2
$\begingroup$

I have a second question about the article "Imperfect Bose Gas with Hard-Sphere Interaction". The authors begins with the sum: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi ^2 a^2 \lambda ^2 }}{{V^2 }}\sum\limits_{} {'\frac{{\langle n_\alpha \rangle \langle n_\gamma \rangle \langle n_\lambda \rangle }}{{\frac{1}{2}\left( {k_\alpha ^2 + k_\beta ^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}\delta \left( {\vec k{}_\alpha + \vec k{}_\beta - \vec k{}_\gamma - \vec k{}_\lambda } \right)} $$ which represents the second order perturbation term of the energy. ${\langle n_\alpha \rangle }$ is: $$\langle n_\alpha \rangle = \sum\limits_{n = 0}^\infty {n\left( {ze^{ - \beta \varepsilon _\alpha } } \right)} ^n /\sum\limits_{n = 0}^\infty {\left( {ze^{ - \beta \varepsilon _\alpha } } \right)} ^n = \frac{{ze^{ - \beta \varepsilon _\alpha } }}{{1 - ze^{ - \beta \varepsilon _\alpha } }} $$ In the sum the terms with a vanishing denominator are omitted. There is also the restriction ${\vec k{}_\alpha \ne \vec k{}_\beta }\ \ $ and ${\vec k{}_\gamma \ne \vec k{}_\lambda }\ \ $. Now the passage that is unclear to me is the passage to the integral: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi a^2 \lambda ^2 }}{{V^2 }}\left( {\frac{{4V^3 }}{{\pi ^3 \lambda ^5 }}} \right)\sum\limits_{i,j,k}^\infty {\frac{{z^{j + k + l} }}{{\left( {j + k + l} \right)^{1/2} \left( {j - k} \right)l}}} \frac{{\partial J}}{{\partial u}} $$ where: $$J = \int\limits_0^\infty {\int\limits_0^\infty {dqdq\frac{{\cosh \left( {upq} \right)}}{{q^2 - p^2 }}} } e^{ - vq^2 - wp^2 } $$ and: $$u = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{2\left( {j - k} \right)l}}{{j + k + l}}} \right) $$ $$v = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l}}{{j + k + l}}} \right) $$ $$w = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l + 4jk}}{{j + k + l}}} \right) $$ $\frac{{\partial J}}{{\partial u}}$ is calculated here: Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ Do you have any suggestion?

$\endgroup$
1
$\begingroup$

Maybe we don't have to do thermal average of n and we need : $$n = \frac{1}{{z^{ - 1} e^{\varepsilon /kT} - 1}} = \sum\limits_{n = 1}^\infty {\left( {ze^{ - \varepsilon /kT} } \right)^l } $$ so: $$\sum\limits_{} {'... = \sum\limits_{i,j,k = 1}^\infty {\sum\limits_{} ' } } z^{ijk} \frac{{e^{ - \frac{{\hbar ^2 k_a^2j }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\gamma ^2k }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\lambda ^2l }}{{2m}}} }}{{\frac{1}{2}\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}} $$ This explain the factor $z^{ijk}$ and the sum. How can we continue?Now we have to transform the sum' in a integral. The density of state depends of $kdk$ so i think we have to evaluate: $$\int\limits_0^\infty {\frac{{e^{ - \frac{{\hbar ^2 k_a^2j }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\gamma ^2k }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\lambda ^2l }}{{2m}}} }}{{\frac{1}{2}\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}k_a k_b k_\gamma k_\lambda } dk_a dk_b dk_\gamma dk_\lambda $$ We have to evaluate two integrals, i think.

$\endgroup$
0
$\begingroup$

I have tried to solve this integral: $$F\left( {a,b,c,d} \right) = \int {\frac{{e^{ - \frac{{\hbar ^2 }}{{2m}}\left( {ak_a^2 + bk_b^2 + ck_\gamma ^2 + dk_\lambda ^2 } \right)} }}{{\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}k_a k_b k_\gamma k_\lambda dk_a dk_b dk_\gamma dk_\lambda } $$ from which we can calculate our integral putting $b=0$. We have: $$\frac{{\partial F}}{{\partial a}} + \frac{{\partial F}}{{\partial b}} - \frac{{\partial F}}{{\partial c}} - \frac{{\partial F}}{{\partial d}} \propto \frac{1}{{abcd}} $$ Now we put: $$i=a+b+c+d$$ $$l=a+b-c-d$$ $$m=a-b$$ $$n=c-d$$ So: $$\frac{\partial }{{\partial a}} = \frac{\partial }{{\partial i}} + \frac{\partial }{{\partial l}} + \frac{\partial }{{\partial m}} $$ $$\frac{\partial }{{\partial b}} = \frac{\partial }{{\partial i}} + \frac{\partial }{{\partial l}} - \frac{\partial }{{\partial m}} $$ $$\frac{\partial }{{\partial c}} = \frac{\partial }{{\partial i}} - \frac{\partial }{{\partial l}} + \frac{\partial }{{\partial n}} $$ $$\frac{\partial }{{\partial d}} = \frac{\partial }{{\partial i}} - \frac{\partial }{{\partial l}} - \frac{\partial }{{\partial n}} $$ $$\frac{\partial }{{\partial a}} + \frac{\partial }{{\partial b}} - \frac{\partial }{{\partial c}} - \frac{\partial }{{\partial d}} = 4\frac{\partial }{{\partial l}} $$ and: $$a=(i+l+2m)/4$$ $$b=(i+l-2m)/4$$ $$c=(i-l+2n)/4$$ $$d=(i-l-2n)/4$$ and the equation becomes: $$\frac{{\partial F'}}{{\partial l}} \propto \frac{1}{{\left( {\frac{{i + l}}{2} + m} \right)\left( {\frac{{i + l}}{2} - m} \right)\left( {\frac{{i - l}}{2} + n} \right)\left( {\frac{{i - l}}{2} - n} \right)}} $$ The integration can be evaluated with derive, and the substitutions gives an expression divergent if we put $b=0$. What is wrong?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.