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In the laboratory we are observing the motion of a particle moving in the positive direction of the axis $x$ of a frame system attached to it. At the initial moment, the particle is in the origin and in a time interval of $2.0$ ns covers a distance of $25$ cm. A spacecraft passes with velocity $v = 0.80c$ along the $x$ direction of the laboratory, in the positive direction, and from it observes the motion of the same particle. Determine the average particle speeds in the two systems of reference. What interval of time and what distance would an observer placed on the spaceship?

Why can't I use the formulas of time expansion and length contraction, but instead I must use the Lorentz transformations?

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    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – user4552 Jun 23 '19 at 20:13
  • $\begingroup$ @BenCrowell Thank you very much. I sincerely do not attend this site also because of a bad appreciation of my questions. I put my soul to write them better. Can you see my previous question: physics.stackexchange.com/questions/487680/…. I use the translator. If there are unclear parts you are allowed to change what you want. Also for any errors. $\endgroup$ – Sebastiano Jun 23 '19 at 20:18
  • $\begingroup$ @G.Smith I am referring to the text of the exercise. I am not interested in what is going on. Rather because I have to use Lorentz's formulas instead of the classic ones of time dilation and length contraction. If my question is not clear you can edit it according to my comments. $\endgroup$ – Sebastiano Jun 23 '19 at 21:23
  • $\begingroup$ @G.Smith What should I write or change to make the question clearer? I use the translator. $\endgroup$ – Sebastiano Jun 23 '19 at 21:30
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    $\begingroup$ Notice that you CAN use the formulas of time expansion and velocity composition instead of Lorentz transformations: see here (second solution) for an example. $\endgroup$ – Intelligenti pauca Jun 24 '19 at 13:42
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Why I can not to use the formulas of time expansion and length contraction and I must applicate, necessary, Lorentz transformations?

The time dilation and length contraction formulas are derived from the Lorentz transform using some specific assumptions. For time dilation the assumption is that the clock is at rest in one of the frames. For length contraction the assumptions are that the endpoints are at rest in one of the frames and that the length is constant.

In this problem you cannot use the length contraction or time dilation formulas because the assumptions are not met. The particle is not at rest in either the lab or rocket frames.

As an aside, I recommend that new students of relativity not use the time dilation and length contraction formulas at all. They are too easy to misapply as in this problem. Just use the Lorentz transform, it will automatically simplify when appropriate, and you will avoid misapplying them when they are not appropriate.

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IMO newcomers (and teachers) in SR should avoid both time dilation and length contraction. I add Lorentz transformations too. Too often they are applied mechanically, without understanding how and why. The most basic instrument in SR is invariance of spacetime interval. It's enough to solve all elementary problems.

So I welcome @Aretino 's comment. There is however a weak point. In Aretino's solution use is made of RTV - relativistic transformation of velocities (I prefer "transformation" to "addition" or "composition" but I can't dwell on that here). Now RTV formula is usually established through Lorentz', so it looks that Aretino's solution couldn't avoid them.

That's not true since RTV can be proved using only invariance of interval (admittedly in a longer way). I believe this proof isn't widely known, although I'm pretty sure that there are books containing it. I'm afraid that reporting it here could be deemed as OT.

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  • $\begingroup$ In the meantime, my sincere thanks for your response, which I vote in favour of. I believe as I do on TeX.SE that any valid answer is undoubtedly voted positive in my humble opinion. Can I get an explanation of the IMO acronyms? It would be very interesting for me to look into the matter in more detail and I would be grateful. The reason for my question is the difference between the solution given by Aretino's solution by comment and the final comment of this soluttion $\endgroup$ – Sebastiano Jun 27 '19 at 16:28
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    $\begingroup$ @Sebastiano As you can read italian, and if nobody else asks for a specific question, I can refer you to this [link](sagredo.eu/articoli/addvel-2.pdf $\endgroup$ – Elio Fabri Jun 28 '19 at 14:35
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    $\begingroup$ @Sebastiano The final comment you cited is not in contradiction with the solution I linked (by the way: it wasn't written by me, you can see here the main page). The reason is explained in Dale's answer. In the solution I linked a third reference frame is introduced (the particle rest frame) acting as a "bridge" between the two given frames. $\endgroup$ – Intelligenti pauca Jun 29 '19 at 20:20
  • $\begingroup$ @Aretino I've seen both solutions. It's true what you wrote. What I do not agree with is the final comment. You can use the transformations of Lorentz and the classic formulas of dilations of time and contractions of lengths making the appropriate assumptions. Writing that it is not possible to use these last relations is misleading. Thank you and my regards. a solution $\endgroup$ – Sebastiano Jun 29 '19 at 20:31
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    $\begingroup$ @Sebastiano I'm afraid there has been some misunderstanding. Yet the link I gave is precise: it points to a paper of mine, entitled Trasformazione relativistica delle velocità. It aims at showing how Einstein's formula can be derived without Lorentz' transformations. $\endgroup$ – Elio Fabri Jun 30 '19 at 10:21
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Let distance be $D=25cm$, you are asking why we can't just use $D'=D/\gamma$ (length contraction) and $t'=\gamma t$ for $t=2ns$ (time dilation) so that we calculate average speed in spaceship's frame as $u'=D'/t'=u/\gamma^2$ where u indicate velocity of particle in the lab frame? If it's your question then it's an easy one, intuitively speaking it's because spaceship moves in $x$ direction as well, same as particle. If spaceship were to move in y axis, you could use time dilation to conclude that speed is $u'_x=u/\gamma$ but now that spaceship moves in x direction, you should consider this fact that spaceship will ascribe his speed to particle too!

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  • $\begingroup$ Surely there is my upvote. Could you please edit your question a little clearer? If a bar is stopped in an inertial $\Sigma'$ system with $v$ speed in the direction of $x$ motion, an observer placed in $\Sigma'$ sees the bar moving. Here I apply the length contraction formulas with the Lorentz factor $\gamma$. Why do you refer to the $y$ direction? $\endgroup$ – Sebastiano Jun 23 '19 at 22:23
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    $\begingroup$ @Sebastiano forget y direction. that was just another scenario. if we drop your typos in the comment, maybe you are asking if we were to analyse scenario in particle's frame, we would only need to use length contraction, why it cant be done in the case of spaceship too? As i said it's because we should consider the relative velocity between particle and spaceship, in the case of laboratory and particle we need not to do that because velocity of lab's frame is zero $\endgroup$ – Paradoxy Jun 23 '19 at 22:36
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Well, firstly, think of how the length contraction formula came. We used $∆t'=0$ in the inverse Lorentz transformation, right? Why did we did that? Because, to measure a length, you have to measure the both end at the same time. But, here the space time coordinate measured at the same time. First coordinate is measured at $t=0$, and second one is at $t=2$ ns. So length contraction is not valid here.

I myself faced the exact same problem at the first days of my learning relativity. I recommend you to go through the derivative of length contraction again.

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  • $\begingroup$ I have not understood: I recommend you to go through the derivative of length contraction again. +1 $\endgroup$ – Sebastiano Jun 23 '19 at 22:35
  • $\begingroup$ @Sebastiano i think he meant "derivation of length contraction" , yeah i also recommend that. $\endgroup$ – Paradoxy Jun 23 '19 at 22:38

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