Fizik Lecture 8 - Metrics Lecture 8 Metrics Objectives: • More on the metric and how it transforms. Reading: Hobson, 2. 8.1 Riemannian Geometry The interval ds 2 = g ?ß dx ? dx ß , is a quadratic function of the coordinate di?erentials. Thisisthede?nitionofRiemannian geometry,ormorecorrectly,pseudoRiemannian geometry to allow for ds 2 < 0. Example 8.1 What are the coe?cients of the metric tensor in 3D Euclidean space for Cartesian, cylindrical polar and spherical polar coordinates? Answer 8.1 The “interval” in Euclidean geometry can be written in Carte sian coordinates as Introducing an obvious notation with x standing for the x coordinate index, etc. ds 2 = dx 2 +dy 2 +dz 2 . The metric tensor’s coe?cients are therefore given by g xx = g yy = g zz = 1, with all others = 0. In cylindrical polars: ds 2 = dr 2 +r 2 d? 2 +dz 2 , 30LECTURE 8. METRICS 31 so g rr = 1, g ?? = r 2 , g zz = 1 and all others = 0. Finally spherical polars: ds 2 = dr 2 +r 2 d? 2 +r 2 sin 2 ?d? 2 , gives g rr = 1, g ?? = r 2 and g ?? = r 2 sin 2 ?. Example 8.2 Calculate the metric tensor in 3D Euclidean space for the coordinates u = x+2y, v = x-y, w = z. Answer 8.2 The inverse transform is easily shown to be x = (u + 2v)/3, y = (u-v)/3, z = w, so dx = 1 3 du+ 2 3 dv, dy = 1 3 du- 1 3 dv, dz = dw, so ds 2 = µ 1 3 du+ 2 3 dv ¶ 2 + µ 1 3 du- 1 3 dv ¶ 2 +dw 2 , = 2 9 du 2 + 5 9 dv 2 + 2 9 dudv+dw 2 . We can immediately write g uu = 2/9, g vv = 5/9, g ww = 1, and g uv = g vu = 1/9 since the metric is symmetric. This metric still describes 3D Euclidean ?at geometry, although not obviously. 8.2 Metric transforms The method of the example is often the easiest way to transform metrics, however using tensor transformations, we can write more compactly: g ? ' ß ' = ?x ? ?x ? ' ?x ? ?x ß ' g ?? . This shows how the components of the metric tensor transform under coor dinate transformations but the underlying geometry does not change. Example 8.3 Use the transformation of g to derive the metric components in cylindrical polars, starting from Cartesian coordinates.LECTURE 8. METRICS 32 Answer 8.3 We must compute terms like ?x/?r, so we need x, y and z in terms of r, ?, z: x = rcos?, y = rsin?, z = z. Find ?x/?r = cos?, ?y/?r = sin?, ?z/?r = 0. Consider the g rr component: g rr = ?x i ?r ?x j ?r g ij , where i and j represent x, y or z. Since g ij = 1 for i = j and 0 otherwise, and since ?z/?r = 0, we are left with: g rr = µ ?x ?r ¶ 2 + µ ?y ?r ¶ 2 = cos 2 ?+sin 2 ? = 1. Similarly g ?? = µ ?x ?? ¶ 2 + µ ?y ?? ¶ 2 = (-rsin?) 2 +(rcos?) 2 = r 2 , and g zz = 1, as expected. This may seem a very di?cult way to deduce a familiar result, but the point is that it transforms a problem for which one otherwise needs to apply intuition and 3D visualisation into a mechanical procedure that is not di?cult – at least in principle – and can even be programmed into a computer. 8.3 First curved-space metric We can now start to look at curved spaces. A very helpful one is the surface of a sphere.LECTURE 8. METRICS 33 Figure: Surfaceofasphereparameterisedbydistancer from a point and azimuthal angle ? The sketch shows the surface “embedded” in 3D. This is a priviledged view that is not always possible. You need to try to imagine that you are actually stuck in the surface with no “height” dimension. Two coordinates are needed to label the surface. e.g. the distance from a point along the surface, r, and the azimuthal angle ?, similar to Euclidean polar coords. The distance AP is given by Rsin?, so a change d? corresponds to distance Rsin?d?. Thus the metric is ds 2 = dr 2 +R 2 sin 2 ?d? 2 . or since r = R?, ds 2 = dr 2 +R 2 sin 2 ³ r R ´ d? 2 . This is the metric of a 2D space of constant curvature. Circumference of circle in this geometry: set dr = 0, integrate over ? C = 2?Rsin r R < 2?r. e.g. On Earth (R = 6370km), circle with r = 10km shorter by 2.6cm than if Earth was ?at. Exactly the same is possible in 3D. i.e we could ?nd that a circle radius r has a circumference < 2?r owing to gravitationally induced curvature. 8.4 2D spaces of constant curvature Can construct metric of the surface of a sphere as follows. First write the equation of a sphere in Euclidean 3D x 2 +y 2 +z 2 = R 2 .LECTURE 8. METRICS 34 If we switch to polars (r,?) in the x–y plane, this becomes r 2 +z 2 = R 2 . In the same terms the Euclidean metric is dl 2 = dr 2 +r 2 d? 2 +dz 2 . But we can use the restriction to a sphere to eliminate dz which implies 2rdr+2zdz = 0, and so dl 2 = dr 2 +r 2 d? 2 + r 2 dr 2 z 2 , which reduces to dl 2 = dr 2 1-r 2 /R 2 +r 2 d? 2 . De?ning curvature k = 1/R 2 , we have dl 2 = dr 2 1-kr 2 +r 2 d? 2 , the metric of a 2D space of constant curvature. k > 0 can be “embedded” in 3D as the surface of a sphere; k < 0 cannot, but it is still a perfectly valid geometry. [A saddle shape has negative curvature over a limited region.] A very similar procedure can be used to construct the spatial part of the metric describing the Universe.