SNe1a data

!

I finally found time to look in more detail at your table which you
repeatedly posted:

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

I wanted to repeat your calculations (including a proper
error analysis). However, it is unclear to me where you got
the numbers in the denominators from (10, 14.5, 13, 22, 20.5,
22, 17.5, 22, 17.5, 14.5, 22).

I tried to find this information with Google, and found out that
you apparently got these numbers somehow from the SN 1995 D, using
data in the paper astro-ph/9810291 by Riess et al. However, it is still
unclear to me how exactly you got these numbers from that paper. Did you
read them off the graph shown for SN 1995 D in fig. 3 somehow?

If yes, why do you have so many different numbers - there are only four
different curves presented there? And: did you take into account what the
figure caption says, namely that B, I and R curves are plotted at an
offset with respect to the V curve?

Sorry if you have addressed all that already somewhere; I was not able to
find this. :-(


***
I must of addressed this in earlier posts but It would take a while
trying to find it.
As it is I will have to tell you just from memory but the process was
this: I used several SN from Reiss to average numbers but I think Steve
insisted I only used 1995D or at least he insisted on only reffering
to it as it was better sampled than most. I did find however that
within the range of low redshift lightcurves there was a certain amount
of variation that 1995D didnt neccesarily fit into well.
Anyways I measured off 1995D delay times for I think it was
peak+1 to peak +2 off of the 4 BVRI *tables* (not template) so I would
have got for example :
12 days for B peak 1-2
22 days for V "
22 days for R "
22 days for I"

I then still had a problem where for instance a high redshift SN
was being compared to restframe where the high redshift -restframe
was inbetween lets say B and V . By this I mean that if 1995D B is 430
and V is 530nm and the high redshift was 480nm restframe obviuosly
it coulnt be compared to either B or V accurately.
I decided the best and fairest way to deal with this would be to
calculate where 480 was proportionally between 430 and 530nm.
And then work out the equivelent in days from 1995D B-V
So 480 is approx 1/2 way between 430-530
So if B was lets say 10 days and V was 22 days I nominally
gave 480 nm as 16 days decay time. Which is wht 496 is 17.5
and 467nm is 14.5
I also found that V R I all seemed to have around the same decay
times of about 22 days for peak 1-2 so anything above 530nm was always
22 days.

One example from 1995D is the following
peak in V is 13.42 and peak+1 is 14.43 day 9783.77
and peak+2 is 15.43 (between 15.35-15.63) day 9793 approx.
thats approx 10 days for peak1-2 in B

Also I think there is a error or typo in my list above for 1998aw.
It should be 22 days.
Hopefully that answers your questions. I welcome your analysis
of those numbers and will accept any corrections you make as long as
I can double check your corrections.

sean wrote:

I finally found time to look in more detail at your table which you
repeatedly posted:

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

A small question about the restframe wavelengths. You probably
got them by dividing the middle wavelength of the I filter
by 1+z, right? If yes, then your value fo 1998eq seems to be
wrong. Should be around 528 nm.


I wanted to repeat your calculations (including a proper
error analysis). However, it is unclear to me where you got
the numbers in the denominators from (10, 14.5, 13, 22, 20.5,
22, 17.5, 22, 17.5, 14.5, 22).

I tried to find this information with Google, and found out that
you apparently got these numbers somehow from the SN 1995 D, using
data in the paper astro-ph/9810291 by Riess et al. However, it is still
unclear to me how exactly you got these numbers from that paper. Did you
read them off the graph shown for SN 1995 D in fig. 3 somehow?

If yes, why do you have so many different numbers - there are only four
different curves presented there? And: did you take into account what the
figure caption says, namely that B, I and R curves are plotted at an
offset with respect to the V curve?

