Bose condensate in 4d - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-17T20:36:03Z https://physics.stackexchange.com/feeds/question/444367 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/444367 3 Bose condensate in 4d Ornella https://physics.stackexchange.com/users/214651 2018-11-30T23:59:36Z 2018-12-01T01:34:00Z <p>Could a boson gas condensate in a hypervolume <span class="math-container">$V$</span> in 4D? How can I find its critical temperature and the heat capacity? In the books it just said volume <span class="math-container">$V$</span>, it does not specify the dimension.</p> <p>My professor asked this and I have no ideia how to start this question. Please help me. It is a statistical mechanics subject at the university.</p> https://physics.stackexchange.com/questions/444367/bose-condensate-in-4d/444383#444383 2 Answer by SuperCiocia for Bose condensate in 4d SuperCiocia https://physics.stackexchange.com/users/37677 2018-12-01T01:28:19Z 2018-12-01T01:34:00Z <p>Yes.</p> <p>Th key thing here is that for non-interacting bosons the mean occupancy of each (single-particle) state <span class="math-container">$j$</span> is given by:</p> <p><span class="math-container">$$f(E_j) = \frac{1}{e^{(E_j - \mu)/kT}-1} .$$</span></p> <p>Now you see that the ground state <span class="math-container">$E_0$</span>, the occupancy is infinity. This is because for the <span class="math-container">$E_0$</span> state the chemical potential <span class="math-container">$\mu$</span> also needs to be zero, in order to guarantee <span class="math-container">$f$</span> to still be positive. Physically, the chemical potential is defined as <span class="math-container">$\partial U/\partial N$</span>, i.e. the energy added when you add one particle to the system. But if you add it to the <span class="math-container">$E=0$</span> state, then the extra energy is 0...</p> <p>Bose-Einstein condensation begins when you saturate the excited states and start macroscopically occupying the ground state, which has infinite occupancy.</p> <p>Below <span class="math-container">$T_c$</span>, <span class="math-container">$f(E_0)$</span> starts blowing up so it does not make sense using the above distribution anymore, since the atoms start amassing into the ground state. <br> So <span class="math-container">$T_c$</span> is extracted from when your total <span class="math-container">$N$</span> is equal to the number of atoms in the excited states, <span class="math-container">$N_{ex} = \int_0 ^{\infty}dE \, g(E) \, f(E)$</span> where <span class="math-container">$g(E)$</span> is the <em>density of states</em>, i.e. the number of states in a given interval <span class="math-container">$[E, E+dE]$</span>. The sum should have been over the <em>states</em>, but I changed it to the <em>energy</em> <span class="math-container">$E$</span> just by introducing this density of states term. </p> <p>The density of states <span class="math-container">$g(E)$</span> scales with the number of dimensions <span class="math-container">$d$</span>. For a free <span class="math-container">$d$</span> dimensional system it goes as <span class="math-container">$g(E) \propto E^{d/2 -1}$</span>, while for <span class="math-container">$d$</span> dimensional harmonic potential it scales as <span class="math-container">$g(E) \propto E^{d-1}$</span>.</p> <p>In general you can write:</p> <p><span class="math-container">$$g(E) \propto E^{\alpha -1},$$</span></p> <p>with <span class="math-container">$\alpha$</span> being the number of degrees of freedom in the system divided by 2. For free particles in <span class="math-container">$d$</span> dimensions, <span class="math-container">$\alpha = d/2$</span>, and for a <span class="math-container">$d$</span> dimensional harmonic potential, the degrees of freedom are <span class="math-container">$2d$</span> (<span class="math-container">$d$</span> translations and <span class="math-container">$d$</span> oscillations) so <span class="math-container">$\alpha = d$</span>. All agree with the above.</p> <p>The integral above can be rewritten as:</p> <p><span class="math-container">$$N_{ex} = \int_0 ^{\infty}dE \, g(E) \, f(E) \propto (k T_c)^{\alpha} \int_0 ^{\infty} dx\frac{x^{\alpha-1}}{e^x - 1}$$</span></p> <p>where I defined <span class="math-container">$x$</span> as <span class="math-container">$E/k T_c$</span>.</p> <p>The intregral</p> <p><span class="math-container">$$\int_0 ^{\infty} dx \frac{x^{\alpha-1}}{e^x - 1} = \Gamma(\alpha) \zeta(\alpha), \qquad \alpha &gt; 1$$</span></p> <p>with <span class="math-container">$\Gamma$</span> being the gamma function, <span class="math-container">$\zeta$</span> being the Riemann zeta function.</p> <p>Which gives you:</p> <p><span class="math-container">$$k T_c \propto \frac{1}{[\Gamma(\alpha) \zeta(\alpha)]^{1/\alpha}}.$$</span></p> <p>To have a BEC transiton, you want <span class="math-container">$T_c \neq 0$</span>, i.e. a non-trivial solution.</p> <p>In free space, <span class="math-container">$d=2,3,4$</span> have <span class="math-container">$\alpha = 1, 3/2, 2$</span>:</p> <p><span class="math-container">$$\begin{array}{ccc} \alpha &amp; \Gamma(\alpha) &amp; \zeta(\alpha) \\ \hline 1/2 &amp; \text{integral does not converge} \\ 1 &amp; 1 &amp; \infty \\ 3/2 &amp; \sqrt{\pi}/2 &amp; 2.612 \\ 2 &amp; 1 &amp; \pi^2/6 \\ \dots &amp; \dots &amp; \dots \end{array}$$</span></p> <p>So in a free system with <span class="math-container">$d = 1,2$</span> the only solution is <span class="math-container">$T_c = 0$</span>, but for <span class="math-container">$d&gt;2$</span>, <span class="math-container">$T_c$</span> is finite. </p> <p>--</p> <p>All the details and numerical factors can be found in books like Pethick &amp; Smith, and Pitaevskii &amp; Stringari.</p>