Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A Starship is going to accelerate from 0 to some final four-velocity, but it cannot accelerate faster than $g_M$, otherwise it will crush the astronauts.

what is the appropiate equation to constraint the movement so the astronauts never feel a gravity higher than $g_M$? for a moment i thought the appropiate relationship was

$$ \left\lvert \frac{d u}{d \tau}\right\rvert \le g_M $$

where the absolute value is of the spatial component of the four-acceleration

But going down this route i get the following:

$$ \lvert u_F \rvert = \int_0^{\tau_F} \left\lvert \frac{d u}{d \tau} \right\rvert\,d \tau \le g_M \int_0^{\tau_F} d \tau = g_M \tau_F $$

where $u_F$ is the spatial component of the final velocity, and $\tau_F$ is the proper time it takes to reach the final velocity. The above gives me:

$$ \tau_F = \frac{ \lvert u_F \rvert }{ g_M } $$

i'm doing some silly mistake, because there are no gamma factors, and i'm getting a finite proper time to reach $\lvert u_F \rvert = c$

share|improve this question
    
Why is starship written with upper case letters, did you have any specific starship in mind? –  NiftyKitty95 Aug 15 '12 at 12:05
add comment

2 Answers

up vote 5 down vote accepted

You mistake is that you use the absolute value "of the spatial components" (your words) of the velocity only. Picking spatial components only is clearly not a Lorentz-covariant procedure, so it cannot calculate the invariant "feelings of the astronauts".

Instead, the right condition is given by the same inequality but $|d u^\mu / d \tau|$ is the length of the four-vector one obtains by differentiating the four-velocity $u^\mu$, where $u_\mu u^\mu = 1$, over the proper time $\tau$. The vector $d u^\mu / d \tau$ is spacelike and perpendicular (according to the Lorentzian metric) to the velocity vector $u^\mu$ itself; but this derivative isn't a purely spatial vector in any inertial system. In a frame in which the spatial components of $u^\mu$ are already nonzero, $d u^\mu / d \tau$ contains a nonzero time component, too.

When you calculate it correctly, the proper time needed to achieve the speed of light is infinite.

The easiest way to calculate it is one that assumes some knowledge of the Lorentzian geometry and how it's analogous to the Euclidean geometry. A uniformly accelerating object in the Euclidean spacetime would produce a circular world line. In the real, Minkowski space, the world line is a hyperbola. The coordinates after proper time $\tau$ may be written in analogy with sines and cosines but they're hyperbolic ones: $$ t = \sinh (\tau/\tau_0),\quad x = \cosh(\tau/\tau_0) $$ Here, $\tau_0$ is a constant depending on the acceleration. Consequently, the speed after proper time $\tau$ is simply the ratio, $$ v = \tanh (\tau/\tau_0) $$ For a small $\tau$, this gets reduced to $\tau/\tau_0$ in the limit and the $\tau$-derivative $1/\tau_0$ should be the (maximum) acceleration $g_M$ so $\tau_0=1/g_M$: $$ v = \tanh (\tau g_M) $$ in the $c=1$ units. You may invert it: $$ \tau = \frac{c}{g_M} {\rm arctanh} (v/c) $$ where I restored the powers of $c$ for your convenience. Note that arctanh of one is infinity. For a small $v/c$, one uses ${\rm arctanh}\, x\approx x$ and the right formula reduces to your nonrelativistic formula from the original question.

share|improve this answer
3  
You might want to look at chapter 6 of Gravitation by Misner, Thorne and Wheeler, or for a quick summary at math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html –  John Rennie Aug 15 '12 at 7:06
    
thanks for your answer Lubos, i have a question regarding your answer: on the final formula where proper time is related to constant acceleration and final $\beta$, i'm wondering if there is some missing constraint between final velocity, acceleration and total distance; for instance if i assume that $\tau \approx 10^4 sec$, that is, less than four hours of proper time, then $\frac{\rm arctanh(\beta)}{g_M} \approx 10^{-4} m^{-1} sec^2$ but i'm not clear how to proceed from that to obtain the required $g_M$ for this shorter travel. This ought to depend on the total traveled distance –  user56771 Aug 16 '12 at 1:29
    
Apologies, I don't understand this question. The original question was about the relationship of the final velocity and the proper time. You may also ask about the relationship of the total distance traveled and the final velocity, or the total distance traveled and the proper time. All these things may be calculated. The proper distance traveled, as sketched already above, is $x=[\cosh(\tau g_M)-1]/g_M$. You may also substitute the formula for $\tau$ in terms of $v$ (final) to get the other relationship. –  Luboš Motl Aug 16 '12 at 6:09
1  
@thanks Lubos, this really helped. –  user56771 Aug 17 '12 at 14:14
add comment

What you're interested in is the proper acceleration, the acceleration as recorded by the accelerometers of the starship, being constrained:

$\alpha \leq g_M$

For the unidirectional case, the coordinate acceleration $a$ and proper acceleration are related by:

$\alpha = \dfrac{d(\gamma v)}{dt} = \gamma ^3 a$

From this, it is easy to see that while the starship passengers feel a constant acceleration, the coordinate acceleration approaches zero as the coordinate speed approaches $c$.

Also, please note that the non-zero space-like components of the four-velocity, $\vec u = \gamma \vec v$, go to infinity as the coordinate speed goes to $c$, i.e., the proper speed does in fact reach $c$ in finite proper time.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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