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# Line 71 | Line 71 | of Notre Dame, Notre Dame, IN 46556}
71   \begin{abstract}
72    This document includes useful relationships for computing the
73    interactions between fields and field gradients and point multipolar
74 <  representations of molecular electrostatics.  We also provide
74 >  representations of molecular electrostatics. We also provide
75    explanatory derivations of a number of relationships used in the
76    main text. This includes the Boltzmann averages of quadrupole
77 <  orientations, and the interaction of a quadrupole with the
77 >  orientations, and the interaction of a quadrupole density with the
78    self-generated field gradient. This last relationship is assumed to
79    be zero in the main text but is explicitly shown to be zero here.
80   \end{abstract}
# Line 82 | Line 82 | One important task in performing out the simulations m
82   \maketitle
83  
84   \section{Generating Uniform Field Gradients}
85 < One important task in performing out the simulations mentioned in the
86 < main text was to generate uniform electric field gradients. We rely
87 < heavily on both the notation and results from Torres del Castillo and
88 < Mend\'{e}z Garido.\cite{Torres-del-Castillo:2006uo} In this work,
89 < tensors were expressed in Cartesian components, using at times a
90 < dyadic notation. This proves quite useful for computer simulations
91 < that make use of toroidal boundary conditions.
85 > One important task in carrying out the simulations mentioned in the
86 > main text was to generate uniform electric field gradients.  To do
87 > this, we relied heavily on both the notation and results from Torres
88 > del Castillo and Mend\'{e}z Garido.\cite{Torres-del-Castillo:2006uo}
89 > In this work, tensors were expressed in Cartesian components, using at
90 > times a dyadic notation. This proves quite useful for computer
91 > simulations that make use of toroidal boundary conditions.
92  
93   An alternative formalism uses the theory of angular momentum and
94   spherical harmonics and is common in standard physics texts such as
# Line 313 | Line 313 | exerted on object $a$ by the electric field is given b
313   ${E}_\gamma$ terms are all evaluated at the center of the object (now
314   a point). Thus later the ${E}_\gamma$ terms can be written using the
315   same (molecular) origin for all point charges in the object. The force
316 < exerted on object $a$ by the electric field is given by,
316 > exerted on object $a$ by the electric field is given by,\cite{Raab:2004ve}
317   \begin{align}
318   F^a_\gamma = \sum_{k \textrm{~in~} a} E_\gamma(\mathbf{r}_k) &=  \sum_{k \textrm{~in~} a} q_k \lbrace E_\gamma + \nabla_\delta E_\gamma r_{k \delta}
319   + \frac {1}{2} \nabla_\delta \nabla_\varepsilon E_\gamma r_{k \delta}
# Line 370 | Line 370 | The dielectric properties of the system  mainly arise
370   \end{table}
371  
372   \section{Boltzmann averages for orientational polarization}
373 < The dielectric properties of the system  mainly arise from two
374 < different ways: i) the applied field distort the charge distributions
375 < so it produces an induced multipolar moment in each molecule; and ii)
376 < the applied field tends to line up originally randomly oriented
377 < molecular moment towards the direction of the applied field. In this
378 < study, we basically focus on the orientational contribution in the
379 < dielectric properties. If we consider a system of molecules in the
380 < presence of external field perturbation, the perturbation experienced
381 < by any molecule will not be only due to external field or field
382 < gradient but also due to the field or field gradient produced by the
383 < all other molecules in the system. In the following subsections
373 > If we consider a collection of molecules in the presence of external
374 > field, the perturbation experienced by any one molecule will include
375 > contributions to the field or field gradient produced by the all other
376 > molecules in the system. In subsections
377   \ref{subsec:boltzAverage-Dipole} and \ref{subsec:boltzAverage-Quad},
378 < we will discuss about the molecular polarization only due to external
379 < field perturbation. The contribution of the field or field gradient
380 < due to all other molecules will be taken into account while
388 < calculating correction factor in the paper.
378 > we discuss the molecular polarization due solely to external field
379 > perturbations.  This illustrates the origins of the polarizability
380 > equations (Eqs. 6, 20, and 21) in the main text.
