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Putting this into the formula that gives the derivative of a vector, we have A = a A b eb + a x x
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We rearranged the order of quantities in the second term for later convenience. Now, recall that any repeated indices that are both up and down in an expression are dummy indices, and so can be relabeled. In the second term, we swap b c to change c ba Ab ec b ca Ac eb , and the expression for the derivative of a
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The expression in parentheses is the covariant derivative of a vector A. We denote the covariant derivative by b Aa : b Aa = Aa + xb
b ca A c
(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 right-hand 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 4-1 Consider polar coordinates. Find the covariant derivative a V a of V = r 2 cos er sin e . SOLUTION 4-1 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 4-2 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 4-2 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 right-hand 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
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