Tensor Calculus in .NET framework

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Tensor Calculus
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To obtain the last line, we used a gbc = 0. We now concentrate on the second term. To use a Ub b Ua = 0 to change the indices, rst we need to lower the index on U a inside the derivative. Remember, we are free to pull gab outside the derivative since a gbc = 0. So we get gbc U a a U c + Ua b U a = gbc U a a U c + Ua b g ac Uc = gbc U a a U c + g ac Ua b Uc = gbc U a a U c + g ac Ua c Ub The rst term is already in the form we need to prove the result. To get the second term in that form, we are going to have to raise the indices on the U s. a We do this and recall that g ab gbc = c , which gives gbc U a a U c + g ac Ua c Ub = gbc U a a U c + g ac gae U e c gbd U d = gbc U a a U c + g ac gae gbd U e c U d
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c = gbc U a a U c + e gbd U e c U d
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= gbc U a a U c + gbd U c c U d To obtain the last line, we used the Kronecker delta to set e c. Now note that c is a dummy index. We change it to a to match the rst term gbc U a a U c + gbd U c c U d = gbc U a a U c + gbd U a a U d Now we focus on the other dummy index d. We re free to change it so we set it equal to c and have gbc U a a U c + gbd U a a U d = gbc U a a U c + gbc U a a U c = gbc (U a a U c + U a a U c ) = 2gbc U a a U c We started by taking the derivative of a constant, which is zero, b (U a Ua ) = 0, and so this result must vanish. We divide the above expression by 2 and get gbc U a a U c = 0
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There are two possibilities: If gbc = 0, then the equation is trivially zero. If it is not zero, we can divide both sides by gbc and get the desired result U a a U c = 0
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The Torsion Tensor
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The torsion tensor is de ned in terms of the connection as T a bc =
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a bc
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(4.13)
It is easy to show that even though the connection is not a tensor, the difference between two connections is a tensor. In general relativity, the torsion tensor is taken to vanish, which means that the connection is symmetric; i.e.,
a bc
(4.14)
In some theories of gravity (known as Einstein-Cartan theories), the torsion tensor does not vanish. We will not study such cases in this book.
The Metric and Christoffel Symbols
In n dimensions there are n 3 functions called Christoffel symbols of the rst kind, abc . In a coordinate basis (see 5) these functions can be derived from the metric tensor using the relationship
gab gbc gca + a b x x xc
(4.15)
More generally,
gab gbc gca + + Cabc + Cacb Cbca a b x x xc
where the Cabc are called commutation coef cients (see 5).
EXAMPLE 4-3 Show that
Tensor Calculus
gab = xc
SOLUTION 4-3 This is a simple problem to solve. We simply use (4.15) and permute indices, which gives
1 2 1 2
gbc gca gab + xa xb xc gab gbc gca + xb xc xa
Adding, we nd 1 2 1 2 gab gbc gca + xa xb xc 1 2 gab gbc gca + xb xc xa
= = =
gca gab gab gbc gbc gca + + + xa xa xb xb xc xc = gca xb
gca 1 2 b 2 x
We can obtain the Christoffel symbols of the second kind (which are usually referred to simply as the Christoffel symbols) by raising an index with the metric
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