At The Origin Of A Uniformly Accelerated One Spatial Dimensional Frame Of Reference.

Let us recall that the SOLT, the definition of the parameter p and an "invariance relationship",

were presented by equas.(26a), (26b), (27) and (28) of the original (link) preliminary paper.

These equations are reproduced below and are renumbered, for convenience,

in the present article.

X = p(x - (1/2)a*t^2) (eq.1a)

T = (p(t^2 - 2(a^2/L^3)x))^0.5 (eq.1b)

where p is given by

p = 1/(1 - (a/L)^3)^(2/3) (eq.2)

The invariance relationship

(X^3 - ((L/2)^3)*T^6)^2 = (x^3 - ((L/2)^3)*t^6)^2 (eq.3)

will now be shown to be satisfied when x = (1/2)at^2.

But first we will recall the informal discussion of the physical interpretation

of the SOLT. The frames of reference r (with coordinates (x,t) ) and

the frame of reference R (with coordinates (X,T)) are both to be

considered as consisting of one spatial dimension and a time

dimension. At the point in space - time

(x, t) = (X, T) = (0, 0) (eq.4)

the R frame is moving with initial velocity = zero and constant

acceleration = a in a direction that is parallel to the x - axis of the

r frame. The parameter L has the physical dimensions of acceleration

and is associated with optical phenomena.

If an observer is standing at the origin of the (X, T) then his spatial

coordinate X is zero. So we have X = 0 and from (eq.1a) it follows,

in that case, that

x = (1/2)at^2 (eq.5)

and also, with X = 0, (eq. 3) reduces to

(((L/2)^3)*T^6)^2 = (x^3 - ((L/2)^3)*t^6)^2 (eq.6)

Equa.(6) may be simplified as follows: let

W = (1/2)LT^2 (eq.7a)

w = (1/2)Lt^2 (eq.7b)

Therefore (eq.6) may be written as

(W^3)^2 = (x^3 - w^3)^2

W^6 = (x^3 - w^3)^2 (eq.8)

Now to verify (eq.8) we start off by squaring both sides of (eq.1b) to obtain

T^2 = p(t^2 - 2(a^2/L^3)x)

now multiply both sides of the above equation by (1/2)L to give us

(1/2)LT^2 = p((1/2)Lt^2 - (a^2/L^2)x)

but x = (1/2)at^2 from (eq.5) so we may write

(1/2)LT^2 = p((1/2)Lt^2 - (a^2/L^2)(1/2)at^2)

(1/2)LT^2 = ((1/2)t^2)p(L - (a^2/L^2)a)

1/2)LT^2 = ((1/2)t^2)p(L - (a^3/L^2))

1/2)LT^2 = ((1/2)Lt^2)p(1 - (a^3/L^3))

Now making use of (eq.7a) and (eq.7b), the equation above becomes

W = wp(1 - (a^3/L^3)) (eq.9)

If we write

A = a/L (eq.10)

then (eq.9) becomes

W = wp(1 - A^3) (eq.11)

By making use of (eq.10), we may also simplify (eq.2) as follows:

p = 1/(1 - A^3)^(2/3) (eq.12)

Cubing both sides of (eq.11) will give us

W^3 = (w^3)(p^3)(1 - A^3)^3 (eq.13)

Making use of (eq.12) we have

p^3 = 1/(1 - A^3)^2

substituting this result into (eq.13) gives us

W^3 = w^3(1 - A^3) so that

W^6 = w^6(1 - A^3)^2 (eq.14)

Recalling (eq.5) and (eq.7b) we have x = (1/2)at^2 and w = (1/2)Lt^2

respectively. We may therefore write

x^3 - w^3 = ((1/2)at^2)^3 - ((1/2)Lt^2)^3

x^3 - w^3 = ((1/2)t^2*(a))^3 - ((1/2)Lt^2)^3

x^3 - w^3 = ((1/2)Lt^2*(a/L))^3 - ((1/2)Lt^2)^3

recalling that we set A = a/L

x^3 - w^3 = ((1/2)Lt^2)^3*A^3 - ((1/2)Lt^2)^3

x^3 - w^3 = ((1/2)Lt^2)^3(A^3 - 1)

x^3 - w^3 = - ((1/2)Lt^2)^3(1 - A^3 )

Now recall that w = (1/2)Lt^2

x^3 - w^3 = - (w)^3(1 - A^3 )

x^3 - w^3 = - w^3(1 - A^3 )

(x^3 - w^3)^2 = w^6(1 - A^3 )^2 (eq.15)

Comparing the right sides of (eq.14) and (eq.15) we are led to the

conclusion that

W^6 = (x^3 - w^3)^2 (eq.16)

as asserted in (eq.8)

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