Sorry if you have addressed all that already somewhere; I was not able to
find this. :-(


***
I must of addressed this in earlier posts but It would take a while
trying to find it.
As it is I will have to tell you just from memory but the process was
this: I used several SN from Reiss to average numbers but I think Steve
insisted I only used 1995D or at least he insisted on only reffering
to it as it was better sampled than most. I did find however that
within the range of low redshift lightcurves there was a certain amount
of variation that 1995D didnt neccesarily fit into well.
Anyways I measured off 1995D delay times for I think it was
peak+1 to peak +2 off of the 4 BVRI *tables* (not template) so I would
have got for example :
12 days for B peak 1-2
22 days for V "
22 days for R "
22 days for I"

I looked up what templates Knop et al. used for the fitting. These
templates were determined by using the data from a large number of
low-z supernovae, instead of only the single one SN1995D. They provide
the data in their table 2. And it turns out that these templates
have only 16.8 days for peak+1 to peak+2 in the V band (I did
not check the R and I band so far)! That obviously makes the
numbers you calculated above much too low - the actual numbers are
about 30% greater!


I then still had a problem where for instance a high redshift SN
was being compared to restframe where the high redshift -restframe
was inbetween lets say B and V . By this I mean that if 1995D B is 430
and V is 530nm and the high redshift was 480nm restframe obviuosly
it coulnt be compared to either B or V accurately.
I decided the best and fairest way to deal with this would be to
calculate where 480 was proportionally between 430 and 530nm.
And then work out the equivelent in days from 1995D B-V
So 480 is approx 1/2 way between 430-530
So if B was lets say 10 days and V was 22 days I nominally
gave 480 nm as 16 days decay time. Which is wht 496 is 17.5
and 467nm is 14.5
I also found that V R I all seemed to have around the same decay
times of about 22 days for peak 1-2 so anything above 530nm was always
22 days.
One example from 1995D is the following
peak in V is 13.42 and peak+1 is 14.43 day 9783.77
and peak+2 is 15.43 (between 15.35-15.63) day 9793 approx.
thats approx 10 days for peak1-2 in B

I am not sure if this linear interpolation is even necessary. From
what I have read so far about the K-correction, it is precisely for
taking this into account. So, if you use K-corrected data, you
don't have to do such an interpolation. See e.g. what Steve Willner
wrote already on Nov. 23rd last year:

"For example, for I-band observations of 1997ek at z=0.86, one would use
a template at 438 nm rest. Since such a template won't in general
exist, one applies a "k-correction" to the observations to account for
the difference between the available template (B, approx 440 nm) and
the rest frame."

(Message ID: )



Also I think there is a error or typo in my list above for 1998aw.
It should be 22 days.

Do you mean 27/22 or 22/20.5?


Hopefully that answers your questions. I welcome your analysis
of those numbers and will accept any corrections you make as long as
I can double check your corrections.

I hope that you agree with me that using a V-band decay time derived
from a lot of SNs makes more sense than using only one SN, even if
that changes your numbers substantially.


Bye,
Bjoern

A small question about the restframe wavelengths. You probably
got them by dividing the middle wavelength of the I filter
by 1+z, right? If yes, then your value fo 1998eq seems to be
wrong. Should be around 528 nm.

Yes. youre right

I looked up what templates Knop et al. used for the fitting. These
templates were determined by using the data from a large number of
low-z supernovae, instead of only the single one SN1995D. They provide
the data in their table 2. And it turns out that these templates
have only 16.8 days for peak+1 to peak+2 in the V band (I did
not check the R and I band so far)! That obviously makes the
numbers you calculated above much too low - the actual numbers are
about 30% greater!

In fact it was Steve who insisted I had to use 1995D as a reference.
I had averaged out a larger sample of SN from Reiss samples but
he said it was prefereable to use just one well sampled SN.

I am not sure if this linear interpolation is even necessary. From
what I have read so far about the K-correction, it is precisely for
taking this into account. So, if you use K-corrected data, you
don't have to do such an interpolation. See e.g. what Steve Willner
wrote already on Nov. 23rd last year:

"For example, for I-band observations of 1997ek at z=0.86, one would use
a template at 438 nm rest. Since such a template won't in general
exist, one applies a "k-correction" to the observations to account for
the difference between the available template (B, approx 440 nm) and
the rest frame."

I have to admit you have confused me here. What do you mean by
linear interpolation? As 1997ek is 438 nm and restframe is 440nm
(actually Reiss B looks more like 430nm to me from his fig 1
transmission response graph) are you saying k correct 438 to
440nm? What was done in the my calculations was assume 438 to be
so close to 440 that they could be compared like for like.
And I`m sure that this is what Steve suggested doing.
Could you elaborate a bit more. What is the `linear
interpolation` I`ve done which isnt neccesary and how
does that effect the overall results of my table of ratios?

