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Please follow this thought experiment:

1) A ball moves one centimeter in one unit of time.

2) A ball disappears. Then, after a unit of time, it reappears one centimeter far away.

By now we don't define the unit of time: it can be an hour like one thousandth of a second.

The difference between a movement (1) and a teleportation (2) is that in the second case the ball does not exist between one unit of time and another, while in the first case it continues to exist. So, if the unit of time is a minute, after half a minute in (1) the ball is half a centimeter away, while in (2) it does not exist.

But if I assume the existence of a minimum unit of time, the two cases coincide, because also in (1) there is no time fraction - no "meanwhile" - in which the ball can be elsewhere.

In short, if there is a minimum unit of time (if time and space are discrete), is the fastest movement equivalent to disappearing and reappearing?

enter image description here

(by 'teleportation' here I intend that the object disappears at location X prior to appearing at location Y)

Francesco D'Isa
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    That's a great question. My understanding is that there is a Planck distance.. but it's defined by continuous time. So I'm not sure teleportation is required? In other words time isn't discrete.. space might be. – Richard Dec 09 '18 at 23:12
  • Your measuring equipment (to know _where_ the ball is) will be affected by the same time interval... That means the ball would appear to move smoothly i.e. there won't be a time interval at which it is "not there. However it will also mean that objects at different velocity will seem to have different _minimal spacial units_. See also https://en.wikipedia.org/wiki/Planck_units and https://en.wikipedia.org/wiki/Zeno%27s_paradoxes – christo183 Dec 10 '18 at 04:27
  • This is what I thought about when I was 15. At that time I said to myself it is a teleportation. Even further, I assume speed only changes the frequency of directed teleportation. But it all boils down to what teleportation means. – rus9384 Dec 10 '18 at 07:20
  • @rus9384 of course, here by teleportation I intend to simply disappear and later reappear elsewhere – Francesco D'Isa Dec 10 '18 at 07:42
  • @christo183 you are right, but let’s suppose for the sake of the argument that I don’t need to measure it, we just know that the balls behave like that – Francesco D'Isa Dec 10 '18 at 07:45
  • Well. You should take a few possibilities. 1) The object disappers at location X and appears at location Y simultaneously. 2) The object disappears at location X prior to appearing at location Y. 3) The object appears at location Y prior to disappearing at location Y. Does it all count for teleportation? – rus9384 Dec 10 '18 at 08:13
  • "measured" being "observed" in the QM sense. - You can have neither distance nor time as discrete, in which case it would be impossible to ask this question. You can have time (exclusive)OR distance discrete, this is where the bizarre observations above would be possible. But if you are looking at real objects, quantum effects could come into play. Lastly you can have both discrete (Planck units), in which case it would be impossible to observe the phenomena. – christo183 Dec 10 '18 at 08:16
  • @rus9384 You are right, thank you. I meant (2); i added it to the question. – Francesco D'Isa Dec 10 '18 at 09:43
  • @christo183 I agree, I think that you should have time AND distance discrete to ask this question. – Francesco D'Isa Dec 10 '18 at 13:05
  • @Richard It's a common misconception to think that the Planck length implies discrete space, which isn't the case. It's just the minimum distance at which two positions can be differentiated. Points A and B which are one Planck length apart can be identified as different from one another, but that doesn't preclude the possibility of a Point C that lies between them and can't be differentiated from either. – Nuclear Hoagie Dec 11 '18 at 16:00

5 Answers5

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Your thought-experiment is excellent. I've also examined this issue in these terms and it is a much clearer approach than Zeno's arguments.

What you have done is realised the paradoxical nature of our usual idea of time, motion and change. On close examination it doesn't work.

This is not news, but the correct solution is a matter of debate. The topic is too difficult for me to say much here but I can recommend that you google the writings on time and change by Hermann Weyl. If I could PM I'd mention an essay of mine on this topic but there's no PMs here.

I'd say you are right, if time is 'grainy' then motion requires 'teleportation' (of some kind). If time is continuous then even greater problems arise. If time is conceptual then they are all solved. If you conduct a literature review you'll notice that time baffles all those who believe it to be metaphysically real.

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If I understood your question correct I think you are making a logical typing error.

Your Question states two core concepts. (1)The Ball disappears and reappears again. (2)Time is (possibly) a discrete entity

Concept (1) suggests a ego-centric view of time. Meaning theres only the "now/present" which is always the case. Occuring differences of the ball therefore only take place if the location changes. Meaning that in the "now" the ball is first in location x0 dissapears then and appears in the same "now" again in location x1.

This difference allows you (if you want) to measure time based on the balls location changes and therefore make time discrete. We could additionally ad another ball/clock inorder to fit your description of the ball not being the time defining element. But just because you can create concept (2) out of concept (1) doesn't mean they are logicaly compatible views of time.

