| Topic: |
Science > Physics |
| User: |
"Patty" |
| Date: |
16 Aug 2005 05:20:34 PM |
| Object: |
HUP nuclear questions |
Hi,
It is said that the electron can't be confined in the
nucleus because the energy requirement would be so much...
around 3.77 GeV from the principle of HUP. But there is
such thing as electron capture where the proton in the
nuclues act with electron and gets transmuted to neutron.
So what's the exact explanation how the electron manage to
get inside the nucleus and bypass HUP??
Another. About QCD and fractional charge. Some argued
that fractional charges cannot exist in a spin 1/2 hbar
particle, regardless of its size, what do you think?
Patty
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| User: "" |
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| Title: Re: HUP nuclear questions |
16 Aug 2005 05:26:15 PM |
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I'm not sure, but what is sure is that all you'll get here is a bunch
of ridiculing, sarcastic answers which do nothing but to destroy the
normal dialectic of information exchange and learning. Quite a shame.
"ya cudda been a contender, sci.physics".
**************
No replies will be read due to the above . Yall have a nice one,
heah????
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| User: "" |
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| Title: Re: HUP nuclear questions |
16 Aug 2005 05:34:21 PM |
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In article <1124230834.221073.123340@g47g2000cwa.googlegroups.com>, "Patty" <partistuds@yahoo.com> writes:
Hi,
It is said that the electron can't be confined in the
nucleus because the energy requirement would be so much...
around 3.77 GeV from the principle of HUP. But there is
such thing as electron capture where the proton in the
nuclues act with electron and gets transmuted to neutron.
So what's the exact explanation how the electron manage to
get inside the nucleus and bypass HUP??
You've to distinguish between "confined in" and "has nonzero
probability to be in". Not the same thing.
An electron confined to some volume has in general a non-zero
probability to interact within any subvolume of said volume, no matter
how small.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "tj Frazir" |
|
| Title: Re: HUP nuclear questions |
16 Aug 2005 06:32:18 PM |
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If you can get the electron cold enouph its too small .
The electron and the neutron are going full speed and cant go any
faster.
The small mass of the electron is a range where it can exchange
frequecy with the ( exchange energy ) with the electron.
The photon is 1 mass the electron is 100 masses and the neutron is 2500
masses and the proton is 3500 masses.
The orbits will change with the mass .
No 2 parts of the atom will touch as the intence wave presure pushes
them apart as gravity and motion hold them in place.
Intence waves and CF pushing out wile gravity pushes in.
Mass is how much energy is where how long.
.
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| User: "Y.Porat" |
|
| Title: Re: HUP nuclear questions |
16 Aug 2005 11:38:41 PM |
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now you see to physicists that know mathematics
flat page mathematics and nothing more
what is left for them is to mumble.
it is not the picture that the nucleus is a small ball
and somewhere far away you have some electrons ruining around
IE
for some reason people cannot detach themselves from the 'solar system
of Bhore
that is not the picture
the nuc and the atom are some continuous entity *continuous* got it??!!
so there is matter all along between the nuc and the last electron
around it
it is a (just stick it to you new mind)
it is a chain of orbitals many orbitals all along
now the electrons are indeed not 'inside the nuc'
that is not their location
they are located at the *periphery* of the nuclear particles
you have the solid massive 'skeleton' of the nuc
then around it attached neutrons that sort of dress that skeleton
all around
and only there more distantly the electrons that are attached to the
so rounding neutrons (yes sometimes at the poles surrounding protons as
well)
and those electrons are extending out wards - not orbiting
*extending out wards!!
so there is no trouble of lack of 'living room ' for those electrons
and they are not 'squeezed' as you have in your false imagination
now since one picture is worth a thousand words
just have a look at my site.
and realize that there is some life in science without QM and even
without
mathematics.
physics thinking is before all that !!
ATB
Y.Porat
----------------------------
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| User: "Uncle Al" |
|
| Title: Re: HUP nuclear questions |
16 Aug 2005 06:49:25 PM |
|
|
Patty wrote:
Hi,
It is said that the electron can't be confined in the
nucleus because the energy requirement would be so much...
around 3.77 GeV from the principle of HUP. But there is
such thing as electron capture where the proton in the
nuclues act with electron and gets transmuted to neutron.
So what's the exact explanation how the electron manage to
get inside the nucleus and bypass HUP??
Another. About QCD and fractional charge. Some argued
that fractional charges cannot exist in a spin 1/2 hbar
particle, regardless of its size, what do you think?
s-orbitals have an anti-node at the nucleus. An electron need not be
confined for electron capture decay, merely overlap enough probability
to make it happen. Having beta-capture isotopes in extreme oxidation
states or encapsulated in C60 alters half-lives by a few percent
compared to the free element.
As for your second query, there are no quarks in protons and
neutrons? Where do the observed geometries of high energy deep
scattering orginate?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
.