Do you mean 27/22 or 22/20.5?

Yes

I hope that you agree with me that using a V-band decay time derived
from a lot of SNs makes more sense than using only one SN, even if
that changes your numbers substantially.

Yes it does make a difference I have to admit.
Although looking at Reiss where the data has both peak and
peak+2 I get these results. Which is about 18.5 average
I suppose Knop used other SN?

1995E V= 19
1995al V= 20
1995ac V= 21
1995ak V= 15
1995bd V= 21
1996X V= 16
1996bl V= 19
1996ba V= 15
1996ai V= 20 ~

Anyways here`s the table using your times from Knop
(R from Knop is 19.5 and V from Knop 16)
1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86*
1998eq (restframe528nm) z=.54 15/116 =1 " 1.54
1997ez (restframe457nm) z=.78 16/12 =1.33 " 1.78
1998as (restframe602nm) z=.35 23/19.5 =1.17 " 1.35
1998aw (restframe565nm) z=.44 27/18 =1.5 " 1.44*
1998ax (restframe542nm) z=.50 30/16 =1.87 " 1.50*
1998ay (restframe496nm) z=.64 20/13 =1.53 " 1.64*
1998ba (restframe569nm) z=.43 24.5/18 =1.36 " 1.43*
1998be (restframe496nm) z=.64 18/14 =1.28 " 1.64
1998bi (restframe467nm) z=.74 15.5/12 =1.29 " 1.74
2000fr (restframe528nm) z=.54 24.5/16 =1.1 " 1.54

Not as good results for me as I would have liked but if you look
at the * where the result favors dilation you still get
a 1/2 1/2 split between dilation and no dilation.
What conclusions can be drawn from that I dont know as it seems
the results arent clear enough either way.
I would still like to resolve the debate around the non dilated
fit debate . Here is a follow up from another of your posts
on this subject.

In other
words you cant find out s until you have done the chi
squared fit.


Or unless I have information from another chi squared fit
which was already done to a similar formula. That's the part
of the argument which you apparently keep missing.


Here is a point I cant agree with you on. When it
is neccesary to do a fit to find s, *that does not* include
using results from a different fit. It cant be done as
the variables I, I max are different for each example.

Why? Because if you think about it,in the fit to find s that
Knop has done ,which you claim to use, it does not use the
data in the same way as the fit to find s where no dilation
is present. And why do I say it uses different data?
Because although the SN data is the same for both
the template data that the function uses to `fit` is
different.
Dont forget because the template has not been dilated, all
the day values of the templates are different between the
two versions. For instance in Knops fit day 5 of the template
is multiplied by s AND z so it becomes something like day 9
lets say. Whereas in the non dilated fit day 5 only gets
multiplied by the (unknown) variable s and not by z
(remember z is always 0) so day 5 becomes maybe only day 6.
And if the data from the SN is being fitted to the dilated
template the dilated version uses day 9 from the SN data whereas
in the undilated fit it compares against day 6 from the SN data.
And those are two different data points from the SN being
compared to the same template day(which is the original day 5)
I admit that I havent had the experience of performing a chi
squared fit in maple and you have I assume , yet it seems
undeniable that by dilating the template one ends up with
a different set of values for all the variables in the
formula compared with a undilated fit. Therefore its not
possible for you to use the same information
from another fit as you say you can.
Unless of course you substitute s with z in all cases
in an undilated fit as you seem to be arguing.
I could accept that but only if you actually have done
the fit to see if in fact the fitting procedure does
give s as z in an undilated fit. And as you admit
you have only *assumed* this to be the outcome. One reason
I doubt this outcome is I have done a few single `sanity check`
fits using the s value Knop got in an undilated fit and
the fit is as good as knops dilated fit. In fact if I
substituted 0 instead I would have got a better fit.
So I believe that only performing a chi squared fit of an
undilated template would resolve this as I can see that clearly
an undilated template can be made to fit the data by simply
moving the peak day back to 50822.
In fact in most of the SN I have tested seperately in single
fits I can see that they can all fit the data with s=0
as well as Knops dilated fits. Unfortunately I dont
have maple to confirm this.
But why dont you accept single fits as tests?
They are not incorrect as long as the data is followed
and as far as I can see, the only flaw in them is that
they arent as good as a `best fit`. And if you can only
criticise them for not being as good as a best fit then
the only conclusion is that best fits would produce results
better than Knops dilated fits (seeing as my single fits
are as good as Knops)
Why dont you double check my single fit? I`ll supply the
neccesary numbers again. At least it will show you that
a single fit can be as good as Knops.
Sean

