Radioactive decay and the myth of randomness
in [Logic]
|
Prev: Can we predict the outcome of a tossed coin?
Next: group of formal differences vs. Grothendieck group
From: tg on 19 Nov 2009 06:38 On Nov 17, 6:45 pm, James Burns <burns...(a)osu.edu> wrote: > tg wrote: > > On Nov 17, 3:00 pm, James Burns <burns...(a)osu.edu> wrote: > > > **** > > >>This is not the sort of situation which we can use to > >>test local reality vs. quantum mechanics using Bell's > >>theorem. > > > Very good. > > >>Those situations involve entangled pairs of > >>particles being measured at macroscopically separated > >>locations. However, I have already explained once why > >>the collapse of local reality in those other situations > >>is still bad news for local reality in this situation. > > > No, I don't think so. 'Bad news' is not a sufficient > > characterization in science or philosophy. > Top Post interjection---please ignore the obviously not; you are now apparently paying attention. > What, people don't speak informally in philosophy, > expecting others to fill in the formalities for themselves? > How odd. They certainly do in the sciences. > > The bad news is that you cannot assume the constraints > of local reality will hold for a decaying atom, because > they have been shown not to hold for entangled pairs of > particles. > > My original justification for extending the implications > of Bell's theorem from entangled pairs of particles to > individual decaying atoms was that this is just how science > works. We generalize to the greatest extent we can when > we theorize, and then use the contradictions from > experiment to drive the next cycle of theorizing. > The maximal generalizing is important because we /want/ > contradictions -- or "new" contradictions, at least. > An argument could be made that, if we are not getting > contradictions of theory from our experiments, then we > are not learning anything we didn't already know. > > I suppose maximal generalization is "merely" a > methodological concern, but that methodology is more firmly > embedded in the sciences than any theory you could name. > > However, it occurs to me that I'm holding this thing the > wrong way around. The tests of various Bell inequalities > do not /validate/ any particular theory. What they do > is /invalidate/ the entire class of hidden-variable theories. > Not bad, but I thought it invalidated the entire class of hidden variable theories *that purport to be consistent with and as complete as QM*. But anyway, I accept that you now see more fully why I don't like the facile use of decay as an example in this discussion. > You think that you may have a hidden-variable > theory that side-steps the usual Bell-centric issues. OMG. I am being so cautious that I use the somewhat awkward phrase "proposed conjecture", and suddenly I have a Theory. Now, you are either a creationist improperly using the term, or, JJ-like, setting up quite the strawman. > > It's /your/ theory, so /you/ get to explain what the > difference is between decaying atoms and entangled > pairs of particles. Your theory seems to be "Let's > hide the outcome of the quantum measurement (the time > of decay of the atom) somewhere inside the atom > and let that determine the time of decay." How does > this generalize to spin measurements of entangled particles? > The obvious way is to say all quantum measurement > outcomes are somewhere in the quantum system, but > then the predictions for entangled pairs will be > bounded by Bell's theorem and definitely inconsistent > with experiment. Do you refuse to generalize? Then > you must explain why your theory applies in one place > and not another. This is /required/ for what you might > consider methodological reasons: if researchers cannot > answer questions like that from your description of > your new theory, then they will have to seek you out > and ask "Does it apply here?" every time they find > themselves in a novel situation. This is unworkable. > > [...] > > > But I thought that's what you just did---you > > translated my problem into the language of QM, > > which I appreciate. We are not looking for > > equations here but clear language and reasoning, > > so we can consider the philosophical implications, > > if any. You have framed this situation as follows: > > > *The lifetime of the particle is dependent on its > > quantum state.