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Any complete standardized SN11 data out there?
(steve)
You could have saved Bjoern and me a lot of work if you had simply asked what the k-correction is instead of guessing. Actually if you look back through the last months posts you can see that I do repeatedly ask for clarification on whether or not the minuet formula is part of the k correction. And I never got a clear answer on that so I had to assume it was until finally now it is clarified to me that the two are seperate. It seems so simple now , but believe me I did ask and no answer was given. But thanks anyways for the info. Here is another example of where I can only `guess` a process yet I can understand well enough the overall concept. Its the details I lack not having done the calculation myself. It is the fitting process. I can tell that from what I understand that when a template is fitted to SN1997ek data there are at least two steps. Heres the formula from Goldhaber. I(t) Imax = fR ((t-tmax)/s(1+z))=b I have no idea what fR is or what a `function `is but I have to assume that if one were to describe in words what happens to the template in this fitting process is that essentially it is normalized to the data peak and divided by 1+z. Looking at fig 1 the template peak is much lower than the data peak so I can only assume that to get the decay of the template from 1.54 onwards to fit the template one doesnt use the averaged day peak value of 4.6 but rather its low end of the error margin is used. That way the HST data and the other decay data fits well and the peak data not so well but still within error margins. That is I assume is called a `best fit`. So is that roughly correct if one had to describe the formula in words? So in words what is happening is that the chi square process calculates all the different possible template lightcurves shapes where the template lies within all the available error bars and `Best fitting` is a complex calculation that produces the optimum lightcurve shape that fits the most data as closely as possible to the middle point of the error margin, With the least amount of datapoints only being included at the outside of their error margins? In other words a bad fit is where the template has its peak at 4.6 but because the template wont be within the error margins of the table data at all from about day 20 onwards this is excluded. Is this roughly correct as a verbal description of the minuet formula? Because if it is then I believe that one can get an even better fit to more of the data than Knops fit if one doesnt include the 1+z transformation in the formula. And the reason why knop doesnt use this result is he presupposes expansion which rules out in his mind using any fitting that doesnt get transformed by 1+z. If this is the case I would critisize him for presupposing expansion in his fitting for the very purpose of PROVING expansion. This would be very unscientific as really he should have done best fits including those that did AND didnt dilate the template timescale by 1+z rather than fits that only included multiplying by 1+z. So thats a fair argument dont you think? Of course it depends on if I have understood the formula well enough. And if I havent can I ask you .. where have I got my verbal description of the minuet formula above wrong? After all, you do say` just ask`. I have no idea where you get this. Are you looking at 1997ek in Figure 1 from the Knop et al. paper? If I draw a horizontal line between the 0.4 tick marks on each side of the graph, it goes directly through the point for the HST reading. The time is clearly before 30 days, but the exact value is hard to read from the graph. I make it about 28 days or so, and this is consistent with the times in Table 11. Yes definitely the graph is hard to read and maybe its not a good idea to base any final judgement from reading the graph.? The problem is its the only way to confirm where the HST reading sits on the template. I also now refer to fig 1 I band 1997ek top right hand graph page 11. I even blow up the graph to 800% and count the pixel heights between 0.4 and 0.5 (8 pixels) The pixel height of the HST reading (5 pixels) . The point of the HST reading is 5 pixels high and its top pixel runs one pixel above 0.4 while its bottom pixel runs 3 pixels below 0.4. That puts as I`ve said before, the HST reading at about 0.38 seeing as the middle pixel of the HST reading is parallel to the 0.38 pixel on the graph. Small change I agree but if you then consider that 1 pixel on the graph works out to 2 days on the time axis and if the middle pixel of the HST reading hits the template 1 pixel later than where the 0.4 pixel hits the template that technically puts the HST reading at 1 pixel or 2 days later than 0.4 on the graph. And in our argument way back on this thread heres your calculations from the table..... (Steves quote..) "For 1997ek, we have from HST data in Table 11 a flux density of 1.54 at day 50846.7 and 0.75 at day 50858.8. This is a decay of 0.78 mag in 12.1 days or 16 days per magnitude. The next measurement is 0.46 at day 50871.9, giving 0.53 mag in 13.1 days or 25 days per magnitude. The error bars are about 15% on the first decay time and 20% on the second one..." And my argument was that if it takes 16 days to decay 1 mag from day 50846.7 and 25 days to decay 1 mag from day 50871 then that means on average every day later than 50846.7 the decay rate increases by 1 day . In that case one can extrapolate that every day before day 50846 if we follow the trend through it will take 1 day less to decay 1 mag. And as the pixels show on the graph the HST reading you refer to is 2 days later(1 pixel later) than the official 0.4 peak +1 reading which means that extraopolating back from the HST reading one gets 14 days (16-2) for a 1 mag decay from 0.4. And thats what I have always said I get from the graphs. But is this too close to call ? Maybe yes. Thats why I went through all 11 to see what I get for 1 mag decay from 0.4 on the graphs. And on average there is no or little time dilation. You can check through yourself and see but I`m sure we`ll get a rough agreement. Even if you lets say get a couple of days extra then me for each SN it will still be in favor n average for no time dilation because in fact 1997ek is the most preferable to a time dilation argument at 1.4 which is midway between 1 and 1.86 .The rest are much closer to 1 than 1997ek. Sean |
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sean wrote:
(steve) You could have saved Bjoern and me a lot of work if you had simply asked what the k-correction is instead of guessing. Actually if you look back through the last months posts you can see that I do repeatedly ask for clarification on whether or not the minuet formula is part of the k correction. There is no "minuet formula". There is a fitting procedure (algorithm) with the title MINUIT. And I never got a clear answer on that so I had to assume it was until finally now it is clarified to me that the two are seperate. It seems so simple now, but believe me I did ask and no answer was given. But thanks anyways for the info. Here is another example of where I can only `guess` a process yet I can understand well enough the overall concept. Its the details I lack not having done the calculation myself. It is the fitting process. I can tell that from what I understand that when a template is fitted to SN1997ek data there are at least two steps. Heres the formula from Goldhaber. I(t) Imax = fR ((t-tmax)/s(1+z))=b You probably mean I(t)/Imax = fR((t-tmax)/s(1+z))+b, which one can obviously also write as I(t) = Imax * ( fR((t-tmax)/w) + b ), w = s(1+z). So one has four free parameters he Imax, tmax, w and b. I have no idea what fR is or what a `function `is A function is simply something which assigns one value to another. In this case, the function fR(t) gives for every time t the magnitude of a SN, normalized so that the peak is at 1. I.e. the function fR describes the known lightcurves of SNs Ia. but I have to assume that if one were to describe in words what happens to the template in this fitting process is that essentially it is normalized to the data peak and divided by 1+z. No, that is not at all what is actually done. What is done is (roughly) that the differences between the measured fluxes (let's call them I_i) and the theoretical magnitudes at the times of measurement t_i are computed, squared and summed up: sum_i (I_i - I(t_i))^2 For this sum, one searches then the minimum by varying the four parameters Imax, tmax, w and b, until the sum is minimal (i.e. until the deviations between the measured fluxes and the theoretical curve becomes minimal). This is done using derivatives an the like - things which you unfortunately probably never heard of, if you don't even know what a function is. :-( If you think there is something wrong with such an approach, please consider that all of this is totally standard stuff, done in *every* are of physics. Oh, and please consider that what I described above is only a rough description - in a more detailed analysis, one also has to take the error margins into account. Looking at fig 1 the template peak is much lower than the data peak so I can only assume that to get the decay of the template from 1.54 onwards to fit the template one doesnt use the averaged day peak value of 4.6 but rather its low end of the error margin is used. That way the HST data and the other decay data fits well and the peak data not so well but still within error margins. That is I assume is called a `best fit`. So is that roughly correct if one had to describe the formula in words? No, sorry. One does not use the "low end of the error margin". One uses all the data and all error margins, and then tries to minimize the sum of the squared differences between the measured data and the theoretical curve, by varying the four free parameters. Also, the formula *is* not the fit. The formula gives the function *with* (or *to*) which the fit is performed. So in words what is happening is that the chi square process calculates all the different possible template lightcurves shapes where the template lies within all the available error bars and `Best fitting` is a complex calculation that produces the optimum lightcurve shape that fits the most data as closely as possible to the middle point of the error margin. With the least amount of datapoints only being included at the outside of their error margins? In other words a bad fit is where the template has its peak at 4.6 but because the template wont be within the error margins of the table data at all from about day 20 onwards this is excluded. Is this roughly correct as a verbal description of the minuet formula? Yes, this is roughly right - with the exception that there still is no "minuet formula". ;-) Because if it is then I believe that one can get an even better fit to more of the data than Knops fit if one doesnt include the 1+z transformation in the formula. And the reason why knop doesnt use this result is he presupposes expansion which rules out in his mind using any fitting that doesnt get transformed by 1+z. No, that makes no sense. s is a completely free parameter. If there were no time dilation in the data, the fit would have *shown* that! It would have yielded the result that s is proportional to 1/(1+z). But what the fit actually showed is that s is (roughly) constant. Steve and you have been through this before; unfortunately, you still have not understood this... :-( [snip more of that] So thats a fair argument dont you think? No. [snip] I have no idea where you get this. Are you looking at 1997ek in Figure 1 from the Knop et al. paper? If I draw a horizontal line between the 0.4 tick marks on each side of the graph, it goes directly through the point for the HST reading. The time is clearly before 30 days, but the exact value is hard to read from the graph. I make it about 28 days or so, and this is consistent with the times in Table 11. Yes definitely the graph is hard to read and maybe its not a good idea to base any final judgement from reading the graph.? The problem is its the only way to confirm where the HST reading sits on the template. Huh? Why? The table data gives you both the time and the magnitude of the reading. I also now refer to fig 1 I band 1997ek top right hand graph page 11. I even blow up the graph to 800% and count the pixel heights between 0.4 and 0.5 (8 pixels) The pixel height of the HST reading (5 pixels) . The point of the HST reading is 5 pixels high and its top pixel runs one pixel above 0.4 while its bottom pixel runs 3 pixels below 0.4. That puts as I`ve said before, the HST reading at about 0.38 seeing as the middle pixel of the HST reading is parallel to the 0.38 pixel on the graph. Well, as you yourself said, reading of the graphs is not very accurate. Did you ever consider the possibility that already in the "printing" of the graphs, some graphical errors crept in? Small change I agree but if you then consider that 1 pixel on the graph works out to 2 days on the time axis Well, so my estimate of an error margin of 2 days was quite good, apparently. ;-) and if the middle pixel of the HST reading hits the template 1 pixel later than where the 0.4 pixel hits the template that technically puts the HST reading at 1 pixel or 2 days later than 0.4 on the graph. And in our argument way back