Relativity Question Fireworks explosion A firecracker explodes at the origin of an inertial reference frame. Then, 2.0 microseconds later, a second firecracker explodes 300m away. Astronauts in a passing rocket measure the distance between the explosions to be 200m. According to the astronauts, how much time elapses between the two explosions?
Okay. My textbook answers this question using spacetime interval consistency which is simple.
I don't understand why the time dilation formula does not work:
$$Δt = \frac {Δτ}{ \sqrt{1-\frac{v^2}{c^2}}}$$
Where v is simply the ratio between 300m and 2 microseconds.
Δτ=?
Δt=2x10^-6 s
 A: If both firecrackers went off at the same spatial point in the Earth frame then you could just use the time dilation formula to get the time interval in the rocket. However they go off at different spatial points and this means you have to do the full calculation. Let's see how this works.
We'll choose our origin so that in the Earth frame the first firecracker goes off at $(0, 0)$, then the second firecracker goes off at $(t, d)$, where $t = 2\mu s$ and $d = 300m$, but let's keep it general for now. We'll choose the coordinate system of the rocket so its origin coincides with the Earth frame, then the point $(0, 0)$ is the same in both frames. We just need to find where the point $(t, d)$ is in the rocket frame.
The Lorentz transformations tell us:
$$\begin{align}
t' &= \gamma \left( t - \frac{vx}{c^2}\right) \\
x' &= \gamma \left( x- vt \right)
\end{align}$$
Actually we are only asked for the time difference in the rocket frame so we just need $t'$, and this is:
$$ t' = \gamma \left( t - \frac{vd}{c^2}\right) $$
Now if the two firecrackers had been in the same place $d$ would be equal to zero and we would just have:
$$ t' = \gamma t$$
So you could just use the usual time dilation equation. It's because $d \ne 0$ that we can't do this.
Later:
The question provides an interesting example of the easy and hard ways to solve problems. I’ll go through the details in case it’s of interest to anyone.
If we choose the origins of the two frames to coincide when the first firecracker goes off, so it’s at $(0, 0)$ in both frames, then in the Earth frame the second firecracker is at $(2\mu s, 300m)$. The question asks if in the rocket frame the second firecracker is at $(t’, 200m)$ then what is the value of $t’$?
The easy way to answer this is to use the invariance of the proper time. The proper time, $\Delta \tau$, between any two spacetime points $(t, x, y, z)$ and $(t + \Delta t, x + \Delta x, y + \Delta y, z + \Delta z)$ is given by:
$$ c^2\Delta \tau^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 $$
The proper time is the same for all observers, so in this case the Earth observer and the rocket observer must calculate the same value for $\tau$. For the Earth observer we have:
$$ \tau_{Earth} = c^2t^2 – d^2 $$
where $t = 2\mu s$ and $d = 300m$, and for the rocket we have :
$$ \tau_{Rocket} = c^2t’^2 – d’^2 $$
where we’re told that $d’ = 200m$. Setting $\tau_{Earth} = \tau_{Rocket}$ we get:
$$ c^2t^2 – d^2 = c^2t’^2 – d’^2 $$
And a quick rearrangement gives:
$$ t’^2 = \frac{ c^2t^2 – d^2 + d’^2}{c^2} $$
And putting in the values for $t$, $d$ and $d’$ we get:
$$ t’ = 1.86 \mu s $$
And this answers the question.
The hard way to do it is to use the Lorentz transformations. These tell us that:
$$ t' = \gamma \left( t - \frac{vd}{c^2}\right) $$
But we don’t know the value of $v$. To get this we have to use the transformation equation for distance:
$$ x' = \gamma \left( x- vt \right) $$
We know that $t = 2\mu s$, $d = 300m$ and $d’ = 200m$, so we can solve for $v$. The reason this is hard is that after much scribbling (and swearing) the result is:
$$ v = \frac{2dt \pm \sqrt{4d^2t^2 - 4(t^2 + \frac{d’^2}{c^2})(d^2 – d’^2) }}{2 ( t^2 + \frac{d’^2}{c^2})} $$
And putting in the values for $t$, $d$ and $d’$ we get two solutions (corresponding to the $+$ and $-$ in the $\pm\sqrt{}$ term) with the values for $v$ of:
$$ v = 2.185 \times 10^8 \text{m/s} \space \text{and} \space 5.148 \times 10^7 \text{m/s} $$
And finally substituting either of these into:
$$ t' = \gamma \left( t - \frac{vd}{c^2}\right) $$
gives us $ t’ = 1.86 \mu s$. So we get the same result as using the simpler method, but after much more pain! Still, using the Lorentz equations does tell us that there are two possible velocities for the rocket.