381  
382   \subsection{Dipoles}
383   \label{subsec:boltzAverage-Dipole}
384   Consider a system of molecules, each with permanent dipole moment
385 < $p_o$. In the absense of external field, thermal agitation orients the
386 < dipoles randomly, reducing the system moment to zero.  External fields
387 < will tend to line up the dipoles in the direction of applied field.
388 < Here we have considered net field from all other molecules is
389 < considered to be zero.  Therefore the total Hamiltonian of each
390 < molecule is,\cite{Jackson98}
385 > $p_o$. In the absense of an external field, thermal agitation orients
386 > the dipoles randomly, and the system moment, $\mathbf{P}$, is zero.
387 > External fields will line up the dipoles in the direction of applied
388 > field.  Here we consider the net field from all other molecules to be
389 > zero.  Therefore the total Hamiltonian acting on each molecule
390 > is,\cite{Jackson98}
391   \begin{equation}
392 < H = H_o - \mathbf{p_o}\cdot \mathbf{E},
392 > H = H_o - \mathbf{p}_o \cdot \mathbf{E},
393   \end{equation}
394   where $H_o$ is a function of the internal coordinates of the molecule.
395 < The Boltzmann average of the dipole moment is given by,
395 > The Boltzmann average of the dipole moment in the direction of the
396 > field is given by,
397   \begin{equation}
398 < \braket{p_{mol}} = \frac{\displaystyle\int d\Omega\; p_o\; cos\theta\;  e^{\frac{p_oE\; cos\theta}{k_B T}}}{\displaystyle\int d\Omega\; e^{\frac{p_oE\;cos\theta}{k_B T}}},
398 > \langle p_{mol} \rangle = \frac{\displaystyle\int p_o \cos\theta
399 >  e^{~p_o E \cos\theta /k_B T}\; d\Omega}{\displaystyle\int  e^{~p_o E \cos\theta/k_B
400 >    T}\; d\Omega},
401   \end{equation}
402 < where $\bf{E}$ is selected along z-axis. If we consider that the
403 < applied field is small, \textit{i.e.} $\frac{p_oE\; cos\theta}{k_B T} << 1$,
402 > where the $z$-axis is taken in the direction of the applied field,
403 > $\bf{E}$ and
404 > $\int d\Omega = \int_0^\pi \sin\theta\; d\theta \int_0^{2\pi} d\phi
405 > \int_0^{2\pi} d\psi$
406 > is an integration over Euler angles describing the orientation of the
407 > molecule.
408 >
409 > If the external fields are small, \textit{i.e.}
410 > $p_oE \cos\theta / k_B T << 1$,
411   \begin{equation}
412 < \braket{p_{mol}}  \approx \frac{1}{3}\frac{{p_o}^2}{k_B T}E,
412 > \langle p_{mol} \rangle \approx \frac{{p_o}^2}{3 k_B T}E,
413   \end{equation}
414 < where $ \alpha_p = \frac{1}{3}\frac{{p_o}^2}{k_B T}$ is a molecular
414 > where $ \alpha_p = \frac{{p_o}^2}{3 k_B T}$ is the molecular
415   polarizability. The orientational polarization depends inversely on
416 < the temperature and applied field must overcome the thermal agitation.
416 > the temperature as the applied field must overcome thermal agitation
417 > to orient the dipoles.
