[<< wikibooks] Plasma Fusion Preface
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

n

i

n

n

=
2.4
∗

10

21

T

3

/

2

n

i

exp
−
(

U

i

k
T

)

{\displaystyle {\frac {n_{i}}{n_{n}}}=2.4*10^{21}{\frac {T^{3/2}}{n_{i}}}\exp -({\frac {U_{i}}{kT}})}
,
where ni is the ion density and nn is the neutral atoms density and Ui is the ionization energy of the gas.
Putting for ordinary air

n

n

=
3
∗

10

25

m

−
3

{\displaystyle n_{n}=3*10^{25}m^{-3}}

T
=
300
K

{\displaystyle T=300K}

U

i

=
14
,
5
e
V
(
n
i
t
r
o
g
e
n
)

{\displaystyle U_{i}=14,5eV(nitrogen)}

gives

n

i

n

n

=

10

−
122

{\displaystyle {\frac {n_{i}}{n_{n}}}=10^{-122}}

which is ridiculously lowThe ionization remains low until Ui 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.

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

m

d
v

d
t

=
q
v
X
B

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

A simple way of solving this equation is to put

v
=

v

0

e

j
w
t

{\displaystyle v=v_{0}e^{jwt}}

The equation then becomes

m
j
w

v

0

e

j
w
t

=
q
v
X
B
=
m
j
w
v

{\displaystyle mjwv_{0}e^{jwt}=qvXB=mjwv}

While only considering the magnitude we get

w

c

=

|

q

|

B

m

{\displaystyle w_{c}={\frac {|q|B}{m}}}

and while v=wr we get

r

L

=

m
v

|

q

|

B

{\displaystyle r_{L}={\frac {mv}{|q|B}}}

where wc is called the cyclotron frequency and rL 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.

== Energy and temperature of a plasma ==
It will later on be shown that the average energy may be written

E

A
V

=

1
4

m

v

2

=

1
2

k
T

{\displaystyle E_{AV}={\frac {1}{4}}mv^{2}={\frac {1}{2}}kT}

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

v
=

2
k
T

m

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

The above energy equation can be derived while using the Maxwellian velocity distribution function

f
(
v
)
=
A
exp
⁡

(
−

m

v

2

/

2

k
T

)

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

where the volume particle density can be calculated using

n
=

∫

−
∞

∞

f
(
v
)
d
v

{\displaystyle n=\int _{-\infty }^{\infty }f(v)dv}

which gives us

A
=
n

m

2
π
k
T

{\displaystyle A=n{\sqrt {\frac {m}{2\pi kT}}}}

What this means is that while the most probable speed is when

m

v

2

2

=
k
T

{\displaystyle {\frac {mv^{2}}{2}}=kT}

there are particles with both lower and higher speeds that has the same temperature.

== Some ITER calculations ==
According to Francis F. Chen, physicists use

k
T
=
e
V

{\displaystyle kT=eV}

to avoid confusion.
Let's state some constants:

k
=
1
,
38
∗

10

−
23

[
J

/

K
]

{\displaystyle k=1,38*10^{-23}[J/K]}

e
=
1
,
6
∗

10

−
19

[
A
s
]

{\displaystyle e=1,6*10^{-19}[As]}

m

p

=
1
,
67
∗

10

−
27

[
k
g
]

{\displaystyle m_{p}=1,67*10^{-27}[kg]}

μ

0

=
4
π
∗

10

−
7

[
V
s

/

A
m
]

