The Direction of geodesics

[Luigi Fortunati]

A geodesic passes from A and also from B.

Is the direction from A to B fully equivalent to the direction from B
to A?

Or can it happen that one of the two directions prevails over the
other?

[Richard Livingston]

In general relativity, the physical (spatial) distance from A to B is always
the same as from B to A. This is not true of the time for light travel
however. If A is higher in a gravitational field than B, the round trip
light travel time is longer for A to B to A (from the point of view of A)
than it is for B to A to B (from the point of view of B). For any single
observer the round trip times are the same either way. That may seem
paradoxical, but it is actually true and consistent.

Rich L.


[[Mod. note -- Referring to Richard's opening sentence, I take it that
he is referring to the metric distance integrated along the geodesic
between points A and B. This is also known as the "geodesic distance".
-- jt]]

[Luigi Fortunati]

Richard Livingston mercoledě 07/09/2022 alle ore 05:12:12 ha scritto:
>> A geodesic passes from A and also from B.
>>
>> Is the direction from A to B fully equivalent to the direction from B
>> to A?
>>
>> Or can it happen that one of the two directions prevails over the
>> other?
>
> In general relativity, the physical (spatial) distance from A to B is always
> the same as from B to A. This is not true of the time for light travel
> however. If A is higher in a gravitational field than B, the round trip
> light travel time is longer for A to B to A (from the point of view of A)
> than it is for B to A to B (from the point of view of B). For any single
> observer the round trip times are the same either way. That may seem
> paradoxical, but it is actually true and consistent.

I absolutely agree with what you wrote.

But how does all this affect the direction of motion that a body
decides to take when it is left free to follow its geodesic?

I simplify with my animation
https://www.geogebra.org/m/zdevssyz
where there is an elevator that (initially) cannot follow its geodesic
since it is constrained at point A.

If we eliminate the constraint (Start) the elevator starts moving
towards B to follow its geodesic which I have highlighted with the blue
dotted line.

This blue line has no limits and has no privileged directions, it is
not a vector that goes in an obligatory direction.

One direction is as good as the other.

But the elevator always goes in one direction only, the one that goes
down, towards point B.

And it never goes up.

What is it that forces the elevator to always get started downwards and
never upwards?

What has the down direction more than the up direction?

In other words, are geodesics open lines or are they vectors with one
direction privileged over the other?

[Stefan Ram]

Luigi Fortunati <fortuna...@gmail.com> writes:

>Richard Livingston mercoledi 07/09/2022 alle ore 05:12:12 ha scritto:
>>In general relativity, the physical (spatial) distance from A to B is always
>>the same as from B to A. This is not true of the time for light travel
>>however. If A is higher in a gravitational field than B, the round trip
>>light travel time is longer for A to B to A (from the point of view of A)
>>than it is for B to A to B (from the point of view of B). For any single
>>observer the round trip times are the same either way. That may seem
>>paradoxical, but it is actually true and consistent.
>I absolutely agree with what you wrote.

Richard wrote about two points in /space/.

In GR, a geodesic usually is a curve in /space-time/.

It's points are not (x, y, z), but (t, x, y, z).
One should not think of its end points as being two points in space!

A geodesic has no direction by itself.

[Luigi Fortunati]

Stefan Ram mercoledě 07/09/2022 alle ore 15:41:17 ha scritto:
> Richard wrote about two points in /space/.
>
> In GR, a geodesic usually is a curve in /space-time/.
>
> It's points are not (x, y, z), but (t, x, y, z).
> One should not think of its end points as being two points in space!

Okay, let's talk about two events in spacetime.

Event A is the lift's departure from point A and event B is the lift's
arrival at point B.

Why does the elevator (when it starts) go down (i.e. it goes from event
A to event B) and doesn't go up (where event B is not there)?

> A geodesic has no direction by itself.

On this I absolutely agree, however the elevator of my animation
https://www.geogebra.org/m/zdevssyz
it goes only and always in one direction and never in the other, why?

[Richard Livingston]

On Wednesday, September 7, 2022 at 2:43:54 PM UTC-5, Luigi Fortunati wrote:

> Richard Livingston mercoledÄ=9B 07/09/2022 alle ore 05:12:12 ha scritto:
> >> A geodesic passes from A and also from B.
> >>
> >> Is the direction from A to B fully equivalent to the direction from B
> >> to A?
> >>
> >> Or can it happen that one of the two directions prevails over the
> >> other?
> >
> > In general relativity, the physical (spatial) distance from A to B is always
> > the same as from B to A. This is not true of the time for light travel
> > however. If A is higher in a gravitational field than B, the round trip
> > light travel time is longer for A to B to A (from the point of view of A)
> > than it is for B to A to B (from the point of view of B). For any single
> > observer the round trip times are the same either way. That may seem
> > paradoxical, but it is actually true and consistent.
> I absolutely agree with what you wrote.
>
> But how does all this affect the direction of motion that a body
> decides to take when it is left free to follow its geodesic?
>

...

> What is it that forces the elevator to always get started downwards and
> never upwards?
>
> What has the down direction more than the up direction?
>
> In other words, are geodesics open lines or are they vectors with one
> direction privileged over the other?

