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Tensor Calculus
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EXAMPLE 4-8 Find the exterior derivative of the following one forms: = e f (r ) dt SOLUTION 4-8 In the rst case, we obtain d = d e f (r ) dt = e f (r ) dr dt = f (r ) e f (r ) dr dt r and = e g(r ) cos ( ) sin ( ) dr
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For the second one form, we get d = d e g(r ) cos ( ) sin ( ) dr = g (r )e g(r ) cos ( ) sin ( ) dr dr e g(r ) sin ( ) sin ( ) d dr + e g(r ) cos ( ) cos ( ) d dr = e g(r ) sin ( ) sin ( ) d dr + e g(r ) cos ( ) cos ( ) d dr = e g(r ) sin ( ) sin ( ) dr d + e g(r ) cos ( ) cos ( ) d dr Note that we used dr dr = 0, and on the last line we used (4.20) to eliminate the minus sign.
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The Lie Derivative
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The Lie derivative is de ned by L V W = V b b W a W b b V a The Lie derivative of a (0, 2) tensor is L V Tab = V c c Sab + Scb a V c + Sac b V c In 8, we will study Killing vectors, which satisfy L K gab = 0 (4.29) (4.28) (4.27)
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Tensor Calculus
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The Absolute Derivative and Geodesics
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Consider a curve parameterized by . The absolute derivative of a vector V is given by DV a dV a = + D d
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a bc u b
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(4.30)
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where u is a tangent vector to the curve. The absolute derivative is of interest in relativity because it allows us to nd geodesics. We will see later that relativity theory tells us that free-falling particles (i.e., particles subject only to the gravitational eld) follow paths de ned by geodesics. A geodesic can be loosely thought of as the shortest distance between two points. If a curve is a geodesic, then a tangent vector u satis es Du a = u a D where the tangent vector is de ned by u a = a Using u a = dx with (4.30), we obtain d d2 x a + d 2
a bc dx a d
(4.31) and is a scalar function of .
dx a dx b dx c = d d d
(4.32)
We can reparameterize the curve and can do so in such a way that = 0. When we use a parameter for which this is the case we call it an af ne parameter. If we call the af ne parameter s, then (4.32) becomes d2 x a + ds 2
dx b dx c =0 ds ds
(4.33)
Geometrically, what this means is that the vector is transported into itself. (see Figure 4-2). That is, given a vector eld u, vectors at nearby points are parallel and have the same length. We can write (4.33) in the more convenient form: d2 x a + ds 2
dx b dx c =0 ds ds
(4.34)
Tensor Calculus
Fig. 4-2. A vector u is parallel transported along a curve if the transported vector is
parallel to the original vector and is of the same length.
The solutions to this equation, which give the coordinates in terms of the af ne parameter, x a = x a (s), are the geodesics or straightest possible curves for the geometry. EXAMPLE 4-9 Find the geodesic equations for cylindrical coordinates. SOLUTION 4-9 The line element for cylindrical coordinates is ds 2 = dr 2 + r 2 d 2 + dz 2 It is easy to show that the only nonzero Christoffel symbols are
r r
= r =
r
Using (4.34), we nd the geodesic equations. First we set the indices a = r, b = , c = to get d2 r r ds 2 d ds