on this thread heres your calculations from the table..... (Steves quote..) "For 1997ek, we have from HST data in Table 11 a flux density of 1.54 at day 50846.7 and 0.75 at day 50858.8. This is a decay of 0.78 mag in 12.1 days or 16 days per magnitude. The next measurement is 0.46 at day 50871.9, giving 0.53 mag in 13.1 days or 25 days per magnitude. The error bars are about 15% on the first decay time and 20% on the second one..." Note that all three points lie quite well on the curve, so using the table data instead of the curve directly is entirely reasonable. And my argument was that if it takes 16 days to decay 1 mag from day 50846.7 and 25 days to decay 1 mag from day 50871 then that means on average every day later than 50846.7 the decay rate increases by 1 day. Sorry, I do not understand what exactly you mean here. A decay rate is measured in one over time (inverse time), so saying that it increases by 1 day makes no sense. Do you mean that the decay *time* increases by 1 day? [snip] Bye, Bjoern |
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From: Bjoern Feuerbacher )
Subject: Any complete standardized SN11 data out there? Oh, and please consider that what I described above is only a rough description - in a more detailed analysis, one also has to take the error margins into account. Thanks for the description. Much appreciated. No, that makes no sense. s is a completely free parameter. If there were no time dilation in the data, the fit would have *shown* that! It would have yielded the result that s is proportional to 1/(1+z). But what the fit actually showed is that s is (roughly) constant. I dont follow this bit about s. I dont think I said remove s or that s isnt roughly constant? If I gave that impression I didnt mean to. All I suggest is that the minuit formula could be redone exactly as Knop does it with a fit to table 11 data except for 1 single change. Instead of the 1+z in the formula, relace that with either 1+0 or a very low z like 1+0.01. Basically I would like to try redoing Knops fit for 1997ek but without a time dilation to see what happens. Am I right in thinking you think that doing this is not allowed? At the worst all that would happen is that there would be no fit . (I wouldnt be taking s out of the formula or change s in any way) Sorry, I do not understand what exactly you mean here. A decay rate is measured in one over time (inverse time), so saying that it increases by 1 day makes no sense. Do you mean that the decay *time* increases by 1 day? Yes. So for each consecutive day the decay time increases by one day so that starting on day 50847 (1 day after the HST 1.54 reading) it will take 17 days to decay 1 mag . And starting from day 50848 it will take 18 days etcetc. THanks for the reading you made of 1997ek peak+1 to peak+2 from the 1997ek template. You got 14 days I believe. Could I possibly interest you in doing maybe 1 more? Like 1998be or 1998as? Just to compare with what I got. Sean |
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sean wrote:
From: Bjoern Feuerbacher ) Subject: Any complete standardized SN11 data out there? Oh, and please consider that what I described above is only a rough description - in a more detailed analysis, one also has to take the error margins into account. Thanks for the description. Much appreciated. For the third time I ask: do you have access to Mathematica or Maple? If yes, I could try to tell you how to do the fits for yourself. No, that makes no sense. s is a completely free parameter. If there were no time dilation in the data, the fit would have *shown* that! It would have yielded the result that s is proportional to 1/(1+z). But what the fit actually showed is that s is (roughly) constant. I dont follow this bit about s. I dont think I said remove s or that s isnt roughly constant? If I gave that impression I didnt mean to. All I suggest is that the minuit formula could be redone exactly as Knop does it with a fit to table 11 data except for 1 single change. Instead of the 1+z in the formula, relace that with either 1+0 or a very low z like 1+0.01. Basically I would like to try redoing Knops fit for 1997ek but without a time dilation to see what happens. Then you would have only s instead of s*(1+z) in the formula, and the result is obvious: the s resulting from the fit would be proportional to 1+z. It simply can't be another way, due to the math of the fitting process! Am I right in thinking you think that doing this is not allowed? It is allowed - but it won't change the result that there is indeed time dilation. At the worst all that would happen is that there would be no fit. (I wouldnt be taking s out of the formula or change s in any way) No, there would indeed be a fit, as long as there is any free parameter for stretching the time axis in the formula. Sorry, I do not understand what exactly you mean here. A decay rate is measured in one over time (inverse time), so saying that it increases by 1 day makes no sense. Do you mean that the decay *time* increases by 1 day? Yes. So for each consecutive day the decay time increases by one day so that starting on day 50847 (1 day after the HST 1.54 reading) it will take 17 days to decay 1 mag . And starting from day 50848 it will take 18 days etc etc. So the "each consecutive day" refers to the day when you start the counting, or what? Well, that sounds strange. I am by no means an expert on SN light curves, but I would have thought that the decay is exponential (because, AFAIK, the light comes from radioactive decay), i.e. it takes the same amount of time to decay by a certain factor no matter where one starts the counting. THanks for the reading you made of 1997ek peak+1 to peak+2 from the 1997ek template. You got 14 days I believe. Could I possibly interest you in doing maybe 1 more? Like 1998be or 1998as? Just to compare with what I got. 1998as: 0.4 at day 34, 0.16 at day 60 (both again with an error margin of +- 2 days), so the difference would be 26 +- 3 days. Bye, Bjoern |
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So the "each consecutive day" refers to the day when you
