Wikipedia:Reference desk/Archives/Science/2024 September 14
| Science desk | ||
|---|---|---|
| < September 13 | << Aug | September | Oct >> | September 15 > |
| Welcome to the Wikipedia Science Reference Desk Archives |
|---|
| The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
September 14
editAtomic electron transition time
editIn Wikipedia article Atomic electron transition it is written that:
"The time scale of a quantum jump has not been measured experimentally",
why ?
Malypaet (talk) 12:07, 14 September 2024 (UTC)
- Because nobody knows how to do it (or if they do, they cannot yet implement it in an experiment). An attosecond is a rather short time. The article lists the shortest laser light pulse created as 43 attoseconds, so that is the shortest time scale that is technically accessible at the moment. But if you have an idea, go for it. --Wrongfilter (talk) 12:35, 14 September 2024 (UTC)
- There may be a more fundamental reason. Such a measurement would necessarily (I think) require two observations, one essentially being that a some time t0 the transition has not yet taken place, the other that at a later time t1 the transition has now occurred. Such observations require an electromagnetic interaction, which will unavoidably disturb the observed system, in particular potentially causing the electron to behave differently. If there is some clever way around this fundamental issue, no one has thought of it. --Lambiam 13:07, 14 September 2024 (UTC)
- Apparently, to date, we do not have the tools precise enough for this measurement. On the other hand, the disturbance of the measurement can be anticipated and circumvented, as in this article measuring the delay of a photoemission by the 2023 Nobel Prize winner L'Huillier:
"The determination of photoemission time delays requires taking into account the measurement process, involving the interaction with a probing infrared field. This contribution can be estimated using a universal formula and is found to account for a substantial fraction of the measured delay."
In the past this delay was considered zero, today it is measured around 10 atoseconds.
So...
Malypaet (talk) 21:45, 14 September 2024 (UTC)- Malypaet, the imminent development of a practical Nuclear clock, which will be able to measure smaller intervals of time than any possible Atomic clock, may soon enable the precision necessary to measure quantum-jump timescales. {The poster formerly known as 87.81.230.195} 94.6.83.137 (talk) 14:36, 17 September 2024 (UTC)
- Apparently, to date, we do not have the tools precise enough for this measurement. On the other hand, the disturbance of the measurement can be anticipated and circumvented, as in this article measuring the delay of a photoemission by the 2023 Nobel Prize winner L'Huillier:
- There may be a more fundamental reason. Such a measurement would necessarily (I think) require two observations, one essentially being that a some time t0 the transition has not yet taken place, the other that at a later time t1 the transition has now occurred. Such observations require an electromagnetic interaction, which will unavoidably disturb the observed system, in particular potentially causing the electron to behave differently. If there is some clever way around this fundamental issue, no one has thought of it. --Lambiam 13:07, 14 September 2024 (UTC)
In traditional quantum theory it's literal a discontinuity, giving rise to the measurement problem. Web search for "speed of quantum decoherence" mentions dechoherence times as short as 10-31 seconds, far below attoseconds. I'm not sure if that's the same as the collapse time. Gravity may be involved, quantized gravity still being a notoriously unsolved problem. 2601:644:8581:75B0:0:0:0:C813 (talk) 00:31, 26 September 2024 (UTC)
How does one deal with equations with incorrect units?
editSpecifically thinking this one, equation 31. It doesn't seem to yield a metric unit when you put in some values. Jo-Jo Eumerus (talk) 16:52, 14 September 2024 (UTC)
- Do you expect us to read and understand the entire paper up to that equation? Maybe you could help us a bit by summarising what the terms stand for, what their units are and why you think the entire equations do not yield "metric" units. --Wrongfilter (talk) 17:06, 14 September 2024 (UTC)
- It should. All the units in the paper are SI. Script-H in the paper should be heat flux (e.g. conductive thru a 2D surface i.e. an ice sheet), which will be in Joules per second per meter^2. Make sure to write down every step carefully, using the values given in the paper, including the table of constants. SamuelRiv (talk) 17:08, 14 September 2024 (UTC)
- The problem is that the first equation after doing that insertion yields (kg^7*m^11)^(1/10) which is obviously wrong. JoJo Eumerus mobile (main talk) 18:48, 14 September 2024 (UTC)
- Ok, I got it to work, to give delta H in meters. The one constant that seemed hidden in the paper was b, but it has the same dimensions as b0 (eqn A2). Also they don't give the Coriolis coefficient explicitly, but that has rad/sec (1/s) units. Apart from that, all the other constants should be in there. SamuelRiv (talk) 19:44, 14 September 2024 (UTC)
- Is it the Coriolis coefficient rather than the rotation frequency? Jo-Jo Eumerus (talk) 09:02, 15 September 2024 (UTC)
- Yes, as defined after Eq. (10) of the paper, f represents the Coriolis coefficient, which is subsequently used in Eq. (31). Nanosci (talk) 16:30, 15 September 2024 (UTC)
- Is it the Coriolis coefficient rather than the rotation frequency? Jo-Jo Eumerus (talk) 09:02, 15 September 2024 (UTC)
- Ok, I got it to work, to give delta H in meters. The one constant that seemed hidden in the paper was b, but it has the same dimensions as b0 (eqn A2). Also they don't give the Coriolis coefficient explicitly, but that has rad/sec (1/s) units. Apart from that, all the other constants should be in there. SamuelRiv (talk) 19:44, 14 September 2024 (UTC)
- The problem is that the first equation after doing that insertion yields (kg^7*m^11)^(1/10) which is obviously wrong. JoJo Eumerus mobile (main talk) 18:48, 14 September 2024 (UTC)