Energy loss in GEANT

The two problems mentioned above have been solved in GEANT 3.15. As far as the precision is concerned, the solution was to revert to the algorithm of GEANT 3.13 every time the relative energy loss in the step ( DESTEP/GEKIN) is smaller than five times the machine precision. This has given good results without loosing the substantial improvements obtained in GEANT 3.14 with the introduction of the stopping range algorithm. As a matter of fact, the above condition happens only the in the case of very small steps or when the $dE/dx$ curve is very flat, and in both cases the algorithm based on the $dE/dx$ tables is a good approximation.

The second problem has been solved by changing the interpolation algorithm, going from a linear to a quadratic interpolation. Instead of assuming a linear relation between energy and stopping range in every energy bin, we assume a quadratic relation of the sort:

The only problem now is the determination of the coefficients a, b and c. To do this we recall that the general formula of the parabola which passes through the points $(x$₁,y₁), (x₂,y₂), (x₃,y₃) is the following:

This allows us to calculate and tabulate the coefficients a, b and c just by replacing $y$_i by ELOW(I) and $x$_i by the corresponding stopping range ELOW(i ). In the routine GRANGI a table of length $3\; x(NEKBIN-1)$

is created:

Q(JINTRP+3*(I-1)	+1)	=	$a(R$I,RI+1,RI+2,EI,EI+1,EI+2)	=	$A$I .3cm
	+2)	=	$0.5\; \hspace{0.17em}\{b(R$I,RI+1,RI+2,EI,EI+1,EI+2)a(RI,RI+1,RI+2,EI,EI+1,EI+2)}	=	$B$I .3cm
	+3)	=	$\{c(R$I,RI+1,RI+2,EI,EI+1,EI+2)a(RI,RI+1,RI+2,EI,EI+1,EI+2)}	=	$C$I .3cm

where $E$_I= ELOW(I) and $R$_I= R(ELOW(I)) . The calculation of the energy loss is now done in the following steps:

1. Evaluate the stopping range for the threshold energy ( STOPC), only once at the beginning of each new particle tracking.
2. From the energy of the particle derive the stopping range by a quadratic interpolation of the range table:

   $R$₀= -B_I+ {A_I| A_I|}&sp;B_I²-( C_I-{E₀A_I})

   The value of I is chosen according to the following. Let i be such that ₀≤ELOW(i+1) :

   $max(i-1,1)$	$if$	$\{E$0-ELOW(i)ELOW(i+1)-ELOW(i)}< 0.7 .3cm
   $min(i,NEKBIN-1)$	$if$	$\{E$0-ELOW(i)ELOW(i+1)-ELOW(i)}≥0.7 .3cm

   $.$

   The value of $0.7$ is an empirical number which minimises the discontinuities of the reconstructed $dE/dx$ curve.

3. Evaluate the stopping range for the particle after the step: $R\text{'}$₀= R₀- step . If this is less than the stopping range of a particle of threshold energy, the particle is terminated as a stopping particle below the energy cut. Otherwise, the final energy is computed as:

   $E\text{'}$₀= A_I(C_I+R'₀(2B_I+R'₀))

4. the energy loss is computed as:

   $\Delta E\; =\; E$₀-E'₀

   This value is then corrected to take into account the energy loss fluctuations (see [PHYS332]).

PHYS100