When Will Fusion Work? Insights from ITER and Expert Opinions

[PAllen]

Thank you for your educational information!

But I'm a stupid guy. When it comes to the birth of a star like our sun I do not understand how gravity can play such an important role.

Protons are positively charged, right?

And equal charge repell, right?

Still protons have obviously bundled up.

Why or more scientifically, how?

It is not enough that protons bundle up, they bundle up so tightly that they start to fuse into Helium.

I don't get this part.

Roger

The pressure at the center of the sun is about 250 billion kg / cm ^2, and this is all due to gravity. Does this help?

Also, why are planets round? One definition of a planet versus and asteroid is a body large enough the gravity overwhelms all possible sources of mechanical rigidity, making the body round. A moon mass collection of diamond crystals will 'collapse' into round mass carbon, overcoming the rigid resistance of diamond. Now, how many times more massive is the sun than the moon?

 

[Drakkith]

Also remember that the Sun has electrons and is not charged overall. The atoms in the gas cloud that initially collapsed to form the Sun didn't repel each other because they were not ionized.

 

[rogerk8]

I know so little and understands so little so maybe I should quit now?

Anyway here is how I see it:

$$F_G=G\frac{m_1m_2}{r^2}[N]$$

$$F_Q=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}[N]$$

where for protons

$$m_p=1,67e^{-27}[kg]$$
$$G=6,67e^{-11}[Nm^2/kg^2]$$
$$q=+e=1,6e^{-19}[As]$$
$$\epsilon_0=8,85e^{-12}[As/Vm]$$

which gives

$$\frac{F_Q}{F_G}=10^{36}$$

This clearly states that, in the beginning, protons could not have bundled up due to gravity while the electromagnetic force is way much higher (to say the least).

So what happened? I see two scenarious:

1) The first particles to bundle up was neutrons and when they bundled up tight enough they somehow mutated into protons which after a while where able to fuse into He_2.

2) Reading your kind answer makes me think that perhaps the first neutral (which is a must here) particles where neutral protons i.e pure H_1 which later fuses into He_2.

Now I will try to answer your question "How many times more massive is the sun than the moon": I have no clue To me the sun is of course massive but it is also gasous like a plasma, right? So, stupid as I am, I would actually consider the moon to be more massive than the sun because it is made of dirt, so to speak. Please, educate me some more here if I'm wrong.

Roger
PS
I kind of know how to write isotopes but I fail using <sup>.

 

[PAllen]

I know so little and understands so little so maybe I should quit now?

Anyway here is how I see it:

$$F_G=G\frac{m_1m_2}{r^2}[N]$$

$$F_Q=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}[N]$$

where for protons

$$m_p=1,67e^{-27}[kg]$$
$$G=6,67e^{-11}[Nm^2/kg^2]$$
$$q=+e=1,6e^{-19}[As]$$
$$\epsilon_0=8,85e^{-12}[As/Vm]$$

which gives

$$\frac{F_Q}{F_G}=10^{36}$$

This clearly states that, in the beginning, protons could not have bundled up due to gravity while the electromagnetic force is way much higher (to say the least).

So far, so good, in the Newtonian sense. However, as Drakkith noted, the sun is electrically neutral, and started from Hydrogen atoms (and other stuff). By the time you need to worry about proton repulsion, you already have the 250 billion kg/cm^2 pressure of neutral matter above to overcome it.

Now, for the more remarkable GR correction to the Newtonian picture (though this is not relevant to the formation of stars). Suppose you put one proton in a cubic meter of vacuum, building this pattern out. According to GR, there would come a point, as you built this outwards, where this framework was within its Schwarzschild radius, despite the ultra-low density. Then, no matter what, the assemblage would collapse to a singularity, no matter what forces applied to the protons. The progress toward the singularity would be exactly mathematically and physically equivalent to the progress of time, so even forces approaching infinite would not be able to stop the collapse. Thus, per GR, enough stuff, however sparse, must collapse - if you have enough of it.

So what happened? I see two scenarious:

1) The first particles to bundle up was neutrons and when they bundled up tight enough they somehow mutated into protons which after a while where able to fuse into He_2.

No, much simpler, it started as Hydrogen atoms.

2) Reading your kind answer makes me think that perhaps the first neutral (which is a must here) particles where neutral protons i.e pure H_1 which later fuses into He_2.

Correct in that the starting point is Hydrogen gas. However, the fusion reaction is not to Helium 2, which would not release energy. It is to Deuterium when the proton-proton interaction is accompanied by emission of a positron and a neutrino. This process releases energy but is very rare. The further steps from here to Helium 4 occur and much higher rates and release much more energy.

Now I will try to answer your question "How many times more massive is the sun than the moon": I have no clue To me the sun is of course massive but it is also gasous like a plasma, right? So, stupid as I am, I would actually consider the moon to be more massive than the sun because it is made of dirt, so to speak. Please, educate me some more here if I'm wrong.

Roger
PS
I kind of know how to write isotopes but I fail using <sup>.

