I have just learnt that an object appears shorter in length when viewed by an observer in relative motion with respect to the object. We can derive this with Lorentz transformation formulas, and that's where my doubt lies. If a rod is moving with velocity $v$ with respect to a frame $S$, and another frame $S'$ is moving in the same direction as the rod with respect to $S$ with the same velocity $v$, then the length measured by an observer in frame $S'$ will be the proper length, as the observer is at rest with respect to the rod. Let $x',y' and\,z'$ be the coordinate axis of $S'$ and $ x,y\,and\, z$ be the coordinate axis of the frame $S$. Therefore, we can write the proper length $L_\circ = x_2' - x_1'$. We can also find the value of length as seen from the frame $S$, i.e., $L=x_2 - x_1$, if we put the values of $x_2'\, and \, x_1'$ from the formula $x'=\gamma(x-vt)$ (Inverse Lorentz Transformation) in $L_\circ=x_2' - x_1'$, as the term containing the variables $t_1 \,and \,t_2$ will cancel out because $t_1=t_2$. The result will come out to be $L_\circ=\gamma L$ or simply $L= \frac{L_\circ}{\gamma}$. My question is that when we put the value of $x$ directly in the formula $L=x_2 - x_1$ as $ x= \gamma(x' + vt')$, and then calculate, the result comes out to be $L=\gamma L_\circ$, quite the opposite of what we got earlier. Is there something that I am doing wrong here?


1 Answer 1


It is important to keep in mind that $S$ can determine the length of the moving rod only when he measures the two end points of the rod at the same time in his reference frame : this is why we take $t_1=t_2$. It is also important to note that $t_1' \neq t_2'$ (which is what you have assumed to get the wrong result : $L=\gamma L_0$) since the clocks co-moving with $S'$ read different times at different $x$ when observed from the frame of $S$ : See $(1)$.

$$t' = \gamma (t-\frac{vx}{c^2}) \tag{1}$$

At each point $x$ in frame $S$, there are two clocks present at every instant : one at rest in $S$ reading time $t$ and the other co-moving with $S'$ (attached to $S'$) reading time $t'$ which happened to pass $x$ at that instant (the co-moving clock changes from one instant to the next). A person at rest in frame $S$ $\textit{observes}$1 all the clocks that are at rest in $S$ to be synchronized and $\textit{observe}$s all the moving clocks to be out of sync (Relativity of Simultaneity).

1 Observation is a technical term here : What the person $\textit{sees}$ is not what s/he $\textit{observes}$. Observation takes the correction of travel-time of light into account.

$$\underline{\textbf{Updated Answer}}$$ I think your confusion most probably stems from a lack of understanding of what the quantities related by the Lorentz transformation actually are. This is going to be lengthy explanation. So, hold tight!

$$\underline{\text{A (Common) Instructive Way to Visualize Inertial Frames}}$$ Let us consider motion only along the $x$-direction for our purposes. You're at rest in frame $S$. Imagine an infinitely long ruler to be affixed to the frame, with its zero mark positioned at your location (your chosen origin). Also, imagine mini robots (please go with me on this) with synchronized clocks stationed at all points on the ruler. These robots have cameras that record everything that happens in their near vicinity. To us, they just look like table clocks affixed to the ruler (very convenient for pictorial purposes) [Refer the figure below]. Therefore, when the frame moves, so does the ruler, and in turn, so do the clocks attached to the ruler.

Fig. 1

What do I mean by synchronized clocks, you ask? Pick any bot and instruct it to shout to you both, the distance ($x$) (by looking at the reading on the stationary ruler) and the reading on its clock ($t$). Let's say, you receive this message when the time on your watch reads $(T )$. What you find is that the quantity $t+({x \over c})$ matches $T$ (here, $c$ is the speed of sound in your frame). If this condition is true for any bot (and as a consequence, every bot), then the clocks on the bots are synchronized with respect to your frame.

Instead of writing this in a lengthy manner, I could have simply said that you $\textit{observe}$ all the bot-clocks to read the same time. What you see isn't what you $\textit{observe}$, for $\textit{observation}$ takes correction of travel time of signals into account. Please keep this visualization in mind for the upcoming parts to make sense.

$$\underline{\text{Lorentz Transformation}}$$

Your frame is $S$ and the frame moving to your right with speed $v$ is $S'$. I'll call the ruler affixed to frame $S$, the $S$-ruler, and the mini-bots affixed to $S$-ruler as $S$-bots (and likewise for $S'$). Let's say the $S$-bot located at $x$ records the reading of the adjacent moving $S'$-bot's position and time (i.e., $(x',t')$) when its own clock reads $t$ [Refer the figure below]. Likewise, you can always imagine the adjacent $S'$-bot to do the same thing. Lorentz transformation just relates these two pairs of quantities : $(x,t)$ and $(x',t')$.

The figure below shows what a person at rest in frame $S$ $\textit{observes}$.

enter image description here

To get to your question in the comments : Why is $t_1' \neq t_2'$? Look at the figure above and imagine the moving rod's end points to be located at $O'$ and $x'$. Do you see that the $S'$-clocks located at $O'$ and $x'$ read different time? That is what I meant when I said that they are not synchronized. The $S$-clocks below read the same time at all points on the $S$-ruler.

The $S'$- clocks are synchronized in frame $S'$ but they are not synchronized in frame $S$. When a person at rest in frame $S$ tries to measure the end points of the rod simultaneously ($t_1=t_2$), he will notice that the $S'$-bots that are next to the rod's end points will read different times ($t_1' \neq t_2'$).

I apologize if I've only confused you further. I know there must be a better way to explain this but this is the picture that I have in mind that, personally, works for me and prevents me from making mistakes.

Let me know in the comments if you have questions / find something that doesn't make sense in my explanation.

  • 3
    $\begingroup$ This is why I maintain that the "concepts" of length contraction and time dilation are just as bad as relativistic mass, pedagogically. They all encourage "lazy thinking" by poorly summarizing the full effects of SR (I call it "gamma slinging"). $\endgroup$
    – m4r35n357
    Commented Nov 29, 2019 at 14:27
  • $\begingroup$ But why is $t_1 \ne t_2$? Can't the observer who is moving with respect to $S$ measure the length of the rod at a particular instant as the rod is at rest with respect to him? $\endgroup$ Commented Nov 29, 2019 at 14:40
  • $\begingroup$ @PrathamYadav Is that a typo? I didn't say $t_1 \neq t_2$. The observer who is stationary at $S'$ will measure the end points of the stationary rod to be located at $x_1'$ and $x_2'$ : It doesn't matter if s/he measures the end points at the same time or at different times as the rod is stationary with respect to him. (contd.) $\endgroup$
    – Ajay Mohan
    Commented Nov 29, 2019 at 14:50
  • $\begingroup$ @AjayMohan I am sorry. It was a typo. Actually I was trying to ask why you had stated $t_1 ' \ne t_2'$, as the observer in S' can observe the two ends of the rod at the same time. $\endgroup$ Commented Nov 29, 2019 at 14:53
  • $\begingroup$ @PrathamYadav I'm updating my answer to give a comprehensive explanation. Stand by. $\endgroup$
    – Ajay Mohan
    Commented Nov 29, 2019 at 15:09

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