3 ms·

If you were to imagine photons having an internal clock, that clock never ticks. That's not the same as light not moving through time. Light does not experience

If you were to imagine photons having an internal clock, that clock never ticks. That's not the same as light not moving through time. Light does not experience proper time. Any statement that combines the word "move" with the claim that it shows how light lacks time has already contradicted its goal. I think it is better to imagine light as line-like objects that extend through time, not point-like objects that move through time, and do away with speaking of light as something that moves, if you wish to express that it lacks a clock (proper time).

fjsolwmv 9y ago
Every particle has a life line (segment). Light particle's life line segment is perpendicular to the time axis, jumping across empty space (from birth/emission to death/absorption) at one constant time. If an atom emits an alpha particle (helium atom) and gamma particle (photon) at the same time, the alpha particle moves through space while its proper time changes (its "imaginary clock" ticks), whereas the gamma particle instantly arrives at its destination somewhere else in the Universe without experiencing any change in time, absorbed by another particle. Since light has no mass, it has no impact on spacetime geometry, so it only exists at emission and absorption. There is no way to detect or affect it between those to moments in its life, so it makes sense to consider them exactly one moment. Lights line segment is a "dotted line" the line segment (exclusive of endpoints) between emission and absorption does not really represent any part of our Universe; it's purely virtual. The only way to affect a photon is to plan ahead and set up an apparatus before it is born. Once the photon exists (relative to any observer), it's already gone (absorbed) by whatever first observed it. (And nothing else can directly observe it. All of us can bounce lots of stuff off the moon to observe it, but only one person can bounce only one thing off a photon , which kills it)
CuriousSkeptic 9y ago
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

fjsolwmv 9y ago

Every particle has a life line (segment). Light particle's life line segment is perpendicular to the time axis, jumping across empty space (from birth/emission to death/absorption) at one constant time. If an atom emits an alpha particle (helium atom) and gamma particle (photon) at the same time, the alpha particle moves through space while its proper time changes (its "imaginary clock" ticks), whereas the gamma particle instantly arrives at its destination somewhere else in the Universe without experiencing any change in time, absorbed by another particle. Since light has no mass, it has no impact on spacetime geometry, so it only exists at emission and absorption. There is no way to detect or affect it between those to moments in its life, so it makes sense to consider them exactly one moment. Lights line segment is a "dotted line" the line segment (exclusive of endpoints) between emission and absorption does not really represent any part of our Universe; it's purely virtual. The only way to affect a photon is to plan ahead and set up an apparatus before it is born. Once the photon exists (relative to any observer), it's already gone (absorbed) by whatever first observed it. (And nothing else can directly observe it. All of us can bounce lots of stuff off the moon to observe it, but only one person can bounce only one thing off a photon , which kills it)

CuriousSkeptic 9y ago

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