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I thought that the answer is yes, a little bit. As the sun runs out of hydrogen, then it it gets more and more contracted, increasing pressure, temperature, and therefore luminosity. This is supported by this answer saying, about main sequence stars, that:

result is that older stars a slightly hotter and brighter than younger stars.

However, Wikipedia says

low- and intermediate-mass stars expand and cool until at about 5,000 K they begin to increase in luminosity in a stage known as the red-giant branch. The transition from the main sequence to the red giant branch is known as the subgiant branch.

Am I right that this cooling which wikipedia mentions happens after the star left the MS, while on the sub-giant branch? (Not cooling, and then entering SGB, on the way to RGB.)

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    $\begingroup$ Your diagram here astronomy.stackexchange.com/questions/44058/31410 shows that the temperature increases during the main sequence and cools during the subgiant phase. $\endgroup$ May 29 at 1:19
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    $\begingroup$ Also David Hammen's answer here astronomy.stackexchange.com/a/44052/31410 says that the sun is growing hotter. $\endgroup$ May 29 at 1:20
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    $\begingroup$ I think that as helium ash accumulates in the core of a main sequence star, this causes the core to slightly contract (due to the lower rate of hydrogen fusion per unit volume), which increases temperatures and pressures, "overcompensating" for the initial drop in fusion rate and thus causing an increase in rate of hydrogen fusion. This causes temperatures in the core to rise (slowly), but pushes the outer layers out, causing the star to expand and the surface temperature to actually drop. But I couldn't quite find a source supporting the last sentence, so I could be wrong. $\endgroup$
    – YiFan
    May 29 at 1:38
  • $\begingroup$ Thank you all for these comments, all of them are useful. $\endgroup$
    – zabop
    May 29 at 8:58
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The answer is no: the Sun cools throughout its main sequence lifetime. This effect is due to the accumulation of helium ash in the core, which decreases the amount of energy released per unit volume. The decreased energy output to counteract gravity means that the core contracts under its own gravity, increasing temperatures and pressures in the core, and hence the rate of fusion. This overcompensates for the drop in fusion rate due to accumulation of helium ash, increasing the energy output of the core.

However, this increased energy output actually causes the outer layers of the star to expand, correspondingly decreasing in temperature. The surface temperature of this star thus actually drops (at least for sufficiently massive stars, I'm not too sure if there is a difference for < 1 solar mass stars).

This is why the bottom part of the main sequence is called the Zero Age Main Sequence (ZAMS): throughout its life, stars on the MS will shift upwards and rightwards on the main sequence, away from the ZAMS. This is referred to as main sequence broadening. See the diagram below, where the "MS" in the evolutionary tracks tells you when the star is still on the main sequence (but not on the ZAMS).

enter image description here

However, it should be noted that this is much less significant than the cooling (in terms of surface temperatures) in the subgiant or red giant phase, which is why it's often ignored or glossed over.

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  • $\begingroup$ Are you sure about this? Wikipedia plot shows the temperature to increase during MS, as pointed out by the comments on the OP. $\endgroup$
    – zabop
    Jun 2 at 7:07
  • $\begingroup$ @zabop Hi, I think the main focus of that plot might be to show the post-MS evolution, hence the apparent inaccuracy in the MS phase. Alternatively, it could be showing the evolutionary track of a low-mass star (less than 1 solar mass)---in the diagram in my answer, also taken from wikipedia, you can see that the temperature does slightly increase for stars of that mass. However I am unfortunately not aware of the mechanism for this. (My guess would be that the atmosphere does not increase in radius as much due to the star being fully convective, so that the cooling is more insignificant.) $\endgroup$
    – YiFan
    Jun 2 at 8:05
  • $\begingroup$ @zabop I will admit however that I am not an expert in this topic as some contributors on the site might be, so there's a chance I might be wrong here. This is my best understanding of the topic though. $\endgroup$
    – YiFan
    Jun 2 at 8:13

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