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August 29

Rounding error in large numbers

At 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)[reply]

Multiplying the numbers is the same as adding the exponents. So

10^{10^{115}}\times 10^{10^{10^{56}}}=10^{\left(10^{115}+10^{10^{56}}\right)}

. And

10^{115}

is negligible when added to

10^{10^{56}}

. --Amble (talk) 17:45, 29 August 2024 (UTC)[reply]

Or does the text try to say that

10^{10^{115}}

is negligible compared to the rounding error in

10^{10^{115}}\times 10^{10^{10^{56}}}

? 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"

10^{10^{115}}

possible combinations, the specific combination that results in a repeat of the Big Bang is bound to occur sometime in the next

10^{10^{10^{56}}}

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)[reply]

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)[reply]

(2^{3})^{2}=(2^{2})^{3}=2^{6}

, but

2^{3^{2}}=2^{(3^{2})}=2^{9}

. Exponentiation goes right to left; follow the braces. I trust you'll forgive my not writing out

10^{115}

. —Tamfang (talk) 18:53, 3 September 2024 (UTC)[reply]

Photons question.

When 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).[reply]

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)[reply]

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)[reply]

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).[reply]

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)[reply]

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)[reply]