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I Stellar Thermal Energy with no Nuclear source

  1. Apr 16, 2018 #1
    Can anyone enlighten me as to why the total thermal energy of a star will increase with time if the star has no nuclear energy source?
     
  2. jcsd
  3. Apr 16, 2018 #2

    Drakkith

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    The star will heat up as it contracts, if that is what you mean. The mechanism is called the Kelvin–Helmholtz mechanism, and it converts gravitational potential energy into thermal energy.
     
  4. Apr 16, 2018 #3
    when big stars run out of hydrogen to burn, fusion stops. This causes the core to cool down. The cooler core doesn’t push up as hard so gravity pulls the rest of the star inwards. The collapsing core then heats up because it’s being pushed on by all of that mass. This causes temperature to skyrocket, leading to helium fusion, then the process repeats up the periodic table until iron.

    If you add pressure to anything, it’ll increase in temperature.
     
  5. Apr 16, 2018 #4
    Most stars don't go all the way to iron though.
    After the CNO elements are produced. it's game over for stars around the mass of the Sun.
    A short puff of a red giant stage, then it's white dwarf time for near eternity unless other objects get involved.
    It should eventually cool to the point where it doesn't significantly radiate at all; a black dwarf.
    Far as I know though, that could take trillions of years.
     
    Last edited: Apr 16, 2018
  6. Apr 17, 2018 #5
    The article only specifies the source of heat - not that the heat should increase.
    Heat does not always increase - only if the star is mostly ideal gas it does.

    Consider a self-gravitating sphere.
    Now it cannot actually have constant density - the density at centre must be bigger than at the surface (because of compressibility)
    Now suppose it shrank in such a way that its density radial distribution remained unchanged. So, its radius decreases 2 times, density at centre increases s3=8 times, and new density at 50 % of new radius is also exactly 8 times of what the old density was at 50 % of old radius, and so for every other percentage of new and old radii.
    What is the central pressure then?
    The surface gravity, and gravity at every depth inside, was increased 4 times.
    The density was increased 8 times as stated.
    So weight of each fluid column of given length was increased 32 times.
    But the length of the column surface to centre was decreased 2 times.
    So in total the central pressure was increased 16 times.
    But as stated before, the central density was increased just 2 times.
    If the contents are ideal gas, by ideal gas law the only way to achieve this is to increase temperature 2 times.

    If the contents are not ideal gas, then for example liquid´s pressure can easily be increased 16 times without appreciable increase of density.
    So, if the interior of a body is appreciably degenerate, there is little reason for temperature to increase on contraction - actually, there is a reason for the body to not contract unless the temperature decreases.

    Bodies whose interiors are largely degenerate, like brown dwarfs, planets and white dwarfs, do produce heat on contraction, but only contract as they cool down (having dissipated the heat produced on contraction). Bodies whose interiors consist of largely ideal gas need to heat up on contraction, and get that heat by Kelvin mechanism.
     
  7. Apr 17, 2018 #6

    Drakkith

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    Well said.
     
  8. Apr 20, 2018 #7

    stefan r

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    I think that is typo. Otherwise the logic is good. If I understood correctly then that line should read:

    Here is link to ideal gas law in case anyone forgot chem.


    That sounds like you are saying the red giant phase comes after helium fuses to carbon. The Asymptotic giant branch is much shorter and does come after helium burns. The red giant phase starts much earlier.

    I find it interesting that the luminosity drops at the time when the helium flash occurs.
     
  9. Apr 21, 2018 #8
    So does a star heat up when it contracts? Or Contracts when it cools down?
     
  10. Apr 21, 2018 #9

    Drakkith

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    Both, actually. Nuclear fusion muddles the issue a bit, but if we imagine a star made out of non-fusing gas then it becomes a bit clearer. In such a case, the star would initially heat up as it contracts, converting gravitational potential energy into heat, which is radiated away from its outer layer. The star heats up as it contracts until the rate of heat loss is equal to the heat generated from the contraction. At this point, the star reaches equilibrium and stops contracting. Well, kind of. It still contracts, but it does so very slowly, with the rate of contraction being limited by the rate of heat loss from its surface.

    The star very slowly contracts for millions of years until its gravitational potential energy is nearly used up, at which point it begins to cool as it finishes contracting. Finally all that's left is a cold ball of inert gas that is in equilibrium with the background radiation.

    Real stars get hot enough to ignite nuclear fusion in there cores, which lengthens the amount of time required for the star to contract fully in addition to causing cycles of expansion and contraction in larger stars.
     
  11. Apr 21, 2018 #10
    How?

    Is it the heat that causing nuclear fusion? Or the fact that gravity is smashing one nucleus into another or a combination of both?



    How can gravitational PE be used up? Is it because of the loss of mass as heat? How much mass would a main sequence star lose in its life time? It would be very very small, right? So how gravitational potential gets used up, or are you saying that it comes into equilibrium with other forces like electromagnetic forces between atoms.
     
  12. Apr 22, 2018 #11

    Drakkith

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    Like I said in my last post, it causes cycles of contraction and expansion and greatly lengthens the amount of time required for the star to contract, in addition to adding in another source of potential energy to consider.

    The heat smashes the nuclei together, while gravity holds it all together under very high pressure.

    The energy radiated away by the star's radiation comes from the combination of the GPE of the star and fusion. In our non-fusing example, all of the energy would come from the GPE. As for 'how', it's because gravity accelerates gas and plasma particles towards the center of the star as they fall, and as they hit the other particles underneath them they transfer some of their kinetic energy. This gets spread out through the star as heat.

    More mass is lost through solar wind, flares, and other similar phenomena than through heat loss. As for how much, I don't know the exact amount offhand.
     
  13. Apr 22, 2018 #12

    stefan r

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    Here is wikipedia on thermal energy and heat:

    Consider what happens when you stick your thumb over the end of a hose or a sink. The flow of water mass through the pipe decreases because you are blocking the flow. You can, however, spray the water much further. The velocity of water leaving the hose is higher. It would take much longer to fill a bucket with the high velocity water. The Mississippi river has a much higher total flow rate than a faucet even though the flow rate per unit area is much lower. Analogies should not be take too far. The surface temperature of a star is a balance of energy radiated away and the flow energy to the surface from within. If you increase the energy then the star expands making it more like the unobstructed pipe. If it loses energy it contracts and becomes more like a sprayer.

    E = mc2 You can calculate mass lost by measuring the luminosity. Converting hydrogen to helium releases less than 1% of the mass.
     
  14. Apr 22, 2018 #13

    Ken G

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    Yes, the internal kinetic energy is always rising the same, degenerate or ideal, but we only call the internal kinetic energy "thermal energy" if it's not too degenerate. If it's highly degenerate, that energy cannot be lost in the form of heat, even when in thermal contact with the coldness of space, so that's the situation where the "thermal energy" can actually drop. Ironically, that's the more intuitive response to losing heat, but requires quantum mechanics to get it!
     
  15. Apr 22, 2018 #14

    JMz

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    Be careful here: The energy lost in the form of photons is very, very small. (Well, 1%-ish.) A little is lost in the form of the stellar wind during its main-sequence phase. The major mass loss for stars near the Sun's size and somewhat larger comes near the end, as it sheds its outer layers with a huge wind during the red-giant phase, leaving the core as a white dwarf. This can easily be 1/2 the total mass of the star.
     
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