Plasma Fusion Preface

Template:Subpages This book is an effort in describing some basic considerations with regard to fusion power and its creation. The focus is, however, not only on fusion power alone but on understanding related physical phenomena such as, for instance, pressure.

Plasmas in nature

The Saha equation states

$\frac{n_i}{n_n}=2.4*10^{21}\frac{T^{3/2}}{n_i}\exp -(\frac{U_i}{kT})$,

where n_i is the ion density and n_n is the neutral atoms density and U_i is the ionization energy of the gas.

Putting for ordinary air

$n_n=3*10^{25}{m}^{-3}$

$T=300K$

$U_i=14,5eV(nitrogen)$

gives

$\frac{n_i}{n_n}=10^{-122}$

which is ridiculously low

The ionization remains low until U_i is only a few times kT.

So there exist no plasmas naturally here on earth, only in astronomical bodies with temperatures of millions of degrees.

Numerical Example

While the pressure inside the Sun, according to below if that is right, seems to be in the order of two and half million atmospheres and the (surface) temperature being some 50 000K, we can use the formula p=nkT to calculate the particle density (n), using this formula I get n as 3,6E29 (which is some 10000 more dense than air), here we can use the Saha Equation to calculate the proportion (n_i/n_n) of the Hydrogen ions (n_i) to that of neutral Hydrogen atoms (n_n) while using the ionization energy (Ui=13,6eV) of Hydrogen as 5,6%.

Unproven ideas regarding sun pressure

The intensity (I) from something omnidirectional is

$I=\frac{P}{4\pi R^2}$

where P is the radiated power and R the distance, this may give rise to a density of

$\rho ={\rho }_0(1-\frac{4\pi R^2}{4\pi b^2})={\rho }_0(1-\frac{R^2}{b^2})$

the mass (m) is then

$M=\int \rho dv=\underset{0}{\overset{b}{\int }}\rho 4\pi R^{2}dR$

where b is the radii of the Sun and R some radii in the Sun, this gives

$M=4\pi {\rho }_0[\frac{R^3}{3}-\frac{R^5}{5b^2}]=4\pi {\rho }_0{R}^3[\frac{1}{3}-\frac{R^2}{5b^2}]$

the pressure is then

$p=\rho gh=\rho gR=\rho \frac{GM}{R^2}R=\rho \frac{GM}{R}$

thus

$p={\rho }_0(1-\frac{R^2}{b^2})G4\pi {\rho }_0{R}^2(\frac{1}{3}-\frac{R^2}{5b^2})$

the mean rho is

$\tilde{\rho }=\frac{M}{V}$

and

$\rho ={\rho }_0(1-\frac{R^2}{b^2})$

thus

$\tilde{\rho }=\frac{1}{b}\underset{0}{\overset{b}{\int }}{\rho }_0(1-\frac{R^2}{b^2})dR$

which gives

$\tilde{\rho }=2/3{\rho }_0$

resulting in

${\rho }_0=\frac{3}{2}\tilde{\rho }=\frac{3M}{2V}$

the pressure is then

$p=(\frac{3M}{2V}{)}^2(1-\frac{R^2}{b^2})G4\pi {\rho }_0{R}^2(\frac{1}{3}-\frac{R^2}{5b^2})$

now we need to check what value R should have, I have calculated below that the function has a minima/maxima (depending on which direction you are going), this point is found by regular derivation and putting the derivate to zero, derivating the formula with regard to R gives

$\frac{2}{3}-\frac{32}{15}(\frac{R}{b}{)}^2+\frac{6}{5}(\frac{R}{b}{)}^4=0$

where we make a change of variable by putting

$x=(\frac{R}{b}{)}^2$

which gives

$x^2-\frac{160}{90}x+\frac{10}{18}=0$

solving this second order equation then gives

$x=\frac{16}{18}-\sqrt{(\frac{16}{18}{)}^2-\frac{10}{18}}$

where only minus in front of the root-expression is possible because otherwise R>b, a numerical solution is

$x=0,89-0,48=0,40$

thus

$\frac{R}{b}=0,63\approx \frac{2}{\pi }$

where I just recognized the value from the DC value of a full wave rectified signal, b(sun) is then 700Mm which gives that at the radii of 446Mm there excist a maxima/minima, Physics Handbook then states

