they were derived from an equivalent error of interpretation, in SR's
case the second postulate...
>
>Make up yout mind.
>Does SR predict fringe shifts, or doesn't it?
>And please remeber that you opinion about the corectness of the theory
>is utterly irrelevant to what it predicts.
It predicts zero fringe shift for the wrong reasons. It uses a mathematical
trick.
>
>> >Now I think it is about time you understand this, Henry.
>> >It shouldn't be hard at all.
>> >So please stop repeating your nonsense.
>
>> The reason I repeat my good sense is because I DO understand!
>
>When you repeat your "wave crest diagram" above, you are repeating
>what never has been disputed, and what nobody ever have failed to
>understand. So what is the bloody point with that?
Don't try to make out you have seen it before. Everybody has been under the
delusion that rain drops, bullets and photons appear diagonal to a moving
observer. THEY DO NOT! They remain vertical.
>
>When you repeat that "the photons are moving vertically at c",
>you repeat what is correct according to one specific theory only,
>and again - nobody has ever failed to understand that
>the "ballistic light theory" states so.
>So what is the bloody point with that?
>
>When you repeat that the predictions of SR and Michelson's ether theory
>are not what they are because you think the theories are wrong,
>then you are repeating a huge logical nonsense.
>So what is the bloody point with that?
How the hell can a theory be right if it is based on a monumental error?
>
>Paul
Paul B. Andersen
>
> On Wed, 12 Sep 2001 16:14:03 +0200, "Paul B. Andersen" <paul.b....@hia.no>
> wrote:
>
> >Henry Wilson wrote:
> >> Paul B. Andersen wrote:
> >> >However, according to "Michelson type ether theory",
> >> >the photons are moving at the speed c IN THE ETHER.
> >> >And the transverse arm is moving at the speed v IN THE ETHER.
> >> >That means that the speed of the photons in the arm frame
> >> >will be sqrt(c^2-v^2), and the time to traverse the transverse arm
> >> >= L/sqrt(c^2-v^2) = L/(c*sqrt(1-v^2/c^2))
> >> >
> >> > /|
> >> > c/ |
> >> > / | sqrt(c^2-v^2)
> >> > /___|
> >> > v
> >> >How the hell can you have missed such an obvious matter?
> >
> >> I know THAT!!
> >
> >Fine. So we agree then.
> No we don't bloody agree! When I said "I know that" I meant I know all about
> your argument.
Stop right there, Henry!
What is it you don't agree about?
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
If it isn't correct, please point out what's wrong.
> >According to Michelson's ether theory, the photons (infinitesimal
> >light elements) move diagonally at c in the ether, and
> >the velocity component along the transverse arm is sqrt(c^2-v^2).
> NO Paul! (tearing hair out).
> A photon is not an 'infinitesimal point'. A photon has length, cross section and
> structure. Otherwise it couldn't contain the information about its source,
> needed to give the correct Doppler shift when it hits something. The double slit
> experiment wouldn't work, either.
> Each infinitesimal point moves along a unique infinitesimally thin diagonal at
> sqrt(c^2+v^2). Just like each minute part of the 'ants' walking up the pole.
This is mumbo jumbo, Henry.
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
> >> >AND PLEASE DON'T REPEAT YOUR IDIOTIC CLAIM THAT THE PHOTONS
> >> >ARE NOT MOVING DIAGONALLY AT C IN THE ETHER FRAME!
> >
> >It is obvious from the context that this is
> >ACCORDING TO MICHELSON'S ETHER THEORY!
> SR IS an aether theory.
> What other justification do you have for claiming that the light beam moves
> diagonally at c. That is just what sound would do in air.
Stick to the issue!
We are discussing what Michelson's ether theory predicts!
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
> >> The bloody photons move vertically at c, in all frames.
> >
> >So you DO repeat your MORONIC claim that according to
> >Michelson's ether theory, the photons are not moving
> >diagonally at c in the ether.
> According to any theory, they are not. Maxwell for instance.
> The whole thing is a huge blunder.
WHAT is a blunder, Henry?
Stick to the issue!
We are discussing what Michelsons ether theory predicts!
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
> Just draw the path of each 'wavecrest' in the
> MMX cross beam and you will see why.
> Surely there is someone else in the world capable of carrying out this simple
> exercise in graphing.
/|
c/ |
/ | sqrt(c^2-v^2)
/___|
v
> Plot the body of an ant walking up the pole in the moving observer's frame. IT
> NEVER ROTATES DIAGONALLY. IT ALWAYS REMAINS VERTICAL!!!
So what?
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
> >You are very inconsistent, Henry.
> >Right above, you said: "I know THAT" to what you now deny.
> I have attended to that.
Meaning what?
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
> >> >And the "round trip speed" of the photons along the parallel
> >> >arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
> >> >and a fringe shifts is predicted.
> >> >How the hell can you have missed such an obvious matter?
> >
> >> The fringe shift is WRONGLY predicted.
> >
> >How so?
> >We agree that according to Michelson's ether theory:
> >the round trip speed of light in the interferometer frame is:
> >along the transverse arm: sqrt(c^2-v^2)
> It varies. The individual rays of the cross beam which meet to create fringes do
> not have exactly the same travel time. They come from different parts of its
> cross section and enter the eyepiece in such a away as to result in no fringe
> change.
And what does this mumbo jumbo mean?
Do you understand that according to Michelson's ether theory,
the infinitesimal light elements (or whatever you prefer to call them)
move diagonally at c in the ether frame, and the velocity component
along the tranverse arm is sqrt(c^2-v^2), or don't you?
> But like I said, there is no aether anyway so why worry about it.
YOU claim that the MMX doesn't falsify Michelson's ether theory.
We are discussing whether or not your claim is correct.
> >along the parallel arm: (c^2-v^2)
> Not if light speed is source dependent over short distances.
Stick to the issue!
We are discussing what Michelson's ether theory predicts.
Do you understand that according to Michelson's ether theory,
the round trip speed along the parallel arm is (c^2-v^2),
or don't you?
> >You are claiming that this doesn't imply fringe shifts
> >when the interferometer is rotated.
> >Show it!
> I'm working on it - along with a few others.
Ah. You are working on it!
So that means that you are not able to support you claim,
but hope that you will be in the future.
But why do you claim something you cannot support now, Henry?
Do you guess first, and try to find a reason for your guess later?
> >> The reason I repeat my good sense is because I DO understand!
> >
> >When you repeat your "wave crest diagram" above, you are repeating
> >what never has been disputed, and what nobody ever have failed to
> >understand. So what is the bloody point with that?
> Don't try to make out you have seen it before. Everybody has been under the
> delusion that rain drops, bullets and photons appear diagonal to a moving
> observer. THEY DO NOT! They remain vertical.
> >
> >When you repeat that "the photons are moving vertically at c",
> >you repeat what is correct according to one specific theory only,
> >and again - nobody has ever failed to understand that
> >the "ballistic light theory" states so.
> >So what is the bloody point with that?
> >
> >When you repeat that the predictions of SR and Michelson's ether theory
> >are not what they are because you think the theories are wrong,
> >then you are repeating a huge logical nonsense.
> >So what is the bloody point with that?
> How the hell can a theory be right if it is based on a monumental error?
Hand waving, Henry.
