Natural Science Forum / Physics / Relativity / November 2007

When we push a car, it pushes back for a period of *time*.
There is no gorilla on the other end of the car and no
viscosous goo under the wheels to explain this reaction
force so we have to attribute the resistance to our force
to massive objects, remote from the car, that the car can't
communicate with faster than the speed of light.

The curvature that is refered to is deeply bound up in
the work spaces used by tensor calculus. Some texts offer
the "rubber sheet diagram" just as a very coarse illustration
for laypeople.  It has little if any formal significance.

I prefer to say:

Consider all the masses in the universe are pulling you
equally in all directions. That is inertia and is isotropic.

Consider now that a nearby planet spoils that isotropy.
That is gravity and your inertial motion will be toward
the planet.

That still doesn't put the curvature where it really exist
in the formalism but it gets it a bit closer. It also puts the
horse in front of the cart as Einstein suggests:

<<  Already Newton recognized that the
law of  inertia is unsatisfactory
in a context so far unmentioned in this
exposition, namely that it gives no
real cause for the special physical
position of the states of motion of the
inertial frames relative to all other
states of motion. It makes the observable
material bodies responsible for the
gravitational behaviour of a material
point, yet indicates no material cause
for the inertial behaviour of the material
point but devises the cause for it
(absolute space or inertial ether). This
is not logically inadmissible although
it is unsatisfactory. For this reason
E. Mach demanded a modification of the
law of inertia in the sense that the
inertia should be interpreted as an
acceleration resistance of the bodies
against one another and not against "space".
This interpretation governs the expectation
that accelerated bodies have concordant
accelerating action in the same
sense on other bodies (acceleration induction).
This interpretation is even more
plausible according to general relativity
which eliminates the distinction between
inertial and gravitational effects.
It amounts to stipulating that, apart
from the arbitrariness governed by the
free choice of coordinates, the
gm v -field shall be completely determined
by the matter. Mach's stipulation is favoured
in general relativity by the circumstance
that acceleration induction in accordance
with the gravitational field equations really
exists, although of such slight intensity
that direct detection by mechanical experiments
is out of the question. >>
http://nobelprize.org/nobel_prizes/physics/laureates/1921/einstein-lecture.html

Sue...

It is spaceTIME that is curved by gravity.  Does this help?

--
Martin Hogbin

Straight lines in curved space are curved lines in straight space.
  http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov

The essential aspect of gravity in GR is that spaceTIME is curved. In
plotting the trajectory of a small object in spaceTIME, its velocity
becomes an angle in the corresponding space-time plane, and naturally
affects the path of the object.

Tom Roberts

This is a good question, especially as applies to whether lab-frame
acceleration of bodies would be different depending on their velocity.  You
can look for my thread "What is acceleration of particles moving transverse
to field of extended planar mass?", but I quote it here (with some extra
notes in square brackets] for convenience:

I was told, that the lab-frame acceleration of a mass moving transverse to
the field around an extended planar mass (with essentially uniform field
over a wide range) is not the same as what I expect from the "accelerating
elevator" (AE) and my original understanding of the equivalence principle.
In the AE, a mass dropped straight down and a bullet fired horizontally hit
the floor at the same time. That means, the same lab acceleration along g
(which looks higher to the bullet due to time dilation.) What I was told
[by Greg Egan, who seems credible]:
g(moving transverse to g) = g(1 + v^2/c^2).
That is supposedly due to the moving body cutting planes of space time
differently than a simple falling body, etc.

[But like gravitomagnetism, this sort of effect violates the equivalence
principle since it is not being subject to being transformed away, even in a
tiny region, by acceleration.]

But what if you accelerate a ring of vast radius R from rest to rapid
rotation, using up its own mass-energy? The total mass-energy per unit ds of
the ring, seen in the lab, stays the same (and we can use discrete points to
avoid stretching.) In my original understanding, the close to 1/r gravity
field near the ring current therefore stays the same value. That avoids free
energy tricks like raising/lowering parallel static mass rings before/after
acceleration of the first ring.

If the acceleration difference is real, then we can speed up the first ring,
get g(new) = g(rest)*(1 + v^2/c^2), lower in some sandwiching static rings,
decelerate the first ring (keeping the energy there for same mass-energy per
unit), then raise the other rings back out and keep the extra energy. It
would be worth 0.36 mg*delta h [with m the mass of either static ring, and g
the gravity of the accelerated ring at the location of making tiny
displacements to get this result] if the main ring got up to 0.6c, etc.  The
other rings or sets of masses go right towards the spinning ring, there's no
way for corrections to fix energy using their own transverse velocity.  Play
with it some, and maybe you'll see that differential values of g cause
problems.

Comments, anyone?

BTW, below are some references regarding this issue.

http://www.mathpages.com/home/kmath530/kmath530.htm
http://arxiv.org/pdf/gr-qc/0503092
http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.2906v1.pdf

Thanks for all the answers, but I still don't quite get it. Maybe if I
ask it differently...

How does the curved spacetime cause an object to accelerate towards an
other object?

Also wondering, I always read that objects always fall as fast on
earth no matter what their mass is (in vacuum), but shouldn't heavier
objects fall faster?
Maybe not measurable because the difference in mass is rather small
compared to the mass of earth but still, if you'd put 2 earths next to
each other they'd go twice as fast towards each other right?

And you won't likely find the answer in mainstream thinking. The
question is much deeper than the math is capable of reaching. A
mathematical answer only leads circularly back to the same questions,
it being the source of the questions to begin with.

Regardless of any explanation that it's spacetime that is curved
rather than space, it still remains that every macroscopic mass is
composed entirely of electromagnetic quanta, none of which shares the
same path as the others. When two "objects" are at rest wrt each other
initially, then it is only thier centers of mass that are at rest wrt
each other, while the particles of which they are composed are moving
at relativistic speeds wrt each other. In light of this, any equations
that might get these "objects" to accelerate toward one another by
bending spacetime, are reduced to nothing more than an alternative
mathematical approach to ordinary forces in flat spacetime. A true
curvature model, would have to be carried out on the microscopic
domain, and then integrated over all of the particles involved to
account for macroscopic effects. The same macroscopic theory (GR)
cannot be used on these particles to obtain consisent results, since
these particles are electromagnetic masses. The laws govering their
interactions are quite different. If OTOH, you were to postulate that
these particles introduced spacetime curvatures, then integration over
the whole would necessarily produce a set of macroscopic curvatures as
well, but of a different nature.

As an example note that two passing point charges exert an inductive
force on each other that is proportaional to their relative
velocity^2, whereas these same particles, when part of electrical
currents in opposing neutral conductors result in a force between the
conductors that is proportional to velocity^1. The microscopic forces
yeild macroscopic forces, but the two are of different natures. One
spacetime curvature cannot account for both, but rather one must be
found to be more fundamental than the other, and in fact the latter is
a purely macroscopic effect.

Wrt point charges (fundamental particles) it is space that is curved
rather than spacetime. Spacetime is purely Euclidean, as prescribed by
special relativity. Because the interactions of these particles is
conforms to special relativity, and because space is the extension of
em fields, then there is no mass on this level, and this makes GR a
superficial treatment of purely special relativistic effects.

So in effect the spacetime curvature of GR, is simply a set of
equations to which the idea of spacetime curvature is attached, though
that curvature is only imaginary, just as the B field is imaginary.
Both superficial realities.