I am trying to intuitively derive length contraction in special relativity using a thought experiment, without relying on Lorentz transformations. My aim is to obtain a derivation similar to how time dilation is derived using the classic light clock thought experiment. However, I have not been able to find or create a similarly intuitive derivation for length contraction.
I understand that deriving length contraction is challenging because it involves two points with different worldlines, the concept of simultaneity, and a precise definition of 'measuring' distance between these points.
Here's the gedankenexperiment I've devised, but for which I haven't found a satisfactory answer:
- A rod with a photon emitter at its center is placed on an 2-dimensional plane. At a certain instant, the emitter sends photons in opposite directions.
- The plane is made of a material that records an imprint when the photons reach the rod's endpoints.
- The rod's length is determined by measuring the distance between the imprinted spots.
When the rod is at rest with respect to the plane, triggering the emitter records two spots at a distance we'll call the proper length of the rod, $L_0$.
Now, consider the rod moving at velocity $v$ with respect to the plane, where $v$ is parallel to the direction defined by the rod's endpoints. What is the distance between the imprinted spots when the photon emitter is triggered?
I expect the answer to be $L_0 \gamma^{-1}$, as length contraction would imply, but I have been unable to derive this result. Below is my attempt:
Suppose the rod is moving to the right and the device trigger event takes place at $(t,x) = (0,0)$ in the rest frame of the plane.
We now find the points at which light reaches each endpoint, which we do by solving a simple system of equations for each point. Let $x_A$ and $x_B$ be the position of the right and left endpoint respectively. Then
For $A$:
\begin{equation}
\left. \begin{aligned}
x_A = c t_A \\
x_A – L/2 &= v t_A
\end{aligned} \right\} \ \rightarrow \ x_A = \frac{L}{2} \frac{1}{1-\beta}
\end{equation}
For $B$:
\begin{equation}
\left. \begin{aligned}
x_B = – c t_B \\
x_B + L/2 &= v t_B
\end{aligned} \right\} \ \rightarrow \ x_B = \frac{L}{2} \frac{1}{1+\beta}
\end{equation}
Where $\beta = v/c$, $t_A$ and $t_B$ are the time it takes for the photons to reach the $A$ and $B$ endpoints respectively, and $L$ is the length of the rod in the plane frame.
Then the difference between $x_A$ and $x_B$ which determines the positions of the spots imprinted on the plane is
\begin{equation}
\Delta x = x_A – x_B = L \gamma^2
\end{equation}
I do not know how to make sense of this.
If I assume ad hoc length contraction of the rod ($L = L_0\gamma^{-1}$) then the result would be $\Delta x = L_0 \gamma$, which doesn't make sense to me. It implies that the spots are further apart when the rod is moving, suggesting some kind of length dilation (if we define length by the distance between spots).
Can anyone help clarify this issue?
Best Answer
What we mean by the rod's length (in the plane frame) is the difference between the end positions of the rod at the same time (in the plane frame). For instance, if both ends emit a flash of light simultaneously (in the plane frame), and a detector on either side of the rod measures the difference between the arrival times of the flashes as $\tau$, you can calculate the rod's length as $c\tau$. This works because the flashes were emitted simultaneously in the plane frame.
I don't quite follow your math, but as has been pointed out, the problem is that in the plane frame, the imprints occur at different times so the distance between them is not the length of the rod. As for what this distance is, note that in the rod frame the distance between the imprints is $L_0$ (the rest length of the rod). Since the imprints are subsequently moving with respect to the rod, the distance between them is length-contracted in the rod frame, so in the plane frame, the imprints are separated by $\gamma L_0$: this is the proper distance.