# Local EPR-experiments with photons in vacuum?

The principle of non-locality states "that an object is influenced directly only by its immediate surroundings." (Wikipedia) When two entangled particles are measured in an EPR experiment, we consider these events as non-local. Generally I agree. But does this also apply to a pair of photons in vacuum? (and if so, why?) From the point of view of any observer's Minkowski diagram, the two spacetime intervals of the lightlike worldlines of the two photons AB and AC are always zero (so-called "empty intervals"), and by consequence, B and C are located in the "immediate surroundings" of A, and the events should not be nonlocal.

For more clarity I replaced "B in immediate surroundings of C" by "B&C in the immediate surroundings of A". In spite of the attribution of a bounty, the question is still unanswered.

As user 12262 pointed out, some comments have disappeared in the meanwhile.

Edit 01/07/14: Up to now, no answer explains why there is the word "interval" in "spacetime interval = 0".

• Moonraker: "why there is the word "interval" in "spacetime interval = 0"" -- Especially because it refers to some pair of events; in general to some pair of distinct events, such as "spacetime interval between events $A$ and $B$", with magnitude $s[ A B ] = 0$ (provided that the distinct events $A$ and $B$ are light-like related to each other). The magnitude $s$ of spacetime intervals is consequently a binary function which takes two events as arguments. – user12262 Jul 1 '14 at 21:43

spacetime intervals of the lightlike worldlines of the two photons [...] in vacuum [...] AB and AC are always zero (so-called "empty intervals"),

Rather "null intervals" (i.e. intervals of zero magnitude).

and by consequence, B is located in the "immediate surroundings" of C

Hardly ...
But first of all we might investigate whether events A and B are "in the "immediate surroundings" of each other", for instance (or similarly: events A and C).

The sketches clearly suggest that events A and B are distinct events. (Which does not contradict the magnitude $s[ A, B ] = 0$. This property of "null intervals" is certainly different from null vectors in the arguably more usual sense of preserving identity when being added to any other vector.)

Event B is "on the future light-cone of" event A. (And event C is "on the future light cone of" event A as well.) In some sense it may therefore be said that in some sense event B "was influenced first hand by" event A (and that event C "was influenced first hand by" event A, too).

But, as both sketches suggest, we should not merely consider the events A, B and C themselves but the geometric relations between photon source (which took part in event A) and the two (separate) detectors (one of which took part in event B, but not in event C; and the other detector vice versa).

Presumably the source and the two detectors were and remained at rest to each other throughout the experiment; accordingly their configutation is characterized by the indicated (non-zero!) distance values. Surely it cannot be said that the source and the detector which took part in event B had been "in the "immediate surroundings" of each other"; neither could be said that the source and the detector which took part in event C had been "in the "immediate surroundings" of each other"; nor that the two detectors had been "in the "immediate surroundings" of each other".

Finally looking again at the relation between events B and C it can be said that

• they're distinct,

• they're spacelike separated, with interval magnitude $\sqrt{ 440 } ~ \text{light years}$, and

• the principal identifiable participants of the described experimental setup who took part in one or the other of these two events (namely the two detectors) were at rest to each other, throughout the experiment, at a distance of $21 ~ \text{light years}$.

These three characterizations are mutually consistent; and either one would seem to indicate that events B and C should not be called "in the "immediate surroundings" of each other".

p.s.

The principle of non-locality states [... http://en.wikipedia.org/wiki/Principle_of_locality ]

Rather, what's stated seems to be called "principle of locality", for what it's worth.

