Author: bugman Date: Thu May 8 19:48:53 2014 New Revision: 23099 URL: http://svn.gna.org/viewcvs/relax?rev=23099&view=rev Log: Added proper punctuation to the B14 dispersion model equations in the manual. Equations should be readable as English sentences and they follow standard punctuation rules. All of the equations in the B14 model section of the dispersion chapter have been updated to follow this. Modified: trunk/docs/latex/dispersion.tex Modified: trunk/docs/latex/dispersion.tex URL: http://svn.gna.org/viewcvs/relax/trunk/docs/latex/dispersion.tex?rev=23099&r1=23098&r2=23099&view=diff ============================================================================== --- trunk/docs/latex/dispersion.tex (original) +++ trunk/docs/latex/dispersion.tex Thu May 8 19:48:53 2014 @@ -568,66 +568,65 @@ \begin{subequations} \begin{align} \Rtwoeff & = \frac{\RtwozeroA + \RtwozeroB + \kex }{2}-\frac{ N_{\textrm{CYC}} }{ \taucpmg } \cosh{}^{-1}(v_{1c}) \nonumber \\ -& \qquad - \frac{1}{\taucpmg}\ln{\left(\frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2 \kAB p_D)\right)} \\ +& \qquad - \frac{1}{\taucpmg}\ln{\left(\frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2 \kAB p_D)\right)} , \\ & = \Rtwoeff^{\textrm{CR72}} - \frac{1}{\taucpmg}\ln{\left( \frac{1+y}{2} + \frac{1-y}{2\sqrt{v_{1c}^2-1}}(v_2 + 2\kAB p_D )\right)} , \end{align} \end{subequations} - where Appendix 1 in \citet{Baldwin2014} list the recipe for exact calculation of $\Rtwoeff$. -Establish the complex free precession Eigenfrequency. +Establish the complex free precession Eigenfrequency with \begin{subequations} \begin{align} - \Delta \Rtwozero & = \RtwozeroA - \RtwozeroB \\ - \alpha_- & = \Delta \Rtwozero + \kAB - \kBA \\ - \zeta & = 2 \dw \alpha_- \\ - \Psi & = \alpha_-^2 + 4 \kAB \kBA - \dw^2 \\ - h_3 &= \frac{1}{\sqrt{2}}\sqrt{ \Psi + \sqrt{\zeta^2 + \Psi^2} } \\ - h_4 &= \frac{1}{\sqrt{2}}\sqrt{ -\Psi + \sqrt{\zeta^2 + \Psi^2} } + \Delta \Rtwozero & = \RtwozeroA - \RtwozeroB , \\ + \alpha_- & = \Delta \Rtwozero + \kAB - \kBA , \\ + \zeta & = 2 \dw \alpha_- , \\ + \Psi & = \alpha_-^2 + 4 \kAB \kBA - \dw^2 , \\ + h_3 &= \frac{1}{\sqrt{2}}\sqrt{ \Psi + \sqrt{\zeta^2 + \Psi^2} } , \\ + h_4 &= \frac{1}{\sqrt{2}}\sqrt{ -\Psi + \sqrt{\zeta^2 + \Psi^2} } . \end{align} \end{subequations} The ground state ensemble evolution frequency $f_{00}$ expressed in separated real and imaginary components, in terms -of definitions $\zeta , \Psi , h_3 , h_4$. +of definitions $\zeta , \Psi , h_3 , h_4$ is \begin{equation} - f_{00} = \frac{1}{2}(\RtwozeroA + \RtwozeroB + \kex) + \frac{1}{2}(\dw - h_4) i -\end{equation} - -Define substutions for 'stay' and 'swap' factors. + f_{00} = \frac{1}{2}(\RtwozeroA + \RtwozeroB + \kex) + \frac{1}{2}(\dw - h_4) i . +\end{equation} + +Define substutions for 'stay' and 'swap' factors are \begin{subequations} \begin{align} - N & = h_3 + h_4 i \\ - NN^* & = h_3^2 + h_42 \\ - F_0 & = (\dw^2 + h_3^2) / NN^* \\ - F_2 & = (\dw^2 - h_4^2) / NN^* \\ - F_1^b & = (\dw + h_4) (\dw - h_3 i) / NN^* \\ - F_1^{a+b} & = (2\dw^2 + \zeta i) / NN^* + N & = h_3 + h_4 i , \\ + NN^* & = h_3^2 + h_42 , \\ + F_0 & = (\dw^2 + h_3^2) / NN^* , \\ + F_2 & = (\dw^2 - h_4^2) / NN^* , \\ + F_1^b & = (\dw + h_4) (\dw - h_3 i) / NN^* , \\ + F_1^{a+b} & = (2\dw^2 + \zeta i) / NN^* . \end{align} \end{subequations} -Weighting factors for frequencies ($E_{0-2}$) emerging from a single CPMG block, ($F_{0-2}$). -Here $\taucpmg = 1 / 4\nucpmg $. +Weighting factors for frequencies ($E_{0-2}$) emerging from a single CPMG block, ($F_{0-2}$), are \begin{subequations} \begin{align} - E_0 & = 2 \taucpmg \cdot h_3 \\ - E2 & = 2 \taucpmg \cdot h_4 \\ - E1 & = (h_3 - h_4 i) \cdot \taucpmg + E_0 & = 2 \taucpmg \cdot h_3 , \\ + E2 & = 2 \taucpmg \cdot h_4 , \\ + E1 & = (h_3 - h_4 i) \cdot \taucpmg . \end{align} \end{subequations} -Final result, with identities to assist efficient matrix exponentiation optimised for numerical calculation. +Here $\taucpmg = 1 / 4\nucpmg $. +Final result, with identities to assist efficient matrix exponentiation optimised for numerical calculation is \begin{subequations} \begin{align} - \nu_{1c} & = F_0 \cosh(E_0) - F_2 \cos(E_2) \\ - \nu_{1s} & = F_0 \sinh(E_0) - F_2 \sin(E_2)i \\ - \nu_{3} & = \sqrt{\nu_{1c}^2 - 1} \\ - \nu_{4} & = F_1^b (-\alpha_- - h_3 ) + F_1^b (\dw - h_4) i \\ - \nu_{5} & =(-\Delta \Rtwozero + \kex + \dw i) \nu_{1s} + 2 (\nu_{4} + \kAB F_1^{a+b}) \sinh(E_1) \\ - y & = \left( \frac{\nu_{1c} - \nu_{3}}{\nu_{1c} + \nu_{3}} \right) ^ {N_{\textrm{CYC}}} \\ - T & = \frac{1}{2}(1 + y) + \frac{(1 - y)\nu_{5}}{2 \nu_{3}N} \\ - \Rtwoeff{}_{\_ \textrm{CR72}} & = \frac{(\RtwozeroA + \RtwozeroB + \kex)}{2} - \frac{N_{\textrm{CYC}}}{\taucpmg} \, \textrm{arcosh}(\, \operatorname{Re}(\nu_{1c}) \, ) \\ - \Rtwoeff{} & = \Rtwoeff{}_{\_ \textrm{CR72}} - \frac{1}{\taucpmg} \log(\operatorname{Re}(T)) + \nu_{1c} & = F_0 \cosh(E_0) - F_2 \cos(E_2) , \\ + \nu_{1s} & = F_0 \sinh(E_0) - F_2 \sin(E_2)i , \\ + \nu_{3} & = \sqrt{\nu_{1c}^2 - 1} , \\ + \nu_{4} & = F_1^b (-\alpha_- - h_3 ) + F_1^b (\dw - h_4) i , \\ + \nu_{5} & =(-\Delta \Rtwozero + \kex + \dw i) \nu_{1s} + 2 (\nu_{4} + \kAB F_1^{a+b}) \sinh(E_1) , \\ + y & = \left( \frac{\nu_{1c} - \nu_{3}}{\nu_{1c} + \nu_{3}} \right) ^ {N_{\textrm{CYC}}} , \\ + T & = \frac{1}{2}(1 + y) + \frac{(1 - y)\nu_{5}}{2 \nu_{3}N} , \\ + \Rtwoeff{}_{\_ \textrm{CR72}} & = \frac{(\RtwozeroA + \RtwozeroB + \kex)}{2} - \frac{N_{\textrm{CYC}}}{\taucpmg} \, \textrm{arcosh}(\, \operatorname{Re}(\nu_{1c}) \, ) , \\ + \Rtwoeff{} & = \Rtwoeff{}_{\_ \textrm{CR72}} - \frac{1}{\taucpmg} \log(\operatorname{Re}(T)) . \end{align} \end{subequations} @@ -636,9 +635,8 @@ The term $p_D$ is based on product of the off diagonal elements in the CPMG propagator, see supplementary Section 3, \citet{Baldwin2014}. It is interesting to consider the region of validity of the Carver Richards result. The two results are equal when the correction is zero, which is true when - \begin{equation} - \sqrt{v_{1c}^2-1} \approx v_2 + 2k_{\textrm{AB}}p_D + \sqrt{v_{1c}^2-1} \approx v_2 + 2k_{\textrm{AB}}p_D . \end{equation} This occurs when $k_{\textrm{AB}}p_D$ tends to zero, and so $v_2=v_3$.