lec29.ppt
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This document discusses the metric tensor, which describes how distances are measured in different coordinate systems. It gives the general formula for the metric tensor gij in terms of partial derivatives of the coordinate transformations. As an example, it derives the specific form of the metric tensor in spherical polar coordinates, obtaining the known result that the line element ds2 is given by (dr)2 + r2(dθ)2 + r2sin2θ(dφ)2.
lec29.ppt
- 1. NPTEL – Physics – Mathematical Physics - 1 Lecture 29 Metric Tensor The rotation of distance is fundamental in all areas of physics and mathematics. The arc length in different coordinate system is represented as, 𝑑𝑠2 = (𝑑𝑥1)2 + (𝑑𝑥2)2 + (𝑑𝑥3)2 = 𝛿𝑖𝑗 𝑑𝑥𝑖𝑑𝑥𝑗 More generally, this can be written as, 𝑑𝑠2 = gij𝑑𝑥𝑖 𝑑𝑥𝑗 In polar coordinates : (𝑥1, 𝑥2) = (𝑟, 𝜃) 𝑑𝑠2 = (𝑑𝑥1)2 + (𝑥1)2(𝑑𝑥2)2 in spherical polar coordinates : (𝑥1, 𝑥2, 𝑥3) = (𝑟, 𝜃, 𝛷) 𝑑𝑠2 = (𝑑𝑥1)2 + (𝑥1)2(𝑑𝑥2)2 + (𝑥1𝑠𝑖𝑛𝑥2)2(𝑑𝑥3)2 etc. thus in general notation, let {𝑥1𝑗} denote a set of Cartesian coordinates and {𝑥𝑗} denote some other coordinates of which {𝑥1𝑗} are functions. Then we have, 𝑑𝑥1𝑖 (1) 𝑑𝑥1𝑖 = 𝑑𝑥𝑗 (sum over j is implicit) thus the element of length (squared) in a 𝑉 𝑛 space is, 𝑑𝑠2 = ∑𝑛 (𝑑𝑥1𝑖)2 = ∑𝑛 𝑖=1 𝑖=1 𝑑𝑥1𝑖 𝑑𝑥1𝑖 = ∑𝑛 (𝜕𝑥 1𝑖 𝜕𝑥 𝑗 𝑑𝑥𝑗 ) (𝜕𝑥 𝑑𝑥𝑘 ) 𝑖=1 1𝑖 𝜕𝑥 𝑘 = (∑𝑛 𝜕𝑥 𝜕 𝑥 1𝑖 𝜕𝑥 𝜕 𝑥 𝑗 𝑖=1 1𝑖 𝑘 ) 𝑑𝑥 𝑑𝑥 𝑗 𝑘 (2) Compose Eq. (1) and (2) 𝑛 𝑔 = ∑ 𝑗 𝑘 𝑖=1 𝜕𝑥1𝑖 𝜕𝑥1 𝑖 𝜕𝑥𝑗 𝜕𝑥 𝑘 𝑔𝑗 𝑘 is called the metric tenk of order (0, 2) which can be proved as follows, 𝑛 Page 15 of 20 Joint initiative of IITs and IISc – Funded by MHRD 𝑔𝑗𝑘 = ∑ 𝜕𝑥̅ 𝑗 𝜕𝑥̅𝑘 𝑖=1 𝜕𝑥1𝑖 𝜕𝑥1𝑖 Using a chain rule,
- 2. NPTEL – Physics – Mathematical Physics - 1 𝑔𝑗𝑘 = ∑ 𝜕𝑥𝑝 𝜕𝑥̅ 𝑗 𝜕𝑥𝑞 𝜕𝑥̅𝑘 𝑖=1 𝜕𝑥1𝑗 𝜕𝑥𝑝 𝜕𝑥1𝑗 𝜕𝑥 𝑞 𝑛 = (∑𝑛 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 1 𝑗 𝜕𝑥 𝜕𝑥 𝜕𝑥̅ 𝜕𝑥̅ 𝜕𝑥̅ 𝜕𝑥̅ 𝑝 𝑞 𝑗 𝑘 𝑗 𝑘 𝑖=1 1𝑗 𝑝 𝑞 ) = 𝑔𝑝𝑞 𝜕𝑥 𝜕𝑥 𝑝 𝑞 𝑔𝑝𝑞 The elements of the metric tensor are invertible which can be extended as, 𝑔𝑗𝑘ℎ𝑘𝑚 = 𝛿𝑗 𝑚 Example Metric tensor in spherical polar coordinates We have, x = rsin𝜃𝑐𝑜𝑠, 𝑦 = 𝑟𝑠𝑖𝑛𝜃𝑠𝑖𝑛, metric tensors are 𝑧 = 𝑟𝑐𝑜𝑠𝜃. The components of the 𝑔𝑟𝑟 (𝑟, 𝜃, ) = (𝜕𝑟 ) + (𝜕𝑟 ) + ( ) 𝜕𝑥 𝜕𝑦 𝜕𝑧 2 2 2 𝜕𝑟 = (sin 𝑐𝑜𝑠)2 + (𝑠𝑖𝑛𝜃𝑠𝑖𝑛)2 + (𝑐𝑜𝑠𝜃)2 = 1 𝑔𝑟𝑟 (𝑟, 𝜃, ) = 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕 𝑧 + + 𝜕𝑟 𝜕𝜃 𝜕𝑟 𝜕𝜃 𝜕𝑟 𝜕𝜃 = 0 (check !!) 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝑔𝜃𝜃 = ( 𝜕𝑟 ) + ( 𝜕𝑟 ) + ( 𝜕𝑟 ) = 𝑟2 𝑔 = ( 𝜕 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 ) + ( ) + ( ) 𝜕 𝜕 𝑑𝑠2 = (𝑑𝑟)2 + 𝑟2(𝑑𝜃)2 + 𝑟2𝑠𝑖𝑛2𝜃(𝑑)2 -a known result Page 16 of 20 Joint initiative of IITs and IISc – Funded by MHRD