Discussion on tensor dimensions

[bcrowell]

Here's my attempt at extending the analysis of #32 from abstract index notation to Cartan notation.

Let vector r and covector $\omega$ be duals of each other, and let r represent a displacement.

Cartan notation:
$r=r^\mu \partial_\mu$
$\omega=\omega_\mu dx^\mu$

Attribute these units to the above in Cartesian coordinates (i.e., r has units of $L^A$, etc.):

(1) A=B+C ... units of $r=r^\mu \partial_\mu$

(2) D=E+F ... units of $\omega=\omega_\mu dx^\mu$

The following two requirements seem clear:

(3) $A+D=2\sigma$ ... because $r \cdot \omega=ds^2$

(4) $D=2\gamma+B$ ... because $\omega_\mu=g_{\mu \nu}r^\nu$

To avoid a clash between Cartan and concrete index notation in a Cartesian coordinate system, it seems like we want the following three conditions:

(5) $F=\xi$ ... We want units of Cartan notation $dx^\mu$ not to clash with units of abstract index notation $dx^a$ in Cartesian coordinates.

(6) $C=-\xi$ ... We want units of the derivative $\partial_\mu$ not to clash with units of abstract index notation $dx^a$ in Cartesian coordinates.

(7) $B=\xi$ ... We want units of concrete components in Cartan notation not to clash with units of abstract index notation $dx^a$ in Cartesian coordinates.

Taking $(\sigma,\gamma,\xi)$ to be fixed and imposing (1)-(6), we find:

$A=\sigma-\gamma-\frac{1}{2}\xi$
$B=\sigma-\gamma+\frac{1}{2}\xi$
$C=-\xi$
$D=\sigma+\gamma+\frac{1}{2}\xi$
$E=\sigma+\gamma-\frac{1}{2}\xi$
$F=\xi$

If we also impose (7), then because $\sigma=\gamma+\xi$, we have

$\xi=0.$

This means that if we want to be able to extend the system to Cartan notation, we have to use Dicke's system $(\sigma,\gamma,\xi)=(1,1,0)$, with unitless coordinates (or some trivial variation such as (2,2,0)). The results are:

A=B=C=F=0, D=E=2

The units of the factors in $r=r^\mu \partial_\mu$ are $L^0=L^0L^0$.

The units of the factors in $\omega=\omega_\mu dx^\mu$ are $L^2=L^2L^0$.

 

[stevendaryl]

Let the units of the factors be $[ds]=L^\sigma$, $[g_{ab}]=L^{2 \gamma}$, and $[dx^a]=[dx^b]=L^\xi$, where $L$ stands for units of length. We then have

$\sigma=\gamma+\xi .$

Since $ds$ is a proper length, I can't understand why anyone would make any choice other than $\sigma = 1$.

 

[bcrowell]

Since $ds$ is a proper length, I can't understand why anyone would make any choice other than $\sigma = 1$.

Terry Tao does choose $\sigma=0$. He doesn't give any detailed justification for his system, but I think it's a natural system to use if you think of unit conversions as being a uniform rescaling of coordinates $(t,x,y,z)\rightarrow(kt,kx,ky,kz)$. Proper time is a scalar, and scalars don't change under such a change of coordinates. As an example, suppose we have a cuckoo clock whose bird comes out at regular time intervals, and we also have the Earth spinning on its axis. If we observe that the cuckoo comes out 24 times during one rotation of the earth, then this is simply a statement about counting events and coincidences of events in spacetime. A change of coordinates is just a renaming of the events, which can't change this relationship. There is nothing in the physical process of measurement that even says which thing, cuckoo or earth, should be considered as the "clock" that defines our "unit" of time.

As discussed in the Dicke paper, there's considerable ambiguity in terms of how to define what it means to change units.

 

[bcrowell]

It turns out that Schouten gives an analysis that's almost identical to what I did in #32. This is in ch. VI of Tensor Analysis for Physicists. His "relative units" are the same as my dynamical units, and his "absolute units" are the same as my complete set of units (kinematic $\times$ dynamical), with $\gamma=1$.