Impact on Physics Asymmetries

The ``true'' asymmetry can be written as

$$A=\frac{n^{\downarrow }-n^{\uparrow }}{n^{\downarrow }+n^{\uparrow }}$$

denotes normalized count rates for the two helicity states (indicated by the arrows). The measured asymmetry, affected by electronics dead-time only, can be written as

$$A_{m}=\frac{(1-DT_{el}^{\downarrow })\cdot n^{\downarrow }-(... ...ot n^{\downarrow }+(1-DT_{el}^{\uparrow })\cdot n^{\uparrow }}$$ (1)

with denoting the electronics dead-time.

Alternatively we can express with A:

$$n^{\downarrow }=\frac{n^{\downarrow }+n^{\uparrow }}{2}(1+A)$$

$$n^{\uparrow }=\frac{n^{\downarrow }+n^{\uparrow }}{2}(1-A)$$

The electronics dead-time is small in our case and can therefore be expressed as proportional to the raw Bar rate:

$$DT_{el}^{\uparrow \downarrow }=k_{el}\cdot b^{\uparrow \downarrow }$$ (2)

The bar rates and can be expressed with the unpolarized rate and the bar rate asymmetry A_b:

$$b^{\uparrow }=b(1+A_{b})$$

$$b^{\downarrow }=b(1-A_{b})$$

Making all the substitutions in eq. leads to

$$A_{m}=\frac{A-k_{el}b(A+A_{b})}{1-k_{el}b(1+A_{b}A)}$$

Since is small we do a Taylor expansion of A_mabout (k_elb=0):

A_m=A-A_b(A²-1)(k_elb)+O[(k_elb)²]

A² is in our case very small and we can further approximate and obtain

$$A_{m}\cong A-A_{b}k_{el}b$$

For the deviation from the true asymmetry we get

$$\Delta A=A_{m}-A\cong -A_{b}k_{el}b$$

For k_el we use the number determined in the above mentioned test measurement: . Using scaler information the average bar rate can be determined as (which is only half of what was always quoted). The size of A_b depends on the sign of the target polarization as mentioned above. Therefore we obtain a deviation from the ``true'' asymmetry which is dependent on the sign of the target polarization:

$$Tpol-:\: A_{b}=-1.134\cdot 10^{-3}\: \Rightarrow \: \Delta A=1.02\cdot 10^{-5}$$

$$Tpol+:\: A_{b}=-7.6\cdot 10^{-5}\: \Rightarrow \: \Delta A=6.84\cdot 10^{-7}$$