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Space Engine
Benjamin Yiwen Faerber
January 23, 2017
K is thermal conductivity, A is surface area dQ/dt is the power transfer, and dT/dx is
the temperature gradient across the pipe.
dQ
dt
= KA
dT
dr
(1)
Reference https://www.physicsforums.com/threads/change-in-temperature-of-water-as-
it-flows-through-a-pipe.450206/
r
r−x
1
r
dr =
Ts
Tw
k2πl
P
dT (2)
ln
r
r − x
=
k2πl
P
(TS − TW ) (3)
1 Special Relativity
In space-time, we use U instead of V . For an object travelling at speed V , a Lorentz
transformation from the rest frame of the object gives
Ua
=
γ
γ #»v
Ua = (γ, −γ #»v )
| ¯U|2
= 1
UaUa
= (γ, −γ #»v )
γ
γ #»v
= γ2
1 − V 2
= 1
Newton: Classically:
F = ma
In Special Relativity, this becomes
fa = maa
1
where
aa
=
dUa
dτ
The 4-acceleration is perpendicular to the 4-velocity:
aaUa
= 0 (4)
¯a ¯U =
d ¯U
dτ
¯U
=
1
2
d ¯U ¯U
dτ
=
1
2
d
dτ
(1) = 0
The Lorentz Force equation. For a particle of charge q
#»
F = q
#»
E + #»v ×
#»
B (5)
Faraday Tensor:
Fab
=




0 Ex Ey Ez
−Ex 0 Bz −By
−Ey −Bz 0 Bx
−Ez By −Bx 0



 (6)
Lorentz Force:
fa
= qUbFba
(7)
f1
= qUbFb1
(8)
f1
= q U0F01
+ U1F11
+ U2F21
+ U3F31
= qγ F01
− VxF11
− VyF21
− VzF31
= qγ (Ex − Vx (0) − Vy (−Bz) − VzBy)
= qγ (Ex + (VyBz − VzBy))
= qγ Ex +
#»
V ×
#»
B
x
f1
, f2
, f3
= qγ
#»
E +
#»
V ×
#»
B (9)
Let inertial frame B move at speed V ˆx with respect to inertial frame A. Suppose in frame
A the magnetic field components vanish. Using the Faraday Tensor, find the magnetic field
components and electric field components in frame B. In general:
qUbFba
= fa
(10)
2
f1
= qγ F01
− vxF11
− vyF21
− vzF31
f2
= qγ F02
− vxF12
− vyF22
− vzF32
f3
= qγ F03
− vxF13
− vyF23
− vzF33
Rest Frame A: v=0 and B=0.
f1
= qγF01
f2
= qγF02
f3
= qγF03
Inertial Frame B moves at speed V ˆx with respect to inertial frame A:
f1
= qγ F01
− vxF11
f2
= qγ F02
− vxF12
f3
= qγ F03
− vxF13
Forces in Frame A and frame B are equal.
FA = FB
qγF01
= qγ F01
− vxF11
qγF02
= qγ F02
− vxF12
qγF03
= qγ F03
− vxF13
vxF12
= 0
vxF13
= 0
vxBz = 0
−vxBy = 0
By = Bz = 0
B = (Bx, 0, 0)
E = (Ex, Ey, Ez)
3
2 Fermi Function
F ( ) =
1
eβ( −µ) + 1
(11)
µ = −kTα (12)
Although these equations are applicable only for electrons in a metal, I only speculate that
the Fermi Dirac distribution might be useful in understanding the temperature relationship
of water particles and quantum mechanical equations.
With the movement of water particles the temperature changes. This is a general
indication that water is a magic fluid! The movement of a water particle through a tube
is comparable to the movement of a train on the rail. For the movement of the train the
laws of special theory of relativity are applicable. Therefore the laws of movement must be
combined with temperature, maybe also with the laws of special theory of relativity. The
theory of special relativity theory can be easily understood, when one sits on a train. One
is then in a moving inertial system which moves relative to the stationary inertial system,
which is the outside environment, and one distinguishes himself through the velocity. The
special relativity theory postulates to draw a linear slope or a straight line in a coordinate
system, which represents the moving inertial system. Alternatively the straight line draws
an angle to the x-axis of the static coordinate system. Actually a very simple theory, which
is not complete, as the speed of light cannot be the maximum velocity. Instead one has
to combine the special theory of relativity with the temperature of a quantum mechanical
system.
1. v (velocity) must be connected to T (Temperature).
2. v is infinite
3. An inertial system which moves on a geodesic, moves above a vacuum, and this vacuum
must be put into relation to the movement of the inertial system.
4
The von-Neumann type interaction
n
cn|φn |Φ0 →
n
|φn |Φn (13)
may be described by the Hamiltonian
H = γ
n
α|φn φn|ˆp (14)
The Hamiltonian should describe an open boundary quantum system. The wave function
of the pointer after time t is
Φn (x, t) = Φ (x − γαnt) (15)
A measurement is ’complete’ when the wave packets Φn for different n are approximately
orthogonal (and therefore equal to the Schmidt states Φn (t).
5

