User talk:PAR

Recent edits

Hello, I was wondering if you could explain why you reverted my edits to Wealth condensation. Marble Weaver (talk) 00:52, 25 May 2010 (UTC)[reply]

Because people who save and invest cannot avoid taxes. It was just a wrong statement. PAR (talk) 00:57, 25 May 2010 (UTC)[reply]

You are incorrect, People who invest most of their money pay a smaller portion of their income in sales tax, because they don't spend most of it. Likewise, people of modest means must spend all their money and pay a sales tax. Marble Weaver (talk) 01:05, 25 May 2010 (UTC)[reply]

Yes, but thats only a tiny piece of the picture. There are other taxes besides sales tax. People who invest and save must pay capital gains taxes on the money they make. If they invest in property, they must pay property taxes, etc., etc. People who invest and save are not avoiding taxes, they are subjecting themselves to different forms of taxation. If you had a lot of money, you would probably want to buy stuff and pay the sales tax, but you would also want to invest and save some of your money to protect you against bad times and as a way to increase your holdings in good times. Thats not tax avoidance, that's common sense. Maybe the point you want to make is that sales tax is regressive since it takes up a larger portion of a poorer persons wealth than a rich person. If a poor person pays five percent of their income on sales tax, saying that a rich person is avoiding taxes by not spending five percent of their income on sales tax is not correct. PAR (talk) 04:15, 25 May 2010 (UTC)[reply]

That's correct, the rich person is subject to different forms of taxation, however he has the potential because he is a rich person too write-off money he lost or couldn't repay, he could just write it off in taxes, whereas a poor person couldn't do that. The poor person is stuck with the bad cheeseburger and never will be able to write it off in their taxes. Marble Weaver (talk) 04:56, 25 May 2010 (UTC)[reply]

The main reason that PAR’s revert was reasonable is that your changed wording implies that the tax savings are the important driving cause of wealth condensation, but that’s clearly wrong. Saving and investment lead to wealth accumulation because the value of investments increases over time as the society prospers and its economy grows. The poorest of the poor pay very little tax, and for good reason – they need all the money they can get just to pay for food. The richest of the rich don’t need to spend much money as a percentage of their wealth, so they invest it instead. This is true whether the tax system is progressive or regressive, whether the rate is 1% or 60%, whether its form is sales tax or property tax or capital gains tax. In other words, the tax differences are a distraction to the main point, and elevating them as you did would make the sentence misleading. –jacobolus (t) 19:10, 25 May 2010 (UTC)[reply]

You are now a Reviewer

Hello. Your account has been granted the "reviewer" userright, allowing you to review other users' edits on certain flagged pages. Pending changes, also known as flagged protection, will be commencing a two-month trial at approximately 23:00, 2010 June 15 (UTC).

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File source problem with File:HallettHouse.pngThank you for uploading File:HallettHouse.png. I noticed that the file's description page currently doesn't specify who created the content, so the copyright status is unclear. If you did not create this file yourself, you will need to specify the owner of the copyright. If you obtained it from a website, please add a link to the website from which it was taken, together with a brief restatement of that website's terms of use of its content. However, if the copyright holder is a party unaffiliated from the website's publisher, that copyright should also be acknowledged. If you have uploaded other files, consider verifying that you have specified sources for those files as well. You can find a list of files you have created in your upload log. Unsourced and untagged images may be deleted one week after they have been tagged per Wikipedia's criteria for speedy deletion, F4. If the image is copyrighted and non-free, the image will be deleted 48 hours after 06:38, 2 September 2010 (UTC) per speedy deletion criterion F7. If you have any questions or are in need of assistance please ask them at the Media copyright questions page. Thank you. Magog the Ogre (talk) 06:38, 2 September 2010 (UTC)[reply]TalkbackYou have new messages Hello, PAR. You have new messages at Magog the Ogre's talk page. You can remove this notice at any time by removing the {{Talkback}} or {{Tb}} template.

Magog the Ogre (talk) 16:40, 2 September 2010 (UTC)[reply]

You have new messages Hello, PAR. You have new messages at Magog the Ogre's talk page. You can remove this notice at any time by removing the {{Talkback}} or {{Tb}} template.

Magog the Ogre (talk) 09:59, 3 September 2010 (UTC)[reply]

