How Does Gravitational Compression Generate Heat in Protostars?

[etotheipi]

Homework Statement

A spherical cloud of interstellar gas (which can be considered to be ideal) collapses under its own gravity. During much of the collapse, the star can be considered to be at the same temperature as its surroundings, $T$. Calculate the heat radiated away during collapse from a radius $r_0$ to $r$.

Relevant Equations

N/A

When I did this the first time I didn't really think too much about it, so I just wrote$$p = \frac{nRT}{V} \implies W = \Delta U - Q = - Q = -\int_{\frac{4}{3}\pi r_0^3}^{\frac{4}{3}\pi r^3} \frac{nRT}{V} dV$$That turned out to be correct, but when I thought about it I didn't understand why this expression would be meaningful. The external pressure to the protostar is presumably zero, since the thing is surrounded by vacuum, so the PV-work on the protostar will be zero. What is really responsible for the generated heat is the work done by gravity/decrease in gravitational self energy, $\Delta U_g = \frac{3GM}{5}\Delta \frac{1}{r}$.

So I wondered how do we actually interpret the thermodynamic result at the top? Furthermore in an actual gravitational collapse the pressure is dependent on the radial coordinate, but in the top it's assumed to be uniform. The resultant pressure force on any spherical shell also won't equal the gravitational force on that shell. Thanks

 

[dRic2]

I maybe wrong, but if you assume the initial and the final states are equilibrium states and the process is reversible (probably assumed here), then thermodynamics tells you that the way you get from one state to the other is irrelevant. So, instead of the complicate problem of integrating the change in gravitational energy, you just solve the equivalent problem of the compression of a gas.

 

[etotheipi]

I maybe wrong, but if you assume the initial and the final states are equilibrium states and the process is reversible (probably assumed here), then thermodynamics tells you that the way you get from one state to the other is irrelevant. So, instead of the complicate problem of integrating the change in gravitational energy, you just solve the equivalent problem of the compression of a gas.

Thanks, that's a good insight. I am aware that heat is not a state function and is then generally path-dependent, but do you say that the heat exchanged for all reversible paths is the same, between two identical states?

 

[dRic2]

Suppose you go from A to B with a reversible process, and then you go back from B to A with a different but still reversible process. $\Delta U_{A, A} = 0$ (since you came back to where you started). This implies $\Delta U_{AB} + \Delta U_{BA} = 0$. Now $\Delta U = \int \delta q - \int pdv$, so the heat and the work exchanged in the two processes can be different but the overall sum has to be the same since you came back to the same point $\Delta U_{A,A} = 0$.

So for example, you can "design" a way to go from A to B such that $\Delta U_{A,B} = \Delta Q$, and then you can imagine a process to go back from B to A such that $\Delta U_{B, A} = - \int pdv$. Overall, since $\Delta U_{A, A} = 0$, you get $\Delta Q - \int pdv = 0$. In this way, instead of calculating the heat exchanged in the first process directly you could calculate the work done in the other process.

 

[LCSphysicist]

I am not sure if helps, but we can get the same results if we imagine such collapse as a reversible process, i don't think is so bad imagine it, if the collapse is too slow.

dS = dq/T

When we have T constant, dS can be written as:

The heat could be seen as an addendum to the entropy change consequence.