Re: SNe1a data
- From: willner@xxxxxxxxxxxxxxx (Steve Willner)
- Date: 9 May 2005 18:21:58 -0500
In article <1115395666.410296.301680@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"sean" <jaymoseley@xxxxxxxxxxx> writes:
[referring to Knop et al. 2004 ApJ 598, 102]
> Maybe in the graph
> the data is shown with k correction whereas in the tables
> its without k correction?
That's my understanding. The K correction should be small for
1997eq.
> (Also I would have thought that when Knop writes ... "Magnitude in the
> observed filter at the peak of the rest-frame B-band lightcurve."
> ..that this would insinuate a K correction has been already applied
> if R max is already in "rest frame B band")
What I think they have done is determined the _observed time_ of the
peak in rest-frame B, then given the observed R or I at that time.
But I agree it isn't exactly clear what they have done about K
corrections, and I'm not about to dig into the paper hard enough to
figure it out. Let's stick to the SNe that have redshifts for which
we know the K correction will be small.
> I tried the same method for another SN,... 1998as and ...
> The calculated normalization value is 1.44 whereas its 1.38 if
> I work it out from the graph artwork.
This is only 4%, so I wouldn't worry. I'm still guessing K
correction, but it really is a guess.
> I can now work out how 0.31 as a weighted average is calculated using
> the formula ((M1 x W1) + (M2 x W2) +etc.... but I was wondering if you
> could describe the calculation of how you got + - 0.18 as the weighted
> error margin?
For a weighted average in general, the weight of the final average
will be the sum of the weights of the measurements being averaged.
In the simplest case, which is probably good enough here, the weight
of any measurement is the reciprocal of the square of the
uncertainty. So take each uncertainty (or "error bar"), square it,
take one over it, and add up the results. Then find one over that
sum and take the square root.
The calculation assumes that the measurements are independent and
that the uncertainties are Gaussian and correct. As I say, that's
probably OK here but might not be OK in other situations.
> Look at the last 3 readings of the observed (*1.06) HST
> readings compared to the expected dilated template (t-day) values..
> JD observed(*1.06) t-day t-flux
> 50846.82 .402 22.96 .41
> 50855.83 .286 29.0 .294
> 50863.83 .233 34.46 .215
> Notice how the 2nd and third last observed HST are decaying at the
> same rate but slightly fainter than the expected values(t-flux)
> predicted by the Knops dilated template but the last HST reading
> is brighter than expected from the template.
The question is the _amount_ of difference. The uncertainties of the
HST fluxes are 0.01 or 0.02. In each case, the measurement agrees
with the prediction within that margin or just a bit more. That
means the data are a good fit to the prediction. The chi-square
makes that quantitative. The result, as you have found yourself for
1997eq, is that the time-dilated template fits the observed data.
(Don't you think Knop et al. might have noticed if their observed
data disagreed with the templates?)
The real point is that observational data always have uncertainties,
and you have to take those into account. Exactly how one does that
depends on what one is trying to do with the data. Here the problem
is a form of hypothesis testing, and chi-square is an appropriate
method.
--
Steve Willner Phone 617-495-7123 swillner@xxxxxxxxxxxxxxx
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.)
.
- References:
- Re: SNe1a data
- From: sean
- Re: SNe1a data
- Prev by Date: Re: length of day
- Next by Date: Re: PLANETS ORBIT THE SUN TO CONSERVE TOTAL ENERGY
- Previous by thread: Re: SNe1a data
- Next by thread: History of Astronomy
- Index(es):
Relevant Pages
|
