Deriving the Equations in Minguzzi's Explicit Time Expression

Introduction

Minguzzi considered what he called the reconstruction problem during acceleration in special relativity. For 1+1 dimensions this requires that the accelerated observer O have an accelerometer that records his acceleration Vs time along his spatial direction. His ansatz is that the acceleration, a, can be expressed as

$a = \frac{d}{d t} γ v = \frac{d}{d t} (\frac{v}{\sqrt{1 - v^{2}}})$

(1.1)

where v is speed it has been assumed that c=1

Instead of using this equation directly to compute the speed Vs time he reverts to the Lorentz invariant value a² by taking the square of the time-like component and subtracting the square of the space-like component and obtains the following result

$- a^{2} = {(\frac{d}{d t} (γ))}^{2} - {(\frac{d}{d t} (γ v))}^{2}$

(1.2)

where $γ = 1 / \sqrt{1 - v^{2}}$ .

Taking the differentials we get:

$a^{2} = \frac{1}{{(1 - v^{2})}^{2}} {(\frac{d v}{d t})}^{2}$

(1.3)

which is quite different from the result that we would get using equation (1.1) which is

$a^{2} = \frac{1}{{(1 - v^{2})}^{3}} {(\frac{d v}{d t})}^{2}$

(1.4)

Solving equation (1.3) for v we obtain:

$\int_{0}^{v} \frac{d v'}{1 - v'^{2}} = \tanh^{- 1} (v) = \int_{0}^{t} a (t') d t'$

(1.5)

and therefore

$v (t) = \tanh [\int_{0}^{t} a (t') d t' + \tanh^{- 1} (v_{0})] \equiv \tanh (θ)$

(1.6)

Then the change in the time like component is just the integral $δ τ = \int_{0}^{t} \cosh [θ (t')] d t'$ and the change in space like component is $δ x = \int_{0}^{t} \sinh [θ (t')] d t'$ .

In order to avoid problems of simultaneity at a large distance, Minguzzi chooses to compute the invariant T²(t) which is the difference between the squares of $δ τ$ and $δ x$ or

$T^{2} = δ τ^{2} - δ x^{2}$

(1.7)

This expression turns out to be very simple since $δ τ^{2} - δ x^{2} = (δ τ - δ x) (δ τ + δ x)$ and we may form the components of this product inside the integrals:

$δ τ - δ x = \int_{0}^{t} {\cosh [θ (t')] - s i n h [θ (t')]} d t' = \int_{0}^{t} \exp [- θ (t')] d t'$

(1.8)

$δ τ + δ x = \int_{0}^{t} {\cosh [θ (t')] + s i n h [θ (t')]} d t' = \int_{0}^{t} \exp [θ (t')] d t'$

(1.9)

and therefore

$T^{2} = \int_{0}^{t} \exp [θ (t')] d t' \int_{0}^{t} \exp [- θ (t')] d t'$

(1.10)

Differential Aging Ratio

We would like to derive an expression for dT/dt from our expression for T².

$2 T \frac{d T}{d t} = \frac{d}{d t} {\int_{0}^{t} \exp [θ (t')] d t' \int_{0}^{t} \exp [- θ (t')] d t'}$

(1.11)

$\begin{array}{l} \frac{d T}{d t} = \frac{1}{2 T} \frac{d}{d t} {\int_{0}^{t} \exp [θ (t')] d t' \int_{0}^{t} \exp [- θ (t')] d t'} = \\ \frac{1}{2 T} {\exp [θ (t)] \int_{0}^{t} \exp [- θ (t')] d t' + \exp [- θ (t)] \int_{0}^{t} \exp [θ (t')] d t'} \end{array}$

(1.12)

$\begin{array}{l} \frac{d T}{d t} = \\ \frac{1}{2} \frac{{\exp [θ (t)] \int_{0}^{t} \exp [- θ (t')] d t' + \exp [- θ (t)] \int_{0}^{t} \exp [θ (t')] d t'}}{\sqrt{(\int_{0}^{t} \exp [- θ (t')] d t') (\int_{0}^{t} \exp [θ (t')] d t')}} \end{array}$

(1.13)

$\frac{d T}{d t} = \frac{1}{2} {\exp [θ (t)] \sqrt{\frac{\int_{0}^{t} \exp [- θ (t')] d t'}{\int_{0}^{t} \exp [θ (t')] d t'}} + \exp [- θ (t)] \sqrt{\frac{\int_{0}^{t} \exp [θ (t')] d t'}{\int_{0}^{t} \exp [- θ (t')] d t'}}}$

(1.14)

To simplify this expression let

$b = \sqrt{\frac{\int_{0}^{t} \exp [- θ (t')] d t'}{\int_{0}^{t} \exp [θ (t')] d t'}}$

(1.15)

and note that $b = \exp [\ln (b)]$ Then equation (1.14) can be expressed as

$\begin{array}{l} \frac{d T}{d t} = \frac{1}{2} {b \exp [θ (t)] + \exp [- θ (t)] / b]} = \\ \frac{1}{2} {\exp [θ (t)] e x p [l n (b)] + \exp [- θ (t)] e x p [- l n (b)]} = \\ \frac{1}{2} {\exp [θ (t) + l n (b)] + \exp [- θ (t) - l n (b)]} \\ = c o s h [θ (t) + l n (b)] \end{array}$

(1.16)

Rewriting equation (1.16) with its full expressions for θ and b we have:

$d T / d t = c o s h [\int_{0}^{t} a (t') d t' + \frac{1}{2} \ln (\frac{\int_{0}^{t} \exp [- \int_{0}^{t'} a (t'') d t''] d t'}{\int_{0}^{t} \exp [\int_{0}^{t'} a (t'') d t''] d t'})]$

(1.17)

T is an invariant so it is important to know what this invariant means. It just means that parties in any inertial reference system that happens to be spatially alongside of the one that is being accelerated will find the same time between the acceleration starting event and any subsequent event.