{"id":31207,"date":"2021-03-28T13:00:58","date_gmt":"2021-03-28T13:00:58","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=31207"},"modified":"2025-01-12T21:26:14","modified_gmt":"2025-01-12T21:26:14","slug":"problemes-de-combinatoire-en-thermodynamique","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/problemes-de-combinatoire-en-thermodynamique\/","title":{"rendered":"Probl\u00e8mes de combinatoire en thermodynamique"},"content":{"rendered":"<style>\np, ul, ol {\ntext-align: justify;\n}\nh1, h2 {\ntext-align: center;\n}\n<\/style>\n<h1>Probl\u00e8mes de combinatoire en thermodynamique<\/h1>\n<p style=\"text-align:center;\"><em>Combien de fa\u00e7ons existe-t-il pour organiser un syst\u00e8me physique compos\u00e9 de millions d&#8217;\u00e9l\u00e9ments ? Dans ce cours, nous aborderons comment les math\u00e9matiques permettent de r\u00e9pondre \u00e0 des questions comme celle-ci dans le contexte de la thermodynamique, depuis la distribution des quanta d&#8217;\u00e9nergie dans les syst\u00e8mes atomiques jusqu&#8217;au calcul des configurations possibles dans des syst\u00e8mes \u00e0 grande \u00e9chelle. \u00c0 l&#8217;aide d&#8217;outils comme la combinatoire, les logarithmes et la formule de Stirling, nous explorerons comment manipuler des nombres extraordinairement grands et r\u00e9soudre des probl\u00e8mes apparemment insurmontables.<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Objectifs d&#8217;apprentissage :<\/strong><br \/>\n\u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol>\n<li><strong>Comprendre<\/strong> comment les probl\u00e8mes de combinatoire s&#8217;appliquent dans le contexte de la thermodynamique, notamment dans l&#8217;organisation des syst\u00e8mes physiques.<\/li>\n<li><strong>Calculer<\/strong> les configurations possibles des syst\u00e8mes atomiques en utilisant des nombres combinatoires.<\/li>\n<li><strong>Appliquer<\/strong> la formule de Stirling pour estimer l&#8217;ordre de grandeur des configurations complexes.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><u>TABLE DES MATI\u00c8RES :<\/u><br \/>\n<a href=\"#1\"><strong>Probl\u00e8mes de combinatoire<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>Probl\u00e8mes avec de grands nombres<\/strong><\/a><br \/>\n<a href=\"#3\">Utilisation des logarithmes et de la formule de Stirling pour le calcul de l&#8217;ordre de grandeur<\/a><br \/>\n<a href=\"#4\"><em>D\u00e9veloppement avec l&#8217;approche simplifi\u00e9e<\/em><\/a><br \/>\n<a href=\"#5\"><em>D\u00e9veloppement avec l&#8217;approche classique<\/em><\/a><br \/>\n<a href=\"#6\"><strong>Exemples de calculs combinatoires et d&#8217;ordre de grandeur<\/strong><\/a><br \/>\n<a href=\"#7\">Cas 1 : Grands factorielles<\/a><br \/>\n<a href=\"#8\">Cas 2 : Grandes combinatoires<\/a>\n<\/p>\n<p>Une question courante dans certaines situations physiques est : combien de fa\u00e7ons diff\u00e9rentes peut-on organiser un syst\u00e8me donn\u00e9 ? Ces probl\u00e8mes combinatoires sont fr\u00e9quents en thermodynamique. Bien qu&#8217;ils paraissent simples au d\u00e9part, ils deviennent complexes lorsqu&#8217;on introduit des nombres extr\u00eamement grands, comme le <strong><a href=\"http:\/\/toposuranos.com\/material\/mol-y-masa-molar\/\" rel=\"noopener\" target=\"_blank\">nombre d&#8217;Avogadro<\/a><\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">N_A<\/span>, qui illustre \u00e0 quel point il peut \u00eatre difficile de travailler avec des grandeurs de cette \u00e9chelle.