plots in "Stable distribution" could use more samplesHi PAR, thank you for the nice plots you made for the article "Stable distribution" and possibly others. It seems that these plots could be improved by sampling more densely on the horizontal axis. For example in File:Levy distributionPDF.png the first derivative of the red curve is clearly not continuous to the eye. File:Levyskew_distributionPDF.png also shows the same problem (purple curve close to x=0). Would you mind fixing this or providing the input files for these plots so that others could do it quickly? Thanks, Spheniscus —Preceding unsigned comment added by Spheniscus (talk • contribs) 16:54, 21 October 2010 (UTC)[reply]Hi - I think maybe your browser or PC display has a problem. Click on the graph in the article, and then click on "Full Resolution" and you will see that the curves are solid. PAR (talk) 17:13, 21 October 2010 (UTC)[reply]Hello again. No, my browser displays PNG files correctly of course. What I mean is that one can clearly see the linear segments from which the curves are made. Look, for example, at the peaks in http://upload.wikimedia.org/wikipedia/commons/0/04/Levyskew_distributionPDF.png or at the purple curve around x=0. If you made these curves with gnuplot, you should issue something like "set samples 1000" before plotting them. Spheniscus (talk) 22:40, 21 October 2010 (UTC)[reply]Hi - ok, I see what you mean, I didn't read your first message closely enough. I will try to find the code used to generate them and fix them. PAR (talk) 22:49, 21 October 2010 (UTC)[reply]ApologizingI have to apologize with you for a rather embarassing matter. My reply in Temperature was a bit curt, because I misinterpreted an expression of yours. My English is perhaps better for technical matters than in everyday language, so I understood that by "edit ... then go from there" you intended something like "make what you want, then go away from there/here, we have discussed at length the whole matter, so what you want here." I know this looks ridicolous, but this is. Pardon me --GianniG46 (talk) 22:48, 24 October 2010 (UTC)[reply] LOL. I have to watch my colloquialisms. Yes, I meant for you to edit, and then to see where those edits would lead us.... I did not take any offense at your answer, no apology needed. PAR (talk) 23:33, 24 October 2010 (UTC)[reply]http://en.wikipedia.org/wiki/File:Lognormal_distribution_PDF.svgThanks for the diagrams. There is a slight mistake in the pdf given in the description, pdf(x,sigma,mu)=((1/(x*sqrt(2*pi*sigma**2))) * exp((-log(x-mu)**2)/(2*sigma**2))) should be pdf(x,sigma,mu)=((1/(x*sqrt(2*pi*sigma**2))) * exp(-(log(x)-mu)**2/(2*sigma**2))) The problem does not appear when mu=0. —Preceding unsigned comment added by 122.107.60.43 (talk) 00:22, 25 October 2010 (UTC)[reply]Birdmounth jointSpeedy deletion nomination of Birdsmouth joint