Paul
Henry Wilson
wrote:
>Henry Wilson wrote:
>>
>> On Wed, 12 Sep 2001 16:14:03 +0200, "Paul B. Andersen" <paul.b....@hia.no>
>> wrote:
>>
>> >Henry Wilson wrote:
>
>> >> >= L/sqrt(c^2-v^2) = L/(c*sqrt(1-v^2/c^2))
>> >> >
>> >> > /|
>> >> > c/ |
>> >> > / | sqrt(c^2-v^2)
>> >> > /___|
>> >> > v
>> >> >How the hell can you have missed such an obvious matter?
>> >
>> >> I know THAT!!
>> >
>> >Fine. So we agree then.
>
>> No we don't bloody agree! When I said "I know that" I meant I know all about
>> your argument.
>
>Stop right there, Henry!
>What is it you don't agree about?
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
>If it isn't correct, please point out what's wrong.
There is only one direction in which photons move at c, that is along their wave
axes. as seen by a moving observer, a transverse beam of light will always
appear vertical. Each element of that beam will follow a diagonal path in the
moving frame. Their velocity along that diagonal is sqrt(c^2+v^2).
>
>> >According to Michelson's ether theory, the photons (infinitesimal
>> >light elements) move diagonally at c in the ether, and
>> >the velocity component along the transverse arm is sqrt(c^2-v^2).
>
>> NO Paul! (tearing hair out).
>> A photon is not an 'infinitesimal point'. A photon has length, cross section and
>> structure. Otherwise it couldn't contain the information about its source,
>> needed to give the correct Doppler shift when it hits something. The double slit
>> experiment wouldn't work, either.
>> Each infinitesimal point moves along a unique infinitesimally thin diagonal at
>> sqrt(c^2+v^2). Just like each minute part of the 'ants' walking up the pole.
>
>This is mumbo jumbo, Henry.
>
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
Paul, do you know anything about draughting. If so, just draw a simple diagram
to illustrate what you are trying to tell me and you will soon see your obvious
error.
>
>> >> >AND PLEASE DON'T REPEAT YOUR IDIOTIC CLAIM THAT THE PHOTONS
>> >> >ARE NOT MOVING DIAGONALLY AT C IN THE ETHER FRAME!
>> >
>> >It is obvious from the context that this is
>> >ACCORDING TO MICHELSON'S ETHER THEORY!
>
>> SR IS an aether theory.
>> What other justification do you have for claiming that the light beam moves
>> diagonally at c. That is just what sound would do in air.
>
>Stick to the issue!
>We are discussing what Michelson's ether theory predicts!
There is no ether and michelson's predictions were wrong anyway.
that's my honest opinion
>
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
Yes Paul, I understand that perfectly.
>
>> >> The bloody photons move vertically at c, in all frames.
>> >
>> >So you DO repeat your MORONIC claim that according to
>> >Michelson's ether theory, the photons are not moving
>> >diagonally at c in the ether.
>
>> According to any theory, they are not. Maxwell for instance.
>> The whole thing is a huge blunder.
>
>WHAT is a blunder, Henry?
>Stick to the issue!
>We are discussing what Michelsons ether theory predicts!
According to Michelson et al, Michelson's ether theory predicted a fringe shift.
According to Wilson et al, Michelson's theory was all up the spout.
>
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
You just asked that. Have you got Altzheimer's?
>
>> Just draw the path of each 'wavecrest' in the
>> MMX cross beam and you will see why.
>> Surely there is someone else in the world capable of carrying out this simple
>> exercise in graphing.
>
> /|
> c/ |
> / | sqrt(c^2-v^2)
> /___|
> v
>
WRONG! What possible reason do you have for saying that it moves at c along that
diagonal?
It doesn't - nor do raindrops or the ants on the pole.
>> Plot the body of an ant walking up the pole in the moving observer's frame. IT
>> NEVER ROTATES DIAGONALLY. IT ALWAYS REMAINS VERTICAL!!!
>
>So what?
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
You really should visit a dementia clinic. That's the third time you have asked
the same question!
>
>> >You are very inconsistent, Henry.
>> >Right above, you said: "I know THAT" to what you now deny.
>
>> I have attended to that.
>
>Meaning what?
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
Are you repeating this in the hope that the truth will go away?
>
according to Michelson's ether theory:
>> >the round trip speed of light in the interferometer frame is:
>> >along the transverse arm: sqrt(c^2-v^2)
>
>> It varies. The individual rays of the cross beam which meet to create fringes do
>> not have exactly the same travel time. They come from different parts of its
>> cross section and enter the eyepiece in such a away as to result in no fringe
>> change.
>
>And what does this mumbo jumbo mean?
>Do you understand that according to Michelson's ether theory,
>the infinitesimal light elements (or whatever you prefer to call them)
>move diagonally at c in the ether frame, and the velocity component
>along the tranverse arm is sqrt(c^2-v^2), or don't you?
Definitely a dementia case.
>
>> But like I said, there is no aether anyway so why worry about it.
>
>YOU claim that the MMX doesn't falsify Michelson's ether theory.
>We are discussing whether or not your claim is correct.
My demo at homestead.com shows why a null result would be expected if there
happened to be an aether. I never implied that there was one!
>
>> >along the parallel arm: (c^2-v^2)
>
>> Not if light speed is source dependent over short distances.
>
>Stick to the issue!
>We are discussing what Michelson's ether theory predicts.
>Do you understand that according to Michelson's ether theory,
>the round trip speed along the parallel arm is (c^2-v^2),
>or don't you?
I undertand that it WRONGLY predicts this, YES!!!
>
>> >You are claiming that this doesn't imply fringe shifts
>> >when the interferometer is rotated.
>> >Show it!
>
>> I'm working on it - along with a few others.
>
>Ah. You are working on it!
>So that means that you are not able to support you claim,
>but hope that you will be in the future.
It isn't a question of supporting my claim. It is a matter of creating something
that a SRian or a 5 yo kid can understand.
>
>But why do you claim something you cannot support now, Henry?
>Do you guess first, and try to find a reason for your guess later?
Paul, I do it just for the sake of annoying you!
>
>
>> >
>> >When you repeat that the predictions of SR and Michelson's ether theory
>> >are not what they are because you think the theories are wrong,
>> >then you are repeating a huge logical nonsense.
>> >So what is the bloody point with that?
>
>> How the hell can a theory be right if it is based on a monumental error?
>
>Hand waving, Henry.
Big hand! Big wave!
>
>Paul
Paul B. Andersen
|However, according to "Michelson type ether theory",
|the photons are moving at the speed c IN THE ETHER.
|And the transverse arm is moving at the speed v IN THE ETHER.
|That means that the speed of the photons in the arm frame
|will be sqrt(c^2-v^2), and the time to traverse the transverse arm
|
| /|
| c/ |
| / | sqrt(c^2-v^2)
| /___|
| v
| How the hell can you have missed such an obvious matter?
|
| arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
| and a fringe shifts is predicted.
> > Henry Wilson wrote:
> >> No we don't bloody agree! When I said "I know that" I meant I know all about
> >> your argument.
> Paul B. Andersen wrote:
> >Stop right there, Henry!
> >What is it you don't agree about?
> >Do you understand that according to Michelson's ether theory,
> >the infinitesimal light elements (or whatever you prefer to call them)
> >move diagonally at c in the ether frame, and the velocity component
> >along the tranverse arm is sqrt(c^2-v^2), or don't you?