The core of the sun has a density of about 150 grams/cm^3, well over 10 times the density of the Earth's core. This is because of the enormous pressure of the overlying layers squeezing an ionized plasma to a density beyond any material we know on earth.

 

[phyzguy]

The core of the sun has a density of about 150 grams/cm^3, well over 10 times the density of the Earth's core. This is because of the enormous pressure of the overlying layers squeezing an ionized plasma to a density beyond any material we know on earth.

There is also the fact that the volume of the sun is over 60 million times larger than the moon.

 

[mfb]

And the fact that Earth plus moon orbit the sun, not the other way round.

WolframAlpha: (mass of sun)/(mass of moon)

Even today, the core of the sun is neutral - the hydrogen is ionized and we have a plasma, but the negative electrons are still hanging around there together with the positive protons and helium nuclei.

 

[rogerk8]

Another fun calculation is:

$$F_G=F_Q$$

or

$$GMm_p=\frac{e^2}{4\pi\epsilon_0}$$

which gives

$$M=2 [Gkg]$$

which of couse is the same as the mass for

$$N_{mp}=10^{36}$$

number of protons.

However, this states that the mass of the rising sun has to exceed 2GKg before any protons may be attracted.

This still makes me believe that the sun began as a bundling up of neutrons which later on gave rise to such a high mass (>2Gkg) and therefore gravitational force that it could attract protons. Now when it did that, the rising sun got charged and due to the sign of charge of electrons they came along too.

As you may have noticed I am considering the simplest form of particle soup here which means that we already have a post Big Bang soup of convenient elementary particles. All of them in a plasma state.

So in the core we now have a plasma of both neutrons, protons and electrons.

But why did protons start to fuse?

I don't understand the concept of pressure for instance.

Repeating the formula here for convenience:

$$p=nkT[J/m^3=N/m^2]$$

Loooking at this formula you can see how increased particle density give rise to an increased pressure.

But what about T?

What determines T?

I don't get it.

Finally, gravitational force is of course higher the closer you get to the centre of gravity.

But what makes the pressure and/or gravitational force start the reaction to fuse?

What happens here with the elementary soup of star-life?

Roger

 

[PAllen]

Are you even reading other people's posts? I think about 5 times it has been explained the sun began with neutral hydrogen gas. The, as it collpased and heated (you can think of this simply as conversion of gravitational potential energy to heat), the center became ionized, but still neutral on average. You than have a neutral plasma at high temperature and pressure (= high density), such that the rare p + p -> deuterium + neutrino + positron can occur (at a low rate per volume).

 

[rogerk8]

I am of course reading other people's posts. But when people come with explanations like the GR I am totally lost and it's no use to even debate. Furthermore, I am trying to understand this my own way if this is alright by you?

Ok, let's say the sun began as a neutral Hydrogen gas. I can easily buy that.

Please explain how gravitational potential energy can be converted to heat. I don't even know what gravitational potential energy is (other than mgh).

And please explain p+p->deuterium (H_2) + neutrino + positron because I find this very interesting mainly due to the fantastic mutation of a proton becoming a neutron (i.e deuterium).

Finally, I thank you for your answer.

Roger

 

[PAllen]

I am of course reading other people's posts. But when people come with explanations like the GR I am totally lost and it's no use to even debate. Furthermore, I am trying to understand this my own way if this is alright by you?

Ok, let's say the sun began as a neutral Hydrogen gas. I can easily buy that.

Please explain how gravitational potential energy can be converted to heat. I don't even know what gravitational potential energy is (other than mgh).

mgh is good enough for the basic idea. You have a large mass of hydrogen in a cube .1 light years on a side. Under the influence of self gravitation, it collapses to a diameter of 1 million miles (appx). That means the average h in mgh, for given volume of hydrogen is about 100 billion miles. The g is varying during the collapse, but it should be easy to imagine that you have an enormous amount of energy per unit compressed volume of hydrogen.

And please explain p+p->deuterium (H_2) + neutrino + positron because I find this very interesting mainly due to the fantastic mutation of a proton becoming a neutron (i.e deuterium).

Normally, a neutron decays via weak interaction (in about 10 minutes if outside of a nucleus) into proton, an electron, and an anti-neutrino. Since a proton is slightly lighter than a neutron, it does not decay (by any standard model processes). However, a proton plus energy, can, with low probability, undergo the 'decay' p -> neutron + neutrino + positron, mediated by the same weak interaction. In the core of the sun, where the high temperature give each proton plenty of KE, and the high density makes collisions likely, once in a blue moon this reaction occurs along with with a collision. When it does, the formation of deuterium releases net energy (not much, but enough to keep things going).

Finally, I thank you for your answer.

Roger

 

[rogerk8]

I sincerelly want to thank you for putting so much time and effort into trying to explain these things to me. I feel honored!

I did not understand much though

So I will have to think about this before I can get back to you with adequate questions.

Take care!

Roger

 

[mfb]

This still makes me believe that the sun began as a bundling up of neutrons which later on gave rise to such a high mass (>2Gkg) and therefore gravitational force that it could attract protons. Now when it did that, the rising sun got charged and due to the sign of charge of electrons they came along too.

