By A. Mous
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QED. 7 Theorem If the force F (x, t) has explicit time dependence, we may still define E = T + V, but dE ∂V = . dt ∂t Proof: Since V depends both on position and on time, we have, F =− ∂V ∂x and, using the chain rule, dV = ∂V ∂V ∂V dx + dt = −F dx + dt. ∂x ∂t ∂t We can no longer write, Wab = b a F dx = − b a 58 dV dx = −Vb + Va dx because the middle step no longer holds. However, we can still define E = T + V. Then, dE d 1 ∂V dx ∂V = ( mx˙ 2 ) + + dt dt 2 ∂x dt ∂t ∂V . = mx¨ ˙ x − (m¨ x)x˙ + ∂t Hence, dE ∂V = .
We conclude, bi = g ij bj . QED. 17 Theorem Theorem: ∂bi /∂q j = ∂bj /∂q i . Proof: This symmetry of pattern merely reflects the fact that the order in which partial derivatives are taken does not matter: ∂bi ∂ ∂r ∂ ∂r ∂bj = j i = i j = . j ∂q ∂q ∂q ∂q ∂q ∂q i QED. Since each of these nine partial derivatives, ∂bi /∂q j are themselves vectors, we can express each as some linear combination of either the unit vectors, reciprocal basis vectors or basis vectors. We choose to express them as a linear combination of basis vectors, ∂bi = Γkij bk .
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. Again, this is a formal result that is essential to our derivation, but not one that we will use as a practical tool. 3 A Better Form for the Covariant Components of Acceleration We now begin to change the form of the covariant components of acceleration to put them into a form that is a practical tool.
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