energy flux inside star's core

[Kane Starkiller]

OK, I'm debating over at st-v-sw and one of the debators claims that power intensity at sun's core is 10^23W/m2. At first I didn't know where did he pulled that figure from but later I realized that he simply used Stefan-Boltzman equation and plugged in the 60 million K(number from TNG episode Half a life).
Now, I already explained to him that if core radiated 10^23W per square meter the sun would exceed it's own gravitational binding energy in less than a minute but they responded with something like "the energy is shooting back and forth in the core" so it's Ok tu use the 10^23W/m2. While the energy does indeed take million years to reach the surface there is no way the core could be radiating 10^23W/m2.
So could anyone here point to how we can calculate the power an object would be recieving when inside the star's core.
This particular discussion involves the episode "Half a life" when Enterprise uses torpedoes to try and revive a sun.

[Wyrm]

The basic flaw in his reasoning is that the Stefan-Boltzmann law equals the total output flux of an object only when there's nothing radiating back. The core is indeed pumping out that flux due to its temperature, but the upper layers (at nearly the same temperature) are pumping in a flux almost equal to that back into the core, less the power that the star pumps out as it shines. The net power output is what we're really interested in.

Anyway, finding the net power output of the core is simple, for a star in the main sequence. Just take the output of the star and divide it by the surface area of the core (for main sequence stars, the core is about 10% the diameter of the star).

Things get messy as the star leaves the main sequence, however. For a pulsating star, the power output of the core oscillates from a net power output that would make the star expand to an output that would let the star contract. But this only happens after the star has gone into the red giant/supergiant stage (which means that Timican's planet is already fried) and initiates helium burning (which in your case, hasn't happened yet because 60 million K is too cool for helium burning).

[Darth Wong]

So he makes no distinction between "net" and "gross"? I'd love to see how this guy does his taxes.

[frogcurry]

Energy output of our sun is 3.83E23 kW, diameter is 1.392E6 km. From a spreadsheet I have which took a lot of its data from Wikipedia so thats probably the source of these numbers.

However I'm not sure that will give you the answer you need. If what Wyrm says is accurate then it is the gross you want I think - thats the energy you will experience in the suns core. The net output is what leaves it, but if you stand inside the sun you'll experience that intensity of energy (assuming his maths is correct).

i.e.

-----------><----------
Energy (core) ------> YOU <------- Energy (a bit further out)
-----------><----------
---> (smallish net transfer as core is that bit hotter).

Thats how I understand it anyway. Net is what leaves the sun, not a measure of the energy bouncing around inside it. You are talking about being inside the suns core, and hence all then energy (including that which is recycling internally) will be affecting you.

Am I being stupid?

[frogcurry]

The arrow drawing didn't come out as expected, I'll redraw it as it looks meaningless:

Deep Core slightly further out

--------> <-----------
--------> Object <-----------
--------> <-----------
---> (net energy transfer outwards)

[Wyrm]

If I may venture a guess to what this is about, is the other fellow trying to prove that Fed torps can withstand being sunbaked like nachos, thereby wank out Trek?

[Kane Starkiller]

If I may venture a guess to what this is about, is the other fellow trying to prove that Fed torps can withstand being sunbaked like nachos, thereby wank out Trek?

In the episode "Half a life" Enterprise used it's torpedoes to try and restart the fusion process in a star that was slowly dying. They are claiming that the torpedoes had to endure the power of 10^23W/m2 while inside the core. Now I know that sun produces about 4*10^26W and since it takes about million years for that energy to be emitted on surface then the sun would contain about 1.26*10^40J. Since sun has a volume of about 1.37*10^27m3 then power density inside it would be about 9TJ/m3. So these calculations would tend to disprove his 10^23W/m2 power intensity obtained by using Stefan-Boltzman formula but I would like to know some exact way I could calculate what power an object submerged in 60 million K core would recieve.

[Darth Wong]

If I may venture a guess to what this is about, is the other fellow trying to prove that Fed torps can withstand being sunbaked like nachos, thereby wank out Trek?

