I proved that, under SR, the only way the impact times could be equal in the embankment frame \(t_1=t_2\) is if the length of the body is \(L=0\). I don't know why you cannot agree to that, considering it's just an inverse Lorentz transform. Since you cannot agree, there is no need for me to take it to the next level. I was going to show you how an arbitrarily small gravitational field would not change that result significantly, but I won't do it now. Thanks for saving me the time I would have wasted on you.
This is impossible, as I already pointed out. Thomas precession cannot know how far it is to the ground. Your counter-rotation would have to apply for a fall of 3 inches, 6 feet or 20 miles. This is a logical contradiction unless the body magically remains parallel to the floor at all times, in all frames (or...you know...I gave the correct explanation that the body does not main rigidity in all frames, and you're the one waving your arms)
Tach does not even understand what he is claiming. The only way the head and feet of the body can impact the floor at the same time in both frames is if both frames share the same interpretation of simultaneity. If that were the case, there could be no length contraction and no time dilation between the two frames either. He seems to think this arises due to a gravitational field, no matter how weak. In other words, he thinks the gravity of a grain of salt imposes Newtonian absolute universal time into relativity. What a maroon.
Gravity i snot the issue here. Neddy Bate is correct in his statement that the body cannot land flat in the embankment. We know that the body lans flat in the train frame. Thus if we were to put clocks on the floor of the train at the points where the head and feet land, and those clocks are synchronized in the train frame, they each will have identical readings at impact. But we also know that these clocks are in motion with respect to the embankment and thus are subject to the relativity of simultaneity and cannot be in sync according to the embankment frame. The reading on the clock where the head impacts and the reading on the clock where the feet impact is invariant between frames. Thus the embankment frame must agree that the two clocks read exactly the same at impact. And since the clocks cannot read exactly the same at a given instant in the embankment frame, the head and feet cannot land at the same instant in the embankment frame. This is true whether the body was falling due to gravity or just traveling downward at a constant velocity. Gravity adds an extra complication if you want to do an in depth analysis, but the existence of gravity is not relevant to the solution of the main issue; If one end hits first in one frame, would the force distribution be different than that in the frame in which it lands flat. The answer to this is in the difference between time-like and space-like intervals. Consider the normal situation of a rod falling at an angle. When the first end hits the shock of impact travels through the rod effecting the rest of the rod. Thus if you pick some point along the rod, when the force caused by the impact of the end reaches it acts to counter the points downward momentum. in turn some of its momentum is transfered back down the rod to the impact point. If the force of this interaction is greater than the strength of the rod, the rod will suffer some permanent deformation. Also, since out point along the rod has given up some of its momentum, when it does finally impact it will be at a lesser speed than the end that hit first. The upshot is that the force of impact is not equally distributed along the length of the rod, Since there is time enough between the initial end hitting and the point along the rod hitting the surface for there to be a causal link between them (the impact at the end has an effect on the impact of the point along the rod), there is a time-like interval between the two events. Now let's consider the situation in the OP. here we hava an object that is falling skewed due to RoS while at same time having a relative sideways motion (in the embankment). For our example, assume that the relative velocity is from left to right. This means the the left end will be lower and hit first. when it hits, the impact force travels through the rod. For simplicities sake, we'll say that it moves at the greatest possible speed, that of light. So, from the embankment frame, the impact moves through the rod at c from left to right. however, since the rod itself is also moving from left to right, the speed that the impact moves with respect to the rod is c-v. if x is the distance of some point of the rod form the end in the train frame, then the time it takes for the impact to propagate form the end to this point is \(\frac{x}{\gamma (c-v)}\) The time delay between the end and the same point is: \(\frac{xv}{\gamma c^2}\) For any value of x or v, the RoS time delay will be shorter than the time needed for the impact to propagate to the same point. So what does this mean? It means that due to the RoS, the delay between the end and some point along the rod hitting the surface is less than the time it takes for the impulse of the end's impact to reach this point. Which means that every point along the rod's length hits the surface without being affected by the impact of the points of the rod that impacted before it (and vice versa) . Since there is no causal link between the impacts at different points of the rod, the interval between these events are space-like. This also means each point of the rod hits with the same force and speed as every other part of the rod, and thus the total force of impact is evenly distributed along the length of the rod, just like it it in the train frame.
