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1

But the acceleration is not a partial derivative! Its a total derivative, $\frac{\mathrm dp}{\mathrm dt}$, with a $\mathrm d$ instead of a $\partial$. Anyway, I guess you might want to read about the Hamilton-Jacobi equation.


4

That formula holds for a simple pendulum of length $L$ in a gravitational field $g$ and released from rest with an amplitude $\theta_0$. Since this system is conservative its mechanical energy is constant and equals the gravitational potential energy when it is released. Setting the zero of potential energy at the fixed point of the pendulum, the mechanical ...


3

The notations $\int f(r)\, dr$ and $\int dr\,f(r)$ are equivalent. The latter is convenient if you have nested integrals, e.g. $$\int_0^\infty dr\int_0^\pi d\theta\,f(r,\theta),$$ so that you can see which integration limit belongs to which integration variable. The notation $d^3r$ is a shorthand for "integrate over 3D space", for example $dx\,dy\,dz$ if ...


8

The delta function $\delta(x)$ has unit area, but the function $\delta(2x)$ is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is $$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$ where the $x_i$ are the roots of $f$. In ...


4

You are confused due to the vectorial form of the equation, so you should write it by components (I will use cartesian coordinates), and I will use $\partial_x = \dfrac{d}{dx}$ for comfort, that being said, your equation can be written as: $$ F=-\int{(dr)}{(\vec{\nabla} \cdot \vec{P}) \vec{E} }=-\int{(dr)}{(\partial_x P_x +\partial_y P_y + \partial_z P_z ) ...


14

This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous. In this context, just to give you one overview, a distribution is a functional on the space of test functions. We define the space of test functions over $\mathbb{R}$ as $\mathcal{D}(\mathbb{R})$ being the ...


10

This is not a peculiar physicist notation oddly enough. The notation allows one to interpret $1/x$ as a distribution (which makes sense since it's being added to the delta distribution on the right hand side of the equation). For a suitable test function $\varphi$, one defines this distribution as $$ \mathrm{pv}(1/x)(\varphi) = \lim_{\epsilon\to ...


2

The parallel axis theorem should still be applicable - but note that it states that the moment of inertia about an arbitrary axis is always greater than the moment of inertia about the axis through the center of mass. So in your case after calculating the moment of inertia through D you have to subtract mr^2 as you move the axis to the center of mass. Did I ...



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