Comment on Which is it Lemmy?
NeatNit@discuss.tchncs.de 6 days ago
The answer is definitely B (see other comments), but I think the more interesting question is, how does it affect the portals themselves and the surfaces they’re attached to?
When the portals are attached to a solid unmovable wall (as in the games), we can pretty much ignore the forces acting upon the wall as they will just be automatically cancelled out by the normal force. But the same does not apply for portals attached to detached panels. Let’s say that instead of a piston pushing down, the portal is attached a free-falling panel with the portal facing down.
We are inclined to assume that as the portal falls onto the cube, it feels no force, as if it fell through air or through a vacuum. But is that really the case? Maybe instead, the falling portal experiences resistance from the cube? Or even, actually, an accelerating force?
Let’s assume conservation of momentum at each end of the portal. That is, in the general case: if the portal (which we assume has no mass) is attached to an object with mass M and velocity V, and “swallows” a cube or other object with mass m and velocity v, and at the end of this interaction the portal-surface object with mass M has velocity V’, then we expect:
M × V’ = (M × V) + (m × v)
Mass is not conserved because the cube was transferred to the other portal.
Solve for V’:
V’ = V + (m/M) v
This is a strange result. The cube’s momentum is imparted into the portal, but the result would be different depending on our reference frame. From a reference frame tracking the cube pre-interaction, v = 0 (by definition) and so V’ = V, the portal-surface object has no change in velocity. But in any other reference frame, it does.
Physics can’t be affected by reference frame (thanks, relativity) so this analysis must be wrong.
… I’ll keep thinking on this.
NeatNit@discuss.tchncs.de 6 days ago
Okay, after thinking about this a while, I’ve decided that conservation of momentum just doesn’t work, unless the portal actually absorbs the mass of every object passing through it, which would probably cause its own set of contradictions.
Instead, I think forces that act through the portal are applied to the portal. What do I mean by this? Well, take this classic case of a cube floating halfway through two floor portals:
(sorry, can’t do better than ASCII art right now)
The system has reached equilibrium, and nothing is moving. Both halves of the cube are being pulled down by gravity, but cancel each other out by applying a normal force on each other right at the portal interface. What I’m claiming is that this normal force is applied to the portals, which ultimately transfer it to the floor. If this entire floor was a scale, it would measure the calibrated weight of the Weighted Storage Cube.
But this is the static analysis. What happens when objects are in motion? To be continued…
NeatNit@discuss.tchncs.de 6 days ago
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.
ThatWeirdGuy1001@lemmy.world 5 days ago
I legitimately love how much effort you put into this.
On a shitpost of all things. It’s just beautiful.