| Topic: |
Science > Physics |
| User: |
"David" |
| Date: |
16 Aug 2005 07:12:41 AM |
| Object: |
Nuclear Binding Energy Calculations |
Hi,
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Is it accurate? What's the formula.
David
.
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|
| User: "Uncle Al" |
|
| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 11:30:41 AM |
|
|
David wrote:
Hi,
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Is it accurate? What's the formula.
http://t2.lanl.gov/data/astro/molnix96/molnix.html
If you are not smart enough to find the citation, what makes you think
you are smart enough to understand it?
--
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: Nuclear Binding Energy Calculations |
16 Aug 2005 07:57:36 PM |
|
|
"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:430214B1.D92377A1@hate.spam.net...
[snip crap]
Hey Bozo!
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: "RP" |
|
| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 11:54:18 PM |
|
|
Androcles wrote:
"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:430214B1.D92377A1@hate.spam.net...
[snip crap]
Hey Bozo!
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.
You've set a=b with your conditions.
It's you who doesn't understand algebra. You can't set a=b, and then
insist that a>b. Your error has been discussed many times, but you still
don't get it.
Richard Perry
"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: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: Nuclear Binding Energy Calculations |
19 Aug 2005 05:22:27 PM |
|
|
"RP" <no_mail_no_spam@yahoo.com> wrote in message
news:Ro-dnRpoffF6X5_eRVn-pw@centurytel.net...
|
|
| Androcles wrote:
|
| > "Uncle Al" <UncleAl0@hate.spam.net> wrote in message
| > news:430214B1.D92377A1@hate.spam.net...
| > [snip crap]
| >
| >
| > Hey Bozo!
| >
| > 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.
|
| You've set a=b with your conditions.
| It's you who doesn't understand algebra. You can't set a=b, and then
| insist that a>b. Your error has been discussed many times, but you
still
| don't get it.
|
| Richard Perry
Assertion carries no weight. Claiming prior disproof carries no weight.
Show the flaw you claim is there.
Androcles.
|
| > "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: "RP" |
|
| Title: Re: Nuclear Binding Energy Calculations |
22 Aug 2005 04:08:01 AM |
|
|
Androcles wrote:
"RP" <no_mail_no_spam@yahoo.com> wrote in message
news:Ro-dnRpoffF6X5_eRVn-pw@centurytel.net...
|
|
| Androcles wrote:
|
| > "Uncle Al" <UncleAl0@hate.spam.net> wrote in message
| > news:430214B1.D92377A1@hate.spam.net...
| > [snip crap]
| >
| >
| > Hey Bozo!
| >
| > 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.
|
| You've set a=b with your conditions.
| It's you who doesn't understand algebra. You can't set a=b, and then
| insist that a>b. Your error has been discussed many times, but you
still
| don't get it.
|
| Richard Perry
Assertion carries no weight. Claiming prior disproof carries no weight.
Show the flaw you claim is there.
½f(a+b) = f(a)
½(a+b) = (a)
a+b = 2(a)
a+b = a+a
b = a
Richard Perry
Androcles.
|
| > "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.
| >
|
.
|
|
|
| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: Nuclear Binding Energy Calculations |
22 Aug 2005 02:48:26 PM |
|
|
"RP" <no_mail_no_spam@yahoo.com> wrote in message
news:RMGdnTMIac1HCJTeRVn-vg@centurytel.net...
|
|
| Androcles wrote:
| > | > 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.
| > |
| > | You've set a=b with your conditions.
| > | It's you who doesn't understand algebra. You can't set a=b, and
then
| > | insist that a>b. Your error has been discussed many times, but you
| > still
| > | don't get it.
| > |
| > | Richard Perry
| >
| > Assertion carries no weight. Claiming prior disproof carries no
weight.
| > Show the flaw you claim is there.
|
| ½f(a+b) = f(a)
| ½(a+b) = (a)
| a+b = 2(a)
| a+b = a+a
| b = a
cos(0) = 1
cos(pi/2) = 0
cos(pi) = -1
cos(3*pi/2) = 0
cos(2*pi) = 1
½ cos(a+b) = cos(a)
a = pi/2
b = 0, a > b.
½ cos(pi/2) = cos(pi/2)
0/2 = 0
Setting a = b as you claim,
a = pi/2, b = pi/2.
½ cos(pi/2 + pi/2 ) = cos(pi/2)
½ cos(pi) = cos(pi/2)
-0.5 = 0 ????
It's me who doesn't understand algebra?
Nope, it's you that doesn't know what a function is,
is a phuckwit, not worth wasting time on.
*plonk*
Androcles.
.
