# Does determinism exclude retrocausality?

• I
Gold Member
Suppose the system under examination is fully deterministic. Does that imply that effects follow causes and not precede them?

For instance, if in this system Alice would respond to event X with A, but, if instead of X event Y would have happened, with B, does that mean she has no choice between A and B in that case?

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Suppose the system under examination is fully deterministic. Does that imply that effects follow causes and not precede them?
Bohmian mechanics is fully deterministic, and posits retrocausality, so I'd say determinism does not imply that.

For instance, if in this system Alice would respond to event X with A, but, if instead of X event Y would have happened, with B, does that mean she has no choice between A and B in that case?
This isn't an example of retro causality.
Alice wants vanilla, and she'd respond to X: vanilla ice cream being presented to her right with the action A: reaching to her right. But if instead Y happens (the vanilla is placed to her left), then she'd respond with B: reaching to her left. Straight cause and effect there, nothing retro.
Retrocausality would be her reaching for the ice cream before she knows where it is going to be placed.

Dale
Mentor
Edit: I have changed my notation to make it consistent with @entropy1

Suppose the system under examination is fully deterministic. Does that imply that effects follow causes and not precede them?

For instance, if in this system Alice would respond to event X with A, but, if instead of X event Y would have happened, with B, does that mean she has no choice between A and B in that case?
For any discussion of causality it is absolutely essential to be clear about what definition you are using for causality. Otherwise you have people arguing who think they are arguing about substance when they are actually just arguing because they are using different definitions. Here are some suggested definitions, for clarity: https://en.wikipedia.org/wiki/Causality

Necessary causes: If x is a necessary cause of y, then the presence of y necessarily implies the prior occurrence of x. The presence of x, however, does not imply that y will occur.

Sufficient causes: If x is a sufficient cause of y, then the presence of x necessarily implies the subsequent occurrence of y. However, another cause z may alternatively cause y. Thus the presence of y does not imply the prior occurrence of x.

So, given the laws of classical physics, and in particular the time reversibility, if we have an initial condition ##\cancel{A} \ Y## then we can apply the laws of physics to calculate a final condition ##B## at any later time. However, we can also start from the final condition ##B## and use the laws of physics to calculate backwards to the initial condition ##\cancel{A} \ Y##. So ##\cancel{A} \ Y## implies ##B## and ##B## implies ##\cancel{A} \ Y##.

Thus we see that ##\cancel{A} \ Y## is both a necessary and a sufficient cause of ##B##. However, by the definitions above ##B## fails to be a necessary cause of ##\cancel{A} \ Y## because ##B## did not occur prior to ##\cancel{A} \ Y##. Similarly, ##B## fails to be a sufficient cause of ##\cancel{A} \ Y## because ##\cancel{A} \ Y## did not occur subsequent to ##B##.

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jim mcnamara and entropy1
Gold Member
Could one argue that in the deterministic setting, if at present B would happen instead of A, the consequence would be that consecutively something different would happen (in general/in the universe) than if A would have happened?

And if B would happen instead of A, something different preceded this event than if A would have happened?

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hutchphd
Thus we see that A is both a necessary and a sufficient cause of B.
And if B would happen instead of A, something different preceded this event than if A would have happened?
OK now I am completely confused. How are these statements compatible?

Dale
Mentor
And if B would happen instead of A, something different preceded this event than if A would have happened?
I think @hutchphd is right. This is not logically possible. You can have both A and B or neither A nor B. But you cannot have A instead of B.

Edit: I am confused here. You can have both A and X or neither A nor X, but you could instead have B and Y.

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Not sure. A is a measurement of spin up, B is down. You can't have both or neither, so it's A instead of B, or B instead of A. This is an uncaused difference. Nothing different 'preceded event B than if A would have happened'.
OK, that's a QM interpretational assertion. There are deterministic interpretations that say B is determined and A is out, due to some preceding state, possibly a hidden one. Something like MWI says both A and B, or neither, so we're more in line with Dale's statement. WF collapse interpretations (like RQM for instance) are not necessarily deterministic, so there is B or A, even if there is no 'dice rolling' in some of them.

