Minimum of a potential function

Question

I'm looking for extremes (minimum) of $$V = \frac{\alpha}{|\vec{r}_1-\vec{r}_2|} + \beta (\vec{r}_1 + \vec{r}_2)\cdot \vec{e}_z$$ where $\vec{r}_i = R(\cos\phi_i\sin\theta_i,\sin\phi_i\sin\theta_i,\cos\theta_i)$. I though that the expansion to spherical harmonics would be the easiest way $$ V(\phi_1,\phi_2,\theta_1,\theta_2) = \frac{\alpha}{R} \sum_{m=0}^{\infty} \frac{4 \pi}{2m + 1} \sum_{l=-m}^{m}Y_{l m}(\phi_1,\theta_1) Y_{l m}^{*}(\phi_2,\theta_2) + \beta R(\cos\theta_1+\cos\theta_2).$$ As you see, I am looking for $\phi_i$ and $\theta_i$ for both vectors.

Answer 1

By direct expansion of the vectors, you could see that $$V({\phi _1},{\theta _1},{\phi _2},{\theta _2}) = \frac{\alpha }{{\sqrt 2 R\sqrt {1 - \cos {\theta _1}\cos {\theta _2} - \cos ({\phi _1} - {\phi _2})\sin {\theta _1}\sin {\theta _2}} }} + \beta R(cos{\theta _1} + cos{\theta _2}) % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabdAfawjabcIcaOiabew9aMnaa % BaaaleaacqaIXaqmaeqaaOGaeiilaWIaeqiUde3aaSbaaSqaaiabig % daXaqabaGccqGGSaalcqaHvpGzdaWgaaWcbaGaeGOmaidabeaakiab % cYcaSiabeI7aXnaaBaaaleaacqaIYaGmaeqaaOGaeiykaKIaeyypa0 % ZaaSaaaeaacqaHXoqyaeaadaGcaaqaaiabikdaYaWcbeaakiabdkfa % snaakaaabaGaeGymaeJaeyOeI0Iagi4yamMaei4Ba8Maei4CamNaeq % iUde3aaSbaaSqaaiabigdaXaqabaGccyGGJbWycqGGVbWBcqGGZbWC % cqaH4oqCdaWgaaWcbaGaeGOmaidabeaakiabgkHiTiGbcogaJjabc+ % gaVjabcohaZjabcIcaOiabew9aMnaaBaaaleaacqaIXaqmaeqaaOGa % eyOeI0Iaeqy1dy2aaSbaaSqaaiabikdaYaqabaGccqGGPaqkcyGGZb % WCcqGGPbqAcqGGUbGBcqaH4oqCdaWgaaWcbaGaeGymaedabeaakiGb % cohaZjabcMgaPjabc6gaUjabeI7aXnaaBaaaleaacqaIYaGmaeqaaa % qabaaaaOGaey4kaSIaeqOSdiMaemOuaiLaeiikaGIaei4yamMaei4B % a8Maei4CamNaeqiUde3aaSbaaSqaaiabigdaXaqabaGccqGHRaWkcq % GGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWgaaWcbaGaeGOmaidabeaa % kiabcMcaPaaa!94EC! $$Now, we shall evaluate the derivative of the potential function with respect to the variables. Again by direct calculation, we could see that:$$\begin{array}{l}\frac{{\partial V({\phi _1},{\theta _1},{\phi _2},{\theta _2})}}{{\partial {\phi _1}}} = - \frac{{\alpha \sin {\theta _1}\sin {\theta _2}\sin ({\phi _1} - {\phi _2})}}{{2\sqrt 2 R{{\left( {1 - \cos {\theta _1}\cos {\theta _2} - \cos ({\phi _1} - {\phi _2})\sin {\theta _1}\sin {\theta _2}} \right)}^{\frac{3}{2}}}}},\\\frac{{\partial V({\phi _1},{\theta _1},{\phi _2},{\theta _2})}}{{\partial {\theta _1}}} = \frac{{\alpha \left( {\sin {\theta _1}\cos {\theta _2} - \cos ({\phi _1} - {\phi _2})\cos {\theta _1}\sin {\theta _2}} \right)}}{{2\sqrt 2 R{{\left( {1 - \cos {\theta _1}\cos {\theta _2} - \cos ({\phi _1} - {\phi _2})\sin {\theta _1}\sin {\theta _2}} \right)}^{\frac{3}{2}}}}} - \beta R\sin {\theta _1},\\\frac{{\partial V({\phi _1},{\theta _1},{\phi _2},{\theta _2})}}{{\partial {\phi _2}}} = \frac{{\alpha \sin {\theta _1}\sin {\theta _2}\sin ({\phi _1} - {\phi _2})}}{{2\sqrt 2 R{{\left( {1 - \cos {\theta _1}\cos {\theta _2} - \cos ({\phi _1} - {\phi _2})\sin {\theta _1}\sin {\theta _2}} \right)}^{\frac{3}{2}}}}},\\\frac{{\partial V({\phi _1},{\theta _1},{\phi _2},{\theta _2})}}{{\partial {\theta _2}}} = - \frac{{\alpha \left( {\cos {\theta _1}\sin {\theta _2} - \cos ({\phi _1} - {\phi _2})\sin {\theta _1}\cos {\theta _2}} \right)}}{{2\sqrt 2 R{{\left( {1 - \cos {\theta _1}\cos {\theta _2} - \cos ({\phi _1} - {\phi _2})\sin {\theta _1}\sin {\theta _2}} \right)}^{\frac{3}{2}}}}} - \beta R\sin {\theta _2}.\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOabaeqabaWaaSaaaeaacqGHciITcqWG % wbGvcqGGOaakcqaHvpGzdaWgaaWcbaGaeGymaedabeaakiabcYcaSi % abeI7aXnaaBaaaleaacqaIXaqmaeqaaOGaeiilaWIaeqy1dy2aaSba % 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Solving these equations is quite straightforward. First consider the case $\sin {\theta _1} = 0$. We have to satisfy only the second and fourth equations. This would lead to $$\left\{ \begin{array}{l}\cos ({\phi _1} - {\phi _2})\sin {\theta _2} = 0\\\left( {\frac{\alpha }{{2\sqrt 2 R{{\left( {1 - \cos {\theta _2}} \right)}^{\frac{3}{2}}}}} + \beta R} \right)\sin {\theta _2} = 0\end{array} \right. \to \left\{ \begin{array}{l}\sin {\theta _2} = 0\\{\rm{or}}\\\cos ({\phi _1} - {\phi _2}) = 0,\cos {\theta _2} = 1 - \frac{1}{2}{\left( {\frac{\alpha }{{{R^2}\beta }}} \right)^{\frac{2}{3}}}\end{array} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaamaaceaaeaqabeaacyGGJbWycqGG % VbWBcqGGZbWCcqGGOaakcqaHvpGzdaWgaaWcbaGaeGymaedabeaaki % abgkHiTiabew9aMnaaBaaaleaacqaIYaGmaeqaaOGaeiykaKIagi4C % amNaeiyAaKMaeiOBa4MaeqiUde3aaSbaaSqaaiabikdaYaqabaGccq % GH9aqpcqaIWaamaeaadaqadaqaamaalaaabaGaeqySdegabaGaeGOm % aiZaaOaaaeaacqaIYaGmaSqabaGccqWGsbGudaqadaqaaiabigdaXi % abgkHiTiGbcogaJjabc+gaVjabcohaZjabeI7aXnaaBaaaleaacqaI % YaGmaeqaaaGccaGLOaGaayzkaaWaaWbaaSqabeaadaWcaaqaaiabio % daZaqaaiabikdaYaaaaaaaaOGaey4kaSIaeqOSdiMaemOuaifacaGL % OaGaayzkaaGagi4CamNaeiyAaKMaeiOBa4MaeqiUde3aaSbaaSqaai % abikdaYaqabaGccqGH9aqpcqaIWaamaaGaay5EaaGaeyOKH46aaiqa % aqaabeqaaiGbcohaZjabcMgaPjabc6gaUjabeI7aXnaaBaaaleaacq % aIYaGmaeqaaOGaeyypa0JaeGimaadabaGaee4Ba8MaeeOCaihabaGa % gi4yamMaei4Ba8Maei4CamNaeiikaGIaeqy1dy2aaSbaaSqaaiabig % daXaqabaGccqGHsislcqaHvpGzdaWgaaWcbaGaeGOmaidabeaakiab % cMcaPiabg2da9iabicdaWiabcYcaSiGbcogaJjabc+gaVjabcohaZj % abeI7aXnaaBaaaleaacqaIYaGmaeqaaOGaeyypa0JaeGymaeJaeyOe % I0YaaSaaaeaacqaIXaqmaeaacqaIYaGmaaWaaeWaaeaadaWcaaqaai % abeg7aHbqaaiabdkfasnaaCaaaleqabaGaeGOmaidaaOGaeqOSdiga % aaGaayjkaiaawMcaamaaCaaaleqabaWaaSaaaeaacqaIYaGmaeaacq % aIZaWmaaaaaaaakiaawUhaaaaa!A8FB! $$ (The case for $\sin {\theta _2} = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiGbcohaZjabcMgaPjabc6gaUjab % eI7aXnaaBaaaleaacqaIYaGmaeqaaOGaeyypa0JaeGimaadaaa!47FF! $ is also very similar to this case and would yield $$\left\{ \begin{array}{l}\sin {\theta _1} = 0\\{\rm{or}}\\\cos ({\phi _1} - {\phi _2}) = 0,\cos {\theta _1} = 1 - \frac{1}{2}{\left( {\frac{\alpha }{{{R^2}\beta }}} \right)^{\frac{2}{3}}}\end{array} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaamaaceaaeaqabeaacyGGZbWCcqGG % PbqAcqGGUbGBcqaH4oqCdaWgaaWcbaGaeGymaedabeaakiabg2da9i % abicdaWaqaaiabb+gaVjabbkhaYbqaaiGbcogaJjabc+gaVjabcoha % ZjabcIcaOiabew9aMnaaBaaaleaacqaIXaqmaeqaaOGaeyOeI0Iaeq % y1dy2aaSbaaSqaaiabikdaYaqabaGccqGGPaqkcqGH9aqpcqaIWaam % cqGGSaalcyGGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWgaaWcbaGaeG % ymaedabeaakiabg2da9iabigdaXiabgkHiTmaalaaabaGaeGymaeda % baGaeGOmaidaamaabmaabaWaaSaaaeaacqaHXoqyaeaacqWGsbGuda % ahaaWcbeqaaiabikdaYaaakiabek7aIbaaaiaawIcacaGLPaaadaah % aaWcbeqaamaalaaabaGaeGOmaidabaGaeG4mamdaaaaaaaGccaGL7b % aaaaa!7095! $$). The final case arises when $\sin ({\phi _1} - {\phi _2}) = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiGbcohaZjabcMgaPjabc6gaUjab % cIcaOiabew9aMnaaBaaaleaacqaIXaqmaeqaaOGaeyOeI0Iaeqy1dy % 2aaSbaaSqaaiabikdaYaqabaGccqGGPaqkcqGH9aqpcqaIWaamaaa!4D9E! $. Under this condition, we