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Hot answers tagged mathematical-physics

3

This is easiest to see in 1D (where the 'volume' becomes an interval $[a,b]$ and the boundary consists of 2 points $a$ and $b$). Poisson's equation becomes a 2nd-order ODE, which means that the full solution has 2 integration constants. Imposing both Dirichlet and Neumann boundary conditions (BCs) would lead to 4 conditions (2 at $a$ and 2 at $b$), which ...

2

First, expand the logarithm into its Taylor series: \begin{align} S&\equiv \sum_{n\ge 1}n\log(1-e^{-nx}) \\ &= -\sum_{n\ge 1} n \sum_{k\ge 1} \frac{e^{-knx}}{k} \\ &= -\sum_{k \ge 1} \frac{1}{k}\sum_{n\ge 1}n\,e^{-nkx} \end{align} To sum the inner series, differentiate the following identity with respect to $\beta$, $$\frac{1}{1-e^{-\beta}}-1=\... 2 I think you just need to read more widely to encounter a broad range of number systems being useful in at least theoretical physics. I'll hyperlink to discussions of applications, but just mention the number systems themselves, as some have multiple applications. There are uses for p-adic numbers, split-complex numbers, dual numbers, quaternions, split-... 2 On page 50, it says we note that for a one-dimensional manifold that is special Kahler, the Ricci scalar is related to the invariant coupling by R+4=2\gamma_\text{inv}^2 and we present a three-dimensional plot of the Ricci scalar in fig. 10 On page 55, it says for large \psi, the Ricci scalar of the moduli space differs from its limiting value by ... 2 OP's question seems mainly spurred by the fact that there are different notions of topology. OP is mentioning general topology, while the topology of Feynman diagrams is described by geometric topology. 1 Well, isn't it the case that any sequence a_n that vanishes as n \to\infty will give rise to the same pressure as the model with no field? Indeed, one can bound$$ e^{-\beta ha_nn^2} Z_{\Lambda_n,\beta,0} \leq Z_{\Lambda_n,\beta,ha_n} \leq e^{\beta ha_nn^2} Z_{\Lambda_n,\beta,0}, $$so that$$ -\beta h a_n + \frac1{n^2}\log Z_{\Lambda_n,\beta,0} \leq \...

1

The short answer is that $\sin^2 (ax) /ax^2$ becomes increasingly localized at zero. The effective domain shrinks like $1/a$ while its value at zero is $a$. Moreover, $\int_{-\infty}^\infty \sin^2 (ax) /ax^2 = \pi$. The rest is math.

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