You may also wish to consider using a Wizard to help you create articles. See the Article Wizard. Thank you.A tag has been placed on Birdsmouth joint, requesting that it be speedily deleted from Wikipedia. This has been done under the criteria for speedy deletion, because it is a very short article that does not provide sufficient context to identify its subject. Please see Wikipedia:Stub for our minimum information standards for short articles. Also please note that articles must be on notable subjects and should provide references to reliable sources that verify their content. Please do not remove the speedy deletion tag yourself. If you plan to expand the article, you can request that administrators wait a while for you to add contextual material. To do this, affix the template {{hang on}} to the article and state your intention on the article's talk page. Feel free to leave a note on my talk page if you have any questions about this. Jlivewell (talk) 16:23, 18 February 2011 (UTC)[reply]Consensus on dashesHi, this is to let everyone who has expressed an interest in the topic that the discussion to arrive at a consensus has been opened at Wikipedia talk:Manual of Style/dash drafting, with discussion taking place at Wikipedia_talk:Manual_of_Style/dash_drafting/discussion. Apologies if you have already commented there, or have seen the discussion and chosen not to comment. Casliber (talk · contribs) 22:57, 6 June 2011 (UTC)[reply]Equilibrium systems are homogeneous?Just read your query under this topic in Thermodynamics:Talk. You were given the wrong answer, which I have just corrected. You are correct: so long as a barrier separating the system allow some communication (I chose adiabatic and flexible), the system can be at equilibrium at different temperatures. (One doesn't need a book to answer this, fortunately.) Geologist (talk) 06:01, 29 September 2011 (UTC)[reply]Nomination for deletion of Template:Non-free standard test imageTemplate:Non-free standard test image has been nominated for deletion. You are invited to comment on the discussion at the template's entry on the Templates for discussion page. SchuminWeb (Talk) 08:19, 22 October 2011 (UTC)[reply]LennaThe copyright notice is in the indicia of the November 1972 issue of Playboy. That copyright applies to derivative works as a matter of American law.—Chowbok ☠ 13:10, 4 November 2011 (UTC)[reply]Ok, good, its a terminology problem. Please see my response on the File talk:Lenna.png page.File:Lenna.png listed for deletionA file that you uploaded or altered, File:Lenna.png, has been listed at Wikipedia:Files for deletion. Please see the discussion to see why this is (you may have to search for the title of the image to find its entry), if you are interested in it not being deleted. Thank you. SchuminWeb (Talk) 14:11, 4 November 2011 (UTC)[reply]Thomson ScatteringUnfortunately I cannot find good explanations of Thomson Scattering anywhere. I saw your diagram and thought you may know a bit about it so I thought I'd ask you directly: http://en.wikipedia.org/wiki/File:Thomson_scattering_geometry.png In this diagram, apparently the scattered electric field is found by taking the incident electric field, at "collision point", and basically "reflecting" it's wave as if there were a mirror in the perpendicular angle. Is this a correct way to look at it? Is this in assumption that the electron is free standing, or part of an atom? Or does it matter? How can you tell in what direction, and at what speed the electron will move, given the "location" and "velocity" of an election, and "location" and "velocity" of the electromagnetic wave? Is there a way to predict at what angle the wave will reflect in? I would assume the something changes because of the interaction of the charge of the electron the electromagnetic wave, no? You are my main source of information! Regards, Ven — Preceding unsigned comment added by 173.3.246.53 (talk) 11:25, 19 November 2011 (UTC)[reply]An electric field makes a positive particle move in the direction of the electric field, so an electron will move in the opposite direction. The incoming wave consists of oscillating electric and magnetic fields, but the magnetic field doesn't contribute hardly at all, its mostly the electric field. Since the electric field is oscillating perpendicular to the direction of the incoming wave, it will make the electron move back and forth perpendicular to the direction of the incoming wave. An oscillating electron emits its own radiation in every direction, but its strongest perpendicular to its motion (ie towards and away from the incoming wave). The pattern of radiation is called dipole radiation. If you are at an angle to the direction of the incoming wave, you will see the radiation, but it won't be as strong as the maximum radiation.So that means that, no, thinking of it as a reflection is not right. Reflection is off of a surface full of electrons, not one electron. The direction the electron will move is perpendicular to the incoming wave, because that's the direction the electric field is oscillating in. The incoming wave is considered to be a plane wave, so it doesn't really have a position, just like a sound wave doesn't really have a position. The bottom line is that the incoming wave makes the electron oscillate, and an oscillating electron emits its own radiation. I am pretty sure it has to be a free electron, not bound to an atom, because the binding forces to an atom are usually much stronger than the force of the oscillating electric field from the incoming radiation. If you haven't already, check out the article Thomson_scattering. PAR (talk) 17:24, 19 November 2011 (UTC)[reply]cumulative DISTRIBUTION functionsIs there any way you can fix the various occurrences of the phrase cumulative density function in various graphics you've uploaded. Apparently these histories are not editable. If the files were deleted and uploaded again with correct edit summaries instead of that phrase, maybe that would do it. As it is, innocent people can be misled to think that those who actually think about these things use that phrase. Michael Hardy (talk) 05:18, 22 November 2011 (UTC)[reply]Sure, I will fix that. I'm not sure what you mean by "uploaded again with correct edit summaries instead of that phrase". Also, can you give me the names of the offending files that you are aware of? PAR (talk) 05:29, 22 November 2011 (UTC)[reply] Here's one. 174.53.163.119 (talk) 14:34, 24 May 2013 (UTC)[reply]Kelvin is not a degreehttp://commons.wikimedia.org/wiki/File:PlanckianLocus.png indicates color temperature is Tc(°K) which is wrong because Kelvin is a unit of absolute temperature and therefore must not be called a degree, e.g., the graph should be Tc(K). P.S. If possible, an SVG version would be good, it's 2012 already. — Preceding unsigned comment added by Niks1024 (talk • contribs) 12:58, 3 January 2012 (UTC)[reply]Done, but as a png. Sorry, I'm not