By A. Mous

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**Example text**

2. We may use the expression which we have just derived for the Christoffel symbols to write, ∂gin ∂gjn ∂gij 1 + − n glk Γkij = glk g kn 2 ∂q j ∂q i ∂q Observe, glk g kn = δln , 35 . so that we can write, 1 ∂gin ∂gjn ∂gij glk Γkij = δln + − n 2 ∂q j ∂q i ∂q = 1 2 ∂gil ∂gjl ∂gij + − ∂q j ∂q i ∂q l . 3. Thus, Al = glk q¨k + 1 2 ∂gil ∂gjl ∂gij + − ∂q j ∂q i ∂q l q˙i q˙j . In principle, the covariant components of the acceleration can be obtained from this expression once the covariant form of the metric tensor is known.

2. We may use the expression which we have just derived for the Christoffel symbols to write, ∂gin ∂gjn ∂gij 1 + − n glk Γkij = glk g kn 2 ∂q j ∂q i ∂q Observe, glk g kn = δln , 35 . so that we can write, 1 ∂gin ∂gjn ∂gij glk Γkij = δln + − n 2 ∂q j ∂q i ∂q = 1 2 ∂gil ∂gjl ∂gij + − ∂q j ∂q i ∂q l . 3. Thus, Al = glk q¨k + 1 2 ∂gil ∂gjl ∂gij + − ∂q j ∂q i ∂q l q˙i q˙j . In principle, the covariant components of the acceleration can be obtained from this expression once the covariant form of the metric tensor is known.

We are left with the result, ∂bi ∂gik ∂gjk ∂gij + − k = 2 j · bk . j i ∂q ∂q ∂q ∂q 32 4. We may now use our assumption that ∂bi = Γlij bl . ∂q j Then, ∂gik ∂gjk ∂gij + − k = 2[Γlij bl ] · bk = 2Γlij (bl · bk ) = 2Γlij glk . j i ∂q ∂q ∂q 5. It is tempting to solve for Γlij by dividing both sides of this expression by 2glk , but one must remember that the repeated index l indicates a sum of terms, not an isolated term. In fact, what we are dealing with here are twenty-seven equations, one for each of the combinations of i, j, and k!