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Agree! This is my favorite crazy pet theory: spacetime is simulated on a flat 3d substrate with constant compute and memory per volume, which runs at a fixed ra

Agree! This is my favorite crazy pet theory: spacetime is simulated on a flat 3d substrate with constant compute and memory per volume, which runs at a fixed rate in time (c) [1], and to account for a high density of mass requiring more compute the simulation is slowed down proportionally. Then, black holes are just the extreme end of this slowing down of the simulation. [1]: You always move at a constant speed: the speed of light. But that speed can be broken down into two component directions which you can trade off: time and space. You typically move near the speed of light in the time direction, meaning your speed through the space direction is comparatively low.

shaded-enmity 5y ago
My crazy pet theory: computational bandwidth is constant for any spacetime coordinate. According to Bekenstein bound the lower the temperature of the system the cheaper it is to flip a bit of information. As time progresses the CMBR gets colder, so it's cheaper to flip a bit, but conversely, as time progresses our Hubble volume gets smaller, limiting our total available energy budget to perform a bit flip.
- goldenkey 5y ago
  None of these theories are crazy at all. Special relativity is directly derivable from energy causing a fixed amount of change (action) per unit time. When the information/computational changes that the energy is allocated, is used for non-local movement, then we consider this to be an aspect of velocity, and the amount of changes leftover for internal changes, which effectively are internal time, is decreased by a ratio equivalent to the standard time dilation formulas. Movement is change...change requires energy allocation, movement diminishes other interactions, hence time dilation, hence special relativity. Wolfram talks about this in his new graph based model of physics. Mass is an emergent metric of the amount of internal changes occurring in an object per unit time. Essentially, causal loops in the graph. Feynman had a toy model called the checkerboard model where the amount of bouncing in an area represented mass. https://youtu.be/xQafZ3CvBUs https://youtu.be/xQafZ3CvBUs
  - jpitz 5y ago
    Movement does not require energy, acceleration does.
    - goldenkey 5y ago
      You're wrong. What you probably meant is that movement does not require external/additional energy. It uses already allocated energy. Once you grasp this, you will understand why action is measured in units of Energy x Time. https://en.wikipedia.org/wiki/Action_(physics) https://en.wikipedia.org/wiki/Action_(physics)
      - jpitz 5y ago
        You do have a way with words. Congratulations.
        goldenkey 5y ago
        An object possessing kinetic energy continues to move unless acted on by an external force/energy. It's literally Newton's first law. It doesn't move for free though. Energy bound to the object is being allocated every Planck second that it moves. That energy could have instead been used for internal interactions, local time for that object, if it were at rest with the same relativistic mass. Time dilation is a consequence of energy having to be divvied up between external and internal changes. Nothing I am saying is controversial. But yes, I am making it much easier for everyone to see that movement is just change of state/information, and that all energy causes a constant change of information per unit time. The universe is computational, not really a surprise...
      - jpitz 5y ago
        Let me clarify my feelings here. >>You're wrong. This is no problem. I'm happy to be wrong, I'm happy to be told I'm wrong. >>What you probably meant is [...] This is problematic to me. I'm a grown adult, I own what I said, and I said precisely what I meant. When it's wrong, it's wrong, and no, I don't want to retcon that in some feeble attempt to save face, and I certainly don't want others to do that for me. Obviously, this is not my field. Is movement and motion here the same thing? Motion is relative, so this energy "use" here is relative - in some reference frames where an object would be observed as at rest, is it still "using" energy? You see these quotes? That's because I don't really understand what "use" means here. An object in motion is in motion as a consequence of acceleration, which requires energy, and certainly possesses unrealized kinetic energy, but it isn't consuming energy. So what does "use" mean here?
whoisburbansky 5y ago
Wouldn't it have to be a 4d substrate for c to be the "time" step-rate of the simulation? What does flat mean in this context?
Strilanc 5y ago
Why would gravity follow an inverse square law in this model? Wouldn't you expect there to be a hard cutoff as you moved away from a busy region, instead of a gradual decrease?
- Cookingboy 5y ago
  It's simply some kind of easing function, and if the universe does have a designer, it's obvious to see that said designer prefers smooth transition and curves for most macroscopic phenomenon. Also by defining the inverse square law you can maintain a cohesive interaction model between different objects and regions. Their sphere of influence (which can be described by gravity waves) will now just be superimposed on top of each other without extra work. Or even crazier, maybe it's a combination of both, maybe there is a hard cutoff after all once you pass the event horizon.
- taneq 5y ago
  Because that's how the simulation is configured?
- R0b0t1 5y ago
  The steps of granularity would likely be the Planck length. His proposal does have some unresolved issues. E.g. distance is quantized but the interactions of those quantized variables produce continuous things that affect other things that are quantized. We don't have enough energy to see the limit yet, but it is possible.
taneq 5y ago
It's because the stronger the forces and the higher the velocity in a cell of the simulation, the more iterations the numerical integrator needs to perform to maintain acceptable accuracy. It's obvious, really. ;) Also the universe isn't actually expanding and the red shift we see in distant stars is just a numerical error due to the enormous number of steps between here and there. :)
grishka 5y ago
> meaning your speed through the space direction is comparatively low But you do move in space at a great speed if you account for the rotation of the earth around the sun, the movement of the solar system in the galaxy, and the movement of the galaxy relative to other galaxies. It's not like there's a coordinate system with an absolute origin somewhere.
- dmd 5y ago
  Leaving aside that this completely ignores how relativity works, even if there were a fixed reference point (e.g., the CMB rest frame), our motion through space relative to that frame is ~ 0.07% of c.