<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Probl\u00e8mes de combinatoire<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/K7_De9wux4A?si=q93_T8xh15EdfJ6B\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Pour comprendre l&#8217;ampleur des probl\u00e8mes impliquant la combinatoire en thermodynamique, consid\u00e9rons l&#8217;exemple suivant :<\/p>\n<h4>Exemple : Combinaisons pour les quanta d&#8217;\u00e9nergie<\/h4>\n<p>Supposons un syst\u00e8me compos\u00e9 de 10 atomes. Chaque atome peut stocker uniquement 1 ou 0 unit\u00e9s d&#8217;\u00e9nergie, appel\u00e9es <strong>quanta d&#8217;\u00e9nergie<\/strong>. Combien de fa\u00e7ons diff\u00e9rentes peut-on distribuer ces quanta si nous avons (a) 10 quanta d&#8217;\u00e9nergie et (b) 5 quanta d&#8217;\u00e9nergie ?<\/p>\n<h5>Solution<\/h5>\n<p>Nous repr\u00e9sentons les atomes comme des espaces disponibles pour stocker un quantum d&#8217;\u00e9nergie. Si un espace est rempli, cela signifie que l&#8217;atome correspondant poss\u00e8de d\u00e9j\u00e0 son quantum d&#8217;\u00e9nergie.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-zJa7H7iMOGY\/YF3QhG94SoI\/AAAAAAAAEuI\/olfdSHfeJOgKea3vHXEAyr5QCJ1ZjmcpgCLcBGAsYHQ\/s839\/cuantodeenergia.PNG\" \n               class=\"aligncenter lazyload\" alt=\"Probl\u00e8mes de combinatoire en thermodynamique\" width=\"400\" height=\"200\" \/><\/center><\/p>\n<p>Pour compter les fa\u00e7ons dont <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> quanta d&#8217;\u00e9nergie sont distribu\u00e9s parmi <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> espaces, nous utilisons le nombre combinatoire :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\binom{n}{k}=\\dfrac{n!}{k!(n-k)!}<\/span><\/span><\/p>\n<p>Ce calcul nous donne le nombre <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> d&#8217;\u00e9tats possibles.<\/p>\n<p><strong>(a)<\/strong> Si 10 quanta sont distribu\u00e9s parmi 10 espaces, il n&#8217;existe qu&#8217;une seule fa\u00e7on de le faire. Ainsi, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega=1<\/span><\/span> :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\Omega = \\binom{10}{10}=\\dfrac{10!}{10!(10-10)!} = \\dfrac{10!}{10!0!} = 1 <\/span><\/span><\/p>\n<p><strong>(b)<\/strong> Pour 5 quanta distribu\u00e9s parmi 10 espaces, les calculs sont les suivants :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\Omega &amp;= \\displaystyle\\binom{10}{5} \\\\ \\\\\n\n&amp;=\\dfrac{10!}{5!(10-5)!} = \\dfrac{10!}{5!\\cdot 5!} \\\\ \\\\\n\n&amp;= \\dfrac{5! \\cdot 6\\cdot 7\\cdot 8 \\cdot 9\\cdot 10}{5! \\cdot 2\\cdot 3\\cdot 4\\cdot 5} \\\\ \\\\\n\n&amp;= \\dfrac{ 7\\cdot 8 \\cdot 9\\cdot 10}{ 4\\cdot 5} = 7\\cdot 2 \\cdot 9 \\cdot 2 = 252\n\n\\end{array}<\/span>\n<p>Ainsi, il existe 252 configurations possibles.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Probl\u00e8mes avec de grands nombres<\/h2>\n<p>Ce que nous avons analys\u00e9 jusqu&#8217;\u00e0 pr\u00e9sent n&#8217;est que le d\u00e9but. Si nous \u00e9tendons le syst\u00e8me du cas (b) \u00e0 100 atomes et 50 quanta, nous obtiendrons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega \\approx 10^{28}<\/span><\/span>. Maintenant, imaginez effectuer le m\u00eame calcul avec une mole d&#8217;atomes ; le r\u00e9sultat serait inconcevable.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Utilisation des logarithmes et de la formule de Stirling pour le calcul de l&#8217;ordre de grandeur<\/h3>\n<p>Lorsque nous voulons estimer une grandeur de la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega = \\binom{n}{k}<\/span><\/span> pour de grandes valeurs de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, en particulier lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, ce qui correspond au cas o\u00f9 les valeurs maximales sont atteintes, il est utile d&#8217;utiliser l&#8217;approximation logarithmique de la formule de Stirling.<\/p>\n<p>Pour manipuler des nombres de cette grandeur, nous pouvons reformuler les calculs en prenant des logarithmes, ce qui donne :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\ln(\\Omega)=\\ln\\left(\\dfrac{n!}{k!(n-k)!