418  
419   \subsection{Quadrupoles}
420   \label{subsec:boltzAverage-Quad}
421 < Consider a system of molecules with permanent quadrupole moment
422 < $q_{\alpha\beta}$. The average quadrupole moment at temperature T in
423 < the presence of uniform applied field gradient is given
424 < by,\cite{AduGyamfi78, AduGyamfi81}
421 > If instead, our system consists of molecules with permanent
422 > \textit{quadrupole} tensor $q_{\alpha\beta}$. The average quadrupole
423 > at temperature $T$ in the presence of uniform applied field gradient
424 > is given by,\cite{AduGyamfi78, AduGyamfi81}
425   \begin{equation}
426 < \braket{q_{\alpha\beta}} \;=\; \frac{\displaystyle\int d\Omega\; e^{-\frac{H}{k_B T}}q_{\alpha\beta}}{\displaystyle\int d\Omega\; e^{-\frac{H}{k_B T}}} \;=\; \frac{\displaystyle\int d\Omega\; e^{\frac{q_{\mu\nu}\;\partial_\nu E_\mu}{k_B T}}q_{\alpha\beta}}{\displaystyle\int d\Omega\; e^{\frac{q_{\mu\nu}\;\partial_\nu E_\mu}{k_B T}}},
426 > \langle q_{\alpha\beta} \rangle \;=\; \frac{\displaystyle\int
427 >  q_{\alpha\beta}\; e^{-H/k_B T}\; d\Omega}{\displaystyle\int
428 >  e^{-H/k_B T}\; d\Omega} \;=\; \frac{\displaystyle\int
429 >  q_{\alpha\beta}\; e^{~q_{\mu\nu}\;\partial_\nu E_\mu /k_B T}\;
430 >  d\Omega}{\displaystyle\int  e^{~q_{\mu\nu}\;\partial_\nu E_\mu /k_B
431 >    T}\; d\Omega },
432   \label{boltzQuad}
433   \end{equation}
434 < where $\int d\Omega = \int_0^{2\pi} \int_0^\pi \int_0^{2\pi}
435 < sin\theta\; d\theta\ d\phi\ d\psi$ is the integration over Euler
436 < angles, $ H = H_o -q_{\mu\nu}\;\partial_\nu E_\mu $ is the energy of
437 < a quadrupole in the gradient of the  
438 < applied field and $ H_o$ is a function of internal coordinates of the molecule. The energy and quadrupole moment can be transformed into body frame using following relation,
439 < \begin{equation}
440 < \begin{split}
441 < &q_{\alpha\beta} = \eta_{\alpha\alpha'}\;\eta_{\beta\beta'}\;{q}^* _{\alpha'\beta'} \\
442 < &H = H_o - q:{\nabla}\mathbf{E} = H_o - q_{\mu\nu}\;\partial_\nu E_\mu = H_o -\eta_{\mu\mu'}\;\eta_{\nu\nu'}\;{q}^*_{\mu'\nu'}\;\partial_\nu E_\mu.
443 < \end{split}
444 < \label{energyQuad}
445 < \end{equation}
434 > where $H = H_o - q_{\mu\nu}\;\partial_\nu E_\mu $ is the energy of a
435 > quadrupole in the gradient of the applied field and $H_o$ is a
436 > function of internal coordinates of the molecule. The energy and
437 > quadrupole moment can be transformed into the body frame using a
438 > rotation matrix $\mathsf{\eta}^{-1}$,
439 > \begin{align}
440 > q_{\alpha\beta} &= \eta_{\alpha\alpha'}\;\eta_{\beta\beta'}\;{q}^* _{\alpha'\beta'} \\
441 > H &= H_o - q:{\nabla}\mathbf{E} \\
442 >  &= H_o - q_{\mu\nu}\;\partial_\nu E_\mu  \\
443 >  &= H_o
444 >    -\eta_{\mu\mu'}\;\eta_{\nu\nu'}\;{q}^*_{\mu'\nu'}\;\partial_\nu
445 >    E_\mu. \label{energyQuad}
446 > \end{align}
447   Here the starred tensors are the components in the body fixed
448 < frame. Substituting equation (\ref{energyQuad}) in the equation (\ref{boltzQuad})
449 < and taking linear terms in the expansion we get,
450 < \begin{equation}
451 < \braket{q_{\alpha\beta}} = \frac{ \int d\Omega \left(1 + \frac{\eta_{\mu\mu'}\;\eta_{\nu\nu'}\;{q}^*_{\mu'\nu'}\;\partial_\nu E_\mu }{k_B T}\right)q_{\alpha\beta}}{ \int d\Omega \left(1 + \frac{\eta_{\mu\mu'}\;\eta_{\nu\nu'}\;{q}^*_{\mu'\nu'}\;\partial_\nu E_\mu }{k_B T}\right)},
448 > frame. Substituting equation (\ref{energyQuad}) in the equation
449 > (\ref{boltzQuad}) and taking linear terms in the expansion we obtain,
450 > \begin{equation}
451 > \braket{q_{\alpha\beta}} = \frac{\displaystyle \int q_{\alpha\beta} \left(1 +
452 >    \frac{\eta_{\mu\mu'}\;\eta_{\nu\nu'}\;{q}^*_{\mu'\nu'}\;\partial_\nu
453 >      E_\mu }{k_B T}\right)\;  d\Omega}{\displaystyle \int \left(1 + \frac{\eta_{\mu\mu'}\;\eta_{\nu\nu'}\;{q}^*_{\mu'\nu'}\;\partial_\nu E_\mu }{k_B T}\right)\; d\Omega}.