There are many geodesics passing through any given point (event) in
space-time. What you are asking is how does a velocity 4-vector evolve.
The velocity 4-vector for a "stationary" object is (c, 0, 0, 0) in flat
space-time. In the vicinity of a large mass M it is (c(1-2M/r)^-1/2, 0,
0, 0) (ref eqn. 20.58 in "Gravity: an introduction to Einsteins general
relativity" by J. B. Hartle, 2003). The most rigorous way to calculate
this is to calculate the covariant derivative in the metric you are
concerned with. For the Schwarzschild metric this gives an acceleration
vector in the radial direction:.

(Revtex code:
a^\alpha = \GAMMA^{\alpha}_{tt} \left( u^t \right)
)

where \GAMMA is a Christoffel symbol, which gives
a = (0, M/r^2, 0, 0)

This is actually the acceleration necessary to hold the object
stationary at this radius. If the object was released it would be in
free fall (i.e. no acceleration in the objects frame). In terms of the
stationary observer the free falling object would accelerate at a = (0,
-M/r^2, 0, 0) i.e. a radial acceleration towards the mass M.

This probably doesn't answer your question (it didn't for me) because
the covariant derivative, while precise and correct, is almost opaque to
intuitive understanding of the physics, which I believe is what you are
asking.

The covariant derivative is related to the concept of parallel
transport. If we introduce a simple idea from quantum mechanics we can
more easily visualize the parallel transport of the particle wave
function in time, and recognize the resulting acceleration towards the
mass.

In relativity a particle with mass m has energy mc^2. In QM this is a
frequency mc^2/h, which I will call f = 1/T. For an object at rest, by
definition the wave function is in phase over some local region of
space. In terms of flat space-time the phase of this wave function
advances by 2pi every time interval T. This happens over the local
region at the same time as the g_{00} metric component is a constant
everywhere in flat space-time.

However near a large mass M, due to the gravitational red shift, not
every location has the same frequency. A region closer to the mass M
will appear to advance in phase more slowly, and a region further out
will advance more quickly. As a result the wave function for a
stationary object of mass m, while initially in phase over some local
area, will gradually become more and more out of phase along a radial
axis. This is entirely because the g_{00} metric component varies along
the r coordinate direction. The change in phase along r is, in QM, the
momentum of the particle, and this momentum gradually increases with
time. It is easy to show that this gives the same answer as the
covariant derivative.

Does that help?

Rich L.

[Stefan Ram]

Luigi Fortunati <fortuna...@gmail.com> writes:
>it goes only and always in one direction and never in the other, why?

I currently can't go on websites, but maybe what is intended
is this: a particle is placed on a geodesic compatible with
that particle. The particle now can move on the geodesic,
but how does it know in which direction as the geodesic has
no preferred direction?

The answer might be: While a geodesic has no preferred
direction, the time coordinate has. The particle moves into
that direction where its time coordinate is growing - assuming
the time coordinate was chosen in such a natural manner that
it grows towards the future of the given universe.

[Luigi Fortunati]

No, unfortunately it doesn't help me.

I can't understand why frequency, wave function, MQ, parallel
transport, etc. can make the down direction prefer over the up
direction.

But thank you all the same for your erudite explanation, a little too
technical for me.

Luigi Fortunati

[Luigi Fortunati]

Stefan Ram venerd=EC 09/09/2022 alle ore 08:30:00 ha scritto:
> I currently can't go on websites, but maybe what is intended
> is this: a particle is placed on a geodesic compatible with
> that particle. The particle now can move on the geodesic,
> but how does it know in which direction as the geodesic has
> no preferred direction?
>
> The answer might be: While a geodesic has no preferred
> direction, the time coordinate has.

The time coordinate is part of the geodesic.

If the time coordinate has a preferred direction, the geodesic also has
a preferred direction.

[[Mod. note --
A useful mental model for a geodesic at a point is motion on the surface
of the Earth. That is, starting at some specified point, we can move
in a specified compass direction (e.g., you might start out moving due
west). If, once moving, we don't turn left and you don't turn right,
our motion will be along a geodesic on the Earth's surface.

Thinking of our starting point again, you could have started moving
in any direction (e.g., instead of bearing 090 degrees = due west, we
could have chosen any other compass bearing). So there are a whole
(infinite) family of possible geodesics passing through that starting
point (one for each possible compass bearing).

If I understand you correctly, you're asking "once a particle is moving,
how does it know to continue moving in that direction?". The answer is
basically conservation of momentum: unless there is some external force
pushing on the particle, it's going to continue moving in the *same*
direction it was already moving in.

In terms of geodesics in relativity (the original context of your question),
it's essential to realise that (as others have noted) the trajectoris of
free particles are geodesics in *spacetime*, not geodesics in *space*.
That means the most useful particle velocity to think about is the
4-velocity, which is *never* zero (you're always moving forward in time),
and corresponding momentum is the 4-momentum, which is also never zero.
(Recall that in relativity, even a zero-rest-mass particle like a photon
still has a nonzero momentum so long as it carries nonzero energy.)
-- jt]]

[Luigi Fortunati]

Luigi Fortunati domenica 11/09/2022 alle ore 11:38:28 ha scritto:
> The time coordinate is part of the geodesic.
>
> If the time coordinate has a preferred direction, the geodesic also has
> a preferred direction.
>
> [[Mod. note --
> A useful mental model for a geodesic at a point is motion on the surface
> of the Earth. That is, starting at some specified point, we can move
> in a specified compass direction (e.g., you might start out moving due
> west). If, once moving, we don't turn left and you don't turn right,
> our motion will be along a geodesic on the Earth's surface.
>
> Thinking of our starting point again, you could have started moving
> in any direction

Exactly, I could have started moving in any direction and, among all, I
would have been forced, finally, to choose only one.

