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c ba ec in .NET framework
c ba ec QR Code Recognizer In VS .NET Using Barcode Control SDK for .NET framework Control to generate, create, read, scan barcode image in VS .NET applications. QR Code Creator In .NET Framework Using Barcode maker for VS .NET Control to generate, create QR image in .NET framework applications. Putting this into the formula that gives the derivative of a vector, we have A = a A b eb + a x x Decoding QR Code 2d Barcode In .NET Using Barcode decoder for .NET Control to read, scan read, scan image in VS .NET applications. Barcode Maker In VS .NET Using Barcode generation for .NET Control to generate, create bar code image in Visual Studio .NET applications. c ba A b
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(4.6) It can be veri ed that this object is a (1, 1) tensor by checking how it transforms. To see how the Christoffel symbols transform, we look at (4.5). Writing this in primed coordinates, we have c ab
ec =
ea xb
d c ed
(4.7) = xd e . xc d
Now, the basis vectors transform according to ec = hand side then becomes
c ab
The left ec =
c ed =
xd ed xc
(4.8) Now let s tackle the righthand side of (4.7). We obtain ea = b ( xb x
m a em ) = xb
xm em xa
xn = b x xn = xn xb
xm em xa
x m em 2xm em + a xn xa x xn
Now we know that
em xn
gives another Christoffel symbol, so we write this as xm 2xm em + a xn xa x
l mn el
ea xn = b xb x
Tensor Calculus
In the rst term inside parentheses, we are going to swap the dummy indices. We change m d and now write the expression as ea xn = b xb x 2xd xm ed + a xn xa x l mn el
In the second term, there is a dummy index l that is associated with the basis vector. We would like to factor out the basis vector and so we change l d, which gives ea xn = b xb x = = xn xb xm 2xd ed + a xn xa x xm 2xd + a xn xa x d d mn ed
(4.9) xn xm xn 2xd + b xb xn xa x xa
We started out with (4.7). Setting the result for the left side of (4.7) given in (4.8) equal to the result found in (4.9) gives us the following: c ab
xd ed = xc
xn xm xn 2xd + b xb xn xa x xa
The basis vector ed appears on both sides, so we drop it and write
c ab
xn 2xd xn xm xd = b + b xc x xn xa x xa
Finally, we obtain the transformation law for the Christoffel symbols by xd dividing both sides through by x c . This gives c ab
xc xn 2xd xc xn xm + d b xd xb xn xa x x xa
(4.10) Now we see why the Christoffel symbols act as correction terms. The rst piece of this can be used to cancel out the extra piece found in the transformation of a partial derivative of a tensor. That way, the covariant derivative transforms as it should, as a (1, 1) tensor. Tensor Calculus
Of course we are going to differentiate other objects besides vectors. The covariant derivative of a one form is given by b a = b a c ab c
(4.11) This suggests the procedure to be used for the covariant derivative of an arbitrary tensor: First take the partial derivative of the tensor, and then add a c ab term for each contravariant index and subtract a c ab term for each covariant index. For example, c T a b = c T a b + c Tab = c Ta b c T ab = c T ab + a d d cd T b
+ a bc T d
ac T db cd T db
cb T ad dc T ad
The covariant derivative of a scalar function is just the partial derivative: a = a (4.12) EXAMPLE 41 Consider polar coordinates. Find the covariant derivative a V a of V = r 2 cos er sin e . SOLUTION 41 The summation convention is in effect. Writing it out explicitly, we have a V a = r V r + V Using (4.6), we have r V r = V = Vr + r V + r cr V c
Vr + r V +
rr V
r V
c c V =
r V
V One can show that the Christoffel symbols for polar coordinates are (Exercise) r = r while all other components are zero, and so we obtain r V r = V = Vr + r V +
r rr V r
Tensor Calculus
r V
Vr r V 1 + Vr r
r V
V The sum then becomes a V a = r V r + V = Vr V 1 + Vr + r r
For V = r 2 cos er sin e , this results in a V a = Vr V 1 + Vr+ = 2r cos + r cos cos = 3r cos cos r r = cos (3r 1) EXAMPLE 42 Suppose that for some vector eld U a , U a Ua is a constant and a Ub b Ua = 0. Show that U a a U b = 0. SOLUTION 42 Since U a Ua is a constant, the derivative must vanish. We compute the derivative of this product as b (U a Ua ) = U a b Ua + Ua b U a Now we use a Ub b Ua = 0 to rewrite the rst term on the righthand side, and then use index raising and lowering with the metric to write Ub = gbc U c , which gives U a b Ua + Ua b U a = U a a Ub + Ua b U a = U a a gbc U c + Ua b U a = U a U c a gbc + gbc U a a U c + Ua b U a = gbc U a a U c + Ua b U a