454   \end{equation}
455 < where $\eta_{\alpha\alpha'}$ is the inverse of the rotation matrix that transforms
456 < the body fixed co-ordinates to the space co-ordinates.
455 > Recall that $\eta_{\alpha\alpha'}$ is the inverse of the rotation
456 > matrix that transforms the body fixed co-ordinates to the space
457 > co-ordinates.
458   % \[\eta_{\alpha\alpha'}
459   % = \left(\begin{array}{ccc}
460   % cos\phi\; cos\psi - cos\theta\; sin\phi\; sin\psi & -cos\theta\; cos\psi\; sin\phi - cos\phi\; sin\psi & sin\theta\; sin\phi \\
# Line 450 | Line 462 | Integration of 1st and 2nd terms in the denominator gi
462   % sin\theta\; sin\psi & -cos\psi\; sin\theta & cos\theta
463   % \end{array} \right).\]
464  
465 < Integration of 1st and 2nd terms in the denominator gives $8 \pi^2$
466 < and $8 \pi^2 /3\;{\nabla}.\mathbf{E}\; Tr(q^*) $ respectively. The
467 < second term vanishes for charge free space, ${\nabla}.\mathbf{E} \; = \; 0$. Similarly integration of the
468 < 1st term in the numerator produces
469 < $8 \pi^2 /3\; Tr(q^*)\delta_{\alpha\beta}$ and the 2nd term produces
470 < $8 \pi^2 /15k_B T (3{q}^*_{\alpha'\beta'}{q}^*_{\beta'\alpha'} -
471 < {q}^*_{\alpha'\alpha'}{q}^*_{\beta'\beta'})\partial_\alpha E_\beta$,
472 < if ${\nabla}.\mathbf{E} \; = \; 0$,
473 < $ \partial_\alpha E_\beta = \partial_\beta E_\alpha$ and
474 < ${q}^*_{\alpha'\beta'}= {q}^*_{\beta'\alpha'}$. Therefore the
475 < Boltzmann average of a quadrupole moment can be written as,
476 <
465 > Integration of the first and second terms in the denominator gives
466 > $8 \pi^2$ and
467 > $8 \pi^2 ({\nabla} \cdot \mathbf{E}) \mathrm{Tr}(q^*) / 3 $
468 > respectively. The second term vanishes for charge free space (where
469 > ${\nabla} \cdot \mathbf{E}=0$). Similarly, integration of the first
470 > term in the numerator produces
471 > $8 \pi^2 \delta_{\alpha\beta} \mathrm{Tr}(q^*) / 3$ while the second
472 > produces
473 > $8 \pi^2 (3{q}^*_{\alpha'\beta'}{q}^*_{\beta'\alpha'} -
474 > {q}^*_{\alpha'\alpha'}{q}^*_{\beta'\beta'})\partial_\alpha E_\beta /
475 > 15 k_B T $.
