tag:blogger.com,1999:blog-8883034.comments2017-06-16T08:47:31.771+02:00atdotdeRobert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.comBlogger513125tag:blogger.com,1999:blog-8883034.post-2892641626435577822017-06-16T08:47:31.771+02:002017-06-16T08:47:31.771+02:00Unfortunately, there is not much written informati...Unfortunately, there is not much written information from the workshop beyond the google doc already linked. But we set up a mailing list on the topic, see https://lists.muc.ccc.de/cgi-bin/mailman/listinfo/iot-haftung and in its archive you can find a number of documents on the topic that have been passed around since.Robert Hellinghttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-71796428055829961082017-06-16T08:17:21.850+02:002017-06-16T08:17:21.850+02:00Can you share with us some information from your w...Can you share with us some information from your workshop. If you already mentioned on your blog then can you please share the link. Thanks.Danis Tanovichttp://trevokandrew.blogspot.com/2017/06/the-internet-of-things-weaves-together-various-devices-to-create-comfort.htmlnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-83010164447583876062017-06-15T19:19:10.511+02:002017-06-15T19:19:10.511+02:00"Is it completely raw (except for been down-s..."Is it completely raw (except for been down-sampled to 4096 Hz)?"<br /><br />That's how I understand the LIGO open science web page. I take that data set as is.<br /><br />"You then mention that since the signal is real, Fourier transform has a phase of 0 or pi for a constant phase<br />and you find out that it is pi. I don't know what this tell us. What frequency does the region of constant phase correspond to?"<br /><br />The phase should be random (as it is outside the low frequency region: The dots come at all colors). What is the frequency? You can do the math yourself (and I am too lazy) but given that is the region of low noise I would expect it to be up to a few 100Hz.<br /><br />" Why do you consider h(t) + time-reversed h(t) (containing the signal) noise? <br />Only h(t) containing the signal+LIGO noise -best fit signal template would be noise?<br />But I don't think you are doing that. So I don't understand how an addition of the original signal+ its time-reversal is noise. Also I don't see a cancellation at other frequencies."<br /><br />The signal has orders of magnitude less power than the noise. So not taking out the signal does not really matter.<br /><br />The effect is strongest for the dataset I mentioned.<br /><br />And your time analysis friend is right: Adding to random noise the time reversal should yield noise that is sqrt(2) times stronger (except at omega=0).Robert Hellinghttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-6461691813910510912017-06-15T18:55:05.503+02:002017-06-15T18:55:05.503+02:00I asked someone who has extensively dealt with tim...I asked someone who has extensively dealt with time-series analysis about this and that person said<br /><br />"Time reversing and subtracting a noisy time series should only cancel out 1 sample (for white noise) and increase the variance (by a factor of 2) for the rest. I see from the plot that the noise is canceled over a much longer time (?) before the variance increases. Could be an effect of colored noise though… I would expect the cancellation to be over the auto-correlation timescale (which is 1 sample for white noise but longer for colored noise + lines). Shantanuhttp://www.blogger.com/profile/16322812456382858228noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-39528952682534453342017-06-15T18:49:41.408+02:002017-06-15T18:49:41.408+02:00There are several things that I have not understoo...There are several things that I have not understood in what you did or why your results are surprising. Following are my questions.<br /><br />o Just to be clear.<br />when you say "raw strain data at 4096 Hz, I presume you mean h(t) timeseries containing GW signal + residual LIGO noise right?<br />Is it completely raw (except for been down-sampled to 4096 Hz)?<br /><br />o You then mention that since the signal is real, Fourier transform has a phase of 0 or pi for a constant phase<br />and you find out that it is pi. I don't know what this tell us. What frequency does the region of constant phase correspond to?<br />(Maybe you can show one plot with frequency in (Hz) on X-axis instead of frequency bin number).<br /><br />o Why do you consider h(t) + time-reversed h(t) (containing the signal) noise? <br />Only h(t) containing the signal+LIGO noise -best fit signal template would be noise?