On 10/7/06, Edward d'Auvergne <edward.dauvergne@xxxxxxxxx> wrote:
That was awesome, thanks for such an informative post. The archive of
your post, located at
https://mail.gna.org/public/relax-users/2006-10/msg00020.html
(Message-id: <481156b20610060807v768437bcw2d58f620652c1a0a@xxxxxxxxxxxxxx>),
should be a great resource for later on when asymmetric CSA tensors
are added to relax. I have a small number of questions/points below.
> Here's my little tutorial on CSA nomenclature. It's one of the most
> annoying things I have found in NMR and I've made it a significant part of
> my thesis to boot! So, I'm going to start from scratch and assume almost no
> previous understanding (it's a large message board, so why not?). First,
> CSA means two things: either 'chemical SHIFT anisotropy' or 'chemical
> SHIELDING anisotropy'. The differences come from, I believe, the discipline
> of NMR you are most familiar with: solution state people always display
> chemical shifts whereas solid state people tend towards the shielding side.
> These two terms are synonymous, and yet opposite each other. Shift tensors
> are always described as delta(ii) and shielding tensors are sigma(ii) where:
>
> delta(ii) = sigma(reference) - sigma(ii)
>
> The three principle values are then organized as del(11) > del(22) > del(33)
> and sig(11) < sig(22) < sig(33), ie del/sig(11) has the largest chemical
> shift value (most downfield) while the del/sig(33) value is the most
> shielded value (most upfield). To make things even more ridiculous, some
> people reverse the sigma organization (sig(11) > sig(22) > sig(33)) where
> del11 is now equivalent to sig33, etc! So, beware! As I said before, I
> prefer the delta organization as I am in solution state and there isn't the
> added sigma confusion.
>
> Now, the CSA tensor that defines the position of a peak in your NMR spectrum
> has both the isotropic and anisotropic component:
>
> delta(ij) = delta(iso) + delta(ani,ij)
>
> In relaxation and RCSA studies, the isotropic part is completely irrelevant
> and therefore we are only interested in the anisotropic part which is
> traceless (SUM[del(ani,ii) from 1to3] = 0) and assumed to be symmetric
> (del(ani,ij) = delta(ani,ji)), although this isn't necessarily true, but for
> molecular symmetry reasons almost always is true. From here on I'm just
> refering to the anisotropic part of the tensor and delta(11) is always
> positive and delta(33) is always negative.
>
> For a number of mathematical reasons, the CSA tensor is often reorganized
> once again in terms of x,y,z such |deltz(zz)| > |delta(yy)| > |delta(xx)|,
> similar to diffusion tensor and alignment tensor definitions. The asymmetry
> of the tensor is defined as:
>
> eta = (delta(xx) - delta(yy))/delta(zz)
>
> and varies from 0 (fully symmetric) to 1 (fully asymmetric). This is the
> parameter that is ubiquitously assymed to be zero in relaxation in order to
> keep the CSA term simple. In the situation where eta ~= 1, del(11) ~=
> del(33) or del(zz) ~= del(yy) and del(22) = del(xx) ~= 0.
Should this say "del(22) - del(iso) = del(xx) ~= 0"?
A few paragraphs up I mentioned that I'd only be refering to the anisotropic part of the tensor, but yet del(22) - diso = del(xx) - diso ~= 0.
And doesn't eta
~= 1 mean that the tensor is approximately axially symmetric about
axis defined by del(xx)? In the Diso, Da, and Dr notation I have used
for the diffusion tensor in relax, a Dr value of one is equivalent to
axial symmetry. A simple change of the axis names (Dx, Dy, and Dz)
will, for the same diffusion tensor, produce a Dr value of zero. It's
all in how you define the axis names. The diffusion tensor Dr
definition in relax is slightly different to the CSA eta definition
though, so eta = 1 may not be axially symmetric.
No, I think you're right. Just in the CSA tensor, the nomenclature isn't concerned with symmetry axes, only the magnitude of the principle components.
> Now the confusing
> part (if this hasn't been confusing enough!): there are many values proposed
> in the literature for the principle values of various nuclei and in some
> cases |del(11)| > |del(33)| and in others |del(33)| > |del(11)| ! Given the
> the CSA is a tensor, the orientation of all 3 components are orthogonal, so
> this seemingly harmless difference represents a 90 degree shift in the
> dominant principle component value in terms of x,y, and z. So, the question
> is, how does this relate to relaxation studies?
