tag:blogger.com,1999:blog-22973357.post116567892366190130..comments2023-09-27T07:44:19.769-04:00Comments on Sabine Hossenfelder: Backreaction: Deformed Special RelativitySabine Hossenfelderhttp://www.blogger.com/profile/06151209308084588985noreply@blogger.comBlogger32125tag:blogger.com,1999:blog-22973357.post-90639304526525188992009-09-21T03:16:42.013-04:002009-09-21T03:16:42.013-04:00Anonymous,
Could you please clarify the followin...Anonymous, <br /><br />Could you please clarify the following: you have a momentum space, you introduce coordinates in it, you cut them off at some finite value. How do you achieve this finite value is the same under all Lorentz transformations?<br /><br />The motivation of the model was not to regularize integrals. One just gets this for free, and I think it's neat. It isn't so Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-84277473229082900012009-09-20T13:54:30.978-04:002009-09-20T13:54:30.978-04:00Bee: "The former: Lorentz invariant RESULT.&q...Bee: "The former: Lorentz invariant RESULT."<br /><br />This RESULT is obtained in a Lorentz-covariant way by preserving the Lorentz symmetry. In your MODEL you deform the Lorentz transformations in order to achieve a (the same???) RESULT. <br /><br />"The latter: Lorentz invariant MODEL based on a non-trivial geometry of momentum space." <br /><br />If the motivation for thisAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-24804852985315999482009-09-20T12:07:39.171-04:002009-09-20T12:07:39.171-04:00Anonymous: The difference I was pointing out is th...Anonymous: The difference I was pointing out is that between there being a prescription to use a cutoff in the momentum space integration that leads to a Lorentz invariant result, and the geometry of momentum space being modified in a Lorentz invariant way such that the integration is finite. The former: Lorentz invariant RESULT. The latter: Lorentz invariant MODEL based on a non-trivial geometrySabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-28466494015626963572009-09-20T09:19:08.119-04:002009-09-20T09:19:08.119-04:00I meant by deforming the Lorentz transformations w...I meant by deforming the Lorentz transformations when this can be easily done in the standard way? What's the advantage?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-87797566777001820632009-09-20T09:15:55.047-04:002009-09-20T09:15:55.047-04:00Bee, you first said: "If there is an upper bo...Bee, you first said: "If there is an upper bound on the integration in momentum space at it appears in loop integrals, it requires a modified transformation behaviour for the momentum of the virtual particle, otherwise the bound will indeed break Lorentz invariance."<br /><br />Now you are admitting that: "There are many ways to use a cutoff to regularize momentum-space integrals Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-108761112454985512009-09-20T08:33:46.894-04:002009-09-20T08:33:46.894-04:00Anonymous: There are many ways to use a cutoff to ...Anonymous: There are many ways to use a cutoff to regularize momentum-space integrals that lead to Lorentz-invariant results. The procedure you are talking about is one. That isn't the same as saying the cut-off introduced in momentum space is invariant under Lorentz-transformation. Note that this is before Wick-rotation.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-81130708040520932562009-09-20T08:21:26.995-04:002009-09-20T08:21:26.995-04:00Bee:"Anonymous: If you introduce a hard cutof...Bee:"Anonymous: If you introduce a hard cutoff into any direction and boost it, it will be at a different value, thus it isn't invariant under (usual) Lorentz trafos. The 4 volume trivially stays invariant, but that isn't the point."<br /><br />When one integrates over the 4-momentum of a virtual particle the cutoff is a Lorentz invariant. It's just a radius of a 4-sphere Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-88709384631113349972009-09-20T03:15:16.112-04:002009-09-20T03:15:16.112-04:00Anonymous: If you introduce a hard cutoff into any...Anonymous: If you introduce a hard cutoff into any direction and boost it, it will be at a different value, thus it isn't invariant under (usual) Lorentz trafos. The 4 volume trivially stays invariant, but that isn't the point.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-78936699854447985112009-09-19T21:47:37.265-04:002009-09-19T21:47:37.265-04:00Bee: "If there is an upper bound on the integ...Bee: "If there is an upper bound on the integration in momentum space at it appears in loop integrals, it requires a modified transformation behaviour for the momentum of the virtual particle, otherwise the bound will indeed break Lorentz invariance."