Timeline for Why don't people use Hamilton's equations for a relativistic free particle?
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29 events
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| Aug 27, 2022 at 15:26 | comment | added | Valter Moretti | Thanks @Qmechanics,... you are a hybrid like me. | |
| Aug 27, 2022 at 14:57 | comment | added | Qmechanic♦ | Hi @Valter Moretti. Some but not all subjects are related to research, teaching & supervision. Formally, I have a Ph.D. in theoretical physics, but with some courses in mathematical physics. | |
| Aug 27, 2022 at 14:25 | comment | added | Valter Moretti | Hi @Qmechanic, there is a thing I would like to ask you. Are these subjects part of your teaching activity? Research activity? Both? I am not able to focus on your research field. You seem to me a theoretical physicists but sometimes you come out with issues which are, at least in my country, more appropriate for mathematical physics. | |
| Aug 27, 2022 at 13:54 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Dec 30, 2021 at 13:02 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Dec 29, 2021 at 23:06 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Oct 22, 2021 at 19:33 | comment | added | Qmechanic♦ | $=\frac{r^{2-d}}{4\pi^{d/2}}\Gamma\Big(\frac{d}{2}-1,\frac{(r\Lambda)^2}{2}\Big)$ $\stackrel{\text{IBP}}{\sim}\frac{\Lambda^{d-4}}{(2\pi)^{d/2}r^2}\exp\left[-\frac{(r\Lambda)^2}{2}\right]$ for $r\Lambda\gg 1$. | |
| Oct 13, 2021 at 18:48 | comment | added | Qmechanic♦ | 3. Case $\frac{m}{\Lambda}\ll r\Lambda$: $\quad G_H({\bf r})\approx\int_0^{\Lambda^{-2}}\!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~\exp\left[-\frac{r^2}{2e}\right]$ $=\int_0^{2\Lambda^{-2}}\!\frac{de}{4}~\frac{1}{(\pi e)^{d/2}}~\exp\left[-\frac{r^2}{e}\right]$ $=\int_{\Lambda^2/2}^{\infty} \!\frac{du}{4\pi^{d/2}}~u^{d/2-2}~e^{-ur^2}$ $=\frac{r^{2-d}}{4\pi^{d/2}}\int_{(r\Lambda)^2/2}^{\infty}\!du~\underbrace{u^{d/2-2}}_{\text{diff}}~\underbrace{e^{-u}}_{\text{int}}$ | |
| Oct 10, 2021 at 20:39 | comment | added | Qmechanic♦ | 1. Case $\frac{m}{\Lambda}>r\Lambda\gg 1 \Rightarrow \Lambda^{-2}> r/m$, i.e. stationary point is inside integration interval. Then $m\gg \Lambda \gg r^{-1}$ dominates $\Lambda$: $\quad G_H({\bf r}) \approx G({\bf r})$ for $mr\gg 1$. 2. Case $ r\Lambda > \frac{m}{\Lambda} \gg 1 \Rightarrow \Lambda^{-2}< r/m$, i.e. stationary point is outside integration interval. $\quad G_H({\bf r}) \stackrel{\text{IBP}}{\sim}\frac{\Lambda^{d-2}}{(2\pi)^{d/2}((r\Lambda)^2-(m/\Lambda)^2)}\exp\left[-\frac{m^2}{2\Lambda^2} -\frac{(r\Lambda)^2}{2}\right]$ for $r\Lambda\gg 1$. | |
