COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT Link to: phsicspages home page. To leave a comment o epot an eo, please use the auilia blog. Refeence: d Inveno, Ra, Intoducing Einstein s Relativit (1992, Ofod Uni Pess. - Section 5.4 and Poblem 5.2. Since one of the main aspects of the definition of a tenso is the wa it tansfoms unde a change in coodinate sstems, it s impotant to conside how such coodinate changes wok. We ll conside two coodinate sstems, one denoted b unpimed smbols i and the othe b pimed smbols i. In geneal, one sstem is a function of the othe one, so we can wite i = i ( (1 whee the inde i uns ove the n dimensions of the manifold (so we have a set of n equations, and the smbol without an inde means the set of all components of, so it s equivalent to (but shote than witing i = i ( 1, 2, 3,..., n (2 Now suppose we want to do an integal ove a potion of the manifold that is bounded b some subsuface in the manifold. As we know fom elementa calculus, the diffeential volume (o aea has a diffeent fom depending on which coodinate sstem we e using. Fo eample, in 3-d ectangula coodinates, the volume element is dddz, while in spheical coodinates it is 2 sinθddθdφ. To see how this woks we can stat with one dimension. If we have an integal in ectangula coodinates such as ˆ 2 1 f(d (3 we can change coodinate sstems if we define = (u. Then we have d = du d du. To tansfom the limits of the integal, we need to invet the definition to get u = u(. Then the integal becomes ˆ u(2 u( 1 f((u d du (4 du 1
COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT 2 Essentiall, this edefines the line element into the u coodinate sstem. In two dimensions, we d stat off with (we ll leave out the limits on the integals since we e eall inteested onl in the aea element: ˆ ˆ f(, dd (5 Now if we want to switch to anothe coodinate sstem, we define u = u(, (6 v = v (, (7 Conside now an elemental ectangle R in the plane. The ectangle has its lowe left cone at the point ( 0, 0 and has dimensions and, so that its aea is. We want to see how this ectangle tansfoms unde the coodinate tansfomation above. The new elemental aea will not necessail be a ectangle, but we can tansfom it point b point to get the new shape. Stating with the lowe left cone, this tansfoms to (u 0,v 0 = [u( 0, 0,v ( 0, 0 ] (8 We can wite the geneal tansfomation as a vecto: (, = u(,î + v (,ĵ (9 Hee, is the tansfomed location of the oiginal point (,, witten with espect to the ectangula basis vectos. The idea now is to conside what happens as and tend to zeo. In this case, the tansfomed vesion of R tends to a paallelogam whose sides ae paallel to the tansfomation of the two sides of R that touch at the point P 0 = ( 0, 0 (the lowe left cone of R we mentioned above. Conside fist the edge of R along the line = 0 (the bottom of the ectangle. We can think of this edge as a tangent to the ectangle at the point P 0. How does this tangent tansfom? Well, the lowe edge of R tansfoms as (, 0 = u(, 0 î + v (, 0 ĵ (10 The tangent along this cuve is then the deivative with espect to, so we get (, 0 = u(, 0î + v (, 0ĵ (11 Thus the tangent along the bottom edge of R at the tansfomed location of P 0 is
COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT 3 (, 0 = =0 u(, 0 î + =0 v (, 0 ĵ (12 =0 B the same agument, the tangent at P 0 along the left edge of R is found b setting = 0 and diffeentiating with espect to, and we get ( 0, = =0 u( 0, î + =0 v ( 0, ĵ (13 =0 B the definition of a deivative, we can wite these tangents in the fom ( 0 +, 0 ( 0, 0 = lim 0 = lim 0 ( 0, 0 + ( 0, 0 (14 (15 The vecto ( 0 +, 0 ( 0, 0 connects the tansfomed lowe left cone of R to the tansfomed lowe ight cone. Similal ( 0, 0 + ( 0, 0 connects the lowe left cone to the uppe left cone. Thus these two vectos define the sides of a paallelogam that, fo ve small and, is a good appoimation to the tansfomed R. In this appoimation, we can wite ( 0 +, 0 ( 0, 0 (16 ( 0, 0 + ( 0, 0 (17 The aea of a paallelogam is A = s 1 s 2 sinθ, whee s 1 and s 2 ae two adjacent sides and θ is the angle between them. If we have two vectos coesponding to the sides, the aea is thus the magnitude of the coss poduct of the vectos. So we get A = (18 Using the equations above, we can wok out this coss poduct. We ll use the notation u u/ to save space. We get = î ĵ ˆk u v 0 u v 0 = (u v v u ˆk (19 The coefficient of ˆk is itself a 2 2 deteminant, and can be witten as
COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT 4 J(, u v u v (20 This is called the Jacobian of the tansfomation. The aea element is thus da = J (,dd (21 Now this is all ve well, but the diffeentials and ae still in the oiginal coodinate sstem. How can we use this esult to tansfom the integal that we began with? The tick is to assume that the tansfomation is invetible, that is, that we can also wite = (u,v (22 = (u,v (23 We can un though the same agument again to get da = J (u,vdudv (24 with That is: ˆ ˆ J (u,v = u u v v (25 ˆ ˆ f(, dd = f [(u,v, (u,v] J (u,v dudv (26 Note that we ve taken the absolute value of J since we e dealing with an aea element, which must be positive. It can also be shown that (the poof would make this post too long the Jacobian satisfies a ve convenient popet: 1 J (u,v = (27 J (, That is, the Jacobian of an invese tansfomation is the ecipocal of the Jacobian of the oiginal tansfomation. The Jacobian genealizes to an numbe of dimensions, so we get, eveting to ou pimed and unpimed coodinates:
COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT 5 J ( = 1 1 1 2 2 1. n 1 1 2 n 2 2. n.... n n 2 n Fo obvious easons, this can be abbeviated to (28 J = a (29 b As a simple eample, conside the tansfomation fom ectangula to pola coodinates in 2-d. Fom the above, the Jacobian we want is J (,θ which equies epessing the old coodinates in tems of the new ones. The tansfomation is = cos θ (30 = sin θ (31 So we have J (,θ = cosθ sinθ sinθ cosθ = (32 Thus the tansfomation of the aea element is dd ddθ (33 Fo the invese tansfomation, we have = 2 + 2 (34 θ = tan 1 (35 so J (, = 2 + 2 2 + 2 / 2 1+(/ 2 1/ 1+(/ 2 Thus J (u,v = 1/J (, as equied. In 3-d, = 1 2 + = 1 2 (36
COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT 6 = sinθ cosφ (37 = sinθ sinφ (38 z = cos θ (39 J (,θ,φ = sinθ cosφ cosθ cosφ sinθ sinφ sinθ sinφ cosθ sinφ sinθ cosφ cosθ sinθ 0 = 2 sinθ (40 Fo the invese: = 2 + 2 + z 2 (41 θ = tan 1 2 + 2 (42 z φ = tan 1 (43 J (,,z = 2 + 2 +z 2 2 + 2 +z 2 / (z 2 + 2 / (z 2 + 2 1+( 2 + 2 /z 2 / 2 1/ 0 1+(/ 2 1+(/ 2 z 2 + 2 +z 2 1+( 2 + 2 /z 2 2 + 2 /z 2 1+( 2 + 2 /z 2 (44 Conveting back to spheical coodinates poves a bit easie. Substituting the above tansfomation equations, along with helps to simplif things. 2 sin 2 θ = 2 + 2 (45 J (,,z = z 3 sinθ 2 sin 2 θ z 3 sinθ 2 sin 2 θ z sinθ 0 (46 The deteminant now comes out to
COORDINATE TRANSFORMATIONS - THE JACOBIAN DETERMINANT 7 J (,,z = ( sinθ 2 sin 2 θ 1 = 4 sin 2 θ z z 3 sinθ ( sinθ 2 sin 2 θ z z 3 sinθ (47 ( (sinθ 2 + z2 2 + ( (sinθ 2 + z2 sinθ 2 sinθ = 2 + 2 (sinθ 4 sin 2 + z2 θ 2 sinθ = 1 ( 2 sinθ + z2 2 sinθ = 1 ( 2 sin 2 θ + z 2 2 2 sinθ = 1 ( 2 + 2 + z 2 2 2 sinθ = 1 2 sinθ PINGBACKS Pingback: Contavaiant tensos Pingback: Covaiant and mied tensos Pingback: Noethe s theoem and consevation laws (48 (49 (50 (51 (52 (53