Re: My New Website
From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 06/22/04
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Date: Tue, 22 Jun 2004 12:12:44 +0200
Y.Porat wrote:
> Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cb6the$rrc$1@news.urz.uni-heidelberg.de>...
>
>>Y.Porat wrote:
>>
>>>Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<caud0e$11l$1@news.urz.uni-heidelberg.de>...
>>>
>>>
>>>>Y.Porat wrote:
>>>
>> elow.
>>
>>>>>>I wonder where you got the 4.5 A from...
>
> ------------------
> i am not sure about it (too lazy to go back to your calculation)
And you don't think that your constant "excuse" of lazyness is rather
impolite?
> but i dont recall (may be i am wrong) that you used what was
> indicated in the quoted sites
I merely used the distance between adjacent atoms. That's what is
measured by X-ray scattering.
And I explained my calculation in great detail, in many small baby
steps, and illustrated by two pictures. How on earth do you manage
to *still* not understand it???
> that 'a'(a ^3) is a cubbe in which the tertaheder is thaking only
> one third of the 'diagonal' of the a cube
> it is not clear to me if it is the planar diagonal
> or the 3d diagonal of the cubic A
> anyway i dont remember that you used that 'fact'
> that one tertahder is only one third of the diagonal
I have no clue what you are talking about.
A tetrahedron is a three-dimensional object; a diagonal is a
one-dimensional object. Hence a tetrahedron can not "take up"
one third of any diagonal. You aren't making any sense.
[snip]
>>>--------------
>>>we cant go on with hundreds of lines per article
>>
>>Why not? You probably have lots of free time, don't you?
>
> -------------
> not exactly i have more 'paying' occupations
You aren't retired yet in your age?
> and sorry my age is making me less motivated and less patiant...
> (i becoming less 'testeronated' than you......)
You are still "motivated" enough to jump into new threads and
make broad pronouncements there. But you aren't motivated enough
to actually invest some effort in substantiating your claims?
>>And, BTW, if my post was too long, why did you leave the part above
>>in it, instead of snipping it, but nevertheless didn't bother to answer
>>my question?
>>
>>So again: where did you get the 4.5 A from?
>
> ----------
> see above
> anyway i satill doubt the meaning of that 1.54
Learn how the experiments are done. Then you can try attacking this number.
> it might be the distance between atoms of different
> layers
Err, *obviously* it is the distance between atoms of different layers!!!
Look at the structure of diamond! Any two adjacent atoms lie in
different layers!!!
> that 1.54/ sin 54 is the distance between atoms
Absolute utter total nonsense.
Care to draw a picture illustrating this? Too lazy, right?
Hint: you can't draw such a picture. This doesn't make sense geometrically.
> as i sayed above while the real distance is actually 1.8 1.9 A
1.54 A/sin(109/2) gives about 1.89 A. However, what you need is
1.94 A. But you don't care about such deviations, right?
[snip]
>>>had it been a direct photograph than no problem
>>>the problem starts while it is not a direct observation but
>>>a 'multy degree of digetion' degree
>>
>>"digestion"??????????
>
> ---------------
> yes we all realised due to this thread and recalled that
> *it is actually not a **direct observation**
I didn't need to recall that. I knew that all the time since I learned
it (about 10 years ago).
> it involves quite
> a series of 'interpretations' of what we 'see directly'
> (interpretations or handlind or processing you chose the right word)
> and that processing migh tlead to mistakes.
Find the actual descriptions of the experiments and point out what
mistakes were done there, please.
>- especially if
> one is 'locked on a concept' ie takes something as granted.
> i might be wrong
You are.
> but as i sayed
> as long as i have different findings and as long
> i didnt folow all those processing i am not convinced.
In other words: If an experimental result contradicts your model,
then the result has to be wrong. No need to look up how it was
actually obtained - it is surely wrong.
>>>i anm not an expert for all theose techniques that lead to those results
>>>sdo i need a lot of time to study it
>>
>>Please do!
>
> -----------
> i am 66..... you are about half my age ....
> motivation problems you see.
That's a totally lame excuse. If you lack motivation to actually
substantiate your claims, then stop making such wild claims!
