Asymmetries at SLC

  • #1
Hello everybody!
I have a question regarding the measure of the asymmetry left-right (##A_{LR}##) at the SLC. Since the beam was not fully polarized the measured asymmetry is different from the theoretical one. The relation which I have found on papers and books is:
$$A_{LR}^{measured} = \frac{|N_L - N_R|}{N_L + N_R} = A_{LR} <P>$$
with ##N_L## the number of events when the beam is left-polarized and mutatis mutandis for ##N_R##.
But what I call a left-polarized beam is: ##<P>e_L + (1-<P>) e_R##
And what I call a right-polarized beam is: ##<P>e_R + (1-<P>) e_L##
With ##e_L## the number of left-polarized electrons and mutatis mutandis for ##e_R##.
If I substitute:
$$A_{LR}^{measured} = \frac{|N_L - N_R|}{N_L + N_R} = \frac{<P>e_L + (1-<P>) e_R - <P>e_R - (1-<P>) e_L}{<P>e_L + (1-<P>) e_R + <P>e_R + (1-<P>) e_L}$$
Doing a bit of algebra I obtain:
$$A_{LR}^{measured} = \frac{e_L - e_R}{e_L + e_R} (2<P> + 1) = A_{LR} (2<P> +1)$$
which is not the relation mentioned above that I find on books...... In which point my reasoning fail?
Thanks in advance!
 

Answers and Replies

  • #2
34,667
10,807
It looks like you just use a different definition of P, plus a sign error somewhere. If both polarizations appear equally often the first equation has < P > = 0 while you call this < P > = 1/2. If the beam is purely polarized then < P > = +- 1 in the first equation while you would use 0 or 1. If you flip the sign of either 2 <P> or +1 then your brackets go from -1 to 1, just like the original equation.
 
  • Like
Likes Aleolomorfo

Related Threads on Asymmetries at SLC

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
21
Views
2K
Replies
5
Views
3K
Replies
9
Views
832
Replies
7
Views
867
Top