Special Relativity with Constant Acceleration

The differential equation that must be solved when we have linear acceleration is:

$d (γ v) = a d t$

(1.1)

where t is earth time and v is relative speed.

$(v \frac{d γ}{d v} + γ) d v = a d t$

(1.2)

$v \frac{d γ}{d v} = \frac{v^{2}}{c^{2} {(1 - \frac{v^{2}}{c^{2}})}^{\frac{3}{2}}}$

(1.3)

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

(1.4)

Integrating with respect to v letting $β = \frac{v}{c}$

$\int_{0}^{β} \frac{v^{2}}{c^{2} {(1 - \frac{v^{2}}{c^{2}})}^{\frac{3}{2}}} d v = \frac{c β}{\sqrt{1 - β^{2}}} - \sin^{- 1} β$

(1.5)

and

$\int_{0}^{β} \frac{d v}{\sqrt{1 - {(\frac{v}{c})}^{2}}} = \sin^{- 1} β$

(1.6)

resulting in the expression:

$\frac{c β}{\sqrt{1 - β^{2}}} = a t$

(1.7)

$c^{2} β^{2} = {(a t)}^{2} (1 - β^{2})$

(1.8)

Solving for β we get:

$β = \frac{a t}{c} \frac{1}{\sqrt{1 + {(\frac{a t}{c})}^{2}}}$

(1.9)

and

$γ = \frac{1}{\sqrt{1 + {(\frac{a t}{c})}^{2}}}$

(1.10)

The rocket time, T, is obtained by integrating γ

$T = \int_{0}^{t_{f}} \frac{d t}{\sqrt{1 + {(\frac{a t}{c})}^{2}}}$

(1.11)

$T = \frac{c}{a} \sinh^{- 1} \frac{a t_{f}}{c}$

(1.12)

and solving for t_f we get:

$t_{f} = \frac{c}{a} \sinh \frac{a T}{c}$

(1.13)

The distance as a function of t_f is:

$d_{f} = \int_{0}^{t_{f}} c β d t = \frac{c^{2}}{a} \int_{0}^{t_{f}} \frac{\frac{a t}{c}}{{\sqrt{1 + (\frac{a t}{c})}}^{2}} d (\frac{a}{c} t)$

(1.14)

Note that:

$\int_{0}^{x_{f}} \frac{x}{\sqrt{1 + x^{2}}} d x = \sqrt{1 + x_{f}^{2}} - 1$

(1.15)

Then $d x = \frac{a}{c} d t$ so the integral (1.14) result is

$d_{f} = \frac{c^{2}}{a} (\sqrt{1 + {(\frac{a t_{f}}{c})}^{2}} - 1)$

(1.16)

which for $\frac{a t_{f}}{c} > > 1$ becomes $d_{f} \approx c t_{f}$ and for $\frac{a t_{f}}{c} < < 1$ becomes $d_{f} = \frac{a t_{f}^{2}}{2}$ .

Solving equation (1.16) for t_f we obtain

$1 + {(\frac{a t_{f}}{c})}^{2} = {(1 + \frac{a d_{f}}{c^{2}})}^{2}$

(1.17)

$t_{f} = \frac{c}{a} \sqrt{{(1 + \frac{a d_{f}}{c^{2}})}^{2} - 1}$

(1.18)

so that for $\frac{a d_{f}}{c^{2}} < < 1$ we have $t_{f} = \sqrt{\frac{2 d_{f}}{a}}$ and for $\frac{a d_{f}}{c^{2}} > > 1$ we have $t_{f} = \frac{d_{f}}{c}$ .

Here is the question: do these results differ from the results where a time difference is obtained by the effective gravitational field which results from the acceleration? I think the latter assumes that the time involved in the acceleration is infinitesimal.