> weven if i do it it will be
> very slow.ie will take time.
I grant you as much time as you want. Is five years enough?
I only ask you to make no claims about crystal structures in the
meantime.
> and btw my access to data is actually poor.
Books on solid state physics are freely available. Even amazon has some.
<http://tinyurl.com/2482n>
>>>in that ocation
>>>thanks to Bjoern and Lothar for bothering and suplying many links
>>>about it
>>>btw are you two sure you folowand undrestand it unequivocally??
>>
>>Yes! Hey, I even learned that already in *school*, in my final year!
>>This essentially was stuff I had to learn for my "Abitur" (the final
>>exam in schools in Germany, which you have to pass in order to be
>>allowed to go to the university)!
>
> ---------
> no no dont be drifted. or else you dont even now realise
> the complexity of things.
I know far more about the "complexity" of crystals than you.
Or have you ever heard terms like "Bravais lattices", "Brillouin cell"
or "reciprocal lattice vectors"?
[snip]
>>Not an expert, but I know the methods good enough to see the faults
>>in your reasoning.
>
> -----------
> we will se later about it
You mean, in some years, when you have actually learned how the methods
are done, and have found the flaws in the experiments or their
interpretations?
> anyway it is clear that your main driving instinct is to 'kill me'
My main "instinct" is to correct your errors. Why do you take that
as a personal attack?
> (not so much the desire to understand nature....)
My desire to understand nature, I leave in my research papers. Here in
the newsgroup, it is not appropriate.
>>>sorry
>>>it should be 1.54 / half of the 109 derfee (sinus)
>>
>>That would give approx. 1.89 A. However, what you would need is 1.94 A!
>>
>>But you don't care for deviations of 2.5%, right?
>
> --------------
> no i am still not Gods friend as you are with absolut self
> confidence ans acuracy....
The results which *I* gave had deviations well below 1%. All
determinations were consistent with each other. But that doesn't bother
you, right?
>>>>In it, one uses a *powder* of the material one wants to examine, i.e. a
>>>>collection of *very* small crystalline particles. Obviously in such a
>>>>powder, the orientation of every individual particle to the X-rays is
>>>>totally random - i.e. one measures at every possible angle *AT ONCE*!!!
>>>><http://journals.iucr.org/iucr-top/comm/cteach/pamphlets/16/node2.html>
>>>>
>>>>In other methods, where one uses a whole crystal instead of a powder,
>>>>the X-ray scattering is done at a *lot* of different angles, and from
>>>>the collection of the measurements at *all* angles, one then determines
>>>>the distances between the layers!!!
>>>
>>>---------------
>>>very nice
>>
>>You didn't understand what I wrote, right?
>
> -------------
> did you ??
Yes. How could I write something I don't understand myself?
> and still in all those indirect observation
> there is th epossibility of error.
Please point out where exactly an error was made. Hint: reading up on
how the experiments are actually done would be helpful for that.
>>>but i am afried non of us realy folowed it to check all
>>>the detailes
>>
>>"follow" in what way? Actually go to the laboratory and look the
>>instruments up, or what do you want?
>
> ---------------
> i am not a university man to go to the laboratory .....
Well, then what do you suggest? What did you mean by "follow it to check
all the details"?
[snip]
>>>the trouble it it is not TRIVIAL.
>>
>>For people who work on that, it *is* trivial!!!!!
>
> -------------
> thats what you think
> too much self confidence is not a good adviser.
Err, this is not self-confidence, this is confidence in *other* people,
if you didn't notice.
>>Hello? Do you withdraw your assertions now about what I did wrong in
>>my calculation?
Hello?
>>>>I don't know what you mean by "extract" here, sorry.
>>>
>>>
>>>i mean that the figure will be avaiable to the reader by just
>>>cliking it
>>
>>Err, simply give a link to the figure??? I don't see the problem, sorry.
>>---------------
>
> in order to make a link it has to be saved on some computer
> right?
Yes.
> if you have a site it is the computer of the site- not my computer
Yes.
> do you say that i can give a link to some place on *my computer*??
No. Save the picture on the same computer where the web site is located.
Then you can access it with the same URL as the web site.
