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Tensor Operations
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We now summarize a few basic algebraic operations that can be carried out with tensors to produce new tensors. These operations basically mirror the types of things you can carry out with vectors. For example, we can add two tensors of the same type to get a new tensor: R ab c = S ab c + T ab c
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It follows that we can subtract two tensors of the same type to get a new tensor of the same type: Qa b = Sa b Ta b a
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We can also multiply a tensor by a scalar a to get a new tensor Sab = aTab Note that in these examples, the placement of indices and number of indices are arbitrary. We are simply providing speci c examples. The only requirement is that all of the tensors in these types of operations have to be of the same type. We can use addition, subtraction, and scalar multiplication to derive the symmetric and antisymmetric parts of a tensor. A tensor is symmetric if Bab = Bba and antisymmetric if Tab = Tba . The symmetric part of a tensor is given by T(ab) = 1 (Tab + Tba ) 2 (3.4)
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and the antisymmetric part of a tensor is T[ab] = 1 (Tab Tba ) 2 (3.5)
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We can extend this to more indices, but we won t worry about that for the time being. Often, the notation is extended to include multiple tensors. For instance, V(a Wb) = 1 (Va Wb + Wb Va ) 2
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Tensors of different types can be multiplied together. If we multiply a tensor of type (m, n) by a tensor of type ( p, q), the result is a tensor of type (m + p, n + q). For example R ab S c de = T abc de Contraction can be used to turn an (m, n) tensor into an (m 1, n 1) tensor. This is done by setting a raised and lowered index equal: Rab = R c acb Remember, repeated indices indicate a sum. (3.6)
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The Kronecker delta can be used to manipulate tensor expressions. Use the following rule: When a raised index in a tensor matches the lowered index in the Kronecker delta, change it to the value of the raised index of the Kronecker delta. This sounds confusing, so we demonstrate it with an example
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a b T bc d = T ac d
Now consider the opposite. When a lowered index in a tensor matches a raised index in the Kronecker delta, set that index to the value of the lowered index of the Kronkecker delta:
c d T ab c = T ab d
EXAMPLE 3-5 Show that if a tensor is symmetric then it is independent of basis. SOLUTION 3-5 We can work this out easily using the tensor transformation properties. Considering Bab = Bba , we work out the left side: Bab =
c d b Bc d =
xc xd Bc d xa xb
For the other side, since we can move the derivatives around, we nd Bba =
d c a Bd c =
xd xc xc xd Bd c = Bd c xb xa xa xb
Equating both terms, it immediately follows that Bc d = Bd c . It is also true that if Bab = Bba , then B cd = B dc . Working this out, B cd = g ca B a d = g ca g db Bab = g ca g db Bba = g ca B d a = B dc
EXAMPLE 3-6 Let T ab be antisymmetric. Show that S[a Tbc] = 1 (Sa Tbc Sb Tac + Sc Tab ) 3
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SOLUTION 3-6 Since T ab is antisymmetric, we know that Tab = Tba . The expression A[abc] is given by A[abc] = Therefore, we nd S[a Tbc] = 1 (Sa Tbc + Sb Tca + Sc Tab Sb Tac Sa Tcb Sc Tba ) 6 1 (Aabc + Abca + Acab Abac Aacb Acba ) 6
Now we use the antisymmetry of T , Tab = Tba , to write S[a Tbc] = 1 (Sa Tbc Sb Tac + Sc Tab Sb Tac + Sa Tbc + Sc Tab ) 6 1 = (2Sa Tbc 2Sb Tac + 2Sc Tab ) 6 1 = (Sa Tbc Sb Tac + Sc Tab ) 3
EXAMPLE 3-7 Let Q ab = Q ba be a symmetric tensor and R ab = R ba be an antisymmetric tensor. Show that Q ab Rab = 0 SOLUTION 3-7 Since R ab = R ba , we can write Rab = Therefore, we have Q ab Rab = 1 1 ab Q (Rab Rba ) = Q ab Rab Q ab Rba 2 2 1 1 (Rab + Rab ) = (Rab Rba ) 2 2
Note that the indices a, b are repeated in both terms. This means they are dummy indices and we are free to change them. In the second term, we make the switch
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