The discussion revolves around the contraction of the stress-energy tensor (SET) in the context of general relativity, specifically exploring the differences between contractions using the metric and those involving one-forms. Participants are examining the implications of these contractions for understanding the scalar form of the Einstein field equations, particularly in relation to perfect fluid models.
Participants do not reach consensus on several points, including the nature of the contraction results (sum vs. difference) and the interpretation of energy density and pressure. The discussion remains unresolved regarding the utility and interpretation of contracting the SET using differentials versus the metric.
There are limitations in the discussion regarding the assumptions made about the definitions of energy density and pressure, as well as the mathematical steps involved in tensor contraction. The varying interpretations of the results highlight the complexity of the topic.
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 x-direction \hat x 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 N/\sqrt{N^2}.