Once you have a discrete time concept like in (2) you don't have an appearing and disappearing ball anymore. Since discrete time rather suggests that the ball is in location x0 at time t0 (x0, t0) and in x1 at t1 (x1, t1). Therefore the balls position at a certain time can be conceptualized as result of a discrete mapping function f(t) = x that mapps a location based on time or vice versa.

This leads to the ball being present in every instance of time in a specific location, contradicting the notion of disapearing and reappearing.

The same case could be made from a ego-centric view when it's always "now" it doesn't make sense to talk about discrete time. F.e. One could argue that since you only have one point in time we couldn't even investigate if it's discrete or not.

I know it seems like a small criticism that could easily be solved by replacing the words disappearing with discrete mapping. However I think that you used two logical exclusive formulations could suggest that you maybe don't use a proper distinction in your concepts and models. This could lead to all kind of confusion.

I hope I didn't missunderstand you.

CaZaNOx
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  • Thank you for your answer! I think you are right, the ball being present in every instance of time in a specific location, contradict the notion of disappearing and reappearing. I find also interesting that this model leads to imagine objects as properties of the time-space and not individual things. – Francesco D'Isa Dec 11 '18 at 08:30
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This is a case where it is useful to separate two views of the world: endurable and perdurable. An "endurable" view of the world is one which views things as snapshots in time. Perdurable views try to assign persistence to objects through time.

As a highly pertinent example, consider the glider from Conway's game of life.

Glider

This is a structure which reproduces itself every 4 generations. We can view this in an endurable way, viewing this as a series of snapshots with a symmetry that appears of period 4. Or we can view it as a perdurable object that is moving diagonally at c/4 (one quarter the maximum velocity of information in Life)

As a general statement, we do not think of gliders as "teleporting," which suggests that the perdurable model we tend to use when thinking of Conway's Game of Life treats this as something other than teleporting.

So you have to decide what teleporting means to you. How does your perdurable model work? Consider these two options:

  • The ball is teleporting on the closed interval [T1, T2] if there exists a time T on the open interval (T1, T2) such that the ball does not exist strictly between A and B
    • In the divisible example you gave, a middle instant, T1.5, is a witness. (Note: I'm relabeling your example such that T1 is the begining and T2 is the end in both divisible and indivisible cases. I think it's more clear that way) In the teleporting sub-case, it is the instant where the object does not exist. In the moving case, the ball exists strictly between A and B at T1.5.
    • In the indivisible point case, there cannot be a time that exists between T1 and T2. Thus the ball is not teleporting in this case -- it must be moving.
  • The ball is teleporting on the closed interval [T1, T2] unless there exists a time T on the open interval (T1, T2) such that the ball exists strictly between A and B at that time and the ball is moving on the closed intervals [T1, T] and [T, T2].
    • In the divisible example you gave, the middle instant, T1.5 is a witness. In the teleporting sub-case, it is the instant where the object does not exist. In the moving case, the ball exists at that middle instant, and it partitions the time into two divisible time periods. Mathematical induction proves this to be moving.
    • In the indivisible point case, there cannot be a time that exists between T1 and T2. Thus the object must be teleporting.

In a continuous world, those two statements are equivalent because I can always subdivide time and space. In a discrete world, those two statements are different statements when there are no times between T1 and T2 (for all statements are assumed true by convention if there are no elements in the set). Thus, if your sense of what "teleport" means depends on this sort of logic, you have to decide which definition you wish to use.