> (that other people will have access to my comuter?
> can i use anothres comuter memory???!!.
Yes. The "memory" of a computer which is hosting web sites.
In my case, I put the picture on a server of the computer center of
my university, which is especially used by university members to host
web pages (and other things).
[snip]
>>>think about a possibility that those bigger gaps are made by *less electrons'
>>>than those that are *in those layers* .....
>>
>>How can electrons make gaps???
>
> --------------
> electrons do not make gapps they made connections
> if less electrons take part in it the distance
> between atoms of carbon i sbigger see later.
I don't see why.
>>>fo rinstance only one of the two electrons in the 'gap direction'
>>>or even one from each side of the gap is 'working'
>>
>>"working"???
>
> ---- doing the connections. whu are you so priky
> try to read thoughts not words.
Err, I am only a normal human. I can't read your thoughts. I can only
read what is written there - i.e. words.
>>>while the other is remaining free (explaines why it is
>>>conducting electricity unlike the diamond
>>
>>This is also nicely explained by standard physics, thank you.
Did you get this?
>>>so may be anothe indication that the electron bind length
>>>is more than 1.57 A ??
>>
>>Huh? I have no clue what you mean here.
>
> ---------------
> as i will try to show yopu later
> the electron length that makes th econnections (bonds)
> is gernrally *not on the shortest line connecting the
> nuclei* !!!
> it happence sideways to that line!!
By "show", you mean "assert", right?
> you asked later about H2 for me it is another example:
>
> the connection in that case is not on the shortest line
> between nucs each H atom is sending an electron
> to one direction
How do you know?
And why should it do that?
> the connection between them is *at the edges of that
> (1.9 distnace) but 1.9 distance sideways
> ie you get a triangle in which two of its sides
> are 1.9 A while the distance between nucs is .....
> on the third line of that triangle a much shorter distance
> i dont remember what it is but much shorter
I.e. you simply choose the angles with which the arms
deviate from the "shortest line connectin the nuclei",
so that the distance comes out right?
As I said elsewhere: you have *lots* of free parameters!
> now at the H2O case the hydrogen is connecter
> to the edges of the oxygen atoms (separated someting like our
> fqamous 109 deg) in that case *the Hydrogen lost its electrons
> and is connected ti the oxygens electrons edges
> alse about 1.9 A from the O nuc.and 109 deg angle between
> those 'arms '
In the water molecule, the angle is *not* 109 degrees. Try again.
[snip]
>>>so let me add some words about it
>>>of course one picture in worth hundreds of words
>>>but thats what is available to me just now
>>>so please Bjoern and Lothar
>>>try to undrstyand my folowing claimes and explanation.
>>>th4 bottom lne of it is :
>>>all those diffrent distances between atoms can be explained
>>>with *just one electron bond length*!!
>>
>>Well, if you think this is possible, do it. You gave only two examples
>>about this in your book, IIRC - what about explaining all the
>>other examples? You can start with H_2, proceed with N_2, O_2 and
>>H_2O, etc.
Hello?
>>>and please who was the crackpot from whom you first heared about it
>>>i can starty it from the data of the elements
>>>Aluminum Argentum and Gold ie their specific weight
>>>in relation to their atomic weight
>>
>>Incomprehensible.
>
> ---------
> not for you at this stage ....
Huh?
>>>but i preffere to concenrtate at this point with another triall:
>>>
>>>ie the single double and tripple bond if carbon
>>>all of them besd on just one tetraheder
>>>in which there is a Carbon atom at its middle
>>>and it sends electron arms to 4 of its corners.
>>
>>What length do you use for the "electron arms",
>
> 1.9 A !
Is that clear now? In your book, you sometimes use 2.15 A, sometimes
1.93 A, etc.
[snip]
>>So far, I could follow you. But I don't understand the calculation
>>you presented on this in your book.
>
> --------------
> i havw a strong recomendation for you:
> do what i did:
> take two iron wires ( i did it from the wire of two paperclipses
> entangle them at their middles
> and strech out the for 'arms' to 109 deg separation between
> all 'arms'
> you got th emodel of the carbon atom with its electrons
> in your hand
> the imaginary cover of those arms is the famous 3d tetraheder
> a nice symetric pyramid
Why should I do that? As I said, I understood the pictures in
your book. The only problem is that I don't see how you get
from the pictures to the calculations you make!