>
> In much of the literature (which assumes eta = 0), the ANISOTROPY of the
> tensor is given as:
>
> DELTAdelta = delta(zz) - (delta(xx)+delta(yy))/2
>
> or, with a little trivial math:
>
> DELTAdelta = (3/2) delta(zz)
>
> What you find in most relaxation equations for R1 and R2 is that the CSA
> constant is:
>
> (2/5) (omega(i) * DELTAdelta)^2 / 3 or equivalently (3/10) (omega(i) *
> delta(zz))^2
>
> (For those of you wondering where 2/5 comes from, I simply prefer to pull it
> out of the extended model free spectral density function.)
>
> Let's keep things simple for now (ie, isotropic diffusiong). If you want to
> include asymmetry, one should really use the following as the CSA constant:
>
> (3/10) omega(i)^2 delta(zz)^2 (1+eta^2/3)
>
> where delta(zz) * sqrt((1+(eta^2)/3)) is the asymmetric anisotropy, or CSAa.
> The CSAa is equivalently:
>
> sqrt((3/2) (del(xx)^2 + del(yy)^2 + del(zz)^2))
>
> So, for eta = 0, DELTAdelta = CSAa, but when eta != 0, CSAa > DELTAdelta.
> Now, how many biomolecules diffuse isotropically? Can I see a show of
> hands? Ok, I don't see a lot of hands, although ubiquitin is sitting in the
> back waving his/her arms wildly :). Here's the kicker, when a molecule
> tumbles anisotropically the asymmetry term of the CSA can NOT be pulled out
> of the spectral density function! The CSA constant for these need to be:
>
> (3/10) omega(i)^2 delta(zz)^2
>
> and the extended model free coefficients contain the asymmetry as well as
> euler angles relating the CSA tensor with the diffusion tensor (but let's
> not go there for right now).
Awww, that's a pity. Just as you were getting into the really
interesting stuff too!
> > Another thing that confuses me is that the dxx eigenvalue is
> > approximately zero ppm? Is that possible? Also, doesn't traceless
> > refer to a tensor in which the sum of the eigenvalues is zero? That
> > would only occur if the isotropic chemical shift is zero ppm. Sorry
> > about my ignorance, but what does 'traceless anisotropic part' and
> > 'anti-symmetric part' of the CSA tensor mean? Does this involve the
> > diagonalisation of the matrix and then the subtraction of the
> > isotropic shift times the identity matrix from the diagonalised CSA
> > matrix?
>
> I was trying to be a little scary, for some reason. The CSA tensor CAN,
> have an anti-symmetric part (delta(ij) != delta(ji)) which leads to terms in
> the correlation functions, f_lm(tau), and their Fourier transforms,
> g_lm(omega), whith l = 1. !! I'm not very good at describing, let alone
> understanding, this but hopefully you know what I mean. It's an interesting
> bit of trivia, but for now let's pretend I never said anything ;)
Said what? I didn't hear a thing!
Excellent. B-)
> > One last question, if the CSA tensor for RNA/DNA base 13C and 15N
> > spins is not rhombic, does this mean that the angle between the unique
> > axis of the CSA tensor and the XH dipolar interaction vector is ~90
> > degrees?
>
> More often than not, yes, but there's really two principle components to the
> tensor that are important. I don't quite understand the rhombic term, but
> I'm assuming it means assymetry (?)
Yep, rhombicity and asymmetry are synonymous.
> > As you can probably see, I am a bit lost. This is probably because of
> > my ignorance of the variety of symbols used to express the same thing
> > and the terminology used in each instance.
>
> That's completely understandable! The whole topic is filled with duplicate
> and triplicate nomenclature! If I had my way, I would chose yet another way
> to describe the CSA tensor that I didn't even discuss that keeps everything
> in terms of the delta(11) , (22), and (33) that I prefer. But that would
> take a lot of derivations from scratch on my part, so I'm leaving that alone
> ... for now ;) One day, I hope to confuse the literature even more!!! Mwa
> ha ha ha!! I think that's a prerequisite for anyone studying CSAs. I hope
> you've enjoyed this mini tutorial of my knowledge on CSA tensors.
It sounds like you have evil plans ;) Thanks again for the tutorial,
it was a very useful summary.
Edward
You're quite welcome. I had fun doing it.
Re: analysis of limited data sets