$M(Sun)=333000m(Earth)$

and

$V(Sun)=\frac{4\pi b_s^3}{3}=1,4\cdot 10^{27}$

the mass of the Earth is unfortunatelly not stated i Physics Handbook but I have little childlishly tried to estimate it by considering the Eart to mostly consist of water, with b(Earth)=6370km I get

$m(Earth)\approx {\rho }_{H_{2}O}\cdot \frac{4\pi b_j^3}{3}=10^{24}$

due to this estimation I get

$M=3,3\cdot 10^{29}$

which gives the mean density of the sun as

$\tilde{\rho }=\frac{M}{V}=238kg/m^3$

which I think is rather low but we should remember that it is a gas/plasma and air is more than 200 times less dense, the constant of gravity is then

$G=6,62\cdot 10^{-}11$

with this value we can now calculate the Sun pressure which gives

$p=7\cdot 10^{9}Pa=70000atm=70ton/c{m}^2$

This is a rather heavy pressure, remember that normal air pressure is like pointing a finger with 1kg of force on for instance a table, it is possible to estimate the temperature of the Sun by using the Stefan-Boltzmann radion law which states that the intensity goes like T^4, this gives

$\frac{T_s^4}{4\pi (AU{)}^2}=\frac{T_j^4}{4\pi R_j^2}$

where AU is the astronomical unit that is the distance from the Sun to the Earth and Rj is the radii of the Earth, using this we get

$T_s=(\frac{AU}{R_j}{)}^{\frac{1}{2}}{T}_j$

which with Tj=300K, Rj=6370km och AU=1,5E11 gives

$T_s=46000K\approx 50000K$

when we now have the pressure and the temperature of the Sun we can calculate the particle density as

$p=nkT$

which gives a particle ion density (n_i) as

$n_i=10^{28}$

which should be compared to the particle density in ordinary air that is around E25, the particle density in the Sun thus seem to be 1000 times more dense, my calculations however says that the density increases with 50% towards the center, I have by the way used the surface temperature that most likely is higher in the core of the Sun, if my formula is correct it is clear that the pressure in the core of the Sun may be some 7 times higher than the above result (just put R=0), using the Saha Equation above and the ionization energy of hydrogen as 13,6eV we can now calculate the ion proportion in the Sun as 11%.

I have now revised my estimated mass of the Earth by looking it up in Wikipedia, it seems lake it is 6 times higher, this actually makes my pressure estimation 36 times higher so the pressure in the Sun should be some 2 500 000atm=2 500 000kg/cm^2!

Basic considerations

When there is a moving particle of charge in a magnetic field, the following equation applies:

$m\frac{dv}{dt}=qvXB$

A simple way of solving this equation is to put

$v=v_0{e}^{jwt}$

The equation then becomes

$mjw{v}_0{e}^{jwt}=qvXB=mjwv$

While only considering the magnitude we get

$w_c=\frac{|q|B}{m}$

and while v=wr we get

$r_L=\frac{mv}{|q|B}$

where w_c is called the cyclotron frequency and r_L is called the Larmor radius.

This means that a particle will gyrate around the lines of force with the cyclotron frequency and the Larmor radius.

This is the most fundamental reason why a plasma can be confined by a magnetic field.

Numerical Example

If we look at electrons and the magnetic flux density of some 5T which I think ITER runs with (the information has unfortunatelly been removed during the Wikipeadia era, I think) we have a cyklotron frequency of 8E11 rad/s and while I remember that the thermal energy of ITER is some 10keV you get with the below energy formula that the speed of the electrons is some 5,6E7 m/s which yields a Larmor-radii of some 70um, if we look at the protons (gyrating the other way around) the cyklotron frequency becomes 4,8E8 rad/s which at the same temperature yields a speed of 1,4E6 m/s which gives a Larmor-radii of 2,9mm, read that the Larmor-radii approximatelly are 0,1mm and 1mm respectively.

Energy and temperature of a plasma

The average energy may be written

$E_{AV}=\frac{1}{2}m{v}^2\approx kT$

where there is an additional kT/2 for each degree of freedom (whatever that means), the speed is then

$v=\sqrt{\frac{2kT}{m}}$

the above energy equation fits the Maxwellian velocity distribution function

$f(v)=A\exp (-\frac{m{v}^2/2}{kT})$

where the constant A can be calculated using

$1=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(v)dv$

which gives us

$A=\sqrt{\frac{m}{2\pi kT}}$

what this means is that

$<kT>=\frac{m{v}^2}{2}$

where <kT> is a mean value, in other words, there are particles that has higher "temperature" than the mean kinetic energy

Numerical Example

The exponential part of the distribution function is 1 maximum and that is at the speed of 0 m/s but it is the area under the curve that represents the probability 1 when all speeds are taken into account, that means that the distribution function has to be integrated over all speeds determining the constant A by putting the integral to 1, A at 300K and electrons is 6,2E-6.

eV and some constants

According to Francis F. Chen, physicists use eV to avoid confusion, kT is however in Joule so it has to be converted using

$kT/q=eV$

Potentials are in Volt but their energy (qU) may be considered being in eV as is.