* > > > If that's correct, then we have a variable that > > determines the lifetime of the particle, correct? > > I would agree with the starred statement, but I want to > be sure you mean the same thing by "lifetime" as I do. > > The lifetime I am talking about is NOT how long any > particular atom lives, from creation to decay. > It is a parameter for a probability distribution > that permits us to calculate the probability of a > certain number of these atoms decaying with a > certain period of time, or the probability of > one particular atom decaying within a certain period > of time, This last is an interesting aspect of the topic. Are you sure? If you can calculate the probability of one particular particle decaying within a certain period of time, then you appear to be agreeing with JJ. Why don't you elaborate on this point. > or many other things. But it will not give > a guarantee of the time the atom will decay. See above. -tg > > I confess, I do not see the connection to your > original idea. > > Jim Burns
From: James Burns on 19 Nov 2009 16:15 tg wrote: > On Nov 17, 6:45 pm, James Burns <burns...(a)osu.edu> wrote: >>tg wrote: >>>On Nov 17, 3:00 pm, James Burns <burns...(a)osu.edu> wrote: >>>>This is not the sort of situation which we can use to >>>>test local reality vs. quantum mechanics using Bell's >>>>theorem. >> >>>Very good. >> >>>>Those situations involve entangled pairs of >>>>particles being measured at macroscopically separated >>>>locations. However, I have already explained once why >>>>the collapse of local reality in those other situations >>>>is still bad news for local reality in this situation. >> >>>No, I don't think so. 'Bad news' is not a sufficient >>>characterization in science or philosophy. > > Top Post interjection---please ignore the obviously not; > you are now apparently paying attention. As near as I can tell, your "Obviously not." was intended to be a meaningless interjection. No, thank you; I can do meaningless on my own. I'm going to go do something else now. I may read any responses from you, I may not. We'll see. I probably won't be responding, though. Good luck with your questioning. I don't think I've asked for much from you, and I've given you quite a lot of my time, actually. I would like to ask a favor in return: Please don't do the You-can't-make-a-clear-statement- of-what-I'm wrong-*about*,-so-you-either-throw-up-a- lot-of-chaff-and-do-a-lot-of-hand-waving,-or-do-the- indignation-dodge thing again. TIA. >>What, people don't speak informally in philosophy, >>expecting others to fill in the formalities for themselves? >>How odd. They certainly do in the sciences. >> >>The bad news is that you cannot assume the constraints >>of local reality will hold for a decaying atom, because >>they have been shown not to hold for entangled pairs of >>particles. > >>My original justification for extending the implications >>of Bell's theorem from entangled pairs of particles to >>individual decaying atoms was that this is just how science >>works. We generalize to the greatest extent we can when >>we theorize, and then use the contradictions from >>experiment to drive the next cycle of theorizing. >>The maximal generalizing is important because we /want/ >>contradictions -- or "new" contradictions, at least. >>An argument could be made that, if we are not getting >>contradictions of theory from our experiments, then we >>are not learning anything we didn't already know. >> >>I suppose maximal generalization is "merely" a >>methodological concern, but that methodology is more firmly >>embedded in the sciences than any theory you could name. >> >>However, it occurs to me that I'm holding this thing the >>wrong way around. The tests of various Bell inequalities >>do not /validate/ any particular theory. What they do >>is /invalidate/ the entire class of hidden-variable theories. > > Not bad, but I thought it invalidated the entire class of > hidden variable theories *that purport to be consistent > with and as complete as QM*. Why would a physicist want a theory that was inconsistent with QM (that is, with experiment) or is not as complete as QM? > But anyway, I accept that > you now see more fully why I don't like the facile use > of decay as an example in this discussion. Uhmmm. I guess you don't see why the facile use of decay is perfectly appropriate in this discussion. Sorry; time's up. You get to work it out on your own now, unless you can entice someone else into having this conversation. >>You think that you may have a hidden-variable >>theory that side-steps the usual Bell-centric issues. > > OMG. I am being so cautious that I use the somewhat awkward phrase > "proposed conjecture", and suddenly