Seperation Between a Rocket and an Emitted Photon

Question

I am studying for an exam on Special Relativity, and I made a toy problem to practice, but I ended up confusing myself. I know something is wrong with my intuition, but can someone please point where I am making the error? Here's the problem:

Consider a rocket traveling with along the x-axis with constant velocity 0.8c. At some time $t'_0=0s$, the rockets starts a clock and emits a photon. At $t'_1=0.6s$ he stops the clock. What is the separation between the rocket and the photon for the rockets reference frames, and for a reference frame at rest?

Here's my work: okay so from the reference frames of the rocket, the rocket appears to be at rest, so the separation would be the distance the photon travels: $ 0.6 \cdot 3 \cdot 10^8 = 1.8 \cdot 10^8m$

From a stationary rest frame we find

$ \gamma = (1-\frac{v^2}{c^2})^\frac{-1}{2}= 1.\overline{6} $

$t_1 = t_1' \cdot γ = 1s$

The rocket is moving at 0.8c so it travels $ 0.8 \cdot 3 \cdot 10^8 = 2.4 \cdot 10^8 $

The photon travels $ 3 \cdot 10^8 m $

So the separation would be $ (3-2.4) \cdot 10^8 = 0.6 \cdot 10^8m. $ If we apply length contraction we get $0.6 \cdot 10^8 \cdot \gamma = 1 \cdot 10^8$ , but this does not agree with the previous calculation of $ 1.8 \cdot 10^8m $ separation.

Can someone help me understand what I'm doing wrong?

(Apologies if this has terrible formatting, I'm writing from a mobile device!)

Answer 1

I'm too tired to do the math, but I can explain how you have gone wrong. As is the case with many of the SR questions on this site, you are overlooking the relativity of simultaneity. The distance between the rocket and the photon after 1 second in the stationary frame is not a contraction of the distance between the rocket and the photon after 0.6 seconds in the rocket frame. In other words, you are not comparing a proper length in the rocket frame with the corresponding contracted length in the stationary frame. And the reason is that 0.6s in the rocket frame is equal to 1.0s in the stationary frame only at the location of the rocket. Owing to the relativity of simultaneity, the time in the rocket frame is out of synch with time in the stationary frame elsewhere along the direction of travel, so when the rocket passenger at 0.6 seconds in the rocket frame asks 'where is the photon now?' she is using a different 'now' to the now that the stationary observer has used in calculating the position of the photon at 1.0 seconds in the stationary frame. If you draw a spacetime diagram you will see that you are not comparing like with like.

Answer 2

I have solved the problem, and if anyone is interested in the solution for the sake of other inexperienced relativists such as myself, I will solve the problem below:

Let us consider two coordinate systems, $x$ and $x'$. Our stationary observer will inhabit x-land, and the rocket ship will be centered at the origin of x' land. At the start of our system, the coordinate systems are almost exactly the same, with the only difference that the $x'$ coordinate system is moving to the right with a velocity of $ 0.8c $.

We can imaging that the system is frozen and the clocks are synchronized at $ t_0 = t'_0 = 0 $. As shown below:

$<-----(x'_0,t'_0)----->$

$<-----(x_0,t_0)----->$

At time $ t_0 = t'_0 = 0 $ the rocket ship we 'unfreeze' the system, both $t'$ and $t$ clocks will begin to run, and the rocket ship emits a photon which travels some distance $x'_1$. We then stop the clock after one second has passed for our rest frame. The situation appears below:

$\ \ \ \ \ \ \ \ \ \ \ \ <-----(x'_0,t'_0)----$$x'_1->$

$<-----(x_0,t_0)----->$

We now have to find a reliable method for transforming between our coordinates. Notice that our stationary observer sees the origin of the $x'$ coordinate system has shifted to the right by a distance of $vt$. Next let us apply a length contraction to the entire $x'$ coordinate scale. Any length $L$ will become $ \frac{L}{\gamma} $, where $\gamma$ for us is $\frac{1}{0.6}$

Now we can calculate that any point, such as $x'_1$ has been shifted by the equation $ x = \frac{x'}{\gamma}+vt$. Rearranging this equation yields $x' = (x - vt)\gamma$.

Now back to the problem we are trying to solve, at this point all we know is that one second has passed for our stationary observer, that is $t_1 = 1s$. From this we can calculate that the the photon will have moved to a position $ x = 3 \cdot 10^8m$.

Using the transformation rules we find the folowing result:

$x' = (3 \cdot 10^8 - 0.8 \cdot 3 \cdot 10^8)\gamma = 1 \cdot 10^8m $

Since our rocket is centered at the origin, the above result is the total separation between the rocket and the photon: $100,000,000$ meters.