equipped to do svg. PAR (talk) 13:40, 3 January 2012 (UTC)[reply]Thank you for your speedy action and contribution to Wikipedia. Niks (talk) 18:49, 21 January 2012 (UTC)[reply]A barnstar for you!The Barnstar of Diligence For keeping track of all the little 2's that wriggle out of sight from the rest of us, even if it means working through the "homework" problem from scratch, such as in the highly visible uncertainty principle example. Teply (talk) 18:19, 15 July 2012 (UTC)[reply]request for adviceDear PAR, if you have a little time to spare, perhaps you would very kindly be willing to have a look at the articles Heat and Temperature and their talk pages? I do not know any likely way of dealing with the recurrent problem that is there. Perhaps you may have some suggestion?Chjoaygame (talk) 12:36, 13 November 2012 (UTC)[reply]Thank you for your very kind and helpful action. The problem, as perhaps you have noticed, is recurrent and persistent. One manifestation of the problem is that the relevant user seems imperceptive of the distinction between thermodynamics and statistical mechanics, and seems to lack the intellectual skills that enable one to learn to perceive distinctions of this kind. He seems resolutely determined to advance his viewpoint, and not open to reason about it.Chjoaygame (talk) 17:03, 13 November 2012 (UTC)[reply]No problem, Chjoaygame, but let's focus on getting it right, not on what's wrong with those who get it wrong. PAR (talk) 05:38, 14 November 2012 (UTC)[reply]Following this, I see you have made a new edit to the article on thermal equilibrium. Good. But it seems that at the same time you did not notice that a recent undo of an edit to a part of the article was made by an editor who shall be nameless. The undo restored a vandalistic edit that had destroyed the very textual sequence of the relevant part of the article. The undo was thus itself a form of vandalism. I did not try to revert it because I did not wish to encourage the nameless editor by responding to him, no matter how irrational and provocative his action. Better, I thought, in the circumstances, to let an error stand in the article than to reward the vandal by responding to him. I suppose the article on thermal equilibrium is one of the less read in the Wikipedia. One might perhaps think that such an error would eventually be picked up and repaired by a neutral person. So far it has not.Chjoaygame (talk) 10:19, 4 December 2012 (UTC)[reply]Thanks for pointing that out, I fixed it. PAR (talk) 14:47, 4 December 2012 (UTC)[reply]Thank you for that. I value your contributions. This instance is easy to prove, but there are so many others like it that I have lost track of them, but in which the proof is not so obvious. I am not one for chasing administrative remedy because I do not know how to do it and because of the burden of proof in subtle questions and because of my personal lack of interest in administrative remedies.Chjoaygame ([[User Chjoaygame (talk) 20:46, 4 December 2012 (UTC)talk:Chjoaygame|talk]]) 19:23, 4 December 2012 (UTC)[reply]Of course I support your present action. I do not think I know the best way to proceed from here right now.Chjoaygame (talk) 07:45, 12 December 2012 (UTC)[reply]Following your line, I am refraining from doing something about Waleswatcher's grammar. The Oxford English Dictionary clearly recognizes such usage as 'pioneer result'; it comes naturally to a native English speaker. As for his "As a radiance ...", the law is a statement not a quantity. So in effect it seems that I have to treat Waleswatcher as a troll who is not a native English speaker.Chjoaygame (talk) 01:04, 17 December 2012 (UTC)[reply]I'd prefer not to talk about editors, but rather about the usefulness of their edits. I think the point you make is minor. PAR (talk) 04:38, 17 December 2012 (UTC)[reply]I would prefer likewise. But even talking about the usefulness of edits sometimes entails feeding the trolls. I agree that the point is minor.Chjoaygame (talk) 04:59, 17 December 2012 (UTC)[reply]Wanna fix your error?You commented on a potential ArbCom case this morning - but you placed your comment out of order. Could you please move it to the bottom of the list AND you will definitely need to provide links/proof of the commentary you made (✉→BWilkins←✎) 13:39, 19 December 2012 (UTC)[reply]Sure - sorry, this is the first one I was involved in. I will fix it. PAR (talk) 15:40, 19 December 2012 (UTC)[reply]No issues - you didn't need to full-out remove. However, ArbCom cases are all about actual linkable evidence that is related to the subject at hand. Schumin's case is about lack of knowledge of a specific deletion criteria, and his use of tools, PLUS an unwillingness to discuss. Incivility in that case is a very minor sidebar, but if you claim it happened, definitely need to be able to prove it. (✉→BWilkins←✎) 16:34, 19 December 2012 (UTC)[reply]PAR, your comment was actually fine as it was and the link you included adequately proves the issue of incivility and lack of assumption of good faith on the part of the subject Admin (both of which have also been raised by others including myself and Steve Baker) as being key elements in his unwillingness to discuss his actions and thus NOT "minor sidebars" but instead integral parts of the total pattern of administrative misconduct raised by the RfA. It is therefore fine to restore (and expand) your original comment which should then be placed after all the existing comments. It is as relevant, supported, and valuable as all the others. Centpacrr (talk) 17:05, 19 December 2012 (UTC)[reply]Centpacrr, thanks for not reading my comments in full before your tirade above. (✉→BWilkins←✎) 17:38, 19 December 2012 (UTC)[reply]I must admit Bwilkins that I am quite puzzled by your comment immediately above. I read your comments in full and what I had to say to PAR was not a "tirade", nor was it either a criticism of anything you said or a failure to assume good faith on your part. If you were to read the link PAR included in his statement you will find it fully supports his statement in that it illustrates demonstrated violations of WP:ADMINACCT by the subject Admin through his long established patters of incivility, failure to respond to the concerns of other editors, and failure to assume good faith on the part of those who disagree with his administrative actions. These same issues (along with others) have also been raised by many others in their statements in both this RfA and the many previous WP processes started regarding this Admin's practices over the years. That being the case they are certainly not "minor sidebars" to the RfA but they are instead central to it. I would therefore urge that instead of accusing me or any other user involved in this or similar matters of "failing to assume the good faith" on your behalf that you should assume good faith on behalf of myself and the others involved and fully familiarize yourself