sean wrote:
A small question about the restframe wavelengths. You probably
got them by dividing the middle wavelength of the I filter
by 1+z, right? If yes, then your value fo 1998eq seems to be
wrong. Should be around 528 nm.


Yes. youre right


I looked up what templates Knop et al. used for the fitting. These
templates were determined by using the data from a large number of
low-z supernovae, instead of only the single one SN1995D. They provide
the data in their table 2. And it turns out that these templates
have only 16.8 days for peak+1 to peak+2 in the V band (I did
not check the R and I band so far)! That obviously makes the
numbers you calculated above much too low - the actual numbers are
about 30% greater!


In fact it was Steve who insisted I had to use 1995D as a reference.

According to his recent post to this thread, he anything but
"insisted" on that. In contrast, he himself said that using templates
obtained from several SNs would be better, and using a single SN
could serve only as a quick consistency check.


I had averaged out a larger sample of SN from Reiss samples but
he said it was prefereable to use just one well sampled SN.

Well, he himself says differently. I suspect that you scrambled
up the averaging, so he suggested a simpler method.



I am not sure if this linear interpolation is even necessary. From
what I have read so far about the K-correction, it is precisely for
taking this into account. So, if you use K-corrected data, you
don't have to do such an interpolation. See e.g. what Steve Willner
wrote already on Nov. 23rd last year:


"For example, for I-band observations of 1997ek at z=0.86, one would use
a template at 438 nm rest. Since such a template won't in general
exist, one applies a "k-correction" to the observations to account for
the difference between the available template (B, approx 440 nm) and
the rest frame."


I have to admit you have confused me here. What do you mean by
linear interpolation?

I explained that in my mail to you.

"linear interpolation" means that if I have two y values (y1,y2) for
different x values (x1,x2) and want to have the y value (y12) at an x
value in between (x12), I simply use the approximation
y12 = y1 + (y2-y1)/(x2-x1) * (x12 - x1)

That's what you have done for the decay times at wavelengths which
do not correspond exactly to a given band (430 nm, 530 nm etc.)



As 1997ek is 438 nm and restframe is 440nm
(actually Reiss B looks more like 430nm to me from his fig 1
transmission response graph)

I already asked which filter you used for getting this 430 nm.
After all, there are three different curves, for three different
filters, in that figure. Care to answer that?


are you saying k correct 438 to 440nm?

I am saying that the K-correction ensures that the light
curve looks *as if* it would have been observed at rest frame
wavelength 440 nm (or 430 nm, if you prefer).



What was done in the my calculations was assume 438 to be
so close to 440 that they could be compared like for like.

O.k., that gives the same result.

I meant mainly the other examples, where the wavelength was
*not* close to 440 nm (or 430 nm, if you prefer).


And I`m sure that this is what Steve suggested doing.

Indeed.


[snip]


Do you mean 27/22 or 22/20.5?


Yes

Do you have *really* severe reading comprehension problems? Or
is this a bad joke? Since when does one answer a question for
"either-or" with a "yes"?

I repeat my question: Do you mean 27/22 or 22/20.5?


I hope that you agree with me that using a V-band decay time derived
from a lot of SNs makes more sense than using only one SN, even if
that changes your numbers substantially.

Yes it does make a difference I have to admit.
Although looking at Reiss where the data has both peak and
peak+2 I get these results. Which is about 18.5 average
I suppose Knop used other SN?

1995E V= 19
1995al V= 20
1995ac V= 21
1995ak V= 15
1995bd V= 21
1996X V= 16
1996bl V= 19
1996ba V= 15
1996ai V= 20

Both Reiss and Knop tell in detail which SNs they used, and how.
Look it up yourself.