• It seems you want to say that a spacetime interval is not an interval. Could you justify your opinion? By the way, Wikipedia provides fundamental information in the article about intervals. en.wikipedia.org/wiki/Interval_%28mathematics%29#Terminology – Moonraker Jun 16 '14 at 18:49
• Moonraker: "It seems you want to say that a spacetime interval is not an interval." -- By "spacetime interval" I mean an ordered pair of events ("in spacetime"); its magnitude (defined as) zero if these two events are light-like related; non-zero positive if space-like related; non-zero negative if time-like related. These conventions also fit the Inverse triangle relation for time-like or light-like pairs; OTOH, en.wikipedia.org/wiki/Interval_%28mathematics%29#Terminology corresponds only loosely – user12262 Jun 16 '14 at 19:51
• p.s. To state that "Inverse triangle relation for time-like or light-like pairs" in a form that I consider better defensible: Given events $H$, $J$, and $K$ such that the intervals $HJ$, $HK$ and $JK$ are all time-like or light-like, i.e. such that $$0 \ge s[ HJ ],$$ $$0 \ge s[ HK ],$$ and $$0 \ge s[ JK ]$$ then the sum of any two (of these negative magnitudes) is larger than (or at least equal to) the remaining third one. – user12262 Jun 16 '14 at 20:11
• user12262: "[...] then the sum of any two (of these negative magnitudes) is larger than (or at least equal to) the remaining third one." -- That's wrong; sorry. Rather: if (without loss of generality) $$0 \ge s[ HJ ] \ge s[ JK ] \ge s[ HK ]$$ then obviously $$s[ HJ ] \ge s[ JK ] + s[ HK ]$$ as well as $$s[ JK ] \ge s[ HJ ] + s[ HK ],$$ but (in some sense inversely) $$s[ HK ] \le s[ HJ ] + s[ JK ].$$ And likewise writing $\tau := -s \ge 0,$ with $$\tau[ HK ] \ge \tau[ HJ ] \ge \tau[ JK ] \ge 0$$ holds $$\tau[ HK ] \ge \tau[ HJ ] + \tau[ JK ],$$ in contrast to a "triangle inequality". – user12262 Jun 18 '14 at 0:23

The issue is that the measurements at $B$ and $C$ are correlated, even if the spacetime interval $BC$ is spacelike.

• Rob, I don't think that this is the issue: The events B and C are correlated because of the entanglement. But in case of electrons (or photons in a medium) this correlation is a non-local phenomenon. - My question is referring to the specificity of the case of photons in vacuum. – Moonraker Jun 7 '14 at 14:24
• I don't understand why you think the presence or absence of a medium matters. Can you elaborate? – rob Jun 7 '14 at 15:05
• The spacetime interval of vacuum light speed c is zero. Photons in a medium are moving at less than c. – Moonraker Jun 7 '14 at 15:33
• @Moonraker: "The spacetime interval of vacuum light speed c is zero." What you are saying (based on visualizations and not reality) means that one can have a (non-zero) speed without the passage of time or without a change in distance. That's actually the most effective way to produce paradoxes like the one you are trying to resolve. – bright magus Jun 17 '14 at 13:26
• @bright magus: I do not understand well, why you are talking about velocity, my question is about two empty intervals (en.wikipedia.org/wiki/Interval_%28mathematics%29#Terminology). – Moonraker Jun 17 '14 at 15:44

In your diagram, the spacetime interval between B and C is $(x_c-x_b)^2-(t_c-t_b)^2 = 21^2-1^2= 440$. So $B$ is not in the immediate surroundings of $C$, there is a spacelike interval between them. Also, the EPR experiment has an entirely local explanation, see:

• There had been several comments submitted here until June 20th, 2014, by the OP (@Moonraker), by myself, and several others. They all seem to have been removed. I'll hereby restate the pertinent comment (follow-up question) by the OP: $$~$$ @alanf: "the [signed square of the magnitude of the] spacetime interval between B and C is [...] $440$. So B is not in the immediate surroundings of C" -- Can a reverse argument be applied to events A and B, for instance? The signed square of the magnitude of spacetime interval $AB$ is $$s^2[ AB ] = 0.$$ Is B therefore in the immediate surroundings of A? – user12262 Jun 24 '14 at 16:21
• No, but it is possible for a signal to get from A to B. It does so by influencing the immediate surroundings at each point along the line AB. There is no path from B to C along which a signal could propagate in this way. – alanf Jun 24 '14 at 16:28
• alanf: "[...] No." -- Agreed. However: If $s^2[ AB ] = 0$ is not suitable to justify that "A and B were in the immediate surroundings of each other" then the argument (put forth in your answer) that $s^2[ BC ] \ne 0$ implies that "B and C were not" isn't really convincing. IMO, better arguments may be made. (Unfortunately the OP didn't really define how to evaluate "being in immediate surroundings" in the first place.) "[...] influencing the immediate surroundings" -- Is "being in immediate surroundings" symmetric and transitive ?? – user12262 Jun 25 '14 at 5:19
• @alanf "[...] No." User 12262 agreed, but I do not understand. If the spacetime interval is 0 (= so-called empty interval), doesn't that mean that there is nothing between A and B?) – Moonraker Jun 28 '14 at 21:27
• @Moonraker: "but I do not understand" -- And I wonder whether you read my answer thoroughly. Anyways ... "If the spacetime interval is 0 [...] doesn't that mean that there's nothing between A and B?" -- So that's a fair question itself: What exactly is between emission event A and reception event B? Between emission event and reception event is: transmission or propagation of the signal. (And between source and "left" detector of your example are 10 light years; and between the source emitting and the "left" detector receiving are 10 years.) – user12262 Jul 2 '14 at 20:41