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| User: "RP" |
|
| Title: Re: Nuclear Binding Energy Calculations |
22 Aug 2005 02:26:49 PM |
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|
Androcles wrote:
"RP" <no_mail_no_spam@yahoo.com> wrote in message
news:RMGdnTMIac1HCJTeRVn-vg@centurytel.net...
|
|
| Androcles wrote:
| > | > 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.
| > |
| > | You've set a=b with your conditions.
| > | It's you who doesn't understand algebra. You can't set a=b, and
then
| > | insist that a>b. Your error has been discussed many times, but you
| > still
| > | don't get it.
| > |
| > | Richard Perry
| >
| > Assertion carries no weight. Claiming prior disproof carries no
weight.
| > Show the flaw you claim is there.
|
| ½f(a+b) = f(a)
| ½(a+b) = (a)
| a+b = 2(a)
| a+b = a+a
| b = a
cos(0) = 1
cos(pi/2) = 0
cos(pi) = -1
cos(3*pi/2) = 0
cos(2*pi) = 1
½ cos(a+b) = cos(a)
a = pi/2
b = 0, a > b.
½ cos(pi/2) = cos(pi/2)
0/2 = 0
Setting a = b as you claim,
a = pi/2, b = pi/2.
½ cos(pi/2 + pi/2 ) = cos(pi/2)
½ cos(pi) = cos(pi/2)
-0.5 = 0 ????
It's me who doesn't understand algebra?
Yep.
Richard Perry
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| User: "Sam Wormley" |
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| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 07:46:05 AM |
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|
David wrote:
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Nuclear Binding Energy
http://www.chem.purdue.edu/gchelp/howtosolveit/Nuclear/nuclear_binding_energy.htm
The difference between the mass of a nucleus and the sum of the
masses of the nucleons of which it is composed is called the
mass defect. Three things need to be known in order to calculate
the mass defect:
o the actual mass of the nucleus,
o the composition of the nucleus (number of protons and of neutrons),
o the masses of a proton and of a neutron.
.
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| User: "David" |
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| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 07:51:43 AM |
|
|
Sam Wormley wrote:
David wrote:
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Nuclear Binding Energy
http://www.chem.purdue.edu/gchelp/howtosolveit/Nuclear/nuclear_binding_energy.htm
The difference between the mass of a nucleus and the sum of the
masses of the nucleons of which it is composed is called the
mass defect. Three things need to be known in order to calculate
the mass defect:
o the actual mass of the nucleus,
o the composition of the nucleus (number of protons and of neutrons),
o the masses of a proton and of a neutron.
Is there no exception??
Can the above calculate all isotopes???
How come some have to spend many years finding a way to modify
Standard Model so that they can calculate it such as Tom Lockyer
who spent over 30 years to try to figure out an equation to get
the nuclear binding energies of each element and isotope??
David
.
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| User: "Michael Moroney" |
|
| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 11:08:33 AM |
|
|
"David" <davidqanta@yahoo.com> writes:
Sam Wormley wrote:
David wrote:
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Nuclear Binding Energy
http://www.chem.purdue.edu/gchelp/howtosolveit/Nuclear/nuclear_binding_energy.htm
The difference between the mass of a nucleus and the sum of the
masses of the nucleons of which it is composed is called the
mass defect. Three things need to be known in order to calculate
the mass defect:
o the actual mass of the nucleus,
o the composition of the nucleus (number of protons and of neutrons),
o the masses of a proton and of a neutron.
Is there no exception??
Can the above calculate all isotopes???
How come some have to spend many years finding a way to modify
Standard Model so that they can calculate it such as Tom Lockyer
who spent over 30 years to try to figure out an equation to get
the nuclear binding energies of each element and isotope??
I know of no exact calculations, but the Weizsaecker Formula which is
based on the Liquid Drop Model of the nucleus is pretty good. It
has several terms, the first is proportional to A, the nucleon number,
a second "surface" term proportional to A^(2/3) is subtracted, a third
coulombic repulsion term proportional to Z^2/(A^(1/3)) is subtracted,
there's another proportional to the difference from the same number of
protons and neutrons (A-2Z) and another term that shows the effect of
nucleon pairing that's positive for even-even nuclei and negative for
odd-odd nuclei. It's not accurate for small A though.
.
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| User: "Y.Porat" |
|
| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 12:14:38 PM |
|
|
Moroney
since you have a copy of my book
you can understand what is the difference between calculating the
overall
binding energy of all sub particles
and..
pin pointing the binding energy at any connection point of Any proton
and neutron
of the nuc.
moreover
i addition to describe or map the exact location of any proton or
neutron
yet i am not sure if you were able to decode my planar schematic
descriptions of the heavy nucs
because it is quit impossible to show it is 3d models for all of them
but had you some of that 3d training you could do it
not to mention the chemical verifications
the nuclear facts that are unprecedentedly explained etc .
so may be just tell them about how far i went compared to others
about the geometric structure of the nuc
i had no pretensions about say spectrum aspects because i never
invested on it.
or may be other aspects
but what did about the geometric detailed structure is unprecedented.
