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Lorentz transform can be written, according to wikipedia, as:

$$ \begin{align} ct' &= \gamma \left( c t - \beta x \right) \\ x' &= \gamma \left( x - \beta ct \right) \\ y' &= y \\ z' &= z. \end{align} $$

I see a complete symmetry between terms ct/ct' and x/x', that is, I can exchange ct with x and ct' with x' obtaining exactly the same equations.

However, books talk about "time dilation" and "space (Lorentz-Fitzgerald) contraction", not about "time and space dilation".

I assume "time" is related to term ct and "space" to x.

Please, could someone help me to see my error or understand the point of view under the expression "space contraction"?

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  • $\begingroup$ I didnt quite undetstand the problem here. Why its not called space-time dilation but called seperately ? $\endgroup$
    – seVenVo1d
    Jan 13, 2019 at 20:09
  • $\begingroup$ @Reign: space contraction vs space dilation. If the equations implies a time dilation, by symetry, they means also space dilation $\endgroup$ Jan 13, 2019 at 20:14

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Although the treatment of $x$ and $t$ in the Lorentz transformation is exactly analogous, the symmetry is broken by the way we measure distance and time. To measure proper time, we allow a clock to exist along some timelike world-line. To measure proper length, we have to carry out a much more complicated process, including verifying that the the measurements at the ends of the measuring rod are simultaneous. To verify this, we need to send signals back and forth.

The clock's world-line is one-dimensional, and an observer can be with the clock at all times. The measuring rod's world-line is two-dimensional, and the observer can only be at one part of the rod at one time.

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To measure the length of a rod in the frame of reference in which it is stationary (the S' frame or "the rod's frame") we place a ruler against it and take readings against the ruler of either end of the rod. It doesn't matter whether or not we take the readings at the same time.

To measure the length of the rod in the laboratory or S frame, in which the rod is whizzing past the ruler we must clearly take simultaneous readings of either end of the rod against the ruler. Thus the events of taking the readings are at (say) ($T, x_1) \text{and} (T, x_2)$.

The separation of these events in the ruler's frame (where the events are not simultaneous, but this doesn't matter) is$$x_{1}'-x_{2}'=\gamma(x_1-\beta cT-[x_2 -\beta cT])= \gamma(x_1-x_2 )$$

So $x_1-x_2 < x_{1}'-x_{2}'.$ Hence the length contraction.

Now consider time dilation. A rocket whizzes through the laboratory (the S frame). Let it emit flashes of light separated by a time interval $T_0\ (=t_{2}'-t_{1}')$ as measured by its own clock. In the clock's frame (the frame in which the clock is stationary) these flashes are in the same place (X'). In the lab frame they are in different places. Using the inverse transform, the time interval between the flashes, as measured in the lab by (synchronised) clocks at the places where the flashes occur, is $$ct_2-ct_1=\gamma(ct_{2}'+\beta X'-[ct_{1}'+\beta X'])=\gamma(ct_{2}'-ct_{1}').$$ Thus $t_2-t_1\ >\ t_{2}'-t_{1}'.$ This is time dilation.

We can now see how the asymmetry arises. For length comparisons, measurements must be at the same time in the lab frame. For time comparisons the place is the same in the 'moving' frame, but different in the lab frame.

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  • $\begingroup$ Thanks a lot for your answer. In order to make this answer the "pinned" (accepted) one, could you add the equations for time dilation, to allow comparison? I think they are something similar to (using $ncT_0$ as a periodic tick clock of period $T_0$ in its frame of reference S') : $(ct',x')=(ncT_0,0) \rightarrow (ct,x)=(\gamma ncT_0,\beta c t)$ thus $ct=\gamma ct' > ct'$. $\endgroup$ Jan 14, 2019 at 9:55
  • $\begingroup$ Done. One trick is to choose the Lorentz transforms that require least algebra; I've seen some inelegant treatments! The first sentence of Ben Crowell's answer is, in my opinion, a very neat summary of why there is the apparent asymmetry. $\endgroup$ Jan 14, 2019 at 11:02
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Even you can derive those 2 equations in terms of each other, they mean different things. Its called time dilation and space contraction because they mean seperate things. Space and time does mot respond to same at relativistic speeds. While lenghts gets shorter (contracts) time passes more slower (dilates)

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  • $\begingroup$ Could you bind last phrase of your answer to the lorentz transform equations. In particular, about the symetry ct<->x ? $\endgroup$ Jan 13, 2019 at 20:40
  • $\begingroup$ @pasabaporaqui I am not sure how to do that. My point was this. In general when we apply $L'=L/ \gamma$ or $t'=t\gamma$ we can see how these components behave. This is called as we know space contraction and time dilation. In Lorentz transformation, the idea is to write the coordinate transformation for another object that moves in relativistic speed. If space-time would dilate at the same time then we could have write, $t'=t\gamma$ and $L'=L'\gamma$. This is all can I do.. $\endgroup$
    – seVenVo1d
    Jan 13, 2019 at 21:33

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