The analogy of electricity and gravity

The analogy of electricity and gravity

Bjerknes Carl Anton
Bjerknes Carl Anton , 1825-1903, Norwegian mathematician and physicist. Tried to explain the electrodynamics by hydrodynamical analogies.

Gravity Electricity
Quantity Relationship Unit Note Quantity Relationship Unit Note
Mass m kg Charge q C = A.s
Density ρ = m/V kg/m3 Volumetric density of charge ρ = q/V C/m3
Current I = m/t kg/s I=konst (v potrubí) Current I = q/t C/s = A I=konst (Kirchhoff's current law)
Current density j = I/s kg/(m2.s) Current density j = I/s C/(m2.s) = A/m2
Potential V = ϰG*m/r m2/s2 Potential V = ϰE*q/r kg.m2/(A.s3)
Intensity of the grav.field E = ϰG*m/r2 m/s2 Intensity of the el.field E = ϰE*q/r2 V/m
Voltage U = ΔV m2/s2 Voltage U = ΔV V
Potential energy ΔW = U.m J = kg*m2/s2 Energy of el.current ΔW = U.q = U.I.t J = V.A.s = kg*m2/s2
Output P = ΔW/t J/s = kg*m2/s3 Output P = ΔW/t = U.I V.A
Grav.induction D = m/r2 kg/m2 El.induction D = q/r2 C/m2 = As/m2
Induction flux φ = U.t = gh.t m2/s Induction flux φ = U.t V.s = Weber = Tesla/m2
Output of hydro turbine P = m/t(v12-v22) = I*Δ(v2) (kg/s)*m2/s2 Battery output (performance) P = W/t = U.I V.A = (C/S)*V
Current in conduit of variable cross-section I = m/t = ρ.V/t = ρ.S.v kg/s Electric current I = q/t = ρ.V/t = ρ.S.v C/s = A
Resistance R = U/I = ght/m m2/(kg.s) Resistance R = U/I V/A = Ω
Resistivity γ m3/(kg.s) Resistivity (1m,φ1mm2) γ Ω.m
Resistance of wire R = γ*l/S m2/(kg.s) Resistance of wire R = γ*l/S Ω
Resistance of conduit R = U/I = γ*Δh/S m2/(kg.s) El. resistance R = γ*l/S Ω
Intensity (?) H kg/(s.m) Intensity of mg.field H A/m
Permeability μ m/kg Permeability μ V.s/(A.m)
Induction (?) B = μ.H 1/s Magnetic induction B = μ.H kg/(s2.A) = Tesla