Multispin Cross‐Correlated Transverse Dipolar NMR Relaxation in Solution

Abstract

In this paper, we want to consider what would be involved in calculating the R₂ relaxation of amide protons in a protein caused by dipolar interactions with nearby protons, for which there are many. NMR textbooks give analytical equations and sometimes derivations for solution NMR relaxation due to dipolar interactions between two spins. There are also closed equations for dipolar interactions between three spins, which include relaxation interference, also known as cross‐correlated cross‐relaxation. We here derive an expression for interference between four spins. For larger systems, such as amide protons in a protein, we develop a local‐field methodology, from which solution relaxation interference can be computed for a basically limitless number of interacting spins.

1. Introduction

Several researchers have been interested in investigating whether proton-proton dipolar interaction in proteins can be utilized to obtain structural [1, 2] or dynamical information [3]. Here, we want to consider what would be involved in calculating the R₂ relaxation of amide protons in a protein caused by dipolar interactions with other protons and, ultimately, compare those with experiment [4]. NMR textbooks give closed equations and sometimes derivations for solution NMR relaxation due to dipole-dipole relaxation between two spins and for chemical shift anisotropy (CSA) relaxation. There are also closed equations for relaxation due to interactions between three dipoles [5] and two dipoles and the CSA [6]. In these cases, relaxation interference, also known as cross-correlated cross-relaxation, has been taken into account. Dipolar-dipolar relaxation interference has been described very early in the history of NMR [7, 8] and has been exploited to achieve line narrowing [9] and structural information [1, 2]. CSA-dipolar cross-correlated relaxation is at the core of the line narrowing in the TROSY experiment [10].

Here, we derive an analytical expression for dipolar interference between four spins. For larger systems, such as amide protons in a protein, we develop a local-field methodology, from which solution relaxation interference can be computed for a basically limitless number of interacting spins.

2. Theory

Let us start by considering the R₂ relaxation of spin I in a rigid molecule with three protons, S-I-Q, subject to fluctuating dipolar interactions IS and IQ. We assume here that I, S, and Q have different chemical shifts and are thus similar, but “unlike.” The first step is to add the relaxation rates due to the IS and IQ dipolar interactions, as given by equation (A.24) of the Appendix. 1 $\begin{array}{}{R}_2^{I\text{total}}=R_2^{\text{IS}}+R_2^{\text{IQ}},\end{array}$ 2 $\begin{array}{}{R}_2^{I\text{total}}=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2\left\{{\left(\frac{{\gamma }_H{\gamma }_H\hslash }{r_{\text{IS}}^3}\right)}^2+{\left(\frac{{\gamma }_H{\gamma }_H\hslash }{r_{\text{IQ}}^3}\right)}^2\right\}\left\{596J\left(0\right)+J\left({\omega }_H\right)+J\left({\omega }_{2H}\right)\right\}.\end{array}$

However, in a rigid molecule, the stochastic fluctuations of the IS and IQ dipolar interactions are 100% correlated and relaxation interference occurs. Relaxation interference is also called cross-correlated cross-relaxation or cross-correlated relaxation.

Interference between two dipolar mechanisms can be qualitatively understood from Figure 1. Here, we consider a linear three-spin system with spin I in the center and spin S and Q equidistant from I.

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In this case, there are two relaxation rates for spin I, one for molecule A, where the dipolar fields of S and Q at the location I cancel independent of orientation with respect to the magnetic field, and another for molecule B, where the dipoles reinforce each other and fluctuate strongly depending on orientation. This is expressed as 3 $\begin{array}{}{R}_2^{I\text{total}}=R_2^{\text{IS}}+R_2^{\text{IQ}}\pm R_{2\text{CC}}^{\text{IS}-\text{IQ}}.\end{array}$

Of course, the dipoles flip by R₁ processes. For macromolecules, R₁ rates are much slower than R₂ rates. Thus, for R₂ cross-correlations, one may assume that orientations of the dipoles remain set during the typical R₂ relaxation time.

If the different dipolar permutations also give rise to different resonances, by either scalar or residual static dipolar coupling, one observes four NMR lines for I. The two outer lines have equal linewidths, and the two inner lines also have equal, but different, linewidths. Which set will be the broader one depends on the molecular geometry and the sign of the (scalar) coupling. If no such coupling exists, one will observe a superposition of a broader and narrower line for I.

In order to obtain quantitative values for the dipolar-dipolar cross-correlation, one starts with expanding the double commutator master equation for R₂ relaxation, as shown in equation (A.16) of the Appendix. We mostly follow a formalism as introduced by [5, 11].

For two R₂ relaxation mechanisms IS and IQ, modulated by the same motions, the relaxation master equation is 4 $\begin{array}{}\frac{d⟨I_{+}⟩}{\text{d}t}=-\frac{1}{2}{\displaystyle \sum _{m=-2}^{m=+2}\left\{⟨\left[\left[I_{+},T_{\text{mIS}}+T_{\text{mIQ}}\right],T_{\text{mIS}}^{†}+T_{\text{mIQ}}^{†}\right]⟩\right\}{j}_m{\omega }_m},\end{array}$ where T_mIS are the tensor operators of the perturbing Hamiltonian, which for dipolar interaction between spins I and S are given by 5a $\begin{array}{}{T}_0=2I_z{S}_z-\frac{1}{2}\left(I_{+}{S}_{-}+I_{-}{S}_{+}\right),\end{array}$ 5b $\begin{array}{}{T}_{\pm 1}=\mp \sqrt{\frac{3}{2}}\left(I_{\pm }{S}_z+I_z{S}_{\pm }\right),\end{array}$ 5c $\begin{array}{}{T}_{\pm 2}=\sqrt{\frac{3}{2}}\left(I_{\pm }{S}_{\pm }\right),\end{array}$ with similar equations for T_mIQ.

j_mω_m are the spectral densities of the molecular motion at the frequencies of the bilinear spin operators including magnitude terms.

