# Total time taken for an accelerating frame in special relativity [duplicate]

How does one derive the equation for the total time traveled in a constantly accelerating frame? I found some help at this question: Special Relativity and Constant Acceleration
But the information pertains to the measured time at a specific time along the way. I have taken the scenario from the question above, but am asking different questions:

A rocket is constantly accelerating at 1g to reach Andromeda (2.4 x 10^22 m) . Assume travel is in only one direction and no external forces act on the rocket, like gravity. The frames are in standard configuration. Halfway there, it begins to constantly decelerate at 1 g and the change takes no time or energy. Find:
- the maximum speed
- the total time of the journey measured in the rocket's frame
- the total time of the journey measured by someone on Earth

I have tried to do some work on the matter and have derived the following equations so far:

$$v_i ' = c^2 - \sqrt{\frac{(c^2-v_i^2)}{\gamma^2(1-v_x\frac{v}{c^2})^2}}$$
Where $v_i'$ is the instantaneous velocity as measured by the rocket, $v_i$ is the instantaneous velocity as measured by Earth, and $v_x$ is the velocity of travel along the x-axis as measured from Earth (the rocket travels in a straight line along the x-axis). And I have:
$$\frac{\gamma(v_i')}{\gamma(v_i)}= \gamma(v)\frac{c^2-v_xv}{c^2}$$ Where $\gamma$ is the Lorentz factor as a function. I am completely stuck here however! I do not know how to move forward with deriving the necessary equations to answer the questions above. I assume it has something to do with the rapidity functions, but I am not entirely well-versed in their use. I do have a working knowledge of calculus, which I assume is necessary.

## marked as duplicate by John Rennie, Emilio Pisanty, Qmechanic♦Aug 28 '13 at 8:53

• Unfortunately, the answer there does not quite explain the derivation or how to calculate the maximum speed achieved. Seems like I will have to get my hands on Gravitation though! – Guy Haley Aug 27 '13 at 12:10
• @GuyHaley Oh do be careful there. MTW (aka Gravitation) is not a textbook - it's an all-encompassing tome that's none too easy for most people to follow at first. Great reference source, but don't expect it to lay things out in the most clear and concise way. – user10851 Aug 27 '13 at 15:47

If $g$ is the constant proper acceleration of the rocket, and $t$ the coordinate time measured by an observer on Earth, then the velocity of the rocket in the Earth frame is (during the first half of the trip) $$v = \frac{gt}{\sqrt{1 + g^2t^2/c^2}},$$ assuming that the initial velocity is zero. See this post for a derivation, and this post for more info about proper acceleration. Integrating this gives us the travelled distance $x$, measured by an observer on Earth: $$x = \frac{c^2}{g}\left(\sqrt{1 + g^2t^2/c^2}-1\right).$$ Insert $x=D/2$ and you find the time needed for half the trip, and the velocity at that time.

The corresponding proper time $\tau$ on board the rocket is found by integrating $$\text{d}\tau = \sqrt{1 - v^2/c^2}\text{d}t = \frac{\text{d}t}{\sqrt{1 + g^2t^2/c^2}},$$ so that, for the first half of the trip, $$\tau = \frac{c}{g}\ln\left(gt/c + \sqrt{1 + g^2t^2/c^2}\right).$$ The equations need some adjustments for the second half of the trip (see also the first link), but the situation is symmetrical.

• +1: My answer is inferior since, as Michael Brown points out in the comments on my answer, interpreting the "acceleration" in the question as the acceleration as measure by an inertial observer doesn't make sense in this case; since Andromeda is far away, it would imply that that rocket's velocity passes that of light at some point along its journey. – joshphysics Aug 27 '13 at 5:17
• Should I consider this to be the working answer then? – Guy Haley Aug 27 '13 at 12:11
• @GuyHaley Well, that's your decision :-) But I'm confident that this is correct. – Pulsar Aug 27 '13 at 19:14
• Well, it certainly was very helpful, and I'll mark it as the answer I need. I was wondering, if you have the time, if you could explain the relation to the derivation of the function $$x^2-c^2t^2 = \frac{c^4}{\alpha^2}$$ I found some information on its derivation at physics.stackexchange.com/questions/41164/… but it is rather unclear. – Guy Haley Aug 27 '13 at 23:02
• The main problem is in the process of integrating $$\phi(u) = \phi(v) + \phi(u'),$$ to $$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{\mathrm{d}}{\mathrm{d}t'}\phi(u')\frac{\mathrm{d}t'}{\mathrm{d}t}$$ and its subsequent rewrite to $$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{1}{c}\gamma^{2}(u)\frac{\mathrm{d}u}{\mathrm{d}t},$$ and $$\gamma^{3}(u')\frac{\mathrm{d}u'}{\mathrm{d}t'} = \gamma^{3}(u)\frac{\mathrm{d}u}{\mathrm{d}t}.$$ – Guy Haley Aug 27 '13 at 23:18

Note. The following holds if the acceleration to which the question refers is acceleration as measured by an inertial observer external to the rocket.

Let $d$ denote the distance to Andromeda. Note that as measured in an inertial frame $S$, the rocket is accelerating at $1 \, g$ for a distance $d/2$, and then decelerating at $1\, g$ for the same distance $d/2$. Determining the maximum speed and travel time $T$ as measured in $S$ is just a regular kinematics problem. Let's assume this has been done. The velocity as a function of time as measured in $S$ will be \begin{align} v(t) =\bigg\{ \begin{array}{cc} gt, & 0<t<T/2\\ -gt,& T/2<t<T \end{array} \end{align} As mentioned in my answer to the your question Special Relativity and Constant Acceleration that you linked above, we can calculate the travel time $T'$ as measured by an observer in the rocket frame $S'$ via the following integral: \begin{align} T' = \int_0^T \frac{dt}{\gamma} = \int_0^{T/2}dt\left(1-\frac{(gt)^2}{c^2}\right)^{1/2}+\int_{T/2}^{T}dt\left(1-\frac{(-gt)^2}{c^2}\right)^{1/2} \end{align} I'll let you take it from here.

• Hmm....I understand the integrals for finding the time in the moving frame, but does regular kinematics work for the regular travel time $T$? Won't there be issues as the rocket approaches the speed of light? I found a somewhat similar question here: physics.stackexchange.com/questions/41164/… and it seems like the answer is a lot more complicated once that is considered (the process of equations are rather poorly explained, which is why I asked this question) especially when calculating the max speed? – Guy Haley Aug 27 '13 at 2:56
• A uniformly accelerating rocket follows a hyperbolic, not parabolic, path in spacetime. So your formula for the velocity is wrong. See the wiki page (more detail for Rindler coordinates). – Michael Brown Aug 27 '13 at 3:17
• @GuyHaley Would you happen to know if the $1\, g$ to which the question refers is the acceleration as measured by the inertial observer? – joshphysics Aug 27 '13 at 3:26
• @MichaelBrown As indicated in your own link, Hyperbolic motion occurs when the object has constant proper acceleration which is related to acceleration as measured by an external inertial observer by a factor of $\gamma^3$. It's not clear to me that the acceleration in the question refers to proper acceleration. – joshphysics Aug 27 '13 at 3:55
• @joshphysics My apologies, I definitely should have made it more clear. The acceleration does, indeed, refer to proper acceleration, which, I believe, would correspond to the acceleration "experienced" by the rocket. Again, my apologies for not making that clear. Thank you for the work for the inertial observer though, it was still very enlightening. – Guy Haley Aug 27 '13 at 4:36