"I don't understand the right explanation so I invent something different" is not a useful way to learn.

 

[rogerk8]

What is wrong with "free thinking" and trying to understand things your own way?

 

[D H]

I know so little and understands so little so maybe I should quit now?

You should quit speculating and instead try to understand what people have posted.

Anyway here is how I see it: ...
This clearly states that, in the beginning, protons could not have bundled up due to gravity while the electromagnetic force is way much higher (to say the least).

That is precisely the kind of speculating you need to stop doing.

You are ignoring pressure, density, and temperature, and you are also ignoring the fact that the Sun is electrically neutral. The gravitational attraction between two protons is not responsible for fusion. Gravity is far too weak a force to overcome electrical repulsion between two protons. Gravitation is nonetheless absolutely essential. While the gravitational force between two protons is exceedingly small, the mutual gravitational interaction amongst the ~10⁵⁷ protons and neutrons in the Sun is extremely large. This is what is responsible for the extremely high pressure at the center of the Sun. The pressure at some point inside the Sun is equal to the weight of all the stuff above that point.

At the center of the Sun, this makes for a pressure of about 250 billion atmospheres, a temperature of about 15 million Kelvin, and a density of about 150 grams/cm³. That density is immense. Even though the Sun's core is a plasma (and hence a gas), it's is about eight times that of solid uranium at the Earth's surface. It is the temperature and density that are ultimately responsible for fusion. The high temperature makes for particles with very high velocities. The high density means *lots* of collisions between those fast moving particles.

 

[Astronuc]

150 grams/cm³ is really impressive as density goes, especially where hydrogen is concerned. That's 150 moles of H/cc, which is ~9 E25 protons/cc, as compared to 1 gm/cc (density of water), or density of air/gas at sea level, or density of a plasma in magnetic confinement, which is on the order of 1e14 H/cc. So the density in the sun's core is about 1e11 to 1e12 times of what we can accomplish in terrestrial systems.

 

[rogerk8]

I thank you all and will comment on your posts individually later in another letter because I haven't slept a single second tonight.

For now, I just want to say this:

I got it!

This morning about 5:30 AM I "woke up" and finally understood the following (in a closed system):

$$E_{tot}=m_pg_sh+\frac{m_pv^2}{2}=m_pg_sh+kT=constant$$

Which means that as h decreases, T increases!

The only minor question I have here is where this Ek=kT comes from. My guess is the Maxwellian velocity distribution where you however can discuss if the distribution is Maxwellian or not.

I have used this energy conservation solution to a complicated problem before. In that case I used it, with your thankful help, to determine the loss of speed for a planet while being submitted to a gravitational sling shot.

It is amazing how often you can use this fundamental law!

Roger
PS
Feel free to correct me if I'm wrong.

 

[rogerk8]

mgh is good enough for the basic idea. You have a large mass of hydrogen in a cube .1 light years on a side. Under the influence of self gravitation, it collapses to a diameter of 1 million miles (appx). That means the average h in mgh, for given volume of hydrogen is about 100 billion miles. The g is varying during the collapse, but it should be easy to imagine that you have an enormous amount of energy per unit compressed volume of hydrogen.

What is easy about that, you mean?

Normally, a neutron decays via weak interaction (in about 10 minutes if outside of a nucleus) into proton, an electron, and an anti-neutrino.

Very interesting information. Do I dare ask why? And neutrinos, are those the kind of particles that can't be stopped or registered by almost any means?

Since a proton is slightly lighter than a neutron, it does not decay (by any standard model processes). However, a proton plus energy, can, with low probability, undergo the 'decay' p -> neutron + neutrino + positron, mediated by the same weak interaction.

I will not ask why, I simply find it interesting. Recognizing that a positron is a positively charged electron.

In the core of the sun, where the high temperature give each proton plenty of KE, and the high density makes collisions likely, once in a blue moon this reaction occurs along with with a collision. When it does, the formation of deuterium releases net energy (not much, but enough to keep things going).

KE means keV, right? And I have never heard about a blue moon. Totally new to me who watches almost every scientific program on the telly. And yet manage to understand so little Anyway what you seem to say is that the "decays" gives rise to both a proton and a neutron among other fantastic particles and they fuse into deuterium and releases net energy. All of this due to high temperature and high density, right?

Thanks for all your help!

Roger

 

[Astronuc]

KE means keV, right? And I have never heard about a blue moon. Totally new to me who watches almost every scientific program on the telly. And yet manage to understand so little Anyway what you seem to say is that the "decays" gives rise to both a proton and a neutron among other fantastic particles and they fuse into deuterium and releases net energy. All of this due to high temperature and high density, right?

Thanks for all your help!

Roger

KE refers to kinetic energy, and kinetic energy of the atoms/molecules in a gas are related to temperature.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c1

Protons, neutrons and electrons were formed along time ago, back at the origin of the universe.