In the expansion, the “diagonal” terms such as 6a $\begin{array}{}{\displaystyle \sum _{m=-2}^{m=+2}\left\{⟨\left[\left[I_{+},T_{\text{mIS}}\right],T_{\text{mIS}}^{†}\right]⟩\right\}{j}_m{\omega }_m},\end{array}$ 6b $\begin{array}{}{\displaystyle \sum _{m=-2}^{m=+2}\left\{⟨\left[\left[I_{+},T_{\text{mIQ}}\right],T_{\text{mIQ}}^{†}\right]⟩\right\}{j}_m{\omega }_m},\end{array}$ and their complex conjugates will give rise to the autocorrelation rates $R_2^{\text{IS}}$ and $R_2^{\text{IQ}}$, as shown in equation (A.22) of the Appendix.

In the cross terms, 7a $\begin{array}{}{\displaystyle \sum _{m=-2}^{m=+2}\left\{⟨\left[\left[I_{+},T_{\text{mIS}}\right],T_{\text{mIQ}}^{†}\right]⟩\right\}{j}_m^{\text{cc}}{\omega }_m},\end{array}$ 7b $\begin{array}{}{\displaystyle \sum }_{m=-2}^{m=+2}\left\{⟨\left[\left[I_{+},T_{\text{mIQ}}\right],T_{\text{mIS}}^{†}\right]⟩\right\}{j}_m^{\text{cc}}{\omega }_m,\end{array}$ and their complex conjugates, only spin operators with exactly the same precession frequency will be relevant, i.e., only the pair (I_zS_z, I_zQ_z) and the four terms I₊S_z, I₋S_z, I₊Q_z, and I₋Q_z.

We obtain the following equation from equations (6a) and (6b): 8 $\begin{array}{}-\frac{d⟨I_{+}⟩}{\text{d}t}=⟨4\left[\left[I_{+},I_z{S}_z\right],I_z{Q}_z\right]⟩j_0^{\text{cc}}\left(0\right)+\frac{3}{2}⟨\left[\left[I_{+},I_{-}{S}_z\right],I_{+}{Q}_z\right]⟩j_1^{\text{cc}}\left({\omega }_I\right),\end{array}$ where the ½ term is dropped because there are two equivalent cross terms, as given in equations (7a) and (7b).

The cross-correlation spectral densities are 9 $\begin{array}{}{j}_m^{\text{cc}}{\omega }_m=-\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)\left(\frac{{\gamma }_I{\gamma }_Q\hslash }{r_{\text{IQ}}^3}\right)P_2\cos \left({\theta }_{\text{IS}-\text{IQ}}\right)J\left({\omega }_m\right),\end{array}$ with, for isotropic motion, 10 $\begin{array}{}J\left({\omega }_m\right)=\frac{2}{5}\frac{{\tau }_c}{1+{\omega }_m^2{\tau }_c^2},\end{array}$ where the P₂cos(θ_IS−IQ) = 1/2(3  cos²(θ_IS−IQ) − 1) term originates from transforming the IQ dipolar vector into the IS frame or vice versa and using the addition theorem of spherical harmonics. Here, θ_IS−IQ is the angle between vectors IS and IQ and τ_c is the (rotational) correlation time.

In contrast to the R₂ autocorrelation, the double commutators drive cross-relaxation 11a $\begin{array}{}-\frac{d⟨I_{+}⟩}{\text{d}t}=⟨4I_{+}{S}_z{Q}_z⟩j_0^{\text{cc}}\left(0\right)+\frac{3}{4}⟨4I_{+}{S}_z{Q}_z⟩j_1^{\text{cc}}\left({\omega }_I\right),\end{array}$ and back 11b $\begin{array}{}-\frac{d⟨4I_{+}{S}_z{Q}_z⟩}{\text{d}t}=⟨I_{+}⟩j_0^{\text{cc}}\left(0\right)+⟨I_{+}⟩j_1^{\text{cc}}\left({\omega }_I\right).\end{array}$

One ends up with a set of coupled differential equations: 12 $\begin{array}{}\frac{d}{\text{d}t}\left(\begin{array}{c}{I}_{+}\\ 4I_{+}{S}_z{Q}_z\end{array}\right)=-\left[\begin{array}{cc}{R}_{2I}^{\text{IP}}& R_{2I}^{\text{CC}}\\ R_{2I}^{\text{CC}}& R_{2I}^{\text{AP}}\end{array}\right]\left(\begin{array}{c}{I}_{+}\\ 4I_{+}{S}_z{Q}_z\end{array}\right).\end{array}$

Here, $R_{2I}^{\text{IP}}$ and $R_{2I}^{\text{AP}}$ are the in-phase (IP) R₂ rate for the <I₊> coherence and antiphase (AP) R₂ rate for the 4 <I₊S_zI_z> coherence. The antiphase terms relax faster because of the extra S_z and Q_zR₁ relaxation. For macromolecules, the difference can be neglected because R₁ is small as compared to R₂, and one can easily diagonalize the rate matrix and obtain relaxation “eigenvectors” 13 $\begin{array}{}\frac{d}{\text{d}t}\left(\begin{array}{c}\left(I_{+}+4I_{+}{S}_z{Q}_z\right)\\ \left(I_{+}-4I_{+}{S}_z{Q}_z\right)\end{array}\right)=-\left[\begin{array}{cc}{R}_2^{\text{IS}}+R_2^{\text{IQ}}+R_{2I}^{\text{CC}}& 0\\ 0& R_2^{\text{IS}}+R_2^{\text{IQ}}-R_{2I}^{\text{CC}}\end{array}\right]\left(\begin{array}{c}\left(I_{+}+4I_{+}{S}_z{Q}_z\right)\\ \left(I_{+}-4I_{+}{S}_z{Q}_z\right)\end{array}\right),\end{array}$ with 14 $\begin{array}{}{R}_{2I}^{\text{CC}}=\frac{1}{4}{\left(\frac{{\mu }_0}{4\pi }\right)}^2\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{IS}^3}\right)\left(\frac{{\gamma }_I{\gamma }_Q\hslash }{r_{IQ}^3}\right)P_2\cos \left({\theta }_{\text{IS}-\text{IQ}}\right)\left\{43J\left(0\right)+J\left({\omega }_I\right)\right\}.\end{array}$

