Can anyone tell me how to solve this problem:

Alpha Centauri is $4.4$ light years away from Earth. What speed $u$ would a spaceship headed towards Alpha centauri had to have in order to last $t' = 10 \text{ years}$ for a passanger onboard?

The only thing i know how to do is to determine the proper time which is the one measured on a spaceship...

$$\boxed{t'\equiv \tau}$$

I don't know how to calculate any other variable if i don't know $\gamma$ or $u$. I think i should get a system of 2 equations and bust $u$ out of there... But i just cant solve this after 4 hours of trying...


I know equations for time dilation, length contraction:

\begin{align} \Delta t &= \gamma \Delta t' \xleftarrow{\text{time dilation}}\\ \Delta x' &= \gamma \Delta x \xleftarrow{\text{length contraction}} \end{align}

and Lorenz transformations:

\begin{align} \Delta x &= \gamma(\Delta x' + u \Delta t')\\ \Delta x' &= \gamma(\Delta x - u \Delta t)\\ \Delta t&= \gamma\left(\Delta t' + \Delta x' \frac{u}{c^2}\right)\\ \Delta t'&= \gamma\left(\Delta t - \Delta x \frac{u}{c^2}\right) \end{align}

  • $\begingroup$ Express time for the passenger in terms of $\gamma$ and $u$. Express $\gamma$ in terms of $u$ and insert back into the time for passenger. Then solve for $u$. It shouldn't be too difficult from there on. $\endgroup$ – SMeznaric Jun 22 '13 at 17:53
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    $\begingroup$ $c^2\tau^2 = c^2t^2 - r^2$ and the information given is all that you need to solve this problem. $\endgroup$ – Alfred Centauri Jun 22 '13 at 17:56
  • $\begingroup$ How did you get this equation? The only equations i know are those in EDIT. Is there a way to bust the solution out of those? $\endgroup$ – 71GA Jun 22 '13 at 18:39
  • $\begingroup$ The transformation equations come from the equation I gave which is just the well known equation for the invariant interval. In other words, it must be the case that: $c^2 \tau^2 = c^2 t^2 - r^2 = c^2 t'^2 - r'^2$ $\endgroup$ – Alfred Centauri Jun 22 '13 at 18:43
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    $\begingroup$ Of course it's possible but, it takes about 10 seconds to do it with the invariant interval equation. Why on earth would you want to do it the hard way? $\endgroup$ – Alfred Centauri Jun 22 '13 at 19:36

How does this connect in any way to my equations?

You know the distance travelled and you know the proper time, $t' = \tau = 10y$.

Using the timelike invariant interval equation, solve for t:

$t^2 = \tau^2 + (\frac{r}{c})^2 = (10y)^2 + (4.4y)^2 = 119.36y^2 \rightarrow t = 10.93y$

$u = \frac{r}{t} = 0.403c$

Since you insist on doing it the hard way, here's one approach:

From your first transformation equation:

$\dfrac{\Delta x}{\Delta t'} = \gamma u = \dfrac{4.4ly}{10y} = 0.44c$

From your 4th transformation equation:

$\gamma \Delta t = \Delta t' + \Delta x \frac{\gamma u}{c^2} = 10y + 4.4y(0.44) = 11.94y$

Combining these results:

$(\gamma u)(\gamma \Delta t) = 5.25ly = \gamma^2 u \Delta t = \gamma^2 \Delta x = \gamma^2 4.4ly$

From which we get:

$\gamma = 1.093$

$\Delta t = \gamma \Delta t' = 1.093 \cdot 10yr = 10.93y$

$u = \gamma u / \gamma = \dfrac{0.44c}{1.093} = 0.403c$

As I attempted to point out in comments, the invariance of the interval is fundamental and useful here. The Lorentz transformations guarantee that:

$(c\Delta t)^2 - (\Delta x)^2 = (c\Delta t')^2 - (\Delta x')^2$

You know that $\Delta x = 4.4ly, \Delta x' = 0$, and $\Delta t' = 10y$ so you just put in what you know and you get $\Delta t$ immediately and the speed $u = \dfrac{\Delta x}{\Delta t}$ immediately follows. There's no need to find $\gamma$ in this problem

