, however, that the unit of t' is c/( (sec). The light speed in the x'-direction is

= c() = (((cm/sec). (19b)

Thus, the light speed remains (. If ( = 2c, Metric (19a) implies that the light speed would be 6 x 1010 cm/sec; and metric (19b) implies that the light speed is 3 x 1010 cm/half-sec.
In the literature, the units of the coordinates are usually not specified. Then, the distinct metric (18a) would be confused with a rescaling of the Minkowski metric. This creates a illusion that all constant metrics were equivalence in physics. This example illustrates also that it is invalid to ?efine" light speed in terms of local Minkowski spaces in a manifold.
Moreover, if the equivalence principle were valid, according to Einstein? approach, one would obtain

c2dT2 = (2dt2 , and (dX2 + dY2 + dZ2) = (dx2 + dy2 + dz2), (20a)

for a resting observer at a point (x0, y0, z0, t0). Eq. (20a) and ds2 = 0 imply that the light speed is

= = 1 (20b)

Eq. (20b) implies, however, that the light speed is c in the local Minkowski coordinate, but is ( ((2c) in the (x, y, z, t) space. On the other hand, since there is no gravitational force for this case, we can have also

x = X, y = Y, and z = Z (21)

Eq. (20b) and eq. (21) absurdly mean that for the same frame of reference, we have different light speeds. This is in disagreement with the principle of uniqueness for a physical measurement.
Example 2, consider the Minkowski flat metric and the transformation, which is a diffeomorphism,

t = C[exp(T/C) - exp(-T/C)]/2, where C = constant. (22a)
Then
ds2 = [exp(T/C) + exp(-T/C)]2dT2 - dx2 - dy2 - dz2, (22b)

is the metric transformed from the Minkowski metric. If metric (22b) is realizable, according to ds2 = 0, the measured light speed would be [exp(T/C) + exp(-T/C)]/2.
From (22b), the Christoffel symbols ((,(( are zeros except (t,tt = (tgtt/2. Then, according to the geodesic equation, the equation of motion for a particle at (x, y, z, T) is

+ (t,tt= 0, and === 0 (23a)
where
(t,tt = {ln[exp(T/C) + exp(-T/C)]}.

It follows eq. (23a) that one obtains, for some constant k

dT/ds = k[exp(T/C) + exp(-T/C)]-1 and dx(/ds = Constant (23b)

Now, consider the case dx/dT = dy/dT = dz/dT = 0. For this case, one has dx/ds = dy/ds = dz/ds = 0 and dx2/dT2 = dy2/dT2 = dz2/dT2 = 0. Thus, in such a ?ree fall", there is no change in the spatial position or acceleration. Physically, this means that such an observer would have the same frame of reference, whether ?ree fall" or not. Thus, he would absurdly have two different light speeds from the same frame of reference. Accordingly, the equivalence principle is not satisfied and metric (22) is not realizable. Note that, from metric (22), there is no acceleration for a static particle.
Nevertheless, some theorists would disregard all these inconsistency because they believe that space-time coordinates have no physical meaning. Therefore, they also disagree with Einstein [2,3] and regard that coordinate light speeds as meaningless.


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