The total R₂ relaxation for proton i, as stated before in equation (3), is then given by 15 $\begin{array}{}{R}_2^{I\text{total}}=R_2^{\text{IS}}+R_2^{\text{IQ}}\pm R_{2\text{CC}}^{\text{IS}-\text{IQ}}.\end{array}$

One notices that ¹H IS-autorelaxation contains a 5J(0) term (equation (2)), while the dipolar cross-correlation has a 4J(0) term (equation (14)). The extra J(0) in autorelaxation is due to the fact that J(0) and J(ω_I − ω_S) are for two protons similar enough for the spectral densities to add (see Appendix), but (ω_I − ω_S) and (ω_I − ω_Q) are different enough for the spin operators to diphase and not cross-correlate. Thus, there will not be a complete cancelation of relaxation, as suggested in the situation of Figure 1. It can still happen, if the differences in distances IS and IQ would make up for the 4/5 term.

If one cannot make the approximation that $R_{2I}^{\text{IP}}$ and $R_{2I}^{\text{AP}}$ are equal, the rates become 16 $\begin{array}{}{R}_2^{I\text{total}}=\frac{1}{2}{R}_{2I}^{\text{IP}}+\frac{1}{2}{R}_{2I}^{\text{AP}}\pm \sqrt{\frac{R_{2I}^{\text{IP}}x{R}_{2I}^{\text{IP}}}{4}-\frac{R_{2I}^{\text{IP}}x{R}_{2I}^{\text{AP}}}{2}+\frac{R_{2I}^{\text{AP}}x{R}_{2I}^{\text{AP}}}{4}+R_{2\text{CC}}^{\text{IS}-\text{IQ}}x{R}_{2\text{CC}}^{\text{IS}-\text{IQ}}}.\end{array}$

Let us now extend this formalism to a 4-spin system, with a central spin I and 3 other spins S, Q, and P in its vicinity.

From the master equation, we arrive at three cross terms: 17 $\begin{array}{c}-\frac{d⟨I_{+}⟩}{\text{d}t}=⟨4\left[\left[I_{+},I_z{S}_z\right],I_z{Q}_z\right]⟩j_0^{\text{cc}}\left(0\right)+\frac{3}{2}⟨\left[\left[I_{+},I_{-}{S}_z\right],I_{+}{Q}_z\right]⟩j_1^{\text{cc}}\left({\omega }_I\right)+⟨4\left[\left[I_{+},I_z{S}_z\right],I_z{P}_z\right]⟩j_0^{\text{cc}}\left(0\right)+\frac{3}{2}⟨\left[\left[I_{+},I_{-}{S}_z\right],I_{+}{P}_z\right]⟩j_1^{\text{cc}}\left({\omega }_I\right)\\ +⟨4\left[\left[I_{+},I_z{Q}_z\right],I_z{P}_z\right]⟩j_0^{\text{cc}}\left(0\right)+\frac{3}{2}⟨\left[\left[I_{+},I_{-}{Q}_z\right],I_{+}{P}_z\right]⟩j_1^{\text{cc}}\left({\omega }_I\right),\end{array}$ leading to terms such as 4I₊S_zQ_z, 4I₊S_zP_z, and 4I₊P_zQ_z, driven by the rates 18a $\begin{array}{}{R}_{2I}^{\text{IS}-\text{IQ}}=\frac{1}{4}{\left(\frac{{\mu }_0}{4\pi }\right)}^2\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{IS}^3}\right)\left(\frac{{\gamma }_I{\gamma }_Q\hslash }{r_{IQ}^3}\right)P_2\cos \left({\theta }_{\text{IS}-\text{IQ}}\right)\left\{43J\left(0\right)+J\left({\omega }_I\right)\right\},\end{array}$ 18b $\begin{array}{}{R}_{2I}^{\text{IS}-\text{IP}}=\frac{1}{4}{\left(\frac{{\mu }_0}{4\pi }\right)}^2\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)\left(\frac{{\gamma }_I{\gamma }_P\hslash }{r_{\text{IP}}^3}\right)P_2\cos \left({\theta }_{\text{IS}-\text{IP}}\right)\left\{43J\left(0\right)+J\left({\omega }_I\right)\right\},\end{array}$ 18c $\begin{array}{}{R}_{2I}^{\text{IP}-\text{IQ}}=\frac{1}{4}{\left(\frac{{\mu }_0}{4\pi }\right)}^2\left(\frac{{\gamma }_I{\gamma }_Q\hslash }{r_{IQ}^3}\right)\left(\frac{{\gamma }_I{\gamma }_P\hslash }{r_{IP}^3}\right)P_2\cos \left({\theta }_{\text{IQ}-\text{IP}}\right)\left\{43J\left(0\right)+J\left({\omega }_I\right)\right\}.\end{array}$

This appears to lead to eight different relaxation rates $\left(R_2^{\text{IS}}+R_2^{\text{IQ}}+R_2^{\text{QS}}\pm R_{2\text{CC}}^{\text{IS}-\text{IQ}}\pm R_{2\text{CC}}^{\text{IS}-\text{IP}}\pm R_{2\text{CC}}^{\text{IP}-\text{IQ}}\right)$. However, that cannot be right; adding one more dipole P to the S-I-Q situation should just give rise to two more rates, four in total. So, one should not stop here and construct the full relaxation matrix.