Let's state some constants:

$k=1,38\cdot 10^{-23}J/K$

$q=1,6\cdot 10^{-19}As$

$m_p=1,67.2623\cdot 10^{-27}kg$

$m_n=1,67.4929\cdot 10^{-27}kg$

$m_e=9,10.9390\cdot 10^{-31}kg$

$h=6,63\cdot 10^{-34}Js$

${\mu }_0=4\pi \cdot 10^{-7}Vs/Am$

${ϵ}_0=8,85\cdot 10^{-12}As/Vm$

Deriving the magnetic flux density of a current loop

B-field from a current loop

From Maxwell's equations we have

$\mathrm{\nabla }\cdot B=0$

which may be rewritten as

$B=\mathrm{\nabla }XA$

where A might be an arbitrary vector.

Using the vector magnetic potential

$A=\frac{{\mu }_0}{4\pi }\underset{v}{\int }\frac{J}{R}dv$

and realising that

$Jdv=JSdl=Idl$

we have from Biot-Savat law

$B=\frac{{\mu }_{0}I}{4\pi }{\oint }_c\frac{dlX{a}_R}{R^2}$

Defining

$dl=bd\varphi a_{\varphi }$

and

$R=a_{z}z-a_{r}b$

and

$dlXR=a_{\varphi }bd\varphi X(a_{z}z-a_{r}b)=a_{r}bzd\varphi +a_z{b}^{2}d\varphi$

and realising that the r-part cancel out we get

$B=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}{a}_z\frac{b^{2}d\varphi }{(z^2+b^2{)}^{3/2}}$

or

$B=\frac{{\mu }_{0}I}{2}\frac{b^2}{(z^2+b^2{)}^{3/2}}=\frac{{\mu }_{0}I}{2}\frac{b^2}{R^3}\hat{z}$

where the dimension of B obviously is

$B\propto \frac{1}{R}$

Numerical Example

If b=2m and I=100A the magnetic flux density (B) at a distance (R) of 1m is only 0,25mT, this is not much compared to major line of force of some 5T, so it is fair to say that the gyrating current do not affect the B-field.

Deriving the electrical field intensity of a charged loop

E-field from a charged loop

One can write the E-field like this

$E=\frac{{\rho }_L}{4\pi {ϵ}_0}\int \frac{d{l}^{\prime }}{R^2}=\frac{{\rho }_L}{4\pi {ϵ}_0}\int \frac{d{l}^{\prime }R}{R^3}$

where R in the denominator is a vector like

$R=-b\hat{r}+z\hat{z}$

and

$d{l}^{\prime }=bd\varphi$

beacause of symmetry there are no r-components which gives us

$R=z\hat{z}$

which may be rewritten as

$E=\frac{{\rho }_L}{4\pi {ϵ}_0}\int \frac{bd\varphi }{z^2}\hat{z}$

or

$E=\frac{{\rho }_L}{4\pi {ϵ}_0}\int \frac{bd\varphi }{R^2}\hat{z}$

and finally

$E=\frac{{\rho }_L}{4\pi {ϵ}_0{R}^2}2\pi b\hat{z}$

where it actually says Q/R^2 which has the same "dimension" as E, then you may observe that E is in the z-direction for charges of positive sign, rho_L may be viewed as (and must not be homogenous)

${\rho }_L=\frac{\sum q}{2\pi b}$

or

${\rho }_L=\frac{\sum q}{2\pi r_L}$

where the dimension of E obviously is

$E\propto \frac{1}{R^2}$

Numerical Example

If we have a Larmor-radii (b) for protons of 2,9mm, the circumference is then able to hold 1,4E22 protons and this gives a total Q of 0,22uC which in turn gives a line charge density (rho_L) of 12,7uC/m. At he distance (R) of one meter we get a electric field intensity (E) of 2,2kV/m which isn't that low (the dielectric strength of air is in the order of 5MV/m) but it is a scenario that never will happen because the particles hits each other all the time.