> in any direction (e.g., instead of bearing 090 degrees = due west, we
> could have chosen any other compass bearing). So there are a whole
> (infinite) family of possible geodesics passing through that starting
> point (one for each possible compass bearing).

That's right, the choice is one of an infinite number of possible
geodesics.

> If I understand you correctly, you're asking "once a particle is moving,
> how does it know to continue moving in that direction?".

No! I am asking how does it know how to "start" to move (along
4-geodesic) and not how to "continue" to move.

Once the particle has moved, its choice (among the infinite possible)
has already been made!

> The answer is basically conservation of momentum

There is no conservation of 4-momentum when the elevator cables break
and the elevator goes from constrained condition to free fall.

> unless there is some external force pushing on the particle, it's going to
> continue moving in the *same* direction it was already moving in.

The elevator where the cables break, does not keep moving in the *same*
4-direction it was moving before.

> In terms of geodesics in relativity (the original context of your question),
> it's essential to realise that (as others have noted) the trajectoris of
> free particles are geodesics in *spacetime*, not geodesics in *space*.
> That means the most useful particle velocity to think about is the
> 4-velocity, which is *never* zero (you're always moving forward in time),
> and corresponding momentum is the 4-momentum, which is also never zero.

Ok, let's talk about 4-momentum.

When the cables break, the elevator does not retain the 4-momentum it
had before but switches from a certain 4-momentum (the one it had when
standing at the floor) to another completely different 4-momentum (the
one it assumes during free fall).

[Stefan Ram]

Luigi Fortunati <fortuna...@gmail.com> writes:
>The time coordinate is part of the geodesic.

When I wrote my last post, I thought of a geodesic as being
a mere subset of space-time, i.e., a set of events like a
line is a set of points, and a mere line has no direction.

But you may be right that it can be seen as having more
structure. Maybe it depends on the details of how a specific
textbook defines it. For the book "Gravitation" by Misner
et al. it's a "curve". I can't find a formal definition for
"curve" in "Gravitation", but it seems to be a continuous
mapping from the real numbers into space-time. So it has a
natural positive direction.

[[Mod. note -- You're right, in relativity a curve is a continuous
mapping from the real numbers (or a closed interval of real numbers)
into spacetime.
-- jt]]

[Richard Livingston]

On Sunday, September 11, 2022 at 6:38:33 PM UTC-5, Luigi Fortunati wrote:
> Stefan Ram venerd=EC 09/09/2022 alle ore 08:30:00 ha scritto:
> > I currently can't go on websites, but maybe what is intended
> > is this: a particle is placed on a geodesic compatible with
> > that particle. The particle now can move on the geodesic,
> > but how does it know in which direction as the geodesic has
> > no preferred direction?
> >
> > The answer might be: While a geodesic has no preferred
> > direction, the time coordinate has.

> The time coordinate is part of the geodesic.
>

> If the time coordinate has a preferred direction, the geodesic also has
> a preferred direction.

...

This is somewhat over simplified, but you can think of the g_00 metric component
as similar to a potential, in that it multiplies the particle energy rather than
adds/subtracts like a true potential. Never the less, the "potential" decreases
as you get closer to the gravitating mass, and thus objects feel a "force"
in that direction.

When you really understand how it works, you will find that the curvature
of the metric that causes gravitational acceleration is the time metric,
g_00, that varies with radius. Ultimately this is what causes the acceleration
to be toward the large mass.

The radial metric, g_11, also varies with radius, but this is not an intrinsic
curvature that causes an acceleration. It does affect orbits, particularly
close to the event horizon, and is the reason there are no stable orbits
w/in 1.5 Schwarzshild radii of the event horizon.

Rich L.

[[Mod. note --

For massive particles the innermost stable *circular* orbit in
Schwarzschild is a bit farther out, at r=6M. There are circular orbits
inside r=6M, but they're unstable.

For massless particles (photons et al), there's only one circular orbit,
at r=3M. It's unstable.
-- jt]]

[Jonathan Thornburg [remove color- to reply]]

Luigi Fortunati <fortuna...@gmail.com> wrote:
[[about a particle]]

> I am asking how does it know how to "start" to move (along
> 4-geodesic) and not how to "continue" to move.

The simple answer is that it's always been moving. When the particle
first came into existence, it was already moving in spacetime, so it
already had a nonzero 4-velocity.

> There is no conservation of 4-momentum when the elevator cables break
> and the elevator goes from constrained condition to free fall.

In special relativity we start with the notion of a worldline, a particle's
path through spacetime. Introducing $(t,x,y,z)$ coordinates, we can write
this as
t = t(tau)
x = x(tau)
y = y(tau)
z = z(tau)
where tau is the particle's proper time, i.e., the time measured by an
ideal clock carried along with the particle.