{\displaystyle \mu _{0}=4\pi *10^{-7}[Vs/Am]}

== Deriving the magnetic flux density of a current loop ==

From Maxwell's equations we have

∇
⋅
B
=
0

{\displaystyle \nabla \cdot B=0}

which may be rewritten as

B
=
∇
X
A

{\displaystyle B=\nabla XA}

where A might be an arbitrary vector.
Using the vector magnetic potential

A
=

μ

0

4
π

∫

v

J
R

d
v

{\displaystyle A={\frac {\mu _{0}}{4\pi }}\int _{v}{{\frac {J}{R}}dv}}

and realising that

J
d
v
=
J
S
d
l
=
I
d
l

{\displaystyle Jdv=JSdl=Idl}

we have from Biot-Savat law

B
=

μ

0

I

4
π

∮

c

d
l
X

a

R

R

2

{\displaystyle B={\frac {\mu _{0}I}{4\pi }}\oint _{c}{\frac {dlXa_{R}}{R^{2}}}}

Defining

d
l
=
b
d
ϕ

a

ϕ

{\displaystyle dl=bd\phi a_{\phi }}

and

R
=

a

z

z
−

a

r

b

{\displaystyle R=a_{z}z-a_{r}b}

and

d
l
X
R
=

a

ϕ

b
d
ϕ
X
(

a

z

z
−

a

r

b
)
=

a

r

b
z
d
ϕ
+

a

z

b

2

d
ϕ

{\displaystyle dlXR=a_{\phi }bd\phi X(a_{z}z-a_{r}b)=a_{r}bzd\phi +a_{z}b^{2}d\phi }

and realising that the r-part cancel out we get

B
=

μ

0

I

4
π

∫

0

2
π

a

z

b

2

d
ϕ

(

z

2

+

b

2

)

3

/

2

{\displaystyle B={\frac {\mu _{0}I}{4\pi }}\int _{0}^{2\pi }a_{z}{\frac {b^{2}d\phi }{(z^{2}+b^{2})^{3/2}}}}

or

B
=

μ

0

I

2

b

2

(

z

2

+

b

2

)

3

/

2

=

μ

0

I

2

b

2

R

3

{\displaystyle 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}}}}

== Drifts in a plasma ==
Using

m

d
v

d
t

=
q
(
E
+
v
X
B
)

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

and putting the left side to zero while taking the cross product with B we obtain

E
X
B
=
B
X
(
v
X
B
)
=
v

B

2

−
B
(
v
⋅
B
)

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

The transverse components of this equation are

v

g
c

=

E
X
B

B

2

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

and the magnitude of this guiding center drift is

v

g
c

=

E
B

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

Realising that

F
=
q
E

{\displaystyle F=qE}

one could set

v

f
o
r
c
e

=

1
q

F
X
B

B

2

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

where F might be

F

E

=
q
E

{\displaystyle F_{E}=qE}

due to an E-field or

F

g

=
m
g

{\displaystyle F_{g}=mg}

due to gravity or

F

c
f

=

a

r

m

v

/

/

2

R

c

{\displaystyle F_{cf}=a_{r}{\frac {mv_{//}^{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

=

E
X
B

B

2

{\displaystyle v_{E}={\frac {EXB}{B^{2}}}}

and the drift due to gravity will be

v

g

=

m
q

g
X
B

B

2

{\displaystyle v_{g}={\frac {m}{q}}{\frac {gXB}{B^{2}}}}

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

v

R

=

1
q

F

c
f

X
B

B

2

=

m

v

/

/

2

q

B

2

R

c

X
B

R

c

2

{\displaystyle v_{R}={\frac {1}{q}}{\frac {F_{cf}XB}{B^{2}}}={\frac {mv_{//}^{2}}{qB^{2}}}{\frac {R_{c}XB}{R_{c}^{2}}}}

It is interesting to note that

|

v

E

|

=

|

E
B

|

{\displaystyle |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

y

=
−

/

+

q

v

p

r

L

2

d
B

d
y

a

y

{\displaystyle F_{y}=-/+{\frac {qv_{p}r_{L}}{2}}{\frac {dB}{dy}}a_{y}}

where vp denotes speed perpendicular to B.
Which put into the force-formula above gives the guiding center drift

v

g
c

=

1
q

F
X
B

B

2

=

1
q

F

y

|

B

|

a

x

=
−

/

+

v

p

r

L

2
B

d
B

d
y

a

x

{\displaystyle v_{gc}={\frac {1}{q}}{\frac {FXB}{B^{2}}}={\frac {1}{q}}{\frac {F_{y}}{|B|}}a_{x}=-/+{\frac {v_{p}r_{L}}{2B}}{\frac {dB}{dy}}a_{x}}

