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> Any given moment in time is represented by a slice across spacetime Is this actually true when we take into account special relativity? I've always struggle

by nilkn 12y ago

> Any given moment in time is represented by a slice across spacetime Is this actually true when we take into account special relativity? I've always struggled with this. I was taught that this sort of thing is not nearly so simple. For instance, if you fix a point in spacetime and have a timelike vector, representing the motion of an observer, then the set of events which that observer will perceive as simultaneous to the fixed point all lie in the plane orthogonal to that timelike vector. Even more bizarrely, because this is a non-Euclidean notion of orthogonal, this orthogonal complement rotates towards the vector as it deviates from the centerline of the future line cone at that point. Given all this, I don't see how one can just say that a slice across spacetime somehow represents a particular moment in time. I thought the whole point of relativity is that there are no absolute time slices in spacetime.

drostie 12y ago
Yes, it's still true when we take special relativity into account. One good perspective on special relativity is that it sort of says, "everybody's right." That is, you abstract your "present" moment to distant positions as a 3d plane in the 4d space; someone else at the same position with a different velocity will have their present too, and those will not be 100% commensurate, but that's OK -- both of you have valid coordinate systems for any calculations you want to accomplish. The family of all the "presents" of a point in spacetime is the set of all points which are spacelike-separated from that point -- i.e. if that point is at the origin of some coordinate system, the complete "relativistic present" is all of those points such that c² t² − x² − y² − z² < 0 Every observer passing through a point agrees upon these points exactly; the only difference is that my t=0 will not correspond to t' = 0 under a Lorentz transform, so that my "simultaneous" at space-separated points is not someone else's "simultaneous" at those points. (Light bubble picture: imagine that the light which shines upon an event in spacetime expands outwards with speed c, forming an expanding bubble of light. Consider two events. They are both "simultaneous" in the sense that there is an observer who thinks that they are simultaneous, if their light bubbles start out topologically disconnected and overlap eventually. They are both "time-ordered" in the other case, if one bubble is inside the other. If two things are objectively time-ordered then they are not objectively space-separated, because the points in the larger bubble correspond to valid inertial trajectories of a spaceship going less than the speed of light -- there are some spaceships which visited both events inertially. Similarly if two things are objectively space-separated then they are not objectively time-ordered; consider someone on the intersection of the two light bubbles seeing both events happen "right now"; there is always a velocity vector such that they will trace the distances back to the origins of the events as equal -- and hence that observer thinks that both happened simultaneously at their different locations.)
- saganus 12y ago
  Wow... that made my mind bend or something. This sounds pretty dense (or maybe the dense one is me...)

by nilkn 12y ago

drostie 12y ago

Yes, it's still true when we take special relativity into account. One good perspective on special relativity is that it sort of says, "everybody's right." That is, you abstract your "present" moment to distant positions as a 3d plane in the 4d space; someone else at the same position with a different velocity will have their present too, and those will not be 100% commensurate, but that's OK -- both of you have valid coordinate systems for any calculations you want to accomplish. The family of all the "presents" of a point in spacetime is the set of all points which are spacelike-separated from that point -- i.e. if that point is at the origin of some coordinate system, the complete "relativistic present" is all of those points such that c² t² − x² − y² − z² < 0 Every observer passing through a point agrees upon these points exactly; the only difference is that my t=0 will not correspond to t' = 0 under a Lorentz transform, so that my "simultaneous" at space-separated points is not someone else's "simultaneous" at those points. (Light bubble picture: imagine that the light which shines upon an event in spacetime expands outwards with speed c, forming an expanding bubble of light. Consider two events. They are both "simultaneous" in the sense that there is an observer who thinks that they are simultaneous, if their light bubbles start out topologically disconnected and overlap eventually. They are both "time-ordered" in the other case, if one bubble is inside the other. If two things are objectively time-ordered then they are not objectively space-separated, because the points in the larger bubble correspond to valid inertial trajectories of a spaceship going less than the speed of light -- there are some spaceships which visited both events inertially. Similarly if two things are objectively space-separated then they are not objectively time-ordered; consider someone on the intersection of the two light bubbles seeing both events happen "right now"; there is always a velocity vector such that they will trace the distances back to the origins of the events as equal -- and hence that observer thinks that both happened simultaneously at their different locations.)

saganus 12y ago
Wow... that made my mind bend or something. This sounds pretty dense (or maybe the dense one is me...)