with the actual issues and facts of the case. The link PAR included fully meets the test of supporting his statement on the matter with regards to incivility and other relevant issues. Centpacrr (talk) 18:39, 19 December 2012 (UTC)[reply]Again, you've misconstrued my entire original messages, and merely expanded your tirade ... ironic, based on your concerns about Schumin. Not worth further discussing, and I hope that PAR actually reads my original message, not your rather messed up interpretation of them. Cheers (✉→BWilkins←✎) 18:54, 19 December 2012 (UTC)[reply]I'm sorry, sir, but I really have no idea what point you are trying to make here. What I said above is not a "tirade" and I have no clue as to why you think it is. You seem to think that the subject Admin's demonstrated pattern of incivility in dealing with others is both a "minor sidebar" in this whole matter, and that it is unsupported by the link PAR included in his statement. Based on my long involvement in this I happen to view these two points differently. Simply because I or any other editor has a different opinion, however, does not in any way constitute a lack of assumption of good faith on my part in anything you said. It is just a simple difference of opinion and nothing more.I do not believe that I have "misconstrued" anything you said just because I see it differently, and I'm sorry if you think my saying so constitutes a "tirade" which is does not. It is simply a difference of opinion which I would appreciate if you would accept that I have stated in good faith as I have gladly done with everything you have said. I really just don't see any reason for you or anyone else to manufacture a controversy over this. You are absolutely entitled to your interpretation of what PAR's statement means as I am entitled mine. I happen to think what he said originally was appropriate and supported by the link he included and you apparently don't. That's it. I hope, however, that he has not been so miffed or intimidated by this artificial kerfuffle that he now decides to not comment at all on the RfA. Centpacrr (talk) 19:37, 19 December 2012 (UTC)[reply]Ok, I am an editor, not a Wikilawyer and I spoke from my gut rather than spending the time to translate my gut feelings into civil language. I was involved in the Lenna dispute along with SteveBaker, and if his comment meets the standards of civility and proper linkage, then my comment is that I agree with and support SteveBaker's comment. Period. That is why I posted after SteveBaker's post, sorry, my error. Would such a comment be acceptable to all? I deleted my post without further comment because I don't want to get neck deep in the Wikilaw swamp, its not my forte. However, as an editor, I do wish to assist in preventing the type abuse addressed by this ArbCom from happening in the future. PAR (talk) 21:27, 19 December 2012 (UTC)[reply]I think your first one was fine, PAR, and this one is fine too. As I said above, I truly have no idea why any of this Wikilawyering kerfuffle over it arose in the first place as I thought where it was placed next to Steve Baker's statement was fine too (I don't see anything that says you can't put a Statement next to one it closely relates to), and in my view is was correctly supported with the link to the Lenna.png discussion. Go for it. Centpacrr (talk) 21:39, 19 December 2012 (UTC)[reply]Centpacrr - Done, and thanks for your help on this. PAR (talk) 22:46, 19 December 2012 (UTC)[reply]Happy to help. PAR, and thank you as well! The Arbs vote is currently 7-0 to accept so I expect this will all be resolved soon. Have a great holiday. Cheers. Centpacrr (talk) 23:10, 19 December 2012 (UTC)[reply]edit to article on thermal equilibriumA section has been deleted from the article on thermal equilibrium.Chjoaygame (talk) 10:21, 31 December 2012 (UTC)[reply]I think that section was not very useful. It did not convey much information and seemed to confuse things rather than clarify anything. Phrases like "Though Planck does not explicitly say so, it seems from the context of his text that he may mean that ..." are REALLY bad. Editors should not be speculating on an article page about what a particular reference may or may not mean. The whole section sounds like the internal dialog of an editor who has not yet reached any firm conclusions. That editor should keep pondering the subject until he reaches some firm conclusions, and then place that on the page - the internal dialog has no place on a Wikipedia page. Please check outWikipedia:Weasel#Unsupported_attributions. PAR (talk) 19:49, 31 December 2012 (UTC)[reply]I have removed the paragraph on Planck's usages.Chjoaygame (talk) 01:31, 1 January 2013 (UTC)[reply]Viscous stress tensor and related topicsHi, indeed I am working on those pages and they are still in a somewhat messy state. In particular the viscous stress tensor article (and the section in viscosity) need work since the only tensor they really define is the strain rate tensor $\nabla v$ . (I have just created an article for it, see strain rate tensor; so the material is now repeated in 3 places...) The viscous stress tensor proper does not seem to be defined anywhere; there is only a formula specific for isotropic Newtonian fluids. But I must stop now; I hope to get back to them again in 20 hours time. All the best, --Jorge Stolfi (talk) 06:10, 17 January 2013 (UTC)[reply]Hi, I see that you have restored viscous stress tensor to its former state. Let me make another try, quite different from (and hopefully better than) the one I botched yesterday. Note that the part of the derivation that pertains to strain rather than stress is (or will soon be) in strain rate tensor. --Jorge Stolfi (talk) 01:58, 18 January 2013 (UTC)[reply]Well, its not exactly its former state, but hopefully improved. It now contains a definition of the viscosity tensor $(c_{ijk\ell })$ . I kind of screwed things up too, because I didn't realize that there was a separate viscous stress tensor page. The only thing I was worried about is that the derivation I put back should definitely not be deleted. As to where it should fit, I'm not sure. Stress and the strain rate are so closely connected, maybe they should not be separated? Also, there are two derivations that are near mirror images of each other, the development outlined in viscous stress tensor and the development outlined in Hooke's law#Isotropic materials on the elasticity tensor. I think that the explanation of the elasticity tensor should be brought into this package of related concepts we are trying to properly present. I'm not sure at this point what is best. I will look at the other articles and try to think of possibilities. PAR (talk) 03:46, 18 January 2013 (UTC)[reply] I have worked some more on both pages. Strain rate tensor now has some pictures, and the derivation you refer to should be all there (unless I lost something by mistake). Viscous stress tensor focuses more on