Re: calorimeter question


"Jeff K deJong" <jdejong@xxxxxxxxxxxxxxxx> wrote in message
news:Pine.LNX.4.58.0601122206270.25930@xxxxxxxxxxxxxxxxxxxxxxxx
Hi all,

I'm really hoping someone out there can help me, I've checked the web my
textbooks and I can't seem to figure this out. When it comes to a sampling
calorimeter we get resolutions of the form

Delta E 20%
------- = -----
E sqrt(E) <--- this E is the absolute energy of the
particle

I can understand that you increase the number of measurements you will
increase the precision(ie 1/SQRT(N)), now the number of measurements you
do will be dependant on the Energy.

Now what I'm trying to figure out is what if your calibration is off, say
you miscalibrate these individual measurements by 10% what would that do
to the above formula will you get simply SQRT(20^2+10^2)=22.4% ?

I believe it is worse than this. The reason is that the calibration error is
a systematic error that shifts the set of measurements in one direction,
rather than just widening the distribution. Perfectly symmetric error
distributions can, in principle, be reduced greatly by taking large numbers
of measurements, but that is not true of a directional error, where you are
not aware whether it is up or down.

I think that, roughly speaking, the calibration error has to be added to the
(presumed) symmetric error.

Any help would be greatly appreciated.

Jeff

--
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Jeffrey K. de Jong,PhD Candidate | " If my human mind cannot
U. of Alberta | acknowledge that all that is, is
Center for Subatomic research | right; yet since what is must be,
Edmonton,AB Canada T6G 2N5 | I will sit amidst the ruins
(F) (780) 492-9658 | and smile"
(W) (780) 492-5017 | 'Last Man'- Mary Shelley
http://csr.phys.ualberta.ca/~jdejong |


.

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