Dear Jacob,
To add one more thing to the many ones from James, despite air scattering noise being flat and easily subtractable over time, that noise wont be the same if you take images with and without the sample. Since some of the photons will be absorbed and diffracted by the crystal, the air noise will be different from what you recorded without the sample (with or without the loop/stuff) and you would like to subtract.
D
-----Original Message-----
From: CCP4 bulletin board [mailto:[log in to unmask]] On Behalf Of James Holton
Sent: 15 January 2015 17:00
To: ccp4bb
Subject: Re: [ccp4bb] X-ray Source Differences (WAS: RE: [ccp4bb] How far does rad dam travel?)
Jacob,
Background subtraction is unfortunately not as forgiving as you think it is. You can subtract the background, but you can't subtract the noise.
This is because noise (by definition) is never the same twice. Yes, the "average" or "true" background under a spot may be flat, but any given observation of it will be noisy, and there is no way to separate the noise that came from the Bragg-scattered photons from the background-scattered photons that hit the same pixel. Each photon is an independent event, after all.
Consider an example: if you see 4 photons in an isolated spot after 1 second and there is zero background then sigma(I) = sqrt(I) = sqrt(4) = 2, and your I/sigma is most likely 2. I write "most likely" because the "true" photon arrival rate (the thing that is proportional to F^2) doesn't have to be 4 photons/s just because you counted four in one observation. The long-term average could easily be something else, such as 3 photons/s, 5 photons/s or even 3.2 photons/s (on average).
Observing 4 photons is not unlikely in all these scenarios. However, if you consider all possible "true" rates, simulate millions of trials and isolate all the instances where you counted 4 photons you will find that the "true" rate of 4.0 photons/s turns up more often than any other, so that's your best guess.
Nevertheless, if your "true" rate really is 4.0 photons/s, then the probability of actually seeing 4 photons is only ~20%! The other 80% of the time you will see something else. 20% of the time you will count 3, there is a 15% chance you will only see 2, and a ~2% chance you will see zero. There is also a 2% chance of seeing 11 photons in 1 second when the "true" rate is really 4. You just don't know. Of course, with more observations you can narrow it down. If you do 100 trials and see an average of 4.0, then you are obviously a lot more confident in the "true" rate of 4 photons/s. But this isn't observing 4 photons, this is observing 400 photons and dividing that number by 100. Because the error in counting 400 photons is sqrt(400)=20, your signal-to-noise is
400/sqrt(400) = 20! This is why multiplicity is a good thing. However, if all you have is one sample of 4 photons your best guess is I = 4 and
sigma(I) = 2.
Now consider the case where there is background. Say the "true rate"
for the background is 10 photons/pixel/s, and for simplicity lets say your 4 photon/s spot lies entirely within one pixel. What is your signal-to-noise now? Well, if you take 100 pixels in the neighborhood around your spot you will count ~1000 photons, giving you an excellent estimate of the "true" background rate: 10 photons/pixel/s with a sigma of sqrt(1000)/100 = 0.3, or a "signal-to-noise" of 31.6 for the estimate of the background level. Looking at the one pixel containing your spot, let's say you saw 14 photons in it, that means you have a "sigma" of 2 from the 4 "spot photons" and a "sigma" of 0.3 from the background photons for a total sigma of sqrt(2^2+0.3^2) = 2.02 and I/sigma = 1.97, right? Wrong.
The reality of the situation is the 14 photons that landed in your spot pixel were independent events from all the photons that landed in the nearby background pixels. Yes, you know that there "should" be 10, but the probability of actually getting 10 exactly is only 12.4%. You are just as likely to see 11 or 9, and there is a 5% chance of 14 background photons hitting the spot area. In that case the Bragg intensity could easily be zero, and the 14 photons you saw were just a random fluctuation from the average of 10. You just don't know! And your error bars should reflect that. The correct value for sigma(I) turns out to be the square root of the TOTAL number of photons that hit the spot area: sqrt(14) = 3.7 and the I/sigma of your 4-photon spot is now 1.07. With 100 background photons/pixel, your I/sigma = 0.4. This is how background degrades your resolution limit.
The bright side of it is that the degradation of I/sigma rises only with the square root of the background level, not the background level itself. As a general rule: it takes 3x the Bragg photons to cut I/sigma in half, 15x the Bragg photons to cut it to 1/4, etc. So, a factor of 2 in background is at worst a 40% hit in I/sigma. Not bad, but not good either.
Now, since the noise from background is proportional to the square root of the background level, the background level is proportional to the illuminated area, and the illuminated area, in turn, is the square of the beam size (for a square beam), the extra noise from a beam-crystal size miss-match is equal to the excess linear dimension of the beam.
That is, if you have a situation where you would get 16 Bragg photons and 48 background photons into a given spot from a 10 micron crystal with a 10 micron beam, that will give you I/sigma = 2. But if you use a
20x20 micron beam you have quadrupled the illuminated area. Assuming your 10 micron crystal is embedded in a block of vitrified water, for the same dose to the crystal (photons/area) you will still get 16 Bragg photons but 192 background photons in the spot area pixels, reducing your I/sigma to 1.1. Most people would rather not do that.
So, yes, smaller beams are better if your crystal is actually small, and it is surrounded by "stuff" of similar thickness, density and elemental composition. You also need to bear in mind the "stuff" that is in the beam path before and after your crystal because this is part of the illuminated volume too. Ideally, you want your crystal sticking out into the air, then any beam size is more or less equivalent because
air/N2 scatters 1000x less than the solid stuff in the loop. The only problem with that is a lot of crystals hate surface tension. This is why I recommend having surrounding "stuff", but no too much. A factor of 2 in volume is a good compromise.