<br /><br />Introducing the upper bound when integrating over the 4-momentum of a virtual particle does not break the Lorentz invariance Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165856298924948172006-12-11T11:58:00.000-05:002006-12-11T11:58:00.000-05:00Dear Anonymous,Thank you for providing the referen...Dear Anonymous,<BR/><BR/>Thank you for providing the references. Indeed, I've read these papers some while ago. I will address it elsewhere, this comment section isn't really the best place for it. Also, it is connected to a paper I'm working on.<BR/><BR/>Regarding your concern that my model doesn't really provide us with any insights, there were several reasons why I found the approach pursued Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165855078736997532006-12-11T11:37:00.000-05:002006-12-11T11:37:00.000-05:00Hi Eugene,Thanks, this is very interesting. It see...Hi Eugene,<BR/><BR/>Thanks, this is very interesting. It seems to me it should be possible to set this in some connection to the formalism I have used. I mean, I have essentially postulated such an behaviour and just parameterized it, but it might be helpful to approach it this way. <BR/><BR/>But I will have to think about it, and get back to you. Best regards,<BR/><BR/>SabineSabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165854661130722482006-12-11T11:31:00.000-05:002006-12-11T11:31:00.000-05:00Dear Arun:Bee,Your comment of 3:03 PM, December 09...Dear Arun:<BR/><BR/><I>Bee,<BR/>Your comment of 3:03 PM, December 09, 2006 is indeed confusing.</I><BR/><BR/>Well, reading it again, I realize it is confusing. I am sorry. What I was trying to explain there is the motivation for 'standard' DSR. This motivation being: one can't boost someone into the super-planckian regime. The motivation for my model is somewhat different. To stay with the Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165811244921334532006-12-10T23:27:00.000-05:002006-12-10T23:27:00.000-05:00Bee,Your comment of 3:03 PM, December 09, 2006 is ...Bee,<BR/>Your comment of 3:03 PM, December 09, 2006 is indeed confusing.<BR/><BR/>Presumably one (desired?) result of the kind of deformation you're looking at is a softening of the high energy behavior of scattering amplitudes? <BR/><BR/>Presumably the difference from what you're trying to do and using a Lorentz-violating momentum cut-off is that in the latter, the cut-off is usually chosen to Arunhttps://www.blogger.com/profile/03451666670728177970noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165801219404827622006-12-10T20:40:00.000-05:002006-12-10T20:40:00.000-05:00hi Bee, How would you describe a particle being in...hi Bee,<BR/><BR/><I> How would you describe a particle being in an interaction, and the dependence of the boost on it? </I><BR/><BR/>Let's consider a simple example of two interacting particles 1 and 2. The generators of the Poincare Lie algebra are<BR/><BR/>P = p_1 + p_2 (total momentum)<BR/>J = j_1 + j_2 (total angular momentum)<BR/>H = h_1 + h_2 + V (total energy)<BR/>K = k_1 + k_2 + W (total Eugene Stefanovichhttps://www.blogger.com/profile/00126106418312618470noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165781998161948442006-12-10T15:19:00.000-05:002006-12-10T15:19:00.000-05:00For references on the velocity, I suggest hep=th/0...For references on the velocity, I suggest hep=th/0304027, hep-th/0207022, and hep-th/0211057.<BR/><BR/>The only possible way that DSR could be physical is indeed in its impact on virtual particles. By choosing a relatively nice-looking regulator in the deformed coordinates, one could potentially get a physical effect. However, doing this is no different from taking ordinary coordinates and Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165775939764627612006-12-10T13:38:00.000-05:002006-12-10T13:38:00.000-05:00Dear Thomas,How can anyone believe that one can ge...Dear Thomas,<BR/><BR/><I>How can anyone believe that one can get new physics just by chosing some wacky new coordinates? It's like claiming that Newtonian mechanics is diffeomorphism invariant just because it can be formulated in curvilinear coordinates. </I><BR/><BR/>This is not the point. The question is what were the appropriate coordinates (observables) from the beginning on, when there are Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165773168903410812006-12-10T12:52:00.000-05:002006-12-10T12:52:00.000-05:00Dear Eugene,This is a very nice picture you have c...Dear Eugene,<BR/><BR/>This is a very nice picture you have chosen :-) Regarding you suggestion, I see some serious conceptual problems. <BR/><BR/><I>Why shouldn't the (Lorentz) boost transformations of the particle's momentum depend on the interaction of this particle with the rest of the system?