# Does the $t = 0$ hypersurface of simultaneity constitutes the universal present?

Does the hypersurface of simultaneity in the diagram below represent the universal present moment?

• There is no such thing as "a present moment" in special relativity and that non-existence continues in general relativity. The only physically relevant events were in your past light cone and will be in your future light cone. The "hypersurface of simultaneity" is completely irrelevant. Think of them as you (hopefully) think of all the beautiful women who will never be your girlfriend. ;-) Jun 8 '16 at 14:51
• That's a lot of big words in your question, but, no. Jun 8 '16 at 15:29
• Define "coexist" in a physical way. Jun 8 '16 at 15:46
• @ChrisDegnen If I can send an instantaneous signal to something, that something is in the present, isn't it? Otherwise I couldn't send an instantaneous signal, only one that moves at some speed - i.e. travels a distance in a non-zero time interval. Jun 8 '16 at 16:36
• The diagram is misleading in that it defines a spacelike curve as a curve contained within the hypersurface of simultaneity. Actually any curve outside the light cone is a spacelike curve. It is also misleading because it shows two timelike worldlines and only one hypersurface, where there should be one for each wordline, illustrating the fact that simultaneity depends on the observer, not on the light cone. Jun 8 '16 at 20:36

The hypersurface of simultaneity does not represent the present. It is just a plane cut through spacetime. If you changed your own state of motion this would tilt the plane by some angle. So the notion of "now" as the hypersurface of simultaneity would depend on your state of motion, which is of course not meaningful. In fact, the notion of "now" itself is meaningless. There is just a meaning of "here and now".

• If we are talking about the universal present moment then there is no motion. Nothing moves in zero time, so frames are irrelevant. We're just talking about the hypersurface at t = 0. Jun 8 '16 at 15:34
• @ChrisDegnen: There is no hypersurface there. The surface is given by the cones, and that's it. What you are doing is to embed the cones in a higher dimensional space that doesn't exist and then you do a scribble in there that has nothing to do with reality. It's obvious why you think that it makes sense: when you draw the same embedding for Galilean relativity, then you end up with a lower half-space for the past and an upper half-space for the future. The dividing surface is the present... in Galilean relativity, which is a different, non-existing universe. Jun 8 '16 at 20:16
• @CuriousOne I think I see this now. The surface is given by the cones because the speed of light isn't just a speed beyond which one can imagine greater speeds on up to 'instantaneous'. The speed of light itself is 'instantaneous' because at the speed of light no time passes. Jun 9 '16 at 10:37
• @ChrisDegnen: Only if you can become light. :-) The latter is, of course, potentially possible. "You" are a fairly classical being, which means that the no-cloning theorem doesn't apply to you. There is no law of nature that says you can't convert yourself into packets of information that can be sent around the universe at the speed of light, to be reconstituted into a copy of yourself somewhere else. Whether that's something a human being would actually consider as a way of traveling their future light cone... that's a matter of taste. Jun 9 '16 at 13:26
• @ChrisDegnen: You found more than you were missing, actually, because in this universe you can actually travel faster than in the one you wanted to be in just yesterday. :-) Jun 9 '16 at 13:46

Quite clearly the answer to this is that no, it does not. In particular, consider two inertial observers moving (in flat spacetime) relative to one another. We know that neither of these two observers is more privileged than the other: the laws of physics are the same for each of them and so on. Yet they will draw different hypersurfaces of simultenaity for the same event. (They will agree, for instance, on the light cone of the event, which is a physically meaningful thing.)

Since there is a continuum of different such surfaces, no one of which it is possible to privilege over any other, none of them can be physically important.

To put this another way: if you believe that some hypersurface defined by $t=0$ for some coordinate $t$ which parameterizes a particular family of timelike geodesics defines a 'universal present' then this is exactly equivalent to picking a particular inertial observer and saying that their reference frame is the one that matters: this particular observer is somehow more important than all the others, because only this observer gets to define this magic 'universal present'. Yet there is no way of identifying such an observer.

• I'm now seeing from here that hypersurfaces are used in analysis of relative motion. Nevertheless I am considering only an instant of time, so there is no motion involved. I think I am taking the hypersurface idea out of its usual context. I'll pick this up later. Jun 8 '16 at 16:03
• @ChrisDegnen probably better not to consider 'instants of time' since such things have no meaning in SR (and therefore none in GR).
– user107153
Jun 8 '16 at 16:38
• Or rather not consider SR or GR in describing the present instant of time. The light cones are relevant only in affirming that the present "at any reasonable distance" is unobservable. Jun 8 '16 at 17:42
• @ChrisDegnen If you take the time interval during motion as zero, the distance is also zero, but the velocity is still non-zero. And since the velocity is what matters in relativistic comparisons, you can't ignore it even for an instant. Jun 8 '16 at 18:08
• @ChrisDegnen There's obviously no problem with considering some vaporous unobservable-even-in-principle gallilean 'present', so long as no-one thinks it's science.
– user107153
Jun 8 '16 at 18:42

As the saying goes, a picture is worth a thousand words. Below is a time-spatial axis diagram of the time and x-axis in two frames. One is at rest relative to the blue, and the other is Lorentz boosted to some velocity. What is the plane of simultaneity is dependent upon the frame that you are on

• I understand this but I'm only asking if all observers can have t = 0 without issue. If there is warp in spacetime like this wormhole then they can't. I image universal t = 0 is fine, but there so much talk about there being no such thing as Now (e.g. Feynman) I was beginning to have my doubts. Jun 8 '16 at 18:29
• The diagram above is exactly showing you how two observers cannot have t=0 without issue. There is two t=0 lines in it, the blue and the red, one for each observer. The only point in spacetime where they agree that it happens now is the one at x=0 and t=0. Only on "here and now" can all observers agree, and only when they meet. There is no universal present, as all others have said. Jun 8 '16 at 20:29

The hypersurface at t = 0 as shown in the diagram is not the observers plane of instantaneous, contemporaneous events constituting the present. The speed of light is as instantaneous as anything gets because at the speed of light no time passes. (Therefore the present is relative.)