So we might think: a cube falling through a floor portal and flying out another floor portal right next to it, kinda looks like how a bouncy ball bounces off the floor, and should apply forces on the floor similar to that - an elastic collision. Unfortunately, this brings us right back to conservation of momentum, which didn’t work out. But maybe I just did it wrong? Maybe, if done more carefully, conservation of momentum can work without being dependent on the reference frame?
I genuinely don’t know yet. Time to bring out the calculus.
We have:
M = the mass attached to the portal (e.g. the floor)
V = the velocity of M
V’ = dV/dt = the acceleration of M (assuming no other forces)
F = MV’ = the force applied on the portal
m’ = dm/dt = the rate of mass entering the portal
v = the velocity of the mass entering the portal
If we assume conservation of momentum (our old nemesis), the force applied on the portal should be the same as the momentum taken from the mass entering the portal:
F = m’v
But now we’re back to the case from before, where the choice of reference frame changes the outcome! We’re not having that. So I hereby decree:
The force applied upon a portal by a mass entering it must be calculated by conservation of momentum in the intertial reference frame of the portal itself at the instant under examination.
This gives us, effectively: F = m’(v-V)
Now THAT’S something we can work with!
Note that this formula also works for the exit portal. In that case, m’ is negative.
A curious result follows.
For the entry portal, m’ is positive, and (v-V) is a vector pointing into the portal. (This must be the case, otherwise the mass would be exiting the portal). Therefore, the force would be pushing the portal into the wall.
For the exit portal, m’ is negative, and (v-V) points out of the portal, therefore the portal again would be pushed into the wall.
The result: whenever anything passes through a portal, both ends always get pushed in the same direction.
So, back to my thought experiment.
If a portal attached to a free-falling panel falls onto a cube, the cube’s entry into it would brake its fall to some extent. However, this wouldn’t be the only force in play.
As I said in my previous comment, forces applied between the two halves of an object going through a portal, get applied to the portal itself. In practice, this mostly means tension and compression forces. In the static analysis from before, it was compression.
As the panel is falling onto the cube, a part of the cube is already out the other side and moving away the portal. If the panel has slowed, this would apply a pulling force (tension) on the part of the cube that still hasn’t made it in. This tension force actually compels the portal, and the panel it’s attached to, to speed up towards the ground.
So, in total, would the panel be slowed, accelerated, or unaffected? Maybe the forces cancel out and it just continues on its normal freefall? I don’t know yet. Would need to do more math.
So we might think: a cube falling through a floor portal and flying out another floor portal right next to it, kinda looks like how a bouncy ball bounces off the floor, and should apply forces on the floor similar to that - an elastic collision. Unfortunately, this brings us right back to conservation of momentum, which didn’t work out. But maybe I just did it wrong? Maybe, if done more carefully, conservation of momentum can work without being dependent on the reference frame?
I genuinely don’t know yet. Time to bring out the calculus.
We have:
M = the mass attached to the portal (e.g. the floor)
V = the velocity of M
V’ = dV/dt = the acceleration of M (assuming no other forces)
F = MV’ = the force applied on the portal
m’ = dm/dt = the rate of mass entering the portal
v = the velocity of the mass entering the portal
If we assume conservation of momentum (our old nemesis), the force applied on the portal should be the same as the momentum taken from the mass entering the portal:
F = m’v
But now we’re back to the case from before, where the choice of reference frame changes the outcome! We’re not having that. So I hereby decree:
The force applied upon a portal by a mass entering it must be calculated by conservation of momentum in the intertial reference frame of the portal itself at the instant under examination.
This gives us, effectively: F = m’(v-V)
Now THAT’S something we can work with!
Note that this formula also works for the exit portal. In that case, m’ is negative.
A curious result follows.
For the entry portal, m’ is positive, and (v-V) is a vector pointing into the portal. (This must be the case, otherwise the mass would be exiting the portal). Therefore, the force would be pushing the portal into the wall.
For the exit portal, m’ is negative, and (v-V) points out of the portal, therefore the portal again would be pushed into the wall.
The result: whenever anything passes through a portal, both ends always get pushed in the same direction.
So, back to my thought experiment.
If a portal attached to a free-falling panel falls onto a cube, the cube’s entry into it would brake its fall to some extent. However, this wouldn’t be the only force in play.
As I said in my previous comment, forces applied between the two halves of an object going through a portal, get applied to the portal itself. In practice, this mostly means tension and compression forces. In the static analysis from before, it was compression.
As the panel is falling onto the cube, a part of the cube is already out the other side and moving away the portal. If the panel has slowed, this would apply a pulling force (tension) on the part of the cube that still hasn’t made it in. This tension force actually compels the portal, and the panel it’s attached to, to speed up towards the ground.
So, in total, would the panel be slowed, accelerated, or unaffected? Maybe the forces cancel out and it just continues on its normal freefall? I don’t know yet. Would need to do more math.