The principle of non-locality states "that an object is influenced directly only by its immediate surroundings." (Wikipedia)

This is a rather colloquial formulation of the principle of locality. In this "definition" the surrounding is meant with respect to space. A better formulation would be that an event $A$ can only be influenced by events that lie within or on the backward light cone of $A$. That means the special relativistic distance between two events must not be spacelike, and the causing event must occur "before" the caused one.

In your example, the distance between $B$ and $C$ is spacelike, so they cannot influence another according to the principle of locality. However, they are both be influenced by event $A$, because $A$ is on the backwards light cone of both $B$ and $C$.

The EPR paradox shows a violation of locality only in the Copenhagen interpretation and in several hidden variable theories. In contrast, locality is not violated in the EPR of the many world interpretation (but that is a different question).

• M.Herzkamp, "In this "definition" the surrounding is meant with respect to space." : Why "spacetime interval" is called "interval"? – Moonraker Jun 17 '14 at 13:56
• For non-negative values the spacetime interval is the proper time an observer experiences, when they move from one event to another. I don't know if there is a meaning like that behind negative values. – M.Herzkamp Jun 17 '14 at 16:52
• M.Herzkamp: "[...] the proper time an observer experiences" ... a.k.a. the duration of that observer ... "[...] from one event to another." -- For a fixed pair of events (say $J$ and $K$) which are light-like separated there are many distinct observers to be considered who all took part in both events $J$ and $K$, but who "went their separate ways" inbetween. Their corresponding durations ("from having the others depart, until having the others join again") are in general different. Of all these durations the maximum is (taken as) magnitude of interval $JK$ (give or take a minus sign.) – user12262 Jun 18 '14 at 20:08
• The magnitude of any space-like interval, say $PQ$, is also related to durations (by Synge's dictum: "For us time [duration] is the only basic measure.") In a flat region, we can consider pairs of observers who remained always at rest to each other, where one took part in event $P$, the other in $Q$. The ping durations of either observer (of any particular such pair) wrt. the other are equal; the value $$c/2 \times T_{ping}$$ is taken as their (mutual) distance. Of all such distance values (of all pairs to be considered) the minimum is (taken as) magnitude of interval $PQ$ (plus/minus). – user12262 Jun 18 '14 at 20:52

Minkowski geometry is not Euclidean geometry.

In Euclidean geometry, the "distance" between any two points can be positive or zero, but not negative or imaginary. In Minkowski geometry, that's not true.

In Euclidean geometry, if the "distance" from A to B is zero, then A and B are actually the same. In Minkowski geometry, that's not true.

In Euclidean geometry, there is the triangle inequality: The "distance" from B to C is less than or equal to the sum of the "distance" from A to B plus the "distance" from B to C. In Minkowski geometry, that's not true. Your question is a perfect example.

People sometimes use words like "interval" and "distance" in Minkowski space but that doesn't mean that you can intuitively think of them as being like a traditional Euclidean-space distance or interval. If this really bothers you, then you should mentally replace the phrase "spacetime interval" with different words, "spacetime shminterval". This will vividly remind you that it's entirely different than the intervals that you're used to from everyday life.