start the counting, or what? Well, that sounds strange. I am by no means an expert on SN light curves, but I would have thought that the decay is exponential (because, AFAIK, the light comes from radioactive decay), i.e. it takes the same amount of time to decay by a certain factor no matter where one starts the counting. Maybe but remember you also read 14 days off the graph which matches my reading of 14.5 and conflictes with Steves 16 days. From: Bjoern Feuerbacher ) For the third time I ask: do you have access to Mathematica or Maple? If yes, I could try to tell you how to do the fits for yourself. No I dont sorry. I`ve never heard of them actually. Then you would have only s instead of s*(1+z) in the formula, and the result is obvious: the s resulting from the fit would be proportional to 1+z. It simply can't be another way, due to the math of the fitting process! This would only be the case if the universe were expanding. BUT if I am correct and there is no time dilation then you would have s as roughly constant and the result a fit of the restframe template to the data. Anyways you havent tried the fit nor have I so it seems only speculation at this point. You may be interested to hear that I plotted the B and V band restframe templates used by Knop and supplied as table data in his and Goldhabers paper. I then scaled the time axis by 1.86 % in fireworks and found they fit his templates very well .Showing a) that he can show that a time dilated template can fit the data and b) that scaling by percentages in graphic softwqare like fireworks is acceptable and the same as a mathematical calculation ofmultiplying each data point in the tables by 1.86. I also multiplied by s and that gives a better fit. I`m sure thatwil make you happy but the next bit will not. Using the undilated restframe b band template (ie before it was time dilated by 1.86) I simply shifted the the template forward by about 10 days on the graph. And the result is that a fit of all the table data within error margins is produced!! In other words although a time dilated b band template can give a fit of the table data, as Knop has done, a non time dilated b band template also gives as good a fit to the data. The only difference is which day one puts the peak of the template. Furthermore I tried doing this with most of the other 11 SN and found that the appropriate B or V band non time dilated templates fit all the data in each SN`s set of table data. I am sure a re-fit of all the SN data with non dilated templates will verify my findings and prove that there is at least as strong an argument for no time dilation as there is for time dilation. And if you dont believe me try a fit of 1997ek with z=1 1998as: 0.4 at day 34, 0.16 at day 60 (both again with an error margin of +- 2 days), so the difference would be 26 +- 3 days. Thanks for that reading. It`s close to my reading of 23 days below in the table so that makes both your 1997ek and 1998as match my table readings reprinted below. (you had 14 days for 1997ek , I had 14.5). How did you get error margins off the template? (It doesnt have error margins.) Although for 1998as:a careful analysis at 800 % on acrobat shows that peak+1 is at the middle pixel between 30 and 40 making it officially 35 for peak+1 and the 2nd last pixel of seven between 50 and 60 making it roughly 58 for peak + 2. Thats then 58-35=23 Exactly as I had found earlier and not 26 as you get. Anyways if nothing else it show that my readings off the templates are roughly correct. And on average the ratios below are much closer to no time dilation than time dilation. Although if you add in error margins I imagine you can argue that the results can weakly support time dilation. Thats OK as long as you realize that within error margins the results also support no time dilation. In fact within error margins all the data supports much more strongly a no time dilation argument. 1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86 1998eq (restframe469nm) z=.54 15/14.5 =1.03 " 1.54 1997ez (restframe457nm) z=.78 16/13 =1.2 " 1.78 1998as (restframe602nm) z=.35 23/22 =1.04 " 1.35 1998aw (restframe565nm) z=.44 27/20.5 =1.3 " 1.44 1998ax (restframe542nm) z=.50 30/22 =1.36 " 1.50 1998ay (restframe496nm) z=.64 20/17.5 =1.2 " 1.64 1998ba (restframe569nm) z=.43 24.5/22 =1.1 " 1.43 1998be (restframe496nm) z=.64 18/17.5 =1.02 " 1.64 1998bi (restframe467nm) z=.74 15.5/14.5=1.1 " 1.74 2000fr (restframe528nm) z=.54 24.5/22 =1.1 " 1.54 Regarding my predictions that GRB`s do not need `host galaxies` and will in many cases have none even in hubble deep field please note that grb041219 may be offering verification of this prediction . It is a bright grb observed as a compact point source suggesting even at limits of observation no underlying host galaxy. If this bears out with follow up observations we will have an example of how grbs will appear too bright for the high redshift to be accomadated by theory. It also emphasises the need for NASA to change the xrt localization procedure and have the UVOT camera search the entire xrt field of view rather than just any candidate galaxies in the field of view. I also wonder if maybe Craig now on the swift team could check the arrival times of grbs from HETE and SWIFT and INTEGRAL to see if they do indeed appear to give time of arrival localizations that do not match observed localizations as I discussed with Craig previously. Sean |
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In article ,
"sean" writes: Using the undilated restframe b band template (ie before it was time dilated by 1.86) I simply shifted the the template forward by about 10 days on the graph. And the result is that a fit of all the table data within error margins is produced!! Claims such as yours are easy to make, but so far no one has demonstrated any such thing. Please show your work in detail. For example, give a table of the day number, the template value, and the data value. Are you quite sure you didn't stretch the magnitude axis? I thought we had agreed that the I-band light curve for 1997ek does not match the B-band curve for 1995D. If that is so, 1997ek cannot match the template unless a time dilation is put in. -- Steve Willner Phone 617-495-7123 Cambridge, MA 02138 USA (Please email your reply if you want to be sure I see it; include a valid Reply-To address to receive an acknowledgement. Commercial email may be sent to your ISP.) |
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From: Steve Willner )