In the episode "Half a life" Enterprise used it's torpedoes to try and restart the fusion process in a star that was slowly dying. They are claiming that the torpedoes had to endure the power of 10^23W/m2 while inside the core.

Let me get this straight: he argues that torpedoes are millions of times tougher than battlecruisers, which get annihilated if they so much as graze the Sun's surface? Why aren't Fed engineers hard at work making starships out of whatever torpedoes are made of, then?

Now I know that sun produces about 4*10^26W and since it takes about million years for that energy to be emitted on surface then the sun would contain about 1.26*10^40J. Since sun has a volume of about 1.37*10^27m3 then power density inside it would be about 9TJ/m3. So these calculations would tend to disprove his 10^23W/m2 power intensity obtained by using Stefan-Boltzman formula but I would like to know some exact way I could calculate what power an object submerged in 60 million K core would recieve.

He's assuming it's a perfect blackbody radiator. Keep in mind that this is what you're doing when you use the Stefan-Boltzmann formula. The correct method to determine solar core power intensity is to estimate the total power output of the star and then divide it by the surface area of the core.

[Wyrm]

They are claiming that the torpedoes had to endure the power of 10^23W/m2 while inside the core. ... Since sun has a volume of about 1.37*10^27m3 then power density inside it would be about 9TJ/m3.

The thing is, that it takes a lot of time for the heat to get anywhere in the Sun, because the photons can only travel a short distance before it bashes into something. Thus, while the intensity absorbed by the torpedo is initially 10^23 W/m^2, but it soon (within a fraction of a second) cools off as the heat is absorbed. Also, realize that 60 million K is just the temperature at core. The temperature falls off from the surface of the core to the outer layers (that's why fusion doesn't occur outside the core!).

Therefore, the pertinent question is not how much the interior of the star radiates, but how much hot stuff does the torpedo actually come "in contact" with. ("In contact" means hot stuff kicking a photon to the torp.)

What you really want is the amount of energy in a thin pancake of material that is in contact with the torpedo for the time it takes to pass. Radiative transport has an approximate speed of 1.3e-5 m/s. That means that for every second the torpedo's skin is in contact with the solar atmosphere, there will be a skin of about 1.3e-5 meters that actually dumps energy into the torpedo (actually, the radiative contribution to the skin decays at a rate proportional to exp(d/{\sigma t}) where \sigma is 1.3e-5 m/s, d is the distance from the torpedo, and t is the time the torpedo is in contact with that skin, but the expected value is \sigma t). Considering how quickly the torpedo gets through the atmosphere, there is obviously not going to be very much heat transfered between the flat pancake and the skin of the torpedo.

Okay, how long does it take for the torpedo to travel a torpedo-length? A torpedo is about 2m long. Knowing how long it takes things to happen on ST, it takes perhaps 20-30 seconds for the torpedo to go from the surface to the outer fringes of the core, during which time it goes 5.564e5 km (the core of the sun is actually 20% the diameter of the sun), which makes for a speed of 1.85e4 and 2.78e4 km/s, so it takes 7.19e-5 to 1.08e-4 s for the torpedo to travel a torpedo-length. This means that the torpedo only has to absorb heat from a skin 1.0e-9 to 1.4e-9 m thick (because that's all the time the stellar atmosphere has to radiate its energy and hit the torpedo's skin).

Because the torpedo is traveling so damned fast, the shockwave forming around the torpedo is hardly larger than the torpedo itself. We'll say that the torpedo is a cylinder of .8 m wide and .5 m tall, so the surface area of the shell is 2.0e-10 to 2.8e-10 m^2. Multiply by the length of the tube (5.564e8 m) and you get 1.1e-1 to 1.4e-1 m^3. Using your energy density of 9 TJ/m^3, we get a total amount of heat absorbed by the torpedo: 1.0 to 1.4 TJ.

