My question pertains to the concept of a constantly accelerating rocket as it approaches the speed of light. The scenario is as follows:

A rocket is constantly accelerating at 1g to reach Andromeda. Assume travel is in only one direction and no external forces act on the rocket e.g. gravity. It's just the rocket. Halfway there it begins to constantly decelerate at 1 g, the change takes no time or energy. Find:
- the maximum speed
- the time measured in the rocket's frame
- the time measured by someone in a stationary frame (Earth).

The biggest issue is figuring out how the time dilation works because of the constantly changing Lorentz factor. How does the relativistic acceleration work in terms of time dilation? Please know that I am looking for the process as in, what does the derivation look like? Just giving the equations without explanation is not the understanding I am looking for.

I have knowledge of mathematics up to Calculus 2 and have a working knowledge of the other standard areas of physics study (except for relativity!).


Let $S$ denote an inertial frame, and let $S'$ denote the rocket frame.

Take, first, the case of zero acceleration where as viewed in $S$, the rocket frame moves at velocity $v$ in a straight line. If a clock that is stationary in the rocket frame measures an amount of time $\Delta t'>0$ between two events, then a clock in the inertial frame $S$ will measure an amount of time \begin{align} \Delta t = \gamma\Delta t' \end{align} where the factor $\gamma$, often called "relativistic gamma" is defined as \begin{align} \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \end{align} and is constant when $S'$ is not accelerating. Now, we could ask,

Is an analogous expression relating time intervals in the two frames valid even when the rocket is accelerating?

Well, the answer to this is a bit tricky. If we try to blindly apply the formula, then we see that there is an ambiguity: which $\gamma$ would we use? Since the rocket frame is accelerating, it's gamma factor is constantly changing. However, if we pick a sufficiently small period of time, then we see that the gamma factor doesn't actually change very much, so we might be tempted to say that in the limit of vanishingly small time intervals $dt'$ and $dt$, we still have the relation \begin{align} dt = \gamma \,dt' \end{align} where here $\gamma$ is the value of gamma the rocket has during this "infinitesimal" time interval. It turns out that this is basically correct. In fact, we can make this more mathematically precise. To determine the time interval $\Delta t'$ between two events measure measured by someone in the rocket, we integrate \begin{align} \Delta t' = \int_{t_1}^{t_2} dt' = \int_{t_1}^{t_2} \frac{dt}{\gamma} \end{align}

Here is also the more mathematically highbrow way of summarizing these facts:

If $x^\mu(\lambda)$ is a parameterized curve in Minkowski space as measured by some inertial observer $S$, then the amount of time $\mathrm{time}(\lambda_1, \lambda_2)$ measured by an observer that is stationary in the rocket between spacetime points $x^\mu(\lambda_1)$ and $x^\mu(\lambda_2)$ along its path is given by \begin{align} \mathrm{time}(\lambda_1, \lambda_2) = \int_{\lambda_1}^{\lambda_2}d\lambda\, \sqrt{\eta_{\mu\nu}\dot x^{\mu}(\lambda)\dot x^\nu(\lambda)} \end{align}

where overdots denote derivatives with respect to $\lambda$, and $\eta_{\mu\nu} = (+1, -1, -1, -1)$ is the Minkowski metric.

I'd like to note that the reason this integration procedure works is an extra physical fact about the way the universe works that does not follow directly, mathematically from the Lorentz transformation, a transformation which holds between inertial frames. In fact, some people often call this fact the clock postulate. In fact, I posted my own question about this a while ago that you might find interesting:

Why do clocks measure arc-length?

You will probably also find this illuminating; John Baez discussing the affects of acceleration on clocks in the context of special relativity:


  • $\begingroup$ The integral could do with a change of variable since $\gamma$ is a function of $v$ $\endgroup$ – John McVirgo Aug 26 '13 at 23:10
  • $\begingroup$ @JohnMcVirgo Not sure what you mean. The notation I use then I integrate $dt/\gamma$ is standard shorthand in physics in the sense that one really means $dt/\gamma(v(t))$ where $\gamma(v) = (1-v^2/c^2)^{-1/2}$. The more mathematically (notationally) precise form the of the integral statement is made in the boxquote below. Are you simply suggesting I be more descriptive with the notation and write $\int_{t_1}^{t_2} \frac{dt}{\gamma(v(t))}$? Is there a more significant error I am missing? $\endgroup$ – joshphysics Aug 26 '13 at 23:20
  • $\begingroup$ The question only asked for clarification on the time dilation, so on second thoughts my suggestion unnecessarily complicate things. I was thinking that $dv/dt = \gamma^3dv'/dt'$. The question says the proper acceleration is constant so $dt = dv/(a'\gamma^3)$ giving an integral you can evaluate in terms of the start and final velocities. $\endgroup$ – John McVirgo Aug 27 '13 at 13:27

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