ATB
Y.Porat
----------------------
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Nuclear Binding Energy Calculations |
17 Aug 2005 04:01:52 AM |
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David wrote:
[snip]
How come some have to spend many years finding a way to modify
Standard Model so that they can calculate it such as Tom Lockyer
who spent over 30 years to try to figure out an equation to get
the nuclear binding energies of each element and isotope??
It's news to me that Tom Lockyer has one single equation. Judging from
what I've seen from him so far, he fumbles around quite a bit in order
to arrive at the numbers he wants to have.
Bye,
Bjoern
.
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| User: "" |
|
| Title: Re: Nuclear Binding Energy Calculations |
17 Aug 2005 11:52:16 AM |
|
|
Bjoern Feuerbacher wrote:
David wrote:
[snip]
How come some have to spend many years finding a way to modify
Standard Model so that they can calculate it such as Tom Lockyer
who spent over 30 years to try to figure out an equation to get
the nuclear binding energies of each element and isotope??
It's news to me that Tom Lockyer has one single equation. Judging from
what I've seen from him so far, he fumbles around quite a bit in order
to arrive at the numbers he wants to have.
Bye,
Bjoern
Bjoern, you fellows refuse to get my book. I use the near field
magnetic moments to calculate the binding energy.
I find there are groups of nucleons that are common to all isotopes,
and that the binding energy is saturated, that is, the nucleons only
"feel" the fields of adjacent nucleons.
This greatly simplified the calculation because one can pre-calculate
group bindings and sum to the total nucleus binding. The only ones
needed seem to be: (p to n), (n to n), (p to p), (p+n) n, (p+n) p,
(p-n) n, (p-n) p, (n+n) p, and (p-n)(p-n).
There is one more caveat, your arrangement of the groupings must get
the experimentally known spin angular momentum of the isotope you
model.
The QVPP method works good enough to tell which nuclei are stable or
unstable against decay.
Here is a page out of the book showing stable Boron11.
http://members.aol.com/thomasl283/Page72.gif
David, the Wormsley reference requires empirical measurement of masses
to get binding energy, and the old liquid drop model mistakenly tried
to make "one equation" fits all. Neither method can account for the
subtle arrangement of nucleons in nuclei.
Only the methods, adopted by QVPP, are the correct approach.
Regards: Tom;
www.amazon.com 0963154664
.
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: Nuclear Binding Energy Calculations |
18 Aug 2005 04:57:00 AM |
|
|
wrote:
Bjoern Feuerbacher wrote:
David wrote:
[snip]
How come some have to spend many years finding a way to modify
Standard Model so that they can calculate it such as Tom Lockyer
who spent over 30 years to try to figure out an equation to get
the nuclear binding energies of each element and isotope??
It's news to me that Tom Lockyer has one single equation. Judging from
what I've seen from him so far, he fumbles around quite a bit in order
to arrive at the numbers he wants to have.
Bye,
Bjoern
Bjoern, you fellows refuse to get my book.
Well, I saw the page from your book you cite again below, and from
that it was quite clear that you have *not* one single equation. But
indeed fumble around quite a bit.
[snip rant]
Here is a page out of the book showing stable Boron11.
http://members.aol.com/thomasl283/Page72.gif
David, the Wormsley reference requires empirical measurement of masses
to get binding energy, and the old liquid drop model mistakenly tried
to make "one equation" fits all.
Err, the model works, so why do you call it "mistaken"?
[snip]
Bye,
Bjoern
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 11:11:39 AM |
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see my reply to David
just to know the overall binding energy does not make you much wiser
you have to know the exact structure and exact location of each
proton and neutron and the binding energy at any connection point
it seems that what i did is so fantastic far away from others
that people even do not start to realize and understand what they see
in my model!1
iow
it is even beyond to their Deming ability.
oh yes
indeed
for the heavy nuclei it is so 3d complicated that i had to invent
a planar abbreviation (or abstraction )graphic system
so it looks as hieroglyphs
but i have in my book the ' Shampoleon stone' IE a key
for translating it to the 3d structure
a bit of it is shown more tangibly at the iron description
IE you can see there how the 3d tangible structure is 'translated' to
the
more abstract schematic graphic code
just have a look at the iron description on two pages.(in two ways of
presentation)
(that is of course not for idiots like David.. but a bit more serious
and intelligent people than him)
ATB
Y.Porat
--------------
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 10:46:00 AM |
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David wrote:
Hi,
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Is it accurate? What's the formula.