}\\right)= \\ln(n!) - \\ln((n-k)!) - \\ln(k!)<\/span><\/span><\/p>\n<p>Cette expression peut \u00eatre travaill\u00e9e en utilisant l&#8217;approximation de Stirling pour le logarithme factoriel. Pour cela, nous avons deux versions possibles : l&#8217;approximation classique et l&#8217;approximation simplifi\u00e9e :<\/p>\n<ul>\n<li><strong>Approximation classique :<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n<\/span><\/span><\/li>\n<li><strong>Approximation simplifi\u00e9e :<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx n\\ln(n) - n<\/span><\/span><\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/p>\n<h4>D\u00e9veloppement avec l&#8217;approche simplifi\u00e9e<\/h4>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/fnt99FeeHPM?si=H6r-5GhO2yqaeaXd\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>En utilisant l&#8217;approche simplifi\u00e9e, nous obtenons les r\u00e9sultats suivants :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\ln(\\Omega) &amp; \\approx n\\ln(n) - n - (n-k)\\ln(n-k) + (n-k) - k\\ln(k) + k \\\\ \\\\\n\n&amp;= n\\ln(n) - (n-k)\\ln(n-k) - k\\ln(k) \\\\ \\\\\n\n&amp;= n\\ln(n) - n\\ln(n-k) + k\\ln(n-k) - k\\ln(k) \\\\ \\\\\n\n&amp;= \\ln\\left[ \\left( \\dfrac{n}{n-k} \\right)^n \\right] + k\\ln\\left( \\dfrac{n-k}{k} \\right) \\\\ \\\\\n\n&amp;= \\ln\\left[ \\dfrac{1}{\\left(1 - \\dfrac{k}{n} \\right)^n} \\right] + k\\ln\\left( \\dfrac{n}{k} - 1 \\right)\n\n\\end{array}<\/span>\n<p>Pour de grandes valeurs de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, nous appliquons la relation suivante :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{n\\to\\infty} \\left(1-\\dfrac{k}{n} \\right)^n = e^{-k} <\/span><\/span><\/p>\n<p>Par cons\u00e9quent :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(\\Omega) \\approx \\ln(e^k) + k\\ln\\left( \\dfrac{n}{k} -1 \\right) = k + k\\ln\\left( \\dfrac{n}{k} -1 \\right) <\/span><\/span><\/p>\n<p>Enfin, en utilisant la conversion en logarithmes d\u00e9cimaux, nous obtenons :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(\\Omega) = \\log(e)\\ln(\\Omega) \\approx k\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{k} - 1 \\right) \\right] <\/span><\/span><\/p>\n<p>Ce qui nous conduit au r\u00e9sultat :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\Omega \\approx 10^{k\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{k} - 1 \\right) \\right]}}<\/span><\/span><\/p>\n<p>Bien que ce r\u00e9sultat ne donne pas la valeur exacte de <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span>, il permet d&#8217;obtenir une estimation du nombre de chiffres n\u00e9cessaires pour le repr\u00e9senter et devient plus pr\u00e9cis \u00e0 mesure que <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> augmente. Avec cette m\u00e9thode, il suffit de calculer ce qui figure dans l&#8217;exposant, une op\u00e9ration r\u00e9alisable avec la plupart des calculatrices.<\/p>\n<p>De plus, cette approche permet d&#8217;estimer rapidement la valeur maximale de <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> pour une grande valeur de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>. En consid\u00e9rant le cas o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, nous obtenons :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Max}\\left(\\Omega\\right) \\approx 10^{\\dfrac{n}{2}\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{n\/2} - 1 \\right) \\right]} = 10^{ n\\log(e)\/2 } <\/span><\/span><\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h4>D\u00e9veloppement avec l&#8217;approche classique<\/h4>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/i3OO4lHV24Q?si=jEPCafXqgVxNYqH-\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Bien que le d\u00e9veloppement avec l&#8217;approche classique offre un r\u00e9sultat plus pr\u00e9cis, il implique des calculs suppl\u00e9mentaires qui, pour de grandes valeurs de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, aboutissent \u00e0 des r\u00e9sultats approximativement \u00e9quivalents \u00e0 ceux de l&#8217;approche simplifi\u00e9e. Ce d\u00e9veloppement r\u00e9utilise plusieurs calculs d\u00e9j\u00e0 r\u00e9alis\u00e9s dans l&#8217;approche simplifi\u00e9e et se pr\u00e9sente comme suit :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(\\Omega) &amp; = &amp; \\ln\\left(\\dfrac{n!