You obviously don't.
How the hell is it possible to be so dense?
> >If it isn't correct, please point out what's wrong.
Henry Wilson wrote:
> There is only one direction in which photons move at c, that is along their wave
> axes. as seen by a moving observer, a transverse beam of light will always
> appear vertical. Each element of that beam will follow a diagonal path in the
> moving frame. Their velocity along that diagonal is sqrt(c^2+v^2).
You are talking nonsense, Henry.
First of all, the frames of interest are the ether frame
and the interferometer frame, moving at v in the ether.
According to Michelson's ether theory, light is moving at
c in the ether frame ONLY. So NO light can move at c in
the interferometer frame (Galilean relativity!).
How the hell is it possible to claim that light
according to Michelson's ether theory is moving
at the velocity sqrt(c^2+v^2) IN THE ETHER FRAME,
when it moves at c per bloody definition?
Sober up, Henry.
This is to stupid even for you!
We know that every infinitesimal light element is moving along
the transverse arm. And since this arm is moving at v in the ether,
every infinitsesimal light element must blatantly obvious have
a horizontal velocity component v IN THE ETHER!
That means that each infinitesimal light element is moving
diagonally in the ether. You DO understand that, don't you?
AND THEY MOVE AT c PER DEFINITION BECAUSE THIS IS MICHELSONS
ETHER THEORY!
Thus:
/|
c/ |
/ | sqrt(c^2-v^2)
/___|
v
There is no way it can be disputed that this is what
MICHELSON'S ETHER THEORY PREDICTS!
> >Stick to the issue!
> >We are discussing what Michelson's ether theory predicts!
=========================================================
> There is no ether and michelson's predictions were wrong anyway.
> that's my honest opinion
Of bloody course we all know by now that Michelson's
predictions were wrong.
THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
But whether a theory is correct or not is utterly irrelevant
to what it predicts. Even YOU must understand that, don't you?
WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
You claim that the MMX do NOT falsify Michelson's ether
theory, because Michelson MISINTERPRETED it.
The only way he could have done so is by doing an error
in the calculation of what his theory predicts.
So Henry, to defend your claim, you must show that
Michelson did an error when calculating the predictions
of MICHELSON'S ETHER THEORY!
And you cannot do that by your moronic repetition of
the fact we all have always known: that "Ballistic light
theory" predicts zero fringe shift!
So Henry - WHAT DOES MICHELSON'S ETHER THEORY PREDICT?
Paul
Henry Wilson
wrote:
>Paul B. Andersen wrote:
Paul, no infinitesimal elements actually move!
The infinitesimally thin diagonal lines, which you, Einstein and the whole
physics establishment often refer to, are the loci or paths followed by
infinitesimally short elements of an infinitesimally narrow light beam WHEN THEY
ARE PLOTTED IN THE FRAME OF A MOVING OBSERVER.
The diagonals are in no way representative of the path of a complete light ray.
In a moving frame, individual photons will appear to move diagonally but with
their long axes always pointing in the original direction.
Paul, imagine - or better still - draw a diagram of a vertical laser beam.
Imagine that you can see it in a smokey atmosphere.
Now, what is its configuration in your frame, if you go past at near c.
Answer, exactly the same. It appears as a vertical light beam, moving away from
you at near c.
IT DOES NOT APPEAR TO BE DIAGONAL.
What is more, there is absolutely no reason for believing that the time taken
for a section of the beam to reach a certain height is in any way dependent on
observer speed.
>
>> >If it isn't correct, please point out what's wrong.
>
>Henry Wilson wrote:
>> There is only one direction in which photons move at c, that is along their wave
>> axes. as seen by a moving observer, a transverse beam of light will always
>> appear vertical. Each element of that beam will follow a diagonal path in the
>> moving frame. Their velocity along that diagonal is sqrt(c^2+v^2).
>
>You are talking nonsense, Henry.
Never done that in my life!
>First of all, the frames of interest are the ether frame
>and the interferometer frame, moving at v in the ether.
>
>According to Michelson's ether theory, light is moving at
>c in the ether frame ONLY. So NO light can move at c in
>the interferometer frame (Galilean relativity!).
>
>How the hell is it possible to claim that light
>according to Michelson's ether theory is moving
>at the velocity sqrt(c^2+v^2) IN THE ETHER FRAME,
>when it moves at c per bloody definition?
Because it doesn't behave like a spherical sound wave in air.
That's how Michelson treated it.
Light in the interferometer is collimated, for a start.
There is no way it can leave the 45 mirror at any angle other than 90.
By what mechanism could it possible appear diagonal?
It is the same as my laser beam example, above. The whole MMX cross beam remains
vertical when plotted in all frames (those moving along the parallel axis, of
course).
So all the past books about the MMX are wrong.
All SR is based on the same incorrect logic and is also therefore wrong!!!
>
>Sober up, Henry.
>This is to stupid even for you!
>
>We know that every infinitesimal light element is moving along
>the transverse arm. And since this arm is moving at v in the ether,
>every infinitsesimal light element must blatantly obvious have
>a horizontal velocity component v IN THE ETHER!
It can be construed that way. But the velocity is obviously not c!
There is no principle or mechanism, maxwellian or otherwise which even suggests
that it might be c.
It is a mistake.
>
>That means that each infinitesimal light element is moving
>diagonally in the ether. You DO understand that, don't you?
>AND THEY MOVE AT c PER DEFINITION BECAUSE THIS IS MICHELSONS
>ETHER THEORY!
>
>Thus:
>
> /|
> c/ |
> / | sqrt(c^2-v^2)
> /___|
> v
>
>There is no way it can be disputed that this is what
>MICHELSON'S ETHER THEORY PREDICTS!
I'm not disputing WHAT Michelson and followers seem to have thought. They have
all been wrong, to this day!
Light doesn't behave like sound in air - aether or no aether!
>
>> >Stick to the issue!
>> >We are discussing what Michelson's ether theory predicts!
> =========================================================
>
>> There is no ether and michelson's predictions were wrong anyway.
>> that's my honest opinion
>
>Of bloody course we all know by now that Michelson's
>predictions were wrong.
>
>THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
But it DIDN"T.
The conclusion was that it DID falsify the theory - but the proof itself was
based on equally false logic!
>
>But whether a theory is correct or not is utterly irrelevant
>to what it predicts. Even YOU must understand that, don't you?
You mean "what it is CLAIMED or WRONGLY INTERPRETED to predict?"
>
>WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
>
>You claim that the MMX do NOT falsify Michelson's ether
>theory, because Michelson MISINTERPRETED it.
Yes. I think I can agree about that..
>
>The only way he could have done so is by doing an error
>in the calculation of what his theory predicts.
>
>So Henry, to defend your claim, you must show that
>Michelson did an error when calculating the predictions
>of MICHELSON'S ETHER THEORY!
No I don't. No need. Michelson already did that. His own experimental evidence
refuted the predictions of his own theory.
>
>And you cannot do that by your moronic repetition of
>the fact we all have always known: that "Ballistic light
>theory" predicts zero fringe shift!
Don't insult me Paul. My comments are much more comprehensive than any primitive
ballistic light theory.
>
>So Henry - WHAT DOES MICHELSON'S ETHER THEORY PREDICT?
It depends which way one looks at the question.
>
>Paul
David Evens
What axes would you be halucinating about?
