On Jul 7, 2013, at 1:44 PM, Ian Tickle <[log in to unmask]> wrote:
>
> On 29 June 2013 01:13, Douglas Theobald <[log in to unmask]>
> wrote:
>
> > I admittedly don't understand TDS well. But I thought it was
> > generally assumed that TDS contributes rather little to the
> > conventional background measurement outside of the spot (so Stout
> > and Jensen tells me :). So I was not even really considering TDS,
> > which I see as a different problem from measuring background (am I
> > mistaken here?). I thought the background we measure (in the area
> > surrounding the spot) mostly came from diffuse solvent scatter, air
> > scatter, loop scatter, etc. If so, then we can just consider Itrue
> > = Ibragg + Itds, and worry about modeling the different components
> > of Itrue at a different stage. And then it would make sense to
> > think about blocking a reflection (say, with a minuscule, precisely
> > positioned beam stop very near the crystal) and measuring the
> > background in the spot where the reflection would hit. That
> > background should be approximated pretty well by Iback, the
> > background around the spot (especially if we move far enough away
> > from the spot so that TDS is negligible there).
>
> Stout & Jensen would not be my first choice to learn about TDS! It's
> a textbook of small-molecule crystallography (I know, it was my main
> textbook during my doctorate on small-molecule structures), and small
> molecules are generally more highly ordered than macromolecules and
> therefore exhibit TDS on a much smaller scale (there are exceptions of
> course). I think what you are talking about is "acoustic mode" TDS
> (so-called because of its relationship with sound transmission through
> a crystal), which peaks under the Bragg spots and is therefore very
> hard to distinguish from it. The other two contributors to TDS that
> are often observed in MX are "optic mode" and "Einstein model". TDS
> arises from correlated motions within the crystal, for acoustic mode
> it's correlated motions of whole unit cells within the lattice, for
> optic mode it's correlations of different parts of a unit cell (e.g.
> correlated domain motions in a protein), and for Einstein model it's
> correlations of the movement of electrons as they are carried along by
> vibrating atoms (an "Einstein solid" is a simple model of a crystal
> proposed by A. Einstein consisting of a collection of independent
> quantised harmonic-isotropic oscillators; I doubt he was aware of its
> relevance to TDS, that came later). Here's an example of TDS:
> http://people.cryst.bbk.ac.uk/~tickle/iucr99/tds2f.gif . The acoustic
> mode gives the haloes around the Bragg spots (but as I said mainly
> coincides with the spots), the optic mode gives the nebulous blobs,
> wisps and streaks that are uncorrelated with the Bragg spots (you can
> make out an inner ring of 14 blobs due to the 7-fold NCS), and the
> Einstein model gives the isotropic uniform greying increasing towards
> the outer edge (makes it look like the diffraction pattern has been
> projected onto a sphere). So I leave you to decide whether TDS
> contributes to the background!
That's all very interesting --- do you have a good ref for TDS where I
can read up on the theory/practice? My protein xtallography books say
even less than S&J about TDS. Anyway, this appears to be a problem
beyond the scope of this present discussion --- in an ideal world we'd
be modeling all the forms of TDS, and Bragg diffraction, and comparing
those predictions to the intensity pattern over the entire detector ---
not just integrating near the reciprocal lattice points. Going on what
you said above, it seems the acoustic component can't really be measured
independently of the Bragg peak, while the optic and Einstein components
can, or least can be estimated pretty well from the intensity around the
Bragg peak (which means we can treat it as "background"). In any case,
I'm going to ignore the TDS complications for now. :)
> As for the blocking beam stop, every part of the crystal (or at least
> every part that's in the beam) contributes to every part of the
> diffraction pattern (i.e. Fourier transform). This means that your
> beam stop would have to mask the whole crystal - any small bit of the
> crystal left unmasked and exposed to the beam would give a complete
> diffraction pattern! That means you wouldn't see anything, not even
> the background!
That's all true, but you can detect peaks independently of one another
on a detector, so obviously there is some minimal distance away from a
crystal where you could completely block any given reflection and
nothing else. Clearly the "reflection stop" would have to be the size of
the crystal (or at least the beam).
