9 votes
Accepted

Asymptotic rate of decay of the integrals

EDIT. After reading @Zarrax's answer, I realized that my original, heuristic guess cannot hold true. Below is a new answer. Performing integration by parts, $$ y_n = \frac{n}{2} \int_{0}^{\infty} x e^...
Sangchul Lee's user avatar
7 votes

Asymptotic rate of decay of the integrals

The $(1 - e^{-{x^2 \over 4}})^n$ factor starts becoming relevant when $e^{-{x^2 \over 4}}$ is about ${1 \over n}$, since then $(1 - e^{-{x^2 \over 4}})^n = (1 - {1 \over n})^n \sim {1 \over e}$. In ...
Zarrax's user avatar
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5 votes

Asymptotic rate of decay of the integrals

$$y_n= \int_{0}^\infty \Big(1-e^{-\frac{x^2}{4}})\Big)^n e^{-x}dx$$ $$y_n=1+ \sum_{k=1}^n (-1)^k \,\binom{n}{k}\int_{0}^\infty e^{-\frac{k x^2}{4}-x}\,dx$$ $$y_n=1+\sqrt \pi \sum_{k=1}^n (-1)^k \,\...
Claude Leibovici's user avatar
4 votes
Accepted

Discrepancy in Results with Self-Adjoint Operator on a Special Hilbert Space in 2D Geometric Algebra

The scalar product corresponding to the quadratic form $\|u\|^2 = u^{\ddagger}u$ is $(u, v) = \mathrm{re}(u^{\ddagger}v)$, where $\mathrm{re}(a + xe_1 + ye_2 + be_{12}) = a$. Self-adjointness of $g$ ...
Vladimir Lysikov's user avatar
2 votes

Why does this trick make the oscillating exponential integral converge?

I guess you encountered this result in the context of a physics topic, where the argument for such a trick is usually omitted or implicit. In fact, two "formalisms" are here merged together ...
Abezhiko's user avatar
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2 votes
Accepted

Inverse Proportionality

The second law does occur often in the sciences. It occurs in the context of conservation, e.g. conserved mass/charge etc. Think of Chemistry: if $x$ is the mass of one reactant and $y$ is the mass of ...
Joshua Tilley's user avatar
2 votes
Accepted

Show that the real K-G equation, $(\Box + m^2)\phi=0$ is the EOM for the action $S=\frac12\int d^4x(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2)$

We will use the following definition of functional differentiation: \begin{equation} \frac{\delta S}{\delta\phi(y)}=\lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(S[\phi+\epsilon\delta^{(4)}(x-y)]-S[\...
QuantumSuperfield's user avatar
2 votes

Show that the real K-G equation, $(\Box + m^2)\phi=0$ is the EOM for the action $S=\frac12\int d^4x(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2)$

Let ${M}$ be the spacetime manifold. Let there be an action functional \begin{align*} {S}[\Phi]=\frac{1}{2}\int_{M}\mathrm{d}^{4}{x}{\,}\sqrt{-\mathrm{det}({g}({x}))}{\,}\bigg({g}^{\mu\nu}({x})\frac{\...
DeVoyd's user avatar
  • 76
2 votes
Accepted

Christoffel symbol on $T^*M$

Leibniz rule is, for $X$ and $Y$ vector fields and $\alpha$ a $1$-form, $$ \nabla_X(\alpha(Y)) = (\nabla_X\alpha)(Y) + \alpha(\nabla_XY). $$ Applying to $\alpha = dx^k$, $X= \partial_i$ and $Y=\...
Didier's user avatar
  • 17.9k
2 votes
Accepted

Show that, in a normal coordinate chart, $\Gamma^{i}_{(x)(jk)}(p)=0$

It should have said $y(p)=(0,\dots, 0)$, so because of this, the $\Gamma^{i}_{(y)\,jk}(p)$ are just numbers, so they come out of all the derivatives, i.e you do not differentiate them at all. Anyway, ...
peek-a-boo's user avatar
  • 50.9k
1 vote
Accepted

Seeking 4x4 Real Matrix Representation of Generators for Clifford Algebra Cl(3,1)

$ \newcommand\Cl{\mathrm{Cl}} $The following was inspired by Proposition 15.20 of Clifford Algebras and the Classical Groups by Ian Porteous. Recall that the gamma matrices are $$ \gamma_0 = \begin{...
Nicholas Todoroff's user avatar
1 vote
Accepted

Definition of $a_{0}$ - what is wrong with my calculations?

It is because $w$ is compactly supported, thus far away the solution looks like that ($\Delta f = 0$ for $|x|>R$). Near the origin $\frac{df}{dr}$ should be bounded (because $f\in C^2$ near the ...
Shuhao Cao's user avatar
  • 18.8k
1 vote

Show that the real K-G equation, $(\Box + m^2)\phi=0$ is the EOM for the action $S=\frac12\int d^4x(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2)$

Step $(A)$. Actually, the action is a functional, that is a linear form defined on a functional vector space. Practically, it means that $S[\phi]$ takes a function $\phi$ as input and nothing else is ...
Abezhiko's user avatar
  • 6,320

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