The Bethe-Weizsaecker formula gives a good approximation. If you want
to achieve better accurary, you have to do ab initio calculations
using model potentials, or equivalent approaches; there is no single
formula which could deal with such complex systems as nuclei.
Bye,
Bjoern
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
16 Aug 2005 12:25:23 PM |
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on this for a change i agree with Feuerbacher
(he has my book so he knows about what i am talking
provided he was able to understand the 3d schemes of the heaviest ones)
now the drop model is dead long ago
and anyone who has my book realizes that it is impossible to
formulate the binding energies just by one formula!!
and again for me the binding energies is not the overall sum of it
that is very primitive knowledge
for me it is at any point of connection (and there are a lot there
my luck was that ... there is a lot of repetitions on those bindings
and surprisingly enough they are *constant* (many types but those
types are constant)
no matter if the nuc is small or big!!
(as one finds in electrons binding energies !!)
(so dont tell me who is expert No 1 for binding energies of the nuc
and a lot more than just binding energies. (:-))
Y.Porat
----------------------------
as i said above the reasons.
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| User: "tj Frazir" |
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| Title: Re: Nuclear Binding Energy Calculations MORON |
16 Aug 2005 10:35:01 PM |
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what a bunch of *****.
Mass is equal its binding energy.
If you want to call it binding energy.
MASS GRAVITY .
as intence wave presure and motion push out ,
gravity pushes in.
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations MORON |
16 Aug 2005 11:03:58 PM |
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mass is Energy divided by C^2
thats all
because E=mC^2
the rest is your *****
no one will agree with you.
just stop drinking!!!....(or drugging ??)
ATB
Y.Porat
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| User: "Chris" |
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| Title: Re: Nuclear Binding Energy Calculations |
17 Aug 2005 05:31:46 PM |
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No it is measured from the mass of the nucleus built then subtract the mass
of all the separate componernts (protons + neutrons) multiply this by the
velocity of light squared and that is the binding energy.
There is no way (yet) of working it out as the way the nucleons bind is not
really known as you cannot see them and so the wavefunctions cannot be
calculated. In an hydrogen molecule ion a quantum mechanical calculation
based on the properties of electrons and assuming things the binding energy
(chemical) holding the two protons and one electron together can be done by
a big computer over hours and hours.
All these calculations are based on measurements of simple systems and
modelling for more complex systems.
There are no absolutes in science, you have to start somewhere with
measurements, you cannot dream up the real universe it is just there for you
to play with.
Chris.
"David" <davidqanta@yahoo.com> wrote in message
news:1124194361.311072.36300@z14g2000cwz.googlegroups.com...
Hi,
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Is it accurate? What's the formula.
David
.
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
18 Aug 2005 12:07:47 AM |
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Chris (no-spam) wrote:
No it is measured from the mass of the nucleus built then subtract the mass
of all the separate componernts (protons + neutrons) multiply this by the
velocity of light squared and that is the binding energy.
There is no way (yet) of working it out as the way the nucleons bind is not
really known as you cannot see them and so the wave functions cannot be
calculated.
some shocking news for you
it is news 12 years old;
in my model (book) i decoded the binding energy
of any nucleid to its neighbors
listen carefully- all along the periodic table
and a lot of other knowledge alike
indeed i did it without a bit of wave functions!!
no need of malfunctions for it
just a need of some gray stuff above your shoulders
and alot of work and not least persistence!!
you can see just some hints for it(if you are sharp eyed-
in my site:
http://www.geocities.com/potat_y/mypage.html
(quote from memory i hope i didnt mistyped it..
ATB
Y.Porat
----------------------
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
18 Aug 2005 12:20:33 AM |
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sorry
it seems that i somehow mistyped it
solet me try again
it is without the http..
www.geocities.com/porat_y/mypage.html
hope now it will work
ATB
Y.Porat
----------------
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| User: "" |
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| Title: Re: Nuclear Binding Energy Calculations |
18 Aug 2005 12:56:33 PM |
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Chris (no-spam) wrote:
No it is measured from the mass of the nucleus built then subtract the mass
of all the separate componernts (protons + neutrons) multiply this by the
velocity of light squared and that is the binding energy.
Yes, Chris, that was my point. With QVPP methods, one can calculate
binding energy between individual nucleons.
I find that the strong force is electromagnetic, and that the near
field magnetic moments of the nucleons far exceed the enormous near
field electric forces.
Here is the equation for the deuteron which is simply a proton bound to
a neutron.