I have a Theory. Now, you are > either a creationist improperly using the term, or, JJ-like, setting > up quite the strawman. If you don't have a theory, then you have nothing worth spending time on. It doesn't have to be a very detailed theory; maybe just "What you guys say, except for this". But it's your <fill in the blank/>. You decide. >>It's /your/ theory, so /you/ get to explain what the >>difference is between decaying atoms and entangled >>pairs of particles. Your theory seems to be "Let's >>hide the outcome of the quantum measurement (the time >>of decay of the atom) somewhere inside the atom >>and let that determine the time of decay." How does >>this generalize to spin measurements of entangled particles? >>The obvious way is to say all quantum measurement >>outcomes are somewhere in the quantum system, but >>then the predictions for entangled pairs will be >>bounded by Bell's theorem and definitely inconsistent >>with experiment. Do you refuse to generalize? Then >>you must explain why your theory applies in one place >>and not another. This is /required/ for what you might >>consider methodological reasons: if researchers cannot >>answer questions like that from your description of >>your new theory, then they will have to seek you out >>and ask "Does it apply here?" every time they find >>themselves in a novel situation. This is unworkable. >> >>[...] >> >> >>>But I thought that's what you just did---you >>>translated my problem into the language of QM, >>>which I appreciate. We are not looking for >>>equations here but clear language and reasoning, >>>so we can consider the philosophical implications, >>>if any. You have framed this situation as follows: >> >>>*The lifetime of the particle is dependent on its >>>quantum state.* >> >>>If that's correct, then we have a variable that >>>determines the lifetime of the particle, correct? >> >>I would agree with the starred statement, but I want to >>be sure you mean the same thing by "lifetime" as I do. >> >>The lifetime I am talking about is NOT how long any >>particular atom lives, from creation to decay. >>It is a parameter for a probability distribution >>that permits us to calculate the probability of a >>certain number of these atoms decaying with a >>certain period of time, or the probability of >>one particular atom decaying within a certain period >>of time, > > This last is an interesting aspect of the topic. > Are you sure? Am I sure what I mean? Yes. I'm not entirely sure that this use of "lifetime" is standard. However, you can bet your life (as I would mine) that the ideas and equations are as standard as anything in science. If you wish to insist upon a JJ-style meaning for "lifetime", that is, a specific time of decay for each atom, then I disagree with the starred sentence above, and I think you would have a hard time finding someone who understands the situation and agrees with that meaning. > If you can calculate the probability > of one particular particle decaying within a certain > period of time, then you appear to be agreeing with JJ. > Why don't you elaborate on this point. However it appears to you, I am not agreeing with JJ. (Not that this would be a bad thing in itself. I'm sure that JJ sometimes gets things right, by chance if no other way.) Anyway, JJ seems to say that an atomic decay would not appear random if we knew more. Giving a probability (except for 0 and 100%) is utterly unlike this lack of randomness that JJ claims. (Not that this matters to JJ's claim. That is about what /would/ be the case /if/ we knew everything.) You ask for elaboration, but I don't know what to give you. Equations, I suppose. Suppose that some radioactive isotope has a /lifetime/ T (my sense of the word). Then, if we know that the atom has NOT YET decayed at time t0, then the probability that it WILL NOT decay before time t1 is p(t1-t0) = exp( -(t1-t0)/T ) If we know that the atom has NOT YET decayed at time t1, then the probability that it WILL decay before time t2 is 1 - p(t2-t1) = 1 - exp( -(t2-t1)/T ) If we know that the atom has NOT YET decayed at time t0, then the probability that it WILL decay between time t1 and time t2 is p(t1-t0)*[1 - p(t2-t1)] = exp( -(t1-t0)/T )*[1 - exp( -(t2-t1)/T )] exp( -(t1-t0)/T ) - exp( -(t2-t0)/T ) If the time interval dt = t2-t1 is small, then the probability that the atom will decay within an interval dt near time t1, after time t0 -- which is the latest we know that the atom has not yet decayed -- (where was I? ... Oh, yeah) that probability is p(t1-t0)*dt/T = exp( -(t1-t0)/T )*dt/T This last seems to be closest to what you want, a specific time of decay, t1. However, all you get is a probability that it decays near any particular time t1. >>or many other things. But it will not give >>a guarantee of the time the atom will decay. > > See above. Answered above. >>I confess, I do not see the connection to your >>original idea. Ditto. Jim Burns