<br />But I don't think you are doing that. So I don't understand how an addition of the original signal+ its time-reversal is noise. Also I don't see a cancellation at other frequencies.<br /><br />oAlso have you repeated the same exercise for Hanford? Do you get the same results?<br />I guess at any rate the best way to resolve this through some toy numerical experiments with injecting a mock signal in mock time series containing white noise (and also doing the same with colored noise)<br />Shantanuhttp://www.blogger.com/profile/16322812456382858228noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-27253106159414197872017-06-14T08:38:31.575+02:002017-06-14T08:38:31.575+02:00Thank you for this post. You have done a great job...Thank you for this post. You have done a great job.Adamhttp://www.top5masala.comnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-7543221399288943852017-06-10T05:40:51.323+02:002017-06-10T05:40:51.323+02:00>> I do not agree that it is out of the que...>> I do not agree that it is out of the question to have a thermometer with a relative velocity in thermal equilibrium with a heat bath at rest.<br /><br />Assume thermometer and system are made from the same type of atoms (to keep the argument simple).<br />The distribution of velocities of those atoms has to be different, otherwise they would not be moving relative to each other in any reasonable sense.<br />But if the velocity distributions are different, they are not in equilibrium.<br /><br />wolfgangnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-54921120596256085512017-06-09T14:59:12.580+02:002017-06-09T14:59:12.580+02:00Rather than commenting here, I updated the origina...Rather than commenting here, I updated the original article in order to be able to write formulas. See above. Robert Hellinghttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-70206314611275330972017-06-09T11:54:57.331+02:002017-06-09T11:54:57.331+02:00This whole story doesn't make sense but in gen...This whole story doesn't make sense but in general, not special, relativity. <br /><br />Another way of realizing what's going on is that, when the temperature is fixed, the energy isn't-it fluctuates. Which means that time translations are not well-defined, therefore Poincaré invariance isn't a global symmetry and it doesn't make sense to ask any question that depends on this assumption. While all these issues may have been discussed in many ways in the past, it's the intuition that's come from Unruh's work, building on Hawking's, that's relevant and clarifies the picture. Once one works within the context of non-inertial frames, there's nothing more to discuss-beyond what could happen if backreaction can't be ignored. While this is an issue for the applications that Hawking has in mind, it isn't for Unruh, where the accelerating system is a probe of the spacetime. Stam Nicolishttp://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-60532755256351557502017-06-09T09:19:31.709+02:002017-06-09T09:19:31.709+02:00You want to know how temperature transforms under ...You want to know how temperature transforms under Lorentz transformations?<br /><br />Let's consider an example. Take an ideal gas. In the rest-frame of the gas (i.e., the frame in which the mean velocity of the gas molecules is zero), the pressure and density are proportional to each other. And the constant of proportionality is (up to a constant factor) the temperature.<br /><br />Now, in that frame, the stress tensor T_{\mu\nu} is diagonal, T_{\mu\nu} = diag(ρ,p,p,p). Hence, the temperature is (up to the aforementioned factor) the ratio of the distinct eigenvalues of T_{\mu\nu}.<br /><br />In any other Lorentz frame, T_{\mu\nu} is, of course, no longer diagonal. But it's still a symmetric tensor. The temperature is defined to transform as a scalar field, which is the ratio of the distinct eigenvalues of T_{\mu\nu} in the local rest-frame of the gas.<br /><br />The generalization to other fluids with other equations of state is straightforward. And the whole story generalizes to General Relativity. See any textbook on cosmology.bloghttps://golem.ph.utexas.edu/~distler/blog/noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-16547271638047848072017-06-09T07:49:37.121+02:002017-06-09T07:49:37.121+02:00I think that one should define the temperature in ...I think that one should define the temperature in equilibrium as the inverse length of the Euclidean time circle. If you start your experiment with a line element<br /><br />ds^2 = -dt^2 + dx^2<br /><br />and posit, in addition, that \tau=it (or is it -it? I always forget) has periodicity<br /><br />\tau \sim \tau+\beta<br /><br />then you can define the temperature as the inverse length of the time circle.