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From a mathematical perspective this means that it is not differentiable. The problem is that you need the discreteness to be able to count states. If you replace the discreteness by something smooth you get something differentiable, but your definition of entropy no longer makes sense. This is just one of the points where mathematicians cringe, but it works ...


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No. In mathematics we prove conjectures, theorems, etc. In science we do not prove a hypothesis. We run experiments to either support or refute the hypothesis, but as time goes on and new evidence becomes available conclusions can swing direction. Proof is an exercise pursued by mathematicians.


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Well asked. Wherever you try to find a continuum in nature, you will fail. An edge is a geometric abstraction, in reality there are atoms in a more or less geometric pattern. Energy is quantised in photons. So matter and energy are quantised. What about time? Don't know. The only thing I know is that we could measure time only in portions. So every event is ...


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It seems to me that there should be a mistake. Let $A^{\mu}(x)=\partial^{\mu} \theta (x)$. Clearly you have $F^{\mu \nu}=0$. Thus, you just have to find a function $\theta(x)$ such that: $$(\partial_{\nu} \partial_{\mu} \theta)(\partial^{\nu} \partial^{\mu} \theta)\neq \frac{1}{2} (\partial_{\nu} \partial^{\nu} \theta)(\partial_{\mu} \partial^{\mu} \theta). ...


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Continuing Dave's answer (and using his notation), a practical way to check for this kind of self-similarity is to draw a log-log plot of $u(x,t)$ as a function of $x$ for different values of $t$. If $u$ is self-similar, the different curves would all all have the same shape and would be related by translations. To see this, define $X \equiv \log x$ and $T ...


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What's going on is that the definition of the scaling variable $\xi=x^{-1/\beta} t$ (note: I prefer to parameterize it this way) defines how to resale the the two variables in a way that generates a scale transformation: $$ u(\lambda^\beta x, \lambda t) = A \lambda^{-\alpha} t^{-\alpha} f( x^{-1/\beta} t ) = \lambda^{-\alpha} u(x,t) $$ by using ...



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