476 > Therefore the Boltzmann average of a quadrupole moment can be written
477 > as,
478   \begin{equation}
479 < \braket{q_{\alpha\beta}}\; = \; \frac{1}{3} Tr(q^*)\;\delta_{\alpha\beta} + \frac{{\bar{q_o}}^2}{15k_BT}\;\partial_\alpha E_\beta,
479 > \langle q_{\alpha\beta} \rangle =  \frac{1}{3} \mathrm{Tr}(q^*)\;\delta_{\alpha\beta} + \frac{{\bar{q_o}}^2}{15k_BT}\;\partial_\alpha E_\beta,
480   \end{equation}
481 < where $ \alpha_q = \frac{{\bar{q_o}}^2}{15k_BT} $ is a molecular quadrupole polarizablity  and  ${\bar{q_o}}^2=
482 < 3{q}^*_{\alpha'\beta'}{q}^*_{\beta'\alpha'}-{q}^*_{\alpha'\alpha'}{q}^*_{\beta'\beta'}$ is a square of the net quadrupole moment of a molecule.
481 > where $\alpha_q = \frac{{\bar{q_o}}^2}{15k_BT} $ is a molecular
482 > quadrupole polarizablity and
483 > ${\bar{q_o}}^2=
484 > 3{q}^*_{\alpha'\beta'}{q}^*_{\beta'\alpha'}-{q}^*_{\alpha'\alpha'}{q}^*_{\beta'\beta'}$
485 > is the square of the net quadrupole moment of a molecule.
486  
487   \section{Gradient of the field due to quadrupolar polarization}
488   \label{singularQuad}
489 < In this section, we will discuss the gradient of the field produced by
490 < quadrupolar polarization. For this purpose, we consider a distribution
491 < of charge ${\rho}(\mathbf r)$ which gives rise to an electric field
492 < $\mathbf{E}(\mathbf r)$ and gradient of the field ${\nabla} \mathbf{E}(\mathbf r)$
493 < throughout space. The total gradient of the electric field over volume
494 < due to the all charges within the sphere of radius $R$ is given by
495 < (cf. Jackson equation 4.14):
489 > In section IV.C of the main text, we stated that for quadrupolar
490 > fluids, the self-contribution to the field gradient vanishes at the
491 > singularity. In this section, we prove this statement.  For this
492 > purpose, we consider a distribution of charge $\rho(\mathbf{r})$ which
493 > gives rise to an electric field $\mathbf{E}(\mathbf{r})$ and gradient
494 > of the field $\nabla\mathbf{E}(\mathbf{r})$ throughout space. The
495 > gradient of the electric field over volume due to the charges within
496 > the sphere of radius $R$ is given by (cf. Ref. \onlinecite{Jackson98},
497 > equation 4.14):
498   \begin{equation}
499 < \int_{r<R} {\nabla}\mathbf{E}\;d^3r = -\int_{r=R} R^2 \mathbf{E}\;\hat{n}\; d\Omega
499 > \int_{r<R} \nabla\mathbf{E} d\mathbf{r} = -\int_{r=R} R^2 \mathbf{E}\;\hat{n}\; d\Omega
500   \label{eq:8}
501   \end{equation}
502   where $d\Omega$ is the solid angle and $\hat{n}$ is the normal vector
503   of the surface of the sphere,
486 $\hat{n} = sin[\theta]cos[\phi]\hat{x} + sin[\theta]sin[\phi]\hat{y} +
487 cos[\theta]\hat{z}$
488 in spherical coordinates.  For the charge density ${\rho}(\mathbf r')$, the
489 total gradient of the electric field can be written as, ~\cite{Jackson98}
504   \begin{equation}
505 < \int_{r<R} {\nabla}\mathbf {E}\; d^3r=-\int_{r=R} R^2\; {\nabla}\Phi\; \hat{n}\; d\Omega  =-\frac{1}{4\pi\;\epsilon_o}\int_{r=R} R^2\; {\nabla}\;\left(\int \frac{\rho(\mathbf r')}{|\mathbf{r}-\mathbf{r}'|}\;d^3r'\right) \hat{n}\; d\Omega