>Paul, imagine - or better still - draw a diagram of a vertical laser beam.
>Imagine that you can see it in a smokey atmosphere.
>
>Now, what is its configuration in your frame, if you go past at near c.
>Answer, exactly the same. It appears as a vertical light beam, moving away from
>you at near c.
>IT DOES NOT APPEAR TO BE DIAGONAL.
This is irrelevant. In order for the light to APEAR verticle, it must
all move DIAGONALLY.
>What is more, there is absolutely no reason for believing that the time taken
>for a section of the beam to reach a certain height is in any way dependent on
>observer speed.
Othen than the fact that it has to move diagonally to reach that
height, and its speed is still invariant.
>>> >If it isn't correct, please point out what's wrong.
>>
>>Henry Wilson wrote:
>>> There is only one direction in which photons move at c, that is along their wave
>>> axes. as seen by a moving observer, a transverse beam of light will always
>>> appear vertical. Each element of that beam will follow a diagonal path in the
>>> moving frame. Their velocity along that diagonal is sqrt(c^2+v^2).
>>
>>You are talking nonsense, Henry.
>
>Never done that in my life!
No, that would be sense you've never talked in your life.
>>First of all, the frames of interest are the ether frame
>>and the interferometer frame, moving at v in the ether.
>>
>>According to Michelson's ether theory, light is moving at
>>c in the ether frame ONLY. So NO light can move at c in
>>the interferometer frame (Galilean relativity!).
>>
>>How the hell is it possible to claim that light
>>according to Michelson's ether theory is moving
>>at the velocity sqrt(c^2+v^2) IN THE ETHER FRAME,
>>when it moves at c per bloody definition?
>
>Because it doesn't behave like a spherical sound wave in air.
>That's how Michelson treated it.
>Light in the interferometer is collimated, for a start.
>There is no way it can leave the 45 mirror at any angle other than 90.
>
>By what mechanism could it possible appear diagonal?
AT LEAST one arm cannot avoid not being in line with ANY relative
motion. The light in that arm MUST follow a diagonal path, or it
would fall out of the arm.
>It is the same as my laser beam example, above. The whole MMX cross beam remains
>vertical when plotted in all frames (those moving along the parallel axis, of
>course).
>So all the past books about the MMX are wrong.
>All SR is based on the same incorrect logic and is also therefore wrong!!!
Yo forgot to mention why looking at the path the light travels, rather
than where the unconnected and unconnectable photons happen to be, is
somehow wrong.
>>Sober up, Henry.
>>This is to stupid even for you!
>>
>>We know that every infinitesimal light element is moving along
>>the transverse arm. And since this arm is moving at v in the ether,
>>every infinitsesimal light element must blatantly obvious have
>>a horizontal velocity component v IN THE ETHER!
>
>It can be construed that way. But the velocity is obviously not c!
>There is no principle or mechanism, maxwellian or otherwise which even suggests
>that it might be c.
>It is a mistake.
Thwn qwhy does Maxwell's equatiuons REQUIRE the total velocity to
ALWAYS be c?
>>That means that each infinitesimal light element is moving
>>diagonally in the ether. You DO understand that, don't you?
>>AND THEY MOVE AT c PER DEFINITION BECAUSE THIS IS MICHELSONS
>>ETHER THEORY!
>>
>>Thus:
>>
>> /|
>> c/ |
>> / | sqrt(c^2-v^2)
>> /___|
>> v
>>
>>There is no way it can be disputed that this is what
>>MICHELSON'S ETHER THEORY PREDICTS!
>
>I'm not disputing WHAT Michelson and followers seem to have thought. They have
>all been wrong, to this day!
>Light doesn't behave like sound in air - aether or no aether!
Light is unable to NOIT behave like sound if there is assumed to be an
aether. Since the predictions of observables are the same in LET
(which assumes aether) and SR (which does not) it is impossible to
avoid the prediction that light will look very much lik sound.
>>> >Stick to the issue!
>>> >We are discussing what Michelson's ether theory predicts!
>> =========================================================
>>
>>> There is no ether and michelson's predictions were wrong anyway.
>>> that's my honest opinion
>>
>>Of bloody course we all know by now that Michelson's
>>predictions were wrong.
>>
>>THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
>
>But it DIDN"T.
>The conclusion was that it DID falsify the theory - but the proof itself was
>based on equally false logic!
Then why have you never found a flaw in the logic?
>>But whether a theory is correct or not is utterly irrelevant
>>to what it predicts. Even YOU must understand that, don't you?
>
>You mean "what it is CLAIMED or WRONGLY INTERPRETED to predict?"
That would be the kind of thing you like to do, yes: Make wrong
claims about what is predicted.
>>WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
>>
>>You claim that the MMX do NOT falsify Michelson's ether
>>theory, because Michelson MISINTERPRETED it.
>
>Yes. I think I can agree about that..
>>
>>The only way he could have done so is by doing an error
>>in the calculation of what his theory predicts.
>>
>>So Henry, to defend your claim, you must show that
>>Michelson did an error when calculating the predictions
>>of MICHELSON'S ETHER THEORY!
>
>No I don't. No need. Michelson already did that. His own experimental evidence
>refuted the predictions of his own theory.
SDo you admit that the MMX DOES falsify the Michelson aether theory.
>>And you cannot do that by your moronic repetition of
>>the fact we all have always known: that "Ballistic light
>>theory" predicts zero fringe shift!
>
>Don't insult me Paul. My comments are much more comprehensive than any primitive
>ballistic light theory.
Yes, you are much more eroneous that any mere balistic light theory.
>>So Henry - WHAT DOES MICHELSON'S ETHER THEORY PREDICT?
>
>It depends which way one looks at the question.
Equations are unable to have the property of producing different
values becaue of things not included in the equations.
Henry Wilson
>On Sun, 16 Sep 2001 04:40:54 GMT, He...@the.edge(Henry Wilson) wrote:
>
>What axes would you be halucinating about?
Evens you are a complete moron and I have killfiled you.
>
>>Paul, imagine - or better still - draw a diagram of a vertical laser beam.
>>Imagine that you can see it in a smokey atmosphere.
>>
>>Now, what is its configuration in your frame, if you go past at near c.
>>Answer, exactly the same. It appears as a vertical light beam, moving away from
>>you at near c.
>>IT DOES NOT APPEAR TO BE DIAGONAL.
>
>This is irrelevant. In order for the light to APEAR verticle, it must
>all move DIAGONALLY.
the file has again been used.
>
>>Because it doesn't behave like a spherical sound wave in air.
>>That's how Michelson treated it.
>>Light in the interferometer is collimated, for a start.
>>There is no way it can leave the 45 mirror at any angle other than 90.
>>
>>By what mechanism could it possible appear diagonal?
>
>AT LEAST one arm cannot avoid not being in line with ANY relative
>motion. The light in that arm MUST follow a diagonal path, or it
>would fall out of the arm.
Draw the bloody thing evens. - or into the file again.
>
>>It is the same as my laser beam example, above. The whole MMX cross beam remains
>>vertical when plotted in all frames (those moving along the parallel axis, of
>>course).
>>So all the past books about the MMX are wrong.
>>All SR is based on the same incorrect logic and is also therefore wrong!!!
>
>Yo forgot to mention why looking at the path the light travels, rather
>than where the unconnected and unconnectable photons happen to be, is
>somehow wrong.