Also, one must take cross-relaxation between the different four-spin terms into account (here only showing some J(0) terms), 19a $\begin{array}{}-\frac{d⟨4I_{+}{S}_z{Q}_z⟩}{\text{d}t}=⟨4\left[\left[4I_{+}{S}_z{Q}_z,I_z{S}_z\right],I_z{P}_z\right]⟩=⟨4I^{+}{Q}_z{P}_z⟩,\end{array}$ 19b $\begin{array}{}-\frac{d⟨4I_{+}{S}_z{Q}_z⟩}{\text{d}t}=⟨4\left[\left[4I_{+}{S}_z{Q}_z,I_z{Q}_z\right],I_z{P}_z\right]⟩=⟨4I_{+}{S}_z{P}_z⟩,\end{array}$ 19c $\begin{array}{}-\frac{d⟨4I_{+}{S}_z{P}_z⟩}{\text{d}t}=⟨4\left[\left[4I_{+}{S}_z{P}_z,I_z{P}_z\right],I_z{Q}_z\right]⟩=⟨4I_{+}{S}_z{Q}_z⟩,\end{array}$ and three reverse processes.

The differential equations in matrix form will now become 20 $\begin{array}{}\frac{d}{\text{d}t}\left(\begin{array}{c}{I}_{+}\\ 4I_{+}{S}_z{Q}_z\\ \begin{array}{c}4I_{+}{S}_z{P}_z\\ 4I_{+}{Q}_z{P}_z\end{array}\end{array}\right)=-\left|\begin{array}{ccc}{R}_{2I}& R_{2\text{cc}}^{\text{SQ}}& \begin{array}{cc}{R}_{2\text{cc}}^{\text{SP}}& R_{2\text{cc}}^{\text{QP}}\end{array}\\ R_{2\text{cc}}^{\text{SQ}}& R_{2I}& \begin{array}{cc}{R}_{2\text{cc}}^{\text{QP}}& R_{2\text{cc}}^{\text{SP}}\end{array}\\ \begin{array}{c}{R}_{2\text{cc}}^{\text{SP}}\\ R_{2\text{cc}}^{\text{QP}}\end{array}& \begin{array}{c}{R}_{2\text{cc}}^{\text{QP}}\\ R_{2\text{cc}}^{\text{SP}}\end{array}& \begin{array}{c}\begin{array}{cc}{R}_{2I}& R_{2\text{cc}}^{\text{SQ}}\end{array}\\ \begin{array}{cc}{R}_{2\text{cc}}^{SQ}& R_{2I}\end{array}\end{array}\end{array}\right|\left(\begin{array}{c}{I}_{+}\\ 4I_{+}{S}_z{Q}_z\\ \begin{array}{c}4I_{+}{S}_z{P}_z\\ 4I_{+}{Q}_z{P}_z\end{array}\end{array}\right).\end{array}$

Here, we are neglecting the differences between R₂ in-phase and antiphase relaxation. This matrix has four eigenvalues, which correspond to what must be physically correct. While the exact eigenvalues of this matrix could potentially be analytically obtained, it becomes harder when differences between R₂ in-phase and antiphase relaxation have to be considered as well. However, because matters will quickly become more complicated when even more spins interact, we will not attempt to derive an analytical solution for the four-spin case here. For instance, in a protein, there are typically at least 20 protons around every amide proton in a 6 Å sphere which would all have to be taken into account. In such a case, one would need to diagonalize a 262144∗262144 matrix (2^20 − 2).

We must take another approach. To arrive at an estimation for the effects in a multiproton spin system, we will start from a “solid-state NMR” point of view. We calculate $B_{\text{loc}\left(i\right)}^{\mathrm{\Omega }}$, the net local magnetic field at center proton i due to the M surrounding protons j [12] for a certain orientation of the molecule in the external magnetic field: 21 $\begin{array}{}{B}_{\text{loc}\left(i\right)}^{\Omega }=\frac{{\mu }_0}{4\pi }{\displaystyle \sum _{j\ne i}^{j=M}\mathbb{D}\left(\frac{{\gamma }_i{\gamma }_j}{r_{ij}^3}\right)P_2\left(\cos {\theta }_{ij}\right).}\end{array}$

Here, θ_ij is the angle between the internuclear vector ij and the magnetic field direction Ω in the molecular frame. $\mathbb{D}$ represents a particular distribution of the signs of the z-components of the dipoles j. If one varies the magnetic field direction according to a sphere distribution and adds the results, one obtains the powder pattern for that particular distribution $\mathbb{D}$. Subsequently, one coadds all powder patterns for different values of $\mathbb{D}$ and normalizes to arrive at the “cross-correlated” dipolar powder pattern for the ¹HN under consideration.

It is the time dependence of B_loc, as caused by molecular motion, that drives the solution NMR dipolar relaxation. The R₂ relaxation is then obtained as the second moment of the (cross-correlated) powder pattern [12]: 22 $\begin{array}{}{R}_{2i}^{\text{solution}}=4{\tau }_c{\displaystyle \sum _{\Omega }{\left(B_{\text{loc}\left(i\right)}-B_{\text{loc}\left.\left(i\right)\right)}^{\Omega }\right)}^2},\end{array}$ where the brackets indicate average overall orientations. In practice, we permute only the signs of the eight closest protons (256 different distributions $\mathbb{D}$) and treat the spins further out with a single random distribution where spins up and down are on average equal. So, each distribution then gives rise to an individual resonance for i with its own R₂ rate. The sum of all of those creates the inhomogeneous sum-line for resonance i.