> You could leave a small hole in the centre for the direct beam and
> that would give you the air scatter contribution, but usually the air
> path is minimal anyway so that's only a very small contribution to the
> total background. But let's say by some magic you were able to
> measure only the background, say Iback". In a separate experiment
> before you rigged up the mask you will have measured Ispot. How does
> that help?: you haven't measured Iback' directly and Iback" will
> differ from Iback' again due to count fluctuations. So I think that's
> a non-starter.
If Iback' and Iback" come from the same process, then one informs the
other. Of course you'd have to account for statistical fluctuations.
This is exactly the same principle behind using Iback to give us
information about Iback' in French and Wilson's method.
> > Ahh, this all seems a bit too philosophical, what's really real and
> > what's not really real. There are of course many different
> > observationally equivalent QM interpretations, not just the one you
> > espouse above (e.g., "the only real quantities are the observables"
> > and "wave function collapse" talk). I won't go down the QM woo road
> > -- next thing we'll confirm Godwin's law and start talking about
> > Nazis ... (Blargh! there I did it :) Anyway, I don't think any of
> > this matters for the practice of probabilistic inference. I can
> > model the background from solvent/air scatter as Poisson (as we're
> > dealing with photons that are well known to have Poisson
> > distributions), and this background adds to the intensity from the
> > coherent scatter of the crystal (which is also from Poissonian
> > photons) --- giving the sum of two Poisson variates, which is itself
> > a Poisson variate. If, OTOH, we can't validly use that model, then
> > I don't see any justification for F&W's method (see below).
> > >
> > > >
> Sorry your allusion to "Godwin's law" is lost on me. Photons only
> have a Poisson distribution when you can count them: QM says it
> meaningless to talk about something you can't observe.
Aw, come on --- QM is a theory, it says no such thing. The claim that
"it's meaningless to talk about something you can't observe" is a
philosophical principle, not science. There are many interpretations of
QM, some involving hidden variables, which are precisely things that
exist that you can't observe. Heck, I'd say all of science is *exactly*
about the existence of things that we only infer and cannot observe
directly. Remember, when you get the readout from a detector, you are
not directly observing photons even then --- you are formally inferring
things that you can't observe. There's a whole chain of theory and
interpretation that gets you from the electronic readout to the claim
that the readout actually represents some number of photons.
> It's like asking which path a particular photon took in the
> double-slit experiment: the best answer we can give (though QM says
> you shouldn't even attempt to answer since it's a meaningless
> question) is that the photon went through both slits simultaneously
> while still remaining as a single particle!
Again, this is your own personal philosophical interpretation of QM ---
QM itself says nothing of the sort. For instance, Bohm's pilot wave
interpretation of QM, which is completely consistent with observation
and QM theory and calculation, states that individual photons *do* go
through one slit or the other. But this is really off point here, I
think --- as I said, I don't want to get into a QM debate.
> As you say all the QM theories differ only in their interpretation
> (i.e. what's going on "behind the scenes"), they all agree about the
> observables. It's meaningless to talk about the measurement error (or
> its distribution) if it's physically impossible to make the
> measurement!
I disagree. Following that logic, we could not talk about the error in
our estiamte of the gravitational force on the Earth from the Moon
(because, if our theories of gravity are correct, the force exerted on
the Earth is the sum of the gravitational pull of all massive objects in
the universe, and it is physically impossible for us to, say, remove the
Sun and then independently measure the force exerted by the Moon).
> It seems to me that the only way we can get at the PDF of Itrue is
> from the PDFs of the observables Ispot & Iback and the prior
> distribution of Itrue. This is precisely what F & W does: I don't see
> what's wrong with that.
I agree completley with that. My point is, however, that F&W implicitly
assumes that the background we measure (in Ib) comes from the same
process as the background under the spot (Ib'). In other words, the
underlying model is:
Is = Ib' + Ij
Ib
where we experimentally measure Is (the spot) and Ib (the background
around the spot), and we assume that both Ib' and Ib come from the same
(Poisson) distribution, and that Ij (a sample from the true spot
intensity J) comes from a Poisson as well.
If that's not the F&W model, then what is?
[snip]
> > > The only function of Is and Ib that's relevant to the joint
> > > distribution of Is and Ib given J and sdJ, P(Is,Ib | J), is the
> > > difference Is-Ib (at least for large Is and Ib: I don't know what
> > > happens if they are small).
