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The minimum mass for the spontaneous collapse of a gas cloud is the Jeans mass. For low temperatures and high densities, this jeans mass is greater than the mass of a typical star. What are the implications of the fact that the Jeans masses of most clouds are larger than the masses of most stars?

Qmechanic
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user1887919
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1 Answers1

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The Jeans mass is given by:

$$ M_J = \left(\frac{5kT}{Gm}\right)^{3/2} \sqrt{\frac{3}{4\pi\rho}} $$

The important point to note if that the Jean's mass is proportional to $T^{3/2}$ and $\rho^{-1/2}$. As the cloud collapses its density increases, but the temperature stays roughly constant. It may seem odd that the temperature stays constant - the reason is that the cloud does not scatter IR light so it can cool efficiently.

The result is that as the cloud collapses the Jean's mass decreases, so the cloud fragments into smaller pieces. These in turn collapse and increase in density, so the Jean's mass falls further and the cloud fragments further. This process only stops when the cloud is thick enough to start scattering light and its temperature starts to rise. At this point the cloud will form a protostar.

I'm not sure the details of the process are well understood (they are not understood by me!) so I don't know at what mass the fragmentation stops. Presumably it is well above a typical stellar mass since only a fraction of the cloud will end up in the star.

John Rennie
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