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.

In article <uloseh$3dbv3$1@dont-email.me>, Luigi Fortunati replied:

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.

--
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currently on the west coast of Canada
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-- Pat Dennis, 2022-04-25

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