20. Potential and Riemann zeta function

 

In the previous article¹, the zeta function has the following relationship.

 

$\zeta(s)=\int x^{s-2}b(x)dx, g(x)=p(1/x-1), |x| \leq 1$

 

If $\zeta_{n}^{s}=\int x^{s-2}b_n(x)dx$,

 

(1) $\zeta_{n}^{s}=\frac{ b_n}{s-(n+1)}+\frac{p}{s-(n+1)}\zeta_{n+1}^{s-(n+1)}$.

 

($b_n=b_n(1), n=0, 1, 2, ...$)

 

Here, $\zeta_n(s)=\zeta_n^s, b_{n+1}(x)=-\frac{x^2}{p}\frac{d(b_{n}(x))}{dx}, b_0(x)=(\frac{1/x-1}{e^{g(x)}-1})'$. $\zeta_{n}^{s}$ or $\zeta_{n}(s)$ is not the same as $\zeta(s)$ in the Riemann zeta function. This is because $\zeta(s)=\int x^{s-2}b_0(x)dx$. When $n=0$, they are same.

 

If upper bound of $\zeta(s)$ is $x$, not $1$, it can be written like $\zeta(s,x)$.

 

Considering $x=\cos(\pi\theta/2)=\beta$,

 

$\theta^2 \approx \sin^2(\pi\theta/2)=1-x^2$ => $1-\zeta(3, x)$.

 

Therefore we can guess that $s$ in the $\zeta(s, x)$ is the number of elementary particles, and $x$ is $\theta=GM/rc^2$. It is notable that even if $b(x)$ is differentiated multiple times, $b_n$ has a finite value at $x=1$. This may be a description of quarks or the elementary particles that make up quarks. At this time, if the mass $M$ is the mass of the particle and is a constant, the gravitational constant can be calculated more accurately. If $e^{g(x)}-1$ is close to $g(x)$ in the macroscopic system, $\zeta(s, x)$ can be seen as representing potential from a particle perspective.

 

Last modified in 2024.01.24

 

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^{1. Refer to Riemann zeta function and quantum mechanics, Bernoulli sequence and Riemann zeta function}