# Tag Info

## New answers tagged integration

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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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For this sort of problem it really helps to go back to the Riemann sum. You don't even need to write it down per se; you just need to think about what your $\delta x$ is in the Riemann sum. So, in this case you're trying to find a "sum of forces" or we might say a "net" force. To find this, it helps a little to think about momentum conservation/Newton's ...

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Well in this case, you could consider a really thin ring of elemental thickness say t, then proceed to form the integral as you have done to that point... Now, you might write dS as tdx , dx being an elemental strip along the circumference of the ring which subtends an angle d(theta) at the center. So you may now rewrite dS as t* R *d(theta) , and proceed ...

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The variational operator $\delta$ satisfies the product rule $$\delta (A \wedge B) = (\delta A)\wedge B + A \wedge (\delta B)$$ To find the variation of $G \wedge \star G$ where $\star G$ is the Hodge dual of $G$, note that $$G \wedge \star G = \frac{1}{p!}G_{\mu_1 \ldots \mu_p} G^{\mu_1 \ldots \mu_p} \sqrt{|g|}d^{D}x$$ From this, it is immediately ...

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In principle, you could use exactly the same steps as are in those notes to derive the full 3D expression. You would have to substitute $\vec{r} = x \hat{\imath} + y \hat{\jmath} + z \hat{k}$ in step 2, and then follow the logic through from there. However, an easier way to find the result for point out of the chosen plane is to exploit the rotational ...

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So that is correct: $$G(x) = \frac{1}{x^2+i\varepsilon^2}$$ The equation $$\Box G(x) = \delta_4(x)$$ is satisfied since $$\lim_{\varepsilon\rightarrow 0} \frac{\varepsilon^2}{(x^2+i\varepsilon^2)^3} = \delta(x^2)$$ Note that $\delta(0) = \frac{1}{\varepsilon^4}$ The issue with $\delta(x^2-x_0^2)$ not being the same as $\delta(x^2+x_0^2)$ is ...

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$f(v)$ is a probability density, so it is convenient to set $$\int_0^{2V{_0}} \mathrm{d}v \, f(v) = 1$$ In your case this will give you an expression for $A$ in terms of $V_0$. Now, e.g. $$\int_{V_0}^{2V{_0}} \mathrm{d}v \, f(v)$$ is the probability that a particle has velocity between $V_0$ and $2V_0$. The average value of any function of $v$ can then ...

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$N$ is equal to the sum of the two integrals you have: $$\begin{eqnarray} N = \int_0^{2 V_0} dv \ f\left(v\right) &=& \int_0^{V_0} dv \ A + \int_{V_0}^{2 V_0} dv \left(-\frac{A}{V_0} v + 2 A\right) \\ &=& A V_0 + \left[ -\frac{A}{2 V_0} v^2 + 2 A V\right]_{V_0}^{2 V_0} \\ &=& \frac{3}{2} A V_0 \\ \end{eqnarray}$$

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The volume element is $(dr)*(rd \phi)*(dz)$. Hence, the extra r in your integrand should be eliminated.

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