curvilinear

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

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

View Full Document Right Arrow Icon
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