}{k!(n-k)!}\\right)= \\ln(n!) - \\ln((n-k)!) - \\ln(k!) \\\\ \\\\\n\n&amp; \\approx &amp; \\color{red}\\dfrac{1}{2}\\ln(2n\\pi)\\color{black} + n\\ln(n) - n \\\\ \\\\\n\n&amp; &amp; \\color{red}-\\dfrac{1}{2}\\ln(2(n-k)\\pi)\\color{black} - (n-k)\\ln(n-k) + (n-k) \\\\ \\\\\n\n&amp; &amp; \\color{red}-\\dfrac{1}{2}\\ln(2k\\pi)\\color{black} - k\\ln(k) + k\n\n\\end{array}<\/span>\n<p>Les parties en rouge repr\u00e9sentent les \u00e9l\u00e9ments suppl\u00e9mentaires pris en compte dans l&#8217;approche classique, tandis que le reste correspond aux r\u00e9sultats d\u00e9j\u00e0 obtenus avec l&#8217;approche simplifi\u00e9e. \u00c0 partir de cela, nous obtenons :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(\\Omega) &amp; \\approx &amp; \\color{red}\\dfrac{1}{2}\\ln\\left( \\dfrac{2n\\pi}{2(n-k)\\pi \\cdot 2k\\pi} \\right)\\color{black} + k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) \\\\ \\\\\n\n&amp; = &amp;  k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right)\n\n\\end{array}<\/span>\n<p>Puis, en appliquant la conversion en logarithmes d\u00e9cimaux :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(\\Omega) = \\log(e)\\ln(\\Omega) \\approx \\log(e) \\left[ k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right) \\right] <\/span><\/span><\/p>\n<p>Et enfin, en passant \u00e0 l&#8217;exponentielle d\u00e9cimale :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\Omega \\approx 10^{\\log(e) \\left[ k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right) \\right]}\n\n<\/span>\n<p>Maintenant, de mani\u00e8re similaire \u00e0 l&#8217;approche simplifi\u00e9e, nous pouvons estimer la valeur maximale de cette grandeur en \u00e9valuant <span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span>. Dans ce cas, le r\u00e9sultat est le suivant :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Max}(\\Omega) &amp;\\approx &amp; 10^{\\log(e) \\left[ \\dfrac{n}{2} + \\dfrac{n}{2}\\ln\\left(\\dfrac{n}{(n\/2)} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2(n\/2)\\pi(n-n\/2)}{n}\\right) \\right]} \\\\ \\\\\n\n&amp; = &amp; 10^{\\log(e) \\left[\\dfrac{n}{2} - \\dfrac{1}{2}\\ln\\left(\\dfrac{n\\pi}{2} \\right) \\right]} = 10^{\\log(e)(n-\\ln(n\\pi\/2))\/2}\n\n\\end{array}<\/span>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Exemples de calculs combinatoires et d&#8217;ordre de grandeur<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/M7NrtICrSzU?si=6UT34333Wp5YwFU1\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p><a name=\"7\"><\/a><\/p>\n<h3>Cas 1 : Grands factorielles<\/h3>\n<p>Estimons l&#8217;ordre de grandeur de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span><\/span>, c&#8217;est-\u00e0-dire le nombre de chiffres n\u00e9cessaires pour \u00e9crire ce nombre.<\/p>\n<h5>Solution<\/h5>\n<p>Pour effectuer ce calcul, nous utilisons la formule de Stirling de la mani\u00e8re suivante :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n\\ln\\left[ \\left(10^{50}\\right)! \\right] &amp;\\approx 10^{50}\\ln\\left(10^{50}\\right) - 10^{50}\\\\ \\\\\n\n&amp;= \\left[\\ln\\left(10^{50}\\right) -1\\right]10^{50}  \\\\ \\\\\n\n&amp;= \\left[50\\ln(10)-1 \\right]10^{50} \\\\ \\\\\n\n\\end{array}\n\n<\/span>\n<p>Ensuite, nous appliquons la conversion en logarithmes d\u00e9cimaux :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\ln\\left[ \\left(10^{50}\\right)! \\right] = \\dfrac{\\log\\left[\\left(10^{50}\\right)!