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2

Since the total mechanical energy is conserved, from which we get $KE_i+PE_i=KE_f+PE_f$, it should be more clear now that $\theta_0$ is the angle at which you raise the pendulum (i.e., that term represents the $PE_i$). The terms on the left side your first centered equation are the final kinetic and potential energies, after some time $t$. In effect, you are ...

2

If you are sure that $f$ is continuous and does not vanish in the integration domain, it is by no means necessary making use of regularization theory of distributions. Consider the initial integral: $$F:= \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y \frac{1}{f(x,y)+\mathrm{i}\epsilon}.$$ It can be re-written as: $$F:= \int_{T} ... 36 It is exactly because we have a factor of \frac 1 2 in the area formula of a triangle. To understand what I'm saying, consider what is the v(t) graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :) 4 You know that you can pull a multiplicative constant out in front of an integral, right?$$\int cf(t)\mathrm{d}t = c\int f(t)\mathrm{d}t$$where f(t) is any function of t, like t^2 or t(2\text{ s} - t) (and c does not depend on t). Units can be part of that constant factor too. In this case, the constant factor is 4\mathrm{\ kg\ m/s^4}. The ... 2 This probably is why it is useful to use variables. If we just let q=4\,{\rm kg\,m/s^4} and h=2\,{\rm s}, then your force is$$ \mathbf{F}=qt\left(h-t\right)\hat{\mathbf{i}} $$We then integrate this over time t,$$ \int_0^t\mathbf{F}\,dt'=q\int_0^tt'\left(h-t'\right)dt'\,\hat{\mathbf{i}} $$we get,$$ ...

2

EDIT The method you want to use is ok, and gives a quick result. Here it is: I=\int\prod d\theta^{*}d\theta\theta_{k}^{*}\theta_{l}exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\int\prod d\theta^{*}d\theta ... 0 Start with \begin{align} \alpha(t) = \frac{d\omega}{dt}(t). \end{align} Integrate both sides from t_i to t_f; \begin{align} \int_{t_i}^{t_f}dt\,\alpha(t) = \int_{t_i}^{t_f}dt\, \frac{d\omega}{dt}(t) = \omega(t_f) - \omega (t_i) \end{align} The second equality on the right follows from the fundamental theorem of calculus which basically says that if ... 1 I don't think that you really understand integration. Let me clear this up for you. In that question there is a rod of length l. You know how to calculate gravitational force between two point masses but not in continuous mass bodies. If you apply the formula to find the gravitational force you don't know what to take the distance as because it is ... 2 First discretize the spacetime, assign a fermion pair \bar{\psi}(i) and \psi(i) at each point i. Then assuming operator \hat{A} is symmetric, hence which can be diagonalized by a unitary operator whose determinant is one, the path integral can be written in the following way:\int \Pi_{i} d\psi(i)d\bar{\psi}(i) e^{i \delta \sum ...

0

$\Longrightarrow$ Four different indices: $(\mu\neq\nu\neq\lambda\neq\sigma)$ $$(\mu\nu|\lambda\sigma)\neq(\mu\lambda|\nu\sigma)\neq(\mu\sigma|\nu\lambda)\hspace{4mm}\Longrightarrow\hspace{4mm}3\cdot\binom{100}{4}$$ $\Longrightarrow$ Three different indices: $(\mu\neq\nu\neq\lambda=\sigma)$  ...

4

The integral you wrote down would simply be computed as follows: \begin{align} \int_\Sigma f\,d\theta\wedge d\phi = \int_0^{2\pi}d\phi\int_0^\pi d\theta f(\theta, \phi) \end{align} You just "erase the wedge." The extra factor of $\sin\theta$ is included if you are integrating a 2-form $\omega$ that is proportional to the volume form; \begin{align} ...

0

Here I want to discuss this answer by Luboš Motl in detail: Basically, you have 4 indices for your wavefunctions, out of which $k=1,\dots,4$ can be different. There are ${n \choose k}$ different possibilities to choose $k$ different values from $n$ choices. If you have 4 different indices, there are 3 possibilities to divide them into pairs. For 3 ...

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