| Oct 10, 2021 at 19:59 | comment | added | Qmechanic♦ | High/heavy modes: $G_H({\bf r})=G({\bf r})-G_L({\bf r})$ $=\int_0^{\Lambda^{-2}}\!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(e)}$ $=\Lambda^{d-2}\int_0^1\!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(\Lambda^{-2}e)}$ $=\frac{r^{2-d}}{(2\pi )^{d/2}}\int_{(r\Lambda)^2}^{\infty}\!\frac{du}{2}~u^{d/2-2}~e^{-S(r^2/u)}$ $=\frac{r^{2-d}}{(2\pi )^{d/2}}\int_{(r\Lambda)^2}^{\infty}\! \frac{du}{2}~\underbrace{\frac{u^{d/2-2}}{1-(mr/u)^2}}_{\text{diff}}~\underbrace{(1-(mr/u)^2)e^{-S(r^2/u)}}_{\text{int}}$. High/heavy action is not natural. | |
| Oct 9, 2021 at 12:05 | comment | added | Qmechanic♦ | $S_{H,2}[\phi_H]=\frac{1}{2}\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}~\widetilde{\phi}_H(-{\bf k})(k^2+m^2) \frac{1}{1-K}\widetilde{\phi}_H({\bf k})$. Low/light modes: $G_L({\bf r}) =\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}\frac{e^{i{\bf k}\cdot{\bf r}}}{k^2+m^2}\underbrace{\exp\left[-\frac{k^2+m^2}{2\Lambda^2}\right]}_{=K}$ $=\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}e^{i{\bf k}\cdot{\bf r}}\int_{\mathbb{R}_+}\!\frac{de}{2}~ \exp\left[-\frac{1}{2}(e+\Lambda^{-2})(k^2+m^2)\right]$ $=\int_{\Lambda^{-2}}^{\infty}\!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(e)}$. | |
| Oct 8, 2021 at 21:50 | comment | added | Qmechanic♦ | Low/light action: $S_L[\phi_L]=S_{L,2}[\phi_L]+\ldots$, where $S_{L,2}[\phi_L] =\frac{1}{2}\int_{\mathbb{R}^d}\!d^dr~\phi_L({\bf r})(-\nabla^2+m^2)\exp\left[\frac{-\nabla^2+m^2}{2\Lambda^2}\right]\phi_L({\bf r})$ $=\frac{1}{2}\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}~\widetilde{\phi}_L(-{\bf k})(k^2+m^2)\underbrace{\exp\left[\frac{k^2+m^2}{2\Lambda^2}\right]}_{=1/K}\widetilde{\phi}_L({\bf k})$. High/heavy action: $S_H[\phi_H]=S_{H,2}[\phi_H]+\ldots$, where | |
| Oct 8, 2021 at 20:59 | comment | added | Qmechanic♦ | Wilsonian effective action is the generator of connected diagrams: $\exp\left\{-\frac{1}{\hbar}W_c[J^H,\phi_L] \right\} := \int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right\}$, where $\sqrt{k_L^2+m^2} < \Lambda \equiv \Lambda_L < \sqrt{k_H^2+m^2}< \Lambda_H$. Note that the cut-off is species dependent $m\ll\Lambda$. Non-$x$-local action terms are exponentially suppressed. Hm. The mass $m$ might run with $\Lambda$. physics.stackexchange.com/q/602474/2451 physics.stackexchange.com/q/254260/2451 | |
| Oct 7, 2021 at 21:46 | comment | added | Qmechanic♦ | Here the action $S(e)=\frac{em^2}{2}+\frac{r^2}{2e}\approx mr$. Stationary point: $0\approx S^{\prime}(e)=\frac{m^2}{2}-\frac{r^2}{2e^2}$, so that $e\approx r/m$. And $S^{\prime\prime}(e)=\frac{r^2}{e^3}\approx \frac{m^3}{r}$. Steepest descent method yields $G({\bf r}) \sim \frac{1}{2}~\frac{1}{(2\pi e)^{d/2}}~\sqrt{\frac{2\pi}{S^{\prime\prime}(e)}}e^{-S(e)} = \frac{1}{(2\pi)^{d/2}}\Big(\frac{m}{r} \Big)^{\frac{d}{2}-1} \sqrt{\frac{\pi}{2mr}}e^{-mr}$ for $mr\gg 1$. Idea: Substitution $e=\exp(u)$ changes integration region to whole $\mathbb{R}$. | |