> now if you totch two of those 'pyramid 'leges'
> the cenroid of those pyramids are at a certain
> distance that can be calculated exacly as dependat of the
> 'arm length' (the 1.9 A)
> now try to atach 3 'arms' each from each pyramid 9each Carbon-
> to make it shorter)
> once you touch 3 arms insted of only two
> you realise that the center of those pyramids
> is becoming closer (than the case of only two arms)
Yes, that is perfectly clear. The question is only *how much closer*
they will become.
Let's see, for a tetrahedron with side length a, the distance of one
corner to the middle, i.e. the arm length, is a*sqrt(6)/4. You
propose an arm length of 1.9 A, hence a=1.9 A * 4/sqrt(6) = 3.10 A.
If two tetrahedrons touch at one corner, i.e. one arm of each is connected
(a single bond), the distance between the centers of the two
tetrahedrons (i.e. the binding length) is then obviously 2*1.9 A = 3.8
A. Disagreement #1 with experimental data.
If two tetrahedrons touch at an edge, i.e. two arms of each are
connected (a double bond), the distance between the centers of the two
tetrahedrons (i.e. the binding length) is two times the height of the
triangle shown here:
<http://www.rzuser.uni-heidelberg.de/~bfeuerba/triangle.jpg>,
with x = 1.9 A. A short calculation then gives for the bond length
a value of 2.19 A.
In your book (page 129), you say that the result is 1.52 A - this is
simply wrong. Apparently you fooled up your geometric calculations! You
use the formula d = sqrt((E/2)^2 + (E/2)^2) here. Could you please
explain me why on earth you think that formula gives the distance
between the two atoms????????????
Also, this is disagreement #2 with experimental data. The experiments
give us a bond length of 1.35 A. Again, you don't bother for a deviation
of even 10%.
If two tetrahedrons touch at a whole face, i.e. three arms of each are
connected (a triple bond), the distance between the centers of the two
tetrahedrons (i.e. the binding length) is given by two times the
distance of the center of a tetrahedron to a face. This distance is
given by a/sqrt(24) = 1.9 A / 3 = 0.63 A, hence the binding length is
1.26 A. You said that it is 1.42 A - apparently you again fooled up your
geometric calculations. And this is disagreement #3 with experimental
data: the experimental value is 1.2 A.
> you can make it(the diastance netween the middles of
> the centoid of the two pyramids ) not only qualitatively but
> quamtitatively as i did it in my book
You fooled up the calculation. Your formula
d = sqrt((E/2)^2 + (E/2)^2)
makes no sense at all.
> as a function of the 1.9 A arm
> you will get at the case of the 'double bond' (only 2 arms touch)
> 1.54
You contradict your own book here. You say that for an arm length
of 2.15 A, one gets a bond length of 1.52 A (BTW, that's wrong - they
right result would be 2.48 A). So obviously, for an arm length of 1.9 A,
one can't get a bond length of 1.54 A!
Please stop lecturing me about geometry. Learn it yourself first!
> while at the case of 3 arms touch (the tripple bond)
> you will get a distance (of columes centroids)
> about 1.42 A
For an arm length of 2.15 A, according to your book and your wrong
calculation, not for an arm length of 1.9 A.
> that fits known data (as much as written in my old books)
The experimental data is:
* double bond: 1.35 A
* triple bond: 1.2 A
That does *not* fit your numbers in any way.
> so no need for comlicated QM arguments and calculations
Hint: the QM calculations reproduce the values I mentioned above
with accuracies lower than 1%.
> just the 3d steromerty that you *not me* ....
> learned at your primery school
On your page 129, you show nicely that you aren't able to do
such elementary geometrical calculations.
> actually inmy opinion to find the centroid of that
> tetraheder and diatnances to another one are not 'soo trivial'
I did it long ago. I told you the results several times, even
in this thread. Have you already forgotten this again?
> it is not 'terribly complicated' but needs time and thinking
> and calculations.
Yes.
Bye,
Bjoern
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