Do you understand what 'infinitesimal' means?
>
>>>Sober up, Henry.
>>>This is to stupid even for you!
>>>
>>>We know that every infinitesimal light element is moving along
>>>the transverse arm. And since this arm is moving at v in the ether,
>>>every infinitsesimal light element must blatantly obvious have
>>>a horizontal velocity component v IN THE ETHER!
>>
>>It can be construed that way. But the velocity is obviously not c!
>>There is no principle or mechanism, maxwellian or otherwise which even suggests
>>that it might be c.
>>It is a mistake.
>
>Thwn qwhy does Maxwell's equatiuons REQUIRE the total velocity to
>ALWAYS be c?
looks like you have been taking typing lessons from tj frazir.
>
>>>That means that each infinitesimal light element is moving
>>>diagonally in the ether. You DO understand that, don't you?
>>>AND THEY MOVE AT c PER DEFINITION BECAUSE THIS IS MICHELSONS
>>>ETHER THEORY!
>>>
>>>Thus:
>>>
>>> /|
>>> c/ |
>>> / | sqrt(c^2-v^2)
>>> /___|
>>> v
>>>
>>>There is no way it can be disputed that this is what
>>>MICHELSON'S ETHER THEORY PREDICTS!
>>
>>I'm not disputing WHAT Michelson and followers seem to have thought. They have
>>all been wrong, to this day!
>>Light doesn't behave like sound in air - aether or no aether!
>
>Light is unable to NOIT behave like sound if there is assumed to be an
>aether. Since the predictions of observables are the same in LET
>(which assumes aether) and SR (which does not) it is impossible to
>avoid the prediction that light will look very much lik sound.
Except that it only travels in straight lines IN THE DIRECTION OF ITS AXIS.
>
>>>THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
>>
>>But it DIDN"T.
>>The conclusion was that it DID falsify the theory - but the proof itself was
>>based on equally false logic!
>
>Then why have you never found a flaw in the logic?
>
>>>But whether a theory is correct or not is utterly irrelevant
>>>to what it predicts. Even YOU must understand that, don't you?
>>
>>You mean "what it is CLAIMED or WRONGLY INTERPRETED to predict?"
>
>That would be the kind of thing you like to do, yes: Make wrong
>claims about what is predicted.
>
>>>WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
>>>
>>>You claim that the MMX do NOT falsify Michelson's ether
>>>theory, because Michelson MISINTERPRETED it.
>>
>>Yes. I think I can agree about that..
>>>
>>>The only way he could have done so is by doing an error
>>>in the calculation of what his theory predicts.
>>>
>>>So Henry, to defend your claim, you must show that
>>>Michelson did an error when calculating the predictions
>>>of MICHELSON'S ETHER THEORY!
>>
>>No I don't. No need. Michelson already did that. His own experimental evidence
>>refuted the predictions of his own theory.
>
>SDo you admit that the MMX DOES falsify the Michelson aether theory.
No. That was a misinterpretation.
>
Paul B. Andersen
Why the hell do you keep repeating this triviality?
In never was disputed.
But the path of every individual photon or light element or what
the hell you prefer to call it is diagonal IN THE ETHER FRAME!
> What is more, there is absolutely no reason for believing that the time taken
> for a section of the beam to reach a certain height is in any way dependent on
> observer speed.
But it sure does depend on the ether speed.
(Or the arm's speed in the ether.)
> >> >If it isn't correct, please point out what's wrong.
> >
> >Henry Wilson wrote:
> >> There is only one direction in which photons move at c, that is along their wave
> >> axes. as seen by a moving observer, a transverse beam of light will always
> >> appear vertical. Each element of that beam will follow a diagonal path in the
> >> moving frame. Their velocity along that diagonal is sqrt(c^2+v^2).
> >
> >You are talking nonsense, Henry.
>
> Never done that in my life!
>
> >First of all, the frames of interest are the ether frame
> >and the interferometer frame, moving at v in the ether.
> >
> >According to Michelson's ether theory, light is moving at
> >c in the ether frame ONLY. So NO light can move at c in
> >the interferometer frame (Galilean relativity!).
> >
> >How the hell is it possible to claim that light
> >according to Michelson's ether theory is moving
> >at the velocity sqrt(c^2+v^2) IN THE ETHER FRAME,
> >when it moves at c per bloody definition?
>
> Because it doesn't behave like a spherical sound wave in air.
> That's how Michelson treated it.
Of bloody course it does exactly that ACCORDING TO MICHELSON'S ETHER
THEORY, and of bloody course Michelson treaded it as such when he
calculated the predictions of MICHELSON'S ETHER THEORY!
> Light in the interferometer is collimated, for a start.
> There is no way it can leave the 45 mirror at any angle other than 90.
In the mirror frame obviously not.
In the the ether frame where the mirror is moving, the angle
will equally obvious NOT be 90 degrees.
> By what mechanism could it possible appear diagonal?
Isn't that blatantly obvious? The mirror is MOVING!
> It is the same as my laser beam example, above. The whole MMX cross beam remains
> vertical when plotted in all frames (those moving along the parallel axis, of
> course).
Why do you keep repeating this obvious triviality?
> So all the past books about the MMX are wrong.
Because they state that the path of each individual photon
(or what the hell etc.) which is moving along the MOVING transverse arm
blatantly obvious is diagonal in the ether frame?
Of course that is correct!
> All SR is based on the same incorrect logic and is also therefore wrong!!!
We are not talking about SR!
We are discussing the predictions of Michelson's ether theory.
The null prediction of SR is obvious.
No reason to bring in any moving observer to see that!
> >Sober up, Henry.
> >This is to stupid even for you!
> >
> >We know that every infinitesimal light element is moving along
> >the transverse arm. And since this arm is moving at v in the ether,
> >every infinitsesimal light element must blatantly obvious have
> >a horizontal velocity component v IN THE ETHER!
>
> It can be construed that way. But the velocity is obviously not c!
> There is no principle or mechanism, maxwellian or otherwise which even suggests
> that it might be c.
> It is a mistake.
Of bloody course the speed IN THE ETHER is c ACCORDING TO
Michelson's ether theory!
> >That means that each infinitesimal light element is moving
> >diagonally in the ether. You DO understand that, don't you?
> >AND THEY MOVE AT c PER DEFINITION BECAUSE THIS IS MICHELSONS
> >ETHER THEORY!
> >
> >Thus:
> >
> > /|
> > c/ |
> > / | sqrt(c^2-v^2)
> > /___|
> > v
> >
> >There is no way it can be disputed that this is what
> >MICHELSON'S ETHER THEORY PREDICTS!
>
> I'm not disputing WHAT Michelson and followers seem to have thought. They have
> all been wrong, to this day!
> Light doesn't behave like sound in air - aether or no aether!
OF BLOODY COURSE IT DOESN'T!
We know that BECAUSE the MMX falsified Michelson's ether theory.
And the MMX falsifies Michelson's ether theory BECAUSE this theory
predicts fringe shift.
Michelson made no error when analyzing what his theory predicted.
He CORRECTLY calculated that his theory predicted fringe shifts.
MICHELSON'S THEORY WAS WRONG.
BUT MICHELSON'S CALCULATION OF WHAT HIS THEORY PREDICTED WAS CORRECT.
THEREFORE THE MMX FALSIFIED HIS THEORY!
