Timeline for Wave packet in curved spacetime
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 20, 2016 at 15:15 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 21, 2016 at 7:20 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 14, 2016 at 16:19 | answer | added | Michael Seifert | timeline score: 3 | |
| Aug 6, 2015 at 5:08 | history | edited | DanielSank | CC BY-SA 3.0 |
title capitalization
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| Aug 6, 2015 at 3:46 | comment | added | Alex Nelson | @user39158 This has been discussed on usenet back in the day, but you may find similar discussion in, say, Aspects of Quantum Field Theory in Curved Spacetime by Stephen A. Fulling...or any introductory book on QFT in curved spacetime. | |
| May 26, 2015 at 9:46 | comment | added | Noix07 | found this link.springer.com/article/10.1007%2FBF02817959#page-1 | |
| May 26, 2015 at 9:36 | comment | added | Noix07 | I don't think you can recover the geodesic just from the field equation | |
| May 26, 2015 at 9:35 | comment | added | Noix07 | Can you give a reference for the first sentence of the post, "it is known that...". I've seen the geodesic equation for a point particle, i.e. an equation on a parameterized curve that has interpretation the space-time position as a function of time. I've also seen the KG field (ok I did not study it in curved space-times) and the duality wave-particle came from quantization of the KG field if I'm not mistaken? So I would say that if you want a geodesic equation you should first recover from the field the interpretation of position, somehow a position observable evaluated on a state, not just | |
| Apr 18, 2015 at 12:01 | history | tweeted | twitter.com/#!/StackPhysics/status/589398252019195907 | ||
| S Jul 9, 2014 at 9:08 | history | suggested | Nick Stauner |
additional tag (you may also want to fix "how can one derived")
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| Jul 9, 2014 at 8:58 | review | Suggested edits | |||
| S Jul 9, 2014 at 9:08 | |||||
| Jun 6, 2014 at 19:05 | answer | added | Robin Ekman | timeline score: 1 | |
| Jun 6, 2014 at 17:59 | comment | added | Valter Moretti | I was referring to that formula for the delta fuction based on Fourier transform. It generally fails since no global coordinates exist. | |
| Jun 6, 2014 at 16:31 | comment | added | user109798 | Yes, curved spacetime. I want to see how particles (wave packets) can realize the geodesic via the equation of motion. | |
| Jun 6, 2014 at 16:17 | comment | added | Valter Moretti | in curved spacetime? | |
| Jun 6, 2014 at 16:15 | comment | added | Zo the Relativist | Well, $\delta(x) = \frac{1}{2\pi}\int e^{ikx}dk$ | |
| Jun 6, 2014 at 16:11 | comment | added | user109798 | Actually, that's how I first did the problem. No, I don't see how one can get it from there, but there might be away to see that by choosing $\Phi \sim e^{ik^\mu x_\nu}$, then if from something like $k^2 = m^2$ condition one gets $k^\mu \nabla_\mu k^\nu = 0$ (this is the geodesic equation, as $k_\mu \sim u_\mu$, in classical limit), then problem solved. | |
| Jun 6, 2014 at 16:04 | comment | added | Zo the Relativist | Hint: assume $\phi = \delta(t(\tau))\delta(x(\tau))\delta(y(\tau))\delta(z(\tau))$ | |
| Jun 6, 2014 at 15:47 | history | asked | user109798 | CC BY-SA 3.0 |