 

[rogerk8]

You are ignoring pressure, density, and temperature, and you are also ignoring the fact that the Sun is electrically neutral. The gravitational attraction between two protons is not responsible for fusion. Gravity is far too weak a force to overcome electrical repulsion between two protons. Gravitation is nonetheless absolutely essential. While the gravitational force between two protons is exceedingly small, the mutual gravitational interaction amongst the ~10⁵⁷ protons and neutrons in the Sun is extremely large. This is what is responsible for the extremely high pressure at the center of the Sun. The pressure at some point inside the Sun is equal to the weight of all the stuff above that point.

1) Very educational!
2) If you read my earlier post you may discover that I have already calculated the Fq/Fg for protons to be 10³⁶ and thus enormous.
3) What has this to do with pressure? To me pressure is nkT.
4) This is very interesting and educational information. I can not really understand it even though it sounds obvious.

At the center of the Sun, this makes for a pressure of about 250 billion atmospheres, a temperature of about 15 million Kelvin, and a density of about 150 grams/cm³. That density is immense. Even though the Sun's core is a plasma (and hence a gas), it's is about eight times that of solid uranium at the Earth's surface. It is the temperature and density that are ultimately responsible for fusion. The high temperature makes for particles with very high velocities. The high density means *lots* of collisions between those fast moving particles.

1) Tells me nothing. It's just a huge number.
2) This is a verification of the figure given for ITER which aims at 10 times the temperature of the Sun for some reason. Feel free to educate me
3) This unfortunatelly also tells me nothing. 150 grams per cubic cm almost sounds tiny to me.
4) This is amazing to say the least! How do I calculate that?
5) Roger that.
6) Incredible good explanation. Thank you!

Roger

 

[rogerk8]

150 grams/cm³ is really impressive as density goes, especially where hydrogen is concerned. That's 150 moles of H/cc, which is ~9 E25 protons/cc, as compared to 1 gm/cc (density of water), or density of air/gas at sea level, or density of a plasma in magnetic confinement, which is on the order of 1e14 H/cc. So the density in the sun's core is about 1e11 to 1e12 times of what we can accomplish in terrestrial systems.

Hi Astronuc!

Above I recently said that I thought 150 grams per cubic cm was almost tiny. I kind of regret that now because I was referring to one cubic cm of something solid. I thus formed my fingers to 1 cm and immagined it to weigh 150 grams. It did actually not sound so impressive coming from the core of the Sun but I kind of forgot the cubic cm being a gas (i.e a plasma)

Take care!

Roger

 

[PAllen]

Hi Astronuc!

Above I recently said that I thought 150 grams per cubic cm was almost tiny. I kind of regret that now because I was referring to one cubic cm of something solid. I thus formed my fingers to 1 cm and immagined it to weigh 150 grams. It did actually not sound so impressive coming from the core of the Sun but I kind of forgot the cubic cm being a gas (i.e a plasma)

Take care!

Roger

It is still 10 times denser than the core of the earth, and also much denser than any terrestrial solid. For example, gold is 19.3 grams / cubic centimeter, lead is 11.3. Note, iron at the surface of the Earth is only 7.9 , but under the pressure in the core, it nearly doubles.

 

[rogerk8]

Amazing!

 

[rogerk8]

It is still 10 times denser than the core of the earth, and also much denser than any terrestrial solid. For example, gold is 19.3 grams / cubic centimeter, lead is 11.3. Note, iron at the surface of the Earth is only 7.9 , but under the pressure in the core, it nearly doubles.

I agree I was very stupid with my estimation. And now that you explain even further I feel even more stupid. Heavy stuff like gold only some 20 grams per cubic cm. 20 grams! And gold is among the most dense material I know. Maybe except for Uranium.

But would you mind explaining the bold part. I simply do not grasp this pressure business.

It also seems like I can't understand that solids can get denser due to pressure.

But by saying this I also kind of understand that they can.

Roger

 

[rogerk8]

By the way, my formula is wrong.

I kind of forgot 1/r^2 in g and more importantly, that h is defined above the surface of the Sun.

Roger

 

[D H]

3) What has this to do with pressure? To me pressure is nkT.

No. Look at the units. Always look at the units! The right hand side has units of energy, not pressure.

The ideal gas law says PV=NRT (chemistry), or PV=nkT (physics). They're the same equation, and in both cases the units are correct. The only difference between the two is whether one uses number of moles or number of molecules. Divide both sides by volume and you'll find that pressure is proportional to the product of density and temperature.

1) Tells me nothing. It's just a huge number.

You need to think when you see a very large number such as 250 billion atmospheres. Think about what it means in terms of pressure and temperature.

2) This is a verification of the figure given for ITER which aims at 10 times the temperature of the Sun for some reason. Feel free to educate me

I'll say more about this below.

3) This unfortunatelly also tells me nothing. 150 grams per cubic cm almost sounds tiny to me.

I very specifically said that this is about eight times the density of solid uranium. You even highlighted that phrase! We are trying very hard to make this understandable to you by relating to things with which you might be familiar. That you highlighted, in bold, what I wrote and then had the audacity to write that this "almost sounds tiny to me" is rather annoying. We have spent a good deal of time responding to your queries. You should respond in kind and try to comprehend what we write.