| Oct 7, 2021 at 10:23 | comment | added | Qmechanic♦ | $G({\bf r}) =\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d} \frac{e^{i{\bf k}\cdot {\bf r}}}{k^2+m^2}$ $=\int_{\mathbb{R}^d}\! \frac{d^dk}{(2\pi)^d} e^{i {\bf k}\cdot {\bf r}} \int_{\mathbb{R}_+}\!\frac{de}{2}~\exp\left[-\frac{e}{2}(k^2+m^2)\right]$ $=\int_{\mathbb{R}_+}\!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(e)}$ $=\frac{1}{(2\pi)^{d/2}}\Big(\frac{m}{r}\Big)^{\frac{d}{2}-1}K_{\frac{d}{2}-1}(mr)$ $\longrightarrow \frac{r^{2-d}}{(d-2){\rm Vol}(\mathbb{S}^{d-1})}$ $=\frac{\Gamma(d/2-1)}{4\pi^{d/2}}r^{2-d}$ for $m\to 0^+$. Here ${\rm Vol}(\mathbb{S}^{d-1}) =2\frac{\pi^{d/2}}{\Gamma(d/2)}$. | |
| Oct 7, 2021 at 7:35 | comment | added | Qmechanic♦ | Notes for later: Greens function for Helmholtz operator $(-\nabla^2+m^2) G({\bf r}) = \delta^d({\bf r})$ in $d$ dimensions (Euclidean signature). | |
| Oct 7, 2021 at 7:24 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Dec 14, 2020 at 14:39 | comment | added | Qmechanic♦ | Notes for later: The KG Lagrangian density ${\cal L}=\frac{1}{2}\left(\pm(\partial\phi)^2 -m^2\phi^2\right)$ has Fourier transform $\widetilde{\cal L}=\frac{1}{2}\left(\pm k^2 -m^2\right)\widetilde{\phi}(k)\widetilde{\phi}(-k)$. From the on-shell EL eq. $\left(\pm k^2 -m^2\right)\widetilde{\phi}(k)\approx 0$ we get the dispersion relation $\pm k^2 \approx m^2$. | |
| Dec 14, 2020 at 14:24 | comment | added | Qmechanic♦ | Notes for later: Here in point mechanics the momentum is in target space. In FT it is of interest to Fourier transform in the world volume. E.g. the Klein-Gordon (KG) Hamiltonian density ${\cal H} =\frac{1}{2}\left( \pi^2+(\nabla\phi)^2 +m^2\phi^2\right)$ gets its square root from a Lorentz covariant square root normalization of creation & annihilation operators. | |
| Dec 12, 2020 at 21:59 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Jul 9, 2019 at 16:25 | comment | added | Qmechanic♦ | Notes for later: $\quad{\rm Re}(i\Delta \tau)>0$; $\quad \int_{\mathbb{R}_+} \!\frac{\mathrm{d}e}{e^{1-\nu}}\exp\left[-ae-\frac{b}{e}\right] ~=~2\left(\frac{b}{a}\right)^{\nu/2} K_{\nu}\left(2\sqrt{ab}\right) $; $\quad \int_{\mathbb{R}_+} \!\frac{\mathrm{d}e}{e^{1+\nu}}\exp\left[-ae-\frac{b}{e}\right] ~=~2\left(\frac{a}{b}\right)^{\nu/2} K_{\nu}\left(2\sqrt{ab}\right) $; $\qquad {\rm Re}(a), {\rm Re}(b)~>~0 $; Integral (6) is a modified Bessel function $K_{1/2}$. | |
| Jul 22, 2017 at 5:38 | vote | accept | Candy Man | ||
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Sep 18, 2016 at 10:24 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Sep 17, 2016 at 21:09 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Sep 17, 2016 at 18:10 | comment | added | Qmechanic♦ | Notes for later: Observation: We Wick rotate in target space $p^0_E=i p^0_M$, while ${\bf p}$ is not Wick rotated, cf. e.g. physics.stackexchange.com/q/275918/2451 , physics.stackexchange.com/q/313921/2451 | |
| Sep 17, 2016 at 15:32 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Apr 22, 2016 at 16:46 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Apr 22, 2016 at 16:35 | history | answered | Qmechanic♦ | CC BY-SA 3.0 |