which can be generalized to

v

∇
B

=
−

/

+

v

p

r

L

2

∇
B
X
B

B

2

{\displaystyle v_{\nabla B}=-/+{\frac {v_{p}r_{L}}{2}}{\frac {\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

c
v

=

v

R

+

v

∇
B

=

m

q

B

2

R
c
X
B

R

c

2

(

v

/

/

2

+

1
2

v

p

2

)

{\displaystyle v_{cv}=v_{R}+v_{\nabla B}={\frac {m}{qB^{2}}}{\frac {RcXB}{Rc^{2}}}(v_{//}^{2}+{\frac {1}{2}}v_{p}^{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

== The plasma as a fluid ==
If we consider a plasma as a fluid we have

m
n
[

d
v

d
t

+
(
v
⋅
∇
)
v
]
=
q
n
(
E
+
v
X
B
)
−
∇
p

{\displaystyle mn[{\frac {dv}{dt}}+(v\cdot \nabla )v]=qn(E+vXB)-\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
=
q
n
[
E
X
B
+
(

v

p

X
B
)
X
B
]
−
∇
p
X
B

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

or

0
=
q
n
[
E
X
B
−

v

p

B

2

]
−
∇
p
X
B

{\displaystyle 0=qn[EXB-v_{p}B^{2}]-\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

=

E
X
B

B

2

−

∇
p
X
B

q
n

B

2

=

v

E

+

v

D

{\displaystyle v_{p}={\frac {EXB}{B^{2}}}-{\frac {\nabla pXB}{qnB^{2}}}=v_{E}+v_{D}}

where the so-called diamagnetic drift is

v

D

=
−

∇
p
X
B

q
n

B

2

{\displaystyle v_{D}=-{\frac {\nabla pXB}{qnB^{2}}}}

where the force is

F

D

=
−

∇
p

n

{\displaystyle F_{D}=-{\frac {\nabla p}{n}}}

meaning the gradient of the pressure

p
=
n
k
T

{\displaystyle p=nkT}

to volume particle density.
For an isoterm plasma we have

∇
p
=
k
T
∇
n

{\displaystyle \nabla p=kT\nabla n}

== Inductance Calculation ==
The definition of inductance, L, is:

N
∫
B
d
S
=
L
I

{\displaystyle N\int BdS=LI}

Using

B
=

μ

0

N
I

l

m

{\displaystyle B=\mu _{0}{\frac {NI}{l_{m}}}}

for a long solenoid
or

B
=

μ

0

N
I

2
R

=

μ

0

N
I

D

{\displaystyle B=\mu _{0}{\frac {NI}{2R}}=\mu _{0}{\frac {NI}{D}}}

for a short solenoid it is clear that the adequate formula depends upon diameter versus length. But in many cases the reality is somewhere in between.
Anyway, the inductance of our short solenoid is:

L
=

N
B
A

I

=

μ

0

N

2

A

2
R

{\displaystyle L={\frac {NBA}{I}}=\mu _{0}{\frac {N^{2}A}{2R}}}

Estimating A to be circular then

L
=

μ

0

N

2

π
R

2

=

μ

0

N

2

π
D

4

{\displaystyle L=\mu _{0}{\frac {N^{2}\pi R}{2}}=\mu _{0}{\frac {N^{2}\pi D}{4}}}

And with N=10 and R=2m this yields

L
=
0
,
4
m
H

{\displaystyle L=0,4mH}

Visualising some ten coils around the tokamak which may be connected in series yields some

L
=
5
m
H

{\displaystyle L=5mH}

and to make things complete

e
=
−
L

d
i

d
t

{\displaystyle e=-L{\frac {di}{dt}}}

This inductance does however only affect power-on. With a smooth onset of voltage (read Amps) the inductance does not matter so much as the resistive losses.