> >
> > So why are we using Is-Ib and not {Is,Ib}? First, note that:
> >
> > P(Is,Ib | J,sdJ,B,sdB) = P(Is | Ib,J,sdJ) P(Ib | B,sdB)
> >
> > since Is and Ib are dependent. I've augmented the notation some, to
> > show the explicit dependence on parameters that will be important
> > later.
> >
> > Note that if you want to work as if Is and Ib are independent, then
> >
> > P(Is,Ib | J,sdJ,B,sdB) = P(Is | J,sdJ) P(Ib | B,sdB)
> >
> > But then you've got the likelihood function P(Is | J,sdJ), which is
> > all you need to find an ML (or Bayesian) estimate of J given data
> > Is. So if Is and Ib are independent, there's no need for F&W at all.
>
> Where did the assumption of independence come from? I don't see how
> they can be independent (if you have a big Ib chances are Is will be
> big too). The implication of having the likelihood function P(Is |
> J,sdJ) is that you don't even need to measure Iback to get an estimate
> of J, all you need is Ispot! That makes no sense.
This is precisely the point I was trying to make, so we agree.
Ib and Is are not independent.
> > But where does P(Is-Ib | J,sdJ) actually come from? Can you derive
> > it? It's not immediately obvious to me how I could predict Is-Ib if
> > all you gave were the values for J and sdJ. In fact, I don't think
> > its possible. What you need is P(Is-Ib | J,sdJ,B,sdB), as I showed
> > above. Now you and F&W say that P(Is-Ib | J,sdJ,B,sdB) is Gaussian,
> > but where exactly does the Gaussian approximation come in?
>
> Sorry I don't see the problem. P(Is-Ib | J,sdJ) just means this is
> conditional on J & sdJ (i.e. _if_ we know J & sdJ _then_ we know Is-Ib
> with the specified probability). I hope we've established that Is-Ib
> is (approximately) Gaussian for reasonably large Is & Ib (from the
> difference of 2 Poissons). So Is-Ib is a sample of the population
> generated by a Gaussian distribution with mean = J and s.d. = sdJ.
I agree that Is-Ib is approximately Gaussian for reasonably large Is and
Ib. The problem is P(Is-Ib | J,sdJ). Why is the mean of this Gaussian
J? Why not J^2, or J/4? How do you derive the Gaussian
P(Is-Ib |J,sdJ)? The claim that the mean of Is-Ib is J requires some
sort of justification --- otherwise I might as well claim that
the mean of Is-Ib is the number of quarters in my pocket.
We both agreed that Is and Ib are dependent, and so that necessarily
means that P(Is-Ib | J,sdJ) cannot be the whole story. Dependence means
that the distribution of Is-Ib will also be a function of B and sdB,
unless you've integrated them out somehow (and even then the form
of the marginalized distribution depends on B and sdB).
> > For example, I've been pushing the model:
> >
> > Is = Ib + Ij
> >
> > where Ib comes from P(Ib|B,sdB) and Ij comes from P(Ij|J,sdJ).
> > Given this model, I can derive F&W (at least one way, there may be
> > others). From theory, both of those distributions are Poisson, and
> > so for large values of B and J, both distributions will approximate
> > a Gaussian. So Is will also be Gaussian, from N(Is | Ib+Ij,sds)
> > (sds^2=sdB^2+sdJ^2). It follows then that Id=Is-Ib will also be
> > Gaussian, from N(Id | J,sdd) (where sdd will be a bit complicated,
> > but it will be larger than sds).
> > > >
> > So it already seems to me that by using Is-Ib, the std dev of J will
> > be larger than it needs to be --- we should be able to do better if
> > we don't subtract the variates. And the way I derived this, the
> > Gaussian approximation is applied to both Ib and Ij, which is
> > exactly where we don't need it --- supposedly F&W applies to weak
> > observations, not strong ones.
>
> This all depends on your assumption that P(Ij|J,sdJ) is Poisson & I'm
> saying that you can't possibly know that (in fact you can use F & W to
> get an estimate & show that it's not). How could you prove it since J
> cannot be observed?
F&W assumes that P(Ij|J,sdJ) is Poisson. If you think F&W doesn't, then
explain how you get the Gaussian P(Is-Ib | J,sdJ). Again, why is the
mean of Is-Ib equal to J? I gave one derivation, based on the model
that P(Ij|J), P(Ib|B), and P(Ib'|B) are all Poisson. If you disagree,
then what is your derivation of the Gaussian P(Is-Ib | J,sdJ)?
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