wrote in message
...
Perion wrote:

I'm curious why many authors
prefer the -1,1,1,1 signiture for the metric rather than 1,-1,-1,-1.
Since
any possible observable event for is either timelike or null (lightlike)
it
seems that it would be more natural for the time component in the
interval
to be positive.

If you're looking at a canonical formulation of gravity, in which
a spacelike hypersurface evolves in time, it's natural to want the
induced spatial metric to be positive. If you're doing Euclidean
quantum gravity, it's easiest to have the ``imaginary time'' metric
positive definite rather than negative definite. On the other hand,
as you say, if you're looking at motion of observers in a spacetime,
it's easiest to have ds be proper time, and if you want to talk to
particle physicists, who usually use a +--- signature, it's easiest
to not have to keep explaining signs.

Yes. This later part is why I prefer the +--- signature.

Thanks for the clarity Steve

Pmb

Tom Roberts wrote in message ...
This is basically units: 4-velocity is the tangent vector to an object's
trajectory. As such one can use any affine parameter along the curve,
and the norm of the 4-vector will depend on which affine parameter one
selects. By requiring its norm to be -1 one is choosing to use the same
units for proper time

Please be very careful here TR, (anyone else and I wouldn't
rebuttal). Given the proper time ds or its integral worldline
S, are invariant, no units can be moved from some specific
set of coordinates to the proper time ds or its integral S.
Numbers ds and S can produce any components
relative to an infinite variety of CS's.

along the trajectory and the coordinates onto
which one projects the 4-velocity.
Naturally for this to make sense one must select units with c=1, but
that was implicit in your original post.

A very simple question, does ds or S have units?
Ken S. Tucker

Tom Roberts

On 10/15/2003 1:32 PM, Ken S. Tucker wrote:
A very simple question, does ds or S have units?

ds is a distance, and must therefore have the units of length. Ditto for
its integral, s.

Whether or not coordinates {x^i} have units depends on one's choice of
coordinates, but the {g_ij} must correspond so ds^2 = g_ij dx^i dx^j has
units (length)^2. Note that different elements of {g_ij} can have
different units (and often do, e.g. in spherical coordinates)....


Tom Roberts

Tom Roberts wrote in message ...
On 10/15/2003 1:32 PM, Ken S. Tucker wrote:
A very simple question, does ds or S have units?

ds is a distance, and must therefore have the units of length. Ditto for
its integral, s.
Whether or not coordinates {x^i} have units depends on one's choice of
coordinates, but the {g_ij} must correspond so ds^2 = g_ij dx^i dx^j has
units (length)^2. Note that different elements of {g_ij} can have
different units (and often do, e.g. in spherical coordinates)....
Tom Roberts

Oct 12, from thread "Observer dependant....", TR says,
(asterisks are ***Kens***)....

"Yes. But what you call "m" is quite clearly the ENERGY of the object (up
to factors of c), as measured by that observer. OF COURSE it is observer
dependent (energy always is), and OF COURSE it is invariant (because
which observer is referenced is independent of coordinates, and because
the observer in question measures a ***specific number for the energy***
, and all numbers are invariants)."

This obviously refers to an invariant like, E^2 = g_uv E^u E^v.
and you call this "a specific number", implying it has no units.
(calling it an *energy* invariant in fine, as it sources the
equation).
A few weeks back you corrected my terminology to say
that the A in A^2 = g_uv A^u A^v isn't called a length,
but instead should be called a norm, and after some study,
I think you are right, because a norm has no units, it is always
a pure number, a scalar invariant.

To support this point, let me provide a breakdown of
vector S, (the arrows mean vectors)...

- - -
S = e_ u * x^u = e^u * x_u
(u=1) = (1/cm) * (cm) = (cm) *(1/cm)
(u=0) = (1/sec)*(sec) = (sec)*(1/sec)

the e_u and e^u are basis vectors.

Of course the units cm and secs are arbituary but
convey the meaning. The units are implicit within the
definition of covariant (like e_u or A_u) and likewise
with contravariant.
Above you point out elements of {g_ij} can have
different units, well agreed. ie.
g_22 = theta^-2 iff dy = theta
then g_22 dy^2 = no units.

Thats why I interjected TR, to remind you what you
taught me :-), of course your comments are invited.
Regards
Ken S. Tucker

"Pmb" wrote in message .. .

Perion wrote:

I'm curious why many authors
prefer the -1,1,1,1 signiture for the metric rather than 1,-1,-1,-1. Since
any possible observable event for is either timelike or null (lightlike) it
seems that it would be more natural for the time component in the interval
to be positive.

If you're looking at a canonical formulation of gravity, in which
a spacelike hypersurface evolves in time, it's natural to want the
induced spatial metric to be positive. If you're doing Euclidean
quantum gravity, it's easiest to have the ``imaginary time'' metric
positive definite rather than negative definite. On the other hand,
as you say, if you're looking at motion of observers in a spacetime,
it's easiest to have ds be proper time, and if you want to talk to
particle physicists, who usually use a +--- signature, it's easiest
to not have to keep explaining signs.