<br /><br />You can carry out whatever coordinate transformation you want. It may be that the parametric length of the time circle will change as a result, but the temperature which is defined as the actual length of the time circle will remain the same since it is defined in a coordinate invariant way.<br /><br />The entire discussion can be stated more covariantly. Given a line element with a timelike Killing vector we can define an equilibrated configuration as one which is periodic in Euclidean time. The temperature would be the inverse length of the thermal circle which is coordinate invariant. <br /><br />So this is a notion of temperature which is well defined. You might wonder what your thermometer will measure when you do the actual experiment. There's some discussion that can be carried out here but before that, let me point out that you'd be using the thermometer in a setting which it wasn't designed for, so it's not clear why one would do that (apart from saying that it's an interesting academic question) and it's not clear why one would want to call the result of the measurement a temperature. <br /><br />If you are in equilibrium and do want to measure the temperature I would suggest measuring the average energy density instead, which is well defined. With that information (and perhaps the energy flux) one can obtain the temperature.Amosnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-86759102920856201672017-06-08T21:11:49.828+02:002017-06-08T21:11:49.828+02:00@Robert
as you already stated there is a lot of l...@Robert<br /><br />as you already stated there is a lot of literature on this, e.g. arxiv.org/abs/1606.02127<br /><br />In order for a "thermometer" to measure the temperature of a system (the rail) both need to be in equilibrium, but moving relative to each other they cannot be, so the measurement is just meaningless.<br /><br />Btw you dont really have to go to the relativistic case, the non-relativistic Doppler effect is sufficient for your "paradox" imho.<br /> wolfganghttp://tsm2.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-23068083089539925122017-06-08T20:00:44.392+02:002017-06-08T20:00:44.392+02:00In the non-relativistic approximation, temperature...In the non-relativistic approximation, temperature is related to the average kinetic energy-only the latter isn't a Lorentz invariant quantity.That's why it isn't useful, in the relativistic context. <br /><br />This presentation: http://www.hartmanhep.net/topics2015/3-Rindler.pdf might, also, be useful.Stam Nicolishttp://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-8424607452169265612017-06-08T15:23:25.057+02:002017-06-08T15:23:25.057+02:00Isn't temperature still going to be a measure ...Isn't temperature still going to be a measure of the average kinetic energy of the constituent atoms of the rail? If they're Boltzmann distributed with mean kT in the rest frame then why wouldn't the temp as seen by the conductor be the mean of the new velocity distribution in which each probability bin has the same probability mass but is now associated with the boosted velocity?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-74278889859305622942017-06-08T13:42:01.069+02:002017-06-08T13:42:01.069+02:00Temperature, in a relativistic context, does mean ...Temperature, in a relativistic context, does mean acceleration. The relation was found by Unruh. So an observer, at equilibrium in T, is uniformly accelerating with an acceleration, a, related to the temperature by T = hbar a/(2π). http://www.scholarpedia.org/article/Unruh_effect<br />One notices that it is a quantum effect-so for a classical object the limit is quite tricky. <br />Stam Nicolishttp://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-60663734846628043432017-06-08T13:32:57.997+02:002017-06-08T13:32:57.997+02:00Stam, I am not sure I understand your point. I don...Stam, I am not sure I understand your point. I don't think temperature implies acceleration, at best it's the other way round. Just for the fun of it, please compute the acceleration corresponding to 300K that I roughly have here in my room. I am sure, this would exceed the 5-7g that my body might be willing to tolerate by many orders of magnitude.Robert Hellinghttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-65125567678043743082017-06-08T13:32:24.503+02:002017-06-08T13:32:24.503+02:00Finite temperature, in a special relativistic cont...Finite temperature, in a special relativistic context, means acceleration. This was the point of Unruh's observation: a uniformly accelerating observer, in Minkowski spacetime, is in equilibrium at a specific temperature; conversely, a relativistic observer, in equilibrium at a fixed temperature, is uniformly accelerating.