Anyways here`s the table using your times from Knop
(R from Knop is 19.5 and V from Knop 16)
1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86*
1998eq (restframe528nm) z=.54 15/116 =1 " 1.54
1997ez (restframe457nm) z=.78 16/12 =1.33 " 1.78
1998as (restframe602nm) z=.35 23/19.5 =1.17 " 1.35
1998aw (restframe565nm) z=.44 27/18 =1.5 " 1.44*
1998ax (restframe542nm) z=.50 30/16 =1.87 " 1.50*
1998ay (restframe496nm) z=.64 20/13 =1.53 " 1.64*
1998ba (restframe569nm) z=.43 24.5/18 =1.36 " 1.43*
1998be (restframe496nm) z=.64 18/14 =1.28 " 1.64
1998bi (restframe467nm) z=.74 15.5/12 =1.29 " 1.74
2000fr (restframe528nm) z=.54 24.5/16 =1.1 " 1.54

Not as good results for me as I would have liked but if you look
at the * where the result favors dilation you still get
a 1/2 1/2 split between dilation and no dilation.

No. You get dilation in all cases except 1998eq. Any factor greater
than 1 *is* time dilation. The time dilation is not as great as
predicted, granted - but there *is* indeed time dilation.

On the other hand, if no time dilation at all happened, the
differences in the decay times could only be due to internal
differences in the SNs. But then there would be no reason at all
why the factor should always be greater than 1! What we would expect
then is that the factor comes out often also lower than 1, and that
on the average, it is exactly one.

So, the results you show above show clearly that there *is* time
dilation (and hence your claims are refuted) - just not so much
as predicted by the theory.



[snip]


In other
words you cant find out s until you have done the chi
squared fit.

Or unless I have information from another chi squared fit
which was already done to a similar formula. That's the part
of the argument which you apparently keep missing.



Here is a point I cant agree with you on. When it
is neccesary to do a fit to find s, *that does not* include
using results from a different fit.

I only point out that when I already *know* the result for a
*different* type of fit to the *same* data, I can predict with
good confidence what the result of the new, not yet performed
fit will be.

What on earth is your problem with that? To me, it looks totally
logical and perfectly sensible. I showed the argument also to my
brother, who is also a scientist, without telling him that it has
anything to do with cosmology, SNs, lightcurves etc. And he agreed
that my argument is 100% right, and did not understand that you have a
problem with it.


It cant be done as
the variables I, I max are different for each example.

Err, we are talking about *one and the same* data set in both
cases, so I and Imax *are* the same in both cases.

You yourself kept talking about only stretching the time scale with
s in your plots. You never mentioned changing Imax!


Why? Because if you think about it, in the fit to find s that
Knop has done, which you claim to use, it does not use the
data in the same way as the fit to find s where no dilation
is present.

I don't understand what exactly you mean with "use the data not
in the same way".

In both cases, the *same* data is fitted. To *different* formulas.


And why do I say it uses different data?
Because although the SN data is the same for both
the template data that the function uses to `fit` is
different.

Err, indeed, that's the entire point. That's why I get s=1.1
in the first, but s=1.21 in the second example: because the function
to which is fit is different.


Dont forget because the template has not been dilated,

I am not sure what you even mean exactly by this constant
mentioning of "dilated template". Do you think that because
a factor 1+z appears in the formula to which is fitted, this
automatically means that the template is dilated, or what?


all the day values of the templates are different between the
two versions. For instance in Knops fit day 5 of the template
is multiplied by s AND z

By an *unknown* s, and a *known* factor 1+z.


so it becomes something like day 9 lets say.

No. It simply becomes 5 * s * (1+z). The s which appears here
is not known until *after* the fit, so you can't simply proclaim that
the result is 9.

What you *could* say is something like: "Let's assume that the
fit tells us that the 5 days have to be stretched to 9 days in
order for the template to match the data".

I.e. here we have:
5 * s * (1+z) = 9,
and hence
s*(1+z) = 1.8


Whereas in the non dilated fit day 5 only gets
multiplied by the (unknown) variable s and not by z
(remember z is always 0) so day 5 becomes maybe only day 6.