Am I totally inconsistent with this line of thinking?

entropy1
Dale
Mentor
A is a measurement of spin up, B is down. You can't have both or neither, so it's A instead of B, or B instead of A.
Where did that come from? That isn’t what we are talking about. B is caused by A so A and B are not what you described. Not to mention that we are talking about completely deterministic systems.

Edit: I am also confused here. B is caused by Y, not A.

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I'm talking about followup discussion on entropy1's post 4, which is A or B, due to some prior state or not, not the scenario of one being the cause of the other.
It did get confusing when the same terms were used to describe A causing B.

But you are apparently talking about the other scenario in post 6, which is A being necessary and sufficient for B, in which case yes, both A,B or neither. I don't think that's what @entropy1 was talking about.

entropy1
Dale
Mentor
I'm talking about followup discussion on entropy1's post 4,
That was still explicitly deterministic, so I still have no idea why you brought in your example. It seems off topic for a thread about deterministic systems.

Gold Member
B is caused by A
If I wrote that, that would be a mistake...
I don't think that's what @entropy1 was talking about.
Thanks for clearing that up. Post #4 is a somewhat separate question from post #1. In fact, I was working toward a more physical matter myself.

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Dale
Mentor
If I wrote that, that would be a mistake...
Ok, so in your notation what is the cause and what is the effect? Please be clear, there has been substantial confusion.

Gold Member
Ok, so in your notation what is the cause and what is the effect? Please be clear, there has been substantial confusion.
In post #1 A and B are events that are caused by resp. causes X and Y. In post #4 I extended this with that A and B cause unspecified effects (consequence) and are caused by unspecified causes (precedence). I left it unspecified deliberately at this point.

Dale
Mentor
In post #1 A and B are events that are caused by resp. causes X and Y. In post #4 I extended this with that A and B cause unspecified effects (consequence) and are caused by unspecified causes (precedence). I left it unspecified deliberately at this point.
Ah, ok, my post above was incorrect then. I have gone back and added edits to correct my usage and make it consistent with your notation.

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Dale
Mentor
Could one argue that in the deterministic setting, if at present B would happen instead of A, the consequence would be that consecutively something different would happen (in general/in the universe) than if A would have happened?
Definitely. If B instead of A at present then previously Y instead of X.

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Gold Member
And if B would happen instead of A, something different preceded this event than if A would have happened?
Edit: I am confused here. You can have both A and X or neither A nor X, but you could instead have B and Y.
Yes, I'm suggesting exactly that: if B would have happened at present, there would be a different preceding cause than if A would have happened (in general, but in a deterministic setting when it is either A or B, think of a quantum measurement, like Halc mentioned). But I'm not sure if this is true.

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Dale
Mentor
Yes, I'm suggesting exactly that: if B would have happened at present, there would be a different preceding cause than if A would have happened (in general, but in a deterministic setting). But I'm not sure if this is true.
Yes, this is true by your above notation and the above definitions. Since X and Y are necessary and sufficient causes of A and B respectively: we therefore know that B at present implies prior occurrence of Y, whereas A at present implies prior occurrence of X. This is by the definition of necessary cause.

Gold Member
Yes, I'm suggesting exactly that: if B would have happened at present, there would be a different preceding cause than if A would have happened
Yes, this is true by your above notation and the above definitions. Since X and Y are necessary and sufficient causes of A and B respectively: we therefore know that B at present implies prior occurrence of Y, whereas A at present implies prior occurrence of X. This is by the definition of necessary cause.
I have to study this; this is brain exercise for me. (necessary cause vs. sufficient cause)

If X is only a sufficient cause for A, X is not mandatory to have happened. X' or X'' could be causes too. Then, X' could be a cause for both B and A. This means that if B happened the preceding cause could be X', and if A happened it could be too. So there would be a single cause, not different ones, and no need for different histories for A and B. Does that make sense?

