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added anamech 3
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@@ -85,3 +85,58 @@ Fuer kleine Schwingungen entwickle $V (x)$ um $x_0 $ (Position des Minimums)
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v (x)= v (x_0 )+1/2 V'' (x_0 ) (x-x_0 )^2 +1/6 V''' (x_0 ) (x-x_0 )^3 +1/24 V'''' (x_0 ) (x-x_0 )^(4) + ...
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Fuer kleine Amplituden sind die letzten Terme zu vernachlaessigen, da $O((x-x_0)^3 )$.
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$
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V (x)=V (delta_(x) )+1/2 V'' (x_0 )delta_(x) ^2 \
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m dot.double(delta)_(n) =m dot.double(x)=-V' (x)=-V'' (x_0 )delta_(x) \
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delta_(x) =x-x_0 ==> dot.double(delta)_(x) =dot.double(x)\
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==> m delta_(x) +V'' (x_0 )delta_(x) = 0, omega^2 _(0) = (V'' (x_0 )) / (m)
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$
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Periode allgemeine (beliebige Amplidtuden) geschlossene ahn, Umkehrpunkte $x_(3) ,x_(2) $
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$
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t=T,x=x_2 ,x_0 = x_3 ,t_0 = 0, x_0 = x_3
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$
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$
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t-t_0 = +- integral_(x)^(x_0 ) d x' (1) / (sqrt(2/m (E-V (x)))) \
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==> T=2 integral_(x_2 )^(x_3 ) d x (1) / (sqrt(2/m (E-V (x))))
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$
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+ $
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V (x)=1/2k x^2
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==> T=(2 pi) / (omega_0 ) , omega_(0) =k/m\
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==> omega_0 != omega_0 (a)
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$
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+ Anharmonischer Fall
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$
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==> omega=omega (a)
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$
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$
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V (x)approx k/2 x^2 +epsilon V_1 (x)+ "weitere Korrekturterme"
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$
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Dann wird dieser Ausdruck fuer das Potential in den fuer die Peiode eingesetzt
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$
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==> T = T_0 epsilon^(0) +I_1 epsilon +I_2 epsilon^2 + O(epsilon^3).
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$
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Das Ziel ist dann $I_1 $ zu berechnen.
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$
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[E-V (x)]^(-1/2) = [E-k/2x^2 -epsilon V_1 (x)]^(-1/2) \
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E=V (a)= [k/2 (a^2 -x^2 )-epsilon (V_1 (x)-V_1 (a))]^(-1/2) \
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= [k/2 (a^2 -x^2 )]^(-1/2) [1-epsilon A]^(-1/2)
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$
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$
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A=(V_1 (x)-V_1 (a)) / (k/2 (-x^2 +a^2 )) \
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(1-u)^(-1/2) approx 1 +1/2 u, "fuer" u "klein"
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$
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#highlight[TODO: refactor the calculations]
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