Claims such as yours are easy to make, but so far no one has demonstrated any such thing. Please show your work in detail. For example, give a table of the day number, the template value, and the data value. Are you quite sure you didn't stretch the magnitude axis? Yes, I did move (though not stretch) the restframe B band scp1997 template without keeping the baseline consistent with the table data baseline. Looking at that now I dont think that you would accept the results. But Im not sure why a stretching of the table data magnitude is unacceptable. It seems that for most of the Knop graphs he has to do that to `normalize` the table data to 1.0 For instance in 1997ek table the peak HST reading is 3.89 but on the graph it reads as a bit less than 1.0 .(I assume to do this Knop divides 3.89 by about 4 to get the table data to fit his template )That essentially is a stretching (although in this a compression)of the mag scale is it not? However for 1997eq the v band undilated template and data do not need to be stretched in magnitude to fit. If I match day 50817 from 1997eq tables with day -5 on the undilated V band template (goldhaber table 2) I get a pretty good match although most are just outside the HST error margin by a very small amount.The comparisons are as follows. The HST readings in the first column are from the tables and include the relevent +- error margin factored in. (ie 50817 is 0.91+-0.3) 1997eq V band template undilated except for s 50817 0.94 .95 out by +0.01 50824 0.88 .99 out by +0.1 50846 0.36 .36 matches 50855 0.25 .22 out by -0.03 50863 0.21 .15 out by -0.05 A rough calculation shows the v band template would only have to be dilated by 1.2 (z=0.2) to fit the table data within error margins if all my numbers are right. Compare that to its redshift which is 1.54. In other words its a good fit undilated and a best fit within error margins by z=1.2 and thus closer to no time dilation. I know you will say... "but the fit above isnt within error margins"...but I notice that quite a few of Knops templates are also not within error margins either by about the same amounts as my 1997eq fit to the undilated v band template above. Ive noted these deviations below so maybe you could check his paper to confirm... The 4th HST reading in 1998ay I band at 0.4 on about day 50 does not fit Knops template. Its off by about .08+- by my calculation. And for 1998ba all the ground based measurements do not fit his template within their error margins at all nor more importantly does the 4th HST reading on day 50947 (out by about 0.1) Also in 1997ez my calculations are that the second HST reading does not match Knops time dilated template within error margins. At 0.77 (0.81 to 0.74 error margins)it is below day 9.53 on the dilated template by about .06 (Day 9.53 is Day 5(0.875) on the undilated scp1997 template (table 2 Goldhaber)5*s*z=9.53) I have also done a visual match of 1998ba to an undilated V band template and it looks pretty good fit but before I do a numbers match from the tables to confirm no time dilation maybe you could look at these problems I notice from Knops calculations ..... For 1998ba I band the 1st HST reading of about 0.925 on the graph is listed in the tables as 5.74. (To normalize from 5.74 from the tables to the graphs 0.925 I divide it by 6.2) Unfortunately the other table datapoints when divided by 6.2 do not match the graph datapoints at all. I found that for the rest of the HST readings one needs to divide by 7.85 to normalize to the graph. Whats happening here? Could you check that out? If this is a mistake on Knops part his template will not fit the peak HST reading from the 1998ba table data. I thought we had agreed that the I-band light curve for 1997ek does not match the B-band curve for 1995D. If that is so, 1997ek cannot match the template unless a time dilation is put in. I never agreed that the I band lightcurve did not match the b band 1995D lightcurve. You agreed with yourself only. In fact I posted to Bjorn proof that it does match 1995 D on dec 5 or thereabouts. Heres the quote from my post... "I plotted out the table data onto a graph. For multiple measurements on the same day like 50817 or 50819 I average out for a day average and plot that as one datapoint.(which Knop also does incidentally) I then have a graph where peak+1 needs to occur at 1 mag less than the highest averaged day peak of 4.65 on day 50817. This I work out to being 1.8. On the graph the only place this can occur with available data is between 50835 and day 50846. Calcula- ting a standard linear decay rate between those two points gives me 1.8 at about day 26. ETcetc for the next reading of peak+2 at 0.74.. ......It also happens that the 12 day decay rate matches very closely the 10 day rate for the restframe low z 1995D lightcurve which I feel strengthens the validity of my methods . Its no accident that they match as the `no time dilation` theory predicts it." Sean |
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sean wrote:
From: Bjoern Feuerbacher ) So the "each consecutive day" refers to the day when you start the counting, or what? Well, that sounds strange. I am by no means an expert on SN light curves, but I would have thought that the decay is exponential (because, AFAIK, the light comes from radioactive decay), i.e. it takes the same amount of time to decay by a certain factor no matter where one starts the counting. Maybe but remember you also read 14 days off the graph which matches my reading of 14.5 and conflictes with Steves 16 days. Err, what on earth do these 14 days have to do with the increase in lightcurve decay time mentioned above? For the third time I ask: do you have access to Mathematica or Maple? If yes, I could try to tell you how to do the fits for yourself. No I dont sorry. I`ve never heard of them actually. A pity... :-( Both are quite good programs for doing standard math tasks. Do you have any other programming experience? Doing a chi squared fit by hand is almost impossible. Then you would have only s instead of s*(1+z) in the formula, and the result is obvious: the s resulting from the fit would be proportional to 1+z. It simply can't be another way, due to the math of the fitting process! This would only be the case if the universe were expanding. *sigh* No. No. No. This is a simple consequence of the *math*. It can not come out another way! This has nothing to do with the actual physics. It simply follows logically that if one uses two completely equivalent fitting procedures for one and the same data, the result will be the same! BUT if I am correct and there is no time dilation then you would have s as roughly constant No. This simply can not happen! That would be a *mathematical* impossibility!!! *Please* try to understand the actual mathematical procedure; then you will see that for yourself! and the result a fit of the restframe template to the data. Anyways you havent tried the fit nor have I so it seems only speculation at this point. No. This is not speculation. This is a *mathematical* *fact*. As sure as the fact that 2 + 2 = 3 + 1. You may be interested to hear that I plotted the B and V band restframe templates used by Knop and supplied as table data in his and Goldhabers paper. Huh? What, exactly, did you plot? Which tables? And how? I then scaled the time axis by 1.86 % in fireworks Huh? Do you mean a factor of 1.86 instead of 1.86 % here? and found they fit his templates very well. Which templates? Showing a) that he can show that a time dilated template can fit