# Tag Info

5

If you formulate the integral in the correct coordinates (i.e. in spherical polar coordinates with the axis along the inter-charge axis), then it reads $$E = \int_0^\infty\!\!\! \int_0^\pi\!\! \int_0^{2\pi} \frac{r-d\cos(\theta)}{\left(r^2+d^2 -2rd\cos(\theta)\right)^{3/2}} \sin(\theta) \,\mathrm d\phi \,\mathrm d\theta \,\mathrm dr ,$$ where $d$ is the ...

1

It is not possible to write $dt =-df/f^2$. It is not possible because time $t$ take on any value from $-\infty$ to $\infty$, whereas the frequency $f$ of a harmonic oscillator is a constant which is determined by parameters of the harmonic oscillator $$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$ But as it was not clearly stated in the post this might be not ...

1

The box operator $\Box = g^{αβ}\nabla_{α}\nabla_{β} = \nabla^{α}\nabla_{α}$ and the two covariant derivatives $\nabla_{μ}\nabla_{ν}$ act on the variation of the inverse metric tensor. You want $δg^{μν}$ to be a multiplying factor so you must integrate by parts twice (because you have two derivatives) to make the derivatives act on $\cfrac{df}{dR}$. The box ...

0

Both are correct as it is just a matter of wording although I would favour change in position and displacement most of the time. Consider one dimensional motion along the x-axis with the unit vector $\hat x$ defining the positive x-direction. A body starts at position $+3 \,\hat x$ with velocity $+6\,\hat x$ and after undergoing constant acceleration ...

1

You’re right. You’ll get the change in position $\Delta x$, which is the displacement. Maybe your teacher is confused about “displacement” ($\Delta x$) vs. “position” ($x$).

0

The crucial concept which is necessary for the integration of differential forms on manifolds is the parametrization of the manifold. Actually, this also needed for "normal integration" over a set $M \in \mathbf R^n$, furthermore vectors tangent to the parameterized manifold are needed. Let's go step by step. 1) Math on manifolds need charts and in general ...

3

A double integral in this context means you are integrating over a surface. The integral here is a double integral because a surface is parametrized by two parameters. I think your confusion lies in the notation. A surface integral is sometimes denoted with two integral symbols, but not always. So the integral in the definition of magnetic flux is no ...

2

But then what does $E_x(r)$ represent? The problem only said that $dE_x$ represents the field component at the point due to a ring $E_x(r)$ is a poor notation here since it certainly looks like the $x$ component of the electric field as a function of the radial coordinate $r$. Further, it is clear from the integral that $E_x = E_x(x)$. Here, $dE_x$ or $... 4 The left hand integral limits should be with respect to the field you get as you add up the contributions to the field due to each charged ring. Therefore, it should start at$0$and end with the final field. i.e. $$\int_0^{E_x}\text dE_x' = \int_0^R{1 \over 4\pi\epsilon_0}{2\pi\sigma rx\,\text dr \over (x^2+r^2)^{3/2}}$$ Trivially, the integral on the left ... 1 This is just complex analysis, in or out of context, so math.SE would've been a better choice, but let me try first: Following from Wikipedia, the only condition to use the Jordan's lemma is if your function is continuous on a semicircular contour ($z=re^{i\theta}:r\in\mathbb{R}^+,\theta\in[0,\pi]$). Since there are 2 poles at$k = \pm i\kappa$we also need ... 3 The fact that$I_2$, in the second form, vanishes is nothing but a direct application of the so-called Riemann-Lebesgue lemma. No computations are necessary. 1 Here is the right way to tackle this integral. We can rewrite: $$\int \frac{d^{2\omega} p}{(2\pi)^{2\omega}} \frac{p_\mu p_\nu}{(p^2)^{s+1}} e^{i p\cdot x} = -\partial_\mu\partial_\nu \int \frac{d^{2\omega} p}{(2\pi)^{2\omega}} \frac{e^{i p\cdot x}}{(p^2)^{s+1}} \tag{4}$$ and then use$(1)\$ to integrate.

0

If you are happy with a numerical recipe then by far the most genuine, easy and error-less method is a pseudo-data approach. Theoretical approach is very often just an approximation (linearization in formulas) and cannot take into account full distributions. Even full correlation matrix contains only a very concentrated information from situation which is ...

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