== Drift considerations in a plasma ==
Getting back to our general B-formula for a short solenoid which is repeated here for convenience

B
=

μ

0

N
I

R

2

2

r

3

{\displaystyle B=\mu _{0}{\frac {NIR^{2}}{2r^{3}}}}

we can see that the magnetic flux density diminishes as

1

/

r

3

{\displaystyle 1/r^{3}}

along the

z
−
a
x
i
s

{\displaystyle z-axis}

which in our case is "almost" equal to the

ϕ
−
a
x
i
s

{\displaystyle \phi -axis}

This however creates a gradient in B but this gradient is mostly along the B-field.
So even though B lessens with distance to the next coil the grad-B drift might be negligible due to the curl of grad-B with B.

== The tokamak current ==

It is preliminary considered that SW1 and SW2 are closed at different times and in such a way that they never are closed at the same time. The voltage source E is preliminary considered stable as a battery.
For the tokamak current, It, we may write

q

/

C
=

R

c
u

i
+
L

d
i

d
t

{\displaystyle q/C=R_{cu}i+L{\frac {di}{dt}}}

[where the charge q stored by C has been converted to an equivalent voltage because of C=Coulomb/Volt=As/Volt]
Putting

i
=
−

d
q

d
t

{\displaystyle i=-{\frac {dq}{dt}}}

(The sign of this one is a bit hard to understand but maybe one can view it like the current coming out of the capacitor is leaving the capacitor, therefore the minus sign.)
and deriving once more gives

L

i
″

+

R

c
u

i
′

+
i

/

C
=
0

{\displaystyle Li''+R_{cu}i'+i/C=0}

or

i
″

+

R

c
u

L

i
′

+

1

L
C

i
=
0

{\displaystyle i''+{\frac {R_{cu}}{L}}i'+{\frac {1}{LC}}i=0}

Putting

b
=

R

c
u

/

2
L

{\displaystyle b=R_{cu}/2L}

and

w
=

1

L
C

{\displaystyle w={\frac {1}{\sqrt {LC}}}}

gives the characteristic equation

(

r

2

+
2
b
r
+

w

2

)
i
=
0

{\displaystyle (r^{2}+2br+w^{2})i=0}

where

r

1
,
2

=
−
b
+

/

−

b

2

−

w

2

{\displaystyle r_{1,2}=-b+/-{\sqrt {b^{2}-w^{2}}}}

And if

b
>
w

{\displaystyle b>w}

the solution may be written

i
(
t
)
=

C

1

e

r

1

t

+

C

2

e

r

2

t

{\displaystyle i(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}}

Using the initial values

i
(
0
)
=
0

{\displaystyle i(0)=0}

and

i
′

(
0
)
=
E

/

L

{\displaystyle i'(0)=E/L}

[This initial value is however a bit hard to understand. But it must come from

E
=
−
L

d
i

d
t

{\displaystyle E=-L{\frac {di}{dt}}}

where E is the capacitor voltage at t=0 and not the induction of the current derivate. In short, E/L forces the current derivate at t=0 in this case and the valid sign comes from the schematic above.]
then

i
(
0
)
=

C

1

+

C

2

==
0

{\displaystyle i(0)=C_{1}+C_{2}==0}

yielding

C

2

=
−

C

1

{\displaystyle C_{2}=-C_{1}}

so now we have

i
(
t
)
=

C

1

(

e

r

1

t

−

e

r

2

t

)

{\displaystyle i(t)=C_{1}(e^{r_{1}t}-e^{r_{2}t})}

Deriving this while putting t=0 yields

i
′

(
0
)
=

C

1

(

r

1

−

r

2

)
==
E

/

L

{\displaystyle i'(0)=C_{1}(r_{1}-r_{2})==E/L}

thus

C

1

=

E

/

L

r

1

−

r

2

{\displaystyle C_{1}={\frac {E/L}{r_{1}-r_{2}}}}

and finally

i
(
t
)
=

E

/

L

r

1

−

r

2

(

e

r

1

t

−

e

r

2

t

)

