Sorry, I don't remember what, if anything, I intended to text.
I'm not expert on how Einstein arrived at his famous field equations. I
know that he insisted on them being tensor equations so that they would
have the same form in all coordinate systems. That may sound like a
mathematical technicality, but it is really to ensure that the things in
the equation, the tensors, could have a physical interpretation. He
also limited himself to second order differentials, probably as a matter
of simplicity. And he excluded torsion, but I don't know why. And of
course he knew it had to reproduce Newtonian gravity in the weak/slow limit.
Brent
On 4/18/2019 7:59 AM, agrayson2...@gmail.com wrote:
On Wednesday, April 17, 2019 at 7:16:45 PM UTC-6, agrays...@gmail.com
wrote:
*I see no new text in this message. AG*
Brent; if you have time, please reproduce the text you intended.
I recall reading that before Einstein published his GR paper, he used
a trial and error method to determine the final field equations (as he
raced for the correct ones in competition with Hilbert, who may have
arrived at them first). So it's hard to imagine a mathematical
methodology which produces them. If you have any articles that attempt
to explain how the field equations are derived, I'd really like to
explore this aspect of GR and get some "satisfaction". I can see how
he arrived at some principles, such as geodesic motion, by applying
the Least Action Principle, or how he might have intuited that
matter/energy effects the geometry of spacetime, but from these
principles it's baffling how he arrived at the field equations.
AG
On Wednesday, April 17, 2019 at 7:00:55 PM UTC-6, Brent wrote:
On 4/17/2019 5:20 PM, agrays...@gmail.com wrote:
On Wednesday, April 17, 2019 at 5:11:55 PM UTC-6, Brent wrote:
On 4/17/2019 12:36 PM, agrays...@gmail.com wrote:
On Wednesday, April 17, 2019 at 1:02:09 PM UTC-6, Brent
wrote:
On 4/17/2019 7:37 AM, agrays...@gmail.com wrote:
On Tuesday, April 16, 2019 at 9:15:40 PM UTC-6,
Brent wrote:
On 4/16/2019 6:14 PM, agrays...@gmail.com wrote:
On Tuesday, April 16, 2019 at 6:39:11 PM
UTC-6, agrays...@gmail.com wrote:
On Tuesday, April 16, 2019 at 6:10:16 PM
UTC-6, Brent wrote:
On 4/16/2019 11:41 AM,
agrays...@gmail.com wrote:
On Monday, April 15, 2019 at 9:26:59
PM UTC-6, Brent wrote:
On 4/15/2019 7:14 PM,
agrays...@gmail.com wrote:
On Friday, April 12, 2019 at
5:48:23 AM UTC-6,
agrays...@gmail.com wrote:
On Thursday, April 11, 2019
at 10:56:08 PM UTC-6, Brent
wrote:
On 4/11/2019 9:33 PM,
agrays...@gmail.com wrote:
On Thursday, April 11,
2019 at 7:12:17 PM
UTC-6, Brent wrote:
On 4/11/2019 4:53
PM,
agrays...@gmail.com
wrote:
On Thursday, April
11, 2019 at
4:37:39 PM UTC-6,
Brent wrote:
On 4/11/2019
1:58 PM,
agrays...@gmail.com
wrote:
He might
have
been
referring
to a
transformation
to a
tangent
space
where
the
metric
tensor
is
diagonalized
and its
derivative
at that
point in
spacetime
is zero.
Does
this
make any
sense?
Sort of.
Yeah, that's
what he's
doing. He's
assuming a
given
coordinate
system and
some
arbitrary
point in a
non-empty
spacetime. So
spacetime has
a non zero
curvature and
the
derivative of
the metric
tensor is
generally
non-zero at
that
arbitrary
point,
however small
we assume the
region around
that point.
But applying
the EEP, we
can transform
to the
tangent space
at that point
to
diagonalize
the metric
tensor and
have its
derivative as
zero at that
point. Does
THIS make
sense? AG
Yep. That's
pretty much
the defining
characteristic
of a
Riemannian space.
Brent
But isn't it weird
that changing
labels on
spacetime points
by transforming
coordinates has
the result of
putting the test
particle in local
free fall, when it
wasn't prior to
the
transformation? AG
It doesn't put it
in free-fall. If
the particle has EM
forces on it, it
will deviate from
the geodesic in the
tangent space
coordinates. The
transformation is
just adapting the
coordinates to the
local free-fall
which removes
gravity as a
force...but not
other forces.
Brent
In both cases, with and
without
non-gravitational
forces acting on test
particle, I assume the
trajectory appears
identical to an
external observer,
before and after
coordinate
transformation to the
tangent plane at some
point; all that's
changed are the labels
of spacetime points. If
this is true, it's
still hard to see why
changing labels can
remove the
gravitational forces.
And what does this buy
us? AG
You're looking at it the
wrong way around. There
never were any
gravitational forces,
just your choice of
coordinate system made
fictitious forces
appear; just like when
you use a merry-go-round
as your reference frame
you get coriolis forces.
If gravity is a fictitious
force produced by the choice
of coordinate system, in its
absence (due to a change in
coordinate system) how does
GR explain motion? Test
particles move on geodesics
in the absence of
non-gravitational forces,
but why do they move at all? AG
Maybe GR assumes motion but
doesn't explain it. AG
The sciences do not try to
explain, they hardly even try to
interpret, they mainly make
models. By a model is meant a
mathematical construct which,
with the addition of certain
verbal interpretations, describes
observed phenomena. The
justification of such a
mathematical construct is solely
and precisely that it is expected
to work.
--—John von Neumann
Another problem is the
inconsistency of the
fictitious gravitational
force, and how the other
forces function; EM, Strong,
and Weak, which apparently
can't be removed by changes
in coordinates systems. AG
It's said that consistency is
the hobgoblin of small minds. I
am merely pointing out the
inconsistency of the
gravitational force with the
other forces. Maybe gravity is
just different. AG
That's one possibility, e.g
entropic gravity.
What is gets you is it
enforces and explains
the equivalence
principle. And of course
Einstein's theory also
correctly predicted the
bending of light,
gravitational waves,
time dilation and the
precession of the
perhelion of Mercury.
I was referring earlier just
to the transformation to the
tangent space; what
specifically does it buy us;
why would we want to execute
this particular
transformation? AG
For one thing, you know the
acceleration due to
non-gravitational forces in this
frame.
*IIUC, the tangent space is a vector
space which has elements with
constant t. So its elements are
linear combinations of t, x, y, and
z. How do you get accelerations from
such sums (even if t is not
constant)? AG*
*
*
So you can transform to it, put
in the accelerations, and
transform back.
*I see no way to put the
accelerations into the tangent space
at any point in spacetime. AG*
The tangent space is just a patch of
Minkowski space. d/t(dx/dt) =
acceleration.
Brent
--
You received this message because you are subscribed to the Google
Groups "Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send
an email to everything-list+unsubscr...@googlegroups.com
<mailto:everything-list+unsubscr...@googlegroups.com>.
To post to this group, send email to everything-list@googlegroups.com
<mailto:everything-list@googlegroups.com>.
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
