Monday, February 26, 2018

general relativity - Dimensional analysis of metric tensor


In the geometry of GR, the metric tensor $g$ can determine the Riemannian connection and curvature tensor by combining the spatial derivatives (w.r.t. the 4d coordinate system) properly.


I am curious about the dimensional analysis of the metric tensor.


According to the geometric picture of GR, the connection as the potential is related with energy and the curvature tensor is related with force strength (with the mass to connect the connection/curvature with energy/force).


Then what's the dimension of the metric tensor? Intuitively it should be dimensionless, but how its first/secondary spatial derivative is related with energy/mass and force/mass=acceleration respectively?


Another observation is from the representation of Lorentz group. Where the rotation/boost is related with $SU(2)$ and $SL(2)$ transformations. If we take the $SU(2)$ or $SL(2)$ as transformations $U$ on quantum states, then they are dimensionless. So the acceleration (boost/time) can be regarded as $$dU/dt=H/\hbar=1/t$$ so we get boost is dimensionless, so time=length and energy=mass (these are normal conclusions since we usually take c=1). The reason that I check the Lorentz group representation is that the general metric tensor is generated from the Minkowski metric by dimensionless operation $GL(4)$, so this seems to confirm that the metric tensor should be dimensionless.


But if we go back to the former analysis, where the spatial derivative of the dimensionless metric tensor gives energy/mass, which is then also dimensionless since energy=mass. So we get the spatial derivative of a dimensionless value is still dimensionless.


There must be something wrong with my deduction. Can anybody help to clarify this?




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