tag:blogger.com,1999:blog-22973357.post3691767716057374766..comments2021-01-28T02:07:34.956-05:00Comments on Sabine Hossenfelder: Backreaction: Well, Actually. 10 Physics Answers.Sabine Hossenfelderhttp://www.blogger.com/profile/06151209308084588985noreply@blogger.comBlogger90125tag:blogger.com,1999:blog-22973357.post-3645051247668492172021-01-08T08:14:17.221-05:002021-01-08T08:14:17.221-05:00Sorry, I screwed up the numbering. 1) should be 1+...Sorry, I screwed up the numbering. 1) should be 1+2) and 2) should be 3). Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-80675502476278748282021-01-08T08:13:13.723-05:002021-01-08T08:13:13.723-05:00Andrei,
1) It's a toy-model. It does not say ...Andrei,<br /><br />1) It's a toy-model. It does not say anything about where the hidden variables come from. The only thing you need to know about them, as I already said, is that they're uniformly distributed in the complex unit disk. (And they're uncorrelated.) <br /><br />I merely suggest the reader thinks about them as the detector's degrees of freedom because I personally Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-57129685583834586442021-01-07T04:39:24.961-05:002021-01-07T04:39:24.961-05:00Sabine,
Let me ask you a few (probably stupid) qu...Sabine,<br /><br />Let me ask you a few (probably stupid) questions about your model.<br /><br />1. About the hidden variables you say:<br /><br />"The reader can think of these variables as encoding the detailed degrees of freedom of the detector"<br /><br />I do not understand how a complex number can encode this kind of information. At least in my understanding, those detailed Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-72763859564289703912021-01-06T06:11:39.828-05:002021-01-06T06:11:39.828-05:00Andrei,
I don't know what it even means for i...Andrei,<br /><br />I don't know what it even means for information to flow. As the title of the paper already says, this is a toy model. It doesn't tell you anything about where the hidden variables come from or what the detectors do in detail. The point I am tryin to convey is that you don't need to know this; the model will give the same average predictions regardless of those Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-17693410951041352132021-01-06T04:47:14.999-05:002021-01-06T04:47:14.999-05:00Sabine,
"At this point I think you are just ...Sabine,<br /><br />"At this point I think you are just fundamentally misunderstanding how the model works to begin with."<br /><br />You are probably right. Not being a physicist I have a hard time understanding this kind of abstract models. If you find some time in the future it would be great to write a blog post about your model and made it more explicit to the non-physicists. What IAndreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-27064882473951118772021-01-05T08:45:21.327-05:002021-01-05T08:45:21.327-05:00Andrei,
"This does not work for measurements...Andrei,<br /><br /><i>"This does not work for measurements performed on the same orientation. Those need to be perfectly anticorrelated, always. If the result itself depends on some random parameter that is only present at each detector you will have failures."</i><br /><br />At this point I think you are just fundamentally misunderstanding how the model works to begin with.<br /><br />Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-42523063328454897642021-01-05T07:08:12.867-05:002021-01-05T07:08:12.867-05:00"If you make those details non-deterministic ..."If you make those details non-deterministic it does not matter for the AVERAGE of the outcome. It matters for the outcome itself, which is no longer determined. Ie, both theories reproduce quantum mechanics. "<br /><br />This does not work for measurements performed on the same orientation. Those need to be perfectly anticorrelated, always. If the result itself depends on some random Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-86070812234195312182021-01-05T06:43:02.525-05:002021-01-05T06:43:02.525-05:00Andrei,
"Sure, but then, if the non-determin...Andrei,<br /><br /><i>"Sure, but then, if the non-deterministic part of the model does not really matter for the experiment you still have a deterministic model. It's like saying that by changing randomly the color of a planet you have a non-deterministic model of gravity"</i><br /><br />If you make those details non-deterministic it does not matter for the AVERAGE of the outcome. Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-67274764681250904662021-01-05T04:18:54.572-05:002021-01-05T04:18:54.572-05:00Sabine,
"I didn't say the setting is non...Sabine,<br /><br />"I didn't say the setting is non-deterministic. I said the position of the settings is not the only degree of freedom of the detector. All the other ones can do whatever you want and be non-deterministic as they please."<br /><br />Sure, but then, if the non-deterministic part of the model does not really matter for the experiment you still have a deterministic Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-70905514678047568732021-01-04T11:47:57.713-05:002021-01-04T11:47:57.713-05:00Andrei,
"For a Bell test with photons, those...Andrei,<br /><br /><i>"For a Bell test with photons, those settings are nothing else but the position of the polarizer. If that position depends on something that is not deterministic (say an electronic device of some sort) you cannot guarantee what it will be at the time of detection. And in this case the model either fails or it requires non-locality to work."</i><br /><br />A) I didnSabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-92059372751971004642021-01-04T11:14:57.452-05:002021-01-04T11:14:57.452-05:00"First, the quote from my paper correctly ref..."First, the quote from my paper correctly refers to the MEASUREMENT SETTINGS. You, in contrast, refer to the "state of the detector". Different thing entirely."<br /><br />For a Bell test with photons, those settings are nothing else but the position of the polarizer. If that position depends on something that is not deterministic (say an electronic device of some sort) you Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-82997673326826326752021-01-04T10:39:44.734-05:002021-01-04T10:39:44.734-05:00Andrei,