I show two calculations, one with the magnetic moments from the
measured binding energy of 2.224573 E6 in equivalent volts, and the
second equation shows the value I get using the original QVPP model
predicted magnetic moment values.
http://members.aol.com/thomasl283/QVPdeuteronbind.jpg
The close agreement between measured (Upmx and Upnmx) and QVPP magnetic
moments (Upm and Unm) was independent proof that the QVPP proton and
neutron models were correct.
( I had the QVPP magnetic moments, from the scaling of the proton
model, for 14 years prior to discovering the way to calculate binding
energy between nucleons.)
I was pleased to see that the measured value gets the same Bpn value
to within about 1 percent.
Notice the error would have required the measured value of binding to
be better than 13.6ppm. So, QVPP may very well be showing us the
correct magnetic moments in Upm and Unm. In fact, in the book I use
QVPP predicted nucleon magnetic moments for binding energy calculations
of complex nuclei, exclusively, with good results.
There is no way (yet) of working it out as the way the nucleons bind is not
really known as you cannot see them and so the wavefunctions cannot be
calculated.
Yes there is, see above reference and this example from the book;
http://members.aol.com/tnlockyer/Page69.gif
The equations and their values are fully shown in the QVPP book.
In an hydrogen molecule ion a quantum mechanical calculation
based on the properties of electrons and assuming things the binding energy
(chemical) holding the two protons and one electron together can be done by
a big computer over hours and hours.
Its not that complicated, Chris. Here is the binding calculations for
the electron to proton using the QVPP models for the electron and
proton.
What Bohr did not know was the binding energy of the electron is a
small mass defect that requires using both the electric moments and the
magnetic moments of the proton and electron in the calculations, as
shown;
http://www.members.aol.com/thomasl283/hydrogen.gif
There are no absolutes in science, you have to start somewhere with
measurements, you cannot dream up the real universe it is just there for you
to play with.
Exactly, that is my quarrel with the QED and QCD theories. They
covered their ignorance by inventing unprecedented things happening in
a dream universe of particles popping in and out of the vacuum and
interfering with their 'point" electron, fractional charges, gluons,
color charges etc. etc.
Do you know that there are (otherwise intelligent) people who actually
believe those fairy tales?
QVPP uses just what the natural particles have, a natural charge of
(+-1e) and the magnetic moments necessary to create the EM binding
energy photon mass defect.
"David" <davidqanta@yahoo.com> wrote in message
news:1124194361.311072.36300@z14g2000cwz.googlegroups.com...
Hi,
Is it possible to calculate the Nuclear Binding Energies
of any element, any isotopes using some kind of formula?
Is it accurate? What's the formula.
Yes, David, see above references and QVPP book for formulae.
Regards: Tom;
www.amazon.com 0963154664
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
19 Aug 2005 01:40:15 AM |
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wrote:
Chris (no-spam) wrote:
No it is measured from the mass of the nucleus built then subtract the mass
of all the separate componernts (protons + neutrons) multiply this by the
velocity of light squared and that is the binding energy.
Yes, Chris, that was my point. With QVPP methods, one can calculate
binding energy between individual nucleons.
----------
OK if your model is so powerful
why dont you go on and solve
all the nuclei of all the periodic table??!!
just go on and do what i did
i solved it all along from mass calculation and binding energy
showing in all nuclei (repeat all of them) the individual
binding energy between any two adjacent nucleids
and just between us so that no one will listen
you have no chance without some new insights that you miss.
unless you will borrow it from me
btw
your 'cube' model how old is it ??
ATB
Y.Porat
---------------------
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| User: "" |
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| Title: Re: Nuclear Binding Energy Calculations |
20 Aug 2005 11:31:18 AM |
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Y.Porat wrote:
tnlockyer@aol.com wrote:
Chris (no-spam) wrote:
snip<
Yes, Chris, that was my point. With QVPP methods, one can calculate
binding energy between individual nucleons.
Y. Porat said:
----------
OK if your model is so powerful
why dont you go on and solve
all the nuclei of all the periodic table??!!
just go on and do what i did
i solved it all along from mass calculation and binding energy
showing in all nuclei (repeat all of them) the individual
binding energy between any two adjacent nucleids
My dear Mr. Porat, are you aware that there are some 105 elements with
2600 known isotopes?
And of the 2600 there are 260 stable isotopes.
Of those 260 stable isotopes, 159 have a spin of zero and 101 stable
isotopes have spin between (1/2) and (9/2)?
Do you still claim your models show the binding energy for all those
isotopes?
Do your models show which isotopes are stable?
Do your models show why some isotopes are unstable, and their decay
particles?
Do your models show the known spin angular momentum of the various
isotopes?
The QVPP methods have been used to show those characteristics of
isotopes up to Sulfur 32.
