Talk:Betelgeuse

Any chance it ALREADY exploded and its flash and shock wave are on the way?

Hate to be a doom-and-BOOM! kinda guy, but hey, I'm a pessimist by nature: since the experts can't for various reasons nail down its correct distance, is it not at least faintly possible that:

(1) it was positioned an even 500 light years out [supposing a number well within the accepted plus-or-minus distance estimates] when (2) it already went supernova 499 years ago in 1525 [three years after Magellan's circumnavigating expedition limped back to Spain, incidentally] and that therefore (3) sometime in 2025 we can expect to watch it instantly flare up, and then many years hence (4) be buffeted (or worse) by the stellar detonation's shock wave?

I hope someone who better understands all this can assure me that there's just NO WAY such a scenario will play out, because I'm losing sleep over this in the meantime.

Thanks in advance for any reassuring words; again, I realize what I'm suggesting is hugely UNLIKELY, but I'd like to know it's instead FLATLY IMPOSSIBLE, given what we already definitely know about this strange star. [signed] FLORIDA BRYAN 2600:1700:B9B0:4500:B410:1586:5924:785D (talk) 05:45, 28 May 2024 (UTC)

Wikipedia, even on its talk pages, is not a forum for this sort of debate; there are many online astronomical communities for this. To answer your question without getting drawn into a discussion, yes, uncertainties in the lifetime remaining to Betelgeuse make it possible that the time to supernova as observed from Earth is less than its light-time distance, and so the explosion may have "already" happened. —BillC ^talk 08:21, 28 May 2024 (UTC)

The supernova is expected to happen about 90,000 years in the future, so keep calm. InTheAstronomy32 (talk) 10:45, 28 May 2024 (UTC)

Much appreciate your explication and grudging admission of the remote possibility of my doomsyear assertion, BillC...as absolutely UNreassuring as it is!; accordingly, can't see how I can relax anytime over the next 90,000 years or so. Hope I'm not met with a shotgun in my face if I come knocking on your repurposed fallout shelter's door. [signed] FLORIDA BRYAN 2600:1700:B9B0:4500:2101:5010:7B23:8644 (talk) 18:11, 28 May 2024 (UTC)

Luminosity margins of error

Joyce et al (2020) contains two values for the (log) luminosity: 4.94+0.06
−0.04 and 4.94+0.10
−0.06. The second one resolves to the values in the starbox: raw value 10^4.94 => 87,096, rounded to 87,100; upper margin 10^5.04 => 109,648 (87,096 + 22,551); lower margin 10^4.88 => 75,858 (87,096 - 11,239). Lithopsian (talk) 15:11, 6 June 2024 (UTC)

This method of calculating errors is different of my own method. My way of calculating errors is a little different: I divide the logarithm error by the logarithm, then multiply by the real value. The step-by-step is as follows:

 

10^4.94 = about 87,096.

 

0.10/4.94 = 0.020242915.
87,096*0.020242915 = 1763.
1763 is the positive error, so let's find the negative error:

 

0.06/4.94 = 0.012145749.
87,096*0.012145749 = 1,058.
1,058 is the negative error, so the error bars are ⁺¹⁷⁶³
₋₁₀₅₈. InTheAstronomy32 (talk) 15:50, 6 June 2024 (UTC)

By "a little different", you mean "wrong", right? You can't divide logarithms and then use that as a proportion of the non-log value. That's not how logarithms work. If you want to approach it in that fashion rather than the somewhat long-winded method I showed for clarity, then the 0.10 (first, positive, margin of error, a difference) already gives the ratio of the upper bound to the mean. So 10^0.10 is 1.259-ish; the upper bound is about 25.9% above the mean. That's the very definition of a logarithm, a difference in the logarithm equates to a ratio of the underlying value. Same for the lower bound, 10^−0.06 is about 0.871, so the lower bound is about 12.9% below the mean. Lithopsian (talk) 16:44, 6 June 2024 (UTC)

Now i know that my method was very wrong, but it appears that some of these errors are in % rather than logarithms. See for example the Arroyo-Torres et al. paper, Table 3 gives the luminosity in watts and in log(L_☉) for AH Scorpii, KW Sgr and UY Sct. I will use AH Scorpii as an example. If we calculate the luminosity logarithm and its errors (5.52±0.26) using your method, it gives 330+270
−150×10³ L_☉. But if we use the luminosity in watts, it give (3.3±8.6)×10³ L_☉. The solar luminosity is equivalent to 3.828×10²⁶ W, and the luminosity of AH Sco is (1.26±0.33)×10³² W. Making the conversions, it give the following values: $${\frac {1.26*10^{32}}{3.828*10^{26}}}=329,153.61$$
$${\frac {0.33*10^{32}}{3.828*10^{26}}}=86,206.9$$. So the rough value should be 329,000±86,000. If we divide 86000 by 329000 it will give 0.26, the same value of the logarithm. InTheAstronomy32 (talk) 23:47, 10 June 2024 (UTC)

Betelgeuse B?

arXiv:2408.09089, not yet accepted for publication but something to keep an eye on. SevenSpheres (talk) 16:57, 20 August 2024 (UTC)

Nice. 21 Andromedae (talk) 17:00, 20 August 2024 (UTC)

Mass of Betelgues

How can beetlejuice have a radius 640 times that of the sun and only 10 to 20 times the mass of the Sun? 2600:8802:B0D:4300:AD95:9A5C:B03A:C3E7 (talk) 06:06, 26 August 2024 (UTC)

That is because Betelgeuse is a red supergiant, a star in the last stages of evolution. See stellar evolution. 21 Andromedae (talk) 15:26, 26 August 2024 (UTC)

Binary companion!

In a paper with a draft on 20 August 2024 “A Buddy for Betelgeuse: Binarity as the Origin of the Long Secondary Period in Orianis“ (arXiv:2408.09089) Goldberg et al propose that Betelgeuse has a binary companion of approximately 1.2 solar masses. Not sure where best to edit this into the article. 2600:1700:466A:A100:F482:9C8C:D5B1:25F4 (talk) 05:18, 30 August 2024 (UTC)