r 2 = 1 (cos2 ) = cos2 cot2 = r2 r 2 sin2 in .NET

Encoder QR Code ISO/IEC18004 in .NET r 2 = 1 (cos2 ) = cos2 cot2 = r2 r 2 sin2

EXAMPLE 2-3 In this example, we consider a ctitious two-dimensional line element given by ds 2 = x 2 dx 2 + 2dxdy dy 2 Write down gab , g ab and then raise and lower indices on Va = (1, 1), W a = (0, 1). SOLUTION 2-3 With lowered indices the metric can be written in matrix form as gab = gx x g yx gx y g yy

We see immediately from the coef cients of dx 2 and dy 2 in the line element that gx x = x 2 and g yy = 1

To obtain the terms with mixed indices, we use the symmetry of the metric to write gx y = g yx and so 2dxdy = dxdy + dydx = gx y dxdy + g yx dydx Therefore, we nd that gx y = g yx = 1 In matrix form, the metric is gab = x2 1 1 1

To obtain the inverse, we use the fact that multiplying the two matrices gives the identity matrix. In other words, gx x g yx gx y g yy x2 1 1 1 = 1 0 0 1

Four equations can be obtained by carrying out ordinary matrix multiplication in the above expression. These are gx x x 2 + gx y = 1 gx x gx y = 0 gx y = gx x g x x x 2 + g yy = 0 g yy = x 2 g x x g yx g yy = 1 In addition, the symmetry of the metric provides the constraint g x y = g yx . Using g x y = g x x in the rst equation, we nd g x x x 2 + g x x = g x x (1 + x 2 ) = 1 1 gx x = 1 + x2 and so gx y = gx x = 1 = g yx 1 + x2

Now with this information in place, we can raise and lower indices as desired. We nd V a = g ab Vb 1 1 (1) + ( 1) = 0 2 1+x 1 + x2 1 x2 1 + x2 V y = g yb Vb = g yx Vx + g yy Vy = (1) ( 1) = =1 1 + x2 1 + x2 1 + x2 V a = (0, 1) V x = g xb Vb = g x x Vx + g x y Vy = In the other case, we have Wa = gab W b 1 1 1 (0) + (1) = 2 2 1+x 1+x 1 + x2 x 2 1 x 2 (1) = W y = g yb W b = g yx W x + g yy W y = (0) + 1 + x2 1 + x2 1 + x2 2 x 1 Wa = , 2 1 + x2 1+x Wx = gxb W b = gx x W x + gx y W y =

Often we raise and lower indices with the metric in a more abstract fashion. The reason for doing this is to get equations in a more desirable form to derive some result, say. We will be seeing more of this later, so it will make more sense as we go along. Right now we will just provide a few examples that show how this works. We ve already seen a bit of this with an ordinary vector X a = g ab X b We can also apply this technique to a vector that is present in a more complicated expression, such as X a Y c = g ab X b Y c