From: tg on 21 Nov 2009 08:11 On Nov 19, 4:15 pm, James Burns <burns...(a)osu.edu> wrote: > tg wrote: > > On Nov 17, 6:45 pm, James Burns <burns...(a)osu.edu> wrote: > >>tg wrote: > >>>On Nov 17, 3:00 pm, James Burns <burns...(a)osu.edu> wrote: ...... .. > > >>However, it occurs to me that I'm holding this thing the > >>wrong way around. The tests of various Bell inequalities > >>do not /validate/ any particular theory. What they do > >>is /invalidate/ the entire class of hidden-variable theories. > > > Not bad, but I thought it invalidated the entire class of > > hidden variable theories *that purport to be consistent > > with and as complete as QM*. > > Why would a physicist want a theory that was inconsistent > with QM (that is, with experiment) or is not as complete > as QM? > > > But anyway, I accept that > > you now see more fully why I don't like the facile use > > of decay as an example in this discussion. > > Uhmmm. I guess you don't see why the facile use of decay > is perfectly appropriate in this discussion. > > Sorry; time's up. You get to work it out on your own now, > unless you can entice someone else into having this > conversation. > > >>You think that you may have a hidden-variable > >>theory that side-steps the usual Bell-centric issues. > > > OMG. I am being so cautious that I use the somewhat awkward phrase > > "proposed conjecture", and suddenly I have a Theory. Now, you are > > either a creationist improperly using the term, or, JJ-like, setting > > up quite the strawman. > > If you don't have a theory, then you have nothing > worth spending time on. It doesn't have to be a very > detailed theory; maybe just "What you guys say, > except for this". > > But it's your <fill in the blank/>. You decide. > > > > >>It's /your/ theory, so /you/ get to explain what the > >>difference is between decaying atoms and entangled > >>pairs of particles. For any lurkers, since JB isn't going to read this, and we don't have to spare his feelings: Nonsense! "Hidden variable" is a concept, not a predictive outcome. The most comprehensive theory is required to make predictions for all cases, so even if we wish to calculate the trajectory of a billiard ball, we should be able to use QM to do it. It would be awkward and difficult, but we could demonstrate that by applying the rules we could indeed make a three-cushion bank shot. Now, it should be obvious to the most naive that there are circumstances where hidden variables exist and we recognize that fact---otherwise, this whole confusion about 'randomness' would never exist. People like JB love to say "randomness in particles is not like randomness in a gas; if we knew everything about the gas molecules we could predict the location and momentum of an individual, but that's not true for particles." Well, if that (gas molecule) isn't a hidden variable situation, what is? And, are gas molecules subject to the rules of QM? Of course, since QM is a comprehensive theory. So we come back to my original question: Let's say I propose that the actual lifetime of a nucleus is determined at the moment of it's creation. How would the world of QM be different if this were the case? Come on, there must be a Bell out there; what would the experiment be like? BTW, this is a serious question even though I am trying to make a point about philosophy and language. There may be such an experiment, but I can't see it. ....... > > >>The lifetime I am talking about is NOT how long any > >>particular atom lives, from creation to decay. > >>It is a parameter for a probability distribution > >>that permits us to calculate the probability of a > >>certain number of these atoms decaying with a > >>certain period of time, or the probability of > >>one particular atom decaying within a certain period > >>of time, > > > This last is an interesting aspect of the topic. > > Are you sure? > > Am I sure what I mean? Yes. > > I'm not entirely sure that this use of "lifetime" > is standard. However, you can bet your life (as I > would mine) that the ideas and equations are as > standard as anything in science. > > If you wish to insist upon a JJ-style meaning for > "lifetime", that