As for why ITER is aiming for a temperature much higher than the 15 million degree temperature at the center of the Sun, I'm going to ask a rhetorical question. Here it is: Per unit volume, what produces more energy, the biological processes in a warm compost pile, or the nuclear fusion at the center of the Sun?

The surprising answer is a warm compost pile.


Almost all of the proton-proton collisions at the energies present in the center of the Sun result in two protons just bouncing off one another. There is no fusion. Only rarely do those collisions result in the production of deuterium. The p-p reaction is by far the slowest link in the p-p chain. On the rare occasion where two protons do combine to form deuterium, the rest of the p-p chain proceeds rather quickly to eventually form helium.

To make fusion worthwhile we have to do a lot (a whole lot) better than creating a very expensive warm compost pile. One way around the problem is to bypass the p-p reaction. That is why ITER is using deuterium and tritium. This is the easiest reaction to create. Even then, the odds are still pretty lousy at 15 million kelvin. Bumping the temperature up an order of magnitude and makes for something that produces a whole lot more energy than a warm compost pile.

 

[the_wolfman]

No. Look at the units. Always look at the units! The right hand side has units of energy, not pressure.

The ideal gas law says PV=NRT (chemistry), or PV=nkT (physics). They're the same equation, and in both cases the units are correct. The only difference between the two is whether one uses number of moles or number of molecules. Divide both sides by volume and you'll find that pressure is proportional to the product of density and temperature.

Actually he is correct P = nkT. This is the standard notation is plasma physics, here n is the number density (with units of inverse volume), k is Boltzmann's constant and T is the temperature. The product KT yields energy and when you multiple by n you get energy per unit volume, which has the same units as pressure.

 

[D H]

Actually he is correct P = nkT.

Ahh. I misunderstood where roger's misunderstanding lies. He was questioning that pressure the result of the weight of all the stuff above.

Roger, you are looking at P=nkT as saying that pressure is caused by the local density and temperature. Don't do that. Look at it the other way around, as saying that local density and temperature are determined by the pressure. The Sun's size is more or less constant. That means the Sun is in equilibrium (hydrostatic equilibrium, to be precise; google that term), which in turn means that the pressure at any point inside the Sun is just enough to buoy the weight of all the stuff above that point.

This fact can be used, for example, to estimate the mass of the Earth's atmosphere. Ignoring variations in terrain and ignoring variations in gravitational acceleration, the weight of the atmosphere must equal the pressure times the Earth's surface area. In other words, $m_{\text{atmos}} \approx 4\pi R^2 P_0/g$, where R is the effective radius of the Earth (6371 km), P₀ is atmospheric pressure at sea level (1 atmosphere), and g is the Earth's mean gravitational acceleration at sea level (9.80665 m/s²). Plugging in the numbers yields a value of 5.27×10¹⁸ kg (link), which is about 2.5% higher than the actual mass of the atmosphere because of those simplifying assumptions. Not bad for a simple estimate.

 

[rogerk8]

Actually he is correct P = nkT. This is the standard notation is plasma physics, here n is the number density (with units of inverse volume), k is Boltzmann's constant and T is the temperature. The product KT yields energy and when you multiple by n you get energy per unit volume, which has the same units as pressure.

Thank you Wolfman!

 

[rogerk8]

I very specifically said that this is about eight times the density of solid uranium. You even highlighted that phrase! We are trying very hard to make this understandable to you by relating to things with which you might be familiar. That you highlighted, in bold, what I wrote and then had the audacity to write that this "almost sounds tiny to me" is rather annoying. We have spent a good deal of time responding to your queries. You should respond in kind and try to comprehend what we write.

Yes, you did. And I found that amazing. But forming a cubic cm with my fingers att telling me that this (gas, however) weighs 150 grams in the core of the Sun did not impress me somehow. But I made a misstake and I'm sorry for that.

Almost all of the proton-proton collisions at the energies present in the center of the Sun result in two protons just bouncing off one another. There is no fusion. Only rarely do those collisions result in the production of deuterium. The p-p reaction is by far the slowest link in the p-p chain. On the rare occasion where two protons do combine to form deuterium, the rest of the p-p chain proceeds rather quickly to eventually form helium.

Sorry, but this does not educate me more than a science program on the telly.

To make fusion worthwhile we have to do a lot (a whole lot) better than creating a very expensive warm compost pile. One way around the problem is to bypass the p-p reaction. That is why ITER is using deuterium and tritium. This is the easiest reaction to create. Even then, the odds are still pretty lousy at 15 million kelvin. Bumping the temperature up an order of magnitude and makes for something that produces a whole lot more energy than a warm compost pile.

I would very much like to understand why and more exactly what happens here.

Finally, I think you are a very good teacher D H!

Thanks for all your help. I really mean that!

Roger

 

[rogerk8]

He was questioning that pressure the result of the weight of all the stuff above.

