Special relativity and Lorentz transformation

I recently learned about special relativity and different reference frames. The conversion from two inertial reference frames is given by Lorentz transformations: $$x' = \frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}$$ and $$x = \frac{x'+ut'}{\sqrt{1-\frac{u^2}{c^2}}}$$. In my question, S' is a reference frame which moves with speed $u$ relative to S, along OX axis. Suppose the observer from S' measures at the same time two points $x'_1$ and $x'_2$. The distance from these 2 points is $x'_2 - x'_1$. In all the books I've seen so far, they calculate the distance in S using the first formula above. However, if we use the second one, we obtain that $$x_2-x_1 = \frac{x'_2-x'_1}{\sqrt{1-\frac{u^2}{c^2}}}$$. We conclude that the length perceived by the observer in S is bigger than the length perceived by the observer in S. However, this is not the correct answer. Why we cannot use the second formula to get the answer?

Thank you!

• What about trasnformation of time coordinate? Mar 22 '15 at 14:41
• For my purposes, it does not matter, beaucase the $ut$ term cancels out in substraction. Mar 22 '15 at 14:45

We conclude that the lenght perceived by the observer in S is bigger than the lenght perceived by the observer in S.

No, that would be an error. Why? The short answer is that the coordinate difference $(x_2 - x_1)$ is not a length in S.

A length is the (spatial) coordinate difference at the same coordinate time.

To be clear, there are two events - event 1 and event 2. In the primed coordinate system, you've stipulated that

$$t'_1 = t'_2$$

thus $\Delta x' = x'_2 - x'_1$ could be the length of an object (to measure the length of an object, the coordinates of the ends of the object must be recorded at the same time).

However, in the unprimed coordinate system

$$t_1 \ne t_2$$

so $\Delta x = x_2 - x_1$ cannot be the length of an object but is, rather, simply the spatial separation of the two events in the unprimed coordinate system which must be larger than the $\Delta x'$.

• What if $t'_1=t'_2=0$? Then we get that also $t_1=t_2=0$. So if the measurement takes place right at the initial time (when the origin of S is the same with the origin in S') $x_2-x_1$ is a length? Mar 22 '15 at 15:23
• The only time this would happen is if u is identically zero then both coordinate frames are in fact the same frame Mar 22 '15 at 17:59
• @SebiSebi, your 2nd sentence in your comment above is not true unless $x'_1 = x'_2 = 0$. This is an elementary result from the Lorentz transformation: $$t = \gamma(t' + \frac{vx'}{c^2})$$ For any $x'\ne 0$ and $t'=0$, $t \ne 0$. Mar 22 '15 at 18:40