at what constant speed should I travel one light-second to make my time and a stationary person's time 1 second off?

  • $\begingroup$ Depends how fast you traveled that distance… $\endgroup$
    – Jon Custer
    Jun 20, 2022 at 23:22
  • $\begingroup$ @joncustor - is it ? if i travel slower it takes me longer to get there ? $\endgroup$
    – commonpike
    Jun 21, 2022 at 7:19
  • $\begingroup$ It took few days for Apollo missions to pass the lightsecond distance. $\endgroup$
    – Poutnik
    Jun 26, 2022 at 9:40
  • 1
    $\begingroup$ I think that the revisions are suitable and recommend reopening $\endgroup$
    – Dale
    Jun 26, 2022 at 12:29

1 Answer 1


It will depend what speed you are travelling at. However, dealing purely with time dilation effects from special relativity, if it takes you $\tau$ amount of time to travel a certain distance as measured in your inertial frame, with an observer at rest wrt to the Earth measuring your velocity as $v$ then that observer will measure the time of travel to be

\begin{equation} t = \gamma\tau \end{equation}

where $\gamma = (1 - v^{2}/c^{2})^{-\frac{1}{2}}$. Therefore, the difference in time measured will generally depend on your speed.

However, since $t = d/v$

\begin{equation} \begin{split} \gamma = (1- d^{2}/c^{2}t^{2})^{-\frac{1}{2}} \implies \\ t = \frac{\tau}{\sqrt{1- d^{2}/c^{2}t^{2}}} \implies \\ t^{2} - \frac{d^{2}}{c^{2}} = \tau^{2} \implies \\ c^{2}t^{2} - d^{2} = c^{2}\tau^{2} \end{split} \end{equation}

where the final equality is an expression for the spacetime invariant, $ s^{2} = c^{2}t^{2} - x^{2}$, which is constant between frames.

  • $\begingroup$ i notice speed is not a variable in the end result, but there does not seem to be a linear relation between t and d - is that right ? $\endgroup$
    – commonpike
    Jun 21, 2022 at 7:22
  • $\begingroup$ So, the final result is that the spacetime interval between two events is the same in all inertial frames of reference meaning that it will necessarily not depend on any observer's velocity. There is certainly a linear result between $t$ and $d$, namely $t = d/v$. However, there will not be a nice expression for $t - \tau$ which is what you are asking about in the original question. $\endgroup$
    – Niall
    Jun 21, 2022 at 7:54
  • $\begingroup$ Sorry, I meant a linear relation between d and Δt $\endgroup$
    – commonpike
    Jun 25, 2022 at 9:45
  • $\begingroup$ Staring at this, isn't it correct that as per the question, d==c ? they have different units, but the numbers are the same ? Would it be possible to write Δt as a function of 𝜏 ? I was assuming Δt was a constant, but from what I am getting, it depends on 𝜏 , meaning, it does depend on the speed I traveled that distance in. $\endgroup$
    – commonpike
    Jun 25, 2022 at 10:00
  • 1
    $\begingroup$ From the penultimate line of the block of equations you would get $t^{2} - t_{year}^{2} = \tau^{2} \implies t^{2} - \tau_{2} = t_{year}^{2} \implies (t - \tau)(t+\tau) = t_{year}^{2} \implies \Delta t(\Delta t + 2\tau) = t_{year}^{2}$. However you arrange this, it's not going to give you a nice expression for $\Delta t$ $\endgroup$
    – Niall
    Jun 26, 2022 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.