"The reason the above fragment is co...Andrei,<br /><br /><i>"The reason the above fragment is correct is that the theory is deterministic. The state of the detector at the time of the measurement is uniquely determined by its past state. Otherwise, this whole mechanism does not work. The model becomes worthless."</i><br /><br />What you say is wrong, once again, and this time for several reasons. <br /><br />First, the Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-49124181440479871112021-01-04T10:29:20.920-05:002021-01-04T10:29:20.920-05:00Let's consider your paper, "A Superdeterm...Let's consider your paper, "A Superdeterministic Toy Model", page 8:<br /><br />"The reason this works is that the dynamical law explicitly depends on the measurement settings at the time of measurement. This may appear as if it goes against the arrow of time, but note that one could express the measurement settings at the time of measurement by an initial state of the detectorAndreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-78780045840459394792021-01-04T10:04:51.858-05:002021-01-04T10:04:51.858-05:00"Well, if you do that your theory would fail ...<i>"Well, if you do that your theory would fail to violate Bell's inequality. "</i><br /><br />That's just bluntly wrong. It doesn't matter whatsoever if you have a random distribution whether you think it's fundamentally random, or random just because you do not know what it is determined by. Same thing for the outcome. Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-88013216258339190092021-01-04T10:03:29.349-05:002021-01-04T10:03:29.349-05:00"A random distribution of the hidden variable...<i>"A random distribution of the hidden variables does not imply indeterminism. It can simply be a property of the initial state."</i><br /><br />Of course it CAN be a property of the initial state. But it doesn't have to.<br /><br /><i>"Non-determinism means that a certain state could evolve into at least two different states."</i><br /><br />Nope, it can as well mean Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-85050870961980152982021-01-04T09:38:22.745-05:002021-01-04T09:38:22.745-05:00Sabine,
"You can make any local, determinist...Sabine,<br /><br />"You can make any local, deterministic hidden variable theory into a non-deterministic one just by claiming that the hidden variables are fundamentally randomly distributed."<br /><br />A random distribution of the hidden variables does not imply indeterminism. It can simply be a property of the initial state. Non-determinism means that a certain state could evolve Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-32325940691882174722021-01-04T08:48:12.370-05:002021-01-04T08:48:12.370-05:00And just to be concrete, you can use for this any ...And just to be concrete, you can use for this any superdeterministic model that has ever been put forward, including my own. Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-42902083217363107832021-01-04T08:44:58.175-05:002021-01-04T08:44:58.175-05:00Andrei, You can make any local, deterministic hidd...Andrei, You can make any local, deterministic hidden variable theory into a non-deterministic one just by claiming that the hidden variables are fundamentally randomly distributed. Of course that's entirely pointless, but it's perfectly possible. What you say is trivially wrong. Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-7087321550267653942021-01-04T08:34:58.150-05:002021-01-04T08:34:58.150-05:00Sabine,
How about showing me a single example of ...Sabine,<br /><br />How about showing me a single example of a local, non-deterministic theory that explains EPR? It should be easy to find. You will not find it. Collapse theories are non-local.<br /><br />I understand that you've lost your patience, fine. You can revise the arguments presented here at some other time if you wish. You should be happy as they prove your superdereministic Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-8189867461262408852021-01-04T08:28:24.849-05:002021-01-04T08:28:24.849-05:00Andrei,
You are not following. This is my last tr...Andrei,<br /><br />You are not following. This is my last try. I was saying: IF YOU ALREADY HAVE A DETERMINISTIC THEORY, THEN.... Bell's theorem shows you so and so. Which exactly means that YOU DO NOT NEED IT TO BE DETERMINISTIC. In contrast to what YOU claimed.<br /><br />I don't know what "GRW mass density theory" is supposed to be, and in any case I don't know how it Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-82049628004330297112021-01-04T08:02:12.743-05:002021-01-04T08:02:12.743-05:00Sabine,
"I didn't say that Bell's th...Sabine,<br /><br />"I didn't say that Bell's theorem assumes determinism."<br /><br />You said:<br /><br />" Bell's theorem shows that if you HAVE a deterministic, local, hidden variables theory that violates the inequality, then that has to violate statistical independence."<br /><br />The word determinism should not be there. This is the correct version:<br /><brAndreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-72794905973252721592021-01-04T07:41:39.520-05:002021-01-04T07:41:39.520-05:00Andrei,
You wrote:
"I disagree that Bell’s ...Andrei,<br /><br />You wrote:<br /><br /><i>"I disagree that Bell’s theorem assumes determinism. It assumes locality, statistical independence and the existence of hidden variables that determine the measurement outcome. Bell’s theorem is perfectly compatible with the assumption that the hidden variables are stochastically determined at the time of emission."</i><br /><br />I didn'tSabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-86845800167990497152021-01-04T07:18:27.011-05:002021-01-04T07:18:27.011-05:00Sabine,
Can you give me an example of spontaneous...Sabine,<br /><br />Can you give me an example of spontaneous collapse model that is local? GRW is non-local.Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-35974962634638579872021-01-04T07:13:24.409-05:002021-01-04T07:13:24.409-05:00Sabine,
I disagree that Bell’s theorem assumes de...Sabine,<br /><br />I disagree that Bell’s theorem assumes determinism. It assumes locality, statistical independence and the existence of hidden variables that determine the measurement outcome. Bell’s theorem is perfectly compatible with the assumption that the hidden variables are stochastically determined at the time of emission. In the paper, "On The Einstein Podolsky Rosen Paradox"Andreihttps://www.blogger.com/profile/05519448415253342448noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-21418716107967682092021-01-04T05:59:50.017-05:002021-01-04T05:59:50.017-05:00Andrei,
Ok, thanks. Bell's theorem doesn'...Andrei,<br /><br />Ok, thanks. Bell's theorem doesn't prove that the theory must be deterministic. I don't know why you think it does. Bell's theorem shows that if you HAVE a deterministic, local, hidden variables theory that violates the inequality, then that has to violate statistical independence. It does not show that this is the only way to do it. Spontaneous collapse models Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.com