# Why shouldn't this be a physical reason for the Riemann Hypothesis to be true? [closed]

By the functional equation for the Riemann $\xi$ function, we know that if $\zeta(\sigma + it) \neq 0$, for real $\sigma, t$ $$\zeta(\sigma + it) \neq \zeta(1-\sigma + it)$$

if $\sigma \neq \frac{1}{2}$.

Due to extensive computational evidence, it is now known that all nontrivial zeros of $\zeta$ with imaginary part $t < 10^{18}$ have real part $\sigma = 1/2$, and together with the above inequality, this tells us that

$$\zeta(\sigma + it) \neq \zeta(1-\sigma + it)$$

for all $\sigma \neq 1/2$ and $t < 10^{18}$.

We also know that $\zeta(s)$ can be interpreted as a partition function of a certain quantum mechanical system in the thermodynamic limit where $s$ is the inverse of complex temperature, therefore the above can have a thermodynamic interpretation of its own: namely for $\sigma \neq 1/2$ and $t<10^{18}$, the thermodynamic properties of the system captured by the partition function are altered by the transformation $\sigma \rightarrow 1-\sigma$.

Note that whatever physical interpretation $\sigma$ has at $s_{1} = \sigma + it_{1}$, it is exactly the same as that at $s_{2} = \sigma + it_{2}$, where $t_{2} > t_{1}$, hence it follows that even at $t_{2} \geq 10^{18}$, the thermodynamic properties of the system captured by the partition function would also altered by the transformation $\sigma \rightarrow 1-\sigma$, so that for $\sigma \neq 1/2$,

$$\zeta(\sigma + it_{2}) \neq \zeta(1-\sigma + it_{2})$$

from which the Riemann Hypothesis follows ?

• I don't quite get what you mean by whatever physical interpretation $\sigma$ has at $s_{1} = \sigma + it_{1}$, it is exactly the same as that at $s_{2} = \sigma + it_{2}$. Could you elaborate how the "whatever" interpretation of $\sigma$ is related to values of $\zeta$ at $s_1$ and $s_2$? Feb 14, 2017 at 11:43
• That a transformation ($\sigma\mapsto 1-\sigma$) is not a symmetry of a system at one point does not mean it does not become a symmetry of the system at another point in "temperature space". Worse, even if it is not a symmetry, it might, by some accident, be the case that the values for the partition function are still equal. Your argument is unclear. Feb 14, 2017 at 12:55
• Feb 14, 2017 at 17:16
• @ACuriousMind, I'm becoming quite curious why you do not want me to attach my real name to this post even in the commentary section ! Because it seems the rules are not against that.
– Q_p
Feb 17, 2017 at 9:02
• @user Theta. You can put your real name in your profile. Feb 23, 2017 at 1:23