• There is no Euclidean geometry used (where?), and it is a fact that a spacetime interval is an interval, with all properties of an interval. In particular, an empty spacetime interval is preventing non-locality. – Moonraker Jan 9 '15 at 18:42
• The distance BC is not important. It is essential that B and C are separated from their respective initial point near A by an empty interval. – Moonraker Jan 9 '15 at 18:46
• When you say "all properties of an interval", I think you mean "all the properties that I intuitively expect an interval to have", but your intuition is wrong because your intuition is based on a lifetime of experience with Euclidean geometry. Well, you're entitled to have the opinion: Anything that deserves to be called an interval must certainly have the following properties...!. But you must understand that other people do not feel this way. That's why I suggest that you mentally replace the word "interval" with "shminterval", a made-up word which is therefore free of your preconceptions. – Steve Byrnes Jan 9 '15 at 20:09
• I recommend you to inform you about intervals, empty intervals, space-time intervals and lightlike intervals – Moonraker Jan 9 '15 at 20:23
• Sometimes words have multiple meanings. You found an article called "Interval (mathematics)" which has a certain definition of "Interval". But you are wrong to conclude that this is the universal and only definition of the word "Interval". The more relevant article is en.wikipedia.org/wiki/Interval , which proves that the word "Interval" has at least 15 different definitions. The word "interval" in the context of special relativity means something different than the the word "interval" in the context of the number line, just as it means something different in music, or in sports. :-D – Steve Byrnes Jan 9 '15 at 21:34

I. our real world, with accelerating space-expansion

In this world, the emitters and receivers cannot keep their distances constant, and you would get a hard-time to calculate and adjust their distances to 'keep' it 10 and 11 light years 'until' the photons reach the receivers.

II. hypotethical world of no space-expansion

OK let's say this world has no space-expansion. The emitter and receivers are constant 10 and 11 light years away.

1. the entangled photon pair is traveling at speed c in opposite space directions towards the receivers. The photons' speed in the time dimension is zero.
2. point A, B, C is stationary in space, and moving at (the magnitude of) speed c in time dimension (that is how the four-vector and our universe is set up)
3. you are correct where you say "B and C are located in the "immediate surroundings" of A" ONLY if you take it in spacetime, in 4D. The four-vektor and the universe is set up so that the spacetime distance of A,B,C will stay constant 0.
4. These events are considered local in spacetime, you are right, if we are using particles with zero rest-mass.

Here I say that entanglement has many explanations, and there are many experiments, but the experiments do not explain HOW. And there is no ACCEPTED explanation of HOW. There is the many-worlds explanation. And there is the block-universe explanation. My explanation is more close to the block-universe explanation (that is the explanation in n° 5).

There are multiple explanations to the experiments on entanglement. But there is no ACCEPTED explanation of HOW. They IMO could all be accepted. I think the many-worlds explanation looks good. But I prefer the block-universe explanation with higher dimensions where the entanglement is and we do not 'see' that higher dimension.

Now let's talk about non-locality and your questions in n° 4. "EPR experiments are currently considered to be non-local" OK you are right where you say that most of the EPR experiments are ACCEPTED by QM theories to be non-local, but only in space (3D). In spacetime (4D) it is different. According to SR, the line that you drew, with "spacetime interval=0" is exactly what SR says, for particles with zero rest-mass, it must be 0 so you are correct, there is no need for the word "interval".

The reason for that is what SR says about photons, they are traveling at speed c in space (3D), but their speed in the time dimension is 0. That is why that is a 45degree line and why on that line as you move, your spacetime interval is constant 0. You are not moving in the time dimension if you move on that line, that is, if you travel at the speed of light, but that is only possible if you have zero rest-mass like a photon.

"EPR experiments are ACCEPTED by QM theories to be non-local" and the reason I said that is because Einstein never accepted that (QM) theory to be complete, and never accepted that it is non-local in the EPR experiments. IMO Einstein was right. Please read en.wikipedia.org/wiki/EPR_paradox and this part: ""Acceptable theories and the experiment According to the present view of the situation, quantum mechanics flatly contradicts Einstein's philosophical postulate that any acceptable physical theory must fulfill "local realism"."

So I think Einstein never agreed on that either. IMO he was right when he said EPR must fulfil local realism. An that QM is not complete as a theory. That is why QM cannot explain HOW entanglement works. So you are correct on locality.