{\displaystyle i(t)={\frac {E/L}{r_{1}-r_{2}}}(e^{r_{1}t}-e^{r_{2}t})}

If we derive this and put it equal to zero in search of maximum, one gets:

r

1

e

r

1

t

=

r

2

e

r

2

t

{\displaystyle r_{1}e^{r_{1}t}=r_{2}e^{r_{2}t}}

or

l
n
(

r

1

r

2

)
=
t
(

r

2

−

r

1

)

{\displaystyle ln({\frac {r_{1}}{r_{2}}})=t(r_{2}-r_{1})}

or

t

m
a
x

=

l
n
(

r

1

/

r

2

)

r

2

−

r

1

{\displaystyle t_{max}={\frac {ln(r_{1}/r_{2})}{r_{2}-r_{1}}}}

The strange thing here is that while r1 needs to be greater than r2 for making the current above positive, the result actually indicates that only if r1 is less than 2,71 times r2, tmax is positive.
Here we could put tmax into i(t) to calculate maximum current. We won't however do that because that is just plain algebra. It is however interesting to view i(t) in another way referring to the definition of r1 and r2 above

i
(
t
)
=

E

/

L

b

2

−

w

2

e

−
b
t

e

b

2

−

w

2

t

−

e

−

b

2

−

w

2

t

2

{\displaystyle i(t)={\frac {E/L}{\sqrt {b^{2}-w^{2}}}}e^{-bt}{\frac {e^{{\sqrt {b^{2}-w^{2}}}t}-e^{-{\sqrt {b^{2}-w^{2}}}t}}{2}}}

To make things complete regarding solutions for second order differential equations we have two more conditions to regard. If

b
=
w

{\displaystyle b=w}

then

i
(
t
)
=
(

C

1

t
+

C

2

)

e

−
b
t

{\displaystyle i(t)=(C_{1}t+C_{2})e^{-bt}}

If

b
<
w

{\displaystyle b
0

{\displaystyle \gamma ={\sqrt {w^{2}-b^{2}}}>0}

and this makes

r

1
,
2

=
−
b
+

/

−
j
γ

{\displaystyle r_{1,2}=-b+/-j\gamma }

which gives the solution

i
(
t
)
=
C

e

−
b
t

s
i
n
(
γ
t
+
α
)

{\displaystyle i(t)=Ce^{-bt}sin(\gamma t+\alpha )}

Here we can see that the current is attenuated sinusoidally by the frequency

γ

{\displaystyle \gamma }
and the "amplitude"

C

e

−
b
t

{\displaystyle Ce^{-bt}}

To summarize, all the above solutions are based on the critical condition that

b
=

R

c
u

/

2
L

1

L
C

=
w

{\displaystyle b=R_{cu}/2L{\frac {1}{\sqrt {LC}}}=w}

where b should be equal or greater than w to yield a stable response.
Finally, let's do calculate Imax just for fun :)

i

m
a
x

=

E

/

L

r

1

−

r

2

(

e

r

1

r

2

−

r

1

l
n
(
r
1

/

r
2
)

−

e

r

2

r

2

−

r

1

l
n
(
r
1

/

r
2
)

)

{\displaystyle i_{max}={\frac {E/L}{r_{1}-r_{2}}}(e^{{\frac {r_{1}}{r_{2}-r_{1}}}ln(r1/r2)}-e^{{\frac {r_{2}}{r_{2}-r_{1}}}ln(r1/r2)})}

Using

e

4
l
n
5

=

5

4

{\displaystyle e^{4ln5}=5^{4}}

we get

i

m
a
x

=

E

/

L

r

1

−

r

2

(
(

r

1

r

2

)

r

1

r

2

−

r

1

−
(

r

1

r

2

)

r

2

r

2

−

r

1

)

{\displaystyle i_{max}={\frac {E/L}{r_{1}-r_{2}}}(({\frac {r_{1}}{r_{2}}})^{\frac {r_{1}}{r_{2}-r_{1}}}-({\frac {r_{1}}{r_{2}}})^{\frac {r_{2}}{r_{2}-r_{1}}})}

