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- %*********************************************************************
- % 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;
- % 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;
- % Let us input very simple example:
- a2(k,k);
- % By the way the next two expressions looks like different ones:
- a2(i,j);
- a2(j,i);
- % But the difference of them has a correct value:
- a2(j,i)-a2(i,j);
- % 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);
- % 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);
- % 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;
- % Of course the contraction symmetric and antisymmetric tensors
- % equal zero:
- a2(i,j)*s2(i,j);
- % By the way, the next example not so trivial for computer...
- a2(i,j)*a2(j,k)*a2(k,i);
- % 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;
- % In the beginning some very simple examples:
- a3(i,k,i);
- a3(i,j,k)*s2(i,j);
- % 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;
- % The next examples demonstrate some calculation with them:
- s3(i,j,k)-s3(i,k,j);
- s3(i,j,k)*a2(i,j);
- a3(i,j,k)*s2(i,j);
- s3(i,j,k)*a3(i,j,k);
- % 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;
- % 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);
- % 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);
- % Let us start with simple example:
- ri(m,n,m,n)-ri(m,n,n,m);
- % Much more complicated example is:
- a2(m,n)*ri(m,n,c,d) + a2(k,l)*ri(c,d,l,k);
- % 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);
- %***************** END OF TEST RUN ************************
- % The total execution time is:
- showtime;
- $END$
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