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

[[about a particle]]

*Post by Luigi Fortunati*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.

*Post by Luigi Fortunati*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.

*Post by Luigi Fortunati*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"]

*Post by Luigi Fortunati**Post by Luigi Fortunati*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.

*Post by Luigi Fortunati*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.