The particle's 4-velocity and 4-acceleration are then defined as
$u = (dt/d\tau, dx/d\tau, dy/d\tau, dz/d\tau)$
and
$a = du/d\tau = (d^2t/d\tau^2, d^2x/d\tau^2, d^2y/d\tau^2, d^2z/d\tau^2)$
Finally, we can define the particle's 4-momentum (assuming the particle
to have constant mass $m$) as
$p = mu = (m dt/d\tau, m dx/d\tau, m dy/d\tau, m dz/d\tau)$

[In general relativity things are a bit messier: we define a worldline
in the same way, but 4-velocity and 4-acceleration are now defined in
terms of "covariant derivatives" or "absolute derivatives", which
effectively add some extra terms which depend on the Christoffel symbols.
4-momentum is still given by $p = mu$.]

See
https://en.wikipedia.org/wiki/World_line
https://en.wikipedia.org/wiki/Four-velocity
https://en.wikipedia.org/wiki/Four-acceleration
https://en.wikipedia.org/wiki/Four-momentum
for more on these concepts.

[Note that some authors -- including those of at least the 4-velocity
and 4-acceleration Wikipedia articles -- use a different convention than
I use; they take the time component of 4-vectors to have an extra factor
of c, i.e., they define
$u = (c dt/d\tau, dx/d\tau, dy/d\tau, dz/d\tau)$
$p = (mc dt^2/d\tau^2, m dx^2/d\tau^2, m dy^2/d\tau^2, m dz^2/d\tau^2)$
I'm omitting these extra factors of c.]

When we say that 4-momentum is conserved, that's shorthand for saying
that a particle's 4-momentum is constant *if* there is no net force
acting on the particle.

To discuss your elevator example, we need to decide how we're going to
model gravity, either
(a) Newtonian mechanics or special relativity, in which we model gravity
as a force acting in flat (Minkowski) spacetime, or
(b) general relativity, in which we model gravity as a manifestation
of spacetime curvature.

> There is no conservation of 4-momentum when the elevator cables break
> and the elevator goes from constrained condition to free fall.

The elevator's movement is always governed by the relativistic version
of Newton's 2nd law, which (taking the elevator's mass $m$ to be constant)
is
a = F_net/m
where a is again the elevator's 4-acceleration, and F_net is the net
4-force acting on the elevator.

In perspective (a) [where gravity is a force in flat spacetime]:
--> Before the cable breaks, there are two forces acting on the elevator:
gravity: F=mg downwards
and
cable tension: F=mg upwards
so F_net = 0, the elevator's 4-momentum p is constant, and the elevator's
4-acceleration is 0.
--> After the cable breaks, there is one force acting on the elevator:
gravity: F=mg downwards
so F_net = mg (pointing down) and hence the elevator's 4-momentum p
changes and the elevator's 4-acceleration is nonzero.

In perspective (b) [where gravity is a manifestation of spacetime curvature]:
--> Before the cable breaks, there is one force acting on the elevator:
cable tension: F=mg upwards
so the elevator's 4-acceleration is nonzero (pointing upwards) and
the elevator's path is not a geodesic. This is consistent with the
elevator remaining stationary (i.e.,
x(tau) = constant
y(tau) = constant
z(tau) = constant
is a solution of Newton's 2nd law) because of the extra
Christoffel-symbol terms in the covariant derivatives in the
definition of 4-acceleration.
--> After the cable breaks, there are no forces acting on it, so it moves
along a geodesic, with zero 4-acceleration. That geodesic path starts
(at the moment the cable breaks) with the 4-velocity of the elevator
being at rest (4-velocity having zero spatial components); at later
times the elevator moves downwards.



I wrote [in the context of *general* relativity, where we model gravity
via spacetime curvature, so gravity is *not* a "force"]

>> unless there is some external force pushing on the particle, it's going to
>> continue moving in the *same* direction it was already moving in.

> The elevator where the cables break, does not keep moving in the *same*
> 4-direction it was moving before.

That's right. The elevator's 4-velocity will change after the cable
breaks. In special relativity we say that that change is due to an
external force (gravity) acting on the elevator. In general relativity
we say that that change is due to geodesic motion in a curved spacetime.
It's the same (changing) 4-velocity either way, we're just describing
it differently.

> Ok, let's talk about 4-momentum.
>
> When the cables break, the elevator does not retain the 4-momentum it
> had before but switches from a certain 4-momentum (the one it had when
> standing at the floor) to another completely different 4-momentum (the
> one it assumes during free fall).

Let's work this out in the context of special relativity.
[Doing this in general relativity would be a bit messier.]

Let's take the time the cable breaks to be t=0, and let's take our (x,y,z)
coordinates to be such that the elevator is at rest at x=y=z=0 before
the cable breaks. Finally, let's orient our (x,y,z) coordinates such
that the external gravitational accelration g points in the -z direction.

To simplify the computation, let's assume that the elevator's 3-velocity
is much less than the speed of light. This is fine for investigating what
happens around the time when the cable breaks. This assumption means that
tau = t, so that the time component of 4-velocity is just 1.