It gets to be a complicated many body problem as the numbers of
nucleons increase. I am an old man, so it will be left to others to
extend the QVPP methods.
and just between us so that no one will listen
you have no chance without some new insights that you miss.
unless you will borrow it from me
btw
your 'cube' model how old is it ??
I have been working on it as a hobby since 1977.
Unfortunately everyone in 1977 had just been committed to the (choke,
gasp) quark model, so VPP was (is) as popular as a skunk at a picnic.
It has just been within the last 6 years that the model has been
extended to nuclei structures.
The quark model has been left behind in the dust, by QVPP.
Regards: Tom:
www.amazon.com 0963154664
.
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
21 Aug 2005 05:11:41 AM |
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wrote:
Y.Porat wrote:
wrote:
Chris (no-spam) wrote:
snip<
Yes, Chris, that was my point. With QVPP methods, one can calculate
binding energy between individual nucleons.
Y. Porat said:
showing in all nuclei (repeat all of them) the individual
binding energy between any two adjacent nucleids
My dear Mr. Porat, are you aware that there are some 105 elements with
2600 known isotopes?
yes so ???
----------
And of the 2600 there are 260 stable isotopes.
Of those 260 stable isotopes, 159 have a spin of zero and 101 stable
isotopes have spin between (1/2) and (9/2)?
Do you still claim your models show the binding energy for all those
isotopes?
not all isotopes
*no need to solve all isotopes solved all the 92
elements
with many of the isotopes
now listen carefully
once you solve all the 92 with many of there isotopes
you cant do it without understanding the common features
of all of them
one you do it it means you have the right *tools* in
your hands
and then it i no problem to decode any isotope
i never though it is needed
because i realized that i am able to do it
for (to be cautious) for most of them.
anyway i never met an isotope or element
that i wanted to decode and got stuck!!
because i have all the tools in my hands.
that is for the mass calculation
chemical verification
beta emitters
the lightest and heaviest possible isotope
nuclear process explanation
unprecedented explanations for instance about the
geologic clock family connections
that t explain why some process are only within some elements
and not outside those elements
for instance why is it that there is transformations
from one element to another only say in the
S Si P elements and not to others that are not in that
family etc etc .
why is it that Pt cannot be transfered to Gold
and many others
IE i have it as if on the palm of my hands- tangible
as you can see only a bit of it in my site.
------------
Do your models show which isotopes are stable?
yes for instance
all the beta emitters are of neutrons that are
connected to the pole particles of he rectangular pipe
you have there a proton or neutron and
another neutron is connected to it
in a very unstable connection
it is the *mechanical instability!!
IE a connection only at one point that is a hinge
all the other stable connections are
*with more than one connection per one sub particle!!
your model have no idea about that crucial aspect
on the other hand
the electron capture occurs mainly
not on the poles of the nuc but around the main body
of it on the neutrons that surround the main skeleton.
and i could go on and on with it
--------------
Do your models show why some isotopes are unstable, and their decay
yes see above
it shows why they are mechanically loose
IE not having enough support by more than one connection
your model does not know it at all
that there are sub particles that have more than one connections
others that have two or three or four connections
in a 3d arrangement that gives them stability.
----------
particles?
Do your models show the known spin angular momentum of the various
isotopes?
no i never invested on it
because it was not important to me
i dont think it has much practical use or meaning
am i wrong on that point ??
--------
The QVPP methods have been used to show those characteristics of
isotopes up to Sulfur 32.
so you have other 60 ahead of you
and do you know that the paradigm of
one electron per one proton is not valid for heavy elements??
may be that is why you cant go on ??!!!that was one of my breakthrough
insights that enabled me to go on
on heavy elements!!
----------
-----------
It gets to be a complicated many body problem as the numbers of
nucleons increase. I am an old man, so it will be left to others to
extend the QVPP methods.
see just above
one of the UNNECESSARY complications is the above
wrong paradigm!!
----------
and just between us so that no one will listen
you have no chance without some new insights that you miss.
unless you will borrow it from me
btw
your 'cube' model how old is it ??
I have been working on it as a hobby since 1977.
did you thought about a cube model right from 1977??
do you know that all the elements under Fluorine
ARE NOT CUBES AND NOT RECTANGULAR PIPES ??
see it in my site
----------
-----------
Unfortunately everyone in 1977 had just been committed to the (choke,
gasp) quark model, so VPP was (is) as popular as a skunk at a picnic.
It has just been within the last 6 years that the model has been
extended to nuclei structures.
my book it copyright exactly as it is today since
1993
IE 12 years !!
and i published an abs tact of it
very similar to my site today
and you probably saw it
i remember even some correspondence about it with you ...
----------
The quark model has been left behind in the dust, by QVPP.
i agree with you about almost all your criticism
about the standard model
the quarks are in the good case only a very partial
story of the real story!!not to mention the lie of the
Gluons etc etc that is shear *shameless* mumbling!!
