Wikipedia:Reference desk/Archives/Science/2024 August 29
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August 29
editRounding error in large numbers
editAt the last point of the tabel or matrix about the far future, there is something like:
"Because the total number of ways in which all the subatomic particles in the observable universe can be combined is
10
10
115
{\displaystyle 10^{10^{115}}},[152][153] a number which, when multiplied by
10
10
10
56
{\displaystyle 10^{10^{10^{56}}}}, disappears into the rounding error,"
My question is: How is the point about the rounding error valid? 2A02:8071:60A0:92E0:30AB:357:41DB:5492 (talk) 16:54, 29 August 2024 (UTC)
- Multiplying the numbers is the same as adding the exponents. So . And is negligible when added to . --Amble (talk) 17:45, 29 August 2024 (UTC)
- Or does the text try to say that is negligible compared to the rounding error in ? The intention is not clear to me. Why should these numbers be multiplied and what do rounding errors have to do with it? The argument should be, I think, that since there are "only" possible combinations, the specific combination that results in a repeat of the Big Bang is bound to occur sometime in the next years. However, this seems to assume that all combinations are about equally likely and ignores the effect of the expansion of the universe. I suspect SYNTH. --Lambiam 19:57, 29 August 2024 (UTC)
- how to calculate 10^{10^{115}}? I thought it's 10^(10×115)? 2A0D:6FC0:8EF:6000:9455:1667:D5E1:3858 (talk) 03:13, 2 September 2024 (UTC)
- , but . Exponentiation goes right to left; follow the braces. I trust you'll forgive my not writing out . —Tamfang (talk) 18:53, 3 September 2024 (UTC)
Photons question.
editWhen an electron collides with positron, and when a proton collides with the anti-proton, both situations they are transformed into 2 photons. Can those 2 photons be distinguished from the 2 situations? That is, can the 2 photons be traced to being formerly an electron or proton? They have different energy of initial states, different total spin? Thanks. 66.99.15.162 (talk) 17:46, 29 August 2024 (UTC).
- See electron–positron annihilation and annihilation. At low energy, electron-positron annihilation will produce two photons, and you can be sure they didn't come from proton-antiproton annihilation because the total energy is less than the rest mass of two protons. The annihilation of a proton and antiproton, or an electron and positron with higher energy, can produce a variety of end states including baryons and weak bosons. A proton-antiproton annihilation to two photons would be a fairly rare process, see [1]. So we're talking about fairly uncommon end states. The high energy electron-positron annihilation and the proton-antiproton annihilation could produce the same types of final states, but with different probabilities, so you can make some statistical inferences, especially if you can observe multiple events. --Amble (talk) 19:15, 29 August 2024 (UTC)
- Proton-antiproton annihilation in a pair of photons is very rare. The far more common outcome is a pair or triple of pions. Ruslik_Zero 20:05, 29 August 2024 (UTC)
And then, photons that were created from an electron moving up/down an orbital or shell, are obviously different than the above mentioned photons? Have different measurable properties? These properties (or just energy) are measured when a photon hits a solid, the energy measured by a photo-multiplier tube? 66.99.15.162 (talk) 20:22, 29 August 2024 (UTC).
- Well these photons have lower energy than those created in annihilation. Also they are usually produced one at a time, rather than a pair or triple. Graeme Bartlett (talk) 00:28, 30 August 2024 (UTC)
- But the concise answer to your question is: no. Once a photon is created, its only unique property is its energy. There is no difference between a 511 keV photon created from positron-electron annihilation or one created from any other source of 511 keV photons. See: indistinguishable particles. PianoDan (talk) 16:04, 30 August 2024 (UTC)