== Supply current ==
This basic part doesn't really need a mathematical derivation because one could easily write

u

c

(
t
)
=
E
(
1
−

e

−
t

/

R
C

)

{\displaystyle u_{c}(t)=E(1-e^{-t/RC})}

It is known that

i
=
C

d
u

d
t

{\displaystyle i=C{\frac {du}{dt}}}
.
The differential equation of first order may be written

E
=
R
i
+

u

c

=
R
C

d
u

d
t

+
u

{\displaystyle E=Ri+u_{c}=RC{\frac {du}{dt}}+u}

Now we know the solution but we could pretend that we doesn't and guess

u

c

(
t
)
=
A

e

−
k
t

+
B

{\displaystyle u_{c}(t)=Ae^{-kt}+B}

Then

u
′

=
−
k
A

e

−
k
t

{\displaystyle u'=-kAe^{-kt}}

Boundary values say that

u
(
0
)
=
0

{\displaystyle u(0)=0}

and

u
′

(
0
)
=
E

/

R
C

{\displaystyle u'(0)=E/RC}

this one is however somewhat tricky but comes from

i
=
C

d
u

d
t

{\displaystyle i=C{\frac {du}{dt}}}

where

i
(
0
)

C

=

E

/

R

C

{\displaystyle {\frac {i(0)}{C}}={\frac {E/R}{C}}}

While there are three unknowns, we need a third condition which is

u
(
∞
)
=
E

{\displaystyle u(\infty )=E}

Using these boundary conditions, one first gets

u
(
0
)
=
A
+
B
==
0
=>
B
=
−
A

{\displaystyle u(0)=A+B==0=>B=-A}

and

u
(
∞
)
=
B
=
E

{\displaystyle u(\infty )=B=E}

Then we have

u
(
t
)
=
E
(
1
−

e

−
k
t

)

{\displaystyle u(t)=E(1-e^{-kt})}

Differentiating this yields

u
′

=
E
k

e

−
k
t

{\displaystyle u'=Eke^{-kt}}

and

u
′

(
0
)
=
E
k
==
E

/

R
C

{\displaystyle u'(0)=Ek==E/RC}

which gives

k
=
1

/

R
C

{\displaystyle k=1/RC}

thus

u

c

(
t
)
=
E
(
1
−

e

−
t

/

R
C

)

{\displaystyle u_{c}(t)=E(1-e^{-t/RC})}

Finally, deriving this and using the capacitor formula for the current gives

i
(
t
)
=
C

d
u

d
t

=

E
R

e

−
t

/

R
C

{\displaystyle i(t)=C{\frac {du}{dt}}={\frac {E}{R}}e^{-t/RC}}

== Standard model ==
electron and positron ("anti-electron")
muon and anti-muon
tau and anti-tauAlong with these comes their neutrino and anti-neutrino which gives six distinct types of particles or:

electron
electron-neutrino
muon
muon-neutrino
tau
tau-neutrinoThe 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.

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)

e
=
∮
E
d
l
=
−

d
ϕ

d
t

=
−
j
w
ϕ
[
V
]

{\displaystyle e=\oint Edl=-{\frac {d\phi }{dt}}=-jw\phi [V]}

states the emf-induction due to magnetic flux change.
And the relationship

N
ϕ
=
L
I

{\displaystyle N\phi =LI}

leads back to my first formulas.
Viewing these equations, one has both B and E 90 degrees out of phase.
Considering the differential version of Faraday's law we have