the stress and relies on the other article for the detailed definition of strain rate. All the best, --Jorge Stolfi (talk) 03:35, 23 January 2013 (UTC)[reply]Your revision todayYou appear to have made a change to the article on Heat without any discussion on the Heat talk page. Would you care to discuss this reversion here? --Damorbel (talk) 20:18, 17 January 2013 (UTC)[reply]No. If you disagree with my restoration of the information that you deleted, please begin the discussion of your deletion on the Heat talk page. PAR (talk) 04:08, 18 January 2013 (UTC)[reply]You write:. . . please begin the discussion of your deletion . . . Oh dear, you seem to have missed it! Here it is :[1] PAR, even if I had made my edit without reasoning it on the talk page, you would not be excused for doing the same thing. ( Wikipedia:BRD_misuse )--Damorbel (talk) 06:55, 18 January 2013 (UTC)[reply]Scalar tensorHi, thanks for the corrections to strain rate tensor. However, I am puzzled by the statement that there is no such thing as "scalar tensor". I have only a very foggy notion of tensor calculus, but I would think that (to mathematicians, if not to physicists) any linear mapping between two tensors is itself a tensor. So multiplication by a scalar too would deserve the title. No? ("Scalar multiple of the Kronecker delta tensor" is quite a mouthful for such a basic concept...) By the way, I just saw in decomposition of spectrum (functional analysis) that function analysts write "T+λ" where T is a functional operator and λ is a complex number; and don't even mention that to their analysts. Why can't physicists commit the same sin? 8-) All the best, --Jorge Stolfi (talk) 14:25, 10 February 2013 (UTC)[reply]Read the Wikipedia page on Tensors very carefully, its a good page.Tensors can be classified with the descriptor (R,D). R is the rank of the tensor, its the number of subscripts you use when using index notation. A is a scalar or, equivalently, a rank 0 tensor. It's represented by a single number. $A_{i}$ is a vector or, equivalently, a rank 1 tensor. Its represented by a column (or row) matrix. $A_{ij}$ is a rank 2 tensor and so on. Stress and strain, for example, are 2nd rank tensors. D is the dimension of the space in which the tensor is defined. A vector is represented by a 1xD matrix. A 2nd rank tensor is represented by a DxD matrix. So stress and strain are (2,3) tensors represented by 3x3 matrices.Notice I said "represented by". A tensor is a physical quantity, it doesn't change when you change coordinate systems. Its representation in a particular coordinate system will change when you change coordinate systems. If I have a vector and a coordinate system where that vector is represented by [1,0] and then I rotate that coordinate system clockwise by 90 degrees, I will now represent that SAME vector by [0,1]. The vector has not changed, but, since you changed coordinate systems, its REPRESENTATION (the matrix you use to describe it) has changed. A matrix is not a tensor, its a representation of a tensor that has no meaning unless you specify the coordinate system you are using. A tensor doesn't need a coordinate system to exist.Scalars are (0,D) tensors and they represent physical quantities that do not change with different coordinate systems. The mass of an object is a scalar, because it doesn't matter what coordinate system you pick, its always has the same value. As long as you have an orthogonal coordinate system, you don't have to worry about "contravarian" and "covariant" representations of tensors. Also, the unit tensor will not change when you change coordinate systems. So, for orthogonal coordinate systems, the representation of the unit tensor doesn't change when you change coordinate systems. But when you get into non-orthogonal coordinate systems, this might not be the case. So its tempting to call the unit tensor a "scalar" because it acts like a scalar for orthogonal coordinate systems, but its not really good. By my understanding, the term "constant tensor" is preferable to "scalar tensor" because saying a tensor is a scalar is saying that a (0,D) tensor is a (2,D) tensor, which is just not true.There are certain properties ("invariants") that tensors have, and we really want to know how to calculate those invariants given a particular matrix representation. A scalar has only one invariant - its value. A vector also has one invariant - its length. To calculate that you take the square root of the sum of the squares of its representative matrix. In index notation, if the vector is represented by the matrix $A_{k}$ , its length is $\sum _{k=1}^{D}A_{k}A_{k}$ or, using Einstein summation $A_{k}A_{k}$ . That length is a scalar. Change coordinate systems, and you will be squaring different numbers and summing, but the result will be the same, because the length is a property of the vector and therefore a property of ANY matrix you use to represent that vector. Rank two tensors have lots of invariant properties. The "trace" of a (2,D) tensor is an invariant, it doesn't depend on coordinate system. If the tensor is represented by the matrix $A_{ij}$ , its trace is $\sum _{k=1}^{D}A_{kk}$ or, using Einstein summation $A_{kk}$ . The determinant is an invariant. The symmetry of a tensor is invariant. An antisymmetric (2,D) tensor is represented by an antisymmetric DxD matrix. A symmetric (2,D) tensor is represented by an symmetric DxD matrix. Change coordinate systems, and the symmetry is not changed, because symmetry is a property of the TENSOR and is reflected in the MATRIX used to represent it. A symmetric (2,D) tensor is equal to the sum of a constant tensor and a traceless symmetric tensor. This is for tensors, its a physical thing, it doesn't depend on the matrix you happen to use to represent them. If you happen to have a particular matrix that represents a symmetric tensor, then it will be equal to a constant matrix plus a traceless symmetric matrix, but those matrices will all change when you change coordinate systems. But their trace and their symmetry won't change, because the tensor itself has its own intrinsic trace and symmetry.This is why I like index notation rather than direct notation. If you have a tensor, in direct notation its given as A and you have to remember that the rank of the tensor and that Tr(A) is invariant. In index notation, its given as $A_{ij}$ and its trace is $A_{kk}$ and you can see right away its a rank 2 tensor and that its trace is invariant because there are no "free indices", only the pair of "dummy indices" k and k. Same for the length of a vector. In direct notation, the vector is B and you have to remember its a vector, and you have to remember its length B.B is invariant. In index notation, the vector is $B_{i}$ and its length is $B_{k}B_{k}$ . You can see right away you are dealing with a vector, and you can see the product is invariant, again because there are no "free indices", only the pair of "dummy indices" k