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Sep 7, 2022, 2:22:46 AMSep 7

to

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?

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?

Sep 7, 2022, 1:12:17 PMSep 7

to

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]]

Sep 7, 2022, 3:43:54 PMSep 7

to

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.
>> 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.

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?

Sep 7, 2022, 11:41:23 PMSep 7

to

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

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 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 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.

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.

Sep 8, 2022, 6:36:03 PMSep 8

to

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.
> 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!

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.

https://www.geogebra.org/m/zdevssyz

it goes only and always in one direction and never in the other, why?

Sep 9, 2022, 3:19:19 AMSep 9

to

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

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.

> 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?

>

...
> >> 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
> 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?

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.

Sep 9, 2022, 4:30:05 PMSep 9

to

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
>it goes only and always in one direction and never in the other, why?

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.

Sep 9, 2022, 4:30:57 PMSep 9

to

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

Sep 11, 2022, 7:38:33 PMSep 11

to

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.
> 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.

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]]

Sep 12, 2022, 1:02:09 PMSep 12

to

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
> 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

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).

geodesics.

> If I understand you correctly, you're asking "once a particle is moving,

> how does it know to continue moving in that direction?".

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

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.

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.

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).

Sep 12, 2022, 3:30:46 PMSep 12

to

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]]

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]]

Sep 12, 2022, 3:36:16 PMSep 12

to

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.

> 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.

...
> 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 --

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]]

Sep 13, 2022, 2:20:07 AMSep 13

to

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

[[about a 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"]

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.

[[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
> 4-geodesic) and not how to "continue" to move.

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.

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.

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
>> 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.

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).

[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.

Sep 13, 2022, 3:03:40 PMSep 13

to

Jonathan Thornburg [remove color- to reply] marted=EC 13/09/2022 alle ore

08:20:03 ha scritto:

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]]

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
> [[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.

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.

Sep 15, 2022, 7:12:42 PMSep 15

to

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?
>> 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.

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]]

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