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Ok I have a simple question.

We consider two frames, one is $F$ which is at rest and one is $F'$ which is moving with $v_0$ in the $+X$ direction with respect to $F$. Suppose a rod is at rest in $F$ and its ends are at $x_A$ and $x_B$. In frame $F'$ its ends are observed to be at $x_A'$ and $x_B'$ respectively. Now Lorentz Transformations give us:

$x_B'− x_A' = γ(x_B − x_A) − βγc(t_B − t_A)$

Assuming $t_b=t_A$ we get:

$x'_B - x'_B = γ(x_B − x_A)$

Now since the rod is at rest in $F$, It should be moving with some velcity $v$ (idk if its $v_o$) in the $-X$ direction with respect to $F$. Now due to Lorentz Contraction $x'_b -x'_A$ should be smaller than $x_b -x_A$ but the equation $x'_B - x'_B = γ(x_B − x_A)$ tells otherwise since $\gamma >0$

What am I missing. I have been bugged on this lately and unable to continue further because of this. Can anyone make this (very)clear to me please?

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You can't assume $t_B=t_A$. The two spacetime events corresponding to the ends of the rod at simultaneous times in frame $F'$ are not simultaneous in frame $F$.

Think about how the times transform (I'm setting $c=1$): $$t_A'-t_B'=\gamma (t_A-t_B) - \beta \gamma (x_A-x_B)=0$$ At the end I'm setting to zero because I want the events to be simultaneous.

We can use this to substitute in your equation:

$$x_A'-x_B'=\gamma (x_A-x_B) - \beta \left(\beta \gamma (x_A-x_B)\right)=\gamma^{-1}(x_A-x_B)$$

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  • $\begingroup$ Whoa, I didn't see that, thanks a lot :) $\endgroup$ – ujwal kumar Dec 8 '18 at 6:33
  • $\begingroup$ No problem, I ran into this same question myself at one point. $\endgroup$ – octonion Dec 8 '18 at 6:34
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@John Rennie offered a link where 18 answers were given to the reality of length contraction. I read some, but it's anyway disconcerting that 113 years after Einstein's paper we are still arguing about its meaning. And not only we, but authoritative books and so on...

Yet it appears to me that something is never (or very seldom) clearly stated. Perhaps this explains why so many beginners (and not only) get confused by dilation and contraction matters. I already examined these points in other posts, in recent days too. But there is always something which requires a better clarification.

The first point is about so-called "coordinate contraction". To me this is an unfortunate definition. It isn't a matter of coordinates, but of reference frames. The difference is very sharp, yet frequently neglected. Coordinates are mathematics, frames are physics. In order to observe (much better, measure) length contraction you need first of all a frame of reference, which is a set of points stationary wrt one another, but also a set of measuring instruments of various sorts. You'll ever do without a set of synchronized clocks.

Then you must have the objects you want to measure distance in that frame. Frequently the objects will be the extremes of a rod, or something like that, but not necessarily. For instance, in the well-known experiment about half-life of muons there is no rod. There are two labs, one up on a mountain, the other at sea level, and counters to measure muons flux. It is important therefore to be prepared to identify the objects we are talking about.

But let's think of a rod, as it gives a more concrete vision of the whole matter. A third thing is however necessary: how do you define the length you are about to measure or to compute? If the rod stands still in your frame, no problem: length is simply a comparison with some standard. Grossly said, you see how many times your standard metre fits between the rod's extremes, and that's all.

But the rod is moving: what is its length? A precise answer is needed, and not always it's given. For instance octonion replied

You can't assume $t_B = t_A$.

He's right, but why? Simply because the understood definition of length of a moving object is another. It's still a definition however.

In order to measure length of a moving rod you have to record positions of its ends at the same time of your frame. Then you will at ease measure the distance between these positions.

The point is that this procedure is not exactly simple. In order to put it in operation you need

  • to have a whole battery of synchronized clocks, ideally an infinite number of them
  • to program clocks to record - each one by itself - the passage times of the rod's ends
  • to search in this collection of data two recordings which report the same time, one for passage of left extreme, another for the right one
  • to get the corresponding coordinates of those clocks and take the difference.

And you're done: this difference is the required length of the moving rod.

Once you know this, you may also do the calculation you tried, but with $t'_B=t'_A$, and you know why :-)

I can't help remarking that I said nothing about a very delicate point: how are clocks synchronized? This was a deliberate omission. This time I preferred to focus attention on a different topic.

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  • $\begingroup$ Thanks a lot :) Gave me a good idea of how it works. $\endgroup$ – ujwal kumar Dec 8 '18 at 17:16

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