Then Newtonian mechancis tells us that the elevator's 3-velocity (the usual
velocity of Newtonian mechanics) is
{ (0 ,0,0) if t <= 0
v(t) = {
{ (-gt,0,0) if t > 0
while the elevator's 4-velocity is
{ (1,0 ,0,0) if t <= 0
u(t) = {
{ (1,-gt,0,0) if t > 0
and hence the elevator's 4-momentum is
{ (m,0 ,0,0) if t <= 0
p(t) = m u(t) = {
{ (m,-mgt,0,0) if t > 0

You can see that v, u, and p are all continuous functions of time.

If we were to analyze the dynamics in general relativity we'd get the same
answers for $v(t)$, $u(t)$, and $p(t)$, but the calculations to get them
would be messier.

[Luigi Fortunati]

Jonathan Thornburg [remove color- to reply] marted=EC 13/09/2022 alle ore
08:20:03 ha scritto:

> Luigi Fortunati <fortuna...@gmail.com> wrote:
> [[about a particle]]
>> I am asking how does it know how to "start" to move (along
>> 4-geodesic) and not how to "continue" to move.
>
> The simple answer is that it's always been moving. When the particle
> first came into existence, it was already moving in spacetime, so it
> already had a nonzero 4-velocity.

Ok, in spacetime the elevator has always been in motion, even when it
is stopped at the floor.

Okay, it's true that his 4-velocity has always been non-zero.

But, when the cables break and the elevator goes into free fall, its
4-velocity changes (i.e. it's a new different 4-velocity that wasn't
there before) or is still the same as it was when it was bound to the
floor?

[[Mod. note -- I think I answered your question in a recent message
which may not have reached you yet:

Let's take the time the cable breaks to be t=0, and let's take our (x,y,z)
coordinates to be such that the elevator is at rest at x=y=z=0 before
the cable breaks. Finally, let's orient our (x,y,z) coordinates such
that the external gravitational accelration g points in the -z direction.

To simplify the computation, let's assume that the elevator's 3-velocity
is much less than the speed of light. This is fine for investigating what
happens around the time when the cable breaks. This assumption means that
tau = t, so that the time component of 4-velocity is just 1.

Then Newtonian mechancis tells us that the elevator's 3-velocity (the usual
velocity of Newtonian mechanics) is
{ (0 ,0,0) if t <= 0
v(t) = {
{ (-gt,0,0) if t > 0
while the elevator's 4-velocity is
{ (1,0 ,0,0) if t <= 0
u(t) = {
{ (1,-gt,0,0) if t > 0
and hence the elevator's 4-momentum is
{ (m,0 ,0,0) if t <= 0
p(t) = m u(t) = {
{ (m,-mgt,0,0) if t > 0

You can see that v, u, and p are all continuous functions of time.

-- jt]]

[Luigi Fortunati]

Richard Livingston luned=EC 12/09/2022 alle ore 07:36:11 ha scritto:
>> The time coordinate is part of the geodesic.
>>
>> If the time coordinate has a preferred direction, the geodesic also has
>> a preferred direction.
> ...
>
> This is somewhat over simplified, but you can think of the g_00 metric
> component as similar to a potential, in that it multiplies the particle
> energy rather than adds/subtracts like a true potential. Never the less, the
> "potential" decreases as you get closer to the gravitating mass, and thus
> objects feel a "force" in that direction.

Why is "force" in quotes?

If you feel it, it is a force without quotes.


[[Mod. note --
In the context of general relativity (GR) you don't feel gravity as
a force.

In GR the apparent downward "force" which I feel right now (sitting in
a chair in a room near the Earth's surface) is understood not as a
gravitational "force", but rather as follows:
(a) An inertial reference frame (IRF) at my location is by
definition in free-fall, and therefore must be accelerating
downwards at 9.8 m/s^2 with respect to the Earth's surface.
(b) Therefore, my chair (which is stationary with respect to the Earth's
surface) is accelerating *upwards* at 9.8 m/s^2 with respect to
an IRF at my location.
(c) Because my chair is accelerating upwards at 9.8 m/s^2 with
respect to an IRF, my chair must push up on my body in order
to accelerate my body upwards at 9.8 m/s^2 with respect to an IRF.
(d) More generally, because a coordinate system fixed to the Earth's
surface is accelerating upwards at 9.8 m/s^2 with respect to an
IRF, every object feels an apparent "force" pointing downwards.
-- jt]]

[Rock Brentwood]

On Wednesday, September 7, 2022 at 1:22:46 AM UTC-5, Luigi Fortunati wrote:

In any metric geometry, around each point is a neighborhood U,
such that between any two points A and B in U
lies a unique geodesic between A and B.
Uniqueness means it's therefore the same regardless of direction.

This even includes non-Riemannian geometries,
like the chronogeometry of Newtonian gravity,
which can be regarded as residing on a light cone
of the 4+1 dimensional pseudo-Riemannian chronogeometry
given by the metric whose line element is

dx^2 + dy^2 + dz^2 + 2 dt du - 2V dt^2

where V = -GM/r (with r^2 = x^2 + y^2 + z^2)
is the gravitational potential per unit mass.

This also works perfectly well for multi-body potentials,
V = -sum_a (-GM_a/|r - r_a|).

Its geodesics comprise the orbital motions of Newtonian gravity
AND the geometrical instantaneous geodesics of space at any instant.

