Using REDUCE in High Energy Physics by A. G. Grozin

By A. G. Grozin

This booklet describes using the symbolic manipulation language decrease in particle physics. There are numerous basic goal arithmetic applications to be had to physicists, together with Mathematica, Maple, and decrease. each one has merits and drawbacks, yet lessen has been discovered to be either strong and handy in fixing a variety of difficulties. This booklet introduces the reader to minimize and demonstrates its application as a mathematical device in physics. the 1st bankruptcy of the booklet describes the decrease approach, together with a few library applications. the next chapters exhibit using lessen in examples from classical mechanics, hydrodynamics, basic relativity and quantum mechanics. the remainder of the publication systematically offers the traditional version of particle physics (QED, susceptible interactions, QCD). a good number of scattering and rot tactics are calculated with decrease. All instance courses from the ebook may be downloaded through web, from http://www.inp.nsk.su/~grozin/book/. The emphasis all through is on studying via labored examples.

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The function solve returns a list of solutions. Its first argument is an equation (or a list of equations), and the second one is an unknown variable (or a list of unknown variables). If the right-hand side of an equation is zero, then it may be omitted together with =. The second argument may be omitted, if the unknown variables are the only variables contained in the equations. Every solution in the list returned by solve is the equality with the unknown variable in the left-hand side, or the list of such equalities in the case of an equation system.

Operator f; x:=xx$ y:=yy$ let f(x)=y; clear x; f(x); f(xx); clear y; f(xx); clear f(xx); yy y x:=xx$ y:=yy$ f(x):=y; f(xx) := yy clear x; f(x); f(xx); clear y; f(xx); clear f(xx); yy yy The statements l e t x= ... and x: = ... both change the value of x (the first one, to the unevaluated right-hand side, while the second one, to its value). Therefore, naturally, any of these statements destroys the result of a previous one. At each moment, only one substitution for a variable x raised to an integer power can exist.

By the way, you can see that an assignment has a value, and may be used in an expression. % Taylor series for exp procedure expl(x,n); begin scalar s,u; s:=l; u:=l; for i:=l:n do « u:=u*x/i; s:=s+u » ; return s end$ on div,revpri; expl(x,10); clear expl; ! ^ 1 l 2 l 3 1 4 2 6 x9 + + x + 24 120 1 fi 720 1 « 7 5040 40320 x10 3628800 362880 3628800 °/0 method using small variables procedure exp2(x); begin scalar s,u,i; s:=l; u:=x; i:=l; repeat « s:=s+u; i:=i+l; u:=u*x/i » until u=0; return s end$ weight x=l$ wtlevel 10$ exp2(x); exp2(x+x~2); clear exp2; .

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