Unproven ideas regarding current/charged loops

I have manually reverted the chapter regarding the magnetic flux density (B) from a current loop because I have become to think it is more profound than I thought because certainly the current revolves in the loop and we get a Bz-field from it BUT current is charges in motion where I just philosophies that for instance protons have the same Larmor-radii which means that there is no obligation that protons will revolve around the line of force beside each other, they may revolve behind each other also like a string of pearls, principally one may view it as the pearl "necklace" perhaps may be jammed with protons where the movement of the protons yields the current and hence the B-field.

My thougth here is that the particles of same mass and charge gyrates along the line of force with the same cyclotron frequency and the same Larmor radii, if one may think that the particles of same charge actually gyrates "behind" eachother, we have two scenarios:

1) The current generated by the gyrating particles around a local Bz-field (aka B_phi from the current which is in the z-direction) gives rise to a B_phi-field, and as far as I think I understand this amplifies the sitting B_phi-field if the charges are positive (and is moving in the local phi-direction).

2) While we have a string of particles around the gyration then we also have a "snapshot" of a static E-field, the E-field does also seem to amplify the sitting E-field in the local phi-direction.

Positively charged particles obviously have a positive Ez component, I'm not sure but according to the Lorentz Force equation

$q(vXB)=m\frac{dv}{dt}$

and using

$v_r=v_{0r}{e}^{jwt}$

and

$v_{\varphi }=v_{0\varphi }{e}^{jwt}$

and recognizing for simplicity that

$v=(0;v_{\varphi };0)$

and

$B=(0;0;B_z)$

the movement kind of have to be

$q{v}_{0\varphi }{e}^{jwt}{B}_z=jwm{v}_{0r}{e}^{jwt}$

where

$e^{jwt}$

may be shorted out and we are left with

$q{v}_{0\varphi }{B}_z=jwm{v}_{0r}$

where 0 only indicates the amplitude, loosing the zeroes we have

$q{v}_{\varphi }{B}_z=jwm{v}_r$

with these (new) directions this actually means that the formula for the cyclotron frequency

$w_c=\frac{qB}{m}$

is wrong, it should be

$w_c=\frac{1}{j}\frac{q{B}_z}{m}\frac{v_{\varphi }}{v_r}=\frac{q{B}_z}{m}\frac{v_{\varphi }}{v_r}\mathrm{\angle }-90\mathrm{deg}$

which means that there has to be a v_r drift for w_c to exist, this means that the Larmor radii will increase with time and is as such not stable, a firm B will not make the particles gyrate around the line of force with the same radii at all time (I am probably wrong here).

Finally it seems like

$0=qE+q(vXB)$

by the loops I am describing actually means the E and the B increases in a similar manner, a constant multiplied with E and B may obviously be of no importance because it can be shorted out, the equation is then stable, disregarding that E and B probably do not increase in the same manner.

Drifts in a plasma

This picture shows how an E-field would interact with a B-field to change the particle orbit.

This picture describes what happens to a particle when the magnetic field is non uniform.

This graph describes the centrifugal drift in a plasma.

Using

$m\frac{dv}{dt}=q(E+vXB)$

and putting it to zero because we are examinating if a plasma can stand still by itself, this gives

$0=q(E+vXB)$

that can be rewritten as

$E=-vXB$

if we then take the cross product with B from the right, we get

$EXB=BX(vXB)$

then we can use the "BAC-CAB" rule which means that

$AXBXC=B(A\cdot C)-C(A\cdot B)$

the evidence for this is rather complicated so I won't prove it here, but

$EXB=BX(vXB)=v{B}^2-B(B\cdot v)$

the transverse components of this equation are

$v_{gc}=\frac{EXB}{B^2}$

and the magnitude of this guiding center drift is

$v_{gc}=\frac{E}{B}$

realising that

$F=qE$

one could set

$v_{force}=\frac{1}{q}\frac{FXB}{B^2}$

where F might be

$F_E=qE$

due to an E-field or

$F_g=mg$

due to gravity or

$F_{cf}=a_r\frac{m{v}_{//}^2}{R_c}$

due to the centrifugal force while a particle is moving along the lines of force, then the drift due to E will be

$v_E=\frac{EXB}{B^2}$

and the drift due to gravity will be

$v_g=\frac{m}{q}\frac{gXB}{B^2}$

and the drift due to a curved B-field will be

$v_R=\frac{1}{q}\frac{F_{cf}XB}{B^2}=\frac{m{v}_{//}^2}{q{B}^2}\frac{R_{c}XB}{R_c^2}$

