My physics book states that in the construction in the figure, if the event C has coordinates x'= 0 and t' = l in the primed frame, then event C has x -coordinate in the unprimed frame:

$$x_C = \frac{l}{\sqrt{1-v^2}}$$ The angle between t and t' should be the same as between x and x' are the same. The slope of the world line of the rear of the rod and the front of the rod are the same. Sorry about my iffy picture. When I try to derive the expression myself I don't get this answer. Using the invariance of the interval at C:

$$-t² + x² = -t'² + x'² = l²$$

Dividing by $x²$ on both sides and rearranging:

$$\frac{x²}{x²} = \frac{l²}{x²} + \frac{t²}{x²} \leftrightarrow$$ $$1 = \frac{l²}{x²}+\frac{1}{v²} \leftrightarrow$$ $$ 1 - \frac{1}{v^2} = \frac{l²}{x²} \leftrightarrow$$ $$x²= \frac{l²}{1-\frac{1}{v²}} \rightarrow$$ $$x = \frac{l}{\sqrt{1-\frac{1}{v²}}}$$

If someone could point me to where I'm going wrong I'd be really grateful. I've been stuck on this for a while. The book is Schutz chapter 1 (Lorentz Contraction).


closed as off-topic by sammy gerbil, Kyle Kanos, Jon Custer, heather, ZeroTheHero Jul 8 '18 at 2:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, Kyle Kanos, Jon Custer, heather, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ In dividing both sides by $x^2$ and going from your second equation to your third equation, why do you assume $\nu = x/t$? $\endgroup$ – Samuel Weir Jul 2 '18 at 21:48
  • $\begingroup$ My book states that the tangent of the angle between the t and t' axis is the velocity of the primed frame. The angle is the same between the x and x'-axis (although that might not be obvious in my figure), so the tangent of this angle should also be the velocity since they are identical. The tangent of the angle is x/t. Oh nonono - it isn' - it is t/x ! Thanks Samuel! $\endgroup$ – user18093 Jul 3 '18 at 18:40
  • $\begingroup$ That solved my problem - if you want write it as an answer and I will accept it, otherwise I can write it out. Thanks! $\endgroup$ – user18093 Jul 3 '18 at 18:46
  • $\begingroup$ I see that the question was put on hold, and I read the guide-lines now. Apart from tagging it 'homework-and-exercises' I don't think there is much to change - the question was more about getting help to find a 'bug' in my calculations than about deeper understanding, so I am fine with it being closed or moved to the correct forum - and really happy for the help! $\endgroup$ – user18093 Jul 8 '18 at 15:41