What you miss here is that the s which appears now is *not* the
same as above. We *still* have to stretch the time axis so
that day 5 becomes day 9 (since the data is the same!. But now we
only have the factor s available for that stretch, since we left
out the factor 1+z (i.e. we set z=0). Hence we now have
5 * s = 9,
and hence
s = 1.8



And if the data from the SN is being fitted to the dilated
template the dilated version uses day 9 from the SN data whereas
in the undilated fit it compares against day 6 from the SN data.

Wrong. You make no sense at all here. This is *not* how fitting
is done!


And those are two different data points from the SN being
compared to the same template day (which is the original day 5)

No. No. No. What you describe here has nothing at all to do with
fitting.


I admit that I havent had the experience of performing a chi
squared fit in maple and you have I assume , yet it seems
undeniable that by dilating the template one ends up with
a different set of values for all the variables in the
formula compared with a undilated fit.

One ends up with a different value for s. That's what I keep telling
you. One *cannot* end up with different values for tmax and Imax,
since these are simply the time and the magnitude of the peak,
and these things don't change merely because I use a different formula
in the fit.


Therefore its not
possible for you to use the same information
from another fit as you say you can.

Wrong, it's entirely possible.


Unless of course you substitute s with z in all cases
in an undilated fit as you seem to be arguing.

No. that's *not* what I argue. Where did you get that strange
idea from?


[snip argument based on false premise]


One reason
I doubt this outcome is I have done a few single `sanity check`
fits using the s value Knop got in an undilated fit and
the fit is as good as knops dilated fit. In fact if I
substituted 0 instead I would have got a better fit.

You can only check if the fit is "as good as" or "better" by comparing
the chi squared values (hint: these could be calculated by hand, no
Maple etc. necessary; see below). You haven't done that, right?


So I believe that only performing a chi squared fit of an
undilated template would resolve this

Please be more clear what you mean. Do you mean doing a fit to
the formula
I = Imax (fR((t-tmax)/s) + b),
or what?


as I can see that clearly
an undilated template can be made to fit the data by simply
moving the peak day back to 50822.

The question if it can be "made to fit" is not relevant. The
question is "Does it have a lower chi squared value, i.e.
is the fit better?"



In fact in most of the SN I have tested seperately in single
fits I can see that they can all fit the data with s=0
as well as Knops dilated fits. Unfortunately I dont
have maple to confirm this.

Not necessary.

You can can calculate chi squared in this way (AFAIK):

chi^2 = sum over all squared deviations between the curve
and the data, divided by the error margins

E.g.. if I have only two data points (t1,I1) and (t2,I2) with
error margins i1 and i2 (i.e. the data is I1 +- i1 and
I2 +- i2), and my curve has the values f1 and f2 for the times
t1 and t2, respectively, I have
chi^2 = ((f1-I1)/i1)^2 + ((f2-I2)/i2)^2


But why dont you accept single fits as tests?
They are not incorrect as long as the data is followed
and as far as I can see, the only flaw in them is that
they arent as good as a `best fit`.

Indeed. That's the whole point!


And if you can only
criticise them for not being as good as a best fit then
the only conclusion is that best fits would produce results
better than Knops dilated fits (seeing as my single fits
are as good as Knops)
Why dont you double check my single fit? I`ll supply the
neccesary numbers again. At least it will show you that
a single fit can be as good as Knops.

Stop claiming that your fits are "as good as" Knops without
actually backing this up (see above).


Bye,
Bjoern

sean, what has happened to our discussion about chi-squared fitting?
You didn't answer in sci.math for quite some time now, and you have
also snipped that part of my post here in sci.astro.

I tried to put all the relevent points on my last response in
sci astro SN1e data thread.

You can can calculate chi squared in this way (AFAIK):

chi^2 = sum over all squared deviations between the curve
and the data, divided by the error margins

E.g.. if I have only two data points (t1,I1) and (t2,I2) with
error margins i1 and i2 (i.e. the data is I1 +- i1 and
I2 +- i2), and my curve* has the values f1 and f2 for the times
t1 and t2, respectively, I have
chi^2 = ((f1-I1)/i1)^2 + ((f2-I2)/i2)^2

It seems like the only way for me to prove that there is no dilation
of SN will be to attempt a manual chi squared fit which I have
not done before, but I will try. You will have to clarify a few points
first.
When you say "my curve*" above are you refering to the day values
from the restframe template? So for instance is f1 also day one
on the template? (so day -1 from the V band template will be f-1
and day 5 will be f5 etc)
And also, how does one decide what day from the table data is t1?
For instance in any SN from Knop all the data is given in Julien days
so how would one assign a JD to t. For instance if I wanted
to do a fit of SN1997ek I-band to the V template from
Knop how do I assign t1 to the data ? Which julien day
from table 11 will become t1 ?