505 > \hat{n} = \sin\theta\cos\phi\; \hat{x} + \sin\theta\sin\phi\; \hat{y} +
506 > \cos\theta\; \hat{z}
507 > \end{equation}
508 > in spherical coordinates.  For the charge density $\rho(\mathbf{r}')$, the
509 > total gradient of the electric field can be written as,\cite{Jackson98}
510 > \begin{equation}
511 > \int_{r<R} {\nabla}\mathbf {E}\; d\mathbf{r}=-\int_{r=R} R^2\; {\nabla}\Phi\; \hat{n}\; d\Omega  =-\frac{1}{4\pi\;\epsilon_o}\int_{r=R} R^2\; {\nabla}\;\left(\int \frac{\rho(\mathbf r')}{|\mathbf{r}-\mathbf{r}'|}\;d\mathbf{r}'\right) \hat{n}\; d\Omega
512   \label{eq:9}
513   \end{equation}
514   The radial function in the equation (\ref{eq:9}) can be expressed in
# Line 500 | Line 520 | If the sphere completely encloses the charge density t
520   If the sphere completely encloses the charge density then $ r_< = r'$ and $r_> = R$. Substituting equation (\ref{eq:10}) into (\ref{eq:9}) we get,
521   \begin{equation}
522   \begin{split}
523 < \int_{r<R} {\nabla}\mathbf{E}\;d^3r &=-\frac{R^2}{\epsilon_o}\int_{r=R} \; {\nabla}\;\left(\int \rho(\mathbf r')\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l}\frac{1}{2l+1}\;\frac{{r'^l}}{{R^{l+1}}}\;{Y^*}_{lm}(\theta', \phi')\;Y_{lm}(\theta, \phi)\;d^3r'\right) \hat{n}\; d\Omega \\
524 < &= -\frac{R^2}{\epsilon_o}\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l}\frac{1}{2l+1}\;\int \rho(\mathbf r')\;{r'^l}\;{Y^*}_{lm}(\theta', \phi')\left(\int_{r=R}\vec{\nabla}\left({R^{-(l+1)}}\;Y_{lm}(\theta, \phi)\right)\hat{n}\; d\Omega \right)d^3r
523 > \int_{r<R} {\nabla}\mathbf{E}\;d\mathbf{r} &=-\frac{R^2}{\epsilon_o}\int_{r=R} \; {\nabla}\;\left(\int \rho(\mathbf r')\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l}\frac{1}{2l+1}\;\frac{{r'^l}}{{R^{l+1}}}\;{Y^*}_{lm}(\theta', \phi')\;Y_{lm}(\theta, \phi)\;d\mathbf{r}'\right) \hat{n}\; d\Omega \\
524 > &= -\frac{R^2}{\epsilon_o}\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l}\frac{1}{2l+1}\;\int \rho(\mathbf r')\;{r'^l}\;{Y^*}_{lm}(\theta', \phi')\left(\int_{r=R}\vec{\nabla}\left({R^{-(l+1)}}\;Y_{lm}(\theta, \phi)\right)\hat{n}\; d\Omega \right)d\mathbf{r}
525   '
526   \end{split}
527   \label{eq:11}
# Line 515 | Line 535 | Using equation (\ref{eq:12}) we get,
535   \end{split}
536   \label{eq:12}
537   \end{equation}
538 < Using equation (\ref{eq:12}) we get,
538 > where $Y_{l,l+1,m}(\theta, \phi)$ is a vector spherical
539 > harmonic.\cite{Arfkan} Using equation (\ref{eq:12}) we get,
540   \begin{equation}
541   {\nabla}\left({R^{-(l+1)}}\;Y_{lm}(\theta, \phi)\right) = [(l+1)(2l+1)]^{1/2}\; Y_{l,l+1,m}(\theta, \phi) \; \frac{1}{R^{l+2}},
542   \label{eq:13}
543   \end{equation}
544 < where $ Y_{l,l+1,m}(\theta, \phi)$ is a vector spherical harmonics \cite{Arfkan}. Using Clebsch-Gorden coefficients $C(l+1, 1, l|m_1,m_2,m) $, equation \ref{eq:14} can be written in spherical harmonics,
544 > Using Clebsch-Gordan coefficients $C(l+1,1,l|m_1,m_2,m)$, the vector
545 > spherical harmonics can be written in terms of spherical harmonics,
546   \begin{equation}
547 < Y_{l,l+1,m}(\theta, \phi) = \sum_{m_1, m_2} C(l+1,1,l|m_1,m_2,m)\; {Y_{l+1}}^{m_1}(\theta,\phi)\; \hat{e}_{m_2}.
