Science

I am not a quantum physicist

I am not a quantum physicist. I do not write this prediction thinking that it is true or a novel deduction on the nature of reality. I write this prediction in order to test my own understanding of quantum physics.

Given all particles are fields:

  1. Fermions are those fields where probability is in the range 0-1 (or possibly -1 to +1, depending on antimatter).
  2. Bosons are those fields where probability can take on any positive or zero value (possibly also any negative value, depending on antimatter).

This “explains” why two fermions cannot occupy the same quantum state, yet bosons can. Inverted quote marks, because this might turn out to not have any explanatory power.

I’m fine with that, just as I’m fine with being wrong. I am not a quantum physicist. I don’t expect to be right. It would be nicer to find I’m wrong rather than not even wrong, but even that’s OK — that’s why I’m writing this down before I see if someone else has already written about this.

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Science, Technology

Railgun notes #2

[Following previous railgun notes, which has been updated with corrections]

Force:
F = B·I·l
B = 1 tesla

I: Current = Voltage / Resistance
l: Length of armature in meters

F = 1 tesla · V/R · l
F = m · a
∴ a = (1 tesla · V/R · l) / m

Using liquid mercury, let cavity be 1cm square, consider section 1cm long:
∴ l = 0.01 m
Resistivity: 961 nΩ·m
∴ Resistance R = ((961 nΩ·m)*0.01m)/(0.01m^2) = 9.6×10^-7 Ω
Volume: 1 millilitre
∴ Mass m = ~13.56 gram = 1.356e-2 kg
∴ a = (1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg)

Let target velocity = Escape velocity = 11200 m/s = 1.12e4 m/s:
Railgun length s = 1/2 · a · t^2
And v = a · t
∴ t = v / a
∴ s = 1/2 · a · (v / a)^2
∴ s = 1/2 · a · v^2 / a^2
∴ s = 1/2 · v^2 / a
∴ s = 1/2 · ((1.12e4 m/s)^2) / ((1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg))

@250V: s = 0.3266 m (matches previous result)

@1V: s = 81.65 m
I = V/R = 1V / 9.6×10^-7 Ω = 1.042e6 A
P = I · V = 1V · 1.042e6 A = 1.042e6 W

Duration between rails:
t = v / a
∴ t = (1.12e4 m/s) / a
∴ t = (1.12e4 m/s) / ( (1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg) )

(Different formula than before, but produces same values)
@1V: t = 0.01458 seconds

Electrical energy usage: E = P · t
@1V: E = 1.042e6 W · 0.01458 seconds = 1.519e4 joules

Kinetic energy: E = 1/2 · m · v^2 = 8.505e5 joules

Kinetic energy out shouldn’t exceed electrical energy used, so something has gone wrong.

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Science, Technology

Railgun notes

Force on the projectile of a railgun:
F = B·I·l
B: Magnetic field
I: Current
l: Length of armature

Current = Voltage / Resistance

Resistivity of seawater:
ρ = 2.00×10^−1 (Ω·m) (because = (Ω/m-length)*(cross-sectional area))

Let cavity be 1cm square, consider section 1cm long:

Volume: 1 millilitre
mass (m): ~1 gram = 1e-3 kg
Cross-section: 1e-4 m^2
Armature length (l): 1e-2 m
Resistance: ((2.00×10^−1 Ω·m)*0.01m)/(0.01m^2) = 0.2 Ω (got that wrong first time! Along with all that followed, which is now updated…)
∴ current (I) = Voltage (V) / 0.2 Ω

Rare earth magnets can be 1 tesla without much difficulty. Assume that here.

F = 1 T · (V/0.2 Ω) · (1e-2 m)

Target velocity: 11.2 km/s = Escape velocity = 11200 m/s
v = at = 11200 m/s
∴ a = (11200 m/s) / t
s = 1/2 · a · t^2
∴ s = 1/2 · ( (11200 m/s) / t ) · t^2
= 1/2 · (11200 m/s) · t
or: t = s / (1/2 · (11200 m/s))
F = ma = (1e-3 kg) · a
∴ a = F / (1e-3 kg)
∴ t = (11200 m/s) / (F / (1e-3 kg))
= (11200 m/s) · (1e-3 kg) / F
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / F
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (V/0.2 Ω) · (1e-2 m) )

Say V = 250 volts:
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (250V/0.2 Ω) · (1e-2 m) ) = 5020m (not ~501760 meters)

Say V = 25,000 volts:
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (25000V/0.2 Ω) · (1e-2 m) ) = 50.2m (not ~5017.6 meters)

Liquid mercury instead of seawater:
Resistivity: 961 nΩ·m = 0.961e-6 Ω·m
Resistance: 9.6e-7 Ω (got this one wrong the first time, too!)
Density: 13.56 times water
F = 1 T · (V/9.6e-7 Ω) · (1e-2 m)
s = 1/2 · (11200 m/s) · (11200 m/s) · (13.56e-3 kg) / ( 1 T · (V/9.6e-7 Ω) · (1e-2 m) )
@250 volts: s = 0.3266 meters (not 3.266m as before correction)
@25kV: s = 3.266 millimetres (not 32.66 millimetres as before)

Power (DC): P = IV where I = V/R,
R = 9.6e-7 Ω
@250 volts: I = 250 / R = 250 V / 9.6e-7 Ω = 2.604e8 amperes (x10 more than before correction)
∴ P = 65.1 gigawatts (x10 than before)
@25kV: I = 25000 / R = 25000 V / 9.6e-7 Ω = 2.604e10 amperes (x10 more than before)
∴ P = 651 terawatts (x10 than before)

Duration between rails:
From t = s / (1/2 · (11200 m/s))
@250 volts:
t = 0.3266 meters / (1/2 · (11200 m/s)) = 5.8321×10^-5 seconds (x10 less than before correction)
@25kV:
t = 3.266 millimetres / (1/2 · (11200 m/s)) = 5.8321×10^-7 seconds (x10 less than before)

Electrical energy usage:
E = P · t
@250 volts:
E = 65.1 gigawatts · 5.8321×10^-5 seconds = 3.797×10^6 joules (unchanged by correction)
@25kV:
E = 651 terawatts · 5.8321×10^-7 seconds = 3.797×10^8 joules (unchanged by correction)
(For reference, 1 litre of aviation turbine fuel is around 3.5e7 joules)

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