3. Verification

We verified the algorithm with a three-atom arrangement (see Table 1) with results in Figure 2. In this figure, the green points were calculated from equations (14) and (15) for dipolar-dipolar cross-correlation, while the drawn black line was calculated using the “solid-state” approach. The results are identical. In red is a relaxation curve calculated from the straight addition of R₂ (IS) and R₂ (IQ) (see equation (1)). It is clear that the cross-correlation cannot be neglected, except when one considers the first part of the curve. Fitting a single exponential against a complete cross-correlated relaxation curve (dashed line) yields an erroneous rate.

Table 1 Coordinates of an arbitrary three-spin arrangement (Å). d_IS = 2 Å, d_IQ = 2.236 Å, and P₂ cos(θ_IS−IQ) = 0.70.

	x	y	z
Center spin (I)	0	0	0
Neighbor (S)	0	2	0
Neighbor (Q)	0	−2	1

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In Figure 3(a), we show calculated R₂ relaxation rates for the HN of Asp45 of the protein GB1. This amide has 21 proton neighbors within a sphere of 6 Å. Here, the difference between relaxation curves with and without cross-correlations is smaller than the example case in Figure 2. From this, one would tend to conclude that many protons cancel each other’s cross-correlations. However, that is not necessarily true; in Figure 3(b), we show calculated R₂ relaxation rates for the HN of Asn35 of the protein GB1. This amide has more (36) proton neighbors within a sphere of 6 Å, yet the difference between relaxation curve with and without cross-correlations is larger than for Asp45. This happens because the R₂ relaxation of Asn35 is dominated by two close-by protons.

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This manuscript was written in anticipation of the interpretation of experimental T_1rho data for protein amide protons, which we hope may contribute to more complete understanding of protein molecular dynamics (E.R.P. Zuiderweg and D.A. Case, in preparation). We conclude from the current analysis that one can still fit a single exponential to the beginning of the T_1rho relaxation curves, even when many protons interact. This result is the same as for a 3-spin network, but that was not previously demonstrated. However, how far down a relaxation curve can go varies from proton to proton environment. Judging from Figure 3(b), one should limit the recording of T_1rho relaxation curves to values larger than 0.8xI₀, where I₀ is the initial value of the decay curve.

4. Description of the Codes

The computer program requires as input a “protonated” PDB file, the radius of the sphere of protons around the amide protons one is interested in, the rotational correlation time, and the spectrometer frequency. Basically, the program consists of four nested loops: amides, protons around amides, permutation of dipole signs of these surrounding protons, and rotation of the magnetic field vector in the molecular frame.

A set of 10 nested loops permutes the dipolar signs of the closest 10 hydrogens (1024 distributions). The more remote hydrogens in the sphere (if any) have their dipolar signs assigned according to a 50% random chance. The local dipolar field at a certain ¹HN due to the surrounding protons in a certain permutation $\mathbb{D}$ of surrounding dipoles is calculated according to equation (20). Here, the program takes the differences between “like” and “unlike” spins (see Appendix) into account. Then, the program calculates, according to equation (22), the R₂ relaxation rate due to that permutation, by rotating the external field direction (the z-axis) through the molecular frame using an isotropic spherical distribution (5000 orientations) ().

The relaxation rate for that permutation $\mathbb{D}$ is then used to compute a R₂ relaxation curve. The computed R₂ relaxation curves for all different permutations $\mathbb{D}$ are then added to obtain the complete R₂ relaxation curve, as shown in Figures 2 and 3.

The program is written in Fortran 90 and contains no references to outside libraries. The source code is available from the author.

Appendix

Appendix

This appendix presents a refresher on dipolar R₂ relaxation, mostly following the formalism, as developed by Goldman [11] and further extended by [13], to help follow the algebra in the main body of the paper.

When the density operator σ is subject to both a time-independent (eigen) Hamiltonian ℋ₀ and a time-dependent perturbing dipolar Hamiltonian ℋ₁(t), its evolution, in units of ℏ, is given by A.1 $\begin{array}{}\frac{d}{\text{d}t}\sigma \left(t\right)=-i\left[{\mathrm{\mathscr{H}}}_0+{\mathrm{\mathscr{H}}}_1\left(t\right),\sigma \left(t\right)\right].\end{array}$

Transforming to the rotating frame of the time-independent Hamiltonian, one obtains A.2 $\begin{array}{}\frac{d}{\text{d}t}\tilde{\sigma }\left(t\right)=-i\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t\right),\tilde{\sigma }\left(t\right)\right],\end{array}$ with the time-dependent Hamiltonian also transformed to the rotating frame. One keeps in mind that this equation describes an ensemble average. First, one imposes on equation (A.2) that the density operator cannot evolve (“relax”) when it was not perturbed in the first place (e.g., by a r.f. pulse). Thus, one makes the substitution A.3 $\begin{array}{}\frac{d}{\text{d}t}\tilde{\sigma }\left(t\right)=-i\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t\right),\tilde{\sigma }\left(0\right)-{\tilde{\sigma }}_{\text{eq}}\right],\end{array}$ where ${\tilde{\sigma }}_{\text{eq}}$ is the density operator at equilibrium, which is not evolving under ℋ₀ (i.e., the usual “high temperature approximation with ${\tilde{\sigma }}_{\text{eq\hspace{0.17em}}}\sim I_z$).