</I><BR/><BR/>How would you define the 'rest of the system' and the interactions with it? In my Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165772342616122042006-12-10T12:39:00.000-05:002006-12-10T12:39:00.000-05:00Dear Anonymous,If you try to constuct a truly phys...Dear Anonymous,<BR/><BR/><I>If you try to constuct a truly physical quantity, like the velocity (to choose an example that was worked out a few years ago), you find that it behaves exactly the same way, obeying ordinary SR, in either theory.</I><BR/><BR/>I'd be happy to see the reference, if you'd be so kind to provide it. I'd have expected this requires knowledge of the formulation in position Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165772053163907912006-12-10T12:34:00.000-05:002006-12-10T12:34:00.000-05:00Clarification to avoid misinterpretation: in my la...Clarification to avoid misinterpretation: in my last comment the sentence <I>the inequality s > sum over mass^2 does not have a Lorentz invariant quantity on the left hand side.</I> refers to that inequality in the DSR scenario.Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165771927830531522006-12-10T12:32:00.000-05:002006-12-10T12:32:00.000-05:00Dear Lubos:Had you read my papers, you would have ...Dear Lubos:<BR/><BR/>Had you read my papers, you would have noticed that I write that the allegedly present threshold corrections in DSR arise from the fact that the inequality s > sum over mass^2 does not have a Lorentz invariant quantity on the left hand side. Which is why I think they either break observer independence (that would indeed be back to a violation of Lorentz invariance), or they Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165764514866481752006-12-10T10:28:00.000-05:002006-12-10T10:28:00.000-05:00How can anyone believe that one can get new physic...How can anyone believe that one can get new physics just by chosing some wacky new coordinates? It's like claiming that Newtonian mechanics is diffeomorphism invariant just because it can be formulated in curvilinear coordinates. <BR/><BR/>But then again, it is not much worse than the quite popular kind of noncommutative geometry, where the configuration space becomes noncommutative because one Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165729746001939082006-12-10T00:49:00.000-05:002006-12-10T00:49:00.000-05:00Bee,if my understanding is correct, in DSR Lorentz...Bee,<BR/><BR/>if my understanding is correct, in DSR Lorentz boost transformations are kinematical (just as in ordinary special relativity): boost transformations of the particle's momentum is exacly the same, no matter whether the particle is free or it interacts with other particles. In this respect, Lorentz boosts are assumed to be similar to space translations and rotations. What do you thinkEugene Stefanovichhttps://www.blogger.com/profile/00126106418312618470noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165726388366880622006-12-09T23:53:00.000-05:002006-12-09T23:53:00.000-05:00I think there is a fairly basic algebraic objectio...I think there is a fairly basic algebraic objection to DSR. DSR is supposed to be constructed as a nonlinear representation of the Lorentz algebra. The Lorentz algebra is just the universal enveloping algebra of so(3,1). When one constructs this eveloping algebra explicitly, as a quotient of the tensor algebra on the generators, it LOOKS like it depends a great deal on how those generators areAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165715905605129622006-12-09T20:58:00.000-05:002006-12-09T20:58:00.000-05:00Dear Bee!"If you think you can easily falsify DSR,...Dear Bee!<BR/><BR/>"If you think you can easily falsify DSR, how about you do it?"<BR/><BR/>Well, I think I have already done it many times, and probably not just me, but feel free to choose to ignore these comments.<BR/><BR/>The precise statements of DSR that have been offered so far are not terribly well-defined and they are not at the level of real research papers, so it is not possible to Luboš Motlhttps://www.blogger.com/profile/17487263983247488359noreply@blogger.comtag:blogger.com,1999:blog-22973357.post-1165713130964936062006-12-09T20:12:00.000-05:002006-12-09T20:12:00.000-05:00Hi Lubos,It's because I think that one can easily ...Hi Lubos,<BR/><BR/><I>It's because I think that one can easily falsify your statement [...] Instead, as the anonymous correctly tried to argue but you ignored him or her, it is only the Lorentz-invariant quantities that can be a subject to general inequalities [...]If there are inequalities that determine different regimes of physics, they refer to invariant scales such as p^2 - for example the Sabine Hossenfelderhttps://www.blogger.com/profile/06151209308084588985noreply@blogger.com