and k. PAR (talk) 18:13, 10 February 2013 (UTC)[reply]Wow, thanks. That does match what I know of tensor calculus. (Although I cannot share your fondness for index notation. Coordinates are valuable but artificial constructs, not physically real; it does not seem "right" to understand the physical concept in terms of them.) The issue of scalar x unit is about conventions, and it seems to have a parallel in programming language design. Some languages (like java) are very fussy about indices; for them a 1xn array is completely different from an n-vector, and a vector with one element is completely different from a real number. Some (like matlab) are totally liberal, and do not make such distinction. Forty years ago, when I was starting a Masters project on computer language design, such choices seemed terribly important, worth spilling blood for. Today I have to use regularly half a dozen different languages, spanning that entire range of views. It gets confusing and tiring at times, but I have grown to accept the use of multiple incompatible notations as a necessary part of life, like rain and getting out of bed in the morning. Indeed, now believe that consistency of notation is useful only within the scope a single stand-alone work (a book, a journal article, or a wikipedia article). Trying to enforce it beyond that scope is not only a waste of time, but actually makes texts harder to write and read. So, I see that function analysts think matlab-wise while physicists think java-wise. 8-) OK, that is fine, no problem with that. All the best, --Jorge Stolfi (talk) 04:21, 11 February 2013 (UTC)[reply]Yes, I agree that being "multinotational" is the best. An argument over notation is like a semantic argument in ordinary language. It doesn't focus on the interesing stuff. But, in defense of the index notation, I think you are still not understanding it completely. In direct notation, you have a symbol, like $\mathbf {A}$ and it represents a vector, no coordinate system implied. What I am saying is that the symbol $A_{k}$ is completely equivalent, it represents a vector, no coordinate system implied. It does not represent the k-th component of the vector in a particular coordinate system. Any similarity to a notation which uses $A_{k}$ as the k-th component in some specific coordinate system is entirely coincidental. As a symbol for the vector, it carries more information than $\mathbf {A}$ , it carries the fact that it is a vector by virtue of the "k". $A_{k}$ , $A_{i}$ , $A_{\alpha }$ , $A_{z}$ all are identically equivalent to each other and to the direct notated $\mathbf {A}$ . Likewise, $A_{k}B_{k}$ is identically equivalent to $\mathbf {A} \cdot \mathbf {B}$ , it's the dot product, independent of coordinate system. Where direct notation uses the dot, index notation uses the matched indices. $A_{k}B_{k}$ does not imply a sum over indices, it implies the dot product. You can see right away that it is a physical scalar, there are no unmatched indices. $A_{i}B_{j}$ is identically equivalent to $\mathbf {A} \mathbf {B}$ , it's the dyadic product (a 2nd rank tensor), independent of coordinate system. Where direct notation uses adjacency, index notation uses the unmatched indices. You can see right away that there are two free indices, making it a 2nd rank tensor. It is not a square matrix. It will be represented by a square matrix once you choose a coordinate system. It will be represented by a different square matrix if you choose another coordinate system.For a dyadic product $(\mathbf {AB} )\cdot \mathbf {C} =\mathbf {A(B} \cdot \mathbf {C} )$ is, I guess, a theorem, but in index notation its $(A_{i}B_{j})C_{j}=A_{i}(B_{j}C_{j})$ but parentheses are not used in the index notation. Its just $A_{i}B_{j}C_{j}$ or $B_{j}A_{i}C_{j}$ or $C_{j}A_{i}B_{j}$ etc., whichever you like. In other words, the theorem is "contained" in the notation. When you start to see how many "theorems" in direct notation are "contained" in the index notation, I think you will see an advantage.To be fair, I am simplifying things. Using all lower case indices I am restricting things to orthogonal coordinate systems. Really, matched indices should be upper and lower (contravariant and covariant). In other words, the dot product is correctly written $A^{k}B_{k}$ . If you get into relativistic electrodynamics, there is no electric field and magnetic field, there is the electromagnetic field and its a 4 dimensional 2nd rank tensor. In relativity, you cannot avoid covariant and contravariant distinctions, and the advantage of index notation becomes obvious. Going to general relativity, it becomes blatantly obvious.I am in no way suggesting that the standard (Ricci) notation be used for the fluid mechanics tensors, that's overkill, but we can at least push the reader towards that standard tensor notation by including index notation and defending it against the mistaken idea that it is coordinate-system-bound. PAR (talk) 05:50, 12 February 2013 (UTC)[reply]two recent editsDear PAR, two recent edits by you (on thermodynamic system and on conservation of energy) have attracted comment from me on the respective talk pages. I think both of them are mistaken and should be undone. I think it fair to ask you to attend to that.Chjoaygame (talk) 21:26, 19 February 2013 (UTC)[reply]Please see the talk page for the conservation of energy article. PAR (talk) 15:11, 20 February 2013 (UTC)[reply]Thank you.Chjoaygame (talk) 19:11, 20 February 2013 (UTC)[reply]Examples of convolutionI saw the wiki page, but I couldn't find any examples using actual numbers evaluating the formula. Could you give some examples of convolution, please? Mathijs Krijzer (talk) 22:22, 9 March 2013 (UTC)[reply] See Talk:Convolution PAR (talk) 05:29, 10 March 2013 (UTC)[reply]DefinitionThe convolution of f and g is written f∗g, using an asterisk or star. It is defined as the integral of the product of the two functions after one is reversed and shifted. As such, it is a particular kind of integral transform:$(f*g)(t)\ \ \,$ ${\stackrel {\mathrm {def} }{=}}\ \int _{-\infty }^{\infty }f(\tau )\,g(t-\tau )\,d\tau$ $=\int _{-\infty }^{\infty }f(t-\tau )\,g(\tau )\,d\tau .$ (commutativity)Domain of definitionThe convolution of two complex-valued functions on R^d $(f*g)(x)=\int _{\mathbf {R} ^{d}}f(y)g(x-y)\,dy$ is well-defined only if f and g decay sufficiently rapidly at infinity in order for the integral to exist. Conditions for the existence of the convolution may be tricky, since a blow-up in g at infinity can be easily offset by sufficiently rapid decay in f. The question of existence thus may involve different conditions on f and g.Circular discrete convolutionWhen a function g_N is periodic, with period N, then for functions, f, such that f∗g_N exists, the convolution is also periodic and identical to: $(f*g_{N})[n]\equiv \sum _{m=0}^{N-1}\left(\sum _{k=-\infty }^{\infty }{f}[m+kN]\right)g_{N}[n-m].\,$ Circular convolution