It is interesting to note that

$|v_E|=|\frac{E}{B}|$

It is harder to derive and explain the drift in a nonuniform B-field where the force may be written

$F=-/+\frac{q{v}_{\perp }{r}_L}{2}\frac{dB}{dy}\hat{z}$

where v_$\perp$ denotes speed perpendicular to B_phi, which put into the force-formula above gives the guiding center drift

$v_{gc}=\frac{1}{q}\frac{F_{z}X{B}_{\varphi }}{B^2}=\frac{1}{q}\frac{F_{z}X{\hat{B}}_{\varphi }}{B}=-\frac{1}{q}\frac{|F|}{|B|}\hat{r}=-/+\frac{v_{\perp }{r}_L}{2B}\frac{dB}{dr}\hat{r}$

where the index just shows the important components, they are still vectors and the cross procuct shows that F_z generates a r-component for the drift due to B being in phi-direction only, however B_phi is attenuated according to 1/r so there exist a B gradient here, the above may be generalized as

$v_{\mathrm{\nabla }B}=-/+\frac{v_{\perp }{r}_L}{2}\frac{\mathrm{\nabla }BXB}{B^2}$

which is the grad-B drift or the drift caused by inhomogeneities in B, it can therefore be shown that the total drift in a curved vacuum field is

$v_{cv}=v_R+v_{\mathrm{\nabla }B}=\frac{m}{q{B}^2}\frac{RcXB}{R{c}^2}(v_{//}^2+\frac{1}{2}{v}_{\perp }^2)$

"It is unfortunate that these drifts add. This means that if one bends a magnetic field into a torus for the purpose of confining a thermonuclear plasma, the particles will drift out of the torous no matter how one juggles the temperature and magnetic fields" –Francis F. Chen

Numerical Example

If we look at the gravitational drift (v_g) and runs the ITER B of 5T we have that for electrons v_g is 1,2E-11 m/s, for protons it is v_g 2E-8 m/s. These are no high velocities but in time the particles will drift out of the plasma no matter how we juggles with the magnetic flux density (B), I am however surprised of the low speeds. On the other hand we have other kind of drifts to consider, this is the most easily calculated.

The plasma as a fluid

If we consider a plasma as a fluid we have

$mn[\frac{dv}{dt}+(v\cdot \mathrm{\nabla })v]=qn(E+vXB)-\mathrm{\nabla }p$

where it can be shown that the two terms to the left may be omitted.

If we then take the cross product with B we have

$0=qn[EXB+(v_{p}XB)XB]-\mathrm{\nabla }pXB$

or

$0=qn[EXB-v_p{B}^2]-\mathrm{\nabla }pXB$

where one term has been deliberately omitted.

Rearranging the above yields the total perpendicular drift in a plasma considered as a fluid

$v_p=\frac{EXB}{B^2}-\frac{\mathrm{\nabla }pXB}{qn{B}^2}=v_E+v_D$

where the so-called diamagnetic drift is

$v_D=-\frac{\mathrm{\nabla }pXB}{qn{B}^2}$

where the force is

$F_D=-\frac{\mathrm{\nabla }p}{n}$

meaning the gradient of the pressure

$p=nkT$

to volume particle density.

For an isoterm plasma we have

$\mathrm{\nabla }p=kT\mathrm{\nabla }n$

Numerical Example

At 300K and a nitrogen density in ordinary air of some E25 the pressure becomes 41kPa which isn't the normal air pressure of some 1E5Pa which makes me think that the density is a bit higher, 2.5E25 makes it right.

The Standard Model

1. electron and positron ("anti-electron")
2. muon and anti-muon
3. tau and anti-tau

Along with these comes their neutrino and anti-neutrino which gives six distinct types of particles or:

1. electron
2. electron-neutrino
3. muon
4. muon-neutrino
5. tau
6. tau-neutrino

The neutrinos are preliminary massless and thus very hard to detect.

The dominant three of these are fundamentals and consist of quarks. For our purposes it is enough to recognize two types of quarks namely the up-quark and the down-quark. This is because a neutron consists of two down-quarks and one up-quark while a proton consists of two up-quarks and one down-quark.

As mentors at PF have explained, a neutron can undergo weak interaction (transmutation) and be converted to a proton releasing an electron and an anti-neutrino. This has to do with the fact that a quark can change its type/flavor. In this case one down-quark "only" has to change to one up-quark to make the change of the particle.

