Gravity and curvature

[Luigi Fortunati]

Gravity is a force and, therefore, is a vector.

Is spacetime curvature a vector? Does it have a direction and a verse?

Luigi Fortunati

[[Mod. note --
In order to answer questions like this, we need to be precise in
our terms. You write that "Gravity is a force". But what do you mean
by "gravity"? Starting with Newtonian mechanics for simplicity,
"gravity" could plausibly mean any of several things:
* gravitational potential energy (which is a scalar in Newtonian mechanics).
* the Newtonian "little g" (which is a 3-vector at any given position
and time
* the *difference* in the Newtonian "little g" between nearby objects
at a given time; this difference is what you can measure about the
gravitational field if you're in a freely falling elevator. This
difference is a 3-vector which depends on the separation between the
nearby objects,
difference = M * separation
where M is a 3x3 matrix and "*" denotes matrix multiplication. This
3x3 matrix M (which is really a rank 2 tensor) provides a complete
description of the local gravitational field at a given position and
time.

In general relativity (GR) things are (not surprisingly) more complicated.
To fully describe spacetime curvature at an event (a point in space, at
a particular time) requires generalizing the 3x3 matrix (rank 2 tensor) M
to the Riemann curvature tensor, which is a 4x4x4x4 4-dimensional matrix
(really a rank 4 tensor), i.e., it's a set of 4x4x4x4 = 256 numbers.
The Riemann curvature tensor has a bunch of symmetries, so it actually
has only 20 independent components.

So in GR, the best answer to your question is that spacetime curvature "is"
the Riemann curvature tensor. This doesn't have a single "direction" any
more than the 3x3 matrix M has a "direction" in Newtonian mechanics.
-- jt]]

[Luigi Fortunati]

Post by Luigi Fortunati
Gravity is a force and, therefore, is a vector.
Is spacetime curvature a vector? Does it have a direction and a verse?
Luigi Fortunati
[[Mod. note --
In order to answer questions like this, we need to be precise in
our terms. You write that "Gravity is a force". But what do you mean
by "gravity"?

By gravity I mean the Newtonian force F directly proportional to the product of the masses m1 and m2 and inversely proportional to the square of their distance.

Post by Luigi Fortunati
Starting with Newtonian mechanics for simplicity,
* gravitational potential energy (which is a scalar in Newtonian mechanics).
* the Newtonian "little g" (which is a 3-vector at any given position
and time
* the *difference* in the Newtonian "little g" between nearby objects
at a given time; this difference is what you can measure about the
gravitational field if you're in a freely falling elevator. This
difference is a 3-vector which depends on the separation between the
nearby objects,
difference = M * separation
where M is a 3x3 matrix and "*" denotes matrix multiplication. This
3x3 matrix M (which is really a rank 2 tensor) provides a complete
description of the local gravitational field at a given position and
time.

Does all this exclude that, in classical mechanics, gravity is a fundamental force and, therefore, is necessarily a vector?

Post by Luigi Fortunati
In general relativity (GR) things are (not surprisingly) more complicated.
To fully describe spacetime curvature at an event (a point in space, at
a particular time) requires generalizing the 3x3 matrix (rank 2 tensor) M
to the Riemann curvature tensor, which is a 4x4x4x4 4-dimensional matrix
(really a rank 4 tensor), i.e., it's a set of 4x4x4x4 = 256 numbers.
The Riemann curvature tensor has a bunch of symmetries, so it actually
has only 20 independent components.
So in GR, the best answer to your question is that spacetime curvature "is"
the Riemann curvature tensor. This doesn't have a single "direction" any
more than the 3x3 matrix M has a "direction" in Newtonian mechanics.

In the Wikipedia entry "Spacetime" there is the figure of the space-time sheet that curves *downward* when we place a mass on it.

If there are many directions in the GR, why does the elastic sheet sink *always and only* downwards, i.e. towards a single direction?

Luigi Fortunati

[Richard Livingston]

...

Post by Luigi Fortunati
By gravity I mean the Newtonian force F directly proportional to the product of the masses m1 and m2 and inversely proportional to the square of their distance.

...

Post by Luigi Fortunati
Does all this exclude that, in classical mechanics, gravity is a fundamental force and, therefore, is necessarily a vector?

...

