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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