This assumes that the torpedo slips easily through the stellar atmosphere (no friction), as well as the stellar atmosphere having uniform energy density (which depends on T, but that involves some rather tedious state equations), so count this as nothing more than a first approximation.

Trektards will probalby find these figures disappointing, though.

[Kane Starkiller]

One of the debators on st-v-sw just "explained" how the star can have 10^23W/m2 power intensity:

Let's draw a (VERY INEXACT) scheme: we divide the star into spherical layers of thichkness l, where l - the distance a photon can go until grabbed byu an atom. Then the outermost layes radiates 10^26 Wt outside and the equal amount inside. The next inner layer radiates 2*10^26 (so IN SUMMARY 10^26 J is transferred to outermost layer) outside, and thus radiates 2*10^26 inside. The third layer has to radiate 3*10^26... the millionth layer has to radiate 1^32(1000000*10^26 J) and so on.

Wonderful isn't it.

[Wyrm]

One of the debators on st-v-sw just "explained" how the star can have 10^23W/m2 power intensity:

<snip absolute crap>

Wonderful isn't it.

"Equilibrium" is obviously not a word that means anything to these people. I think you can quash these dim bulbs on your own, by showing that the net radiation pressure on those thin shells overwhelms the gravitational force holding it to the star. Therefore, the star is shedding mass--which it obviously isn't in the show. (We know what a star shedding mass looks like; that ain't it.)

[Darth Wong]

Do these imbeciles not realize that according to their logic, the core of our own Sun should have a power flux of 3E21 W/m^2? As a result of this absurdly overestimated power output, the core (which is roughly 1/4 the width of the Sun, hence it has a surface area of roughly 4E17 m^2) should have a power output of 1.2E39, which is trillions of times its actual power output.

[Kane Starkiller]

Do these imbeciles not realize that according to their logic, the core of our own Sun should have a power flux of 3E21 W/m^2? As a result of this absurdly overestimated power output, the core (which is roughly 1/4 the width of the Sun, hence it has a surface area of roughly 4E17 m^2) should have a power output of 1.2E39, which is trillions of times its actual power output.

No no you don't understand. Here is the explanation:

This is correct. But what do you imply by this is not . You can give somebody $1000 million times and get $999.99 back each time. You only need to have $11000 for this. If each time you do this, you get $0,01, You can actually never have more than &1000 and still the "money flux" would be one Billion dollars.! Sun matter doesn't have "that much" energy, but it gets and gives it quickly. If a PT enters sun, matter near torpedo would give its energy to torpedo very fast, but it would get this energy replenished by the matter further away and that - by matter eeven further away -- and so on. Given that PT moves, the flux will be never much lower than suggested by the model.

See it all makes perfect sense now.

[Wyrm]

Here is the explanation:

This is correct. But what do you imply by this is not . You can give somebody $1000 [a] million times and get $999.99 back each time. You only need to have $11000 for this. If each time you do this, you get $0,01, You can actually never have more than &1000 and still the "money flux" would be one Billion dollars.! Sun matter doesn't have "that much" energy, but it gets and gives it quickly. If a PT enters sun, matter near torpedo would give its energy to torpedo very fast, but it would get this energy replenished by the matter further away and that - by matter eeven further away -- and so on. Given that PT moves, the flux will be never much lower than suggested by the model.

See it all makes perfect sense now.

I've explained elsewhere that for the amount of time the torpedo is in contact with a thin pancake of material, the only material able to actually give up energy to the torpedo is a thin hoop on the order of a micrometer thick, radius-wise. Thus, over its entire trip, the photon torpedo only comes in thermal contact with about a tenth of a cubic meter of hot gas.

The explaination SailorSaturnmoron13 gives only works insofar as the money swapers have enough time to get the money to you. If you leave before a dollar of the seventh person with $11000 has time to work through the chain, you're only going to get $66000 tops, even though $7 billion is "changing hands." Time is everything.

The gas inside a star is quite opaque, and as such the photon torpedo would be well-insulated from the brunt of the star's heat. That's what really kills the explanation.