Yes thanks for the clarification. But there are still units involved even though it's length over length I still consider those units because it relates to physical measurements.vanhees71 said:It's very convenient to normalize it to 1. The four-velocity of a massive particle is thus
$$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
which by definition of proper time is normalized to 1,
$$g_{\mu \nu} u^{\mu} u^{\nu}=1,$$
when using the west-coast convention for the pseudo-metric, i.e., the signature (1,-1,-1,-1).
Another even more convenient choice is to use natural units with ##\hbar=c=k_{\text{B}}=1## (and maybe even ##G=1##, which then makes everything measured in dimensionless quantities, i.e., Planck units).
dsaun777 said:Yes thanks for the clarification. But there are still units involved even though it's length over length I still consider those units because it relates to physical measurements.
If you are doing calculations on paper in an abstract manner to teach a class, I would say that you are not using any units. But if you are using applied physics to solve a real engineering problem it would be useful, I think, to include units even if they cancel and become dimensionless. If anything to show where unitless quantities get their respective derivation and meaning.robphy said:So, would you say the following?
In Euclidean geometry, the unit-vector along the [itex]x[/itex]-direction [itex]\hat x[/itex] is not dimensionless… but has units of m/m … or, if I were describing the direction of a force in classical physics, that unit vector along the force has units N/N or maybe [itex]N/\sqrt{N^2}[/itex].