Is the metric dimensionful?

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Ken S. Tucker - 28 Feb 2007 18:40 GMT

Hi Dr. Francis, studied your post.

On Feb 27, 11:55 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:

> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>

> >Hi Dr. Most and all.
>
> >On Feb 27, 1:44 am, Andreas Most <Andreas.M...@nospam.de> wrote:
> >> Ken S. Tucker wrote:
> >> > Oh No wrote:
> >> > Ken>>Well, g_00 is 1/sec^2,
>
> >> >> ds^2 = g_mu_nu dx^mu dx^nu
> >> >> I would be inclined to think ds is measured in secs, as is dx, which
> >> >> makes g_00 dimensionless.
>
> >> > We should carefully discuss that.
> >> > "ds" is invariant so it can't have relativistic units,
> >> > which seconds (or minutes etc...) is.
>
> >> Invariance has nothing to do with the units of ds.
>
> >By definition "ds" is a scalar, and therefore invariant,
> >(a scalar, is unitless, it's a pure number).
>
> Andreas is right. Scalar quantities, or invariant quantities are not
> unitless. cf mass in classical mechanics.

A kilogram mass is a rest relatively to Paris, per ISU standards.
Relatively to a moving FoR or one in a different gravitational
potential that ISU standard Kilogram will have a differing
inertia. The use of the word "mass" scientifically refers
to the ISU standard reference kilogram in France, and
that mass is certainly relativisitic (not scalar, not invariant).

> >> Invariance simply means that ds^2 is independent from the choice
> >> of coordinates in which you calculate the distance.
> >> The coordinate transformation takes care of the unit conversion
> >> if necessary. You just have to agree on what units to use for ds.
>
> >I do understand what you are imagining, however
> >you'll need to prove, or reference to a proof that the
> >Kronecker Delta is a dimensionless {1,0}, in the
> >scheme you select.
>
> Kronecker delta is defined in mathematics without any reference to
> units. Look it up with google.

Certainly, see pgs 21-24 (html version)...
http://72.14.253.104/search?q=cache:TM3H8ZIcLPQJ:www.grc.nasa.gov/WWW/K-12//Numb
ers/Math/documents/Tensors_TM2002211716.pdf+tensors+%2B+nasa&hl=en&ct=clnk&cd=1

Please note on pg 24, the Kronecker delta is
given by a scalar product of "unit" vectors.
As I understand it, those unit vectors may be expressed
as inches, centimeters, seconds, years, and General
Covariance permits the free choice of units, as well
as the choice of the CS.

> >I previously posted a proof,
> >as to why g_00 = 1/time^2.
>
> and I already corrected it. Had your post found me moderating it would
> not have gone through.

Ohno, I'm being sent back to "Shut-up and Calculate" :-).

Seriously though, It is foundational (per SPF charter ) to
consider alternative interpretations to the meaning of an
arbituary "norm" like AB = g_uv A^u B^v, especially applied
to ds^2=g_uv dx^u dx^v.

It is my impression mathematicians and phycists differ
on that interpretation. I think our discussion may bring
that difference into relief.
Regards
Ken S. Tucker

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Ken S. Tucker - 28 Feb 2007 19:37 GMT

> Hi Dr. Francis, studied your post.
>
[quoted text clipped - 71 lines]
> Regards
> Ken S. Tucker

> ======================================= MODERATOR'S COMMENT:
> Kronecker delta_ij=1 if i=j and =0 if i=/=j, no units involved; please correct >your misunderstanding, because any other definition will lead to confusion >without any chance to gain in understanding

What you're quoting is the result of the condition
that the axes are defined to be "independant", given
by the partial diff,

&x^u / &x^v = {0,1}

(Fred, see Spiegel's pg. 179, Problem 11.)

In addition to which, in this ref,

see pgs 21-24 (html version)...
http://72.14.253.104/search?q=cache:TM3H8ZIcLPQJ:www.grc.nasa.gov/WWW...

Please note on pg 24, the Kronecker delta is
given by a scalar product of "unit" vectors.

Kronecker = e^u.e_v = {0,1}

For example let the unit time vector "e^0" be "seconds",
(generally e^0 = unit time), then,

Kronecker = e^0.e_0 = 1 , with NO units, therefore,

(e^0 = time)*(e_0 = 1/time) = 1.

>From that we form, (Fred, see Spiegel's pg 148, problem 17)

g_00 = e_0.e_0 == (1/time^2 dimensionally).

To Moderator, how do you *physically* define
the basis vector "e^0"?

Regards
Ken
PS: I think this discussion important.

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Natural Science Forum / Physics / General Physics / February 2007

Is the metric dimensionful?