is, a specific time of decay for each > atom, then I disagree with the starred sentence above, > and I think you would have a hard time finding someone > who understands the situation and agrees with that > meaning. > > > If you can calculate the probability > > of one particular particle decaying within a certain > > period of time, then you appear to be agreeing with JJ. > > Why don't you elaborate on this point. > > However it appears to you, I am not agreeing with JJ. > (Not that this would be a bad thing in itself. > I'm sure that JJ sometimes gets things right, > by chance if no other way.) Anyway, JJ seems to > say that an atomic decay would not appear random > if we knew more. Giving a probability (except for > 0 and 100%) is utterly unlike this lack of randomness > that JJ claims. (Not that this matters to JJ's claim. > That is about what /would/ be the case /if/ we knew > everything.) > > You ask for elaboration, but I don't know what to > give you. Equations, I suppose. > > Suppose that some radioactive isotope has a > /lifetime/ T (my sense of the word). > > Then, if we know that the atom has NOT YET decayed > at time t0, then the probability that it WILL NOT > decay before time t1 is > p(t1-t0) = exp( -(t1-t0)/T ) > > If we know that the atom has NOT YET decayed > at time t1, then the probability that it WILL > decay before time t2 is > 1 - p(t2-t1) = 1 - exp( -(t2-t1)/T ) > > If we know that the atom has NOT YET decayed > at time t0, then the probability that it WILL > decay between time t1 and time t2 is > p(t1-t0)*[1 - p(t2-t1)] = > exp( -(t1-t0)/T )*[1 - exp( -(t2-t1)/T )] > exp( -(t1-t0)/T ) - exp( -(t2-t0)/T ) > > If the time interval dt = t2-t1 is small, then > the probability that the atom will decay within > an interval dt near time t1, after time t0 -- > which is the latest we know that the atom has not > yet decayed -- (where was I? ... Oh, yeah) that > probability is > p(t1-t0)*dt/T = exp( -(t1-t0)/T )*dt/T > > This last seems to be closest to what you want, > a specific time of decay, t1. However, all you get > is a probability that it decays near any particular > time t1. > > >>or many other things. But it will not give > >>a guarantee of the time the atom will decay. > > > See above. > > Answered above. > > >>I confess, I do not see the connection to your > >>original idea. > Here, dear readers-other-than-JB, we have what seems to be an outright error in reasoning, or maybe awkward phrasing. Perhaps a statistician can help out. Let's say nucleus A has a .5 probability of decaying to B in 1 sec. And let's say this takes place in a box which becomes transparent every second, so we can see whether we have an A or B in it. Certainly, at time t0, we can calculate a probability of what we will find in the box at t1, say 5 sec. But JB says "if we know that the atom has NOT YET (his caps) decayed at t1......". As I understand it, if we see A in the box at some t1, however many seconds that is, then the probability that we will see A in the box at the next second is still .5 . -tg
From: Zinnic on 21 Nov 2009 09:42 On Nov 21, 7:11 am, tg <tgdenn...(a)earthlink.net> wrote: > On Nov 19, 4:15 pm, James Burns <burns...(a)osu.edu> wrote: > > > tg wrote: > > > On Nov 17, 6:45 pm, James Burns <burns...(a)osu.edu> wrote: > > >>tg wrote: > > >>>On Nov 17, 3:00 pm, James Burns <burns...(a)osu.edu> wrote: > > ..... > . > > > > > > > > > >>However, it occurs to me that I'm holding this thing the > > >>wrong way around. The tests of various Bell inequalities > > >>do not /validate/ any particular theory. What they do > > >>is /invalidate/ the entire class of hidden-variable theories. > > > > Not bad, but I thought it invalidated the entire class of > > > hidden variable theories *that purport to be consistent > > > with and as complete as QM*. > > > Why would a physicist want a theory that was inconsistent > > with QM (that is, with experiment) or is not as complete > > as QM? > > > > But anyway, I accept that > > > you now see more fully why I don't like the facile use > > > of decay as an example in this discussion. > > > Uhmmm. I guess you don't see why the facile use of decay > > is perfectly appropriate in this discussion. > > > Sorry; time's up. You get to work it out on your own now, > > unless you can entice someone else into having this > > conversation. > > > >>You think that you may have a hidden-variable > > >>theory that side-steps the usual Bell-centric issues. > > > > OMG. I am being so cautious that I use the somewhat awkward phrase > > > "proposed