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6. Incompatibility of the Galilean Transformation to the Equivalence Principle
It will be shown that a Galilean transformation, which is unrealizable, is incompatible with the equivalence principle. Some theorists, however, considered incorrectly that the Galilean transformation would lead to a space-time coordinate system. The root of their problem is that they mistaken the existence of the tetrad as equivalent to a satisfaction of the equivalence principle. They do not understand that a satisfaction of this principle requires the geodesic of ree fall" must be valid in physics.
Consider the Galilean transformation from (x, y, z, t) to the K' coordinates,
t = t', x = x', y = y', and z = z' - vt', (24a)
where v is a constant. Eq. (24a) transforms metric (10a) to another constant Lorentz metric
ds2 = ((c - v)dt' + dz' ( ((c + v)dt' - dz'( - dx'2 - dy'2, (24b)
Metric (24b) is a special case of a space with an indefinite metric. Then, for light rays in the z'-direction, ds2 = 0 would imply at any point the light speeds were
dz'/dt' = c + v, or dz'/dt' = -c + v. (25)
Clearly, ight speed" (25) violates coordinate relativistic causality (i.e. no cause event can propagate faster than the velocity of light in a vacuum). Thus, metric (24b) is not physically realizable, and those in (25) are not coordinate light velocities.
Moreover, according to the geodesic equation (2), metric (24b) implies d2x'(/ds2 = 0, and thus
dx'(/ds = constant. where x'( (= x', y', z', or t') (26a)
at any point. Now, according to metric (24), consider the case of a ree fall" at (x'0, y'0, z'0, t'0)
dx'/ds = dy'/ds = dz'/ds = 0, and dt'/ds = (c2 - v2)-1/2 (26b)
Since there is no acceleration or motion, such a ree falling" observer P' carries with him the frame of reference K'. Since a ree fall" does not automatically obtain a local Minkowski space, point 2) of the equivalence principle is violated. Also, for observer P', according eq. (1) the measured light speed is c, but according (24b) the light speed in the x-direction is (c2 - v2)1/2. This inconsistency also implies that point 3) of the equivalence principle is not satisfied in K'.
Nevertheless, mathematics ensures the existence of a local Minkowski space, which can be obtained by choosing first the path of a particle to be the time coordinate and then the other three space coordinates by using orthogonality. According to condition (26b), the time coordinate would remain the same dt'. But, the coordinate dz' is not orthogonal to dt'. Now, let us work out the local orthogonal tetrad of P', whose direction vP' is (0, 0, 0, dt'). Then, the orthonormal vectors of the tetrad are
a1 = (1, 0, 0, 0), a2 = (0, 1, 0, 0), a3 = (0, 0, (, (), and bp' = (0, 0, 0, () (27a)
where
( = ( -1, ( = -( v/c2, and ( = (c2 - v2)-1/2.
The corresponding transformations is as follows:
dt' = ( (dT - v/c2 dZ) , dz' = ( -1dZ, dx' = dX, and dy' = dY. (27b)
Thus, (dx', dy', dz') and (dX, dY, dZ) share the same frame of reference since there is no acceleration. But, there is a space measurement change in the z-direction. Metric (24b) does not satisfy point 2) of the equivalence principle since there is no physical cause for transformation (27b). In relativity, such a physical transformation happens only when there is relative motion or acceleration. But, P' is rest at K'. Thus, (27b) illustrates also that geodesic (26) does not represent a physical free fall. |
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