<br /><br />Therefore the resolution of the apparent paradox regarding the Doppler shift is the usual one: a uniformly accelerating frame isn't an inertial frame. Stam Nicolishttp://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-69980108019209331132017-06-08T13:28:58.863+02:002017-06-08T13:28:58.863+02:00This comment has been removed by the author.Stam Nicolishttp://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-31229580217252946482017-05-17T11:21:48.136+02:002017-05-17T11:21:48.136+02:00This comment has been removed by a blog administrator.ash grayhttp://www.blogger.com/profile/18114571726683468753noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-47366437273331964022017-02-08T18:29:57.494+01:002017-02-08T18:29:57.494+01:00Nice! I did a very similar exercise in Java some y...Nice! I did a very similar exercise in Java some years ago... https://github.com/jugglingcats/naturepuzzle. The bulk of the code is in Runner.java.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-34480530812464788592017-01-02T15:15:39.173+01:002017-01-02T15:15:39.173+01:00No guts no glory!No guts no glory!Pascal Kwantenhttp://www.blogger.com/profile/01001468331902587067noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-41415682647591509052016-11-18T09:25:33.929+01:002016-11-18T09:25:33.929+01:00nun hab ich's gefunden:
http://www.inetbib.de/...nun hab ich's gefunden:<br />http://www.inetbib.de/listenarchiv/msg40972.html<br /><br />nach dieser "2. Zwischenauswertung" wurden keine weiteren Ergebnisse publiziert.Bernhard Mittermaiernoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-39497142282822545692016-11-17T17:10:49.259+01:002016-11-17T17:10:49.259+01:001. Erfahrungen unseres eigenen Verlags
2. https://...1. Erfahrungen unseres eigenen Verlags<br />2. https://www.inetbib.de/listenarchiv/msg59075.html<br />3. die dort erwähnte frühere Diskussion. Der Verlag Eugen Ulmer hatte ca. 2009 im Nachgang zu den Diskussionen eine Untersuchung gemacht; auf die schnelle habe ich die Ergebnisse nicht gefunden. Jedenfalls keine siginfikante Auswirkung hinsichtlich des Umsatzes (weder positiv noch negativ)Bernhard Mittermaiernoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-79278072777616916442016-11-17T14:24:40.210+01:002016-11-17T14:24:40.210+01:00"Dem Absatz des gedruckten Werkes schadet das..."Dem Absatz des gedruckten Werkes schadet das nicht, wie Untersuchungen gezeigt haben."<br /><br />Oh, das hätte ich anders erwartet. Wo kann ich das nachlesen?Robert Hellinghttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-82861790920510912692016-11-17T14:00:51.869+01:002016-11-17T14:00:51.869+01:00Chapeau!
I take the liberty of adding my own Lett...Chapeau! <br />I take the liberty of adding my own Letter to the editor below.<br />Kind regards,<br />Bernhard Mittermaier<br />Forschungszentrum Jülich<br /><br />Das postfaktische Zeitalter hat nun auch Einzug in deutsche Professorenstuben gehalten. Prof. Frei hätte sich z.B. auf den Webseiten seiner eigenen Universitätsbibliothek sehr leicht darüber informieren können, was Open Access ist. Ob er den Gastkommentar „Goldener Zugang“ ohne Kenntnis der Fakten oder aber in bewusster Verdrehung selbiger geschrieben hat, mag sein Geheimnis bleiben. Falsch ist z.B. die grundlegende Prämisse, dass Open Access der Gegensatz zum gedruckten Buch sei. Tatsächlich bedeutet Open Access den freien Zugang zu elektronischer Literatur, idealerweise mit der Möglichkeit der Nachnutzung, beispielsweise aufgrund einer CC-BY Lizenz. Es stellt damit einen Gegensatz zu einem Zugang zu elektronischer Literatur nur gegen Bezahlung dar. Wenn Herr Frei das gedruckte Buch als adäquates Medium erachtet, dann möge er gerne so publizieren. Er kann aber gleichzeitig eine elektronische Fassung des Buches im Internet im Open Access zur Verfügung stellen, um damit auch Rezipienten seiner Werke zu bedienen, die selbst eine andere Arbeitsweise an den Tag legen und lieber mit elektronischen Texten arbeiten. Dem Absatz des gedruckten Werkes schadet das nicht, wie Untersuchungen gezeigt haben. Um nur noch zwei der weiteren Fehler herauszugreifen: „Chemical Abstracts“ sind keine Fachzeitschrift sondern eine Datenbank; die explodierenden Preise der „Online-Publikationen“ sind kein Kennzeichen von „online“, sondern betreffen gedruckte und elektronische Ausgaben von Fachzeitschriften gleichermaßen.Bernhard Mittermaiernoreply@blogger.com