One uses the Hausdorff expansion to integrate equation (A.3): A.4 $\begin{array}{}\tilde{\sigma }\left(t\right)=\tilde{\sigma }\left(0\right)-i{\displaystyle {\int }_0^t\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t\prime \right),\left.\tilde{\sigma }\left(0\right)-{\tilde{\sigma }}_{\text{eq}}\right]\right]\text{d}{t}^{\prime }}-{\displaystyle {\int }_{\text{0}}^t{\displaystyle {\int }_0^{t\prime }\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t^{\prime }\right),\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t^{″}\right),\tilde{\sigma }\left(0\right)-{\tilde{\sigma }}_{\text{eq}}\right]\right]\text{d}{t}^{\prime }\text{d}{t}^{″}+\dots }}.\end{array}$

Since the dipolar Hamiltonian has zero average in solution, the second term in equation (A.4) vanishes. The fourth and higher-order terms also vanish because the perturbing Hamiltonian causes only small changes of the density operator, allowing the expansion to converge rapidly. One thus obtains the formal relaxation equation: A.5 $\begin{array}{}\tilde{\sigma }\left(t\right)=\tilde{\sigma }\left(0\right)-{\displaystyle {\int }_0^t{\displaystyle {\int }_t^{t\prime }\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t^{\prime }\right),\left[\tilde{{\mathrm{\mathscr{H}}}_1}\left(t^{″}\right),\tilde{\sigma }\left(0\right)-{\tilde{\sigma }}_{\text{eq}}\right]\right]\text{d}{t}^{\prime }\text{d}{t}^{″}}}.\end{array}$

The perturbing Hamiltonian fluctuates not only by the fluctuating dipole-dipole interactions between spins I caused by random molecular motion captured in the terms $F_m^{\text{DD}}\left(t\right)$ but also by frequencies of the spin operators I, S, and I₊S₋ (ωI, ωS, ωI−ωS, etc.).

It can be abbreviated as A.6 $\begin{array}{}\tilde{{\mathrm{\mathscr{H}}}_{\text{IS}}^{\text{DD}}}\left(t\right)={\displaystyle \sum _{m=-2}^{m=2}{F}_m^{\text{DD}}\left(t\right)\widetilde{T_{\text{mIS}}}}.\end{array}$

T_mIS are the tensor operators of the perturbing Hamiltonian, which for dipolar interaction between spins I and S are given by A.7a $\begin{array}{}{T}_0=2I_z{S}_z-\frac{1}{2}\left(I_{+}{S}_{-}+I_{-}{S}_{+}\right),\end{array}$ A.7b $\begin{array}{}{T}_{\pm 1}=\mp \sqrt{\frac{3}{2}}\left(I_{\pm }{S}_z+I_z{S}_{\pm }\right),\end{array}$ A.7c $\begin{array}{}{T}_{\pm 2}=\sqrt{\frac{3}{2}}\left(I_{\pm }{S}_{\pm }\right).\end{array}$

On expanding, this becomes A.8 $\begin{array}{c}\tilde{{\mathrm{\mathscr{H}}}_{\text{IS}}^{\text{DD}}}\left(t\right)=+2F_0^{\text{IS}}\left(t\right)I_z{S}_z-\frac{1}{2}{F}_0^{\text{IS}}\left(t\right)\left(I_{+}{S}_{-}+I_{-}{S}_{+}\right)e^{i\left({\omega }_I-{\omega }_S\right)t}+\sqrt{\frac{3}{2}}{F}_1^{\text{IS}}\left(t\right)\left(I_{+}{S}_z\right)e^{i\left({\omega }_I\right)t}-\sqrt{\frac{3}{2}}{F}_1^{\text{IS}}\left(t\right)\left(I_{-}{S}_z\right)e^{i\left({\omega }_I\right)t}\\ +\sqrt{\frac{3}{2}}{F}_1^{\text{IS}}\left(t\right)\left(I_z{S}_{+}\right)e^{i\left({\omega }_S\right)t}-\sqrt{\frac{3}{2}}{F}_1^{\text{IS}}\left(t\right)\left(I_z{S}_{-}\right)e^{i\left({\omega }_S\right)t}\\ +\sqrt{\frac{3}{2}}{F}_2^{\text{IS}}\left(t\right)\left(I_{+}{S}_{+}\right)e^{i\left({\omega }_I+{\omega }_S\right)t}+\sqrt{\frac{3}{2}}{F}_2^{\text{IS}}\left(t\right)\left(I_{-}{S}_{-}\right)e^{i\left({\omega }_I+{\omega }_S\right)t},\end{array}$ with A.9a $\begin{array}{}{F}_0^{\text{IS}}\left(t\right)=-\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)\frac{31{\cos }^2\theta \left(t\right)-}{2}=F_0^{\text{IS}\ast }\left(t\right),\end{array}$ A.9b $\begin{array}{}{F}_{\pm 1}^{\text{IS}}\left(t\right)=\pm \left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)\sqrt{\frac{3}{2}}\sin \theta \left(t\right)\cos \theta \left(t\right)e^{\pm i\phi \left(t\right)}=-F_{\mp 1}^{\text{IS}\ast }\left(t\right),\end{array}$ A.9c $\begin{array}{}{F}_{\pm 2}^{\text{IS}}\left(t\right)=-\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)\sqrt{\frac{3}{8}}{\sin }^2\theta \left(t\right)e^{\pm 2i\phi \left(t\right)}=F_{\mp 2}^{\text{IS}\ast }\left(t\right).\end{array}$

Here, μ₀ is the permittivity of vacuum, ℏ is the Planck’s constant divided by 2π, γ′s are the gyromagnetic ratios, and r_IS is the distance between the dipoles. The angles θ(t) and φ(t) are fluctuating polar angles between the IS vector and the direction of the external magnetic field.

