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In principle, it is very simple and straightforward. The problem is to map out the region where the integer filling state is the ground state. Suppose you have $L$ sites. Take $N=L$ particles, find its ground state energy, which is denoted as $E_g(L)$. Note that here the Hamiltonian does not contain the $\mu $ term. Do it again for $N=L+1$, the ground ...


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In hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume ...


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An alternative to the conservative formalism of the Euler is the primitive form: $$ \frac{\partial\rho}{\partial t}+\nabla\cdot\rho\mathbf v=0 \\ \rho\frac{\partial\mathbf v}{\partial t}+\rho\mathbf v\cdot\nabla\mathbf v+\left(\gamma-1\right)\nabla\left(\rho e\right)=0 \\ \frac{\partial e}{\partial t}+\mathbf v\cdot\nabla ...


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Yes, these equations exist and can be derived from the partition function in JGab's answer. The internal energy per spin is: $$u(\beta) = - \frac{\partial}{\partial \beta} \left( \ln(2) + \frac{1}{8 \pi^2} \int_{0}^{2\pi} \mathrm{d} q_1 \int_{0}^{2\pi} \mathrm{d} q_2 \\\ln \left[ \big( 1 - \sinh(2 \beta J) \big)^2 + \sinh(2 \beta J) \left( 2 - \cos q_1 - ...


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I think you have a problem with double counting. The Ising Hamiltonian is $$H = - J \sum_{\langle i,j \rangle} S_i S_j$$ where this strange sum notation means to sum over all bonds between neighbouring spins. It is the bond that matters, so you should not count it twice (for $ij$ and $ji$). Actually you can get the energy difference of a spin flip as ...


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Even though your integration method seems wrong, like David Hammen pointed out, the results look do not look wrong. The problem rely lies with the way you define your ellipse. Your definition for the semi-major axis, $a$, and eccentricity, $e$, are correct. The way you define the semi-minor axis, $b$, probably is not, which is defined as: $$ b = a \sqrt{1 - ...


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Here's your key error: var dx = Add(Mul(gv , dt*dt) , Mul(o.v, dt)); // position changes In math, that's $\Delta \vec x = \vec g \Delta t^2 + \vec v \Delta t$. That's incorrect. Assuming a constant acceleration $\vec g$, the change in position over some time interval $\Delta t$ is given by $\Delta x = \frac 1 2 \vec g \Delta t^2 + \vec v \Delta t$. That ...



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