• Is it possible to measure or calculate the total entropy of the Sun?
  • Assuming it changes over time, what are its current first and second derivatives w.r.t. time?
  • What is our prediction on its asymptotic behavior (barring possible collisions with other bodies)?
  • $\begingroup$ Entropy is: $S=k_B \ln \Omega$ Where the $k_B$ is Boltzmann's constant and the number of microstates $\Omega$ is: $1.19\times10^{57}$ The entropy of the Sun is $S=1.81\times 10^{21} \,\text m^2 \,\text {kg} /\text s^2 \,\text K $ $\endgroup$
    – user30274
    Sep 28 '13 at 22:34
  • 1
    $\begingroup$ Its easy given the temprature and density profiles, to calculate the entropy due to the underlying uncertainty in Motion of the particles. It would be interesting, to calculate the entropy increase due to Fussion, I hope some one can demonstrate how to do that. $\endgroup$
    – Prathyush
    Sep 29 '13 at 10:37

For an ideal gas, you can always relate the entropy to the pressure and density via the relation* $$ S(r)\simeq P(r)\rho^{-\gamma}(r) $$ where $\gamma=5/3$ is the adiabatic index. Reasonable (and simple) estimates exist for the density and pressure profiles (e.g., $\rho(r)\simeq\rho_c\left(1-r/R\right)$ where $R$ is radius of the star and $\rho_c$ is the central density).

I do not know that there are functional forms of $P(r,t)$ and $\rho(r,t)$, so I do not know how to answer the questions about the time-derivatives of $S$. Perhaps there is a stellar structure code online that you can modify to give that for you?

*This definition stems from the Stackur-Tetrode relation of the entropy density, $$ s=C_v\ln\left(P\rho^{-\gamma}\right)+{\rm const} $$ but eliminates the "clunky" $C_v$ and ${\rm const}$ terms.


The entropy of the sun is roughly $10^{35} \mathrm{J} /\mathrm{K}$ (see http://dx.doi.org/10.1103/PhysRevD.7.2333 where the entropy of the sun is given as $10^{42}\mathrm{erg}°\mathrm{K}^{-1}$)

It can be calculated using Boltzmanns law ( $S = k_B \ln{W} )$ where $W$ is the disorder parameter of the sun (depends on the number of atoms in the sun).

I did not get you second question.

  • 1
    $\begingroup$ Could you please explain how this estimated value is derived? I rephrased my second question. Is it clear now? $\endgroup$ Sep 28 '13 at 22:39
  • $\begingroup$ On second thought, how can we quantify about the entropy generated due to fusion inside the sun?? $\endgroup$
    – mcodesmart
    Sep 28 '13 at 22:39
  • 5
    $\begingroup$ Huh? I've never heard of $W$ being referred to as a "disorder parameter" before - it's usually called the phase space volume, though I've also seen people call it the "multiplicity", the "probability" (even though it isn't) and the "index of probability of phase". But also, I can't imagine how you would calculate $W$ other than by working out the entropy and doing $W=e^{S/k_B}$. $\endgroup$
    – Nathaniel
    Sep 29 '13 at 9:17
  • 1
    $\begingroup$ This can't be right. See Nathaniel's comment. If we roughly consider the Sun to output 4 000 000 tonnes of energy a second at, say, 6000K (i.e. it's a black body at this temperature), it's increasing the entropy of the World outside it by $\Delta S = \Delta Q / T = 4000000\mathrm{t}\times 1000 \mathrm{kg\,t^{-1}} \times 10^{17} \mathrm{J\,kg^{-1}} / 6000\mathrm{K} = 7\times 10^{22} \mathrm{J\,K^{-1}\,s^{-1}}$. At this rate, it would take about $10^{12}\mathrm{s}$ to make up your figure. That's only about thirty thousand years, i.e. just a bit older than human agriculture. $\endgroup$ Sep 29 '13 at 10:45
  • $\begingroup$ What about the entropy generation? Is that quantifiable and is it significant? $\endgroup$
    – mcodesmart
    Sep 29 '13 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.