547 > Y_{l,l+1,m}(\theta, \phi) = \sum_{m_1, m_2} C(l+1,1,l|m_1,m_2,m)\; Y_{l+1}^{m_1}(\theta,\phi)\; \hat{e}_{m_2}.
548   \label{eq:14}
549   \end{equation}
550   Here $\hat{e}_{m_2}$ is a spherical tensor of rank 1 which can be expressed
# Line 533 | Line 555 | The normal vector $\hat{n} $ is then expressed in term
555   \end{equation}
556   The normal vector $\hat{n} $ is then expressed in terms of spherical tensor of rank 1 as shown in below,
557   \begin{equation}
558 < \hat{n} = \sqrt{\frac{4\pi}{3}}\left(-{Y_1}^{-1}{\hat{e}}_1 -{Y_1}^{1}{\hat{e}}_{-1} + {Y_1}^{0}{\hat{e}}_0 \right).
558 > \hat{n} = \sqrt{\frac{4\pi}{3}}\left(-Y_1^{-1}{\hat{e}}_1 - Y_1^{1}{\hat{e}}_{-1} + Y_1^{0}{\hat{e}}_0 \right).
559   \label{eq:16}
560   \end{equation}
561   The surface integral of the product of $\hat{n}$ and
562 < ${Y_{l+1}}^{m_1}(\theta, \phi)$ gives,
562 > $Y_{l+1}^{m_1}(\theta, \phi)$ gives,
563   \begin{equation}
564   \begin{split}
565 < \int \hat{n}\;{Y_{l+1}}^{m_1}\;d\Omega &= \int \sqrt{\frac{4\pi}{3}}\left(-{Y_1}^{-1}{\hat{e}}_1 -{Y_1}^{1}{\hat{e}}_{-1} + {Y_1}^{0}{\hat{e}}_0 \right)\;{Y_{l+1}}^{m_1}\; d\Omega \\
566 < &=  \int \sqrt{\frac{4\pi}{3}}\left({{Y_1}^{1}}^* {\hat{e}}_1 +{{Y_1}^{-1}}^* {\hat{e}}_{-1} + {{Y_1}^{0}}^* {\hat{e}}_0 \right)\;{Y_{l+1}}^{m_1}\; d\Omega \\
565 > \int \hat{n}\;Y_{l+1}^{m_1}\;d\Omega &= \int \sqrt{\frac{4\pi}{3}}\left(-Y_1^{-1}{\hat{e}}_1 -Y_1^{1}{\hat{e}}_{-1} + Y_1^{0}{\hat{e}}_0 \right)\;Y_{l+1}^{m_1}\; d\Omega \\
566 > &=  \int \sqrt{\frac{4\pi}{3}}\left({Y_1^{1}}^* {\hat{e}}_1 +{Y_1^{-1}}^* {\hat{e}}_{-1} + {Y_1^{0}}^* {\hat{e}}_0 \right)\;Y_{l+1}^{m_1}\; d\Omega \\
567   &=   \sqrt{\frac{4\pi}{3}}\left({\delta}_{l+1, 1}\;{\delta}_{1, m_1}\;{\hat{e}}_1 + {\delta}_{l+1, 1}\;{\delta}_{-1, m_1}\;{\hat{e}}_{-1}+ {\delta}_{l+1, 1}\;{\delta}_{0, m_1} \;{\hat{e}}_0\right),
568   \end{split}
569   \label{eq:17}
570   \end{equation}
571 < where ${Y_{l}}^{-m} = (-1)^m\;{{Y_{l}}^{m}}^* $ and
572 < $ \int {{Y_{l}}^{m}}^*\;{Y_{l'}}^{m'}\;d\Omega =
571 > where $Y_{l}^{-m} = (-1)^m\;{Y_{l}^{m}}^* $ and
572 > $ \int {Y_{l}^{m}}^* Y_{l'}^{m'}\;d\Omega =
573   \delta_{ll'}\delta_{mm'} $.
