I wonder if it is possible for causality to be held, but not determinism. Does any real-world example exist? If yes, please explain how does that satisfy causality and not determinism. (for example, is the act of measurement on quantum systems an answer?)

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    $\begingroup$ I think you're going to have to define what you mean by "causality" first. Is a process causal if we know all causes for all possible outcomes (and in particular, we know that all possible outcomes have a particular cause) while still not being able to predict which particular outcome will happen? $\endgroup$ – probably_someone Dec 19 '19 at 14:41
  • $\begingroup$ Wow, you helped to me to see that I already know the answer! It was just in my definition! But maybe my definition is not the one that others mean... $\endgroup$ – seyed sepehr mousavi Dec 19 '19 at 17:21
  • $\begingroup$ Do you see now what I have been telling you in comments in your two previous questions? In what sense do you use the word 'causality'? It is impossible to answer your questions if you do not make that clear. See for example this answer of mine $\endgroup$ – Stéphane Rollandin Dec 19 '19 at 22:47
  • $\begingroup$ Thank you for it. You are all right! $\endgroup$ – seyed sepehr mousavi Dec 20 '19 at 7:16

Yes, quantum measurement is random (with no local hidden variables), but does not violate causality. A photon cannot be detected, unless it was emitted and given enough time to propagate, however the point where it is detected cannot be predicted.

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  • $\begingroup$ So it does not necessary for one cause to ONLY cause one outcome, but it can be two or more outcomes (that would be called truly random). Right? $\endgroup$ – seyed sepehr mousavi Dec 19 '19 at 17:25
  • $\begingroup$ @seyedsepehrmousavi "Random" means unpredictable (not because we don't know how to predict, but because it cannot be predicted in principle regardless of what we know). So the number of outcomes has nothing to do with them being random. If you have 10 predictable outcomes, none of them is random, but if any of them is unpredictable, then it is random. $\endgroup$ – safesphere Dec 19 '19 at 17:50
  • $\begingroup$ I think you misunderstood my statement. You are absolutely right, but what you said is not a contrast to what I said. This is what I meant: if a cause A can have two effects B and C, then how can we predict if B or C can happen if A is the ONLY cause existing in nature to have the effects B and C. $\endgroup$ – seyed sepehr mousavi Dec 19 '19 at 18:06
  • $\begingroup$ If the type of the outcome is predictable, then conceptually we can predict it. If it is not predictable in principle, then the choice between B or C is random. $\endgroup$ – safesphere Dec 19 '19 at 20:20
  • $\begingroup$ The outcome of experiments on Bell's inequality simply don't require this. They do require that if there are hidden variable they are non-local and if physics is fully local there are no hidden variables. Also admitted is neither strong locality nor realism. My personal prejudice likewise is in favor of locality without hidden variables, but you shouldn't try to present it resolved because it simply isn't. $\endgroup$ – dmckee --- ex-moderator kitten Dec 20 '19 at 0:11

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