For the spherical system one has that sin cos cos

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Unformatted text preview: 3 sin(θ) cos(θ)/r 0 g→ g→ g→ (15) 0 0 1 Note again that the components have been expressed in the standard orthonormal Cartesian basis. For the spherical system one has that sin(ϕ) cos(θ) cos(ϕ) cos(θ)/r g 1 → sin(ϕ) sin(θ) g 2 → cos(ϕ) sin(θ)/r cos(ϕ) − sin(ϕ)/r (16) − sin(θ)/r sin(ϕ) g 3 → cos(θ)/r sin(ϕ) . 0 4 Gradient of a Scalar Function Consider a scalar function f . Its gradient is given as ∇f . This can be converted through the use of the chain rule into curvilinear coordinates as: ∇f = ∂f i ∂f ∂zk ∂f ∂f = e = k i ei = k g k . i ∂x ∂x ∂z ∂x ∂z (17) Typically, however, results are expressed using the physical basis vectors and not the natural basis vectors. For our two coordinates systems we have upon expansion: ∂f er + ∂r ∂f er + ∇f = ∂r ∇f = 5 1 ∂f ∂f eθ + ez r ∂θ ∂z 1 ∂f 1 ∂f eϕ + eθ r ∂ϕ r sin(ϕ) ∂θ (18) (19) Gradient of a Vector To compute the gradient of a vector expressed in curvilinear coordinates we need to be able to compute the gradient of the basis vectors as they are 4 functions of position (unlike in the Cartesian case). Importantly we will need to know the derivatives ∂ ∂ xk ∂ 2 xk ∂ gi = j i ek = j i ek . ∂zj ∂z ∂z ∂z ∂z (20) The components of these vectors are usually expressed in the dual basis as Γk = g k · ij ∂ gi , ∂zj (21) where Γk is called the Christoffel symbol. For the cylindrical coordinate ij system all of the Christoffel symbols are zero except Γ1 = −...
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