Homework Help: Homework - time dilation problem (Lorenz factor is unknown)

1. Jun 22, 2013

71GA

1. The problem statement, all variables and given/known data
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 $t' = 10$ years for a passanger onboard?

2. Relevant equations
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}

3. The attempt at a solution
The only thing i know how to do is to determine the proper time which is the one measured on a spaceship...
\begin{aligned}
\boxed{t'\equiv \tau}
\end{aligned}
By using the equations above i simply can't calculate any other variable because $\gamma$ is unknown and is present in all 6 equations...

2. Jun 22, 2013

Solkar

Right.
And the Lorentz-γ solemnly depends on your "u", and you've given data for Δt', Δx, and Δt = Δx/u to be matched.
So simply substitute and solve for u.

Last edited: Jun 22, 2013
3. Jun 22, 2013

71GA

Ok so i am sure that:

$$u = \frac{\Delta x}{\Delta t}$$

and this holds for an observer in a coordinate system $xy$. In this system only the observers clock isn't moving so this means that his clock is measuring the proper time $\Delta t = \tau$ from this it follows that all other times are longer (holds also for the time measured by the clock in a moving $x'y'$ frame): $\Delta t' = \gamma \Delta t$. If i use this in the equation above i get:

\begin{align}
u = \frac{\Delta x}{\Delta t} = \frac{\Delta x \gamma}{\Delta t'} \xrightarrow{\text{i use only the RHS}} u &= \frac{\Delta x \gamma}{\Delta t'}\\
u&=\frac{\Delta x}{\Delta t' \sqrt{1 - u^2 / c^2}}\\
u^2&=\frac{\Delta x^2}{(\Delta t')^2 (1 - u^2/c^2)}\\
u^2 - \frac{u^4}{c^2} &= \frac{\Delta x^2}{(\Delta t')^2}
\end{align}

How can i now solve this?

4. Jun 22, 2013

Solkar

Sorry, it should by a minus,
$$\Delta t'= \gamma\left(\Delta t - \Delta x \frac{u}{c^2}\right)$$

Now do yourself a favor, set c:=1, and continue
$$\Delta t'= \gamma\left( \Delta x/u - \Delta x\, u\right) = \gamma\left( \Delta x \frac{1-u^2}{u}\right)$$

Now expand γ and see what cancels out.

5. Jun 22, 2013

Staff: Mentor

$$\Delta x = \gamma(\Delta x' + u \Delta t')$$

The x' coordinate of the spaceship is constant, so Δx' = 0. Δx = 4.4 light years, and Δt' = 10 years. And, you already noted that γ is a function of u/c.

6. Jun 23, 2013

71GA

Thank you. If i set point in a origin this is easier :)

Can anyone tell me which time here is the proper time and why.

7. Jun 23, 2013

Staff: Mentor

The proper time is the time measured by the guy on the spaceship who is physically present at both events (spaceship leaves, spaceship arrives), and whose coordinate is not changing during the trip.

8. Jun 24, 2013

Thanks.