We can rewrite the integral, shortening L ( q, q ˙ ) to L for convenience, as: = δ q d d t ∂ ∂ q ˙ L ( q, q ˙ ) + δ q ˙ ∂ ∂ q ˙ L ( q, q ˙ ) ,Ī simple application of the product rule d d t ( f g ) = f ˙ g + f g ˙ which allows us to substitute Keeping in mind that δ q ˙ = d d t δ q and noting thatĭ d t ( δ q ∂ ∂ q ˙ L ( q, q ˙ ) ) L ( q + δ q, q ˙ + δ q ˙ ) = L ( q, q ˙ ) + δ q ∂ ∂ q L ( q, q ˙ ) + δ q ˙ ∂ ∂ q ˙ L ( q, q ˙ ) + O ( δ q 2 ) + O ( δ q ˙ 2 )Īnd since we make little error by discarding higher-order terms in δ q and δ q ˙, we have How are we to simplify this mess? We are considering to the path, which suggests a Taylor series expansion of L ( q + δ q, q ˙ + δ q ˙ ) about ( q, q ˙ ): This point, call it t = t 0, could be a minimum or a maximum, so in the usual calculus of a single variable we’d proceed by taking the second derivative, f ′′ ( t 0 ), and seeing if it’s positive or negative to see whether the function has a minimum or a maximum at t 0, respectively.įormally, we write the condition that produce no change in S as δ S = 0. Of course, since the whole point is to consider δ t ≠ 0, once we neglect terms O ( δ t 2 ) this is just the point where f ′ ( t ) = 0. This is where f ( t + δ t ) ≈ f ( t ) taking a Taylor series expansion of f ( t ) at t, we findį ( t + δ t ) = f ( t ) + δ t f ′ ( t ) + O ( δ t 2 ) = f ( t ) , We wish to find the path described by L, passing through a point q ( a ) at t = a and through q ( b ) at t = b, for which the quantity S is a minimum, for which in the path produce no first-order change in S, which we’ll call a “ stationary point.” This is directly analogous to the idea that for a function f ( t ), the minimum can be found where δ t produce no first-order change in f ( t ).
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |