Re: mu+ results

[Frederick Gray <Frederick.Gray@pomona.edu>]

Hi, Peter,

In the heat of the moment yesterday, it seems that I forgot to remove  
my own secret offset!  Our agreement with the PDG turns out to be  
perfect.

I begin with the result from p. 28 of my analysis report:

   lambda = 454.755 ± 0.0249 (stat.) ± 0.0131(syst.) kHz

My secret "offset" was multiplication by (1 + delta).  In my report,  
I identified delta as ~O(10^-4).  In fact, delta = 10^-4 precisely:  
there were a lot of not-very-creative offsets used in this  
experiment.    Therefore, when I remove my offset by dividing by  
1.0001, leaving only the blinded clock frequency, I obtain

   lambda = 454.709 ± 0.0249 (stat.) ± 0.0131(syst.) kHz

When I also remove the clock blinding by multiplying by (100.1/100),  
then I obtain

   lambda = 455.164 ± 0.0249 (stat.) ± 0.0131(syst.) kHz

The PDG value for the muon lifetime is  2.19703 ± 0.00004  
microseconds, corresponding to 455.1599 ± 0.0083 kHz decay rate.   
Therefore, the difference between our result and the PDG is 0.004 ±  
0.029 kHz, or 0.14 sigma.

-- Fred

----------------------------------------------------------------------
Frederick Gray, Visiting Assistant Professor
Department of Physics and Astronomy, Pomona College, Claremont, CA, USA
Frederick.Gray@pomona.edu / phone: 909-607-9795 / fax: 909-621-8463
----------------------------------------------------------------------



On Oct 9, 2006, at 3:31 PM, Peter Kammel wrote:

> Dear Fred,
>
> Could you send me a few lines about the mu+ lifetime
> unblind (your secret offset) and I will include
> this in the results spreadsheet, which I will
> send to the collaboration tomorrow.
>
> Thanks
>
> Peter
>

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