You may be hard-pressed, in this case, to account for
what the corresponding geodesic *distances* are,
since everything is zero on a light cone!
All the geodesics for Newtonian gravity
are null curves in 4+1 dimensions.
But it's the same in both directions.

This may be compared to the Schwarzschild metric
which is equivalently given as light cones
in the 4+1 dimensional chronogeometry
whose metric has the following line element

dx^2 + dy^2 + dz^2 + 2 dt du - 2V dt^2
+ alpha du^2 - 2 alpha V dr^2/(1 + 2 alpha V)

where dr^2 = (x dx + y dy + z dz)^2/r^2 and alpha = (1/c)^2.

This is a deformation of the Newtonian chronogeometry,
from alpha = 0 to alpha > 0.
Here, however, V = -GM/r is only for one body, not many.
There may exist similar deformations for the many-body cases.

The geodesic distances on the light cones
in the (4+1)-dimensional chronogeometry are all 0,
since all the curves on the light cones are null.
The light cones, however (unlike the Newtonian case)
comprise metric geometries in their own right -
that is: the Schwarzschild chronogeometries,
whose the metric yields - for time-like curves -
a proper time equal to s = t + alpha u.
It's the same in both directions.

In the alpha = 0 case, which is Newtonian gravity,
the (3+1)-dimensional chronogeometry continues to have
s = t + alpha u as an invariant ... which is now just s = t.
So, it can still be taken as a metric of sorts for time-like curves.
Ordinary time is proper time for Newtonian gravity.

[Luigi Fortunati]

Luigi Fortunati il 15/09/2022 10:12:38 ha scritto:

> Richard Livingston lunedi 12/09/2022 alle ore 07:36:11 ha scritto:
>>> The time coordinate is part of the geodesic.
>>>
>>> If the time coordinate has a preferred direction, the geodesic also has
>>> a preferred direction.
>> ...
>>
>> This is somewhat over simplified, but you can think of the g_00 metric
>> component as similar to a potential, in that it multiplies the particle
>> energy rather than adds/subtracts like a true potential. Never the less,
>> the "potential" decreases as you get closer to the gravitating mass, and
>> thus objects feel a "force" in that direction.
>
> Why is "force" in quotes?
>
> If you feel it, it is a force without quotes.
>
> [[Mod. note --
> In the context of general relativity (GR) you don't feel gravity as
> a force.
>
> In GR the apparent downward "force" which I feel right now (sitting in
> a chair in a room near the Earth's surface) is understood not as a
> gravitational "force", but rather as follows:
> (a) An inertial reference frame (IRF) at my location is by

> definition in free-fall...

But is it really like that?

Instead of considering who is on the Earth's surface, consider person A
in the cubicle at the center of the Earth watching elevator B fall
freely.

Person A says: I am in an inertial reference frame and, therefore,
elevator B (which is accelerating towards me) is an accelerated
reference frame.

The person in elevator B sees cubicle A at the center of the Earth and
says: I am in an inertial reference frame and, therefore, cubicle A
(which is accelerating towards me) is an accelerated reference frame.

Who is right?
1) Only A is right
2) Only B is right
3) They are both right
4) They are both wrong
5) A is a little more right than B
6) B is a little more right than A.

For me, (4) is correct because neither of the two reference frame are
totally inertial and (5) because the tidal forces of the cubicle are
infinitely smaller than those (already negligible) of the elevator.

And for you?

Luigi Fortunati

[Tom Roberts]

On 1/1/24 2:18 AM, Luigi Fortunati wrote:
> [...] But is it really like that?

We humans have no hope of knowing "what is really happening". The best
we can do is formulate models of what is happening, apply them to
various physical situations, and improve them -- this is called science.
Guessing "what is really happening" is the province of theology,
astrology, and other arcane wastes of time.

> [... confused writing]

If you want to understand General Relativity, you must STUDY it. Posting
random questions in a newsgroup is outrageously inadequate. Get a good
book and STUDY. You have never posted a question that could not be
answered by yourself if you had STUDIED.

Note in particular you must distinguish between various types of
acceleration: proper acceleration, coordinate acceleration in different
coordinate systems, and 4-acceleration. By just using the term
"acceleration", and failing to distinguish among the different types,
the PUNS destroy any argument you make.

Hint: observer A at the center of the earth would have zero
proper acceleration, as the earth is in freefall [@]. Observer
B on the surface has a proper acceleration of 9.8m/s^2 upward,
due to the force exerted on them by the surface. If either
observer extends their coordinates to the other [#], each would
say the other observer has zero acceleration in their
coordinates.

[@] This is really an approximation in GR, valid only
for "small" objects. Compared to the sun, the earth is
indeed small, and this is an excellent approximation.

[#] extending such coordinates is problematical,
so just assume it can be done without distortions.

Note that "all physics is local" [Einstein and others], so these various
descriptions of accelerations of distant objects never appear in the
laws of physics. They are useful only for personal amusement.

Tom Roberts

[Luigi Fortunati]

Tom Roberts il 02/01/2024 09:17:26 ha scritto:
> Note that "all physics is local" [Einstein and others], so these various
> descriptions of accelerations of distant objects never appear in the
> laws of physics.

No distant objects.