It has also been explained how a proton can be changed to a neutron in a similar manner.

This is the basic reason for all those protons at the birth of a star like our Sun can generate neutrons and thus Deuterium to actually start the fusion process to Helium.

If I am allowed to think anything, I think that these theories are just fancy particles.

Numerical Example

According to Physics Handbook the muon has a mass of 106MeV/c^2, the tauon has a mass of 1807MeV/c^2. 1Mev of energy means in Joule 1,6E-13J which gives the mass of the muon as 1,7E-11J/c^2 and the mass of the tauon as 2,9E-19J/c^2 which yield their masses as 1,9E-30kg and 3,2E-29kg respectively. The smallest particle I know is the electron with a mass of some 1E-30 so the muon weighs about two electron masses and the tauon some 32 electron masses.

Radiation particles

1) Beta-particle (electron)

2) Alpha-particle (ordinary Helium_4 nuclei)

3) Gamma-rays (high energetic photons emitted from the nuclei)

4) X-rays (slightly lower energetic photons emitted when electrons are decelerated or accelerated)

Numerical Example

Gamma-rays start with an energy of 10keV, this means a wavelength of 1Å, ordinary visible light is between 10Å and 100Å.

Bohr model derivation

It has been proven that

$n\lambda =2\pi r$

which means that the length of the electron orbit has to be an integer number of times the wavelength.

With the use of the de Broglie wavelength

$\lambda =h/p$

and

$\hslash =h/2\pi$

the above equation may be rewritten as

$pr=mvr=n\hslash$

Referring to the basic force relationship where the centrifugal force is equal to the electromagnetic force we may write

$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$

where

$k=\frac{1}{4\pi {ϵ}_0}$

Solving for v yields

$v=\sqrt{\frac{k{e}^2}{mr}}$

Integrating the electromagnetic force gives the potential energy as

$E_p=-\frac{k{e}^2}{r}$

The kinetic energy may as usual be written

$E_k=\frac{m{v}^2}{2}$

Adding Ep with Ek with the use of the expression for v above then yields

$E_{tot}=E_p/2=-\frac{k{e}^2}{2r}$

Now,

$mrv=mr\sqrt{\frac{k{e}^2}{mr}}=\sqrt{k{e}^{2}mr}==n\hslash$

Solving for r yields

$r=\frac{n^2{\hslash }^2}{k{e}^{2}m}$

For n=1 this is called the Bohr Radius and for Hydrogen it can be shown that this is some 0,5Å.

Using this equation and the above expression for speed gives

$v=\frac{k{e}^2}{n\hslash }$

which shows how speed is discretely depended on shell number (n).

For optional atom you may view k as kA where A is the atom number (this is however not true in real life).

Numerical Example

If a jump is done from n=1 to n=2, the energy difference is radiated with a photon of hf=10,2eV which means a wavelength of 0,122Å.

Proton-proton fusion

These statements are cited from

1) Protons fuse

2) One proton is transmuted into one neutron forming Deuterium (releasing one positron and an electron-neutrino), which gives

$H_1^2$

3) Deuterium fuses with another proton (which also releases gamma-rays), which gives

$H{e}_2^3$

4) Two of the resulting Helium_3 nuclei fuse, which gives

$2xH{e}_2^3=H{e}_2^4+2p$

5) An Alpha particle (Helium_4) forms with the energetic release of two protons to complete the process.

A fun quote by Arthur Eddington:

"I am aware that many critics consider the stars are not hot enough. The critics lay themselves open to an obvious retort; we tell them to go and find a hotter place."

Numerical Example

If the gain in energy might be written Eb(after)-Eb(before) some Eb(He3)-Eb(H)=7,8-2,2=5,5MeV is gained when He3 is created.

Unproven ideas regarding proton-proton fusion

I think the proton-proton fusion should read

$p+p+e+e->p+n^{-}+\mathrm{\Delta }{E}_b$

here we have mass conservation due to n>~p+2e but we have a negatively charged neutron, so to speak, we could however rewrite this as

$H_1^1+H_1^1=H_1^{2-}+\mathrm{\Delta }{E}_b$

but by definition there exist no neutral atoms in a (hot) plasma.