> >
> >> >Stick to the issue!
> >> >We are discussing what Michelson's ether theory predicts!
> > =========================================================
> >
> >> There is no ether and michelson's predictions were wrong anyway.
> >> that's my honest opinion
> >
> >Of bloody course we all know by now that Michelson's
> >predictions were wrong.
> >
> >THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
>
> But it DIDN"T.
> The conclusion was that it DID falsify the theory - but the proof itself was
> based on equally false logic!
I think I have had enough of this idiocy now, Henry.
You are saying that Michelson's calculations of what his theory
predicted was wrong because the theory was wrong, and thus the fact
that the predictions are wrong does not falsify the theory!
How the hell is it possible to fail to understand how utterly
ridiculous this is?
> >But whether a theory is correct or not is utterly irrelevant
> >to what it predicts. Even YOU must understand that, don't you?
>
> You mean "what it is CLAIMED or WRONGLY INTERPRETED to predict?"
So you don't understand it.
> >WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
> >
> >You claim that the MMX do NOT falsify Michelson's ether
> >theory, because Michelson MISINTERPRETED it.
>
> Yes. I think I can agree about that..
> >
> >The only way he could have done so is by doing an error
> >in the calculation of what his theory predicts.
> >
> >So Henry, to defend your claim, you must show that
> >Michelson did an error when calculating the predictions
> >of MICHELSON'S ETHER THEORY!
>
> No I don't. No need. Michelson already did that. His own experimental evidence
> refuted the predictions of his own theory.
See!!!!
I must be talking to an idiot!
There is no point in going on.
You have obviously no idea of basic logic.
I give you up!
Paul
Henry Wilson
wrote:
>Henry Wilson wrote:
>>
>>
>> The diagonals are in no way representative of the path of a complete light ray.
>> In a moving frame, individual photons will appear to move diagonally but with
>> their long axes always pointing in the original direction.
>>
>> Paul, imagine - or better still - draw a diagram of a vertical laser beam.
>> Imagine that you can see it in a smokey atmosphere.
>>
>> Now, what is its configuration in your frame, if you go past at near c.
>> Answer, exactly the same. It appears as a vertical light beam, moving away from
>> you at near c.
>> IT DOES NOT APPEAR TO BE DIAGONAL.
>
>Why the hell do you keep repeating this triviality?
>In never was disputed.
>But the path of every individual photon or light element or what
>the hell you prefer to call it is diagonal IN THE ETHER FRAME!
Why do you continue to regard light as being similar to sound?
One obvious difference between a transverse and a longitudinal wave it that the
former has a wave axis and travels only in that direction. There is no evidence
that a transverse wave in any medium can disperse and travel at a fixed speed in
any other directions.
I don't think anyone has succeded in actually creating a transverse wave in air.
It would be quickly dampened or become turbulent.
>
>> What is more, there is absolutely no reason for believing that the time taken
>> for a section of the beam to reach a certain height is in any way dependent on
>> observer speed.
>
>But it sure does depend on the ether speed.
>(Or the arm's speed in the ether.)
You are comparing it with sound again. Where is your evidence or logic here?
>
>>
>> >First of all, the frames of interest are the ether frame
>> >and the interferometer frame, moving at v in the ether.
>> >
>> >According to Michelson's ether theory, light is moving at
>> >c in the ether frame ONLY. So NO light can move at c in
>> >the interferometer frame (Galilean relativity!).
>> >
>> >How the hell is it possible to claim that light
>> >according to Michelson's ether theory is moving
>> >at the velocity sqrt(c^2+v^2) IN THE ETHER FRAME,
>> >when it moves at c per bloody definition?
>>
>> Because it doesn't behave like a spherical sound wave in air.
>> That's how Michelson treated it.
>
>Of bloody course it does exactly that ACCORDING TO MICHELSON'S ETHER
>THEORY, and of bloody course Michelson treaded it as such when he
>calculated the predictions of MICHELSON'S ETHER THEORY!
Which is one reason I am not interested in Michelson's aether. It is obviously
wrong.
>
>> Light in the interferometer is collimated, for a start.
>> There is no way it can leave the 45 mirror at any angle other than 90.
>
>In the mirror frame obviously not.
>In the the ether frame where the mirror is moving, the angle
>will equally obvious NOT be 90 degrees.
Did you recall my laser example. To a moving observer, the beam always appears
vertical, even though it is moving sideways..
>
>> By what mechanism could it possible appear diagonal?
>
>Isn't that blatantly obvious? The mirror is MOVING!
Here we go again.
It is NOT THE BEAM ITSELF which moves diagonally.
Infinitesimal elements follow a diagonal path - each one different - but they do
not constitute light and are not moving at c.
The same applies to a boat crossing a flowing river whilst always pointing
perpendicular to the bank. Each section of the boat moves along its own zero
thickness diagonal at velocity sqrt(Vr^2+Vb^2) relative to the bank.
You see that this velocity can range to infinity, for any constant forward boat
speed.
Why do you (and Michelson) think light should behave any differently from this?
>
>> It is the same as my laser beam example, above. The whole MMX cross beam remains
>> vertical when plotted in all frames (those moving along the parallel axis, of
>> course).
>
>Why do you keep repeating this obvious triviality?
Why do you trying to infer that the most important things are trivial?
The beam remains vertical an always traverses a fixed distance in that direction
in the same time.
Michelson, you and all the others are wrong.
>
>> So all the past books about the MMX are wrong.
>
>Because they state that the path of each individual photon
>(or what the hell etc.) which is moving along the MOVING transverse arm
>blatantly obvious is diagonal in the ether frame?
>Of course that is correct!
It is NOT correct to assume that anything moves at VELOCITY C in that diagonal
direction.
Nobody has ever measured the one way velocity of light along its wave axis, let
alone in a direction oblique to that!!! - or even the TWLS for that matter.
So you and your boys have been living in fairyland.
>
>> All SR is based on the same incorrect logic and is also therefore wrong!!!
>
>We are not talking about SR!
>We are discussing the predictions of Michelson's ether theory.
>The null prediction of SR is obvious.
>No reason to bring in any moving observer to see that!
Of course SR will predict the null result. It is based on Michelson's aether
theory. That is, it also assumes that light moves diagonally at c when seen by a
moving observer.
Naturally everything cancels out nicely so the the time taken for light to
travel along both beams is the same.
BUT that doesn't mean the theory matches reality.
THERE IS NO LENGTH CONTRACTION. I have proved that and you agree.
There can only be a physical length contraction if there is absolute space.
>
>>
>> It can be construed that way. But the velocity is obviously not c!
>> There is no principle or mechanism, maxwellian or otherwise which even suggests
>> that it might be c.
>> It is a mistake.
>
>Of bloody course the speed IN THE ETHER is c ACCORDING TO
>Michelson's ether theory!
Stuff Michelson - and Einstein, who said the same thing. They are clearly
wrong, WRONG!
>
>> >That means that each infinitesimal light element is moving
>> >diagonally in the ether. You DO understand that, don't you?
>> >AND THEY MOVE AT c PER DEFINITION BECAUSE THIS IS MICHELSONS
>> >ETHER THEORY!
>> >
>> >Thus:
>> >
>> > /|
>> > c/ |
>> > / | sqrt(c^2-v^2)
>> > /___|
>> > v
>> >
>> >There is no way it can be disputed that this is what
>> >MICHELSON'S ETHER THEORY PREDICTS!
