Dear Sir:
I am concerned with a conformal-Cartesian metric of the form:
guv = (a)* (1,1,1,-1)
where “a” is a time varying number between 0 and 1
....Putting this into the Maxima computer program I can
compute the Curvature-Scalar as:
(3a (att) - 3 (at)^2) / ( 2 a^3 )
where subscript "t" indicates the derivative with respect to time
att can be ignored, while at is on the order of 3x10-8/sec
and a = 0.8
....The above scalar curvature is dimensionless because
it was computed from a dimensionless metric.
.... Can you tell me how to restore “MKS” units to this
expression for the scalar curvature so that it will read
In inverse meters, as a curvature should in MKS units?
(I assume this involves inserting “c” appropriately
somewhere in the expression but I don’t know where)
and it seems to come out in dimensions of m-2
instead of m-1 which has me baffled, although I have
seen comments about this fact before?
Sincerely yours, George Hammond MS physics

On 1/3/22 1:19 PM, George Hammond wrote:
> Dear Sir:
> I am concerned with a conformal-Cartesian metric of the form:
> guv = (a)* (1,1,1,-1)
> where “a” is a time varying number between 0 and 1

I presume you mean diag(1,1,1,-1), and a(t) is dimensionless. So if
{x,y,z,t} are all given in meters, the result is sort-of in MKS because
the distance ds computed from these metric components will be in meters
-- it's only "sort-of" because t in meters is not MKS. So to use MKS the
metric components should be:

g_uv = a(t) diag(1,1,1,-c)

where this is for coordinates (x,y,z,t) with units (m,m,m,s).

Note: this is the simplest and most straightforward correction to your
post; other methods of assigning units to coordinates and metric
components are possible.

Note: the presence of c does not imply any relationship to
electrodynamics, it is just a units conversion.

> ...Putting this into the Maxima computer program I can
> compute the Curvature-Scalar as:
> (3a (att) - 3 (at)^2) / ( 2 a^3 )
> where subscript "t" indicates the derivative with respect to time
> att can be ignored, while at is on the order of 3x10-8/sec
> and a = 0.8
> ...The above scalar curvature is dimensionless because
> it was computed from a dimensionless metric.

This is wrong, as that calculation involves coordinate derivatives of
the metric components, and the coordinates have units -- att has units
of inverse meters^2, as does (at)^2 (when t has units of meters, see above).

> ... Can you tell me how to restore “MKS” units to this
> expression for the scalar curvature so that it will read
> In inverse meters, as a curvature should in MKS units?

Use the correct MKS metric components (above), and remember that in MKS
the coordinates have units. Note the Riemann, Weyl, and Ricci curvature
tensors, and the Ricci scalar all have units meters^-2 (when using the
metric components above).

> (I assume this involves inserting “c” appropriately
> somewhere in the expression but I don’t know where)

It is more complicated than that, involving using the correct metric
components for MKS, and remembering that in MKS the coordinates have units.

Tom Roberts