shaded-enmity 5y ago

My crazy pet theory: computational bandwidth is constant for any spacetime coordinate. According to Bekenstein bound the lower the temperature of the system the cheaper it is to flip a bit of information. As time progresses the CMBR gets colder, so it's cheaper to flip a bit, but conversely, as time progresses our Hubble volume gets smaller, limiting our total available energy budget to perform a bit flip.

goldenkey 5y ago
None of these theories are crazy at all. Special relativity is directly derivable from energy causing a fixed amount of change (action) per unit time. When the information/computational changes that the energy is allocated, is used for non-local movement, then we consider this to be an aspect of velocity, and the amount of changes leftover for internal changes, which effectively are internal time, is decreased by a ratio equivalent to the standard time dilation formulas. Movement is change...change requires energy allocation, movement diminishes other interactions, hence time dilation, hence special relativity. Wolfram talks about this in his new graph based model of physics. Mass is an emergent metric of the amount of internal changes occurring in an object per unit time. Essentially, causal loops in the graph. Feynman had a toy model called the checkerboard model where the amount of bouncing in an area represented mass. https://youtu.be/xQafZ3CvBUs https://youtu.be/xQafZ3CvBUs
- jpitz 5y ago
  Movement does not require energy, acceleration does.
  - goldenkey 5y ago
    You're wrong. What you probably meant is that movement does not require external/additional energy. It uses already allocated energy. Once you grasp this, you will understand why action is measured in units of Energy x Time. https://en.wikipedia.org/wiki/Action_(physics) https://en.wikipedia.org/wiki/Action_(physics)
    - jpitz 5y ago
      You do have a way with words. Congratulations.
      - goldenkey 5y ago
        An object possessing kinetic energy continues to move unless acted on by an external force/energy. It's literally Newton's first law. It doesn't move for free though. Energy bound to the object is being allocated every Planck second that it moves. That energy could have instead been used for internal interactions, local time for that object, if it were at rest with the same relativistic mass. Time dilation is a consequence of energy having to be divvied up between external and internal changes. Nothing I am saying is controversial. But yes, I am making it much easier for everyone to see that movement is just change of state/information, and that all energy causes a constant change of information per unit time. The universe is computational, not really a surprise...
    - jpitz 5y ago
      Let me clarify my feelings here. >>You're wrong. This is no problem. I'm happy to be wrong, I'm happy to be told I'm wrong. >>What you probably meant is [...] This is problematic to me. I'm a grown adult, I own what I said, and I said precisely what I meant. When it's wrong, it's wrong, and no, I don't want to retcon that in some feeble attempt to save face, and I certainly don't want others to do that for me. Obviously, this is not my field. Is movement and motion here the same thing? Motion is relative, so this energy "use" here is relative - in some reference frames where an object would be observed as at rest, is it still "using" energy? You see these quotes? That's because I don't really understand what "use" means here. An object in motion is in motion as a consequence of acceleration, which requires energy, and certainly possesses unrealized kinetic energy, but it isn't consuming energy. So what does "use" mean here?

whoisburbansky 5y ago

Wouldn't it have to be a 4d substrate for c to be the "time" step-rate of the simulation? What does flat mean in this context?

Strilanc 5y ago

Why would gravity follow an inverse square law in this model? Wouldn't you expect there to be a hard cutoff as you moved away from a busy region, instead of a gradual decrease?

Cookingboy 5y ago
It's simply some kind of easing function, and if the universe does have a designer, it's obvious to see that said designer prefers smooth transition and curves for most macroscopic phenomenon. Also by defining the inverse square law you can maintain a cohesive interaction model between different objects and regions. Their sphere of influence (which can be described by gravity waves) will now just be superimposed on top of each other without extra work. Or even crazier, maybe it's a combination of both, maybe there is a hard cutoff after all once you pass the event horizon.
taneq 5y ago
Because that's how the simulation is configured?
R0b0t1 5y ago
The steps of granularity would likely be the Planck length. His proposal does have some unresolved issues. E.g. distance is quantized but the interactions of those quantized variables produce continuous things that affect other things that are quantized. We don't have enough energy to see the limit yet, but it is possible.

taneq 5y ago

It's because the stronger the forces and the higher the velocity in a cell of the simulation, the more iterations the numerical integrator needs to perform to maintain acceptable accuracy. It's obvious, really. ;) Also the universe isn't actually expanding and the red shift we see in distant stars is just a numerical error due to the enormous number of steps between here and there. :)

grishka 5y ago

> meaning your speed through the space direction is comparatively low But you do move in space at a great speed if you account for the rotation of the earth around the sun, the movement of the solar system in the galaxy, and the movement of the galaxy relative to other galaxies. It's not like there's a coordinate system with an absolute origin somewhere.

dmd 5y ago
Leaving aside that this completely ignores how relativity works, even if there were a fixed reference point (e.g., the CMB rest frame), our motion through space relative to that frame is ~ 0.07% of c.