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Consider a frame of reference S' moving with a velocity $v$ along x-axis with respect to a frame of reference S.

Using Lorentz transformation, $x'=\gamma(x-vt)$, where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

$\implies \Delta x'=\gamma\Delta x\tag1$

Using Inverse Lorentz transformation, $x=\gamma(x'+vt')$

$\implies \Delta x=\gamma\Delta x'\tag2$

However, equations $(1)$ and $(2)$ are simultaneously true only if $\gamma=1$. This further leads to 2 different results for length contraction.

$$L'=\Delta x'=\gamma L$$ and $$L=\Delta x=\gamma L'$$

Please tell what's wrong in this.

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  • $\begingroup$ I didn't follow. Set $t^{'}=\gamma(t-xv/c^{2})=0$ so the frames coincide and solve for $t$. Then plug ${t}$ into $x^{'}=\gamma(-vt+x)$ to obtain $x^{'}$ in terms of $\gamma$ and $x$. $\endgroup$ Sep 18, 2019 at 7:50

3 Answers 3

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When you make a measurement, you need to be clear about what reference frame you are using.

If Alice is on Earth watching Bob go past in a rocket, then in Alice's reference frame the rocket appears contracted in its direction of motion.

But for Bob, it's Alice who is doing the moving and so in Bob's reference frame it's Alice who is length-contracted.

Different reference frames, different measurements. As you say, the only situation where they'd all agree on the length of an object (in the direction of travel) is when the two reference frames are not moving with respect to each other, i.e. if $\gamma=1$. But in general they won't agree on lengths if their reference frames are in motion.

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The “length of an object” is not a well-defined concept, because extended objects are problematic in relativity. Indeed, an object is viewed as a set of simultaneous events. But simultaneity depends on the reference frame, so that the identity of the parts of an object depend on the frame. So it is best to stick to events.

Two events separated by $L$ in their rest frame, and simultaneous in that frame

$$x_1=0,\quad x_2=L,\quad t_1=0,\quad t_2=0$$

leads to

$$x_1'=0,\quad x_2'=\gamma L,\quad t_1'=0,\quad t_2'= -\gamma v L/c^2$$

On the other hand, 2 events separated by $L$ in their own rest frame, but occurring simultaneously in the moving rest frame

$$x_1=0,\quad x_2=L,\quad t_1'=0,\quad t_2'=0$$

leads to

$$x_1'=0,\quad x_2=L/\gamma$$ which is the Lorentz contraction.

The former corresponds, say, to two flashes emitted at the front and back end of a rocket, simultaneously as viewed by the rocket's captain. These are observed from a system at rest as being further apart than on the rocket.

The latter corresponds, say, to two photographs of the rocket, taken from the ground, at 2 different positions.

The two results you compute are both meaningful and correct in appropriate circumstances. But they are never both applicable, so no contradiction arises.

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What's wrong is that you are playing around with some special relativity notions without fully adopting the idea that you have to think in $4$ dimensions.

Let's take a simpler example. Suppose we can only grasp $1$-dimension and we are looking at a $1$-dimensional object, a line segment. This line segment will have a length. Consider that this simplified Universe has $2$ dimensions. Imagine that the line segment is rotated in that additional dimension with an arbitrary angle. What will we perceive then? We will then see a shorter line segment than before. So, is the line segment shorter? Well, for all practical purposes, it is measurably shorter for the $1$-dimensional beings, but in $2$-dimensions what really happens is that the line segment has been rotated, while its length in $2$ dimensional space remained the same.

So, you can take this example and make an analogy with the Universe we do live in. Instead of $1$ dimension we can perceive $3$ spatial dimensions. Instead of $2$ dimensions our Universe (at least in SR) has $4$ dimensions. And instead of rotations we have Lorentz transformations, which are a special kind of rotations.

And to actually answer your question, let's go back to our simplified Universe. Bare in mind that you cannot rotate along with the line-segment. You are constrained to move along one single direction. You can only perceive the segment projection onto your world-line. This is equivalent to saying that in the actual Universe you cannot move relative to yourself. You are always in your rest frame (proper frame). Someone else from the same $2$-dimensional Universe which has the world-line parallel to the rotated segment will measure its "full length". But if the line-segment is rotated back to its original position, you will be to one to measure its "full-length", and the other observer will measure its length as being shorter. What actually happens is that the segment-line length in $2$-dimensions is not changing, but its length (as viewed by different $1$-dimensional observers) has different values depending on how much it is rotated with respect to the observer's world-line.

An observer in the $2$-dimensional Universe will always see that the objects that are parallel to its word-line have "full-length". And the observer will always see the objects that are rotated relative to its world-line as being shorter than they really are when they are measured in a world-line parallel to them. But the idea is that there is no special world-line, there is no direction more preferable than the others.

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