Regards
Y.Porat
------------------
Regards: Tom:
www.amazon.com 0963154664
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| User: "" |
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| Title: Re: Nuclear Binding Energy Calculations |
21 Aug 2005 01:07:44 PM |
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Y.Porat wrote:
tnlockyer@aol.com wrote:
Y.Porat wrote:
tnlockyer@aol.com wrote:
Chris (no-spam) wrote:
snip<
Yes, Chris, that was my point. With QVPP methods, one can calculate
binding energy between individual nucleons.
Y. Porat said:
showing in all nuclei (repeat all of them) the individual
binding energy between any two adjacent nucleids
My dear Mr. Porat, are you aware that there are some 105 elements with
2600 known isotopes?
yes so ???
----------
Those isotopes are just elements, with extra neutrons in their
structures.
The added neutrons can completely restructure the nucleus, as witnessed
in their altered spin states. For example He4 has a spin of zero, but
He5 has a spin of -3/2 by adding only one neutron to the He4.
That requires a drastic restructuring, of the nucleon placements, to
go from zero to -3/2 spin.
And of the 2600 there are 260 stable isotopes.
Of those 260 stable isotopes, 159 have a spin of zero and 101 stable
isotopes have spin between (1/2) and (9/2)?
Do you still claim your models show the binding energy for all those
isotopes?
not all isotopes
*no need to solve all isotopes solved all the 92
elements
with many of the isotopes
now listen carefully
once you solve all the 92 with many of there isotopes
you cant do it without understanding the common features
of all of them
once you do it means you have the right *tools* in
your hands
and then it is no problem to decode any isotope
i never though it is needed
because i realized that i am able to do it
for (to be cautious) for most of them.
anyway i never met an isotope or element
that i wanted to decode and got stuck!!
because i have all the tools in my hands.
that is for the mass calculation
chemical verification
beta emitters
the lightest and heaviest possible isotope
nuclear process explanation
unprecedented explanations for instance about the
geologic clock family connections
that explain why some processes are only within some elements
and not outside those elements
for instance why is it that there is transformations
from one element to another only say in the
S Si P elements and not to others that are not in that
family etc etc .
why is it that Pt cannot be transfered to Gold
and many others
IE have it as if on the palm of my hands- tangible
as you can see only a bit of it in my site.
------------
You sure have done a lot of thinking about the elements, and have
apparently found some interesting relationships.
My approach was to work with the binding energy and spin states. If
your work cannot tell the subtle difference between the spin and
binding energy of isotopes, then your understanding is too narrow, in
my view.
Do your models show which isotopes are stable?
yes for instance
all the beta emitters are of neutrons that are
connected to the pole particles of he rectangular pipe
you have there a proton or neutron and
another neutron is connected to it
in a very unstable connection
it is the *mechanical instability!!
IE a connection only at one point that is a hinge
all the other stable connections are
*with more than one connection per one sub particle!!
your model have no idea about that crucial aspect
Nope, Porat, QVPP does show the nucleon binding and it must be
orthogonal, on account of the orthogonal electric and magnetic forces.
Beta decay occurs in certain nuclei only when the neutron turning into
a proton, or the proton changing into a neutron will result in a lower
energy state, not because of connection hinges.
on the other hand
the electron capture occurs mainly
not on the poles of the nuc but around the main body
of it on the neutrons that surround the main skeleton.
Nope, electron capture , by the proton occurs when a proton capturing
an electron results in a lower energy state. Sometimes this results in
an isomeric state (i.e. the same number of protons and neutrons but not
the final configuration. See for example:
http://members.aol.com/tnlockyer/Solarcy.gif
That is why that it is primarily important to be able to model nuclei
for spins and binding energy in order to properly analyze unstable
nuclei.
Do your models show why some isotopes are unstable, and their decay particles?
yes see above
it shows why they are mechanically loose
IE not having enough support by more than one connection
your model does not know it at all
Porat, it is not a matter of loose connections. The connections are
strong in any case, it is just that nature finally gets around to
figuring out that the lower energy state has not been achieved, in
certain decaying nuclei.
that there are sub particles that have more than one connections
others that have two or three or four connections
in a 3d arrangement that gives them stability.
Porat, a neutron must change into a proton in (B-) decays, and a proton
must change into a neutron in (EC) or the competing (B+) decays.
Connections cannot give that mechanism, beta decays require changes in
the neutron or proton structures themselves.
Do your models show the known spin angular momentum of the various
isotopes?
no i never invested on it
because it was not important to me
i dont think it has much practical use or meaning
am i wrong on that point ??
