By Stone M.
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Then, looking back at the derivation of the time-independence of the first integral, we see that if L does depend on time, we instead have dE ∂L =− . 115) so that − dE d J˙ · A d3x = = (Field Energy) − dt dt ˙ + J˙ · A J·A d3 x. 3. LAGRANGIAN MECHANICS 27 ˙ we find Thus, cancelling the duplicated term and using E = −A, d (Field Energy) = − dt J · E d3 x. 117) Now J · (−E) d3x is the rate at which the power source driving the current is doing work against the field. The result is therefore physically sensible.
Is somewhat Obtaining and interpreting the last equation, involving h, trickier. It comes from the variation of φ on the upper boundary. The variation of S due to δφ is δS = ρ0 ∂ ∂ δφ + ∂t ∂x δφ ∂φ ∂x + ∂ ∂y δφ ∂φ ∂y − δφ ∇2 φ dtdxdy. 144) The first three terms in the integrand constitute the three-dimensional divergence ∇ · (δφ Φ) where, listing components in the order t, x, y, Φ = 1, ∂φ ∂φ . 145) The integrated-out part on the upper surface is therefore Here, the outward normal is n= 1+ ∂h ∂t 2 + ∂h ∂x −1/2 2 − ∂h ∂h ,− ,1 , ∂t ∂x (Φ · n)δφ d|S|.
It comes from the variation of φ on the upper boundary. The variation of S due to δφ is δS = ρ0 ∂ ∂ δφ + ∂t ∂x δφ ∂φ ∂x + ∂ ∂y δφ ∂φ ∂y − δφ ∇2 φ dtdxdy. 144) The first three terms in the integrand constitute the three-dimensional divergence ∇ · (δφ Φ) where, listing components in the order t, x, y, Φ = 1, ∂φ ∂φ . 145) The integrated-out part on the upper surface is therefore Here, the outward normal is n= 1+ ∂h ∂t 2 + ∂h ∂x −1/2 2 − ∂h ∂h ,− ,1 , ∂t ∂x (Φ · n)δφ d|S|. 146) and the element of area d|S| = 1+ ∂h ∂t 2 + ∂h ∂x 2 1/2 dtdx.