The two little men (at the center of the Earth and in the elevator in
free fall) pass in front of each other in acceleration (when
approaching) and in deceleration (when moving away) and, therefore,
they are not far away at all: they are 'next to each other.

During their meeting, they cannot perceive the force of gravity but
they can easily measure their mutual gravitational acceleration (even
in 4 or more dimensions).

Luigi Fortunati


[[Mod. note -- A corollary of what Tom Roberts wrote is that our modern
conceptions of physics involve *local* equations.

In the situation you describe, the two little men's relative velocity
when they meet (for later reference, I'll call this velocity V) is the
integrated result of different *local* accelerations with respect to
the Earth over their past trajectories. When the two little men meet,
their relative velocity V is effectively an "initial condition" for
their dynamics right around the meeting -- the relative velocity could
just as well have come from (say) the "free fall" elevator starting out
at rest with respect to the Earth a kilometer away, and firing a rocket
to accelerate to the velocity V with respect to the Earth.

In contrast, the two little men can make a direct *local* measurement
of their differential gravitational acceleration, which is a reflection
of the local gravitational field and is unaffected by initial conditions.
-- jt]]

[Luigi Fortunati]

Luigi Fortunati il 03/01/2024 03:49:09 ha scritto:
> Tom Roberts il 02/01/2024 09:17:26 ha scritto:
>> Note that "all physics is local" [Einstein and others], so these various
>> descriptions of accelerations of distant objects never appear in the
>> laws of physics.
>
> No distant objects.
>
> The two little men (at the center of the Earth and in the elevator in
> free fall) pass in front of each other in acceleration (when
> approaching) and in deceleration (when moving away) and, therefore,
> they are not far away at all: they are 'next to each other.
>
> During their meeting, they cannot perceive the force of gravity but
> they can easily measure their mutual gravitational acceleration (even
> in 4 or more dimensions).
>
> Luigi Fortunati
>
> [[Mod. note -- A corollary of what Tom Roberts wrote is that our modern
> conceptions of physics involve *local* equations.

If the two little men send each other radar signals, each of them
*measures* the acceleration of the approaching other with their own
light and their own watch, throughout the entire journey from the
surface of the Earth to its center.

Therefore, they are local measurements (carried out in your own
reference) and not at a distance.

And there is no clock synchronization to do: each of them only uses
their own.

Luigi Fortunati.


[[Mod. note -- Radar is not a local measurement: it sends a light/radio
signal out to a distant target, and some of that signal is reflected and
(a tiny fraction of it) returns to the sender. "Local" means *everything*
(including any light/radio signals) fits in an elevator.
-- jt]]

[Luigi Fortunati]

Luigi Fortunati il 04/01/2024 16:49:56 ha scritto:
> [[Mod. note -- Radar is not a local measurement: it sends a light/radio signal out to a distant target, and some of that signal is reflected and (a tiny fraction of it) returns to the sender. "Local" means *everything* (including any light/radio signals) fits in an elevator. -- jt]]

Ok, radar is not a local measurement because it measures distances between distant bodies but they certainly serve to establish whether two bodies are moving away or closer to each other.

In my drawing
https://www.geogebra.org/m/nnghqpg6
there is spaceship A stopped at the Earth's surface, spaceship B stopped at the center of the Earth and spaceship C in free fall.

I study Newton's physics and I learn that the two spaceships A and B are inertial reference systems because they are not accelerating anywhere and this convinces me because, in fact, neither of the 2 spaceships is accelerating.

I also learn that spaceship C (in free fall) is an accelerated reference system because it is accelerating and this also convinces me.

Then I study Einstein's physics and learn that only spaceship B maintains its inertial status, while spaceship A (despite being stationary) becomes an accelerated reference system and spaceship C (despite accelerating) becomes a reference system inertial.

And all this generates an inconsistency: how can the 2 spaceships A and B be one accelerated and the other inertial if they never move away or get closer to each other?

And again: how can the 2 spaceships C and B both be inertial reference systems if they are accelerating towards each other?

Luigi Fortunati

[Jonathan Thornburg [remove -color to reply]]

Luigi Fortunati <fortuna...@gmail.com> wrote:
> In my drawing
> https://www.geogebra.org/m/nnghqpg6
> there is spaceship A stopped at the Earth's surface, spaceship B stopped at the center of the Earth and spaceship C in free fall.
>
> I study Newton's physics and I learn that the two spaceships A
> and B are inertial reference systems because they are not accelerating
> anywhere and this convinces me because, in fact, neither of the 2
> spaceships is accelerating.
>
> I also learn that spaceship C (in free fall) is an accelerated reference
> system because it is accelerating and this also convinces me.
>
> Then I study Einstein's physics and learn that only spaceship B maintains
> its inertial status, while spaceship A (despite being stationary) becomes
> an accelerated reference system and spaceship C (despite accelerating)
> becomes a reference system inertial.
>
> And all this generates an inconsistency: how can the 2 spaceships A and B
> be one accelerated and the other inertial if they never move away or get
> closer to each other?

The answer is that in general relativity, inertial reference frames in a
curved spacetime (i.e., one that's not completely empty) are only *local*.
Spacetime curvature (a.k.a "gravitation") is what prevents us from extending
local inertial reference frames to larger ("global") sizes. So in this
case, the fact that the A-B distance doesn't change despite B being inertial
and A not being inertial is a reflection of spacetime curvature (caused by
the Earth's mass).