--------
Yes, Porat, you are wrong, see above example. In the example the
Be8i has a spin of +2 like the parent B8, but the unstable Be8 has a
spin of zero and decays into two alpha particles.
The QVPP methods have been used to show those characteristics of
isotopes up to Sulfur 32.
so you have other 60 ahead of you
and do you know that the paradigm of
one electron per one proton is not valid for heavy elements??
may be that is why you cant go on ??!!!that was one of my breakthrough
insights that enabled me to go on
on heavy elements!!
----------
What led you to that conclusion?
They have measured the electron binding energies for heavy elements,
but I suppose it is possible that after you remove a few electrons,
that the nucleus could capture electrons from the environment, that
then require higher energy to remove, as experiments show.
It gets to be a complicated many body problem as the numbers of
nucleons increase. I am an old man, so it will be left to others to
extend the QVPP methods.
snip<
btw
your 'cube' model how old is it ??
I have been working on it as a hobby since 1977.
did you thought about a cube model right from 1977??
Yes, here is the 1977 scaling of the nested cube structure for the
proton and neutron models.
http://members.aol.com/tnlockyer/VPPsprdb.gif
It has just been within the last 6 years that the VPP model has been
extended to nuclei structures.
The quark model has been left behind in the dust, by QVPP.
Porat said;
i agree with you about almost all your criticism
about the standard model
the quarks are in the good case only a very partial
story of the real story!!not to mention the lie of the
Gluons etc etc that is shear *shameless* mumbling!!
Regards: Tom:
www.amazon.com 0963154664
.
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
22 Aug 2005 03:15:40 AM |
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wrote:
Y.Porat wrote:
wrote:
Y.Porat wrote:
wrote:
My dear Mr. Porat, are you aware that there are some 105 elements with
2600 known isotopes?
yes so ???
----------
Those isotopes are just elements, with extra neutrons in their
structures.
thank you very much for telling me that
yet you could get a little lesson from me about how isotopes are
created
and what is their structure
that Will answer you former question about how many isotopes i decoded:
see for instance the iron description in my site
see for instance the Bismuth of lead nuclei
have you ever dreamed to see the structure of the Lead nuclei??
so go there and see
now you see there along those long schematic figures
Numbers like 207 208 209 ---- 212 etc etc
do you know what that is ?? no ??
so listen carefully
there are the locations in which neutron no 207 212 etc etc are
located
got it man??
can you appreciate what you see or it is not convenient for you to
understand it
and that is only the tip of the iceberg of my model
those 2o7 208 etc etc
is the location of neutron that is No 207 in the order of Euclid's
that gives the whole nuclei its ISOTOPE NAME lead 207 etc etc
so what you see there is not just abstract mumbling of isotopes and how
they are
created
it is the exact location of neutron no 207 that gives it its name
the same with the beta emitters and their exact location
it is not just abs tact mumbling it is
th map of the nuc
got it ?? surely not
it is too shocking for you.
so you are wrong while you say that the proton or neuron have to
change themselves completely!!
that is the QM nonsense
in order of a proton to change to a neutron or vice versa
the change is only at their *periphery* got it
one of them can stay exactly in its position on the nuc
*and be changed* no need for a complete 'reshuffling of the structure'
and please dont teach me about what is a stable connection or not a
stable connection you have not the faintest real idea what is really
going on there
because you have only a poor understanding of it
compared to me you are stumbling in the darkness
my model is as tangible as a geographic map !and explained things you
never
your basic model is far form reality
you didnt answered my question about all the elements below fluorine
that are not rectangular pipes at all
you didnt answer why you are stuck at element No 32
while there should not be a reason for it had you a real good model
you dint answer my remark about the wrong paradigm about
one electron for one proton in heavy elements
do you agree with me on that or not ??
knowing spin is a marginal importance for practical use of it
and i forgot to tell you that i decoded and brought an explanations
and calculations about the
reason for a very hight angular momentum of many isotopes
it seems Thai you didnt deal with it at all
you didnt noticed my achievements about being able to predict
what transformation of elements is possible and what is not possible-
in heaver elements than the first ten
etc etc
so
dont judge my model of which you have no clue about.
because if you do it
i will tell what is my opinion about your model....
ATB
Y.Porat
-------------
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| User: "Autymn D. C." |
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| Title: Re: Nuclear Binding Energy Calculations |
22 Aug 2005 12:55:29 PM |
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Porat! >:(
.
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| User: "Y.Porat" |
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| Title: Re: Nuclear Binding Energy Calculations |
22 Aug 2005 11:18:37 PM |
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Auty did you saw your psychiatrist to day?
what is your real identity anyway?
who sent you ?? who is your boss ??
ATB
Y.Porat
----------------
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