# For simplicity consider the stress tensor its

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Unformatted text preview: r 1 1 uθ,r r (uθ cot(ϕ) + uθ,ϕ ) r sin(ϕ) uθ,θ + ur /r + uϕ cot(ϕ)/r The components of the symmetric part of this tensor give the strain, ε, (in the physical basis as) ￿ ￿ ￿ ￿ 1 1 ur,r 1 1 ur,ϕ + r(uϕ /r),r ur,θ + r(uθ /r),r 2r 2 r sin(ϕ) ￿ ￿ (30) 11 1 1 (uϕ,ϕ + ur ) u + r sin(ϕ) uϕ,θ r 2 r θ,ϕ 1 sym. u + ur /r + uϕ cot(ϕ)/r r sin(ϕ) θ,θ 6 Gradient and Divergence of a Tensor The basic procedure for ﬁnding the gradient and divergence of a tensor follows exactly as we did above. For simplicity consider the stress tensor σ . Its gradient is given by ￿ ∂σ ij ∂ gj ∂ ￿ ij ∂g σ gi ⊗ gj = g i ⊗ g j + σ ij i ⊗ g j + σ ij g i ⊗ (31) ∂x ∂x ∂x ∂x ∂ gj ∂ gi ∂σ ij g i ⊗ g j ⊗ g k + σ ij k ⊗ g j ⊗ g k + σ ij g i ⊗ k ⊗ g k (32) = ∂k ∂ z￿ ∂z ￿ z ij ∂σ = + +σ lj Γi + σ il Γj g i ⊗ g j ⊗ g k . (33) lk lk k ∂z ∇σ = The divergence is obtained by contract...
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## This note was uploaded on 04/11/2013 for the course PHYS 105 taught by Professor Edgarknobloch during the Spring '10 term at University of California, Berkeley.

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