>>
>> I'm not disputing WHAT Michelson and followers seem to have thought. They have
>> all been wrong, to this day!
>> Light doesn't behave like sound in air - aether or no aether!
>
>OF BLOODY COURSE IT DOESN'T!
>We know that BECAUSE the MMX falsified Michelson's ether theory.
>And the MMX falsifies Michelson's ether theory BECAUSE this theory
>predicts fringe shift.
>Michelson made no error when analyzing what his theory predicted.
>He CORRECTLY calculated that his theory predicted fringe shifts.
>
>MICHELSON'S THEORY WAS WRONG.
>BUT MICHELSON'S CALCULATION OF WHAT HIS THEORY PREDICTED WAS CORRECT.
>THEREFORE THE MMX FALSIFIED HIS THEORY!
Well, the fact that a misinterpreted experiment falsifies an incorrect theory
hardly gives credibility to another theory which correctly explains a phenomenon
of the aforesaid incorrect theory, does it, Paul?
>
>> >THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
>>
>> But it DIDN"T.
>> The conclusion was that it DID falsify the theory - but the proof itself was
>> based on equally false logic!
>
>I think I have had enough of this idiocy now, Henry.
Where is the problem? I have made a strightforward statement of fact.
>
>You are saying that Michelson's calculations of what his theory
>predicted was wrong because the theory was wrong, and thus the fact
>that the predictions are wrong does not falsify the theory!
NO. I'm saying that the prediction of a non-null result was an incorrect
conclusion derived from the misinterpretation of experimental analysis
completely independent of his incorrect aether theory.
>
>How the hell is it possible to fail to understand how utterly
>ridiculous this is?
Not at all.
I have shown many times why the experiment SHOULD give a null result EVEN IN A
MICHELSON AETHER.
>
>
>> >But whether a theory is correct or not is utterly irrelevant
>> >to what it predicts. Even YOU must understand that, don't you?
>>
>> You mean "what it is CLAIMED or WRONGLY INTERPRETED to predict?"
>
>So you don't understand it.
>
>> >WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
>> >
>> >You claim that the MMX do NOT falsify Michelson's ether
>> >theory, because Michelson MISINTERPRETED it.
>>
>> Yes. I think I can agree about that..
>> >
>> >The only way he could have done so is by doing an error
>> >in the calculation of what his theory predicts.
>> >
>> >So Henry, to defend your claim, you must show that
>> >Michelson did an error when calculating the predictions
>> >of MICHELSON'S ETHER THEORY!
>>
>> No I don't. No need. Michelson already did that. His own experimental evidence
>> refuted the predictions of his own theory.
>
>See!!!!
>
>I must be talking to an idiot!
OK I'll modify that. I was a bit hasty there.
I have already shown that error.
>
>There is no point in going on.
>You have obviously no idea of basic logic.
>I give you up!
>
>Paul
you can't give up now Paul.
You are on the verge of seeing the light. You know deep down that something is
wrong with SR but your faith prevents you leaping over to the othe side.
PaulDanaher
news:3BA67E64...@hia.no...
<snipped>
>
> I must be talking to an idiot!
>
> There is no point in going on.
> You have obviously no idea of basic logic.
> I give you up!
>
> Paul
But it was a truly heroic attempt, Paul.
Paul B. Andersen
|However, according to "Michelson type ether theory",
|the photons are moving at the speed c IN THE ETHER.
|And the transverse arm is moving at the speed v IN THE ETHER.
|That means that the speed of the photons in the arm frame
|will be sqrt(c^2-v^2), and the time to traverse the transverse arm
| = L/sqrt(c^2-v^2) = L/(c*sqrt(1-v^2/c^2))
|
| c/ |
| / | sqrt(c^2-v^2)
| /___|
| v
|
| And the "round trip speed" of the photons along the parallel
| arm is (c^2-v^2). Thus the "round trip speed" is anisotropic,
| and a fringe shifts is predicted.
| How the hell can you have missed such an obvious matter?
The issue is:
Is the above a correct description of what MICHELSON'S ETHER THEORY predicts?
Henry Wilson wrote:
>
> On Tue, 18 Sep 2001 00:51:16 +0200, "Paul B. Andersen" <paul.b....@hia.no>
> wrote:
>
> >Henry Wilson wrote:
> >>
> >>
> >> The diagonals are in no way representative of the path of a complete light ray.
> >> In a moving frame, individual photons will appear to move diagonally but with
> >> their long axes always pointing in the original direction.
> >>
> >> Paul, imagine - or better still - draw a diagram of a vertical laser beam.
> >> Imagine that you can see it in a smokey atmosphere.
> >>
> >> Now, what is its configuration in your frame, if you go past at near c.
> >> Answer, exactly the same. It appears as a vertical light beam, moving away from
> >> you at near c.
> >> IT DOES NOT APPEAR TO BE DIAGONAL.
> >
> >Why the hell do you keep repeating this triviality?
> >In never was disputed.
> >But the path of every individual photon or light element or what
> >the hell you prefer to call it is diagonal IN THE ETHER FRAME!
>
> Why do you continue to regard light as being similar to sound?
Because according to MICHELSON'S ETHER THEORY, it is.
> One obvious difference between a transverse and a longitudinal wave it that the
> former has a wave axis and travels only in that direction. There is no evidence
> that a transverse wave in any medium can disperse and travel at a fixed speed in
> any other directions.
> I don't think anyone has succeded in actually creating a transverse wave in air.
> It would be quickly dampened or become turbulent.
The difference between transversal and longitudinal waves
is irrelevant in this context.
What IS relevant is that medium waves (+Galilean relativity) moves
in a direction perpendicular to the planes of equal phase IN THE MEDIUM FRAME ONLY.
> >> What is more, there is absolutely no reason for believing that the time taken
> >> for a section of the beam to reach a certain height is in any way dependent on
> >> observer speed.
> >
> >But it sure does depend on the ether speed.
> >(Or the arm's speed in the ether.)
>
> You are comparing it with sound again. Where is your evidence or logic here?
Michelson's ether theory.
[..]
> >> Because it doesn't behave like a spherical sound wave in air.
> >> That's how Michelson treated it.
> >
> >Of bloody course it does exactly that ACCORDING TO MICHELSON'S ETHER
> >THEORY, and of bloody course Michelson treaded it as such when he
> >calculated the predictions of MICHELSON'S ETHER THEORY!
>
> Which is one reason I am not interested in Michelson's aether. It is obviously
> wrong.
SEE!!!!!
> >> Light in the interferometer is collimated, for a start.
> >> There is no way it can leave the 45 mirror at any angle other than 90.
> >
> >In the mirror frame obviously not.
> >In the the ether frame where the mirror is moving, the angle
> >will equally obvious NOT be 90 degrees.
>
> Did you recall my laser example. To a moving observer, the beam always appears
> vertical, even though it is moving sideways..
Irrelevancy repeated over and over ..
> >> By what mechanism could it possible appear diagonal?
> >
> >Isn't that blatantly obvious? The mirror is MOVING!
>
> Here we go again.
>
> It is NOT THE BEAM ITSELF which moves diagonally.
> Infinitesimal elements follow a diagonal path - each one different - but they do
> not constitute light and are not moving at c.
... and over and over again.
