Luigi Fortunati <***@gmail.com> wrote:
[[about a particle]]
Post by Luigi FortunatiI 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 FortunatiThere 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 FortunatiThere 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 FortunatiPost by Luigi Fortunatiunless 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 FortunatiOk, 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.