574   Non-vanishing values of equation \ref{eq:17} require $l = 0$,
575   therefore the value of $ m = 0 $. Since the values of $ m_1$ are -1,
# Line 556 | Line 578 | modified,
578   modified,
579   \begin{equation}
580   \begin{split}
581 < \int_{r<R} {\nabla}\mathbf{E}\;d^3r = &- \sqrt{\frac{4\pi}{{3}}}\;\frac{1}{\epsilon_o}\int \rho(r')\;{Y^*}_{00}(\theta', \phi')[ C(1, 1, 0|-1,1,0)\;{\hat{e}_{-1}}{\hat{e}_{1}}\\  &+ C(1, 1, 0|-1,1,0)\;{\hat{e}_{1}}{\hat{e}_{-1}}+C(
582 < 1, 1, 0|0,0,0)\;{\hat{e}_{0}}{\hat{e}_{0}} ]\; d^3r' \\
583 < &= -\sqrt{\frac{4\pi}{{3}}}\;\frac{1}{\epsilon_o}\int \rho(r')\;d^3r'\left({\hat{e}_{-1}}{\hat{e}_{1}}+{\hat{e}_{1}}{\hat{e}_{-1}}-{\hat{e}_{0}}{\hat{e}_{0}}\right)\\
584 < &= - \sqrt{\frac{4\pi}{{3}}}\;\frac{1}{\epsilon_o}\;C_{total}\;\left({\hat{e}_{-1}}{\hat{e}_{1}}+{\hat{e}_{1}}{\hat{e}_{-1}}-{\hat{e}_{0}}{\hat{e}_{0}}\right).
581 > \int_{r<R} {\nabla}\mathbf{E}\;d\mathbf{r} = &- \sqrt{\frac{4\pi}{{3}}}\;\frac{1}{\epsilon_o}\int \rho(r')\;{Y^*}_{00}(\theta', \phi')[ C(1, 1, 0|-1,1,0)\;{\hat{e}_{-1}}{\hat{e}_{1}}\\  &+ C(1, 1, 0|-1,1,0)\;{\hat{e}_{1}}{\hat{e}_{-1}}+C(
582 > 1, 1, 0|0,0,0)\;{\hat{e}_{0}}{\hat{e}_{0}} ]\; d\mathbf{r}' \\
583 > &= -\sqrt{\frac{4\pi}{{3}}}\;\frac{1}{\epsilon_o}\int \rho(r')\;d\mathbf{r}'\left({\hat{e}_{-1}}{\hat{e}_{1}}+{\hat{e}_{1}}{\hat{e}_{-1}}-{\hat{e}_{0}}{\hat{e}_{0}}\right)\\
584 > &= - \sqrt{\frac{4\pi}{{3}}}\;\frac{1}{\epsilon_o}\;C_\mathrm{total}\;\left({\hat{e}_{-1}}{\hat{e}_{1}}+{\hat{e}_{1}}{\hat{e}_{-1}}-{\hat{e}_{0}}{\hat{e}_{0}}\right).
585   \end{split}
586   \label{eq:19}
587   \end{equation}
588 < In the last step, the charge density was integrated over the sphere, yielding a total charge $\mathrm{C_total}$.Equation (\ref{eq:19}) gives the total gradient of the field over a sphere due to the distribution of the charges.
589 < For quadrupolar fluids the total charge within a sphere is zero, therefore
590 < $ \int_{r<R} {\nabla}\mathbf{E}\;d^3r = 0 $.  Hence the quadrupolar
588 > In the last step, the charge density was integrated over the sphere,
589 > yielding a total charge $C_\mathrm{total}$.Equation (\ref{eq:19})
590 > gives the total gradient of the field over a sphere due to the
591 > distribution of the charges.  For quadrupolar fluids the total charge
592 > within a sphere is zero, therefore
593 > $ \int_{r<R} {\nabla}\mathbf{E}\;d\mathbf{r} = 0 $.  Hence the quadrupolar
594   polarization produces zero net gradient of the field inside the
595   sphere.
596  
572 \section{Geometric Factors for Two Embedded Point Charges}
573
597   \bibliography{dielectric_new}
598   \end{document}
599   %

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