  • $\begingroup$ When you say "first transformation equation", do you mean $\Delta t = \gamma \Delta t'$. When you say "4th transformation equation", do you mean $\Delta t' = \gamma \left(\Delta t -\Delta x \frac{u}{c^2}\right)$? $\endgroup$ – 71GA Jun 22 '13 at 22:04
  • $\begingroup$ @71GA, you list four transformation equations after the dilation equations. The first and last of those four are the equations I refer to. $\endgroup$ – Alfred Centauri Jun 22 '13 at 22:14
  • $\begingroup$ I did manage to reproduce this and get the results, but i still don't understand, how to solve this types of problems. I know, that first thing i have to do is to put myself (an observer) in a coordinate system $xy$ or $x'y'$. (A) If i put myself in a $xy$ the time $\Delta t'=10$ years is not the proper time as clock measuring it is moving relative to me (in this case $\Delta t$ is the proper time). (B) If i set myself in a coordinate system $x'y'$ clock measuring time $\Delta t'=10$ doesn't move acording to me - so this time $\Delta t'$ is the proper time... Correct me if i am wrong. $\endgroup$ – 71GA Jun 23 '13 at 6:54
  • $\begingroup$ I forgot to mention that in think that if in case (A) relative speed was $+u$ it is $-u$ in case of (B) $\endgroup$ – 71GA Jun 23 '13 at 6:59

Let me elaborate a bit on Alfred's answer, but note that his answer is correct and should be accepted and this is just a commentary.

When doing problems like this I suggest identifying spacetime events that you can compare in the two frames. In the Earth frame we can choose the two events to be the moment the rocket passes Earth and the moment it passes Alpha Centauri. If we choose our co-ordinate system such that the first event is (0, 0), i.e. $t = 0$ and $x = 0$, then the second event is ($d/v$, $d$) where $d$ is the distance to Alpha Centauri and $v$ is the relative velocity of Earth and the rocket.

So what do these two events look like in the rest frame of the rocket? Well we can choose the origin of the rocket frame to make (0, 0) the point is passes Earth. In the rocket's rest frame it's at rest (obviously!) so the second event is just ($t'$, 0), where $t'$ is the time measured on the clocks in the rocket. We're given $d$ and $t'$, and to solve the problem we have to calculate $v$.

The easy way to do this is to note that the proper time, $\tau$, is given by:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$

and that $\tau$ is an invarient i.e. all observers will measure the same value of $\tau$.

Start in the Earth frame. $dt$ is the time taken to reach Alpha Centauri, $d/v$, and $dx$ is the distance to Alpha Centauri, $d$ ($dy$ and $dz$ are zero because we take the rocket to be travelling along the $x$ axis). So:

$$ c^2d\tau^2 = c^2d^2/v^2 - d^2 $$

In the rocket's rest frame $dx$ is zero, because the rocket is at rest, so the expression for $\tau$ is simply:

$$ c^2d\tau^2 = c^2t'^2 $$

If we set the two expressions for $c^2d\tau^2$ equal we get:

$$ c^2t'^2 = c^2d^2/v^2 - d^2 $$

To get this in a more familiar form we note that $d = vt$ and substituting this in the above equation gives:

$$ \begin{align} c^2t'^2 &= c^2v^2t^2/v^2 - v^2t^2 \\ &= c^2t^2 - v^2t^2 \\ &= t^2 \left( c^2 - v^2 \right) \\ &= c^2t^2 \left( 1 - \frac{v^2}{c^2} \right) \end{align} $$


$$ t' = t \sqrt{1 - \frac{v^2}{c^2}} $$

that is:

$$ t' = \frac{t}{\gamma} $$

So to answer your question you take the above equation and replace $t$ by $d/v$ and rearrange to get:

$$ u = \sqrt{\frac{d^2}{t'^2 + d^2/c^2}} $$

and this gives $u = 0.403c$ as Alfred said.

The point of all this is that to answer the question you just need the definition of $d\tau$ and the fact that it is an invarient. Beginners to SR tend to throw factors of $\gamma$ around and get themselves thoroughly confused as a result. If you choose your pairs of spacetime points carefully and calculate the proper time then you can't go wrong.


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