In terms of equations (A.6), (A.7a), (A.7b), and (A.7c), equation (A.5) now becomes A.10 $\begin{array}{}\tilde{\sigma }\left(t\right)-\tilde{\sigma }\left(0\right)={\displaystyle \sum _{m,n=-2}^{m,n=2}\left[T_m\left[T_n^{†},\left(\tilde{\sigma }\left(0\right)-{\tilde{\sigma }}_{\text{EQ}}\right)\right]\right]\times {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{t\prime }{F}_m^{\text{IS}}\left(t^{\prime }\right)F_n^{\text{IS}\ast }\left(t^{″}\right)e^{i\left({\omega }_{m}t\prime -{\omega }_n{t}^{″}\right)t}\text{d}{t}^{″}\text{d}{t}^{\prime }}}}.\end{array}$

Here, one recognizes ${\displaystyle {\int }_0^t{F}_m^{\text{IS}}\left(t^{\prime }\right)F_n^{\text{IS}\ast }\left(t^{″}\right)\text{d}{t}^{\prime }}$ as a time correlation function (AC(τ)) of $F_m^{\text{IS}}\left(t^{\prime }\right)$ (make the substitution tʺ = tʹ + τ). One makes the common assumption that time correlation functions decay exponentially: A.11 $\begin{array}{}\text{AC}\left(\tau \right)=\text{AC}\left(0\right)e^{-\left(\tau \right)/{\tau }_c},\end{array}$ with a single rotational correlation time τ_c in case of isotropic motion

The second integral in equation (A.10) is the Fourier transform of that time correlation function, which takes the shape of a Lorentzian: A.12 $\begin{array}{}J\left(\omega \right)=\frac{2}{5}\frac{{\tau }_c}{1+{\omega }^2{\tau }_c^2}.\end{array}$

From the 25 terms in equation (A.10), only the ones with equal frequencies $e^{i\left({\omega }_{m}t\prime -{\omega }_m{t}^{″}\right)t}$ contribute. This is called the Redfield kite, named after the author of one of the earliest papers on NMR relaxation [14]. It suffices to just take the five diagonal terms into account and drop the second summation.

One arrives at the “master equation” of relaxation [11] A.13 $\begin{array}{}\frac{d⟨\tilde{\sigma }⟩}{\text{d}t}=-\frac{1}{2}{\displaystyle \sum }_{m=-2}^{m=+2}\left[T_{\text{mIS}},\left[T_{\text{mIS}}^{†},\left(\tilde{\sigma }\left(0\right)-{\tilde{\sigma }}_{\text{EQ}}\right)\right]\right]j_m,\end{array}$ with A.14 $\begin{array}{}{j}_m=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2{\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)}^{2}J\left({\omega }_m\right)=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2{\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{\text{IS}}^3}\right)}^2\frac{2}{5}\frac{{\tau }_c}{1+{\omega }_m^2{\tau }_c^2},\end{array}$ assuming isotropic motion.

It is interesting to consider the time development of an observable Q: A.15 $\begin{array}{}\frac{d⟨Q⟩}{\text{d}t}=\text{Trace}\left\{Q\frac{d⟨\tilde{\sigma }⟩}{\text{d}t}\right\}.\end{array}$

The master equation for the time development of an observable Q is derived from (A.12) using commutation algebra and no new assumptions: A.16 $\begin{array}{}\frac{d⟨Q⟩}{\text{d}t}=-\frac{1}{2}{\displaystyle \sum _{m=-2}^{m=+2}\left\{⟨\left[\left[Q,T_{\text{mIS}}\right],T_{\text{mIS}}^{†}\right]⟩-{⟨\left[\left[Q,T_{\text{mIS}}\right],T_{\text{mIS}}^{†}\right]⟩}_{\text{eq}}\right\}{j}_{mm}}.\end{array}$

Now, we are finally ready to derive the R₂ relaxation rate for <I₊>. We start by setting the equilibrium term to zero: A.17 $\begin{array}{}\frac{d⟨I_{+}⟩}{\text{d}t}=-\frac{1}{2}{\displaystyle \sum _{m=-2}^{m=+2}⟨\left[\left[I_{+},T_{\text{mIS}}\right],T_{\text{mIS}}^{†}\right]⟩j_{mm}}.\end{array}$

Expanding equation (A.15), one obtains A18 $\begin{array}{}-\frac{d⟨I_{+}⟩}{\text{d}t}=\frac{1}{2}\left\{A+B+B^{\prime }+C+C^{\prime }+D+D^{\prime }+E+E^{\prime }\right\},\end{array}$ with the terms A.19 $\begin{array}{c}A=4⟨\left[\left[I_{+},I_z{S}_z\right],I_z{S}_z\right]⟩j_0\left(0\right)=⟨I_{+}⟩j_0\left(0\right),B=+\frac{1}{4}⟨\left[\left[I_{+},I_{+}{S}_{-}\right],I_{-}{S}_{+}\right]⟩j_0\left({\omega }_I-{\omega }_S\right)=0,\\ B\prime =+\frac{1}{4}⟨\left[\left[I_{+},I_{-}{S}_{+}\right],I_{+}{S}_{-}\right]⟩j_0\left({\omega }_I-{\omega }_S\right)=\frac{1}{4}⟨I_{+}⟩j_0\left({\omega }_I-{\omega }_S\right),\\ C=+\frac{3}{2}⟨\left[\left[I_{+},I_{+}{S}_z\right],I_{-}{S}_z\right]⟩j_1\left({\omega }_I\right)=0,\\ C\prime =+\frac{3}{2}⟨\left[\left[I_{+},I_{-}{S}_z\right],I_{+}{S}_z\right]⟩j_1\left({\omega }_I\right)=\frac{3}{4}⟨I_{+}⟩j_1\left({\omega }_I\right),\\ D=+\frac{3}{2}⟨\left[\left[I_{+},I_z{S}_{+}\right],I_z{S}_{-}\right]⟩j_1\left({\omega }_S\right)=\frac{3}{4}⟨I_{+}⟩j_1\left({\omega }_S\right),\\ D\prime =+\frac{3}{2}⟨\left[\left[I_{+},I_z{S}_{-}\right],I_z{S}_{+}\right]⟩j_1\left({\omega }_S\right)=\frac{3}{4}⟨I_{+}⟩j_1\left({\omega }_S\right),\\ E=+\frac{3}{2}⟨\left[\left[I_{+},I_{+}{S}_{+}\right],I_{-}{S}_{-}\right]⟩j_2\left({\omega }_I+{\omega }_S\right)=0,\\ E\prime =+\frac{3}{2}⟨\left[\left[I_{+},I_{-}{S}_{-}\right],I_{+}{S}_{+}\right]⟩j_2\left({\omega }_I+{\omega }_S\right)=\frac{3}{2}⟨I_{+}⟩j_2\left({\omega }_I+{\omega }_S\right).\end{array}$