conjecture", and suddenly I have a Theory. Now, you are > > > either a creationist improperly using the term, or, JJ-like, setting > > > up quite the strawman. > > > If you don't have a theory, then you have nothing > > worth spending time on. It doesn't have to be a very > > detailed theory; maybe just "What you guys say, > > except for this". > > > But it's your <fill in the blank/>. You decide. > > > >>It's /your/ theory, so /you/ get to explain what the > > >>difference is between decaying atoms and entangled > > >>pairs of particles. > > For any lurkers, since JB isn't going to read this, and we don't have > to spare his feelings: > > Nonsense! "Hidden variable" is a concept, not a predictive outcome. > > The most comprehensive theory is required to make predictions for all > cases, so even if we wish to calculate the trajectory of a billiard > ball, we should be able to use QM to do it. It would be awkward and > difficult, but we could demonstrate that by applying the rules we > could indeed make a three-cushion bank shot. > > Now, it should be obvious to the most naive that there are > circumstances where hidden variables exist and we recognize that > fact---otherwise, this whole confusion about 'randomness' would never > exist. People like JB love to say "randomness in particles is not > like randomness in a gas; if we knew everything about the gas > molecules we could predict the location and momentum of an individual, > but that's not true for particles." Well, if that (gas molecule) isn't > a hidden variable situation, what is? And, are gas molecules subject > to the rules of QM? Of course, since QM is a comprehensive theory. > > So we come back to my original question: Let's say I propose that the > actual lifetime of a nucleus is determined at the moment of it's > creation. How would the world of QM be different if this were the > case? Come on, there must be a Bell out there; what would the > experiment be like? > > BTW, this is a serious question even though I am trying to make a > point about philosophy and language. There may be such an experiment, > but I can't see it. > > ...... > > > > > > > > > >>The lifetime I am talking about is NOT how long any > > >>particular atom lives, from creation to decay. > > >>It is a parameter for a probability distribution > > >>that permits us to calculate the probability of a > > >>certain number of these atoms decaying with a > > >>certain period of time, or the probability of > > >>one particular atom decaying within a certain period > > >>of time, > > > > This last is an interesting aspect of the topic. > > > Are you sure? > > > Am I sure what I mean? Yes. Snip > Here, dear readers-other-than-JB, we have what seems to be an outright > error in reasoning, or maybe awkward phrasing. Perhaps a statistician > can help out. > > Let's say nucleus A has a .5 probability of decaying to B in 1 sec. > And let's say this takes place in a box which becomes transparent > every second, so we can see whether we have an A or B in it. > Certainly, at time t0, we can calculate a probability of what we will > find in the box at t1, say 5 sec. But JB says "if we know that the > atom has NOT YET (his caps) decayed at t1......". > > As I understand it, if we see A in the box at some t1, however many > seconds that is, then the probability that we will see A in the box at > the next second is still .5 . Probability is present in neither past nor future. The flip of a coin demonstrates that you are correct. :-)
From: John Jones on 21 Nov 2009 21:28
David C. Ullrich wrote: > On Wed, 18 Nov 2009 04:10:57 +0000, John Jones > <jonescardiff(a)btinternet.com> wrote: > >> Marshall wrote: >>> On Nov 15, 11:45 am, John Jones <jonescard...(a)btinternet.com> wrote: >>>> Quantum mechanics employs everyday terms to support its mathematical >>>> structure. My complaint, a valid one, is that these terms are no longer >>>> employed with their standard meanings, thus making Quantum theory >>>> meaningfully vacuous. >>> That's obviously bullshit. >> It's obviously NOT. It's self-evident. Look at it! am I talking to Mr. >> stupido? If you describe something in non-meaningful terms then it is >> meaningfully vacuous. Comprende? > > Sometimes you seem stupid. "no longer employed with their > standard meanings" is not the same as "non-meaningful". The whole point of the concept of 'non-standard meanings' is that their signs can be used in two ways - 1) non-meaningfully, and 2) in the case of mathematics, as an ellipsis for syntactical, meaning-void, procedures. |