Cort Ammon
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  • Thank you for your answer, your example is perfect: I also think that the main issue is about endurance/perdurance. I'm not sure I got your points, are they not the same? – Francesco D'Isa Dec 11 '18 at 08:19
  • @FrancescoD'Isa They are better thought of as orthogonal. Think of an animated movie of a person walking. The endurable view sees only a series of individual frames, each one with a person in a different pose. Its your mind that provides the perdurable view that it is a single person that is walking through time. Generally speaking, the endurable views are easier to get agreement on. Perdurable views are much more complicated. The most famous issue that arises is [the ship of Theseus](https://en.wikipedia.org/wiki/Ship_of_Theseus) which questions when an entity ceases to be that entity. – Cort Ammon Dec 11 '18 at 15:26
  • Another fun example of these issues arising is quick change magic, like [this Japanese mask illusion](https://www.youtube.com/watch?v=taGqwxC2dpU). In this case, we know intuitively that there is one (perdurable) view that is "true" in which there is a single actor that exists continuously. However, through the illusion, they give us the impression that one character ends and another begins, which creates a dissonant perdurable view. Of course, being magic, we are forbidden from seeing a complete enough endurable view, which would show us the snapshots we want to see to prove... – Cort Ammon Dec 11 '18 at 15:30
  • ... that this is just an illusion, and indeed there is only one actor all along. Indeed we "know" that there are actually multiple masks hidden somewhere in each (endurable) snapshot, but through skill and art, they give the impression that there may be one mask which is simply changing (a perdurable way of thinking) – Cort Ammon Dec 11 '18 at 15:30
  • Ahh, and I found my math error in my points. I knew I was going to do that. I tripple checked it, but did the same mistake each time. I've ammended the second point to be different from the first. (I've gotten too used to applying the rules for converting universal and existential quantifiers that I kept doing it to preserve meaning when the whole point of those bullets was to make the points mean different things!) – Cort Ammon Dec 11 '18 at 15:48
  • Thank you for your suggestions! I get the first point but not the second. Since there's no T between T1 and T2, the ball In T1 and T2 exists in A or B, thus it should always move, not teleport. – Francesco D'Isa Dec 11 '18 at 17:42
  • But if there is no point inbetween where it is strictly between A and B. – Cort Ammon Dec 12 '18 at 05:20
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    Ahh. I may see the hole I dug. "Strictly" is a term in mathematics which excludes the edge cases. If x is strictly between A and B, that means A < x < B, as opposed to merely between, which often includes A <= x <= B – Cort Ammon Dec 12 '18 at 15:18
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The theory of particles forced on us by Schroedinger's wave equation, is that particles actually have to extend through all of space, and only shift so as to tend to be centered in different places, they don't actually leave one space when they enter another, they are just less in that old space and more in that new space. Or in String Theory terms a string's vibrational energy is centered somewhere, but its ends can sneak a considerable distance away, and then the energy can shift rather suddenly up or down the string. So the string 'moves' without really moving.

If matter itself has this pervasive continuity, whether space and time are discrete or not is not so much of an issue. Archibald Wheeler among others was totally ready to declare them cellular, with the realization that this would not present any more bizarre a situation than we have already been forced to accept. (His motivating data turned out not to be real, but he had much of the detail worked out while he thought it was.)

Basically, consider the process of anti-aliasing in computer graphics. A thing can be located between two pixels, and part of its color will be in one of them and part in the other, as it moves from one to the other it will fade from the previous one and intensify in the new one. No real discontinuity, despite total discontinuity... If matter is wavelike, it has to be capable of this trick. So no, it would not disappear in transition, it would fade over across the boundary largely as if there were no separation. (But not entirely, as uncertainty would have to apply separately to parts divided by the cells. So we would still see some effects.)

  • Thank you for your answer! These theories and their interpretations are not indisputable, but you gave me useful and interesting insights. Just one little note: with anti-aliasing some pixels around the borders of the figures are shaded to create the illusion of smooth curves, but it's just an illusion, every pixel keeps being a square https://www.displayninja.com/best-anti-aliasing-mode/ – Francesco D'Isa Dec 11 '18 at 20:44
  • @FrancescoD'Isa Right, and in this case every cell of the spacetime would keep being a cell. And no, it is not necessarily around the edge of the figure. if something is very far away, it might contribute to the color of only one pixel, and that might be its only representation. When it passes from one pixel to the other, it does just what I said, disappearing from the one it is seen in and contributing more to the one where it will be seen. The analogy is apt, but obviously don't take things too literally. –  Dec 11 '18 at 20:48
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Is the space discrete ?

From metaphysical point of view, time is a point of view. To have moments T1, T2 and T3, you need to have something to distinguish between them. In everyday life we use all sort of clocks and other time-measuring instruments. Even if nothing changes in your system (ball doesn't move), clock hand (clock display, whatever you use) would move or change. So, in order to have T1, T2 and T3 you already implicitly define time as discrete - you noticed new position of clock hand and you have new time moment.

What about space ? In T1 ball was at position P1. In T2 ball was at position P2 - ball moved. Is there an infinite number of moments between T1 and T2, such that T is between T1 and T2 ? It doesn't matter, you didn't notice any of these moments. You only have T1 and T2, therefore you only have position P1 and P2. If you consider T2-T1 to be minimal unit of time, question of ball position between these two moments becomes superficial - ball could not exist in a moment T between T1 and T2 , therefore it does not have any position between P1 and P2 . Therefore - if you consider time to be discrete, space also must be discrete and vice versa.