Two protons could in theory however fuse to

$H{e}_2^2$

but this isotope does not exist (according to Physics Handbook), the simplest Helium isotope is

$H{e}_2^3$

which again uses a neutron. I think that neutrons are some kind of core glue, there are however more scientific theories of this which is called the Yukawa Force (or Potential) but it sounds like a convenient fabrication.

On the other hand, the ratio between gravitational force and Coulombian repelling force for protons is around

$10^{-36}$

so the gravitational force is not much use for holding a couple of protons together, but maybe the secret lies in the neutrons?

There are isotopes of all matter but they can seldom get rid of more than a few neutrons to keep being stable.

Physics Handbook states

$p=1,67.2623\cdot 10^{-27}kg$

$n=1,67.4929\cdot 10^{-27}kg$

$e=9,10.9390\cdot 10^{-31}kg$

which gives

$n-p-e=1,53m_e$

this is the simplest process to create a neutron out of a proton and an electron, here we at least have neutral charge but the mass differs with 1,53m_e.

We may perhaps write this process as

$n-p-e=1,53m_e{c}^2$

or

$n-p-e=780keV$

according to Einstein.

Now, if energy and mass is interchangeable how can we grab 780keV from a process that have not even started?

Thermal energy is closely related to particle speed, so how can we convert this "speed" into mass?

I think that this is not possible, the Einstein relation above is just theoretical, in other words are we stuck with the fact that the process differs 1,53m_e in mass.

I think that the key here is to be able to manufacture a neutron, which I now think may be made by p+e but the mass difference confuses me while at the same time Physics Handbook have the mass of the elementary particles specified to the sixth decimal.

How easy is it to accurately measure elementary particles with the tiny mass of around E-27kg?

And the original proton-proton fusion theory above, including special particles as positrons and neutrinos, has this been verified?

Another question one may ask is, how?

Now what is binding energy (Eb) really? I think that it is the potential energy of the particle but I am probably wrong, if however I am right we may write

$E_b(p)=V(p)=V(n)[eV]$

and

$R(p)=R(n)\approx 10^{-15}m$

R(e) can be estimated by

$(\frac{1}{1880}{)}^{1/3}\cdot R(p)\approx 10^{-16}m$

while the density of elementary particles are the same, computed for a proton it is approximatelly

$\frac{-27}{3\cdot -15}=10^{18}kg/m^3$

which is huge to say the least!

Getting back to the binding energy of the proton we perhaps have

$V(p)=\frac{-19}{-10-15}=1MeV=10^{-13}J$

that is from the formula

$V=\frac{Q}{4\pi {ϵ}_{0}R}$

if then this energy equals kT to enable thermal electron penetration of the proton potential barrier, then we have

$T=10^{10}K$

which also is huge to say the least, due to the fact that the electron is somewhat 10 times smaller, its potential energy is ten times higher, now we have

$1+1+10+10=1+1+\mathrm{\Delta }{E}_b$

which gives a gain of 20MeV for one pair of proton fusion, a fun analogy is

$20MeV\cdot 10^{-19}J=20\cdot 10^{-13}Ws\approx 20\cdot 10^{-16}Wh\approx 10^{-15}Wh/particle$

Now, the volume density of ordinary air is at 1atm approximately

$n_{air}\approx 10^{25}/m^3$

so if all particles within a cubic meter will fusion, we have +25 fusions or +25-15=+10Wh, this transforms as

$E=10^{10}Wh=10GWh$

But this is at 1atm...

Finally, we may compute the particle velocities for +10K as approximatelly

$m{v}^2=10^{-13}J$

which gives

$v_e=\sqrt{\frac{-13}{-31}}=9$

thus the speed of the electron to penetrate a proton must be in the order of 1E9 m/s, checking the speed requirement for the proton we get

$v_p=\sqrt{\frac{-13}{-27}}=7$

thus the speed of the proton to merge into another proton must be in the order of 1E7 m/s.

The electron speed seems impossible due to higher than the speed of light but there seem to be a twist that is called the Maxwellian distribution function which seems to state that the kT is kind of average when compared to Ek (I have however always thought that Ek was the average value) which means that the temperature can be rather higher than Ek, so to speak.

This is probably the only hope there is to make fusion work on the planet.

One last rambling before signing off, I believe that the neutron creation will have to be done in two steps, the first step is

$p+e+625keV=pe=H_1^F$

where 625keV is the estimated energy for enabling the electron to penetrate the proton potential barrier, I have called the result "Fused Hydrogen", the other step is

$H_1^F+780keV=n$

where 780keV is the equivalent Einstein mass for the lacking of 1,53me to yield the neutron mass.

