In order to improve my naive understandings, I need more than just "no". I need the "why" too. Please can I have the why? Do you mean we can't simplify by placing it all in a very weak, even vanishingly small GR field? Can't we ignore a very very weak GR field this way and go on with Pete's exercise using just SR? Can you say why not?
What is "weak"? How weak is "weak"? How do you solve the problem when it isn't 'weak"? Why always hack at problems instead of solving them? Ponder on this.
Basically, you're right. I don't know enough about GR to mathematically express the conditions under which it's equivalent to SR, but qualitatively, it's in the small gravity/acceleration limit. Like you implied in parenthesis, if gravity is the source of the initial motion of the rod, but its effect is weak compared to whatever large velocity is making you use relativity in the first place, special relativity is a good approximation. Tach's objections are getting more and more obstructionist, so I'd suggest you just follow Pete's arguments if you want to improve your understanding. In fact, if you really want to hone your understanding of relativity, go through his post 182 and try to derive it for yourself. It's not too advanced an exercise, and you might even be able to tell Pete where his mistakes are (if any), since Tach is refusing to point them out unless Pete starts a new thread.
I see you snipped the 'missing' lines from your post. \(y_A''(t'')=-ut'\) and \(y_B''(t'')=-ut'\) have to be transformed to \(S''\) coordinates. t' at \(A''\) does not equal t' at \(B''\) for fixed t''.
Because immediately jumping into the most complex version of a problem, without building up from simplified versions, makes it hard to catch errors and alienates non-experts.
No, your stuff is as ridiculous as before, let me ask you again, what is \(-ut'-(-ut')\). Hint : \(0\) Even better, what is \(y"_A(t")-y"_A(t")\)? You understand why I can't take your math seriously, do you?
As "weak" a curvature as can be effectively ignored. Why not wait to solve the "non weak" formalism after we do the weak case? Is there something stopping us from doing the weak case first and then continuing with the significant GR case afterwards?
But the problem exists even in the absence of curvature, in a uniform gravitational field. The point is that in the presence of any field, the rod is accelerated. The platform and the train frame are non-inertial so you can't use Lorentz transforms. I pointed that out multiple times. You haven't answered any of the questions.
Thankyou. I need to understand a little more of what is involved logically in either formalism case before I can improve my naive understandings by reading back with what I will have learned following these latter stages of this discussion to the end, for both cases I hope. Very confusing so far. I just don't understand why we can't take it one step at a time and do the weak GR case first. But I will persevere and come back to read again over the next week.
Ok. I promise to make no more posts on this thread about the dumbed-down scenario, although I can't speak for Pete. But as for your objections to post 151, where are they? You say you made them, but at this point, I just don't believe you. Point me to the post(s). Or repeat the arguments. Something. Because right now, I can only see two replies you've made: 1. I don't specify my reference frames, which is objectively false. 2. The Thomas precession solution is right, and my post conflicts with it, therefore my post is wrong. Neither of these are remotely responsive. If I missed any other replies, I apologize; please point me to them.
I have not the level of understanding to answer questions at this stage. I am here to improve on my naive understandings. That is why I asked you the "why" of it when I outlined my confusion and my naive understandings around the GR objection. Why not do it with ignorably insignificant weak GR field first and then get on to the GR included solution?
Irrelevant. We want the angle of the rod in \(S''\) at fixed \(t''\), not \(t'\). Irrelevant. What is \(y"_B(t")-y"_A(t")\)? Note that they are functions of \(t''\), not \(t'\). If you start from functions of t', remember to transform to functions of \(t''\) before you answer.
Not irrelevant at all , look at what you wrote. Do I need to keep rubbing your nose in your errors? How many mistakes do you see? I know : none.
Going around and around. The whole reason for using the weak insignificant GR field is so that we can do it in effectively SR. You can get to the GR inclusive case after can't you?
Sorry, Tach, but I just think you're mistaken. The error you point out is not an error at all. Work through the transformation from \(S'\) to \(S''\): \(\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\ &= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right) \end{align}\)Do you agree with this result? If not, where is the mistake? \(\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right) \end{align}\)Do you agree with this result? If not, where is the mistake?
Yes you can. If a problem is too much to grasp in its entirety, the best approach is often to take a simplifying approximation and work up from there.
