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The purpose of this calculation is to express
Newtonian force, acceleration, and mass in
terms of Quantum Mechanics.
c speedof light=
mv p=
h
λ
=
c h
f
=
p momentum=
E Energy=
ΔpΔx
hbar
2

t time=
c
1
μ0 ε0
=
ΔEΔt
hbar
2
 x length=
m mass=
ΔE
ΔpΔx
Δt
= Δ mv( )
Δx
Δt
= c
Δx
Δt
= v=
h planksconstant=
ΔE Δmcc= hbar planksconstant normalized( )=
v velocity=
E
m
1
μ0 ε0
= c=
a acceleration=
F force= ma=
E
m
1
μ0 ε0 
= c
2
=
μ permeability=
ε permittivity=
ΔE
Δm
μ0 ε0 
=
hbar
2Δt
= because ΔE Δmcc=
hbar
2Δt
=
Δt
hbar μ0 ε0 
2Δm
= (solve for ∆t)
μ0 ε0
2Δt Δm
hbar
Δv
Δv






=
μ0 ε0
2Δm
hbar a


=
2Δm
hbar
F

m







=
2Δm
2
hbar F


= NOTE: ∆t/∆v = 1/a
a = F/m
F
 2Δm
2
hbar μ0 ε0 
= therefore and a
 2Δm
μ0 ε0 hbar 
=
m
μ0 ε0  hbar F


2
=
As a check, lets see if F=ma...
m a
μ0 ε0  hbar F


2








2Δm
hbar μ0 ε0 






=
μ0 ε0  hbar F


2
2Δm
μ0 ε0  hbar

2Δm
μ0 ε0  hbar
=
m a
2 F

Δm
hbar μ0 ε0 
=
F

 
2
2F

Δm
2
hbar μ0 ε0 
= square both sides
F
 2Δm
2
hbar μ0 ε0 
= divide both side by "F"
So what exactly is this all good for??? Suppose you have a plasma like that found in a fusion reactor...
Also suppose that the relative permeability and permittivity are functions of this plasma's state at any
given delta t....
t .01 0.1 25
some_function_of_plasma_1 t( )
sin 1.2t( ) t
3
t
2
 t 2
2 10( )
4

some_function_of_plasma_2 t( )
sin t( ) t
3
t
2
 t 2
2 10( )
4

εr some_function_of_plasma_1
μr some_function_of_plasma_2
0 10 20
0.1
0.05
0
0.05
0.1
εr t( )
μr t( )
t
then ∆t = t2 - t1 on the graph
Δt
hbar μr μ0 εr ε0 
2Δm
=

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Quantum mass calculation

  • 1. The purpose of this calculation is to express Newtonian force, acceleration, and mass in terms of Quantum Mechanics. c speedof light= mv p= h λ = c h f = p momentum= E Energy= ΔpΔx hbar 2  t time= c 1 μ0 ε0 = ΔEΔt hbar 2  x length= m mass= ΔE ΔpΔx Δt = Δ mv( ) Δx Δt = c Δx Δt = v= h planksconstant= ΔE Δmcc= hbar planksconstant normalized( )= v velocity= E m 1 μ0 ε0 = c= a acceleration= F force= ma= E m 1 μ0 ε0  = c 2 = μ permeability= ε permittivity= ΔE Δm μ0 ε0  = hbar 2Δt = because ΔE Δmcc= hbar 2Δt = Δt hbar μ0 ε0  2Δm = (solve for ∆t) μ0 ε0 2Δt Δm hbar Δv Δv       = μ0 ε0 2Δm hbar a   = 2Δm hbar F  m        = 2Δm 2 hbar F   = NOTE: ∆t/∆v = 1/a a = F/m F  2Δm 2 hbar μ0 ε0  = therefore and a  2Δm μ0 ε0 hbar  = m μ0 ε0  hbar F   2 =
  • 2. As a check, lets see if F=ma... m a μ0 ε0  hbar F   2         2Δm hbar μ0 ε0        = μ0 ε0  hbar F   2 2Δm μ0 ε0  hbar  2Δm μ0 ε0  hbar = m a 2 F  Δm hbar μ0 ε0  = F    2 2F  Δm 2 hbar μ0 ε0  = square both sides F  2Δm 2 hbar μ0 ε0  = divide both side by "F" So what exactly is this all good for??? Suppose you have a plasma like that found in a fusion reactor... Also suppose that the relative permeability and permittivity are functions of this plasma's state at any given delta t.... t .01 0.1 25 some_function_of_plasma_1 t( ) sin 1.2t( ) t 3 t 2  t 2 2 10( ) 4  some_function_of_plasma_2 t( ) sin t( ) t 3 t 2  t 2 2 10( ) 4  εr some_function_of_plasma_1 μr some_function_of_plasma_2 0 10 20 0.1 0.05 0 0.05 0.1 εr t( ) μr t( ) t then ∆t = t2 - t1 on the graph Δt hbar μr μ0 εr ε0  2Δm =