∇
X
E
=
−

d
B

d
t

{\displaystyle \nabla XE=-{\frac {dB}{dt}}}

which also states the direction of it all.
But we all know that for induction to happen the moving conductor has to cut the lines of force.
So this is by definition a TEM-wave.
The speed of the (circulating) charge need however to be non-constant (otherwise no induction can be made) which means that we have to accelerate the charge by for instance heat.
Considering

m
v

r

B

=
ℏ

{\displaystyle mvr_{B}=\hbar }

from below we note that v is constant within the Bohr radius.
So the only way of increasing the speed of the electron is to move it up to another shell.
In other words, linear thermal radiation cannot be achieved by heating.
Planck's law of radiation must be due to another phenomenon, which probably is vibration of the nuclei and/or electron.
Or perhaps the Bohr model is just too simple?
Apart from the Bohr restriction, the conclusion must be that every accelerated charge like above give rise to TEM.
A free accelerated or decelerated charge might suit even better for this reasoning. In this case it is more obvious that there are no spectral lines when it comes to thermal radiation and this is because of the "linear" speed states while adding kT.

== Bohr model derivation ==
It has been proven that

n
λ
=
2
π
r

{\displaystyle 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

λ
=
h

/

p

{\displaystyle \lambda =h/p}

and

ℏ
=
h

/

2
π

{\displaystyle \hbar =h/2\pi }

the above equation may be rewritten as

p
r
=
m
v
r
=
n
ℏ

{\displaystyle pr=mvr=n\hbar }

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

m

v

2

r

=

k

e

2

r

2

{\displaystyle {\frac {mv^{2}}{r}}={\frac {ke^{2}}{r^{2}}}}

where

k
=

1

4
π

ϵ

0

{\displaystyle k={\frac {1}{4\pi \epsilon _{0}}}}

Solving for v yields

v
=

k

e

2

m
r

{\displaystyle v={\sqrt {\frac {ke^{2}}{mr}}}}

Integrating the electromagnetic force gives the potential energy as

E

p

=
−

k

e

2

r

{\displaystyle E_{p}=-{\frac {ke^{2}}{r}}}

The kinetic energy may as usual be written

E

k

=

m

v

2

2

{\displaystyle E_{k}={\frac {mv^{2}}{2}}}

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

E

t
o
t

=

E

p

/

2
=
−

k

e

2

2
r

{\displaystyle E_{tot}=E_{p}/2=-{\frac {ke^{2}}{2r}}}

Now,

m
r
v
=
m
r

k

e

2

m
r

=

k

e

2

m
r

==
n
ℏ

{\displaystyle mrv=mr{\sqrt {\frac {ke^{2}}{mr}}}={\sqrt {ke^{2}mr}}==n\hbar }

Solving for r yields

r
=

n

2

ℏ

2

k

e

2

m

{\displaystyle r={\frac {n^{2}\hbar ^{2}}{ke^{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
=

k

e

2

n
ℏ

{\displaystyle v={\frac {ke^{2}}{n\hbar }}}

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).

== Proton-proton fusion ==
These statements are cited from1) Protons fuse
2) One proton is transmuted into one neutron forming Deuterium (releasing one positron and an electron-neutrino).
3) Deuterium fuses with another proton (which also releases gamma-rays)
4) Two of the resulting Helium_3 nuclei fuse
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."

== Pressure in practice ==
Normal air pressure (1atm) is

1
a
t
m
=

10

5

P
a
=

10

5

N

/

m

2

=

10

4

k
g

/

m

2

=
1
k
g

/

c

m

2

{\displaystyle 1atm=10^{5}Pa=10^{5}N/m^{2}=10^{4}kg/m^{2}=1kg/cm^{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
ρ

v

2

{\displaystyle 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

−
ρ
g
h

{\displaystyle 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,

ρ

{\displaystyle \rho }
is not linear above some 5 km where

p
=

p

0

e

−

m
g
h

k
T

{\displaystyle 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.

== Plasma pressure ==
From the Ideal Gas law we have

p
=

N

m
o
l

V

R
T
=

N
V

k
T
=
n
k
T

{\displaystyle 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

Q
=
Δ
U
+
W

{\displaystyle Q=\Delta U+W}

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
=
K
E
+
P
E

{\displaystyle U=KE+PE}

Where KE is the kinetic energy and PE is the potential energy.

Bohr model
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html