Below are the planes of zero phase \ (hereafter called "wave fronts")
of the light in the beam drawn in the ether frame at an instant.
Their paths in the ether are marked with numbers.
\
1 \
1 2 \
1 2 3 \
1 2 3 4 \
1 2 3 4 5 \
1 2 3 4 5 6 \
-> v
Of course the wavefront elements are aligned along a vertical line,
and of course this line will stay vertical.
But every one of them is moving diagonally in the ether.
And every one of them is moving at c IN THE ETHER.
How the hell can you fail to understand something as obvious as this?
> The same applies to a boat crossing a flowing river whilst always pointing
> perpendicular to the bank. Each section of the boat moves along its own zero
> thickness diagonal at velocity sqrt(Vr^2+Vb^2) relative to the bank.
And you don't see why this is not analogous to the above?
If the water in the river is analogous to the ether, and a line
perpendicular to the river banks is analogous to the arm (moving at vr
relative TO THE WATER) then answer the following questions:
1. In what direction will the boat have to point TO FOLLOW THE LINE
PERPENDICULAR TO THE RIVER BANKS?
2. What will the speed of the boat be relative to the bank?
Think just a tiny bit, Henry.
You will see that the drawing I made above applies to this
case as well. Each boat in the "beam of boats" will have to point
in a direction perpendicular to the wavefronts in my drawing.
> You see that this velocity can range to infinity, for any constant forward boat
> speed.
This is to bloody stupid.
Why the hell don't you think a bit before writing?
> Why do you (and Michelson) think light should behave any differently from this?
It doesn't.
Of course photons (or whatever...) NOT MOVING ALONG THE ARM
can move at HIGHER speed than c RELATIVE TO THE ARM.
But photons (or whatever ..) going in other directions than along
the arm are not part of the beam along the arm.
How the hell can you miss that, Henry?
You are obviously very confused.
> >> It is the same as my laser beam example, above. The whole MMX cross beam remains
> >> vertical when plotted in all frames (those moving along the parallel axis, of
> >> course).
> >
> >Why do you keep repeating this obvious triviality?
>
> Why do you trying to infer that the most important things are trivial?
> The beam remains vertical an always traverses a fixed distance in that direction
> in the same time.
It is trivial and obvious that the arm and thus "the beam as a whole"
remains vertical.
It is plain wrong that each photon (or what the hell..) are moving
at c RELATIVE TO THE ARM, because it is by bloody definition in the
theory we are calculating the predictions of moving at c relative
TO THE ETHER.
How the hell is it possible to keep missing this?
> Michelson, you and all the others are wrong.
> >
> >> So all the past books about the MMX are wrong.
> >
> >Because they state that the path of each individual photon
> >(or what the hell etc.) which is moving along the MOVING transverse arm
> >blatantly obvious is diagonal in the ether frame?
> >Of course that is correct!
>
> It is NOT correct to assume that anything moves at VELOCITY C in that diagonal
> direction.
> Nobody has ever measured the one way velocity of light along its wave axis, let
> alone in a direction oblique to that!!! - or even the TWLS for that matter.
> So you and your boys have been living in fairyland.
SEE!!!!
> >> It can be construed that way. But the velocity is obviously not c!
> >> There is no principle or mechanism, maxwellian or otherwise which even suggests
> >> that it might be c.
> >> It is a mistake.
> >
> >Of bloody course the speed IN THE ETHER is c ACCORDING TO
> >Michelson's ether theory!
>
> Stuff Michelson - and Einstein, who said the same thing. They are clearly
> wrong, WRONG!
SEE!!!!
Why do you state a queue of stupidities like this Henry?
1. A "misinterpreted experiment" cannot falsify anything.
It is a fact that the MMX falsifies Michelson's ether theory because
Michelson interpreted it correctly!
2. Of bloody course the fact that the MMX falsifies Michelson's theory
doesn't "give credibility" to another theory.
It is the fact that the MMX FAILS to falsify SR that give
credibilty to SR.
> >> >THAT'S WHY THE MMX FALSIFIED MICHELSON ETHER THEORY!
> >>
> >> But it DIDN"T.
> >> The conclusion was that it DID falsify the theory - but the proof itself was
> >> based on equally false logic!
> >
> >I think I have had enough of this idiocy now, Henry.
>
> Where is the problem? I have made a strightforward statement of fact.
If you don't see the idiocy in insisting that an experiment fails
to falsify a theory BECAUSE THE THEORY IS WRONG,
I can't help you!
> >You are saying that Michelson's calculations of what his theory
> >predicted was wrong because the theory was wrong, and thus the fact
> >that the predictions are wrong does not falsify the theory!
>
> NO. I'm saying that the prediction of a non-null result was an incorrect
> conclusion derived from the misinterpretation of experimental analysis
> completely independent of his incorrect aether theory.
No, Henry.
Every time I wrote SEE!!! above, you used exactly that logic.
> >How the hell is it possible to fail to understand how utterly
> >ridiculous this is?
>
> Not at all.
> I have shown many times why the experiment SHOULD give a null result EVEN IN A
> MICHELSON AETHER.
No, Henry. You never have, because you over and over again
insist that the photons (or what the hell...) are moving at c
RELATIVE TO THE ARM which is moving at v in the ether,
thus insisting that the photons have the speed sqrt(c^2+v^2)
relative to the ether, which OBVIOUSLY is contrary to Michelsons
ether theory.
That is, you are using "ballistic light theory" to calculate
the predictions of Michelson's ether theory.
That is plain idiocy!
> >> >WE ARE DISCUSSING THE PREDICTIONS OF MICHELSON'S ETHER THEORY!
> >> >
> >> >You claim that the MMX do NOT falsify Michelson's ether
> >> >theory, because Michelson MISINTERPRETED it.
> >>
> >> Yes. I think I can agree about that..
> >> >
> >> >The only way he could have done so is by doing an error
> >> >in the calculation of what his theory predicts.
> >> >
> >> >So Henry, to defend your claim, you must show that
> >> >Michelson did an error when calculating the predictions
> >> >of MICHELSON'S ETHER THEORY!
> >>
> >> No I don't. No need. Michelson already did that. His own experimental evidence
> >> refuted the predictions of his own theory.
> >
> >See!!!!
> >
> >I must be talking to an idiot!
>
> OK I'll modify that. I was a bit hasty there.
> I have already shown that error.
No, you haven't.
You have OTOH repeated your idiotic claim over and over in this very posting.
> >There is no point in going on.
> >You have obviously no idea of basic logic.
> >I give you up!
> >
> >Paul
> you can't give up now Paul.
> You are on the verge of seeing the light. You know deep down that something is
> wrong with SR but your faith prevents you leaping over to the othe side.
This is irrelevant to SR.
We are still discussing what the predictions of Michelson's ether
theory are.
And if you still haven't realized you gigantic blunder when
figuring out what it is, you probably never will.
But now I am utterly fed up with repeating the same
basic, trivial, obvious arguments over and over again!
How the hell is it possible to screw up something as simple as this?
Paul
Luc Bourhis
> "Paul B. Andersen" wrote in message news
>> I must be talking to an idiot!
>> There is no point in going on. You have obviously no idea of basic
>> logic. I give you up!
>> Paul
> But it was a truly heroic attempt, Paul.
I'll second that ! And I suggest we bookmark this thread for immoderate
future uses ...
--
Luc J. Bourhis