Post by Luigi Fortunati
If there are many directions in the GR, why does the elastic sheet sink *always and only* downwards, i.e. towards a single direction?
Luigi Fortunati

As mentioned by the moderator, our current understanding of gravity is considerably more complex and subtle than Newton's concept. Experimental evidence has confirmed all this. In particular, thinking of gravity only as a force is missing some important phenomena.

The current understanding is that space time is not Euclidean everywhere, but can be curved in various ways. The geometry of space-time is described by the metric tensor, which is a 4x4 tensor with 10 unique parameters. The ways that space-time can be distorted is more complex than can be described by a simple vector. However, with the metric tensor one can calculate how objects move from point to point via the covariant derivative. This is an extension of the ordinary derivative to a non-Euclidean space-time. One of these derivatives can calculate what we commonly experience as the "force of gravity", and that is a vector with direction and magnitude. But that is only one of several gravitational effects that we now understand.

To put that more clearly, gravity is not just an attractive force. There are a multitude of effects we now understand to be due to gravity.

In general relativity gravity can manifest in ways other than a simple attractive force. The gravitational red shift is one. Actually, what we regard as a "force" is due to the variation in this red shift with radial position. Geometrically, there is a curvature of the time metric, g_00, in the radial direction. But GR also shows that the radial metric varies in the radial direction. This is a different kind of curvature and it does not result in a "force", but it does affect the paths that free objects follow. As an example. if you could somehow turn off the g_00 curvature (i.e. make g_00 = 1 everywhere), then objects could orbit a black hole at the event horizon at any speed without falling in, and without experiencing anything you would call a force. That is, an object in a circular orbit will behave as if it is going in a straight line, which geometrically it is, even though it returns to the same place every orbit. This is not a force, yet it is a gravitational effect.

Rich L.

[Luigi Fortunati]

Post by Richard Livingston
In general relativity gravity can manifest in ways other than a simple
attractive force. The gravitational red shift is one. Actually, what we
regard as a "force" is due to the variation in this red shift with radial
position.

Gravitational redshift is related to radial direction.
Direction is a property of vectors and, therefore, gravity appears in
vector form also in General Relativity.
If it were a scalar it would have no connection to a specific
direction.

Luigi Fortunati

[Jonathan Thornburg [remove -color to reply]]

Post by Luigi Fortunati
In the Wikipedia entry "Spacetime" there is the figure of the
space-time sheet that curves *downward* when we place a mass on it.
If there are many directions in the GR, why does the elastic sheet
sink *always and only* downwards, i.e. towards a single direction?

As you point out, the rubber-sheet analogy has some serious conceptual
problems. My personal opinion is that this analogy is actively misleading,
and doesn't contrube to any useful understanding of general relativity
(see also <http://xkcd.com/895/>).

Ideally, forget you ever saw the rubber-sheet analogy. :)

<<following text copied from a 2019 posting of mine in this newsgroup>>

In my opinion a better analogy is to consider horizontal motion on the
Earth's surface (which we can treat as being spherical for present
purposes).

Suppose you have two nearby ships on the Earth's (ocean) surface, whose
courses are initially parallel (i.e., d/dt of the distance between them
is zero), and that both ships move along "locally straight" lines, i.e.,
symmetrical hulls and rudders not deflected left or right. Then each
ship will travel along a great circle on the Earth's surface. This
implies that the ships paths will *converge* (i.e., d^2/dt^2 of the
distance between them must be negative).

You could attribute this convergence-of-paths to a mysterious
"Newtonian gravity" ship-to-ship attraction, but it's informative to
instead regard it as a manifestation of the Earth's surface having
"intrinsic curvature", i.e., *not* satisfying the axioms of Euclidean
geometry.
[Here the adjective "intrinsic" means that we're treating
the curvature as an attribute of the 2-dimensional surface
itself; not as an artifact of an embedding in a 3-dimensional
world.]

As supporting evidence for this interpretation, note that the
ship-to-ship acceleration (i.e., d^2/dt^2 of the distance between
nearby ships at the moment they're some standard distance apart) is
universal, independent of the properties of this ship. [This is
analogous to the universality (at a given place and time) of the
free-fall acceleration of test masses in Newtonian gravitation,
regardless of the properties of the free-falling test mass.]

Moreover, we observe that a "locally straight" route [= that taken
by ships & aircraft] from (say) Paris (France) to Vancouver (Canada)
curves far to the North of the origin & destination cities, typically
passing over central to northern Greenland. If we didn't know about
the curvature of the Earth's surface, we might explain this by saying
that there is an equator-attracting "action at a distance" force on
the Earth's surfce, causing ship and aircraft trajectories to be
concave towards the equator for the same reason that a thrown ball's
path is concave towards the ground.

