From: Paola Sala (paola.sala@cern.ch)
Date: Thu Sep 13 2007 - 14:51:54 CEST
Hi
I would just like to remind that also in the case of kinetic
energy the output sum will not be equal to the projectile kinetic
energy. This because of two reasons : kinetic energy can be converted to
mass and vice-versa, (example: pair production by photons, or pion
production in nucleon-nucleon interactions) and part of the kinetic
energy is dissipated in the target.
The only way to check energy conservation is to take into account ALL
processes, all partecipants, all secondaries, and work on total energy.
Again, this is done internally in FLUKA at each step.
For what concerns built-in scoring: double differential distributions
(energy/angle and much more) can be obtained with built in scoring
( USRBDX or USRYIELD). There exists also the possibility to set
additional cuts through the fluscw.f user routine ( that is activated by
the USERWEIG card.)
Going to the detail of your question: the instruction is correct except
for the case of heavy ions ( A>4 ): depending on the situation and on
the type of transport chosen, they can be flagged with
JTRACK=-2
or
JTRACK <-6
In this last case, EKIN = ETRACK - AMNHEA (-JTRACK)
in the case jtrack=2 , the mass is in AM(Jtrack) as for other particles
.. if you use USRBDX or USRYIELD it is enough to ask for kinetic energy,
the program performs all the tests...
Paola
On Wed, 2007-09-12 at 18:02 +0200, Konstantin Batkov wrote:
> Hi Paola,
>
> Thank you for this explanation. I would also like to make clear the
> case of the KINETIC energy.
> My question is: Is it correct to calculate the sum of the kinetic
> energy of secondaries using this code in mgdraw.f?
> ENTRY BXDRAW ( ICODE, MREG, NEWREG, XSCO, YSCO, ZSCO )
> if (MREG.eq.3 .and .NEWREG.eq.2) then
> sum = sum + ETRACK - AM(JTRACK)
> end if
> RETURN
>
> My problem is that I am studying different functions of energy/polar
> angles of different secondaries using many different cuts, so I just
> can't utilize the built-in facilities of FLUKA.
>
> Thank you,
>
> Konstantin
>
>
>
> 2007/9/12, Paola Sala <paola.sala@cern.ch>:
> Hi
>
> If you compare total energies, you have to take into account
> also
> the total energy of the target nucleus in the initial state,
> and the
> total energy of the nuclear fragments in the final state,
> including the
> residual nucleus that will NOT exit from the target. This is
> indeed
> what fluka checks internally at each interaction.
>
> for instance, if you take a reaction like
> n,A --> 2n,A'
> where A=(A,Z) is the target nucleus and A'==(A-1,Z) is the
> residual
> nucleus, if you count only the exiting particles the balance
> will be
> 1 GeV in input ( few MeV kinetic energy are enough)
> 2 GeV in output (twice the neutron mass)
> Of course this is balanced by the mass difference between the
> target
> and residual nuclei ( I leave out binding energies....)
>
> even more striking : in a neutrino quasielastic interaction
> a few MeV projectile will give you a 1 GeV nucleon in the
> final state.
>
>
> Ciao
> Paola
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