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C:\Book\Booktex\C1s4.DVI 108 1.4 DERIVATIVE OF A TENSOR § In this section we develop some additional operations associated with tensors. Historically, one of the basic problems of the tensor calculus was to try and find a tensor quantity which is a function of the metric ∂g ∂2g tensor g and some of its derivatives ij , ij , .... A solution of this problem is the fourth order ij ∂xm ∂xm∂xn Riemann Christoffel tensor Rijkl to be developed shortly. In order to understand how this tensor was arrived at, we must first develop some preliminary relationships involving Christoffel symbols. Christoffel Symbols Let us consider the metric tensor gij which we know satisfies the transformation law ∂xa ∂xb g = gab . αβ ∂xα ∂xβ Define the quantity c a b 2 a b a 2 b ∂gαβ ∂gab ∂x ∂x ∂x ∂ x ∂x ∂x ∂ x (α, β, γ)= = + gab + gab ∂xγ ∂xc ∂xγ ∂xα ∂xβ ∂xα∂xγ ∂xβ ∂xα ∂xβ∂xγ 1 and form the combination of terms [(α, β, γ)+(β,γ,α) (γ,α,β)] to obtain the result 2 − a b c b 2 a 1 ∂gαβ ∂gβγ ∂gγα 1 ∂gab ∂gbc ∂gca ∂x ∂x ∂x ∂x ∂ x + = + + gab . (1.4.1) 2 ∂xγ ∂xα − ∂xβ 2 ∂xc ∂xa − ∂xb ∂xα ∂xβ ∂xγ ∂xβ ∂xα∂xγ In this equation the combination of derivatives occurring inside the brackets is called a Christoffel symbol of the first kind and is defined by the notation 1 ∂g ∂g ∂g [ac, b]=[ca, b]= ab + bc ac . (1.4.2) 2 ∂xc ∂xa − ∂xb The equation (1.4.1) defines the transformation for a Christoffel symbol of the first kind and can be expressed as ∂xa ∂xb ∂xc ∂2xa ∂xb [αγ,β]=[ac, b] + gab . (1.4.3) ∂xα ∂xβ ∂xγ ∂xα∂xγ ∂xβ Observe that the Christoffel symbol of the first kind [ac, b] does not transform like a tensor. However, it is symmetric in the indices a and c. At this time it is convenient to use the equation (1.4.3) to develop an expression for the second derivative term which occurs in that equation as this second derivative term arises in some of our future considerations. ∂xβ To solve for this second derivative we can multiply equation (1.4.3) by gde and simplify the result to the ∂xd form ∂2xe ∂xa ∂xc ∂xβ = gde[ac, d] + [αγ,β] gde. (1.4.4) ∂xα∂xγ − ∂xα ∂xγ ∂xd ∂xd ∂xe The transformation gde = gλµ allows us to express the equation (1.4.4) in the form ∂xλ ∂xµ ∂2xe ∂xa ∂xc ∂xe = gde[ac, d] + gβµ[αγ,β] . (1.4.5) ∂xα∂xγ − ∂xα ∂xγ ∂xµ 109 Define the Christoffel symbol of the second kind as i i 1 ∂g ∂g ∂g = = giα[jk,α]= giα kα + jα jk . (1.4.6) jk kj 2 ∂xj ∂xk − ∂xα This Christoffel symbol of the second kind is symmetric in the indices j and k and from equation (1.4.5) we see that it satisfies the transformation law µ ∂xe e ∂xa ∂xc ∂2xe = + . (1.4.7) αγ ∂xµ ac ∂xα ∂xγ ∂xα∂xγ Observe that the Christoffel symbol of the second kind does not transform like a tensor quantity. We can use the relation defined by equation (1.4.7) to express the second derivative of the transformation equations in terms of the Christoffel symbols of the second kind. At times it will be convenient to represent the Christoffel symbols with a subscript to indicate the metric from which they are calculated. Thus, an alternative notation for i is the notation i . jk jk g EXAMPLE 1.4-1. (Christoffel symbols) Solve for the Christoffel symbol of the first kind in terms of the Christoffel symbol of the second kind. Solution: By the definition from equation (1.4.6) we have i = giα[jk,α]. jk We multiply this equation by gβi and find i g = δα[jk,α]=[jk,β] βi jk β and so i 1 N [jk,α]=gαi = gα + + gαN . jk 1 jk ··· jk EXAMPLE 1.4-2. (Christoffel symbols of first kind) Derive formulas to find the Christoffel symbols of the first kind in a generalized orthogonal coordinate system with metric coefficients 2 gij =0 for i = j and g = h ,i=1, 2, 3 6 (i)(i) (i) where i is not summed. Solution: In an orthogonal coordinate system where gij =0fori = j we observe that 6 1 ∂g ∂g ∂g [ab, c]= ac + bc ab . (1.4.8) 2 ∂xb ∂xa − ∂xc Here there are 33 = 27 quantities to calculate. We consider the following cases: 110 CASE I Let a = b = c = i, then