Yes, there is such a thing as big crystals with a "sweet spot" that can only be accessed with a small beam and there is definitely a lot of excitement about that. What I keep wondering is: what went wrong in the rest of that crystal?
Brighter beams are better for getting your experiment over more quickly, but if you're attenuating then those extra photons are going to waste. Faster detectors can help with this, but if they are too fast they will start picking up high-frequency noise in the beam. This is a problem for anomalous, but not for resolution. 5% error when you are only counting 4 photons is a minor consideration. Background is the primary enemy of resolution. After disorder, of course!
As for where to put resources, I try not to think about which aspect of structural biology should be cut in favor of some other part because I think structural biology as a whole is important! Especially when you consider how much money is wasted on <insert random political statement
here>.
Detectors can always be better, but at the moment low-end noise levels and speed are not limiting factors. The challenges are either 1) detecting weak spots (aka resolution) or 2) accurately measuring small differences between strong spots (aka anomalous differences). 1) is limited by pixel count and 2) by calibration. I say pixel count because larger active areas are always better for background reduction (inverse square law), but only if your spots take up more than a few pixels. If your spots are all smaller than a pixel then your pixels are too big.
Colin Nave (JSR, 2014) has calculated that the ideal MX detector would have about 1e9 pixels in it. Only problem with that is the going rate for a pixel these days is ~$0.25 each. For anomalous, the biggest problem with detectors is calibration, which is a lot harder to deal with than you might think. The best evidence of this fact is that if you simulate data with every kind of noise you can think of you still get low-resolution R-meas values of ~0.5% (Holton et al, FEBS 2014, Diederichs, 2009). I have never seen a real dataset like that.
Nevertheless, if you count 1,000,000 photons, the sigma of that count is 1000, or 0.1% error. Something else is getting in the way. Unfortunate really, because if we could routinely get R-meas = 0.1% we would never need to make metal derivatives again.
-James Holton
MAD Scientist
On 1/8/2015 9:47 AM, Keller, Jacob wrote:
> Yes, this is great info and thoughts. What I still do not understand, however, is why the noise from air/loop scattering is so bad--why not make sure only the top of the Gaussian is engulfing the crystal, and the tails can hit air or loop? Isn't the air scattering noise easily subtractable, being essentially flat over time, whereas uneven illumination of the crystal is highly difficult to correct?
>
> Also, in light of these considerations, it would seem to me a much better use of resources not to make brighter and smaller beams but instead concentrate on making better low-intensity big beam profiles (top-hats?) and lower-noise, faster detectors (like Pilatus and the new ADSC).
>
> Jacob
>
> -----Original Message-----
> From: James Holton [mailto:[log in to unmask]]
> Sent: Tuesday, December 30, 2014 3:57 PM
> To: Keller, Jacob;[log in to unmask]
> Subject: Re: [ccp4bb] How far does rad dam travel?
>
>
> Yes, bigger is okay, and perhaps a little better if you consider the effects of beam/crystal vibration and two sharp-edged boundaries dancing over each other. But bigger is better only to a point. That point is when the illuminated area of non-good-stuff is about equal to the area of the good stuff. This is because the total background noise is equal to the square root of the number of photons and equal volumes of any given "stuff" (good or non-good) yield about the same number of background-scattered photons. So, since you're taking the square root, completely eliminating the non-good-stuff only buys you a gain of 40% in total noise. Given that other sources of noise come into play when the beam and crystal are exactly matched (flicker), 40% is a reasonable compromise. This is why I recommend loop sizes that are about 40% bigger than the crystal itself. Much less risk of surface-tension injury, and the air around the loop scatters 1000x less than the non-crystal stuff in the
> loop: effectively defining the "beam size".
>
> As for what beam profiles look like at different beamlines, there are some sobering mug-shots in this paper:
> http://dx.doi.org/10.1107/S0909049511008235
>
> Some interesting quirks in a few of them, but in general optimally focused beams are Gaussian. Almost by definition! (central limit theorem and all that). It is when you "de-focus" that things get really embarrassing. X-ray mirrors all have a "fingerprint" in the de-focused region that leads to striations and other distortions. The technology is improving, but good solutions for "de focusing" are still not widely available. Perhaps because they are hard to fund.
>
> Genuine top-hat beams are rare, but there are a few of them. Petra-III is particularly proud of theirs. But top-hats are usually defined by collimation of a Gaussian and the more x-rays you have hitting the back of the aperture the more difficult it is to control the background generated by the collimator. If you can see the shadow of your pin on the detector, then you know there is a significant amount of "background" that is coming from upstream of your crystal! My solution is to collimate at roughly the FWHM. This chops off the tails and gives you a tolerably "flat" beam in the middle.
>
> How much more intense is the peak than the tails? Well, at the FWHM,
> the intensity is, well, half of that at the center. At twice that
> distance from the center, you are down to 6.2%. The equation is
> exp(-log(16)*(x/hwhm)**2) where "hwhm" is 1/2 of the FHWM.
>
> HTH!
>
> -James Holton
> MAD Scientist
>
> On 12/30/2014 12:10 PM, Keller, Jacob wrote:
>>> Yes, it gets complicated, doesn't it? This is why I generally
>>> recommend
>> trying to use a beam that matches your crystal size.
>>
>> ...or is bigger, right? Diffuse scattering, yes, but more even illumination might be worth it?
>>
>> Generally, James, I have a question: what is the nature of the intensity cross-sections at most beamlines--are they usually Gaussian, or are some flatter? Or I guess, if Gaussian, how much more intense is the peak than the tails?
>>
>> JPK
>>
>>
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