Unidades Relativistas
Relativistic units are used in Theory of Relativity, consist of taking the speed of light as the dimensionless constant 1, ie
c = 300.000 km/s ≈ 1
Now all the physical formulas can be passed to relativistic units.
Pass a non-relativistic formula units relativistic units is easy, simply replace all factors of c for 1, for example
E = mc2 is, in relativistc units
E = m (equivalence of mass and energy).
Converse is a bit trickier, but not too much, simply complete the units, always remembering that the units of c are [m].[s]-1, for example
E = m
E = m <=> [kg][m]2[s]-2 = m
Then need a [m]2[s]-2 to complete, so the formula becomes
E = mc2
Where these units come from? The answer come by consider Minkowski space, note that
v = (Δ x2 + Δ y2 + Δ z2) / Δt2
if we want v = c = 1, then
(Δ x2 + Δ y2 + Δ z2) / Δt2 = 1
- Δt2 + Δ x2 + Δ y2 + Δ z2 = 0
Taking infinitesimal increments we obtain The Minkowski metric .
Relativistic units are used in Theory of Relativity, consist of taking the speed of light as the dimensionless constant 1, ie
c = 300.000 km/s ≈ 1
Now all the physical formulas can be passed to relativistic units.
Pass a non-relativistic formula units relativistic units is easy, simply replace all factors of c for 1, for example
E = mc2 is, in relativistc units
E = m (equivalence of mass and energy).
Converse is a bit trickier, but not too much, simply complete the units, always remembering that the units of c are [m].[s]-1, for example
E = m
E = m <=> [kg][m]2[s]-2 = m
Then need a [m]2[s]-2 to complete, so the formula becomes
E = mc2
Where these units come from? The answer come by consider Minkowski space, note that
v = (Δ x2 + Δ y2 + Δ z2) / Δt2
if we want v = c = 1, then
(Δ x2 + Δ y2 + Δ z2) / Δt2 = 1
- Δt2 + Δ x2 + Δ y2 + Δ z2 = 0
Taking infinitesimal increments we obtain The Minkowski metric .