Excellent, Janus. Your wave propagation analysis is equivalent to my rigidity explanation. We can all agree that Tach was again bluffing when he said that the mathematical formalism, which he supposedly possessed, would prove his point.
But this is NOT an SR problem, it isn't an exercise for beginners. You are trying desperately to reduce the problem to your (low) level of understanding. Because you can't apply the Lorentz transforms, this is a GR problem and both frames, the car and the platform are accelerated so, you can't apply Lorentz transforms. You won't because you can't.
...only in your reductionist understanding. The presence of the gravitational field nullifies your attempt at reducing the problem to a RoS problem. GR, the proper theory for treating this problem, has different concepts of time dilation, length contraction and RoS than SR. You are attempting to force the SR formalism because this is your limited level of understanding. It is "moron", and it applies to you.
It doesn't have to "know" if the solution is independent of height. Since you cannot produce any mathematical formalism, you wouldn't and couldn't understand this simple concept.
No, the math says that the right end will disconnect first, such that it hits the ground first (in the absence of all other effects). This is what happens when you make claims without doing any calculations, like RJBeery. Correction, right to left, the right end would hit first (if you ignore all other effects). This is what happens when you tell stories instead of doing the calculations. Besides, the force of impact doesn't travel at c, it travels (roughly) at the speed of sound. You mean, \(s-v\) where s is the sound speed. I very much doubt it, you are inventing some very weird relativistic form of materials theory. This makes no sense, the propagation of force speed is not dependent of the relative motion of frames, you are inventing a very weird form of relativistic theory of materials. At the very worst, you could argue that , if the speed of force propagation is \(s\) in the car frame, then it will be \(\frac{s-v}{1+vs/c^2}\) in the ground frame. Even this is reaching, since, best of my knowledge, there is no relativistic formulation of materials theory. Why don't you correct all these errors and start again?
Tach, we've already established that you're a poseur and a bluffer when it comes to claiming that others "need only to work out the math" to see your point. You're simply wrong (again) and your string of responses above is nothing but meltdown behavior.
Who is "we"? You, the biggest fraud of this forum, the person that can't solve a simple exercise if your life depended on it? Tell you what: I just deconstructed Janus' argument, can you correct his mathematical errors. Let's see it, fraudster.
No, fraud , you were very quick to claim that Janus' solution agrees with your "proof", now that I have pointed out that there are quite a few errors in his solution, how do you answer that? Do you even understand the objections? Are you ever going to admit that you are a fraud, pretty much like Farsight? The questions are rhetorical.
Wrong. This can be easily shown by the following: We know that the two ends of the body are released at the same time in the train frame. This can be accomplished by placing a light source at the midpoint between the two ends and having each end release when the light reaches it. Thus the light reaches both ends at the same time and both ends are released simultaneously. In the embankment frame, the train and body are moving from left to right. Now it is obvious that the light traveling to the left towards the left end which is moving to the right will strike the left end before the light traveling right towards the right end will reach the right end which is also moving to the right. Thus in the embankment frame it is the left end that will release first and thus it is the left end that hits first.
Err, let's try again, with math this time: \(t'=\gamma(t-vx/c^2)\) \(dt'=\gamma(dt-vdx/c^2)\) \(dt=0\) \(dt'=\gamma(-vdx/c^2)\) \(dx=x_{right}-x_{left} >0\) Therefore: \(t'_{right}-t'_{left}=dt' <0\) i.e. the right end falls first. There is also a physical explanation for this: in the car frame , the two ends fall simultaneously, this is why, the platform observer must see the end farthest from him (the right end) fall first, not the other way around. You appear to be a good person, genuinely interested in finding a solution. We can work together to plug the holes in your current solution. The biggest ones are in your attempt at creating a relativistic theory of elasticity. I have an even bigger objection: the attempt at solving this problem using SR (and RoS, to make things even worse) is ill founded, the presence of the gravitational field makes this problem a candidate for GR, not SR.
If the train is moving from left to right I would hope that it's obvious that the left side will be released first. I'm also curious how you would claim that EITHER side can be released first yet both sides hit the floor simultaneously if the floor was negligibly close to the hanging body.
No more free lessons for the forum fraud, RJ. You will need to learn the Lorentz transforms all by yourself. In the frame of the train both ends are released simultaneously, did you sleep through the intro again, RJ? You make up strawmen and you beat them up. You can keep your strawmen.
