Sun Aug 18 16:14:28 2002 run on Windows
%*********************************************************************
% ATENSOR TEST RUN.
%
% V.A.Ilyin & A.P.Kryukov
% E-mail: ilyin@theory.npi.msu.su
% kryukov@theory.npi.msu.su
%
% Nucl. Phys. Inst., Moscow State Univ.
% 119899 Moscow, RUSSIA
%*********************************************************************
% First of all we have to load the ATENSOR program using the one of the
% following command:
% 1) in "atensor.red"$ % If we load source code
% 2) load atensor$ % If we load binary (compiled) code.
load atensor;
% To control of total execution time clear timer:
showtime;
Time: 0 ms
% Switch on the switch TIME to control of executing time
% for each statement.
%on time$
% Let us introduce the antisymmetric tensor of the second order.
tensor a2;
% The antisymmetric property can be expressed as:
tsym a2(i,j)+a2(j,i);
% The K-basis that span K subspace is:
kbasis a2;
a2(i,j) + a2(j,i)
1
% Let us input very simple example:
a2(k,k);
0
% By the way the next two expressions looks like different ones:
a2(i,j);
a2(i,j)
a2(j,i);
a2(j,i)
% But the difference of them has a correct value:
a2(j,i)-a2(i,j);
2*a2(j,i)
% Next examples. For this purpose we introduce 3 abstract
% vectors - v1,v2,v3:
tensor v1,v2,v3;
% The following expression equal zero:
a2(i,j)*v1(i)*v1(j);
0
% It is interest that the result is consequence of the equivalence
% of the name of tensors.
% While the next one - not:
a2(i,j)*v1(i)*v2(j);
a2(i,j)*v1(i)*v2(j)
% Well. Let us introduce the symmetric tensor of the second order.
tensor s2;
tsym s2(i,j)-s2(j,i);
% Their K-basis look like for a2 excepted sign:
kbasis s2;
s2(j,i) + (-1)*s2(i,j)
1
% Of course the contraction symmetric and antisymmetric tensors
% equal zero:
a2(i,j)*s2(i,j);
0
% By the way, the next example not so trivial for computer...
a2(i,j)*a2(j,k)*a2(k,i);
0
% Much more interesting examples we can demonstrate with the
% the tensor higher order. For example full antisymmetric tensor
% of the third order:
tensor a3;
% The antisymmetric property we can introduce through the
% permutation of the two first indices:
tsym a3(i,j,k)+a3(j,i,k);
% And the cyclic permutation all of them:
tsym a3(i,j,k)-a3(j,k,i);
% The K basis of a3 consist of 5 vectors:
kbasis a3;
a3(k,i,j) + a3(j,i,k)
a3(k,j,i) + (-1)*a3(j,i,k)
a3(i,k,j) + (-1)*a3(j,i,k)
a3(i,j,k) + a3(j,i,k)
a3(j,k,i) + a3(j,i,k)
5
% In the beginning some very simple examples:
a3(i,k,i);
0
a3(i,j,k)*s2(i,j);
0
% The full symmetric tensor of the third order may be introduce
% by the similar way:
tensor s3;
tsym s3(i,j,k)-s3(j,i,k);
tsym s3(i,j,k)-s3(j,k,i);
kbasis s3;
s3(k,j,i) + (-1)*s3(i,j,k)
s3(k,i,j) + (-1)*s3(i,j,k)
s3(j,k,i) + (-1)*s3(i,j,k)
s3(j,i,k) + (-1)*s3(i,j,k)
s3(i,k,j) + (-1)*s3(i,j,k)
5
% The next examples demonstrate some calculation with them:
s3(i,j,k)-s3(i,k,j);
0
s3(i,j,k)*a2(i,j);
0
a3(i,j,k)*s2(i,j);
0
s3(i,j,k)*a3(i,j,k);
0
% Now we consider very important physical case - Rieman tensor:
tensor ri;
% It has the antisymmetric property with respect to the permutation
% of the first two indices:
tsym ri(i,j,k,l) + ri(j,i,k,l);
% It has the antisymmetric property with respect to the permutation
% of the second two indices:
tsym ri(i,j,k,l) + ri(i,j,l,k);
% And the triple term identity with cyclic permutation the
% third of them:
tsym ri(i,j,k,l) + ri(i,k,l,j) + ri(i,l,j,k);
% The corresponding K basis consist of 22(!) vectors:
kbasis ri;
ri(l,k,i,j) + (-1)*ri(j,i,k,l)
ri(l,k,j,i) + ri(j,i,k,l)
ri(l,i,k,j) + (-1)*ri(j,k,i,l)
ri(l,i,j,k) + ri(j,k,i,l)
ri(l,j,k,i) + (-1)*ri(j,k,i,l) + ri(j,i,k,l)
ri(l,j,i,k) + ri(j,k,i,l) + (-1)*ri(j,i,k,l)
ri(k,l,i,j) + ri(j,i,k,l)
ri(k,l,j,i) + (-1)*ri(j,i,k,l)
ri(k,i,l,j) + (-1)*ri(j,k,i,l) + ri(j,i,k,l)
ri(k,i,j,l) + ri(j,k,i,l) + (-1)*ri(j,i,k,l)
ri(k,j,l,i) + (-1)*ri(j,k,i,l)
ri(k,j,i,l) + ri(j,k,i,l)
ri(i,l,k,j) + ri(j,k,i,l)
ri(i,l,j,k) + (-1)*ri(j,k,i,l)
ri(i,k,l,j) + ri(j,k,i,l) + (-1)*ri(j,i,k,l)
ri(i,k,j,l) + (-1)*ri(j,k,i,l) + ri(j,i,k,l)
ri(i,j,l,k) + (-1)*ri(j,i,k,l)
ri(i,j,k,l) + ri(j,i,k,l)
ri(j,l,k,i) + ri(j,k,i,l) + (-1)*ri(j,i,k,l)
ri(j,l,i,k) + (-1)*ri(j,k,i,l) + ri(j,i,k,l)
ri(j,k,l,i) + ri(j,k,i,l)
ri(j,i,l,k) + ri(j,i,k,l)
22
% So we get the answer for any expressions with 3 and more terms of
% Rieman tensors with not more then 2 terms. For example:
ri(i,j,k,l)+ri(j,k,l,i)+ri(k,l,i,j)+ri(l,i,j,k);
(-2)*ri(l,j,i,k) + 4*ri(l,i,j,k)
% This three identities leads us to very important symmetry property with
% respect to exchange of pairs indices:
ri(i,j,k,l)-ri(k,l,i,j);
0
% Let us start with simple example:
ri(m,n,m,n)-ri(m,n,n,m);
2*ri(m,n,m,n)
% Much more complicated example is:
a2(m,n)*ri(m,n,c,d) + a2(k,l)*ri(c,d,l,k);
0
% The answer is trivial but not so simple to obtain one.
% The dimension of the full space is 6! = 720.
% The K basis consists of 690 vectors (to reduce output we
% commented the last statement):
%kbasis ri(a2);
% One else nontrivial examples with Riemann tensors:
(ri(i,j,k,l)-ri(i,k,j,l))*a2(i,j);
a2(i,j)*ri(i,j,k,l)
---------------------
2
%***************** END OF TEST RUN ************************
% The total execution time is:
showtime;
Time: 354177 ms plus GC time: 4226 ms
$
END$
Time for test: 354247 ms, plus GC time: 4226 ms