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The premise of the question is false. $\nu$ is never exactly 0. It tends toward 0 as $\mu_c$ decreases. However, it is always non-zero and positive. Hence, the value of $CV$ depends on the other variables in its definition.


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Let $I$ denote our integral: $$ I=\int{{e}^{x^2}(1+erfi(x))} dx $$ Using IBP: $$ u=(1+erfi(x))\quad dv={e}^{x^2}dx\\ du=\frac{2}{\sqrt{\pi}} {e}^{x^2}dx \quad v=erfi(x)\frac{\sqrt{\pi}}{2} $$ You get: $$ I=(1+erfi(x))erfi(x)\frac{\sqrt{\pi}}{2}-\int{erfi(x) {e}^{x^2}dx}\\ I=(1+erfi(x))erfi(x)\frac{\sqrt{\pi}}{2}-\int{erfi(x){e}^{x^2}-{e}^{x^2}dx} ...


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I don't think there is any universal "intuition" to tap into, aside from that which comes from practice. You perhaps need to explore different physics texts in the electromagnetic department. I for example loathed Jackson as a learning text: it is comprehensive and useful as a reference for refreshing knowledge, but not good at conveying it. Volume 2 of the ...


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TL;DR: Substitution inside the delta function yields a Jacobian factor $$ \tag{1} \delta(f(v))~=~ \sum_{v_{(0)},f(v_{(0)})=0 }\frac{1}{| f^{\prime}(v_{(0)})|} \delta(v-v_{(0)}). $$ Here the sum is over the zeroes $v_{(0)}$ of the function $f(v)$. Let us for simplicity consider velocity $v$ rather than momentum $p=mv$. So energy conservation $$\tag{2} ...


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For the second case you want to divide by the force, not the velocity. You are basically computing what fraction of the time you spend at a particular point in phase space. However what you have is a probability density. So $P(z,p)$ is something you multiply by a volume in phase space to get a probability. A correct way to get it would be to consider a ...


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Hint : In the Figure below what is the height of your infinitesimal cylinder ??? (a) The red one ($d\mathbf{h}$) ??? (b) The blue one ($d\mathbf{s}$) ???


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IMHO, the notation $\int_a^b\mathrm{d}x\,f(x)$ is much cleaner than $\int_a^b f(x)\,\mathrm{d}x$, because the integration variable ($x$) and its associated integral range $(\int_a^b$) are kept together. This is particularly important in lengthy and multi-dimensional integrals. Consider $$ \Upsilon_{pq}(k)= \int_0^\infty\mathrm{d}x ...


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Besides the reasons listed in Lubos Motl's answer, here is another reason for the $\int \!dx ~f(x)$ notation: By writing the integral sign $\int_a^b$ and $dx$ next to each other in multiple nested integrations, it becomes more easy to trace which limits belong to which integration. This becomes particularly handy when changing the orders of integration. ...


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It's not just QFT literature. Physicists, especially adult research physicists, find this notation sensible and popular – even though it may be more popular among particle physicists than elsewhere. Formally, $dx\,f(x)$ is a product of two factors and $\int$ is a form of a sum. Because product is commutative, it doesn't hurt when the order is interchanged. ...


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How did you integrate acceleration to get velocity? Note that $\Delta v = \int_{t_i}^{t_f} a(t) dt$ But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v ...


3

To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information. Since they give you the solution and you just have to ...


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I have solved my problem with some careful head scratching and reading. It turns out I misread the paragraph after Eq. (2.9). This is NOT the integration formula, just a statement of the boundary condition value for P as R -> inf and U -> constant. It provides the first two values needed for the rest of the inwards integration. The correct formula to use ...



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