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Spaceengine2

  • 1. Space Engine Benjamin Yiwen Faerber January 23, 2017 K is thermal conductivity, A is surface area dQ/dt is the power transfer, and dT/dx is the temperature gradient across the pipe. dQ dt = KA dT dr (1) Reference https://www.physicsforums.com/threads/change-in-temperature-of-water-as- it-flows-through-a-pipe.450206/ r r−x 1 r dr = Ts Tw k2πl P dT (2) ln r r − x = k2πl P (TS − TW ) (3) 1 Special Relativity In space-time, we use U instead of V . For an object travelling at speed V , a Lorentz transformation from the rest frame of the object gives Ua = γ γ #»v Ua = (γ, −γ #»v ) | ¯U|2 = 1 UaUa = (γ, −γ #»v ) γ γ #»v = γ2 1 − V 2 = 1 Newton: Classically: F = ma In Special Relativity, this becomes fa = maa 1
  • 2. where aa = dUa dτ The 4-acceleration is perpendicular to the 4-velocity: aaUa = 0 (4) ¯a ¯U = d ¯U dτ ¯U = 1 2 d ¯U ¯U dτ = 1 2 d dτ (1) = 0 The Lorentz Force equation. For a particle of charge q #» F = q #» E + #»v × #» B (5) Faraday Tensor: Fab =     0 Ex Ey Ez −Ex 0 Bz −By −Ey −Bz 0 Bx −Ez By −Bx 0     (6) Lorentz Force: fa = qUbFba (7) f1 = qUbFb1 (8) f1 = q U0F01 + U1F11 + U2F21 + U3F31 = qγ F01 − VxF11 − VyF21 − VzF31 = qγ (Ex − Vx (0) − Vy (−Bz) − VzBy) = qγ (Ex + (VyBz − VzBy)) = qγ Ex + #» V × #» B x f1 , f2 , f3 = qγ #» E + #» V × #» B (9) Let inertial frame B move at speed V ˆx with respect to inertial frame A. Suppose in frame A the magnetic field components vanish. Using the Faraday Tensor, find the magnetic field components and electric field components in frame B. In general: qUbFba = fa (10) 2
  • 3. f1 = qγ F01 − vxF11 − vyF21 − vzF31 f2 = qγ F02 − vxF12 − vyF22 − vzF32 f3 = qγ F03 − vxF13 − vyF23 − vzF33 Rest Frame A: v=0 and B=0. f1 = qγF01 f2 = qγF02 f3 = qγF03 Inertial Frame B moves at speed V ˆx with respect to inertial frame A: f1 = qγ F01 − vxF11 f2 = qγ F02 − vxF12 f3 = qγ F03 − vxF13 Forces in Frame A and frame B are equal. FA = FB qγF01 = qγ F01 − vxF11 qγF02 = qγ F02 − vxF12 qγF03 = qγ F03 − vxF13 vxF12 = 0 vxF13 = 0 vxBz = 0 −vxBy = 0 By = Bz = 0 B = (Bx, 0, 0) E = (Ex, Ey, Ez) 3
  • 4. 2 Fermi Function F ( ) = 1 eβ( −µ) + 1 (11) µ = −kTα (12) Although these equations are applicable only for electrons in a metal, I only speculate that the Fermi Dirac distribution might be useful in understanding the temperature relationship of water particles and quantum mechanical equations. With the movement of water particles the temperature changes. This is a general indication that water is a magic fluid! The movement of a water particle through a tube is comparable to the movement of a train on the rail. For the movement of the train the laws of special theory of relativity are applicable. Therefore the laws of movement must be combined with temperature, maybe also with the laws of special theory of relativity. The theory of special relativity theory can be easily understood, when one sits on a train. One is then in a moving inertial system which moves relative to the stationary inertial system, which is the outside environment, and one distinguishes himself through the velocity. The special relativity theory postulates to draw a linear slope or a straight line in a coordinate system, which represents the moving inertial system. Alternatively the straight line draws an angle to the x-axis of the static coordinate system. Actually a very simple theory, which is not complete, as the speed of light cannot be the maximum velocity. Instead one has to combine the special theory of relativity with the temperature of a quantum mechanical system. 1. v (velocity) must be connected to T (Temperature). 2. v is infinite 3. An inertial system which moves on a geodesic, moves above a vacuum, and this vacuum must be put into relation to the movement of the inertial system. 4
  • 5. The von-Neumann type interaction n cn|φn |Φ0 → n |φn |Φn (13) may be described by the Hamiltonian H = γ n α|φn φn|ˆp (14) The Hamiltonian should describe an open boundary quantum system. The wave function of the pointer after time t is Φn (x, t) = Φ (x − γαnt) (15) A measurement is ’complete’ when the wave packets Φn for different n are approximately orthogonal (and therefore equal to the Schmidt states Φn (t). 5