Suppose you were to fix MSun and RSun – what would happen to the central temperature if the composition is (magically) transformed to pure helium (i.e., X = 0, Y = 1)?
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$\begingroup$ Soooo....break the rules of physics and then provide an answer using the remaining (unbroken) rules of physics. $\endgroup$– user226006Commented Jan 21, 2020 at 21:35
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2$\begingroup$ @StudyStudy you need to be clear what you mean. Having a star made purely of He does not violate any physical laws. Demanding that it has a radius of $1R_{\odot}$ does. $\endgroup$– ProfRobCommented Jan 21, 2020 at 22:08
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$\begingroup$ Seems one could run this in MESA for a lark. $\endgroup$– Anders SandbergCommented Jan 21, 2020 at 22:30
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$\begingroup$ Related article: forbes.com/sites/briankoberlein/2016/01/02/…. And this related answer : physics.stackexchange.com/questions/480972/… $\endgroup$– user226006Commented Jan 21, 2020 at 22:33
1 Answer
If you had a $1M_{\odot}$ star made of pure He, then it would not have a radius of $1R_{\odot}$. Assuming that it could find a way to cool, then it would continue to contract in size until its core become hot enough to ignite the triple alpha nuclear reaction that starts to transform the He into Carbon.
I am unaware of any numerical models which hypothesize pure He stars, but one can do a back of the envelope calculation:
Using the virial theorem and assuming an ideal gas it can be shown that the radius at which the core becomes hot enough to ignite He is given by $$ R_{3\alpha} \simeq \left(\frac{G \mu m_u}{5k_B}\right) \left( \frac{M}{T_{3\alpha}} \right) = 0.06 \left(\frac{M}{M_{\odot}}\right) \left( \frac{T_{3\alpha}}{10^8\ {\rm K}}\right)^{-1}\ R_{\odot}, \tag*{(1)}$$ where $T_{3\alpha} \simeq 10^8$ K is the approximate ignition temperature.
So a pure He core-burning star of $1M_{\odot}$ would be much smaller than the Sun and its core temperature would reach $\sim 10^8$ K. The He burning would stabilise the star at approximately this radius and core temperature for a considerable length of time (of order a billion years)
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$\begingroup$ Thanks for your comment, I should have been more explicit in mine above. $\endgroup$– user226006Commented Jan 21, 2020 at 22:35