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| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: HUP nuclear questions |
16 Aug 2005 07:56:06 PM |
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"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:43027B84.5CF4A188@hate.spam.net...
[snip crap]
Hey Moron!
Let's see if you know any high school algebra.
Given:
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
Need a reference, phuckwit? No need, you've quoted it yourself before,
stoooopid.
Doubling both sides:
tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,t+x'/(c-v))
Taking out the t for 3:00pm on a Friday afternoon:
tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
Synchronize clocks at t = 0, we remove tau(0,0,0,0)+
tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
Taking coordinate x' as infinitessimally small, as he says,
you not quite realizing x' is both a coordinate and a distance,
he does that to differentiate, so we leave the distance alone,
dx/dt = x/t anyway with a constant velocity.
tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))
Removing the superflous coordinates, all zero:
tau(x'/(c-v)+x'/(c+v)) = 2 * tau(x'/(c-v))
Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity
tau(a+b) = 2*tau(a)
Renaming tau as f,
f(a+b) = 2f(a) or
½f(a+b) = f(a)
Now tell me that's a linear function, a > b.
"In the first place it is clear that the equations must be linear
on account of the properties of homogeneity which we attribute to
space and time." -- Albert Phuckwit/Huckster Einstein.
In the second place tau is not a linear function. -- Androcles.
In the third place there are no coordinates to transform.
In the fourth place you've been had! (and not by me either)
Hahahahahahaha!!
Stoopid Schwartz is a phuckwit and she's been had!
Androcles.
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| User: "Uncle Al" |
|
| Title: Re: HUP nuclear questions |
16 Aug 2005 08:47:05 PM |
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Androcles wrote:
"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:43027B84.5CF4A188@hate.spam.net...
[snip crap]
Hahahahahahaha!!
Stoopid Schwartz is a phuckwit and she's been had!
Androcles.
Res ipsa loquitur. (Hey stooopid Androcles - you've lost the game
again. Get a Project Head Start slum bunny to explain it to you.
Maybe a Darkie can start you out with doing The Threes.)
s-orbitals have an anti-node at the nucleus. An electron need not be
confined for electron capture decay, merely overlap enough probability
to make it happen. Having beta-capture isotopes in extreme oxidation
states or encapsulated in C60 alters half-lives by a few percent
compared to the free element.
As for your second query, there are no quarks in protons and
neutrons? Where do the observed geometries of high energy deep
scattering orginate?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
.
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: HUP nuclear questions |
16 Aug 2005 10:18:44 PM |
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"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:43029719.F4735A85@hate.spam.net...
[snip crap]
Schwartz can't handle schoolgirl algebra and thinks she's won something.
You really are a buffoon, kraut *****.
Androcles
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| User: "Uncle Al" |
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| Title: Re: HUP nuclear questions |
17 Aug 2005 10:38:24 AM |
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Androcles wrote:
"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:43029719.F4735A85@hate.spam.net...
[snip crap]
Schwartz can't handle schoolgirl algebra and thinks she's won something.
You really are a buffoon, kraut *****.
Androcles
Res ipsa loquitur. (Hey stooopid Androcles - you've lost the game
again. Get a Project Head Start slum bunny to explain it to you.
Maybe a Darkie can start you out with doing The Threes.)
s-orbitals have an anti-node at the nucleus. An electron need not be
confined for electron capture decay, merely overlap enough probability
to make it happen. Having beta-capture isotopes in extreme oxidation
states or encapsulated in C60 alters half-lives by a few percent
compared to the free element.
As for your second query, there are no quarks in protons and
neutrons? Where do the observed geometries of high energy deep
scattering orginate?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
.
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| User: "" |
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| Title: Re: HUP nuclear questions |
16 Aug 2005 10:24:16 PM |
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Hey! If I never read what you guys have to say then, by the Copenhagen
Interpretation, you guys don't exist!!!!!!
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| User: "The Ghost In The Machine" |
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| Title: Re: HUP nuclear questions |
16 Aug 2005 11:52:05 PM |
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In sci.physics, Androcles
<Androcles@MyPlace.org>
wrote
on Wed, 17 Aug 2005 00:56:06 GMT
<GWvMe.7884$Wq4.6545@fe1.news.blueyonder.co.uk>:
"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:43027B84.5CF4A188@hate.spam.net...
[snip crap]
Hey Moron!
Let's see if you know any high school algebra.
Given:
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
Need a reference, phuckwit? No need, you've quoted it yourself before,
stoooopid.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
is as good as any. Section 3. Unfortunately the equations
are unnumbered. Note that x' is defined to be x-vt (and
is also *re*defined later on in the section), and that
tau is a function. It is not clear within this section
whether that function is in fact linear with respect to
its arguments.