rs.29
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  • Thank you for your answer. You are right, space should be discrete as well - but the question remains. – Francesco D'Isa Dec 10 '18 at 09:45
  • @FrancescoD'Isa You defined teleportation as ball not existing between T1 and T2. In discrete time (and you set up your system as discrete-time system ) there is no moment T between T1 and T2 where ball would not exist. Your definition of teleportation is superficial in discrete time if T2-T1 equals your minimum unit of time. – rs.29 Dec 10 '18 at 18:46
  • yes, but we have to accept that we have a space where something disappear (a) and then reappear (b). That objects are sort of properties of the spacetime. – Francesco D'Isa Dec 10 '18 at 19:35
  • @FrancescoD'Isa It does not disappear/reappear because you (or anybody else) could not detect system without the object . Coordinates of the object are T1,P1 and T2,P2 . That's about it. If you introduce T between T1 and T2, then your premise about minimal unit of time no longer stands. Make up your mind ;) – rs.29 Dec 10 '18 at 20:04
  • That none can't detect it doesn't mean that it doesn't happen (maybe - this is another puzzle ;)) by the way I can correctly describe this situation as: the object disappear in T1,P1 and the object appears in T2,P2. No T between T1,T2. – Francesco D'Isa Dec 10 '18 at 20:12
  • @FrancescoD'Isa It doesn't happen because none and no instrument could detect it, even theoretically. You are making a mistake of implicitly creating super-universe from which it is possible to observe teleportation in your original universe, because it has different rules then your original universe. That is whole another ball game. – rs.29 Dec 11 '18 at 05:26
  • Your discrete space argument does not hold. You say that space is discrete because the moving ball only exists at times T1 and T2, taking positions P1 and P2 with no intermediate states in between. A ball moving at half the speed will exists at T1 and T2, but at positions P1 and P1.5. A ball moving at half *that* speed exists at positions P1 and P1.25. Even if time is discrete, space doesn't have to be, as you can show that there are indeed infinite positions between P1 and P2. – Nuclear Hoagie Dec 11 '18 at 15:54
  • @NuclearWang That doesn't change the fact that space is discrete. Ball exists at positions P1 and P1.5 (or P1.25, P1.1 etc ...) , but it doesn't exist between those positions. There is infinite number of real numbers between any two given real numbers, but our ball , depending on its speed, exists only in two points. – rs.29 Dec 12 '18 at 07:59
  • @rs.29 That implies time is discrete, not space. You seem to agree that the ball can take *any* position between P1 and P1.5 (P1.1, P1.01, P1.001, P1.0001...). If the space is infinitely divisible, it's not discrete. A discrete space would imply that the ball can only exist at particular locations that fall on a grid. There are an infinite number of real points between two real numbers, and the ball can exist at any of them. The idea that it might only take certain positions and nothing in between relates to the nature of time, but not space. – Nuclear Hoagie Dec 12 '18 at 13:50
  • @NuclearWang Ball could take exactly one position out of many, and this is exactly the reason why the space is discrete. Opposite of discrete space would be continuous space , where position of ball is continuous function p=f(t) . Discrete space does not mean that space has limited number of positions in a given volume . It simply means that available positions are isolated points . https://en.wikipedia.org/wiki/Isolated_point – rs.29 Dec 12 '18 at 18:34
  • @rs.29 Your link only proves my point - the very first sentence states that "x is called an isolated point of a subset S if x is an element of S but there exists a neighborhood of x which does not contain any other points of S". If there are infinite points in a limited space (which you agree is the case), there *is no neighborhood of x which does not contain other points of S*! For any size neighborhood, even as the size approaches 0, I can find a neighboring point. – Nuclear Hoagie Dec 12 '18 at 19:13
  • @NuclearWang You forgot that we have minimal unit of time in this universe. Therefore, any speed that is not zero would give us point that is isolated from previous point (and zero speed would give us exactly one point) . Infinite number of points in limited space does not mean that they are not isolated from each other . – rs.29 Dec 12 '18 at 19:22
  • @rs.29 Yes, it means exactly that. There is no way to fit an infinite number of isolated points in a finite space - there will always be another point nearby, regardless of how narrowly you define "nearby". I encourage you to read the link you posted more closely. I am finished with this discussion. – Nuclear Hoagie Dec 12 '18 at 19:29
  • @NuclearWang Yes, there is :) You lack basic mathematical knowledge for this type of discussion. I'll give you an example. In our universe , minimal movement of ball is defined as p=v*t , where v is speed of ball, and t is our fixed minimal time. v could take any number (infinite number of combinations) . But for any v>0, there is v*t/2, v*t/3, v*t/4 ... As you can see, whole another level of infinity :) This is why positions in our universe are isolated points. – rs.29 Dec 13 '18 at 03:56
  • @NuclearWang Also, there is whole another discussion how would you even measure speed (as derivative of time and space) in a universe with minimal units of time. For example, would Heisenberg's principle still apply ? https://en.wikipedia.org/wiki/Uncertainty_principle – rs.29 Dec 13 '18 at 04:03