Re: GR ?


"Tom Roberts" <tjroberts@xxxxxxxxxx> wrote in message
news:iawCe.95$lX2.75@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| Significant Zero wrote:
| > "Tom Roberts" <tjroberts@xxxxxxxxxx> wrote in message
| > news:XFiCe.4226$Ih7.2317@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | In modern physics, energy is the conserved Noether current
corresponding
| > | to a time translation. In systems without time translation invariance,
| > | energy is not conserved, and loses much of its usefulness.
| >
| > A point I am interested in Tom, thanks. So do you think that in GR time
| > translation is invariant ?
|
| In GR, any manifold that has a timelike Killing vector will have
| invariance for time translation along that vector (that's what "Killing
| vector" means [named after Prof. Killing, of course]). Manifolds without
| a timelike Killing vector do not.
|
| A "Killing vector" is necessarily a vector field on the
| manifold, but the word "field" is usually omitted.

I was really referring to time invariance or variance as a function of the
physical reality that a manifold was describing but this could easily
degenerate into a debate on manifolds that might last till the cosmos
freezes over so I'm not going down that route thank you{:-) Due you concur
with the experimental facts that time and length are variant ? If so do you
agree that to maintain the measured constancy of c locally their is a need
for time and length to be changed in a proportionate manner so this apparent
fact is maintained ? i.e A cubic meter of the vacuum state between galaxies
has different characteristics to a cubic meter just outside an event horizon
but in each place sol will be locally measured as constant but may not be
the same relatively ?

|
|
| > and can you conceptualise a system in which time
| > and distance proportionate variance produced a conserved energy
situation ?
| > and would this have any relevance to GR ?
|
| I have no idea what you are asking ("time and distance proportionate
| variance" means nothing to me).

See above
|
| Note that any manifold with a timelike Killing vector is called
| "stationary" -- in essence as long as you use that Killing vector as a
| time coordinate, then nothing changes over time (nothing moves). Here's
| a short and incomplete list of such manifolds:
| Minkowski spacetime, everywhere
| Schwarzschild spacetime, exterior to the horizon
| Kerr geometry, external to the horizon
| ... there are others
|
| As we observe objects moving in the world we inhabit, any manifold
| modeling our world has no timelike Killing vector.
|
|
| > But can a *Langrangian (or *Hamiltonian) always be considered as being
| > applied to a continuous symmetry in GR.?
|
| Obviously you don't know what a Lagrangian is.

Maybe not in your terms but Lagrangians seem to be describing the actions of
particles in systems not explaining what energy is, which was the point of
my original post and although I agree that explanations of the actions of
objects is important and necessary it does not address what energy *is* as
my original post was trying to do.
..
| That's too complicated
| for me to attempt to explain here -- find a good book and study it. But
| in a nutshell, the Lagrangian of a system expresses the "action" of the
| system as it evolves, and the principle of least action states that
| variations of the Lagrangian around the system's actual dynamical path
| must be zero. Using the calculus of variations can then yield
| differential equations for the system known as its "equations of
| motion"; they can be solved for the trajectories of the various
| components of the system. In Newtonian mechanics this can yield the
| trajectory of a cannonball, for instance. In GR the equation of motion
| is the Einstein field equation (originally discovered another way by
| Einstein).
|
| The "continuous symmetries" discussed here, and which are the subject of
| Noether's theorem, are SYMMETRIES OF THE LAGRANGIAN. That is, if energy
| is to be conserved in a given system then the Lagrangian for that system
| must not change under a time translation -- the Lagrangian must be
| independent of time.
|
|
| > Its certainly confusing to me as it
| > appears that a Langrangian is based on Pythagoras's and Euclidean
geometry
| > which seem incompatible with GR.
|
| In Newtonian mechanics, Euclidean geometry is implicitly used, and
| geometrical terms do not appear in the Lagrangian. In GR, the Lagrangian
| explicitly includes the Ricci scalar, which is a geometrical term. This
| is why GR needs no prior geometry (geometry implicitly assumed and
| forever fixed) -- in GR the geometry of the manifold is dynamical, as
| are the trajectories and interactions of objects.

Fine but are you letting the metric follow the geometry or are you using a
flat Euclidean geometry and metric and referring GR's geometry to this ?
--
Significant Zero E-field = Electric field, M-field =Magnetic field, two
unbound field effects
http://home.freeuk.com/paulps/
Maybe updates. (1-(1/(1/3))^2)/(1 + (1/(1/3))^2) = - 0.08 = FTL ? -p<+p or
(m*-v)<(m*+v) or (m*-c^2)<(m*+c^2) =g?


|
|
| Tom Roberts tjroberts@xxxxxxxxxx


.

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