Doubling both sides:
tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,t+x'/(c-v))
OK.
Taking out the t for 3:00pm on a Friday afternoon:
tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
Careful.
Since you've yet to prove tau(a+b) = tau(a) + tau(b), this
operation cannot be done without very careful considerations.
However, it turns out one can in fact make that assumption.
Synchronize clocks at t = 0, we remove tau(0,0,0,0)+
tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
Coordinate system redefinition. OK.
Taking coordinate x' as infinitessimally small, as he says,
you not quite realizing x' is both a coordinate and a distance,
he does that to differentiate, so we leave the distance alone,
dx/dt = x/t anyway with a constant velocity.
tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))
Hm. You are now on record as stating x-vt is "infinitessimally small".
This also means that
tau(0,0,0,x'/(c-v)+x'/(c+v)) = tau(0,0,0,0) = 0 unless v is *very*
close to c, which leads to some interesting issues.
Removing the superflous coordinates, all zero:
tau(x'/(c-v)+x'/(c+v)) = 2 * tau(x'/(c-v))
Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity
tau(a+b) = 2*tau(a)
Renaming tau as f,
f(a+b) = 2f(a) or
½f(a+b) = f(a)
Now tell me that's a linear function, a > b.
It is. You've stopped time. You've also made the interesting
substitution x' = 0 and for some reason totally neglected
to simply state that tau(0,0,0,0) = 2 * tau(0,0,0,0) = 0,
and that a and b are both zero (within a certain tolerance).
If a=0, b=0, f(0) = 0, then (1/2)f(a+b) = f(a).
I'm not certain about Einstein's methods but yours need work.
"In the first place it is clear that the equations must be linear
on account of the properties of homogeneity which we attribute to
space and time." -- Albert Phuckwit/Huckster Einstein.
In the second place tau is not a linear function. -- Androcles.
In the third place there are no coordinates to transform.
In the fourth place you've been had! (and not by me either)
Hahahahahahaha!!
Stoopid Schwartz is a phuckwit and she's been had!
I for one think Uncle Al is smarter than that. :-)
Androcles.
--
#191,
It's still legal to go .sigless.
.
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| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: HUP nuclear questions |
18 Aug 2005 05:11:57 PM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message news:1mpat2-urq.ln1@sirius.tg00suus7038.net...
| In sci.physics, Androcles
| <Androcles@MyPlace.org>
| wrote
| on Wed, 17 Aug 2005 00:56:06 GMT
| <GWvMe.7884$Wq4.6545@fe1.news.blueyonder.co.uk>:
| >
| > "Uncle Al" <UncleAl0@hate.spam.net> wrote in message
| > news:43027B84.5CF4A188@hate.spam.net...
| >
| > [snip crap]
| >
| > Hey Moron!
| >
| > Let's see if you know any high school algebra.
| >
| > Given:
| > ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] =
tau(x',0,0,t+x'/(c-v))
| >
| > Need a reference, phuckwit? No need, you've quoted it yourself
before,
| > stoooopid.
|
| http://www.fourmilab.ch/etexts/einstein/specrel/www/
|
| is as good as any. Section 3. Unfortunately the equations
| are unnumbered. Note that x' is defined to be x-vt (and
| is also *re*defined later on in the section), and that
| tau is a function. It is not clear within this section
| whether that function is in fact linear with respect to
| its arguments.
|
| >
| > Doubling both sides:
| > tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 *
tau(x',0,0,t+x'/(c-v))
|
| OK.
|
| >
| > Taking out the t for 3:00pm on a Friday afternoon:
| >
| > tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
|
| Careful.
|
| Since you've yet to prove tau(a+b) = tau(a) + tau(b), this
| operation cannot be done without very careful considerations.
|
| However, it turns out one can in fact make that assumption.
It's the same value for emission, reception and reflection,
a constant offset from an arbitrary zero, and can be removed carelessly
and without further consideration.
It's intent was to obfuscate and it makes no further appearance in the
derivation of the cuckoo transformations.
I've no intention of proving tau(a+b) = tau(a) + tau(b).
|
| >
| > Synchronize clocks at t = 0, we remove tau(0,0,0,0)+
| >
| > tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))
|
| Coordinate system redefinition. OK.
Synchronization definition, tau(0) = 0.
|
| >
| > Taking coordinate x' as infinitessimally small, as he says,
| > you not quite realizing x' is both a coordinate and a distance,
| > he does that to differentiate, so we leave the distance alone,
| > dx/dt = x/t anyway with a constant velocity.
| >
| > tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))
|
| Hm. You are now on record as stating x-vt is "infinitessimally
small".
"Hence, if x' be chosen infinitesimally small," --- Einstein.