have $$\left\{ \begin{array}{l}\frac{{\alpha \left( {\sin {\theta _1}\cos {\theta _2} - \cos {\theta _1}\sin {\theta _2}} \right)}}{{2\sqrt 2 R{{\left( {1 - \cos {\theta _1}\cos {\theta _2} - \sin {\theta _1}\sin {\theta _2}} \right)}^{\frac{3}{2}}}}} - \beta R\sin {\theta _1} = 0\\\frac{{\alpha \left( {\cos {\theta _1}\sin {\theta _2} - \sin {\theta _1}\cos {\theta _2}} \right)}}{{2\sqrt 2 R{{\left( {1 - \cos {\theta _1}\cos {\theta _2} - \sin {\theta _1}\sin {\theta _2}} \right)}^{\frac{3}{2}}}}} + \beta R\sin {\theta _2} = 0\end{array} \right. \to \sin {\theta _1} + \sin {\theta _2} = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaamaaceaaeaqabeaadaWcaaqaaiab % eg7aHnaabmaabaGagi4CamNaeiyAaKMaeiOBa4MaeqiUde3aaSbaaS % qaaiabigdaXaqabaGccyGGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWg % aaWcbaGaeGOmaidabeaakiabgkHiTiGbcogaJjabc+gaVjabcohaZj % abeI7aXnaaBaaaleaacqaIXaqmaeqaaOGagi4CamNaeiyAaKMaeiOB % a4MaeqiUde3aaSbaaSqaaiabikdaYaqabaaakiaawIcacaGLPaaaae % aacqaIYaGmdaGcaaqaaiabikdaYaWcbeaakiabdkfasnaabmaabaGa % eGymaeJaeyOeI0Iagi4yamMaei4Ba8Maei4CamNaeqiUde3aaSbaaS % qaaiabigdaXaqabaGccyGGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWg % aaWcbaGaeGOmaidabeaakiabgkHiTiGbcohaZjabcMgaPjabc6gaUj % abeI7aXnaaBaaaleaacqaIXaqmaeqaaOGagi4CamNaeiyAaKMaeiOB % a4MaeqiUde3aaSbaaSqaaiabikdaYaqabaaakiaawIcacaGLPaaada % ahaaWcbeqaamaalaaabaGaeG4mamdabaGaeGOmaidaaaaaaaGccqGH % sislcqaHYoGycqWGsbGucyGGZbWCcqGGPbqAcqGGUbGBcqaH4oqCda % WgaaWcbaGaeGymaedabeaakiabg2da9iabicdaWaqaamaalaaabaGa % eqySde2aaeWaaeaacyGGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWgaa % WcbaGaeGymaedabeaakiGbcohaZjabcMgaPjabc6gaUjabeI7aXnaa % BaaaleaacqaIYaGmaeqaaOGaeyOeI0Iagi4CamNaeiyAaKMaeiOBa4 % MaeqiUde3aaSbaaSqaaiabigdaXaqabaGccyGGJbWycqGGVbWBcqGG % ZbWCcqaH4oqCdaWgaaWcbaGaeGOmaidabeaaaOGaayjkaiaawMcaaa % qaaiabikdaYmaakaaabaGaeGOmaidaleqaaOGaemOuai1aaeWaaeaa % cqaIXaqmcqGHsislcyGGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWgaa % WcbaGaeGymaedabeaakiGbcogaJjabc+gaVjabcohaZjabeI7aXnaa % BaaaleaacqaIYaGmaeqaaOGaeyOeI0Iagi4CamNaeiyAaKMaeiOBa4 % MaeqiUde3aaSbaaSqaaiabigdaXaqabaGccyGGZbWCcqGGPbqAcqGG % UbGBcqaH4oqCdaWgaaWcbaGaeGOmaidabeaaaOGaayjkaiaawMcaam % aaCaaaleqabaWaaSaaaeaacqaIZaWmaeaacqaIYaGmaaaaaaaakiab % gUcaRiabek7aIjabdkfasjGbcohaZjabcMgaPjabc6gaUjabeI7aXn % aaBaaaleaacqaIYaGmaeqaaOGaeyypa0JaeGimaadaaiaawUhaaiab % gkziUkGbcohaZjabcMgaPjabc6gaUjabeI7aXnaaBaaaleaacqaIXa % qmaeqaaOGaey4kaSIagi4CamNaeiyAaKMaeiOBa4MaeqiUde3aaSba % aSqaaiabikdaYaqabaGccqGH9aqpcqaIWaamaaa!F878! $$Note that since ${\theta _1},{\theta _2} \in [0,\pi ) % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabeI7aXnaaBaaaleaacqaIXaqm % aeqaaOGaeiilaWIaeqiUde3aaSbaaSqaaiabikdaYaqabaGccqGHii % IZcqGGBbWwcqaIWaamcqGGSaalcqaHapaCcqGGPaqkaaa!4CC0! $, the last equation seizes to yield any solution. Summarizing, the solutions are: $$\begin{array}{l}1)\sin {\theta _1} = 0,\sin {\theta _2} = 0,\\2)\sin {\theta _1} = 0,\cos {\theta _2} = 1 - \frac{1}{2}{\left( {\frac{\alpha }{{{R^2}\beta }}} \right)^{\frac{3}{2}}},\cos ({\phi _1} - {\phi _2}) = 0,\\3)\sin {\theta _2} = 0,\cos {\theta _1} = 1 - \frac{1}{2}{\left( {\frac{\alpha }{{{R^2}\beta }}} \right)^{\frac{3}{2}}},\cos ({\phi _1} - {\phi _2}) = 0.\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOabaeqabaGaeGymaeJaeiykaKIagi4C % amNaeiyAaKMaeiOBa4MaeqiUde3aaSbaaSqaaiabigdaXaqabaGccq % GH9aqpcqaIWaamcqGGSaalcyGGZbWCcqGGPbqAcqGGUbGBcqaH4oqC % daWgaaWcbaGaeGOmaidabeaakiabg2da9iabicdaWiabcYcaSaqaai % abikdaYiabcMcaPiGbcohaZjabcMgaPjabc6gaUjabeI7aXnaaBaaa % leaacqaIXaqmaeqaaOGaeyypa0JaeGimaaJaeiilaWIagi4yamMaei % 4Ba8Maei4CamNaeqiUde3aaSbaaSqaaiabikdaYaqabaGccqGH9aqp % cqaIXaqmcqGHsisldaWcaaqaaiabigdaXaqaaiabikdaYaaadaqada % qaamaalaaabaGaeqySdegabaGaemOuai1aaWbaaSqabeaacqaIYaGm % aaGccqaHYoGyaaaacaGLOaGaayzkaaWaaWbaaSqabeaadaWcaaqaai % abiodaZaqaaiabikdaYaaaaaGccqGGSaalcyGGJbWycqGGVbWBcqGG % ZbWCcqGGOaakcqaHvpGzdaWgaaWcbaGaeGymaedabeaakiabgkHiTi % abew9aMnaaBaaaleaacqaIYaGmaeqaaOGaeiykaKIaeyypa0JaeGim % aaJaeiilaWcabaGaeG4mamJaeiykaKIagi4CamNaeiyAaKMaeiOBa4 % MaeqiUde3aaSbaaSqaaiabikdaYaqabaGccqGH9aqpcqaIWaamcqGG % SaalcyGGJbWycqGGVbWBcqGGZbWCcqaH4oqCdaWgaaWcbaGaeGymae % dabeaakiabg2da9iabigdaXiabgkHiTmaalaaabaGaeGymaedabaGa % eGOmaidaamaabmaabaWaaSaaaeaacqaHXoqyaeaacqWGsbGudaahaa % WcbeqaaiabikdaYaaakiabek7aIbaaaiaawIcacaGLPaaadaahaaWc % beqaamaalaaabaGaeG4mamdabaGaeGOmaidaaaaakiabcYcaSiGbco % gaJjabc+gaVjabcohaZjabcIcaOiabew9aMnaaBaaaleaacqaIXaqm % aeqaaOGaeyOeI0Iaeqy1dy2aaSbaaSqaaiabikdaYaqabaGccqGGPa % qkcqGH9aqpcqaIWaamcqGGUaGlaaaa!B6F9! $$