the equation (1.4.8) simplifies to 1 ∂g [ab, c]=[ii, i]= ii (no summation on i). (1.4.9) 2 ∂xi From this equation we can calculate any of the Christoffel symbols [11, 1], [22, 2], or [33, 3]. CASE II Let a = b = i = c, then the equation (1.4.8) simplifies to the form 6 1 ∂g [ab, c]=[ii, c]= ii (no summation on i and i = c). (1.4.10) −2 ∂xc 6 since, gic =0fori = c. This equation shows how we may calculate any of the six Christoffel symbols 6 [11, 2], [11, 3], [22, 1], [22, 3], [33, 1], [33, 2]. CASE III Let a = c = i = b, and noting that gib =0fori = b, it can be verified that the equation (1.4.8) 6 6 simplifies to the form 1 ∂g [ab, c]=[ib, i]=[bi, i]= ii (no summation on i and i = b). (1.4.11) 2 ∂xb 6 From this equation we can calculate any of the twelve Christoffel symbols [12, 1] = [21, 1] [31, 3] = [13, 3] [32, 3] = [23, 3] [21, 2] = [12, 2] [13, 1] = [31, 1] [23, 2] = [32, 2] CASE IV Let a = b = c and show that the equation (1.4.8) reduces to 6 6 [ab, c]=0, (a = b = c.) 6 6 This represents the six Christoffel symbols [12, 3] = [21, 3] = [23, 1] = [32, 1] = [31, 2] = [13, 2] = 0. From the Cases I,II,III,IV all twenty seven Christoffel symbols of the first kind can be determined. In practice, only the nonzero Christoffel symbols are listed. EXAMPLE 1.4-3. (Christoffel symbols of the first kind)Find the nonzero Christoffel symbols of the first kind in cylindrical coordinates. Solution: From the results of example 1.4-2 we find that for x1 = r, x2 = θ, x3 = z and 1 2 2 g11 =1,g22 =(x ) = r ,g33 =1 the nonzero Christoffel symbols of the first kind in cylindrical coordinates are: 1 ∂g [22, 1] = 22 = x1 = r −2 ∂x1 − − 1 ∂g [21, 2] = [12, 2] = 22 = x1 = r. 2 ∂x1 111 EXAMPLE 1.4-4. (Christoffel symbols of the second kind) Find formulas for the calculation of the Christoffel symbols of the second kind in a generalized orthogonal coordinate system with metric coefficients 2 gij =0 for i = j and g = h ,i=1, 2, 3 6 (i)(i) (i) where i is not summed. Solution: By definition we have i = gim[jk,m]=gi1[jk,1] + gi2[jk,2] + gi3[jk,3] (1.4.12) jk By hypothesis the coordinate system is orthogonal and so 1 gij =0 for i = j and gii = i not summed. 6 gii The only nonzero term in the equation (1.4.12) occurs when m = i and consequently i [jk,i] = gii[jk,i]= no summation on i.(1.4.13) jk g ii We can now consider the four cases considered in the example 1.4-2. CASE I Let j = k = i and show i [ii, i] 1 ∂g 1 ∂ = = ii = ln g no summation on i.(1.4.14) ii g 2g ∂xi 2 ∂xi ii ii ii CASE II Let k = j = i and show 6 i [jj,i] 1 ∂gjj = = − no summation on i or j.(1.4.15) jj g 2g ∂xi ii ii CASE III Let i = j = k and verify that 6 j j [jk,j] 1 ∂g 1 ∂ = = = jj = ln g no summation on i or j. (1.4.16) jk kj g 2g ∂xk 2 ∂xk jj jj jj CASE IV For the case i = j = k we find 6 6 i [jk,i] = =0,i= j = k no summation on i. jk g 6 6 ii The above cases represent all 27 terms. 112 EXAMPLE 1.4-5. (Notation) In the case of cylindrical coordinates we can use the above relations and find the nonzero Christoffel symbols of the second kind: 1 1 ∂g = 22 = x1 = r 22 −2g ∂x1 − − 11 2 2 1 ∂g 1 1 = = 22 = = 12 21 2g ∂x1 x1 r 22 Note 1: The notation for the above Christoffel symbols are based upon the assumption that x1 = r, x2 = θ and x3 = z. However, in tensor calculus the choice of the coordinates can be arbitrary. We could just as well have defined x1 = z,x2 = r and x3 = θ. In this latter case, the numbering system of the Christoffel symbols changes. To avoid confusion, an alternate method of writing the Christoffel symbols is to use coordinates in place of the integers 1,2 and 3. For example, in cylindrical coordinates we can write θ θ 1 r = = and = r. rθ θr r θθ − If we define x1 = r, x2 = θ, x3 = z, then the nonzero Christoffel symbols are written as 2 2 1 1 = = and = r. 12 21 r 22 − In contrast, if we define x1 = z,x2 = r, x3 = θ, then the nonzero Christoffel symbols are written 3 3 1 2 = = and = r.
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