3 ms·
Your Law of Optimization reminds me of the (often popularly ignored) denominator in the famous e = m c^2 equation. I believe the full version has a (1 - (v/c)^
by jjtheblunt 5y ago
Your Law of Optimization reminds me of the (often popularly ignored) denominator in the famous e = m c^2 equation.
I believe the full version has a (1 - (v/c)^2) denominator on the right side; I think the Lorentz transformations motivate this.
The upshot matches your observation in spirit: the closer you get to the optimum (the celeritas/speed of light) the more energy you require.
- JadeNB 5y ago> Your Law of Optimization reminds me of the (often popularly ignored) denominator in the famous e = m c^2 equation. > I believe the full version has a (1 - (v/c)^2) denominator on the right side …. I think the question is what you mean by `m`. If it's inertial mass, then I believe no correction is necessary. The point is that the inertial mass `m` is the rest mass `m_0` divided by `\sqrt{1 - (v/c)^2}`, as you say.
- jjtheblunt 5y agoyes!
- breck 5y agoIronic the man who wrote `make the irreducible basic elements as simple and as few as possible without having to surrender the adequate representation of a single datum of experience.` is most tagged with `e = mc^2`.