Sean

sean wrote:
sean, what has happened to our discussion about chi-squared fitting?
You didn't answer in sci.math for quite some time now, and you have
also snipped that part of my post here in sci.astro.


I tried to put all the relevent points on my last response in
sci astro SN1e data thread.


You can can calculate chi squared in this way (AFAIK):


chi^2 = sum over all squared deviations between the curve
and the data, divided by the error margins


E.g.. if I have only two data points (t1,I1) and (t2,I2) with
error margins i1 and i2 (i.e. the data is I1 +- i1 and
I2 +- i2), and my curve* has the values f1 and f2 for the times
t1 and t2, respectively, I have
chi^2 = ((f1-I1)/i1)^2 + ((f2-I2)/i2)^2


It seems like the only way for me to prove that there is no dilation
of SN will be to attempt a manual chi squared fit which I have
not done before, but I will try.

If you really want to do all the work...

On the other hand, you could just accept all the arguments presented
to the contrary:
1) Using an "undilated template", i.e. using only s instead of
s*(1+z) in the formula would give the same result of time dilation.
If you are interested, I could try to demonstrate that with a (very
simplified) example of an actual fit: not to SN data, but to data
made up for illustration, and not to the actual formula, but to
a simpler formula.
2) Even the table you keep bringing up supports "time dilation"
(although not as much as required by theory) more than "no time
dilation at all".
3) Steve's detailed analysis which showed time dilation clearly for
one example.



You will have to clarify a few points first.
When you say "my curve*" above are you refering to the day values
from the restframe template?

I am referring to the curve you draw through the data points. I.e.
a curve which has been in general already been stretched, translated
etc., not the original template.

You said that you tried to draw a curve through the data points
without using the factor 1+z, and the resulting curve matched the
data points nicely. I mean *that* sort of curve.


So for instance is f1 also day one on the template?

f1 in my formula above is the value of the curve you draw through
the points for time t1, where you can choose t1 to be any time
for which you have data (I1).


(so day -1 from the V band template will be f-1
and day 5 will be f5 etc)

In general, no, since you will need a stretch of the time axis,
even without an assumed time dilation.


And also, how does one decide what day from the table data is t1?
For instance in any SN from Knop all the data is given in Julien
days so how would one assign a JD to t. For instance if I wanted
to do a fit of SN1997ek I-band to the V template from
Knop how do I assign t1 to the data ? Which julien day
from table 11 will become t1 ?

No tables needed here. Simply use the curves you already have
drawn.



Bye,
Bjoern

In article ,
(sean) writes:
1997ek (restframe438nm) z=.86 14.5/10 =1.45should be1.86
....
As it is I will have to tell you just from memory but the process was
this: I used several SN from Reiss to average numbers but I think Steve
insisted I only used 1995D or at least he insisted on only reffering
to it as it was better sampled than most. I did find however that
within the range of low redshift lightcurves there was a certain amount
of variation that 1995D didnt neccesarily fit into well.
Anyways I measured off 1995D delay times for I think it was
peak+1 to peak +2 off of the 4 BVRI *tables* (not template)

My analysis of 1997ek is in sci.astro.research message-ID
and some followups to it. I
don't "insist" on using a single SN rather than templates (which as
Bjoern says are better), but using a single low-redshift SN as a
template suffices to show obvious time dilation.

1997ek is convenient because it has a high redshift, and there will
be little K-correction between its I-band magnitudes and B-band
magnitudes at low redshift. The paper by Knop et al. does a full
analysis for several SNe and shows time dilation for all SNe in their
sample. (They would have seen different results for 's' in Table 3
if there were no time dilation.)

If someone claims to obtain a different answer, the burden is on that
person to show his work with full details.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
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