On 1/13/2012 11:32 PM, arka chakraborty wrote:
Hi all,
I would like to ask some questions regarding this thread..
1) What is exactly meant by "Fourier transformed electron density"?- according
to my knowldege performing a fourier transform on the
electron density gives you the structure factor back. So, how does it related to
what Prof. James H called "non-lattice-convoluted
pattern"? It will be really nice if somebody can explain the thing in a "
decoded" language?!
And also any articles focusing on the concepts discussed in the entire thread
will be very helpful
The real question is "The Fourier transform of what?". All of our
usual software takes the electron density of the unit cell and Fourier
transforms it assuming that the unit cell is repeated endlessly throughout
the universe. This results in non-zero values only at integral values
of h, k, and l. If, instead, you assume there is only one unit cell
present in the x-ray beam you will get a continuous function for the
intensity of scattering. You can use the convolution theorem to show
that the value of these two functions are equal at the integral values
(except for a scale factor).
This thread has been about the relationship between the features
one would observe in this continuous function (if it could be measured)
and the electron density. If you could measure the continuous function
you can directly calculate the electron density - there is no phase
problem. The paper I referred to earlier gives the details.
Dale Tronrud
Regards,
ARKO
On Sat, Jan 14, 2012 at 12:42 AM, Dale Tronrud <det...@uoxray.uoregon.edu
<mailto:det...@uoxray.uoregon.edu>> wrote:
I think you have to be a little more clear as to what you mean
by an "electron density map". If you mean our usual maps that we
calculate all the time the Patterson map is just the usual Patterson
map. It also repeats to infinity, with the infinitely long Patterson
vectors (infinitely high frequency components) being required to
create the Bragg peaks. If you mean an electron density map of a
single object with finite bounds your Patterson map will also have
finite bounds, just with twice the radius.
The Patterson boundary is not a sharp drop-off because there aren't
as many long vectors as short ones, but the distribution depends on
the exact shape of your object. Once you have a Patterson map that
has an isolated edge (no cross-vectors) back calculating the original
object is pretty easy. (Miao, et al, Annu. Rev. Phys. Chem. 2008,
59:387-410)
Dale Tronrud
On 01/13/12 10:54, Jacob Keller wrote:
> I am trying to think, then, what would the Patterson map of a
> Fourier-transformed electron density map look like? Would you get the
> shape/outline of the object, then a sharp drop-off, presumably? Is
> this used to orient molecules in single-particle FEL diffraction
> experiments?
>
> JPK
>
> On Fri, Jan 13, 2012 at 12:33 PM, Dale Tronrud
> <det...@uoxray.uoregon.edu <mailto:det...@uoxray.uoregon.edu>> wrote:
>>
>>
>> On 01/13/12 09:53, Jacob Keller wrote:
>>> No, I meant the non-lattice-convoluted pattern--the pattern arising
>>> from the Fourier-transformed electron density map--which would
>>> necessarily become more complicated with larger molecular size, as
>>> there is more information to encode. I think this will manifest in
>>> what James H called a smaller "grain size."
>>
>> I've been thinking about these matters recently and had a nifty
>> insight about exactly this matter. (While this idea is new to me
>> I doubt it is new for others.)
>>
>> The lower limit to the size of the features in one of these
>> "scattergrams" is indicated by the scattergram's highest frequency
>> Fourier component. Its Fourier transform is the Patterson map.
>> While we usually think of the Patterson map as describing interatomic
>> vectors, it is also the frequency space for the diffraction pattern.
>> For a noncrystalline object the highest frequency component corresponds
>> to the longest Patterson vector or, in other words, the diameter of
>> the object! The bigger the object, the higher the highest frequency
>> of the scattergram, and the smaller its features.
>>
>> Dale Tronrud
>>
>>>
>>> JPK
>>>
>>> On Fri, Jan 13, 2012 at 11:41 AM, Yuri Pompeu <yuri.pom...@ufl.edu
<mailto:yuri.pom...@ufl.edu>> wrote:
>>>> to echo Tim's question:
>>>> If by pattern you mean the position of the spots on the film, I dont
think they would change based on the complexity of
the macromolecule being studied. As far I know it, the position of the
spots are dictated by the reciprocal lattice points
>>>> (therefore the real crystal lattice) (no?)
>>>> The intensity will, obviously, vary dramatically...
>>>> ps. Very interesting (cool) images James!!!
>>>
>>>
>>>
>
>
>
--
/ARKA CHAKRABORTY/
/CAS in Crystallography and Biophysics/
/University of Madras/
/Chennai,India/
