1.2.2 Definition of a Cartesian Tensor an Entity T Which Has Components

1.2.2 Definition of a Cartesian tensor

An entity T which has components Tijk... (n indices) relative to a rectangular Cartesian basis {ei } and transforms like

TQQQTijk′ ...= ip jq kr... pqr ... (1.2.6) under a change of basis eeiiijj→ ′ = Q e where Q ≡ (Qij ) is a proper orthogonal matrix, is called a Cartesian tensor of order n and denoted CT(n).

Examples

(a) The tensor T in Section 1.2.1 has components that transform according to (1.2.5) and is a CT(2).

(b) The vector v in Section 1.1.2 has components that transform according to (1.1.35) and is a CT(1).

(d) The Krönecker delta:

δδij′′′==ee i•• j()()QQ ip e p jq e q = QQ ip jq pq

and therefore is a CT(2). Furthermore, δδij′ ===QQ ip jq pq QQ ip jp δ ij , i.e., the components of the identity tensor I are unaffected by a change of basis. In fact I is a member of a special class of Cartesian tensors, called isotropic tensors.

See Problems 1.2.1-1.2.5.

October 15, 2007 1.2.2-1 Problem 1.2.1 If S is a CT(3) and T is a CT(2) with components STijk, lm respectively with respect to the basis {ei }, slow that STijk lm are the components of a CT(5). Generalize this result to the situation where S is a CT(m) and T is a CT(n). (An invariant notation to represent such a product is introduced in Section 1.2.3.)

Deduce that USTvSTijk≡= ijp kp, i ijk jk are respectively the components of a CT(3) and a CT(l).

Problem 1.2.2 If T is a CT(2) and Tn0= for arbitrary vectors n show that T0= (the zero second-order tensor), that is Tij = 0 with respect to an arbitrary basis. (This result was used in the derivation of equation (1.2.5).)

Problem 1.2.3 If S and T are each CT(n)’s and α, β are scalars, prove that αβST+ is also a CT(n). This shows that Cartesian tensors of the same order may be added, but no meaning can be attached to the sum of tensors of different orders.

± 1 Problem 1.2.4 Show that Aijkl=±2 (δ ikδδδ jl il jk ) are the components of a CT(4) and prove +− that AAijkk==δ ij,0 ijkk .

Problem 1.2.5 If Eij are the components of a CT(2) and λ, μ are scalars, deduce that

TEEij=+2μ ijλδ kk ij are also the components of a CT(2), and show that

TEkk=+(3λμ 2 ) kk , 2 TEij ij=+2(),μλ EE ij ij E kk 22 TTij ij=++4(34)().μλλμ E ij E ij E kk

If Eij and Tij are defined by

11 EEij=− ij33 E kkδ ij,, TT ij =− ij T kkδ ij

deduce that ETkk== kk0, T ij = 2μ E ij , and hence show that

1 TEij ij=+ TE ij ij3 T kk E ll ,

1 2 EEij ij==− EE ij ij EE ij ij3 () E kk .

Problem 1.2.6 If T is an arbitrary CT(2) and I is the identity tensor on , deduce that TTI= . Hence use the representation (1.2.10) for I to show that T has the representation (1.2.9).

The decomposition Tne= Tnik k i may be used.

October 15, 2007 1.2.2-2