Recalling the ships-moving-on-the-Earth's-surface analogy for general
relativity which I posted on 2023-12-21, Luigi's question is analogous to
the following situation: We have ships P and Q (each with perfectly
symmetrical hull forms) moving on the (idealised, spherical and entirely
oceanic) Earth's surface. P moves westward at fixed speed (relative to
the Earth's rotating surface -- the fact that the Earth rotates is irrelevant
here) along the equator. Q moves westward at half of P's speed at a *fixed
latitude* of 60 degrees north latitude. Initially Q is directly north of P.

By symmetry we can see that P's rudder must be centered in order to stay
on the equator. By what about Q? If Q (initially moving westward at
latidude 60 degrees north) were to center its rudder, it would follow a
great circle, so it wouldn't stay at latitude 60 degrees north: its latitude
would start decreasing. We thus see that in order to *stay* at latitude
60 degrees north, Q must have its rudder set for a steady turn to the right.

(This may be easier to follow if you imagine yourself in space looking
down at the earth from above the north pole: Q is now going in a circular
path around the north pole, at a constant latitude of 60 degrees north.
Since the cosine of 60 degrees is 0.5, this path has a circumference of
1/2 of the Earth's equatorial circumference.)

Luigi's question is now analogous to this: How is it that the distance
between P and Q stays fixed (it's equal to the distance between the
equator and the circle of constant latitude 60 degrees north, which is
~6700 km measured along the Earth's surface), despite
(a) P steaming along with rudder centered while Q is continually holding
right rudder in order to stay at that fixed latitude of 60 degrees
north, and
(b) Q only steaming at half the speed of P?
In this context, the answer is obvious: the Earth's surface is curved,
i.e., it doesn't follow the axioms of Euclidean geometry. It's only our
intuition about flat surfaces that gives the mistaken impression that
two ships steaming at different speeds, one with rudder centerd and the
other with rudder set for a steady turn, can't stay the same distance
apart.



One final point to ponder: Returning to the three spaceships, let's
imagine (gedanken) reducing the Earth's density to zero. This means that
(ignoring the gravitational fields/spacetime curvature from the rest of
the universe) all three of the spaceships would be local inertial reference
frames. And, because we'd no longer have the spacetime curvature of the
Earth's mass, we could in fact extend those local inertial reference
frames to a single *global* inertial reference frame spanning the entire
Earth. In a flat spacetime you can do that, but in a curved spacetime
inertial reference frames are only local.

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] il 09/01/2024 09:37:45 ha
scritto:

> Recalling the ships-moving-on-the-Earth's-surface analogy for general
> relativity which I posted on 2023-12-21, Luigi's question is analogous to
> the following situation: We have ships P and Q (each with perfectly
> symmetrical hull forms) moving on the (idealised, spherical and entirely
> oceanic) Earth's surface. P moves westward at fixed speed (relative to
> the Earth's rotating surface -- the fact that the Earth rotates is irrelevant
> here) along the equator. Q moves westward at half of P's speed at a *fixed
> latitude* of 60 degrees north latitude. Initially Q is directly north of P.
>
> By symmetry we can see that P's rudder must be centered in order to stay
> on the equator. By what about Q? If Q (initially moving westward at
> latidude 60 degrees north) were to center its rudder, it would follow a
> great circle, so it wouldn't stay at latitude 60 degrees north: its latitude
> would start decreasing. We thus see that in order to *stay* at latitude
> 60 degrees north, Q must have its rudder set for a steady turn to the right.

Ok, it's true: your ship Q, which moves while remaining at 60 degrees
north latitude, is an accelerated reference frame because there is the
force of the rudder that makes it deviate from its geodesic and your
ship P is an inertial reference frame because there is no force that
makes it deviate from its equatorial geodesic.

And although the two ships are one accelerated and the other is not,
their distance does not change over time.

And, so far, we agree.

But I don't agree with your implication that it corresponds
(1) your ship Q to my elevator A (stopped on the floor)
(2) your ship P to my elevator B (stationary at the center of the
Earth)
(3) the force of the rudder, which curves the trajectory of the ship Q,
to the force of the cables that keep my elevator A at the floor
because these correspondences are incorrect.

And in fact, if the rudder of your ship Q breaks, it moves forward
inertially and no one forces it to necessarily go towards ship P until
it hits it (so much so that it doesn't hit it!).

Instead, if the cables of my elevator A break, its free fall is not as
free to go wherever it happens but *must* necessarily head towards
elevator B until it hits it and this is not inertial behavior!

Luigi Fortunati

[[Mod. note -- The analogy isn't precise.

Notably, the ships are moving in 2 spatial dimensions (the Earth's
surface), whereas the elevators are (as I understand it) constrained
to move in a tunnel drilled radially through the Earth, and so are
constrained to move ONLY along that radial direction.

But this problem is (gedanken) easy to solve: we can arrange for the
ships P and Q to always have equal longitudes. (For example, we could
have P continuously broadcast radio messages giving its current longitude,
and hvae Q receive these messages and continually adjust its speed so
as to match P's longitude.) The result is that if Q were to center its
rudder, then yes, after a time P and Q would collide.
-- jt]]