# What is wrong in this false derivation of length "dilation"?

Suppose an object measures $$L$$ in moving frame $$S'$$. This is measured at the same time so $$\Delta t'=0$$ :

$$\Delta x = \gamma(\Delta x'+v\Delta t') = \gamma \Delta x' = \gamma L$$
Since $$\gamma \gt 1$$, the same object is dilated in the rest frame $$S$$.
This is clearly wrong as we know that the object is actually contracted. What am I doing wrong?

Before we start, I find that a lot of confusion with Special Relativity can be cleared up if we use a standard convention, so I'm going to use all primed'' quantities to represent quantities measured in the $$S^\prime$$ frame, and all unprimed quantities to represent the same quantities in the $$S$$ frame. (In other words, the quantity $$L$$ that the OP uses in their question is what I will refer to as $$L^\prime$$. I apologise for this, but I find that it makes my answer easier to understand.)

Let me also write down the Lorentz Transformations:

\begin{aligned} &\text{(A)}\quad\Delta x^\prime = \gamma \left(\Delta x - v \Delta t\right)\\ &\text{(B)}\quad \Delta t^\prime = \gamma \left( \Delta t - \frac{v}{c^2}\Delta x\right)\\ \\ &\text{(C)}\quad\Delta x = \gamma \left(\Delta x^\prime + v \Delta t^\prime \right)\\ &\text{(D)}\quad \Delta t = \gamma \left( \Delta t^\prime + \frac{v}{c^2}\Delta x^\prime \right)\\ \end{aligned} \label{LT}

And lastly, let's actually make the definition of a length clear. For an observer sitting in $$S^\prime$$, since the object is at rest with respect to him, its length $$L^\prime$$ is simply the difference in the coordinates, irrespective of when $$x_B^\prime$$ and $$x_A^\prime$$ are measured. He could measure $$x_B^\prime$$, have a coffee, and then measure $$x_A^\prime$$ and the difference would give him the length. However, for an observer sitting in $$S$$, since the object is moving with respect to her, both the endpoints $$x_B$$ and $$x_A$$ need to be measured simultaneously in her frame of reference ($$S$$) in order for the difference to be the length $$L$$. (In other words, if she has a coffee between measuring $$x_B$$ and $$x_A$$, the object would have moved between measurements!) So, we have

$$L^\prime = x_B^\prime - x_A^\prime |_\text{ for any \Delta t^\prime}$$ $$L = x_B - x_A |_\text{ only when \Delta t=0}$$

If you understand this, the rest of the answer is quite simple. Let us consider, as you have, that the object we are measuring is at rest in the frame $$S^\prime$$, and its length is being measured both from $$S$$ (in which it is moving to the right with a velocity $$v$$) and $$S^\prime$$ in which it is at rest.

The observer in $$S$$ requires to measure the endpoints of the object simultaneously in her frame of reference, as otherwise the object would move between measurements. In other words, for $$(x_B - x_A)$$ to be the length, we require that $$\Delta t = t_B - t_A = 0$$. Note: we are not placing any condition on $$\Delta t^\prime$$. It may not be (and isn't!) zero. Two observers, moving at some velocity $$v$$ relative to each other will not agree on simultaneous events.

Thus, we need to find a relation between $$\Delta x$$ and $$\Delta x^\prime$$, when $$\Delta t=0$$. The mistake you've made in your argument is that you're relating $$\Delta x$$ and $$\Delta x^\prime$$ when $$\Delta t^\prime=0$$. So, the mistake comes when you say that $$\Delta x|_{\Delta t^\prime = 0} = L$$, the length measured in $$S$$.

We refer to transformations above, and see that (A) is the transformation we should use, as it relates these quantities.

\begin{equation*} \begin{aligned} \Delta x^\prime &= \gamma \left(\Delta x - v \Delta t\right)\\ \Delta x^\prime|_{\Delta t = 0} &= \gamma \left(\Delta x|_{\Delta t =0} - v \Delta t|_{\Delta t = 0}\right)\\ \\ L^\prime &= \gamma L \end{aligned} \end{equation*}

Thus, the length that an observer measures when she is at rest with respect to the object (i.e. sitting in $$S^\prime$$) $$L^\prime$$ is always greater than $$L$$, since, as you point out, $$\gamma > 1$$. Thus, an observer sitting in $$S$$, with respect to whom the object is moving at a constant velocity will measure a length $$L$$ which is shorter: lengths contract!

• Thank you so much for the detailed explanation. I now get how to measure length of a moving object. However, I'm still a bit confused as the equation $$\Delta x = \gamma (\Delta x' + v\Delta t')$$ seems to suggest that $\Delta x \gt \Delta x'$ as $\gamma \gt 1$. $\Delta x$ and $\Delta x'$ also represent the length of object right? @Philip Cherian Dec 28 '19 at 19:59
• Looks you've addressed this already Note: we are not placing any condition on Δt′. It may not be (and isn't!) zero. Two observers, moving at some velocity v relative to each other will not agree on simultaneous events. I haven't gotten till simultaneity yet... Probably it will make more sense then.. Thanks again:) Dec 28 '19 at 20:06

"This is measured at the same time so Δ𝑡′=0" This requirement of simultaneity (of measurement of positions of front and back of rod) is needed only in frames in which the rod is moving. So your S' frame is a frame in which the rod is moving, and your S frame (in which the front and back measurements are not simultaneous) is the frame in which the rod is at rest, because simultaneity of measurement isn't needed in this case.

So your $$\Delta x$$ is $$L$$ (or $$L_0$$) and your $$\Delta x'$$ is the length of the rod in the frame in which it is moving. Since $$\gamma>0$$ this length is clearly contracted!