I think that this must happen in steps because we humans cannot increase the temperature of controlled fusion that fast which means that the first process will happen first, then the process approximatelly needs another 10^10K to finish.

Thus, there might be a pe-particle which I never have heard of, this particle has neutral charge like the neutron but differs in mass which might be fixed by deaccelerating the electron more than is needed for proton penetration.

I am wondering how a neutron becomes a neutron because we humans can't just inject the right amount of mass/energy to hit the exact mass that a neutron has, it has to "know" it somehow so I think that particle physics is not that far from genetics.

I will sign off here and pick up my studies in Field and Wave Electromagnetics by David K. Cheng, when I am finished with that book I will begin to seriously study Plasma Physics and Controlled Fusion by Fransis F. Chen, the reference given below.

Pressure in practice

Normal air pressure (1atm) is

$1atm=10^{5}Pa=10^{5}N/m^2=10^{4}kg/m^2=1kg/c{m}^2$

This only means that we humans have adapted to 1 kg/cm² and nothing else (except that it all implies an actual atmosphere).

Water depth aside we may also create a pressure difference by moving an object in a fluid:

$p_k=1/2\rho v^2$

This equation says that as soon as we have a fluid we will create a pressure on it simply by moving it.

While we do not feel one whole kg/cm² we feel as little increase as 1 meter under water (+1hg/cm²).

And we only have to dive a couple of 10 m below the water surface before we get drunk due to nitrogen "poisoning" which is the reason why scuba divers breath Helium instead of Oxygen at these depths.

The pressure at the deepest part of our sea is about 1000 atm, but this is only felt if we as humans (needing 1 atm) would want to visit that place (which some have done in spite of all). The vessel hull will have to withstand the above pressure equal to an elephant standing on a dime.

The barometric formula

$p=p_0-\rho gh$

reflects the air pressure at different heights (p₀ being 1 atm)

This formula is approximately accurate up to some 10 km (where it actually equals 0).

Anyway, $\rho$ is not linear above some 5 km where

$p=p_0{e}^{-\frac{mgh}{kT}}$

should be used instead (m simply is the molecular weight).

The atmosphere is not uniform. There are four districtive layers or spheres (defined by temperature):

4) Thermosphere (80 km-Karman Line)

3) Mesosphere (50–80 km)

2) Stratosphere (10–50 km)

1) Troposphere (<10 km)

Where the Karman line is 100 km, specified as the height at which a vessel needs to fly as fast as orbital speed to keep height.

Orbital speed means the speed where the centrifugal force equals the gravitational force.

The atmosphere is thus as high as 100 km.

Numerical Example

The atmosphere is then some 100km high and you get an additional pressure of 1atm for each 10's of meter in depth, at 1000m depths it sums up to 100atm of overpressure which means 100kg/cm^2 or 10 tons/dm^2 which may resemble the foot of an elephant.

Plasma pressure

From the Ideal Gas law we have

$p=\frac{N_{mol}}{V}RT=\frac{N}{V}kT=nkT$

where n is the (particle) density.

Work to the gas may be defined as the increasement of the PV-product because then temperature and thus Ek increases.

Work done by the gas may be defined as the decreasement of the PV-product because then temperature decreases.

The work divided by N gives the work done to, or made by, one single molecule, which in turn gives the temperature and thus speed of that single molecule.

The first law of thermodynamics seems to be

$dQ=dU+dW=dU+pdV=N\frac{3}{2}kdT+pdV$

Where Q is the total energy, U the internal energy and W is the work which is positive if work is done by the gas or negative if work is done on the gas.

The internal energy is defined by

$U=NE{k}_p=N\frac{3}{2}kT$

Where N is the number of particles and Ekp is the kinetic energy of each particle, in a closed system dQ must be zero because heat is neither coming in nor coming out, if then dV is positive because the gas is doing work the change of the internal energy (dU) must be negative.

Numerical Example

If we have a pressure of some 1atm in a 1dm^3 large box of air we have some 3E25 particles per cubic meter (which means 3E22 particles in the box), if now the temperature (T) increases from these 300K to 400K and we obviously have dV=0 then dQ is 62kJ which is the heat that has to be injected.

Sources

1. David K. Cheng, Field and Wave Electromagnetics
2. Francis F. Chen, Plasma Physics and Controlled Fusion
3. Jan Petersson, Matematisk Analys, Del 2
4. http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html