I actually still don't understand this. Even though it sounds kind of obvious.

Roger, you are looking at P=nkT as saying that pressure is caused by the local density and temperature. Don't do that. Look at it the other way around, as saying that local density and temperature are determined by the pressure. The Sun's size is more or less constant. That means the Sun is in equilibrium (hydrostatic equilibrium, to be precise; google that term), which in turn means that the pressure at any point inside the Sun is just enough to buoy the weight of all the stuff above that point.

1) This is very educational. Thanks! Now I only need to understand pressure (beyond nkT...).
2) This sound reasonable but I still don't understand it being a gas and all. Could you perhaps show me a formula that proves this? Would be very helpful, thanks.

This fact can be used, for example, to estimate the mass of the Earth's atmosphere. Ignoring variations in terrain and ignoring variations in gravitational acceleration, the weight of the atmosphere must equal the pressure times the Earth's surface area. In other words, $m_{\text{atmos}} \approx 4\pi R^2 P_0/g$, where R is the effective radius of the Earth (6371 km), P₀ is atmospheric pressure at sea level (1 atmosphere), and g is the Earth's mean gravitational acceleration at sea level (9.80665 m/s²). Plugging in the numbers yields a value of 5.27×10¹⁸ kg (link), which is about 2.5% higher than the actual mass of the atmosphere because of those simplifying assumptions. Not bad for a simple estimate.

Very interesting calculation. I will contemplate this methode of thinking.

Roger
PS
Wait a minute. Considering a certain internal radius, r, of f.i the Sun. The area is then as you say above i.e

$$A=4\pi r^2$$

Moving further into the Sun, the area decreases according to the above. The mass increases "some" due to more particles above this new point but/and these particles are now distributed over a smaller area. Which of course means that the pressure (N/m^2) is increasing (both with regard to increased mass and lesser area).

With the exception of the actual "above" mass increase, I think I got it now!

Or do I?

By the way, I would call this gravitational pressure.

 

[Astronuc]

Moving further into the Sun, the area decreases according to the above. The mass increases "some" due to more particles above this new point but/and these particles are now distributed over a smaller area. Which of course means that the pressure (N/m^2) is increasing (both with regard to increased mass and lesser area).

With the exception of the actual "above" mass increase, I think I got it now!

Or do I?

By the way, I would call this gravitational pressure.

Moving radially into sphere, the area decreases. The pressure is determined by the area and weight of the atmosphere above. The weight is due to the mass being pulled to the center of gravity, by gravity.

A fundamental property of all main sequence stars is thermal equilibrium. Thermal equilibrium is the liberation of energy from the interior of the star balanced by the energy released at the surface of the star. The energy released by a main sequence star is produced by hydrogen burning in its core (the fusion of 4H into 4He).

Another fundamental property of a main sequence star evolution is hydrostatic equilibrium. Hydrostatic equilibrium reflects the required pressure in the core of a star to support the weight of the outer plasma layers. The heat produced from hydrogen in the core burning supports this outward pressure upon the outer plasma layers.

. . . .

http://www.umich.edu/~gs265/star.htm

Hydrostatic equilibirum means the interior pressure balances the exterior pressure, or the outward force of the plasma supports the inward force of the plasma in the atmosphere above a given point. The plasma pressure comes from the kinetic energy (and momentum) of the nuclei, electrons and photons, just the way atmospheric pressure is due to the energy/momentum of gas molecules in the Earth's atmosphere. The energy comes from the fusion reactions going on in the core and atmosphere of the sun.

D H has provided a fair amount of information on the sun and the p-p fusion process. Although fusion on Earth is often explained as the process going on in the sun, it really isn't. Stars like the sun use p-p (proton-proton) fusion, with some fusion occurring by the CNO-cycle. The conditions under which such fusion occurs are well beyond the capability of anything man-made. I indicated that the density of solar (stellar) plasmas is many orders of magnitude greater than a terrestrial plasma.

http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html
http://csep10.phys.utk.edu/astr162/lect/energy/ppchain.html
http://csep10.phys.utk.edu/astr162/lect/energy/cno.html

Terrestrial fusion plasmas use d+t fusion, because it is easier, but has the disadvantage that most of the energy is released to the 14.1 MeV neutron, which means it has to slow down and be captured in some blanket, which is heated and the heat is then passed to a working fluid, which is then used in some thermodynamic (e.g., Brayton, Stirling or Rankine) cycle to produce electrical energy. Other concepts include d+d fusion, which requires slighly higher temperatures to be optimal, or d+He³ fusion, which is aneutronic, but requires even higher temperatures. He³ is unfortunately rather rare on earth, although a small amount is produced by decay of tritium.

It would help of one did some homework to educate oneself on the significance of the information others have presented. 150 g/cc is a substantial density, and even more so for hydrogen than say for Iridium (density = 22.65 g/cc). Compressing hydrogen (or any substance) to that density requires high pressure.