the data and b) that scaling by percentages in graphic softwqare like fireworks is acceptable and the same as a mathematical calculation of multiplying each data point in the tables by 1.86. This will give you only a *visual* comparison. That is *much* worse than an actual chi squared fit, which does not only give you the *best* fit of the curve to *all* data, but also tells you *quantitatively* how good the fit is! I also multiplied by s and that gives a better fit. By which s??? I`m sure that wil make you happy but the next bit will not. Using the undilated restframe b band template (ie before it was time dilated by 1.86) I simply shifted the the template forward by about 10 days on the graph. And the result is that a fit of all the table data within error margins is produced!! In other words although a time dilated b band template can give a fit of the table data, as Knop has done, a non time dilated b band template also gives as good a fit to the data. The only difference is which day one puts the peak of the template. All done by mere visual comparison instead of an actual quantitative mathematical analysis and hence quite worthless. Furthermore I tried doing this with most of the other 11 SN and found that the appropriate B or V band non time dilated templates fit all the data in each SN`s set of table data. I am sure a re-fit of all the SN data with non dilated templates will verify my findings and prove that there is at least as strong an argument for no time dilation as there is for time dilation. And if you dont believe me try a fit of 1997ek with z=1 Please come back when you have more than just visual comparisons. 1998as: 0.4 at day 34, 0.16 at day 60 (both again with an error margin of +- 2 days), so the difference would be 26 +- 3 days. Thanks for that reading. It`s close to my reading of 23 days below in the table so that makes both your 1997ek and 1998as match my table readings reprinted below. (you had 14 days for 1997ek , I had 14.5). How did you get error margins off the template? (It doesnt have error margins.) Err, simply estimating how well one can read the data off the graph. Since the pictures are not very sharp, an estimated error margin of 2 days for reading the data off looks quite sensible. Although for 1998as:a careful analysis at 800 % on acrobat shows that peak+1 is at the middle pixel between 30 and 40 making it officially 35 for peak+1 and the 2nd last pixel of seven between 50 and 60 making it roughly 58 for peak + 2. Thats then 58-35=23 Exactly as I had found earlier and not 26 as you get. Err, did you miss my comment where I pointed out that you can't even be sure that the pictures are accurate, that there could be problems with the printing of them already? And *please* use error margins! Anyways if nothing else it show that my readings off the templates are roughly correct. And on average the ratios below are much closer to no time dilation than time dilation. Although if you add in error margins I imagine you can argue that the results can weakly support time dilation. Knop et al did an actual mathematical analysis, Goldhaber did do an even stronger one. Both showed that there is *strong* support for time dilation. Your hand wavy methods are still far away from challenging that. Thats OK as long as you realize that within error margins the results also support no time dilation. In fact within error margins all the data supports much more strongly a no time dilation argument. 1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86 1998eq (restframe469nm) z=.54 15/14.5 =1.03 " 1.54 1997ez (restframe457nm) z=.78 16/13 =1.2 " 1.78 1998as (restframe602nm) z=.35 23/22 =1.04 " 1.35 According to my reading, (26 +- 3)/22 = 1.18 +- 0.14. Actually, the error has to be greater, since the 22 days also have an error margin. 1998aw (restframe565nm) z=.44 27/20.5 =1.3 " 1.44 1998ax (restframe542nm) z=.50 30/22 =1.36 " 1.50 1998ay (restframe496nm) z=.64 20/17.5 =1.2 " 1.64 1998ba (restframe569nm) z=.43 24.5/22 =1.1 " 1.43 1998be (restframe496nm) z=.64 18/17.5 =1.02 " 1.64 1998bi (restframe467nm) z=.74 15.5/14.5=1.1 " 1.74 2000fr (restframe528nm) z=.54 24.5/22 =1.1 " 1.54 [snip discussion of GRBs - irrelevant for this thread] Bye, Bjoern |
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Bjoern Feuerbacher -
heidelberg.de) wrote.. Err, what on earth do these 14 days have to do with the increase in lightcurve decay time mentioned above? As I mentioned in my earlier post I calculated that if the decay were linear then using Steves calculations I could calculate `back` and get 14 days for a 1 mag decay from peak +1. As it happens both you and I also `read` around 14 days for a 1 mag decay from the graphs. So if you dont like me suggesting that the decay is linear and gives 14 days then ignore that and just use the 14 days that you and I read off the graph. Same result either way. Doing a chi squared fit by hand is almost impossible. Maybe for a best fit but one can easily do the calculations for 1 single fit to see if that works. What I have done is visually found a close fit using graphic software and noted the new peak day for the restframe undilated template. I then multiply each day value from the scp 1997 table by the s value (for that SN) and see if this new calculated value falls within or close to the error margins from the high redshift SN data tables. I did this for a couple and posted them to Steve but repost them here as Steve is unable to accept the fact that a non dilated template fits as well as a time dilated template to the available data. Notice how he fails to respond to the points I raise about Knop templates not fitting and he also fails to respond to the proof that in 1997eq it can be shown that a non dilated template fits as well as the dilated templates. Furthermore if he or you can confirm what the peak HST reading is for 1998ba I can show a good fit , using numbers , and possibly better than the dilated template that Knop uses. Here is the relevent part of my post to Steve ... If I match day 50817 from 1997eq tables with day -5 on the undilated V band template (goldhaber table 2) I get a pretty good match although most are just outside the HST error margin by a very small amount.The comparisons are as follows. The HST readings in the first column are from the tables and include the relevent +- error margin factored in. (ie 50817 is 0.91+-0.3) The second column under `V band` is the template day times s with day 0 on the template matching day 50822 from the tables 1997eq V band template (undilated except for s) 50817 0.94 .95 out by +0.01 50824 0.88 .99 out by +0.1 50846 0.36 .36 matches 50855 0.25 .22 out by -0.03 50863 0.21 .15 out by -0.05 A rough calculation shows