Ethan and I apparently agree that anomalous scattering is "normal" and Friedel's Law is just an approximation. I'll presume that your "unique" is assuming otherwise and your 62,500 reflections only include half of reciprocal space. The full sphere of data would include 125,000 reflections. Since the cube root of 125,000 is 50 you get a range of indices from -25 to +25 which would give you 2 A resolution, which is still far from your hope of 1 A.
For your test case of 1 A resolution with 50 A cell lengths you want your indices to run from -50 to +50 giving a box of reflections in reciprocal space 101 spots wide in each direction and a total of 101^3 = 1,030,301 reflections. (or 515,150.5 reflections for your Friedel unique with the "half" reflection being the F000 which would then be purely real valued.) Assuming you can fit your structure factors into 16 bits (You had better not have many more than 10,000 atoms if you don't want your F000 to overflow.) the information content will be 1,030,301 * 2 * 16 bits (The "2" because they are complex.) giving 32,969,632 bits. If you spread this same amount of information across real space you will have 1,030,301 complex density values in a 50x50x50 A space giving a sampling rate along each axis of 101 samples/unit cell. Complex density values? The real part of the density is what we call the electron density and the imaginary part we call the anomalous density. If there is no anomalous scattering then Friedel's Law holds and the number of unique reflections is cut in half and the density values are purely real valued - The information content in both spaces is cut in half and they remain equal. By sampling your unit cell with 101 samples their rate is half that of the wavelength of the highest frequency reflection. (e.q. a sampling rate of 0.5 A for 1 A resolution data) This is, of course, the Nyquist Theorem which states that you have to sample at twice the frequency of the highest resolution Fourier coefficient. This is exactly how an FFT works. It allocates the memory required to store the structure factors and it returns the map in that same array - The number of bytes is unchanged. It also guarantees that the calculation is reversible as no information is lost in either direction. So, why does your blocky image look so bad? First you have sampled too coarsely. You should have twice the sampling rate in each direction. The next point is more subtle. You are displaying each voxel as a block. This is not correct. The sharp lines that occur at the boundaries between the blocks is a high frequency feature which is not consistent with a 1 A resolution image. Your sample points should be displayed at discrete points since they are not the average density within a block but the value of the density at one specific point. What is the density of the map between the sampled points? The Fourier series provides all the information needed to calculate them and you can calculate values for as fine a sampling rate as you like, just remember that you are not adding any more information because these new points are correlated with each other. If you have only the samples of a map and want to calculate Fourier coefficients there are many sets of Fourier coefficients that will reproduce the sampled points equally well. We specify a unique solution in the FFT by defining that all reflections of resolution higher than 1 A must be identically equal to zero. When you calculate a map from a set of coefficients that only go to 1 A resolution this is guaranteed. When you are calculating coefficients from any old map you had better ensure that the map you are sampling does not contain information of a higher resolution than twice your sampling rate. This is a problem when calculating Fcalc from an atomic model. You calculate a map from the model and FFT it, but you can't sample that map at 1/2 the resolution of your interest. You must sample that map much more finely because an atomic model implies Fourier coefficients of very high resolution. (Otherwise phase extension would be impossible) This problem was discussed in detail in Lynn Ten Eyck's 1976 paper on Fcalc FFT's but is often forgotten. Gerard Bricogne's papers on NCS averaging from the 1970's also discusses these matters in great depth. In summary, your blocky picture (even with double sampling) is not a valid representation because it is not blurry like a 1 A resolution map should be. To create an accurate image you need to oversample the map sufficiently to prevent the human eye from detecting aliasing artifacts such as the straight lines visible in your blocky picture. This requires very fine sampling because the eye is very sensitive to straight lines. When using a map for any purpose other than FFTing you will need to oversample the map by some amount to prevent aliasing artifacts and the amount of oversampling will depend on what you are doing to the map (Again, see Gerard's papers.) Such an oversampled map will have many more voxels but no more information because the density values are correlated. Dale Tronrud On 11/9/2017 4:10 PM, Keller, Jacob wrote: > Dear Crystallographers, > > > > I have been considering a thought-experiment of sorts for a while, and > wonder what you will think about it: > > > > Consider a diffraction data set which contains 62,500 unique reflections > from a 50 x 50 x 50 Angstrom unit cell, with each intensity measured > perfectly with 16-bit depth. (I am not sure what resolution this > corresponds to, but it would be quite high even in p1, I think--probably > beyond 1.0 Angstrom?). Thus, there are 62,500 x 16 bits (125 KB) of > information in this alone, and there is an HKL index associated with > each intensity, so that I suppose contains information as well. One > could throw in phases at 16-bit as well, and get a total of 250 KB for > this dataset. > > > > Now consider an parallel (equivalent?) data set, but this time instead > of reflection intensities you have a real space voxel map of the same 50 > x 50 x 50 unit cell consisting of 125,000 voxels, each of which has a > 16-bit electron density value, and an associated xyz index analogous to > the hkl above. That makes a total of 250 KB, with each voxel a 1 > Angstrom cube. It seems to me this level of graininess would be really > hard to interpret, especially for a static picture of a protein > structure. (see attached: top is a ~1 Ang/pixel down-sampled version of > the image below). > > > > Or, if we wanted smaller voxels still, let’s say by half, we would have > to reduce the bit depth to 2 bits. But this would still only yield > half-Angstrom voxels, each with only four possible electron density values. > > > > Is this comparison apt? Off the cuff, I cannot see how a 50 x 50 pixel > image corresponds at all to the way our maps look, especially at around > 1 Ang resolution. Please, if you can shoot down the analogy, do. > > > > Assuming that it is apt, however: is this a possible way to see the > power of all of our Bayesian modelling? Could one use our modelling > tools on such a grainy picture and arrive at similar results? > > > > Are our data sets really this poor in information, and we just model the > heck out of them, as perhaps evidenced by our scarily low > data:parameters ratios? > > > > My underlying motivation in this thought experiment is to illustrate the > richness in information (and poorness of modelling) that one achieves in > fluorescence microscopic imaging. If crystallography is any measure of > the power of modelling, one could really go to town on some of these > terabyte 5D functional data sets we see around here at Janelia (and on > YouTube). > > > > What do you think? > > > > Jacob Keller > > > > +++++++++++++++++++++++++++++++++++++++++++++++++ > > Jacob Pearson Keller > > Research Scientist / Looger Lab > > HHMI Janelia Research Campus > > 19700 Helix Dr, Ashburn, VA 20147 > > (571)209-4000 x3159 > > +++++++++++++++++++++++++++++++++++++++++++++++++ > > >
