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= sin cos ex + sin sin e y + cos ez
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= r cos cos ex + r cos sin e y r sin ez
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(4.4)
= r sin sin ex + r sin cos e y Now let s see what happens when we differentiate these basis vectors. As an example, we compute the derivatives of er with respect to each of the spherical coordinates. Remember, the cartesian basis vectors ex , e y , ez are constant, so we don t have to worry about them. Proceeding, we have er = sin cos ex + sin sin e y + cos ez = 0 r r
Tensor Calculus
Our rst attempt hasn t yielded anything suspicious. But let s compute the derivative. This one gives er = sin cos ex + sin sin e y + cos ez = cos cos ex + cos sin e y sin ez 1 = e r Now that s a bit more interesting. Instead of computing the derivative and getting zero, we nd another basis vector, scaled by 1/r. Let s go further and proceed by computing the derivative with respect to . In this case, we get er = sin cos ex + sin sin e y + cos ez = sin sin ex + sin cos e y 1 = e r Again, we ve arrived at another basis vector. It turns out that when differentiating basis vectors, there is a general relationship that gives the derivative of a basis vector in terms of a weighted sum. The sum is just an expansion in terms of the basis vectors with weighting coef cients denoted by a bc . ea = xb
c ab ec
(4.5)
The a bc are functions of the coordinates, as we saw in the examples we ve calculated so far. Looking at the results we got above, we can identify the results in the following way: er 1 = e r er 1 = e r
r
1 r 1 r
Tensor Calculus
Consider another derivative e = r sin sin ex + r sin cos e y = r cos sin ex + r cos cos e y = cos r sin ex + r cos e y = cos r sin sin ex + r sin cos e y sin = cot e
Comparison with (4.5) leads us to conclude that
= cot
The coef cient functions we have derived here are known as Christoffel symbols or an af ne connection. Basically, these quantities represent correction terms. A derivative operator needs to differentiate a tensor and give a result that is another tensor. In particular, the derivative of a tensor eld that has valence m should give a tensor n m eld of valence n+1 . We have seen one reason why we need the correction terms: outside of ordinary cartesian coordinates, the derivative of a vector is going to involve derivatives of the basis vectors as well. Another reason for introducing the Christoffel symbols is that the partial derivative of a tensor is not a tensor. First let s remind ourselves how the components of a vector transform: Xa = Keeping this in mind, we have c X a = = xc xc xa b X xb xd xd xa b X xb xa b X xb
xd xc xd
= xa b X xb
Tensor Calculus
x d x a X b xd 2xa Xb + c xc xd xb x xb xd
Now how does a (1, 1) tensor transform It does so like this: Ta b = xa xd c T d xc xb
That s the kind of transformation we got from the partial derivative above, in the second term on the last line: c X a = xd 2xa xd xa Xb Xb + c xc xd xb x xb xd
But the rst term leaves us out of luck as far as getting another tensor thus the need for a correction term. Let s go back and look at the formula we had for the derivative of a vector A: A = a Ab eb + Ab a (eb ) a x x x Now let s use (4.5) to rewrite the second term. This gives Ab (eb ) = Ab xa
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