the v band template would only have to be dilated by 1.2 (z=0.2) to fit the table data within error margins if all my numbers are right. Compare that to its redshift which is 1.54. In other words its a good fit undilated and a best fit within error margins by z=1.2 and thus closer to no time dilation. I know you will say... "but the fit above isnt within error margins"...but I notice that quite a few of Knops template fits are also not within the observed data error margins by about the same amounts as my 1997eq fit to the undilated v band template above. Ive noted these deviations below so maybe you could check his paper to confirm... The 4th HST reading in 1998ay I band at 0.4 on about day 50 does not fit Knops template. Its off by about .08+- by my calculation. And for 1998ba all the ground based measurements do not fit his template within their error margins at all nor more importantly does the 4th HST reading on day 50947 (out by about 0.1) Also in 1997ez my calculations are that the second HST reading does not match Knops time dilated template within error margins. At 0.77 (0.81 to 0.74 error margins)it is below day 9.53 on the dilated template by about .06 (Day 9.53 is Day 5(0.875) on the undilated scp1997 template (table 2 Goldhaber) 5*s*z=9.53) I have also done a visual match of 1998ba to an undilated V band template and it looks pretty good fit but before I do a numbers match from the tables to confirm no time dilation maybe you could look at these problems I notice from Knops calculations ..... For 1998ba I band the 1st HST reading of about 0.925 on the graph is listed in the tables as 5.74. (To normalize from 5.74 from the tables to the graphs 0.925 I divide it by 6.2) Unfortunately the other table datapoints when divided by 6.2 do not match the graph datapoints at all. I found that for the rest of the HST readings one needs to divide by 7.85 to normalize to the graph. Whats happening here? Could you check that out? If this is a mistake on Knops part his template will not fit at all near the peak HST reading from the 1998ba table data. Then you would have only s instead of s*(1+z) in the formula,and the result is obvious: the s resulting from the fit would be proportional to 1+z. It simply can't be another way, due to the math of the fitting process! This would only be the case if the universe were expanding. *sigh* No. No. No. Sorry, I wasnt making myself clear enough. Yes in a calculation for a non time dilated universe s would be proportional to 1+z But because z is always 0 in the calculations then 1+z is always written as 1+0 for every high redshift SN. So you end up with s not being proportional to redshift in a non time dilated universe simply because in the maths the variable redshift z is always stated as 0 to show that the restframe template is *not being time dilated* It couldnt be more simple to understand. Here it is again... To prove a non time dilated universe one DOES NOT DILATE THE RESTFRAME TEMPLATE by z when comparing to a observed high redshift SN. Understand now? If you dont, then tell me: Why would I dilate the restframe template by z to show that the observed high redshift SN in question wasnt dilated? Or maybe I should jus ask you to try the calculations yourself and replace z with 0 to see what you get. I also multiplied by s and that gives a better fit. .By which s??? This is a silly question. By whatever the s value is for that particular SN. What else did you think I meant? 1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86 1998eq (restframe469nm) z=.54 15/14.5 =1.03 " 1.54 1997ez (restframe457nm) z=.78 16/13 =1.2 " 1.78 1998as (restframe602nm) z=.35 23/22 =1.04 " 1.35 According to my reading, (26 +- 3)/22 = 1.18 +- 0.14. Actually, the error has to be greater, since the 22 days also have an error margin. As I have said already a more accurate reading of the graph will give you 23 days = 1.04(+- 0.14.) Thats almost 1 which is much closer to no time dilation than time dilation even with error margins. But if you insist on the 26 days then your above numbers (1.18+- 0.14) are still much closer to 1 than 1.35 within error margins so you seem to be ignoring the fact that your own readings off the graph support a no time dilation argument better than a time dilation argument. Not only that but as your two supplied readings out of the available 11 are very close to my 2 from the 11 one can presume that my other nine readings will also be roughly correct even if you were to double check those. And I have shown all 11 on average support no time dilation more strongly than time dilation within error margins. 1998aw (restframe565nm) z=.44 27/20.5 =1.3 " 1.44 1998ax (restframe542nm) z=.50 30/22 =1.36 " 1.50 1998ay (restframe496nm) z=.64 20/17.5 =1.2 " 1.64 1998ba (restframe569nm) z=.43 24.5/22 =1.1 " 1.43 1998be (restframe496nm) z=.64 18/17.5 =1.02 " 1.64 1998bi (restframe467nm) z=.74 15.5/14.5=1.1 " 1.74 2000fr (restframe528nm) z=.54 24.5/22 =1.1 " 1.54 Sean |
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sean wrote:
sean, as long as you 1) insist that visual comparisons are as valid as an actual mathematical chi squared fit, and 2) refuse to do a proper error analysis for your readings off the graphs, this discussion is quite pointless. Bjoern Feuerbacher - heidelberg.de) wrote.. Err, what on earth do these 14 days have to do with the increase in lightcurve decay time mentioned above? As I mentioned in my earlier post I calculated that if the decay were linear then using Steves calculations I could calculate `back` and get 14 days for a 1 mag decay from peak +1. I remember your calculation - but IIRC it did *not* use a *linear* decay. Also, what justifies the assumption of a linear decay? [snip] Doing a chi squared fit by hand is almost impossible. Maybe for a best fit but one can easily do the calculations for 1 single fit to see if that works. Show your work. What I have done is visually found a close fit using graphic software and noted the new peak day for the restframe undilated template. Visual fits are in no way as rigorous as an actual calculation. I then multiply each day value from the scp 1997 table by the s value (for that SN) and see if this new calculated value falls within or close to the error margins from the high redshift SN data tables. I did this for a couple and posted them to Steve but repost them here as Steve is unable to accept the fact that a non dilated template fits as well as a time dilated template to the available data. By a *visual* comparison only. Notice how he fails to respond to the points I raise about Knop templates not fitting and he also fails to respond to the proof that in 1997eq it can be shown that a non dilated template fits as well