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?
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.
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.
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 alwaysI absolutely agree with what you wrote.
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.
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?
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 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
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.
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 preferredThe time coordinate is part of the geodesic.
direction, the time coordinate has.
If the time coordinate has a preferred direction, the geodesic also has
a preferred direction.
The time coordinate is part of the geodesic.
I am asking how does it know how to "start" to move (along
4-geodesic) and not how to "continue" to move.
There is no conservation of 4-momentum when the elevator cables break
and the elevator goes from constrained condition to free fall.
There is no conservation of 4-momentum when the elevator cables break
and the elevator goes from constrained condition to free fall.
Before the cable breaks, there are two forces acting on the elevator:gravity: F=mg downwards
After the cable breaks, there is one force acting on the elevator:gravity: F=mg downwards
Before the cable breaks, there is one force acting on the elevator:cable tension: F=mg upwards
After the cable breaks, there are no forces acting on it, so it movesalong a geodesic, with zero 4-acceleration. That geodesic path starts
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.
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).
Luigi Fortunati <fortunati.luigi@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.
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.
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?
[...] But is it really like that?
[... confused writing]
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?
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