3
$\begingroup$

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:

  1. 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.
  2. The plane is made of a material that records an imprint when the photons reach the rod's endpoints.
  3. 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?

$\endgroup$
18
  • 1
    $\begingroup$ A "Gedankenexperiment" is simply an educational tool that tries to make learning a fact easier. The problem with the technique for the purposes of special relativity is that all the SR Gedankenexperimente that I have seen are way more complicated than the dozen lines of high school algebra that it takes to derive the Lorentz transformation from first principles. So why bother? $\endgroup$ Commented Apr 20, 2023 at 22:52
  • 2
    $\begingroup$ Events simultaneous in the rod frame are not simultaneous in the plane frame. So you cannot get the length in the plane frame with this method. $\endgroup$
    – BowlOfRed
    Commented Apr 20, 2023 at 23:37
  • 3
    $\begingroup$ @FlatterMann I couldn't disagree more with the point of view that the best way to learn special relativity is to treat it as a bunch of algebra without grappling with the physical implications illuminated by thought experiments like this one. $\endgroup$
    – Andrew
    Commented Apr 20, 2023 at 23:50
  • 4
    $\begingroup$ @Andrew It's worth stressing that in my experience I was confused to no end with the baroque approaches that FlatterMann berates. It was only until I started thinking in terms of Minkowski diagrams did things start to click. Adopting more "modern techniques" doesn't necessarily mean letting go of "physical implications," although I admit striking the right balance may be a challenge. $\endgroup$ Commented Apr 21, 2023 at 0:29
  • 3
    $\begingroup$ @MaximalIdeal I think you misunderstand me. I also find the geometric formalism more intuitive and powerful than the "high school algebra" approach to special relativity. However, I strongly believe the primary goal of a physics student learning physics is to build physical intuition, which comes not from learning to manipulate formalism, but by applying the theory to calculate the consequences in concrete scenarios, especially when those consequences run counter to "everyday" intuition. $\endgroup$
    – Andrew
    Commented Apr 21, 2023 at 17:16

4 Answers 4

4
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ That was helpful! Your definition for the length of the moving rod in a moving reference frame is what I was after. I now see that, in absence of the correct insight, I tricked myself into thinking that the imprints would actually be representative of the length of the rod in any reference frame. I find now that it is a nice counterintuitive result that the separation of the imprints is greater than the proper length of the rod (I imagined the opposite). Thanks! $\endgroup$ Commented Apr 21, 2023 at 0:04
3
$\begingroup$

In your thought experiment, let's say $F_{R}$ is the reference frame of the rod and $F_{P}$ is the reference frame of the background plane.

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.

I get stuck in circular reasoning. If I assume ad hoc that $L = L_0\gamma^{-1}$, 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).

In $F_{P}$, the left endpoint gets marked first, and then the right endpoint gets marked later. This is clearly not what you want, and the distance between the two marks will not be representative of the rod's length in any reference frame.

If you insist on your thought experiment, you have to mark the endpoints at the same instant with respect to simultaneity of $F_{P}$. This is the challenge.


Here is how I would proceed to create a thought experiment. I would take the light clock thought experiment and flip the light clock 90 degrees so that it is longitudinal to its motion.

So to put it another way, suppose we have a rod moving to the right with velocity $v$ in $F_{P}$. At time $t=0$ in $F_{P}$, the left endpoint is at $(t, x) = (0, 0)$ and the right endpoint is at $(t, x) = (0, L)$. At $t=0$, suppose the left endpoint triggers and releases light to the right. The light signal reaches the right endpoint, gets reflected, and goes back to the left endpoint. The total time elapsed with respect to $F_{P}$'s time is $$ \Delta t = \frac{L}{c-v} + \frac{L}{c+v} = \frac{2cL}{c^{2}-v^{2}} = \frac{2L}{c}\gamma^{2}. $$ If we permit ourselves to assume time dilation, then we know that the amount of time that elapses in $F_{R}$'s frame is $$ \Delta t' = \Delta t/\gamma. $$ But reasoning in $F_{R}$ also reveals that the time elapsed (of the total roundtrip of the light) is $$ \Delta t' = \frac{2L_{0}}{c} $$ where $L_{0}$ is the proper length of the rod. Then by putting the equations together we find $L_{0} = L\gamma$, or equivalently $L = L_{0}/\gamma$.

$\endgroup$
1
  • $\begingroup$ Great! I tried myself the longitudinal lightclock thought experiment as well but did not come up with the answer. Very helpful. I think that your answer and @Puk answer below perfectly settle the question. Thanks :) $\endgroup$ Commented Apr 21, 2023 at 0:24
3
$\begingroup$

In the frame of the rod, the two photons will reach the two ends at the same time, and the distance between them will be $L$. In the frame of the rod, it's the plane which is contracted, so the two events will occur at a distance greater or equal to $L$ in the frame of the plane.

As @BowlOfRed points out, there are errors in your equations for $x_A$ and $x_B$:

  1. In the frame of the plane, the ends of the rod will not be at $-L/2$ and $+L/2$ at the time the photons are emitted
  2. I think you also don't get signs right, e.g. for $v=0$, $t_A = t_B$, but then $x_A = ct_A = ct_B = x_B$ and then $|x_B-x_A| = 0$ instead of expected $L$.
$\endgroup$
3
  • $\begingroup$ You are correct about the minus sign, I miss-typed it on the equation for $x_B$. I just corrected it. Regarding the ends of the rod not being at $-L/2$ and $+L/2$, I guess that may be the error. If that's the case, I do not know how to approach the problem. Thanks for the answer. $\endgroup$ Commented Apr 20, 2023 at 23:50
  • 1
    $\begingroup$ @PereRosselló I think there is a problem in the first pair of the equations, $x_A = ct_A$ and $x_B = ct_B$ (see my edited post). But I agree, it's the second equation that's more tricky. I guess you either allow yourself to assume there's Lorentz contraction, otherwise you need to derive it yourself. $\endgroup$ Commented Apr 21, 2023 at 0:02
  • $\begingroup$ True that, another minus sign missing... it was a typing error (corrected). However, the derivation for the separation of the imprints is still sound, although as you and others pointed out, not representative of the contracted length of the moving rod. $\endgroup$ Commented Apr 21, 2023 at 0:28
3
$\begingroup$

The scheme you are using will not get the answer because you did not take into account a hidden important part of doing these things: You can only compare things when there is a coincidence of stuff.

That is, you need to let the light come back to some point, and only then can you say, hey, you have something easy to compare.

The solution is to put a mirror on one end of the rod, and shoot a pulse of light from the other end. In the rest frame, the time it takes for the pulse to go there and back is obviously deduced as being twice the lightlength of the rod.

What you do to derive the length contraction, is to shoot a pulse at the moving far end of a rod at the same time and place as when the near end of a rod just passes you. Then you know that the far end will be even farther away when the pulse finally reaches the mirror and begins to return. By logical reasoning, we would have to do a correction to remove this excess length, which you can do because you assume the fixed relative velocity is known (because you can work out the relative velocity by repeated measurements after this crucial first experiment is done). After that calculation, you should obtain the Lorentz contracted length of the rod.

And then you can scrutinise what the moving observer would think it is doing to their rod, and the time dilation factor can also be used to figure out that there is a Lorentz contraction on the position axis of the moving observer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.