At STP, the density of hydrogen gas is 0.0000899 gm/cc. In liquid form, the density is about 0.07 gm/cc, and even solid, it's density is only about 0.088 gm/cc.
http://hyperphysics.phy-astr.gsu.edu/hbase/pertab/h.html

 

[rogerk8]

Thank you very much Astronuc. Very educational and interesting!

Actually I want to thank all of you guys that is helping me so far in my quest for some serious understanding of plasma physics.

But I hope you can bear with me that I will not stop until you close this thread. I simply like it too much. By saying this I also mean that you really do not have to answer at all (unless it perhas is fun ). Just let me educate myself based on the facts you so kindly have given as well as the basic growing understanding I am slowly acquireing.

It would however be nice if you correct me when I'm totally wrong. And perhaps answer a serious question sometimes even though I will read all of the links above that you so kindly have supplied.

Before starting reading the links (and reread all your posts) I wish to do two things.

1) Present my new estimation of gravitational pressure
2) Summarize of what I think I understand.

Hope this is ok.

My estimation of gravitational pressure:

Consider a radius, r, inside the Sun. Let's say the average density on the outside of this bourder/surface is n_o and on the inside it is n_i.

Now,

$$g==\frac{MG}{r^2}=\frac{n_iV_iG}{r^2}≈rn_i$$
$$A≈r^2$$
$$V≈r^3$$
$$m_o≈N_o=V_on_o≈(R-r)^3n_o$$
$$p==\frac{gm_o}{A}≈\frac{rn_i(R-r)^3n_o}{r^2}=n_in_o\frac{(R-r)^3}{r}$$

which may be rewritten as

$$p≈n_in_oR^2\frac{(1-\frac{r}{R})^3}{r/R}≈\frac{(1-\frac{r}{R})^3}{r/R}$$

where n_i and n_o has been considered constant for small changes in r.

Now,

$$\frac{p(r/R=0,4)}{p(r/R=0,5)}=2,16$$

which seem to show that gravitational pressure is indeed increasing inwards.

Now I kind of think that all matter, f.i a soccer ball made of pure iron, has higher pressure in the core, even if g is very small here.

A fun thing to notice, if my calculations are right, is that for r=0, there is a singularity. In my layman thoughts I do however think this is just math and not physics. Even though I kind of think that for this infinitismally small core, the mass must approach infinity. Or you may see it like the formula simply is not defined here. Or I'm totally wrong

As always, you may correct me if I'm wrong.

Roger
PS
I fail finding the Tex code for tilde ("proportional to").

 

[rogerk8]

Hi!

I have now collected the most interesting statements from you guys. I am attaching a "understanding grade" of 0-5 at each statement. Feel free to educate me if this grade is less than 3.