Post by Luigi Fortunati
From a "Newtonian" perspective, this equator-attracting force can

acclerate ships and airplanes "down" towards the equator, and thus
is a real physical force (it can do work in the Newtonian sense).

Post by Luigi Fortunati
From a curved-space perspective, we instead see that these ship and

aircraft simply follow great circles (geodesics).

Ok, that's enough of the analogy. I can think of 2 major ways in
which this analogy fails:
1. Gravitation is actually a phenomenon of curved *spacetime*, not just
of curved *space*. It's this that allows a very weak gravitational
field like that of the Earth, to curve a thrown ball's path so
strongly (from 45-degrees up to 45-degrees down in just a second
or so).
[One of Wheeler's books has a nice sketch showing a
ball's path in 3-D, with axes x, y, and c*time, making
it clear that in a 1-second flight, the thrown ball
has travelled some small number of meters in x and y,
but also 300,000 km in c*time, so the actual *curvature*
of its path in spacetime is very small.]
2. In our analogy, the Earth's surface has *extrinsic* curvature: it's
curved (does not satisfy Euclid's axioms of plane geometry) due to
the way it is embedded as a 2-dimensional surface in a 3-dimensional
world. In contrast, we do not believe that the curvature of
4-dimensional spacetime is due to it's actually being embedded in
some 5-dimensional "hyper-spacetime"; rather, spacetime curvature
is seen as *intrinsic* to spacetime itself.

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] il 21/12/2023 09:19:38 ha=20

Post by Jonathan Thornburg [remove -color to reply]
In my opinion a better analogy is to consider horizontal motion on the
Earth's surface (which we can treat as being spherical for present
purposes).
Suppose you have two nearby ships on the Earth's (ocean) surface, whose
courses are initially parallel (i.e., d/dt of the distance between them
is zero), and that both ships move along "locally straight" lines, i.e.=

,

Post by Jonathan Thornburg [remove -color to reply]
symmetrical hulls and rudders not deflected left or right. Then each
ship will travel along a great circle on the Earth's surface. This
implies that the ships paths will *converge* (i.e., d^2/dt^2 of the
distance between them must be negative).

If there are 2 ships at the equator initially traveling south (and if=20
the rudder is always kept on starboard), they will end up meeting at=20
the south pole and if they initially travel north, they will end up=20
meeting in a completely different place (the pole north).

Instead, if there are 2 elevators whose cables break while they are=20
going up or down, they will always end up meeting in the same place:=20
the center of the Earth!

This happens because there is no force that changes the direction of=20
motion of the 2 ships but the force that reverses the direction of=20
travel of the 2 elevators that are going up when the cables break is=20
there!

Therefore, the comparison between the two conditions is not at all=20
correct, because in one case the forces are there and in the other they=20
are not.

Luigi Fortunati

[Jonathan Thornburg [remove -color to reply]]

In a moderator's note a few days ago, I wrote:
| Starting with Newtonian mechanics for simplicity,
| "gravity" could plausibly mean any of several things:
| * gravitational potential energy (which is a scalar in Newtonian mechanics).
| * the Newtonian "little g" (which is a 3-vector at any given position
| and time
| * the *difference* in the Newtonian "little g" between nearby objects
| at a given time; this difference is what you can measure about the
| gravitational field if you're in a freely falling elevator. This
| difference is a 3-vector which depends on the separation between the
| nearby objects,
| difference = M * separation
| where M is a 3x3 matrix and "*" denotes matrix multiplication. This
| 3x3 matrix M (which is really a rank 2 tensor) provides a complete
| description of the local gravitational field at a given position and
| time.

Post by Luigi Fortunati
Does all this exclude that, in classical mechanics, gravity is a
fundamental force and, therefore, is necessarily a vector?

To respond to this, I want to elaborate on my 3rd bullet point quoted
above. (This is from the perspective of Newtonian mechanics and
gravitation.)

Suppose we're in a freely-falling elevator. What can we measure
about the local gravitational field?

Since our elevator is "freely-falling", if we measure the (Newtonian)
gravitational acceleration ("little g") at our elevator's center of
mass, we'll find the gravitational acceleration to be *zero* there.

