Discussion on tensor dimensions

[Ben Niehoff]

Let me add a bit more. Let's continue with my convention that the metric tensor $g$ has units of [L]^2, and that the coordinates $x^\mu$ have units of [L]. Then one has that $dx^\mu$ has units of [L], and $\partial/\partial x^\mu$ has units of [L]^(-1). Since

$$g = g_{\mu \nu} dx^\mu dx^\nu,$$
it follows that the components $g_{\mu\nu}$ are dimensionless. Notice that the exterior derivative operator $d$ is dimensionless (as $dx^\mu$ is just the exterior derivative of the function $x^\mu : M \to \mathbb{R}^n$).

If we write the metric in orthonormal frames

$$g = \delta_{ab} \, e^a \otimes e^b,$$
then it follows that the frames $e^a$ have dimension [L] (the Kronecker delta is always dimensionless). The connection 1-forms defined by

$$d e + \omega \wedge e = 0$$
are then dimensionless, and the curvature 2-form

$$\Omega \equiv d \omega +\omega \wedge \omega$$
is also dimensionless. Recall that the curvature 2-form is valued in a $\mathfrak{gl}(n)$ bundle over M. Choosing the coordinate basis for this bundle, we have that the components of the curvature 2-form are simply (the components of) the Riemann tensor:

$$\Omega^\mu{}_\nu \equiv \frac12 R^\mu{}_{\nu \rho \sigma} \, dx^\rho \wedge dx^\sigma.$$
Because the $dx^\mu$ have units [L], it follows that the components $R^\mu{}_{\nu \rho \sigma}$ have units [L]^(-2). This ought to look familiar.

I am happy to call this "Cartan notation", because these are the methods I typically use. But the consistency of unit conventions is an entirely separate issue from notation. Different notations are merely different ways to say the same thing. Given an abstract-index expression where the constituents are tensors, then I should be free to interpret it as a concrete-index expression in a basis of my choosing (including non-coordinate bases).

The difference is that an abstract-index tensor is defined as an object with indices that transform in a specified way, whereas the index-free definition of a tensor is a map between vector spaces which is function-linear in its arguments. These definitions are equivalent.Now, Bcrowell, let me ask you this. In your convention, you have chosen that the metric tensor to be unitless, and you have chosen that the coordinates, and hence the basis 1-forms $dx^\mu$, are unitless. It therefore follows, as far as I can tell, that the partial derivative operators $\partial / \partial x^\mu$ are unitless. In your convention, therefore, what are the units of the Riemann tensor? I think you might be surprised.

 

[bcrowell]

Here are a couple of other people's discussions of this sort of thing:

Dicke, "Mach's principle and invariance under transformation of units," Phys Rev 125 (1962) 2163 -- https://indico.kfki.hu/event/232/material/0/1.pdf

Terry Tao, https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/

Dicke has a nice discussion of how one can give differing descriptions based on one's definitions and assumptions, after which he goes ahead and picks the convention he wants. As in his treatment, consider this equation

$ds^2 = g_{ab}dx^a dx^b ,$

where nothing that follows depends on whether one considers these to be abstract or concrete indices. [EDIT: Changed my mind about this. It really needs to be abstract index notation. In concrete index notation, $dx^a$ would be an infinitesimal change in a coordinate, so we couldn't attribute units to it generically.] Generalizing his treatment, let's make no other assumption than that when something is written in multiplicative notation, the units multiply. 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

$\sigma=\gamma+\xi .$

Some possible choices of these constants are:

$(1,1,0)$ - Dicke

$(0,\mp 1,\pm 1)$ - Tao

$(1,0,1)$ - what I've been advocating in this thread

Raising and lowering indices to form a tensor of rank $(r,s)$ makes units vary in proportion to $L^{\gamma(s-r)}$. In general, you can assign a physical quantity units $L^u$ that are a product of two factors, a "kinematical" or purely geometrical factor $L^k$, where $k=\gamma(s-r)$, and a dynamical factor $L^d\ldots$, which can depend on what kind of quantity it is, and where the ... indicates that if your system of units has more than just one base unit, those can be in there as well. Dicke uses units with $\hbar=c=1$, for example, so there's only one base unit, and mass has units of inverse length and $d_\text{mass}=-1$.

Although the units check out as long as the u exponents add up on both sides of the equation, it's also necessary for both the k's and the d's to add up, independently. To check the k's it suffices, in non-Cartan index notation, to check that the indices are written grammatically. (This fails when you mix in Cartan notation, e.g., $v=v^\mu\partial_\mu$ is notated as if it has (r,s)=(0,0), but it's actually a notation for a (1,0) vector.)

 

[bcrowell]

I do not agree with this because I do not agree with g(v,v) having different units from g(p,p) being a problem. The assignment of dimensions to the metric tensor is based on the difference in dimensions between the object a linear operator acts on and the object it spits out. In that sense, it is perfectly compatible with g(v,v) differing in units by mass squared from g(p,p). The dimension of g(x,y) is [g][x][y], not [g].

I think the issue here is simply that we have a customary, convenient procedure for checking units, and it's based on the idea that the units of a product are the product of the units of the factors. In index-free notation, there are a lot of cases where things get notated non-multiplicatively.

Let me add a bit more. Let's continue with my convention that [...] the coordinates $x^\mu$ have units of [L]. [...] the components $g_{\mu\nu}$ are dimensionless

So if I'm understanding you correctly, then in the notation I described in #32, you're saying you want $(\sigma,\gamma,\xi)=(1,0,1)$, which is the same convention I've been advocating.

In your convention, you have chosen that the metric tensor to be unitless, and you have chosen that the coordinates, and hence the basis 1-forms $dx^\mu$, are unitless.

No, I think you must have misunderstood me. I never said that I thought coordinates should be unitless. Of course coordinates can have any arbitrary mixture of units that we like, as in spherical coordinates.

 

[bcrowell]

I am happy to call this "Cartan notation", because these are the methods I typically use. But the consistency of unit conventions is an entirely separate issue from notation. Different notations are merely different ways to say the same thing. Given an abstract-index expression where the constituents are tensors, then I should be free to interpret it as a concrete-index expression in a basis of my choosing (including non-coordinate bases).

Of course all these notations are isomorphic ways of expressing the same ideas. However, (a) they may clash if you try to mix them, and (b) they may be inconvenient for certain purposes. There can be unpleasant clashes, for example, when you mix index and Cartan notation. Index-free notation can be inconvenient if you want to check units, since it uses a lot of non-multiplicative notation.

 

[bcrowell]

A velocity vector ought to have units of velocity. [...] the components $V^\mu$ have units of length. What exactly do you think is wrong about that?

[...] the coordinates $x^\mu$ have units of [L]. Then one has that $dx^\mu$ has units of [L][...] the components $g_{\mu\nu}$ are dimensionless.

I don't think these statements are logically consistent. We have

$ds^2=g_{\mu\nu}dx^\mu dx^\nu,$

and based on your statements above, the units on the right-hand side (in Cartesian coordinates) are $(L^0)(L^1)(L^1)$, so $[ds]=L$. Then since

$V^\mu = \frac{dx^\mu}{ds},$

and you say that $[dx^\mu]=L$, it follows that $V^\mu$ is unitless.

 

[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 $\gamma=1$.