Tach, I don't understand why you're unwilling to address the dumbed down scenario. It's completely within the domain of SR, well-defined, easily understood, and easily analysed. Let me know when you're ready to tackle it: The ends of the rod hit the floor simultaneously in the train frame. The ends of the rod do not hit the floor simultaneously in the platform frame.
This is wrong. Here's the correct analysis (for a rod moving inertially to hit the floor of the train in flat spacetime). I'm renaming the frames a little, since I'm starting from the rod frame and progressing to the platform frame. (It's not actually necessary to begin with the rod frame, since the worldines of the rod are perfectly well defined in the train frame, but this is more complete.) \(S = \) rest frame of the rod \(S' = \) rest frame of the train \(S'' = \) rest frame of the platform \(v = (0, u)\) is the velocity of \(S'\) relative to \(S\) \(v' = (V, 0)\) is the velocity of \(S''\) relative to \(S'\) \(\gamma = 1/\sqrt{1-u^2/c^2}\) \(\gamma' = 1/\sqrt{1-V^2/c^2}\) The rod has two ends, A and B. In the rod frame, we place them at (x,y) = (0,0) and (1,0) respectively. In \(S\): \(\begin{align} A &= (x_A, \ y_A) \\ &= (0, \ 0) \end{align}\) \(\begin{align} B &= (x_B, \ y_B) \\ &= (1, \ 0) \end{align}\) The angle of the rod with the x-axis at time t is: \(\begin{align} \tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\ &= 0 \end{align}\) In \(S'\): \(\begin{align} A' &= (x_A'(t'), \ y_A'(t')) \\ &= (x_A, \ \frac{y_A}{\gamma} - ut') \\ &= (0, \ -ut') \end{align}\) \(\begin{align} B' &= (x_B'(t'), \ y_B'(t')) \\ &= (x_B, \ \frac{y_B}{\gamma} - ut') \\ &= (1, \ -ut') \end{align}\) The angle of the rod with the x-axis at time t' is: \(\begin{align} \tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\ &= 0 \end{align}\) In \(S''\): \(\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\ &= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right) \end{align}\) \(\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right) \end{align}\) The angle of the rod with the x-axis at time t'' is: \(\begin{align} \tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ &= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\ &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}\) Your Thomas rotation answer is wrong, as demonstrated in (edit) [post=3062610]post 157[/post].
I think you meant a different post number. When I go to post #160 I see this: http://www.sciforums.com/showthread.php?134380-Relativity-paradox&p=3062627&viewfull=1#post3062627 I grabbed that URL by clicking on the post # and then copying and pasting the URL. I would hope that method of linking to a post is more reliable than calling out the post number, but I am not sure. For example, if a moderator deletes some off-topic posts, both methods might be subject to the same re-numbering problems. But hopefully the link method would survive intact. Oh, and good luck explaining anything to Tach. You're going to need it.
I think your counterpoint depends on an invalid mix of two frames of reference when making your assumption (I highlight your assumption in bold type above). You agreed with me earlier that the platform observer would see the train also crooked as well as the rod, so the train floor and the platform surface cannot both be level parallel with each other at any stage for the platform observer unless you mix both frames like that? That is why your counterpoint to Fednis48 confuses me. My naive understanding of SR analysis practice is that casually mixing frames like that is forbidden? Or are you trying to use some sort of absolute "universal frame" or something like that in your counterpoint assumptions?
Because it is unphysical, you are trying to push under the rug the very gravitational field that gets the whole scenario in motion. The dumbed down scenario is, the real scenario is NOT.
The space of SR is Euclidian, so I disagree. This line of thinking is what makes the rod distorted in the platform frame but not in the car frame. You are doing the same exact thing as Pete is doing, stubbornly trying to apply SR to a problem that is a GR problem. GiGo.
No, I am simply trying to point out that one gets absurd results when trying to force the wrong formalism (SR, in this case) on a problem that requires GR in order to solve.
Last I checked \(x"_B=\gamma'(V)(x'_B-Vt')\) \(x"_A=\gamma'(V)(x'_A-Vt')\) Are you inventing new Lorentz transforms? Not that it matters, since you are stubbornly stuck on solving the unphysical , dumbed down version of the actual problem. Like I said to you and to your Neddy Bate sidekick, GiGo.
Simple, you insist on having the rod moving at constant speed when any 9-th grader can tell you it is accelerated.
We want \(x''\) in terms of \(t''\): \(x_B' = \gamma'(x_B'' + Vt'') \\ \frac{x_B'}{\gamma} = x_B'' + Vt'' \\ x_B'' = \frac{x_B'}{\gamma} - Vt''\)
Err, you mean \(x_B' = \gamma'(x_B'' + Vt'') \\ \frac{x'_B}{\gamma'} = x_B'' + Vt"\) I am having a hard time taking you seriously.
But the gravitational field is present. It is what it makes the rod FALL. Even a 9-th grader knows that.
In the dumbed down scenario, there is no gravitational field. The rod is simply moving at constant speed toward the floor.
You seem intent on solving the dumbed down scenario, that is unphysical. I have no interest in unphysical scenarios.