In the end it's an arbitrary choice, and the textbooks are split
almost evenly. The only place it makes a difference is if you aren't
careful about matching your definition of spinors to your choice
of metric signature; then the sign of the metric can matter. (See
S. Carlip and C. DeWitt-Morette, ``Where the Sign of the Metric
Makes a Difference,'' Phys. Rev. Lett.60 (1988) 1599.)
Steve Carlip

Yes. This later part is why I prefer the +--- signature.
Thanks for the clarity Steve
Pmb

Easy Pmb, Dr. Carlip does not define the signature.
Please read and enjoy...

((Wish that was online, the prison library cancelled
Phys. Rev.in 1987 :-))
Recently I've become very uneasy about using
either signature (+---) or (-+++) as neither is regarded
as correct, and so maybe both are wrong.

IMHO these signatures are obsolete, and should
be replaced by signature (++++) with the inclusion
of a nonorthogonal component g_i0 = -v.
In discussions with the fellows in this group,
certain definitions were needed.
Kindly view the post entitled,

"Mass, Energy and Lorentz Transform Summary (kst)"

part 5 is relevent to this thread.

Regards Ken S. Tucker

PS: snippable, please note in the referenced part
the *collapse* of the interval definition excludes g_i4
components, and thereby eliminates the Kerr metric.
The Kerr metric encoded rather awkwardly, rotational
information. The collapse of the interval thereby requires
a suitable metric replacement to represent rotation and
this information (Imo) is represented within the non-
orthogonal antisymmetrical g_ij, better termed a_ij.

I respect the view that the metrics of g_uv are
defined by measurements (surveys) of the effects
on light rays. The most famous is the blue-shift of
a photon as it moves toward a gravitating mass
and is consistent with the metric g_44.
This one is rather easy as it does not involve
any spatial direction, something a bit tougher...

Consider the case of the Earth in revolution
about the Sun. It's well known the apparent
location of the Sun is positionally altered by
aberration effects as observed by the Terrestrials.
Equally obvious, if the revolution of Earth was to
reverse direction, terrestrials would find an equal
but opposite alteration of position, due to
aberration effects.
If we intend to accept the metric is defined
by light rays, then both of these observations
must be true.

Given that these revolutions occur in the XY
plane then this information is encoded in the
metric components a12 = -a21

(the field is quite weak and doesn't warrant
a specific a_12 or a^12 difference yet)

Now we can define these antisymmetrical metrics
by the aberrated path of light by using,

a12 = x/r * dy/cdt - y/r * dx/cdt
a21 = y/r * dx/cdt - x/r * dy/cdt

and see that a12 = -a21.

For example, set the Sun at the origin and Earth at
x=r , y=0, dx/dr=0, V=dy/dr to find

a12 = V/c and a21 = -V/c

and these metrics precisely define and predict the
the aberration of light rays, given opposite system
rotations, (along the Sun Earth axis).

Another reason to replace a Kerr metric by an anti-
symmetrical a{ij) comes from magnetic force.
In the above, relative rotation/revolution of the
sun and earth accounted for aberration. Well,
attach a charge to earth and find,

F{ij} = constant * a{ij}

where F{ij} is the magnetic field.

This should be expected, because the aberration
equation governs light rays by a{ij} and these are
the same as magnetic field effects described by

F{ij} = constant * a{ij}

which is just another form of EM radiation,
and aberration.

Finally, summarizing,

Signature (g_uv) = (++++),

And nonorthogonal dynamic spacetime metrics
(g_i0) = -V

And antisymmetrical (rotational) metrics
(a_ij) = - (a_ji)

prove to be consistent.

If what I said is true, can we discuss the complicated
stuff?
KST

(Ken S. Tucker) wrote in message . com...
"Pmb" wrote in message .. .

Perion wrote:

I'm curious why many authors
prefer the -1,1,1,1 signiture for the metric rather than 1,-1,-1,-1. Since
any possible observable event for is either timelike or null (lightlike) it
seems that it would be more natural for the time component in the interval
to be positive.

If you're looking at a canonical formulation of gravity, in which
a spacelike hypersurface evolves in time, it's natural to want the
induced spatial metric to be positive. If you're doing Euclidean
quantum gravity, it's easiest to have the ``imaginary time'' metric
positive definite rather than negative definite. On the other hand,
as you say, if you're looking at motion of observers in a spacetime,
it's easiest to have ds be proper time, and if you want to talk to
particle physicists, who usually use a +--- signature, it's easiest
to not have to keep explaining signs.

In the end it's an arbitrary choice, and the textbooks are split
almost evenly. The only place it makes a difference is if you aren't
careful about matching your definition of spinors to your choice
of metric signature; then the sign of the metric can matter. (See
S. Carlip and C. DeWitt-Morette, ``Where the Sign of the Metric
Makes a Difference,'' Phys. Rev. Lett.60 (1988) 1599.)
Steve Carlip

Yes. This later part is why I prefer the +--- signature.
Thanks for the clarity Steve
Pmb

Easy Pmb, Dr. Carlip does not define the signature.

I know he doesn't. But he's very clear when he explains the reason for
various usage of a given signiture. All I said was that his later part
is why *I* choose what I use.

Pete


Please read and enjoy...

((Wish that was online, the prison library cancelled
Phys. Rev.in 1987 :-))

:-D


Pmb