But then, if there is some rule that A and B can't happen both, and can't happen neither, the choice between A and B would be random, right?

If X would be sufficient cause for A, and X happens, does that mean that A happens?

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Dale
Mentor
If X is only a sufficient cause for A, X is not mandatory to have happened. X' or X'' could be causes too. Then, X' could be a cause for both B and A. This means that if B happened the preceding cause could be X', and if A happened it could be too. So there would be a single cause, not different ones, and no need for different histories for A and B.
Does that make sense?
It does make sense for a sufficient cause. However, in classical physics all causes are necessary and sufficient causes.

If X would be sufficient cause for A, and X happens, does that mean that A happens?
Yes. If X is a sufficient cause for A and X happens then subsequently A happens.

Gold Member
A different view on the matter is the following:

Provided the system (universe) is deterministic, if I respond differently (to some event), the event (cause) was different (because if the cause was the same, the response would be the same, right?).

For example: if I respond with A, the cause was X, but if I respond with B, there would have to have been a different cause Y.

These causes (X and Y) would have to be necessary causes.

In this example A and B are mutually exclusive (one of them happens at t0).

If X and Y would on the other hand be sufficient causes, then some different cause Z could cause A as well as B. If A and B are mutually exclusive, does that mean that the respons to Z (A or B) is random?

Is that correct?

And also, is it so that if X is necessary cause of A, that occurence of A implies occurence of X, and that if X is sufficient cause of A, that occurence of X implies occurence of A?

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Stephen Tashi
Determinism or non-determinism is only well defined, if we have a definition of "states".

The use of notation like "A" and "B" to denote phenomena is somewhat ambiguous. I think most contributors to this thread are using notation like "A" to denote a set of different possible phenomena.

For example, "A" could denote "It rains at my house on Monday morning". There are many different phenomena that are described by "It rains at my house on Monday morning". The can differ in details like where the rain drops land and what the closing value of the Dow Jones Industrial Average is on that day.

A different use of notation like "A" is to denote a unique event, as opposed to a set of events.

Either interpretation of notation leads to a host of problems!

entropy1
Dale
Mentor
If X and Y would on the other hand be sufficient causes, then some different cause Z could cause A as well as B. If A and B are mutually exclusive, does that mean that the respons to Z (A or B) is random?
This is why, in a deterministic setting, all causes must be both necessary and sufficient.

entropy1
Gold Member
Is it true that if X is necessary cause of A, that occurence of A implies occurence of X, and that if X is sufficient cause of A, that occurence of X implies occurence of A?
And since I chose this title for the thread, does in the first case (X is necessary a condition for A, occurence of A implies occurence of X) mean that A could be a cause for X (retrocausally)?

Stephen Tashi
A different view on the matter is the following:

Provided the system (universe) is deterministic, if I respond differently (to some event), the event (cause) was different (because if the cause was the same, the response would be the same, right?).
If the universe is deterministic there is no "if" about how you respond. You respond deterministically. So, for your idea to be coherent, we have to imagine that your are using the term "some event" to denote a set of phenomena and that the history of universe contains several examples where phenomena of that type occurred. Likewise, what you call your "response" must also described by at least two mutually exclusive sets of phenomena ##R_1, R_2##. Otherwise the history of the universe contains no examples where you responded "differently".

However, if "some event" ##S## can be followed by two different sets of events ##R_1, R_2## then some of the events ##S,R_1,R_2## are not "states" of the universe. (In a deterministic process, a given "state" is always leads to a unique later "state". That is the definition of "state" and "deterministic".)

You are attempting to discuss cause-and-effect properties for sets of events that are not states. This might be possible, but I think it's tricky!

Gold Member
@Stephen Tashi

If we have state of the universe X at t=0, then you say, that, under determinism, there is only a single possible state of the universe at t=1, say A, right?

So I am suggesting that if at t=1, if we don't find the state of the universe to be A, but, say, B, that X wasn't the state of the universe at t=0, but some other state Y.

Does that make sense?