Re: My New Website

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 06/07/04 

Date: Mon, 07 Jun 2004 12:45:37 +0200

Y.Porat wrote:
> Lothar Brendel <l.no.spam.brendel@uni-duisburg.de> wrote in message news:<c9n93k$be7$1@a1-hrz.uni-duisburg.de>...
>
>>Y.Porat wrote:
>>
>>[...]
>>
>>
>>>let me try and make some order in definitions:
>>>in the structure that Bjoern quoted we see
>>>'speres' connected with 'bars'
>>
>>[...]
>>
>>
>>>so i actually hinted above about the definition problem
>>>so lets define it for having a common language:
>>>
>>>i called
>>>'the latice unit volume' -
>>
>>The "lattice unit _volume_" doesn't tell us anything about the "lattice
>>unit's" _shape_.
>>
>>
>>
>>>*the avrage maximum volume that belongs to --
>>>*one* !!! 'sphere
>>
>>What do you mean by "average maximum volume"? The "average volume that
>>belongs to one sphere" is simply the system's volume divided by the
>>number of spheres, we don't even need a lattice structure for that.
>>
>>The nice thing about a lattice, though, is that this volume is _not_ an
>>average value, it is the same for every sphere.
>>
>>
>>
>>>not to the
>>>
>>>not to the nice rectangles or cubes or exagonal
>>>structures that *contain many!! spheres* just one!
>>
>>The primitive unit cell of the diamond structure is, as I explained a
>>rhombohedron containing two atoms, the latter being called the "base".
>>If you insist on a cell with only _one_ atom, you are free to divide the
>>rhombohedron into two halves and to assign to each half one of the two
>>atoms. But then you have _two_ "unit cells" (being congruent at most).
>>
>>
>>
>>>so i think now we can have a common language
>>>unless someone will introduce another name but !!.....
>>>to the same physical entity!
>>
>>I prefer the commonly used notations "conventional/primitive unit cell"
>>instead of your vague defintion of "lattice unit volume".
>>
>>
>>
>>>and my orriginakl question can be answered only
>>>based on specific weight as Bjoern sayed and!! .....
>>
>>Sure, if you know the _mass-density_ (rather than specific weight) and
>>the mass of an atom, you know the volume "belonging" to one atom. And if
>>you further know that there is an underlying lattice structure, you know
>>that this volume is in a way not only an average value. And if you
>>further know _which_ lattice structure it is, you can even tell about
>>possible _shapes_.
>>
>>
>>
>>>x ray observations of all the possible relevant angles
>>>of the crystal
>>>and measurments od the different distances that are needed
>>
>>That's already done, the diamond structure is well known since a long time.
>>
>>
>>
>>>to define my 'latice unit volume as i defined it above
>>>acording to my findings
>>
>>The unit cell of the diamond structure is well known, too. No need to
>>define a new one.
>>
>>
>>
>>>i ahd a surprise to realise that there are
>>>*more than one atom per such sphere*
>>
>>If by "sphere" you refer still to same thing as above, i.e. "what can be
>>seen in the structure quoted by Bjoern" (besides the bars), than your
>>"realisation" is _wrong_. There are drawn 8 spheres (8/8+6/2+4/1) per
>>conventional unit cell, each representing _one_ atom. The same is true
>>for the primitive unit cell: 2 spheres <=> 2 atoms
>
> -----------------------------
> just give me a break with all your 'well known'data
> we have to examine if it is 'well known

How do you plan to examine that? You don't have access to
the professional journals where all that stuff was published.

> the only way to do it is byfinding how many atoms per cubic centimeter
> are in it (the diamond crystal) and ......

*sigh* For the 10th time: 1.77 * 10^23 atoms per cm^3.

> find *in a different way !!! ie through the x ray data

Already done decades ago.

> *how many 'joints' (!) (ie those points from which those 'bars stretch out)
> so how many joints per cn^2 are there* (from the x ray data.

Since the "bars" are not material things, but merely drawn there in
the pictures for illustration, your definition of "joint" doesn't
make sense for a real-world crystal.

> not from your parroting about the 'well known'
> but from your or the calculation that we will do togethere -
> calculation that we will check it here!!

Calculate what?

> (it will be more than one atom per 'joint' !!) thats my prediction.

On what do you base this prediction?

Can you explain the obtained X-ray data based on your prediction?

> and that will lead to a new unknown problem.

Why do you act as if your prediction had already been established
as a fact?

> --------------------
> Y.Porat
> -----------------------
>
>

You forgot to point out that you are a crackpot.

Bye,
Bjoern