If we're restricted to *only* measurements made inside the elevator,
without looking outside, we have no way of knowing the gravitational
acceleration which would be measured if the elevator were at it's
current position but weren't freely falling.

However, we can say a bit more about inside-the-elevator measurements:
if the elevator is of nonzero size (but still small, in a sense I'll
make more precise shortly), then we can measure the gravitational
acceleration at different locations in the elevator. In particular,
let's set up a (non-rotating) Cartesian coordinate system (x,y,z),
with (0,0,0) at our elevator's center of mass. Then the 3 coordinate
components of the measured-in-the-elevator gravitational acceleration
will vary with position like this:
g_x(x,y,z) = M_xx x + M_xy y + M_xz z + O(x^2+y^2+z^2)
g_y(x,y,z) = M_yx x + M_yy y + M_yz z + O(x^2+y^2+z^2)
g_z(x,y,z) = M_zx x + M_zy y + M_zz z + O(x^2+y^2+z^2)

Now I'm going to assume that the elevator is small enough that the
O(x^2+y^2+z^2) terms can be neglected.

Given this assumption, the 3x3 matrix M
[M is really the matrix of coordinate components of
a rank 2 tensor, but we don't need that level of
sophistication here]
tells us *everything* there is to know about the (Newtonian) gravitational
field in the elevator. That is, the matrix M is a complete description
of the gravitational field in the elevator.

So, from the perspective of a freely falling elevator, one could
reasonably say that "gravity" isn't a vector; rather, gravity is a
3x3 matrix (really a rank 2 tensor).

This freely-falling-elevator perspective is a particularly useful
one if we want to move to general relativity, but I won't go there
in this article.

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] il 21/12/2023 09:19:38 ha

Post by Jonathan Thornburg [remove -color to reply]

Post by Luigi Fortunati
Starting with Newtonian mechanics for simplicity,
* gravitational potential energy (which is a scalar in Newtonian mechanics).
* the Newtonian "little g" (which is a 3-vector at any given position
and time
* the *difference* in the Newtonian "little g" between nearby objects
at a given time; this difference is what you can measure about the
gravitational field if you're in a freely falling elevator. This
difference is a 3-vector which depends on the separation between the
nearby objects,
difference = M * separation
where M is a 3x3 matrix and "*" denotes matrix multiplication. This
3x3 matrix M (which is really a rank 2 tensor) provides a complete
description of the local gravitational field at a given position and
time.
Does all this exclude that, in classical mechanics, gravity is a
fundamental force and, therefore, is necessarily a vector?

To respond to this, I want to elaborate on my 3rd bullet point quoted
above. (This is from the perspective of Newtonian mechanics and
gravitation.)
Suppose we're in a freely-falling elevator. What can we measure
about the local gravitational field?

I will gladly answer you if you specify which gravitational field you
are talking to me about.

In the free falling elevator there are 3 masses and 3 gravitational
fields.

There are (1) the mass of the elevator, (2) that of the body inside it
and (3) that of the Earth outside it.

The 3 gravitational fields are:

- that between the mass of the elevator and the mass of the body inside
it

- that between the mass of the Earth and the mass of the elevator

- that between the mass of the Earth and the mass of the body inside
the elevator.

Which of the 3 gravitational fields are you talking about?

Luigi Fortunati

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] il 21/12/2023 09:19:38 ha

Post by Jonathan Thornburg [remove -color to reply]
Since our elevator is "freely-falling", if we measure the (Newtonian)
gravitational acceleration ("little g") at our elevator's center of
mass, we'll find the gravitational acceleration to be *zero* there.

Attention! This acceleration you are talking about is the one between
the center of mass of the elevator and the elevator itself and not the
gravitational acceleration between the Earth and the elevator (which is
never zero).

The center of mass of the elevator does not accelerate relative to the
elevator but accelerates relative to the Earth (gravity is between the
Earth and the entire elevator and not between the elevator and its
center).

Post by Jonathan Thornburg [remove -color to reply]
If we're restricted to *only* measurements made inside the elevator,
without looking outside, we have no way of knowing the gravitational
acceleration which would be measured if the elevator were at it's
current position but weren't freely falling.

It's obvious! If we cannot look outside we do not see the acceleration
between the Earth and the elevator but this does not mean that *this*
acceleration is not there.

We must not confuse the gravitational acceleration (practically
non-existent) between the elevator and the bodies inside it and that
between the Earth and the entire elevator (which is always there, even
if we cannot measure it from the inside).

Luigi Fortunati