Email or username:

Password:

Forgot your password?
Zeno's posts Post Back to profile
Zeno Rogue

Portals, but this time it is not about immersive visualization, but about gravity!

Gravity always tries to get you down. But sometimes it will send you on a path through a portal. Even if that portal is above you! You cannot use portals to violate conservation of energy here.

The "potion of gravision" shows how the gravitational potential is distributed.

This viz reuses some code from the "non-Euclidean platformer" (youtube.com/watch?v=0riQhP_oZl). Unfortunately, this game concept (using hyperbolic geometry to create a metroidvania with exponential space but small radius) wants gravity to be the same everywhere, and that is not compatible with the laplacian of the gravitational potential being zero (which is what the portal gravity viz assumes). (Rogue-style platfomer from x.com/ZenoRogue/status/1757432)

This is inspired by this video by Sabine Hoffenfelder: youtube.com/watch?v=cox7481IE6

8 June at 22:38 | Open on mathstodon.xyz
3 comments
optozorax

@zenorogue how do you simulate teleportation of a gravity? Recently I have a thought that it might be done using teleportation of gravitons as a model of a gravity. And also, portals must be two-sided, for symmetry.

9 June at 11:08 | Open on mathstodon.xyz
Zeno Rogue

@optozorax We use a grid approximation (as seen in "gravision"). Each square has some value of potential. Without portal, it would linearly depend on the 'y' coordinate.

To take the portals into account, we assume that the Laplacian should equal 0. Which in discrete terms means that each value of potential should equal the average of values on 4 adjacent squares. So we iteratively replace potential on every tile with this average until it stabilizes. The actual grid is much larger than the room, and there is a boundary condition on the edges (defined as if there was no portal).

Adjacent squares of course take portals into account. And the portals are indeed two-sided (for example, the adjacent square 'below' the square above the blue portal will be below the orange portal).

@optozorax We use a grid approximation (as seen in "gravision"). Each square has some value of potential. Without portal, it would linearly depend on the 'y' coordinate.

To take the portals into account, we assume that the Laplacian should equal 0. Which in discrete terms means that each value of potential should equal the average of values on 4 adjacent squares. So we iteratively replace potential on every tile with this average until it stabilizes. The actual grid is much larger than the room,...

13 June at 15:29 | Open on mathstodon.xyz
optozorax

@zenorogue after some talking to gpt, couldn't understand theoretic basis of your approach, but it seems easy to program. I will program in some future, and will compare it to my idea with gravitons. Thanks!

16 June at 22:29 | Open on mathstodon.xyz
Go Up