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I don't see any problems up until maybe the very very last step (where you lost some constants at the very least and I can't tell what you were trying to do or why you think there is a problem). So you were at: $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx.$$ I'm not sure if you then tried a u-substitution or just found an ...


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Why is that? Where am I getting wrong? The proper approach is to note that the integrand for the $x$ component is an odd function of $x$. But first, recall that $$\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_{0}^a f(x)dx = \int_a^0 f(-x)dx + \int_{0}^a f(x)dx$$ Now, for an odd function, $$f(-x) = -f(x)$$ thus $$\int_a^0 f(-x)dx + \int_{0}^a ...


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Completeness of a space is a metric notion (in the most usual definitions). In the case at hand means that each Cauchy sequence of Hilbert space vectors converges in the topology inherited by the inner product structure (that also defines naturally a metric). The other fact, i.e. that each vector can be written as a linear combination of a particular set of ...


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Math is a tool for describing stuff in e.g. physics. I could say "Energy conservation means that any change in internal energy must be exactly the same as all heat gained but withdrawn the work done". Or I could say: $$\Delta U=Q-W$$ It is even hard to express in words without using words like "equal to", "added", "plus" and "minus". Math should be ...


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I think it is interesting to pose this question in the setting of a very simple hypothetical universe. Suppose that the universe consisted, for example, exclusively of perfectly hard spheres that move according to classical physics and collide with each other elastically. (Really, the specific rules of our toy universe don't matter for what I will say, as ...


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The existence of undecidable statements does not mean that we cannot hypothetically manage, adding suitable consistent axioms, to determine the truth of all physically relevant statements. Suppose you have a theory $L$ with a meaningful undecidable statement $A$, such that both $L+A$ and $L+\bar{A}$ (I use bar for negation) are consistent. Then you can ...



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