Main article: Circular convolutionWhen a function g_T is periodic, with period T, then for functions, f, such that f∗g_T exists, the convolution is also periodic and identical to: $(f*g_{T})(t)\equiv \int _{t_{0}}^{t_{0}+T}\left[\sum _{k=-\infty }^{\infty }f(\tau +kT)\right]g_{T}(t-\tau )\,d\tau ,$ where t_o is an arbitrary choice. The summation is called a periodic summation of the function f.Discrete convolutionFor complex-valued functions f, g defined on the set Z of integers, the discrete convolution of f and g is given by:$(f*g)[n]\ {\stackrel {\mathrm {def} }{=}}\ \sum _{m=-\infty }^{\infty }f[m]\,g[n-m]$ $=\sum _{m=-\infty }^{\infty }f[n-m]\,g[m].$ (commutativity)When multiplying two polynomials, the coefficients of the product are given by the convolution of the original coefficient sequences, extended with zeros where necessary to avoid undefined terms; this is known as the Cauchy product of the coefficients of the two polynomials.Discussion on HeatPAR you did not contribute to the discussion with Waleswatcher on heat. In this discussion I argued that heat not proportional to the difference of temperature but to temperature alone, i.e. material at 0K has only ground state energy and above 0K the material has motion (heat) that gives it a temperature >0K. This difference from 0K may be at the core of your argument, if so it should be in the article. --Damorbel (talk) 06:31, 13 March 2013 (UTC)[reply] To make myself clearer, perhaps, I could have an explanation of your reversal of my edit. In my view your note (here)" heat is not a quantity, it is a process " needs amplification since energy is something which is measured in Joules i.e. a quantity of heat is stated in Joules. Similarly heat has a quality measured by temperature, a quality that it possesses at the particle level via the Boltzmann constant. I suggest these matters are sufficiently important to be mentioned in the opening section of the Heat article. --Damorbel (talk) 17:11, 13 March 2013 (UTC)[reply]This discussion belongs on the Talk:Heat page. I won't respond here.PAR (talk) 20:30, 13 March 2013 (UTC)[reply]It's the reversal of my edit I'm complaining about. So far you have managed to do that without any comment on the talk pages, am I supposed to wait for you to respond? --Damorbel (talk) 21:36, 13 March 2013 (UTC)[reply]WP:HUSH PAR (talk) 23:05, 13 March 2013 (UTC)[reply][2] --Damorbel (talk) 06:48, 14 March 2013 (UTC) Dear PAR :[reply] Looking over some material you have written above, I believe you will be interested in some research work I have done recently. During last six years or so, I have been playing with stress and strain tensors etc. etc. If you are interested, watch these two videos http://www.youtube.com/watch?v=wSXjxgt6uXs and http://www.youtube.com/watch?v=k3zPaKv5ImE. The book by Saad is based on the assumption that strains are small. Once you try to use strain definition other than engineering strain it gets messy. It seems we may have similar interests. Keep in touch.Rajen Merchant (talk) 19:47, 15 March 2013 (UTC)[reply]File:HallettHouse.png missing description detailsDear uploader: The media file you uploaded as: File:HallettHouse.png is missing a description and/or other details on its image description page. If possible, please add this information. This will help other editors make better use of the image, and it will be more informative to readers. If the information is not provided, the image may eventually be proposed for deletion, a situation which is not desirable, and which can easily be avoided. If you have any questions, please see Help:Image page. Thank you. Theo's Little Bot (error?) 00:54, 13 April 2013 (UTC)[reply]I have added Author:unknown to the information template. PAR (talk) 01:07, 13 April 2013 (UTC)[reply]More Frank Lmbert nonsenseThe dispute you participated in on distorted "definitions" of entropy has surfaced again on the article entropy (energy dispersal). I found a reliable source to back up the contention that the whole idea is merely a hobby horse that has not been widely accepted in the scientific community. http://www.rsc.org/chemistryworld/2013/02/entropy-second-law-arieh-ben-naim 98.109.238.95 (talk) 14:27, 29 June 2013 (UTC)[reply]