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Now I'm on record for quoting Einstein, as I always was,
"Taking coordinate x' as infinitessimally small, as he says" --Androcles
| This also means that
|
| tau(0,0,0,x'/(c-v)+x'/(c+v)) = tau(0,0,0,0) = 0 unless v is *very*
| close to c, which leads to some interesting issues.
Not at all. The mirror can be a light-year away, v can be zero,
and it will take 2 years for the round trip of the light.
tau(0,0,0,1+1) = tau(0,0,0,0) (your equation) is incorrect.
tau(0,0,0,1+1) = 2 * tau(0,0,0,1) (my equation) is correct.
There are no interesting issues here.
|
| >
| > Removing the superflous coordinates, all zero:
| >
| > tau(x'/(c-v)+x'/(c+v)) = 2 * tau(x'/(c-v))
| >
| > Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity
| >
| > tau(a+b) = 2*tau(a)
| >
| > Renaming tau as f,
| >
| > f(a+b) = 2f(a) or
| >
| > ½f(a+b) = f(a)
| >
| > Now tell me that's a linear function, a > b.
|
| It is.
I specifically stated a > b.
½f(1+0) = f(1), a = 1 > b = 0
½f(2+1) = f(2), a = 3 > b = 1
Now tell me that's a linear function, a > b.
| You've stopped time.
Not at all. The mirror can be a light-year away, v can be zero,
and it will take 2 years for the round trip of the light.
tau(0,0,0,1+1) = tau(0,0,0,0) (your equation) is incorrect.
tau(0,0,0,1+1) = 2 * tau(0,0,0,1) (my equation) is correct.
YOU (or Einstein) stopped time, not I.
| You've also made the interesting
| substitution x' = 0 and for some reason totally neglected
| to simply state that tau(0,0,0,0) = 2 * tau(0,0,0,0) = 0,
| and that a and b are both zero (within a certain tolerance).
Yes, but not a totally neglected reason. I left the distance x'
intact, and I can reverse another ruler so that 12 (inches) is beside 0.
That cannot change the physics, but it does change the coordinate
system.
Reversed ruler.
½[tau(x',0,0,0)+tau(x',0,0,b+a] = tau(0,0,0,b)
Note the 'b' on the RHS.
| If a=0, b=0, f(0) = 0, then (1/2)f(a+b) = f(a).
|
| I'm not certain about Einstein's methods but yours need work.
I cannot anticipate your objections or anybody else's, but I can
answer them.
Phuckwit Auntie Alice the bigot isn't interested in physics, she's
interested in promoting Phuckwit Alice the bigot, much as
Albert Phuckwit Einstein the huckster was interested in promoting
Albert Phuckwit Einstein the huckster.
|
| >
| > "In the first place it is clear that the equations must be linear
| > on account of the properties of homogeneity which we attribute to
| > space and time." -- Albert Phuckwit/Huckster Einstein.
| >
| > In the second place tau is not a linear function. -- Androcles.
| >
| > In the third place there are no coordinates to transform.
| >
| > In the fourth place you've been had! (and not by me either)
| >
| > Hahahahahahaha!!
| >
| > Stoopid Schwartz is a phuckwit and she's been had!
|
| I for one think Uncle Al is smarter than that. :-)
|
| >
| > Androcles.
| >
|
|
| --
| #191,
| It's still legal to go .sigless.
.
|
|
|
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|
| User: "Sam Wormley" |
|
| Title: Re: HUP nuclear questions |
16 Aug 2005 05:43:51 PM |
|
|
Patty wrote:
Hi,
It is said that the electron can't be confined in the
nucleus because the energy requirement would be so much...
around 3.77 GeV from the principle of HUP. But there is
such thing as electron capture where the proton in the
nuclues act with electron and gets transmuted to neutron.
So what's the exact explanation how the electron manage to
get inside the nucleus and bypass HUP??
There is a non zero probability that an electron will be
in the nucleus.
The Probability Distributions for the Hydrogen Atom
http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html
.
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|
| User: "Bjoern Feuerbacher" |
|
| Title: Re: HUP nuclear questions |
17 Aug 2005 03:50:15 AM |
|
|
Patty wrote:
Hi,
It is said that the electron can't be confined in the
nucleus because the energy requirement would be so much...
around 3.77 GeV from the principle of HUP. But there is
such thing as electron capture where the proton in the
nuclues act with electron and gets transmuted to neutron.
So what's the exact explanation how the electron manage to
get inside the nucleus and bypass HUP??
Quite simple: the HUP argument is about an electron which would
be *confined to* the nucleus, i.e. bound in it. OTOH, in these
reactions the electron only "goes into" the nucleus for a short
time and then "reacts" with the proton. Afterwards, there is no
electron left in the nucleus, so no problem with the HUP.
Another. About QCD and fractional charge. Some argued
that fractional charges cannot exist in a spin 1/2 hbar
particle, regardless of its size, what do you think?
Who argues that?
Bye,
Bjoern
.
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