1) The higher temperatures (for DT fuel, my note) are needed to counter the lower pressure in the reactor, with a higher required power density [mfb], 3 (power density, not understood)
2) Because the power output of a single cubic meter of solar core material (i.e. ordinary hydrogen, proton-proton fusion) is roughly on par with a toaster oven [mheslep], 1 (is it really that bad?)
3) Pressure is limited by the magnetic fields (in a Tokamak, my note)- a higher temperature does not allow to increase pressure, so the volume density will go down. As the interaction probability rises quickly with temperature, this still leads to a higher fusion rate [mfb], 1 (why not higher pressure? And fusion rate, how is that defined?)
4) The sun, including the core, is electrically neutral. Both ions and electrons exist under very height pressures at the core.[Drakkith], 4 (electrically neutral sounds convenient, though)
5) The magnetic confinement in a tokamak is completely different from the gravitational "confinement" in the sun.[mfb], 5
6) One definition of a planet versus and asteroid is a body large enough the gravity overwhelms all possible sources of mechanical rigidity, making the body round [PAllen], 5
7) Also remember that the Sun has electrons and is not charged overall. The atoms in the gas cloud that initially collapsed to form the Sun didn't repel each other because they were not ionized.[Drakkith], 4 (the post Big Bang soup should have been be more elementary, I believe)
8) By the time you need to worry about proton repulsion (in the Sun, my note), you already have the 250 billion kg/cm^2 pressure of neutral matter above to overcome it.[PAllen], 3 (why suddenly not neutral? What happens under high pressure?)
9) Correct in that the starting point (for the Sun, my note) is Hydrogen gas. However, the fusion reaction is not to Helium 2, which would not release energy. It is to Deuterium when the proton-proton interaction is accompanied by emission of a positron and a neutrino. This process releases energy but is very rare.[PAllen], 1 (but very interesting).
10) The core of the sun has a density of about 150 grams/cm^3, well over 10 times the density of the Earth's core. This is because of the enormous pressure of the overlying layers squeezing an ionized plasma to a density beyond any material we know on earth. [PAllen], 5
11) Even today, the core of the sun is neutral - the hydrogen is ionized and we have a plasma, but the negative electrons are still hanging around there together with the positive protons and helium nuclei. [mfb], 5
12) as it (the neutral Hydrogen, my note) collpased and heated (you can think of this simply as conversion of gravitational potential energy to heat), the center became ionized, but still neutral on average. You than have a neutral plasma at high temperature and pressure (= high density), such that the rare p + p -> deuterium + neutrino + positron can occur (at a low rate per volume). [PAllen], 2 (why ionized?)
13) p + p -> deuterium + neutrino + positron [PAllen], 0 (but extremely interesting)
14) The g (in a .1 lightyears wide cube of Hydrogen, my note) is varying during the collapse, but it should be easy to imagine that you have an enormous amount of energy per unit compressed volume of hydrogen. [PAllen], 2 (no, it's not easy)
15) Normally, a neutron decays via weak interaction (in about 10 minutes if outside of a nucleus) into proton, an electron, and an anti-neutrino [PAllen], 0 (but extremely interesting)
16) Since a proton is slightly lighter than a neutron, it does not decay (by any standard model processes). However, a proton plus energy, can, with low probability, undergo the 'decay' p -> neutron + neutrino + positron, mediated by the same weak interaction [PAllen], 0 (but extremely interesting)
17) In the core of the sun, where the high temperature give each proton plenty of KE, and the high density makes collisions likely, once in a blue moon this reaction occurs along with with a collision. When it does, the formation of deuterium releases net energy (not much, but enough to keep things going). [PAllen], 1
18) You are ignoring pressure, density, and temperature, and you are also ignoring the fact that the Sun is electrically neutral. [D H], 4 (electrically neutral sounds as convenient as the useful law of conservation of energy)
19) While the gravitational force between two protons is exceedingly small, the mutual gravitational interaction amongst the ~10^57 protons and neutrons in the Sun is extremely large. This is what is responsible for the extremely high pressure at the center of the Sun [D H], 1 (what mutual gravitational interaction? And how can this give a high pressure?)
20) The pressure at some point inside the Sun is equal to the weight of all the stuff above that point. [D H], 5
21) Even though the Sun's core is a plasma (and hence a gas), it's is about eight times that of solid uranium at the Earth's surface [D H], 5 (even more amazing now )
22) It is the temperature and density that are ultimately responsible for fusion [D H], 5
23) The high temperature makes for particles with very high velocities [D H], 5
24) The high density means *lots* of collisions between those fast moving particles. [D H], 5
25) You need to think when you see a very large number such as 250 billion atmospheres. Think about what it means in terms of pressure and temperature [D H], 3 (still just a huge number )
26) Roger, you are looking at P=nkT as saying that pressure is caused by the local density and temperature. Don't do that. Look at it the other way around, as saying that local density and temperature are determined by the pressure [D H], 5 (only true in stars though)
27) Moving radially into sphere, the area decreases. The pressure is determined by the area and weight of the atmosphere above. The weight is due to the mass being pulled to the center of gravity, by gravity. [Astronuc], 5
28) Terrestrial fusion plasmas use d+t fusion, because it is easier, but has the disadvantage that most of the energy is released to the 14.1 MeV neutron, which means it has to slow down and be captured in some blanket, which is heated and the heat is then passed to a working fluid, which is then used in some thermodynamic [Astronuc], 3 (please explain "easier")

Roger

 

[PAllen]

Why not use this as a list of questions to read about on your own, from links provided in this thread, or books you can take out of the library?

I don't think I am alone in saying much of this thread has been a case of 'education against active resistance'. There is the sense of 'force me to understand despite my resistance'. I have no more interest in this.

To pick one example out of many, you attach 0 understanding to the idea of p + energy -> neutron + neutrino + positron is a rare but possible reaction. Why not find an introductory book on particle physics, even one meant for the educated layperson, that covers weak, strong, and EM processes? Not only don't you want to do this, you want us to force you to learn this material! Such a book would cover both why this reaction is possible and why it is expected to be rare. To be able to compute how rare, you need a graduate level course. But to understand why it is possible but rare, a good fairly elementary text should suffice.

 

[rogerk8]

Please forgive me, but I thought this was a forum where you can ask questions and thereby cut some corners with regard to how much "useless" information there is out there.

It's almost like when you are discussing things with a person who owns a smartphone. He immediatelly and happilly pulls it out when the answer to the question seems googleable.

But what happens? Well he googles up the information but while he is doing this he gives me lots of unrelative and uninteresting information and I bet that all of this never takes less than ten minutes.

So I am just trying to cut these corners, because I know there is lots of information about this stuff we are discussing here. But really, I have refused to confess this before because I think I know so little and is kind of embarassed about it, but I do hold a degree in Master of Science in Electronical Engineering even though I graduated 96. Which is 18 years ago, if you do the math.

I do however remember very little, as you have noticed, but I am really TIRED of reading books!

And I LOVE chatting with you guys!

So I simply want to refresh my "forgotten" knowledge and understand.

And I can not understand why it seems to be so hard for you to kindly help me.

Don't you like helping people, or what is wrong with you?

While I'm at it, I feel that scientifically skilled people like you are the modern worlds priesthood.

Another thing, D H "bragged" about knowing five decimals in g by heart. I know five decimals in pi by heart, but who cares!

Regards, Roger