curvilinear

# This gives i vi g v gi gi vi i x x x vi gi g gk

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r , 22 Γ2 = Γ2 = 12 21 1 . r (22) For the spherical coordinate system we have all of the Christoﬀel symbols are zero except Γ1 22 = −r sin2 (ϕ) = −r , Γ1 33 Γ2 = Γ2 = 1 , 12 21 r Γ2 33 Γ3 = Γ3 = 1 , 13 31 r Γ3 = Γ3 = cot(ϕ) 32 23 = − sin(θ) cos(ϕ) (23) We can now consider taking the gradient of a vector. This gives ∂ ￿ i ￿ ∂ vi ∂g v gi = gi + vi i ∂x ∂x ∂x ∂ vi ∂ gi = g ⊗ gk + vi k ⊗ gk ∂zk i ∂ ￿i ￿z ∂v + +v j Γi g i ⊗ g k . = jk ∂zk ∇v = (24) (25) (26) For the cylindrical coordinate system we can expand this result to determine the needed components of the gradient. When expressed in terms of the physical basis we ﬁnd that ∇u is given by ur,r 1 (ur,θ − uθ ) ur,z r uθ,r 1 (uθ,θ + ur ) uθ,z . (27) r 1 uz,r u uz,z r z,θ 5 The components of the symmetric part of this tensor give the strain, ε, (in the physical basis as) ￿ ￿1 ur,r 1 1 ur,θ + r(uθ /r),r (ur,z + uz,r ) 2r 2 1 1 (uθ,θ + ur ) (uθ,z + 1 uz,θ ) . (28) r 2 r sym. uz,z For the spherical coordinate system we can also expand this result to determine the needed components of the gradient. When expressed in terms of the physical basis we ﬁnd that ∇u is given by 1 1 (ur,ϕ − uϕ ) (ur,θ − uθ sin(ϕ)) ur,r r r sin(ϕ) u 1 1 (uϕ,ϕ + ur ) − 1 uθ cot(ϕ) + r sin(ϕ) uϕ,θ (29) ϕ,r r...
View Full Document

## This note was uploaded on 04/11/2013 for the course PHYS 105 taught by Professor Edgarknobloch during the Spring '10 term at Berkeley.

Ask a homework question - tutors are online