1. The receiver at B will never receive anything, since before the 2nd photon would get there, the 1st photon going to C is already absorbed, it's wavefunction collapses, that (since they are entangled) collapses (instantenously) the wavefunction of the 2nd photon too. (This is the part what is not describable easily, since according to SR they cannot communicate instantaneously, but that only means that they cannot communicate in space instantenously. But in this case in fact, they are also entangled in 4D spacetime or some higher dimensions(I have no information on this theory, but would be nice to know). Though they are separated in space (3D), they are still entangled, it might be an entanglement in higher dimensions. This is one way to describe how the entangled pair can 'communicate' and collapse their common wavefunction just by absorbing one of the photons.)
2. Now please note that if you would use non-entangled photons, so that they would be shot from the same spacetime position, then the receiver at B would also receive the 2nd (non-entangled) photon. Because in this case they will not have a common wavefunction.
3. Let me really clarify what it means that A,B,C are local. It means that as B is 11 light years from A, nobody could travel faster then as information from A would reach B. Now to move from A to B you, as you have rest-mass, would need to travel slower then speed c in space, you would leg behind the photon (let's say you started with the photon) but that means you need to move in the time dimension too. Now as you would move in space from A to B, your space coordinates would change just as much that it will change your time coordinate exactly as much (the travel will take as much time) so that when you get to B, your movement in space and your movement in time will give the result that in spacetime position you are not anymore as when you were in A. But the photon in different, it moves in space at speed c, and does not move in the time dimension. A person from B would see the photon coming, and would see you legging behind and getting older faster then he is on B. In spacetime you did not move only if you moved with the photon. If you move with the photon, the person in B will see you get older the same speed as he is getting older. You moved in space, and you had to move in time, but the spacetime coordinates will show with the four vektor that your spacetime interval that you took is 0.

4. "Up to now, no answer explains why there is the word "interval" in "spacetime interval = 0"." To answer your question, the reason to use that is that if you use particles with non zero rest-mass, they will move in space (3D) with a speed less then c, and they will move in the time dimension with a nonzero speed. Their spacetime interval thus will be nonzero too. On your image, the line that has spacetime interval=0 is only for photons, since they are moving at speed c in space, and are stationary in the time dimension. Using the word interval is based on the four-vector, and the 4D 'distance'measurement based on that. For particles with zero rest-mass, this will be constant zero and you are correct, in case of photons, there is no reason using the word 'interval' because it is constant 0. But for particles with rest-mass it is the spacetime interval that describes their 'distance' in 4D. Now to answer why in a usual case we use the word interval, is just because if you would travel from A to B, and you would start at the same time as the photon, your spacetime interval would be calculated based on the four vektor, and in a case you would leg behind the photon. By the 'time' the photon would get to B, you would still be behind, and you would need more time to get to B. A person at B looking at you from B traveling from A to B would see you getting older on your way faster then he is. So the spacetime interval would be non-zero. But if you, traveling from A to B would travel at the speed of light, the person at B would see you getting older just as fast as he is. You would be moving in his spacetime position, your spacetime interval would be constant 0. Your space interval would be decreasing as you would move closer to B, but your spacetime interval would not change, because your time interval would change accordingly.

• You are understanding my question by far better than the other answers. +1 for n° 4, but this answer is a bit too simple, if you are agreeing with me on locality, you should provide some more information because EPR experiments are currently considered to be non-local (e.g. do you think that everybody is wrong? What might be the reason why this massless-particle-exception is nowhere to be found?) -In contrast, don't care about my edit of 01/07/14. This question is only for those who do not agree. – Moonraker Nov 30 '16 at 10:17
• Regarding n° 5: You did not get the whole reasoning: When both particles are at A, they are only nearby, they are not exactly at the same point. What I say is that for massless particles, their respective end points B and C correspond exactly to their respective initial points at A which are very close one to each other (= local) but I do not say that they must be considered to be at the same place. Two particles can never be at the same place, but this is just sufficient for locality. – Moonraker Nov 30 '16 at 10:18
• Regarding n° 5 You are right when you say this "which are very close one to each other (= local) " So we could do this, and shoot them from very close (but not the same spacetime coordinate), and still they would stay local in spacetime ONLY. So that one is correct. – Árpád Szendrei Nov 30 '16 at 17:33
• Regarding n° 5 "but I do not say that they must be considered to be at the same place."You are correct, they do not HAVE to be shot from the same spacetime position, but in the case of what you write, is entangled photons, they will be created (there are some cases when you create entangled particles separately but I have no information on that, would be nice to know) at the same spacetime position. That is why I used this example where they are created at the same spacetime position and shot from there. – Árpád Szendrei Nov 30 '16 at 17:33
• Now here starts the misunderstanding: "Two particles can never be at the same place" Photons are bosons, they do not obey the Pauli exclusion principle. An infinite number of photons can be at the same place. ", but this is just sufficient for locality." Here you are right. But I used the example you gave, so entangled photons, and then those are usually shot from the same place. – Árpád Szendrei Nov 30 '16 at 17:36