For the expansion of the commutators, one uses A.20a $\begin{array}{}\left[I_{+},I_z\right]=-I_{+};\left[I_{-},I_z\right]=I_{-};\left[I_{+},I_{-}\right]=2I_z,\end{array}$ and the product rules A.20b $\begin{array}{}{I}_z{I}_z=\frac{1}{4}E;I_{+}{I}_{-}=\frac{1}{2}E+I_z;I_{-}{I}_{+}=\frac{1}{2}E-I_z,\end{array}$ A.20c $\begin{array}{}{I}_z{I}_{+}=-I_{+}{I}_z=\frac{1}{2}{I}_{+};I_z{I}_{-}=-I_{-}{I}_z=-\frac{1}{2}{I}_{-},\end{array}$ with E being the unit matrix.

One finds that all double commutators in equation (A.19) result in terms linear in I₊, yielding an uncoupled differential equation A.21 $\begin{array}{}-\frac{d⟨I_{+}⟩}{\text{d}t}=R_2^I⟨I_{+}⟩,\end{array}$ with A.22 $\begin{array}{}{R}_2^I=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2{\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{IS}^3}\right)}^2\left\{4366J\left(0\right)+J\left({\omega }_I\right)+\left({\omega }_S\right)+J\left({\omega }_I-{\omega }_S\right)+J\left({\omega }_I+{\omega }_S\right)\right\}.\end{array}$

For two protons with different chemical shifts, one makes the substitutions A.23 $\begin{array}{}J\left({\omega }_I-{\omega }_S\right)\cong J\left(0\right);J\left({\omega }_S\right)\cong J\left({\omega }_I\right)\cong J\left({\omega }_H\right),\end{array}$ which yields A.24 $\begin{array}{}{R}_2^I=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2{\left(\frac{{\gamma }_H{\gamma }_H\hslash }{r_{IS}^3}\right)}^2\left\{\left\{596J\left(0\right)+J\left({\omega }_H\right)+J\left(2{\omega }_H\right)\right.\right\}.\end{array}$

This is called the “unlike” proton-proton dipolar relaxation equation.

In the case of two protons I and Iʹ with identical chemical shifts, or when the chemical shifts are forced to be equal during a spin lock, the cross terms between different coherence orders in the dipolar Hamiltonian cannot be neglected.

In particular, two cross terms between the single quantum terms C and D of equation (A.19) enter in the relaxation master equation, giving rise to cross-relaxation to the other spin: A.25 $\begin{array}{}\left[\left[I_{+},I_{-}{I}_z^{\prime }\right],I_z{I}_{+}^{\prime }\right]=2\left[I_z{I}_z^{\prime },I_z{I}_{+}^{\prime }\right]=\frac{1}{2}\left[I_z^{\prime },I_{+}^{\prime }\right]=\frac{1}{2}{I}_{+}^{\prime }.\end{array}$

Similarly, the frequencies of J(0) and zero-quantum terms in this case are precisely equal, so that two cross terms between terms A and B of equation (A.17) will occur as well: A.26 $\begin{array}{}-\left[\left[I_{+},I_z{I}_z^{\prime }\right],I_{-}{I}_{+}^{\prime }\right]=\left[I_{+}{I}_z^{\prime },I_{z-}{I}_{+}^{\prime }\right]=\frac{1}{2}{I}_{+}^{\prime }.\end{array}$

Capturing both transfers into one equation yields A.27 $\begin{array}{}-\frac{d⟨I_{+}⟩}{\text{d}t}=\frac{1}{2}\left\{j_0\left(0\right)+\frac{3}{2}{j}_1\left({\omega }_H\right)\right\}⟨I_{+}^{\prime }⟩\equiv {\sigma }_{\text{ROE}}⟨I_{+}^{\prime }⟩,\end{array}$ A.28 $\begin{array}{}{\sigma }_{\text{ROE}}=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2{\left(\frac{{\gamma }_I{\gamma }_S\hslash }{r_{IS}^3}\right)}^2\left\{46J\left(0\right)+\left({\omega }_I\right)\right\}.\end{array}$

The cross-relaxation rate in equation (A.28) is called the rotating frame Overhauser effect and was first discovered and quantified by [15] using the name “camel-spin.”

Since the I and I′ chemical shifts are identical, one may add equation (A.28) to equation (A.24) to obtain A29 $\begin{array}{}{R}_2^I=\frac{1}{8}{\left(\frac{{\mu }_0}{4\pi }\right)}^2{\left(\frac{{\gamma }_H{\gamma }_H\hslash }{r_{IS}^3}\right)}^2\left\{\left\{9156J\left(0\right)+J\left({\omega }_H\right)+J\left(2{\omega }_H\right)\right.\right\}.\end{array}$

This equation holds two protons with identical chemical shifts, or spinlocked. It is called the “like” spin relaxation equation. This has led to confusion. A better nomenclature would be to call this the identical spin relaxation equation.

Data Availability

No experimental data were used in this work.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The authors thank Professor Arno Kentgens at the University of Nijmegen for discussions and encouragement.

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