\\right]}{\\log{e}}\n\n<\/span>\n<p>Par cons\u00e9quent :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\log\\left[ \\left(10^{50}\\right)! \\right] \\approx \\log(e)\\left[50\\ln(10)-1 \\right]10^{50}\n\n<\/span>\n<p>Enfin, en passant \u00e0 l&#8217;exponentielle d\u00e9cimale, nous obtenons :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\left(10^{50}\\right)!  \\approx  10^{\\log(e)\\left[50\\ln(10)-1 \\right]10^{50}} = 10^{49,5657 \\cdot 10^{50}}\n\n<\/span>\n<p>L&#8217;exposant au-dessus de 10 repr\u00e9sente l&#8217;ordre de grandeur, fournissant une estimation du nombre de chiffres que poss\u00e8de le nombre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span><\/span>.<\/p>\n<p><a name=\"8\"><\/a><\/p>\n<h3>Cas 2 : Grandes combinatoires<\/h3>\n<p>Une maison moyenne poss\u00e8de environ 12 interrupteurs, qui peuvent \u00eatre allum\u00e9s ou \u00e9teints. En moyenne, chaque maison abrite 4 personnes. Si une ville compte 5 millions d&#8217;habitants, combien de fa\u00e7ons possibles y a-t-il pour que la moiti\u00e9 des interrupteurs de la ville soient allum\u00e9s ?<\/p>\n<h5>Solution<\/h5>\n<p>Le nombre <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> total d&#8217;interrupteurs dans la ville est calcul\u00e9 comme suit :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\nn &amp;=&amp;\\dfrac{\\text{habitants dans la ville}}{\\text{personnes par maison}} \\times \\text{interrupteurs par maison} \\\\ \\\\\n\n&amp;=&amp; \\dfrac{5\\cdot 10^6}{4}\\cdot 12 = 15\\cdot 10^6\n\n\\end{array}\n\n<\/span>\n<p>L&#8217;\u00e9tat macro form\u00e9 par tous les micro-\u00e9tats dans lesquels la moiti\u00e9 des interrupteurs sont allum\u00e9s correspond \u00e0 l&#8217;\u00e9tat macro avec le plus grand nombre de configurations possibles. En d\u00e9signant ce nombre maximal comme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max}<\/span><\/span>, nous pouvons obtenir les estimations suivantes selon chaque m\u00e9thode :<\/p>\n<ul>\n<ol><strong>Estimation classique :<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max} = 10^{\\log(e)\\left[15\\cdot10^6 - \\ln\\left(15\\pi\\cdot10^6 \/ 2 \\right) \\right]\/2} \\approx 10^{6.514.413,542}<\/span><\/span><\/ol>\n<ol><strong>Estimation simplifi\u00e9e :<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max} = 10^{\\log(e)\\left[15\\cdot10^6 \\right]\/2} \\approx 10^{6.514.417,229}<\/span><\/span><\/ol>\n<\/ul>\n<p>Bien qu&#8217;il y ait une diff\u00e9rence d&#8217;environ 4 ordres de grandeur entre les deux estimations (ce qui peut sembler beaucoup), cela devient n\u00e9gligeable en comparaison avec les plus de 6,5 millions d&#8217;ordres de grandeur.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Probl\u00e8mes de combinatoire en thermodynamique Combien de fa\u00e7ons existe-t-il pour organiser un syst\u00e8me physique compos\u00e9 de millions d&#8217;\u00e9l\u00e9ments ? Dans ce cours, nous aborderons comment les math\u00e9matiques permettent de r\u00e9pondre \u00e0 des questions comme celle-ci dans le contexte de la thermodynamique, depuis la distribution des quanta d&#8217;\u00e9nergie dans les syst\u00e8mes atomiques jusqu&#8217;au calcul des configurations [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":30534,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":2,"footnotes":""},"categories":[647,931],"tags":[],"class_list":["post-31207","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-physique","category-thermodynamique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Probl\u00e8mes de combinatoire en thermodynamique - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Dans ce cours, nous examinerons une 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