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How can I calculate the proper time of an accelerating frame of reference until it reaches a certain velocity relative to the stationary frame (before acceleration)?

For example, I depart from Earth on a spaceship under constant acceleration of 10 m/s2. How long do I need to keep accelerating (as measured on my clock), until an Earthbound observer sees me moving at .87 c relative to him? Or .99995 c? What would be the Earthbound frame time measured from start until the engine cutoff in these cases?

In classical mechanics, 10 m/s2 acceleration over 1000 seconds gives the same final speed relative to start as 100 m/s2 acceleration over 100 seconds, but I'm not sure if the same holds in special relativity and my speed vs. Earth would be the same for example with 10 m/s2 over 30 000 000 seconds as with 300 m/s2 over 1 000 000 seconds (acceleration and time measured using travelling clock).

Edit: To keep it simple, I'm only interested in special relativity point of view, not how the acceleration itself also affects the clock. I believe that with small values such as 1g the acceleration-induced time dilation can be ignored.

Tondo PX
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1 Answers1

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As seen in this lecture, the expression of velocity in terms of (constant) proper acceleration $\alpha$ is given by

$$v(\tau) = \frac{\alpha\tau}{\sqrt{1 + \frac{\alpha^2\tau^2}{c^2}}}$$

Where $v$ is the proper velocity and $\tau$ the proper time. If we set $v = \beta c$ for $0 \leq \beta \leq 1$ a percentage of the speed of light

$$\beta c\sqrt{1 + \frac{\alpha^2\tau^2}{c^2}} = \alpha\tau$$

Squaring both sides and rearranging we find

$$\tau = \frac{\beta c}{\alpha\sqrt{1 - \beta^2}}$$

If $\tau$ is your proper time, then we can find the proper time $t$ of an observer at rest with respect to you as

$$t = \gamma\tau$$

Given $\alpha = 10 \frac{\text{m}}{\text{s}^2}$ and $c = 3 \times 10^8\frac{\text{m}}{\text{s}}$, the solutions are:

  • For $\beta = 0.87$

$$\tau = \frac{0.87 \times 3 \times 10^8}{10 \times \sqrt{1 - 0.87^2}} = 52935622.91 \text{s}$$ $$t = \gamma \tau = 2.028 \times 52935622.91 = 107363225.0122\text{s} $$

  • For $\beta = 0.99995$

$$\tau = \frac{0.99995 \times 3 \times 10^8}{10 \times \sqrt{1 - 0.99995^2}} = 2999887498.8280416 \text{s}$$ $$t = \gamma \tau = 100.0013 \times 2999887498.8280416 = 299992499812.4827\text{s} $$

Antoniou
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