as the dilated templates. He simply is fed up with your attitude that fits which you do visually have as much validity as an actual chi squared calculation. Furthermore if he or you can confirm what the peak HST reading is for 1998ba I can show a good fit , using numbers, and possibly better than the dilated template that Knop uses. As long as you continue claiming that your visual fits have as much (or even more) validitiy than Knops et al. and Reiss' et al. actual mathematical statistical analysis, I see no point in this. Here is the relevent part of my post to Steve ... If I match day 50817 from 1997eq tables with day -5 on the undilated V band template (goldhaber table 2) I get a pretty good match although most are just outside the HST error margin by a very small amount.The comparisons are as follows. The HST readings in the first column are from the tables and include the relevent +- error margin factored in. (ie 50817 is 0.91+-0.3) The second column under `V band` is the template day times s with day 0 on the template matching day 50822 from the tables 1997eq V band template (undilated except for s) 50817 0.94 .95 out by +0.01 50824 0.88 .99 out by +0.1 50846 0.36 .36 matches 50855 0.25 .22 out by -0.03 50863 0.21 .15 out by -0.05 Nice. 5 data points for a single SN. Why on earth do you think this proves anything? A rough calculation shows the v band template would only have to be dilated by 1.2 (z=0.2) to fit the table data within error margins if all my numbers are right. Compare that to its redshift which is 1.54. In other words its a good fit undilated and a best fit within error margins by z=1.2 and thus closer to no time dilation. I know you will say... "but the fit above isnt within error margins"...but I notice that quite a few of Knops template fits are also not within the observed data error margins by about the same amounts as my 1997eq fit to the undilated v band template above. I told you how to do a chi squared fit in principle: calculate the deviations between the theoretical curve and the data for *all* data points, squared them and add them up. Search for the minimum of the obtained number. As long as you don't have done this, you have no basis for claiming that your fit is better than the one uses by Knop et al. [snip more of this] Then you would have only s instead of s*(1+z) in the formula,and the result is obvious: the s resulting from the fit would be proportional to 1+z. It simply can't be another way, due to the math of the fitting process! This would only be the case if the universe were expanding. *sigh* No. No. No. Sorry, I wasnt making myself clear enough. Yes in a calculation for a non time dilated universe s would be proportional to 1+z. But because z is always 0 in the calculations then 1+z is always written as 1+0 for every high redshift SN. *sigh* No. No. No. You *still* have not understood the actual method. Try again. So you end up with s not being proportional to redshift in a non time dilated universe The *data* shows time dilation, no matter if you include a factor 1+z in the fit or not. It is a *mathematical* *fact* that both methods (using s or s(1+z)) will give the same result. How often do we have to repeat this until you understand it? simply because in the maths the variable redshift z is always stated as 0 You make no sense at all. [snip] To prove a non time dilated universe one DOES NOT DILATE THE RESTFRAME TEMPLATE by z when comparing to a observed high redshift SN. Understand now? Yes. I've understood all along what you want to do. But you *still* fail to understand: when doing the chi squared analysis with s only instead of s(1+z), the resulting s will *come out of the mathematical analysis to be proportional to 1+z*. This is an unavoidable *mathematical* *fact*. If you dont, then tell me: Why would I dilate the restframe template by z to show that the observed high redshift SN in question wasnt dilated? Huh? Sorry, I don't understand the question. Or maybe I should jus ask you to try the calculations yourself and replace z with 0 to see what you get. Not necessary. The outcome is a *mathematical* *fact*. I also multiplied by s and that gives a better fit. By which s??? This is a silly question. No, not at all. By whatever the s value is for that particular SN. What else did you think I meant? I did not know. I would not have asked if I knew! 1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86 1998eq (restframe469nm) z=.54 15/14.5 =1.03 " 1.54 1997ez (restframe457nm) z=.78 16/13 =1.2 " 1.78 1998as (restframe602nm) z=.35 23/22 =1.04 " 1.35 I notice that you *still* do not bother to give error margins. According to my reading, (26 +- 3)/22 = 1.18 +- 0.14. Actually, the error has to be greater, since the 22 days also have an error margin. As I have said already a more accurate reading of the graph will give you 23 days = 1.04(+- 0.14.) Why do you think you are in the position to judge if yours or my reading is more accurate? I *also* used a good magnification to read off the graph. Thats almost 1 which is much closer to no time dilation than time dilation even with error margins. But if you insist on the 26 days then your above numbers (1.18+- 0.14) are still much closer to 1 than 1.35 within error margins So you conveniently ignore my remark that the error margin actually has to greater? so you seem to be ignoring the fact that your own readings off the graph support a no time dilation argument better than a time dilation argument. For this one single example, yes. So what? We are talking about a quite big amount of data here. Even if you manage to show this for all 11 of Knop's SNs (and I strongly doubt that), you still have not addressed all the other SNs used by previous researchers. Anyway, as long as you refuse to do a proper error analysis, such a discussion is moot. Not only that but as your two supplied readings out of the available 11 are very close to my 2 from the 11 one can presume that my other nine readings will also be roughly correct even if you were to double check those. That's a rather strange jumping to conclusions. Because we agree *roughly* on two, I have to agree on all the other nine, too? And I have shown all 11 on average support no time dilation more strongly than time dilation within error margins. 1998aw (restframe565nm) z=.44 27/20.5 =1.3 " 1.44 1998ax (restframe542nm) z=.50 30/22 =1.36 " 1.50 1998ay (restframe496nm) z=.64 20/17.5 =1.2 " 1.64 1998ba (restframe569nm) z=.43 24.5/22 =1.1 " 1.43 1998be (restframe496nm) z=.64 18/17.5 =1.02 " 1.64 1998bi (restframe467nm) z=.74 15.5/14.5=1.1 " 1.74 2000fr (restframe528nm) z=.54 24.5/22 =1.1 " 1.54 Show the error margins. Bye, Bjoern |
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