The mathematics of clock rates

Suppose that you buy an alarm clock and then discover that the clock loses ten minutes every day. How would Newton or Galileo have expressed the time of that clock mathematically?
They would have used the Galilean transformation equations twice, once for the rate of a correct clock and once for the rate of the incorrect slower clock. This would show how the rate of the slower clock relates to the rate of a correct clock. Suppose that t is the time of a correct clock in frame of reference S.

x'=x-vt
y'=y
z'=z
t'=t

t'=t shows that the time of the correct clock is being used in both frames of reference. v is the velocity of S' relative to S according to the time of the correct clock.
Since t' is already defined to be equal to t, we cannot use that variable to show the time of the incorrect clock. We have to use a different variable to show the time of that clock.

x' = x - (v2)(t2)
y'=y
z'=z
(t2)' = (t2)

The equation for time shows that the time of the incorrect clock is being used in both frames of reference. (v2) is the velocity of S' relative to S according to the time of the incorrect clock.
Now we apply this same principle to atomic or scientific time. The difference between this and the incorrect clock equations is that scientific time is correct time. In other words, if we are considering incorrect time, the speedometer of a car will agree with a clock that shows correct time, not with a clock that shows incorrect time, while in scientific time, a second is based on a certain number of transitions of a cesium atom, so a slower or faster cesium clock is still correct. In other words the speed of S' as observed from S' agrees with a clock in S', not with a clock in S. This is a different interpretation than the one being used by scientists, which says that the speed between frames of reference S and S' is the same measured from either frame of reference. We use two sets of Galilean transformation equations as with the example of the incorrect clock, but they are applied differently.

x'=x-vt
y'=y
z'=z
t'=t

t is the time of a clock in S. t'=t shows that the time of that clock is being used in both frames of reference. v is the velocity of S' relative to S according to the time of the clock in S.

x = x' - (v2)(t2)'
y = y'
z = z'
(t2) = (t2)'

(t2)' is the time of the clock in S', which could be faster or slower than the clock in S, depending on conditions of motion, gravitation, etc. (v2) is the velocity of S relative to S' according to the time of the clock in S'.
Now we apply the results of the Michelson-Morley experiment to these two sets of transformation equations. We know from that experiment that light is traveling at a rate of c in both frames of reference. In other words, the time of a clock in S shows light to be traveling at c, and a clock in S' shows light to be traveling at c. We have already determined by experiment that the two clocks have different rates, just as Einstein predicts in his theory. We can show the constancy of the speed of light by two equations, x = ct and x' = c(t2)'. But if we consider a photon going the opposite direction, the equations would be x = - ct and x' = - c(t2)' because the photon would have a negative velocity relitive to S and S'. We are going to resolve this difficulty the same way Lorentz did, by keeping the speed of light squared wherever it occurs in the equations, but we will solve the equations using velocity of light as shown by the variable w. So our two equations are now x=wt and x' = w(t2)'.

x'=x-vt
w(t2)' = wt - vt
(t2)' = t - vt/w = t - vx/w^2 = (t-vx/c^2)

We can now show why the Lorentz equations work when adding velocities using scientific time.

u=dx/dt
u'=dx'/dt'

x' = (x-vt)gamma
dx' = (dx - v dt)gamma
= dt(dx/dt - v)gamma
= dt(u-v)gamma

t' = (t-vx/c^2)gamma
dt' = (dt - v dx / c^2)gamma
= dt(1 - v dx/dt /c^2)gamma
= dt(1 - vu/c^2)gamma


u' = dx'/dt' = (u-v)/(1-vu/c^2)

Now we figure the same velocities as obtained from our two sets of Galilean transformation equations.

u=dx/dt
u' = dx'/dt'

x'=x-vt
dx' = dx-vdt
= dt(dx/dt - v) = dt(u-v)

(t2)' = (t-vx/c^2)
d(t2)' = (dt - v dx/c^2)
d(t2)' = dt(1 - v dx/dt / c^2) = dt(1 - vu/c^2)

u' = dx'/d(t2)' = (u-v)/(1-vu/c^2)

So we get the same addition of velocites using scientific time if we use the Galilean transformation equations twice keeping the velocity of light constant as we do with the Lorentz equations.