3 ms·

But still, that line, in a minkowski diagram, represents a boundary, so it still seems more intuitive to me to eliminate the area beyond it completely from the

But still, that line, in a minkowski diagram, represents a boundary, so it still seems more intuitive to me to eliminate the area beyond it completely from the diagram.

ajkjk 9y ago
For a single observer that area is effectively invisible; they cannot be affected or affect things in it (outside of their forward/backward light-cone). However, other events can still exist in there. And their lightcones may intersect at a later point in time, so they are not totally uninteracting. Point (x1, t1) may never be able to affect (x2, t2), but it can always affect (x2, t2 + <waiting for a while>).
nabla9 9y ago
That boundary is in 45 degree angle for a reason. Spacetime interval is lightlike when spacetime interval is zero (Δs²=0). Spacetime interval is zero when Δr²=c²Δt². In other words light moves exactly as much in time (t) and space (r). If you want to see 'lights point of view' in spacetime intervals and not in the block universe model, every reachable spacetime location is zero distance away. The whole cone, not just the boundary is just one dot (light can travel into any point inside the lightcone using mirrors) and outsides are removed. . edit: the whole eqution is Δs² = Δr² - c²Δt². If we write all coordinates x,y,z,t open, the distance is Δs² = Δx² + Δy² + Δz² - c²Δt². Otherwise nice euclidean distance measure is screwed by that minus sign before time that makes spacetime hyperbolic.
jacobolus 9y ago
It’s not that the points outside the light cone don’t exist. It’s just that each point represents a “spacelike displacement” rather than a “timelike displacement”. If you like, you can think of any point outside the boundary as happening simultaneously to the spacetime point at the vertex of the cone, if you just pick the right reference frame.
QAPereo 9y ago
A causal boundary, but not a real boundary. That is to say, events separated by that line can never meet, never interact, never observe each other, but they can still exist. For a “real world” example see: https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates
- fjsolwmv 9y ago
  They exist in the sense that the uncountable real numbers exist. The reals you can name are like your light cone. By symmetry it seems that the other reals/ spacetime points should exist, but there's no way to prove they actually do if you're not already "nearby". :-)

But still, that line, in a minkowski diagram, represents a boundary, so it still seems more intuitive to me to eliminate the area beyond it completely from the diagram.

ajkjk 9y ago

For a single observer that area is effectively invisible; they cannot be affected or affect things in it (outside of their forward/backward light-cone). However, other events can still exist in there. And their lightcones may intersect at a later point in time, so they are not totally uninteracting. Point (x1, t1) may never be able to affect (x2, t2), but it can always affect (x2, t2 + <waiting for a while>).

nabla9 9y ago

That boundary is in 45 degree angle for a reason. Spacetime interval is lightlike when spacetime interval is zero (Δs²=0). Spacetime interval is zero when Δr²=c²Δt². In other words light moves exactly as much in time (t) and space (r). If you want to see 'lights point of view' in spacetime intervals and not in the block universe model, every reachable spacetime location is zero distance away. The whole cone, not just the boundary is just one dot (light can travel into any point inside the lightcone using mirrors) and outsides are removed. . edit: the whole eqution is Δs² = Δr² - c²Δt². If we write all coordinates x,y,z,t open, the distance is Δs² = Δx² + Δy² + Δz² - c²Δt². Otherwise nice euclidean distance measure is screwed by that minus sign before time that makes spacetime hyperbolic.

jacobolus 9y ago

It’s not that the points outside the light cone don’t exist. It’s just that each point represents a “spacelike displacement” rather than a “timelike displacement”. If you like, you can think of any point outside the boundary as happening simultaneously to the spacetime point at the vertex of the cone, if you just pick the right reference frame.

QAPereo 9y ago

A causal boundary, but not a real boundary. That is to say, events separated by that line can never meet, never interact, never observe each other, but they can still exist. For a “real world” example see: https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

fjsolwmv 9y ago
They exist in the sense that the uncountable real numbers exist. The reals you can name are like your light cone. By symmetry it seems that the other reals/ spacetime points should exist, but there's no way to prove they actually do if you're not already "nearby". :-)