Discussion in 'Physics & Math' started by renislaj, Apr 18, 2013.

1. ### Fednis48Registered Senior Member

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I don't know if this is a typo, but why does the y coordinate of $B''$ have an A as its subscript? Is that the mistake?

3. ### PeteIt's not rocket surgeryModerator

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Thanks, missed that typo.

5. ### TachBannedBanned

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There are more. Bottom line, accordong to what he wrote $tan(\theta")=\frac{-ut'-(-ut')}{....}=0$

Why don't you two guys take all this thrash into another thread? You can also take what Neddy Bate posted in the same thread, so we have it all together.

7. ### Neddy BateValued Senior Member

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In 1920, Einstein explained relativity of simultaneity by using a train and a platform. (Chapter 9 of his book, Relativity, the Special and General Theory)

http://www.bartleby.com/173/9.html

It appears that Tach would have argued with Einstein on the grounds that he was using a "dumbed down" and "unphysical" version of the real problem. Tach would have told Einstein that he should have considered the gravitational field that holds the train on the tracks. When Einstein explained to Tach that the point of the thought experiment was to demonstrate that events which are simultaneous in one frame are not necessarily simultaneous in the other frame, (such as the ends of the body impacting the floor simultaneously in the train frame, but not impacting simultaneously in the platform frame), Tach would have told Einstein that he has created a paradox, and that SR was never meant to be used in that way, (because gravity must always be considered).

Poor Albert.

8. ### PeteIt's not rocket surgeryModerator

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The bottom line is this:
\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}​

Have you worked through the transformation of $A'$ and $B'$ 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?

When I do, I hope you explain why the scenario is unphysical. I'd hate to waste time on a flawed scenario.

9. ### PeteIt's not rocket surgeryModerator

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...duouble...

10. ### Fednis48Registered Senior Member

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725
I made a new thread for the dumbed-down scenario. Partly so we can get past the debate over whether a new thread is needed and do some actual physics, and partly in the hopes that Tach will address my post 151* now that the math disagreements have been moved elsewhere.

*Tach said he already did respond to post 151, so I'll also point to post 231 detailing why I don't think that's true. It's too easy for posts to get lost in the clutter when a debate is happening quickly.

11. ### brucepValued Senior Member

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First of all Marcus explained this in the beginning of the thread. Spacetime events are not frame dependent. Spacetime events are invariant. The originator of the thread claims a paradox exists because the spacetime event, body falls, is frame dependent. Not true.

12. ### Fednis48Registered Senior Member

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I guess I don't quite follow you. By "body falls", do you mean the rod falling? If so, that's not a spacetime event. A spacetime event is something that happens at a particular point in spacetime; it has zero extent in space and time. The bar has length and it falls at a finite rate, so its falling is extensive in both space and time.

13. ### brucepValued Senior Member

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You taking a dump is a spacetime event. If you actually conducted this experiment your analysis would include the effect of gravity. Read what Markus wrote down. It's the second post in the thread you linked for Tach's edification. Spacetime events are invariant, not frame dependent. If the ends of the rods hit the floor simultaneously in one frame they hit the floor simultaneously in all frames. Your analysis concludes that spacetime events are frame dependent and that an objects natural path through the gravitational field is frame dependent. Nonsense.

14. ### Fednis48Registered Senior Member

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How can you just assert that without deriving it? One of the hallmarks of relativity is that things can happen simultaneously in one frame but not in another. If you want to show mathematically that the two ends of the rod hit at the same time, head on over to the new thread (I linked to it four posts up) where we're examining Pete's argument to the contrary. But you definitely can't prove it just by saying spacetime events are frame-independent, because the two ends of the rod hitting the ground are NOT the same spacetime event. They're spatially separated.

Oh, and regarding Marcus' second post you pointed me to? In post 151, I show why his description leads to absurd scenarios. I'm still waiting for Tach to respond to said post, but you're welcome to take a crack at it.

15. ### PeteIt's not rocket surgeryModerator

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There is some question over whether this problem can be modeled with SR. It can, if we take advantage of the equivalence principle and choose our reference frames carefully.

(As per the convention established in the thread, the body is now a rod, the track is now a platform.)
According to the equivalence principle, we can't tell whether the rod is pulled to the floor by gravity, or if the floor is accelerating upward to meet the rod (eg the train and platform are on an accelerating spacecraft).

Since the two situations are equivalent, we can analyse either situation and be confident that the results apply equally to the other.

If there is no gravitational field and the train and platform are accelerating, then once the rod is free of the wires its rest frame is inertial, and can safely be modeled by SR.
In the rod rest frame (call it $S$), the train is accelerating straight upward, and the platform is moving horizontally as well as accelerating upward.
We can identify another inertial frame (call it $S'$), moving horizontally relative to the rod rest frame, in which the platform is accelerating directly upward.

In $S$, the wires are cut simultaneously, the two ends of the rod remain at the same level as the floor rises up to meet them, and they hit the floor simultaneously.
In $S'$, the wires are not simultaneously, the two ends of the rod are at different levels as the floor rises up to meet them, and they do not hit the floor simultaneously.

So, that much of the OP is correct.

But this part is wrong:
The impact of the floor meeting the rod is still spread out along the rod.
When the floor hits the head end of the rod, this impact is not felt at the tail end until after the tail end has met the floor.
The tail end of the rod is accelerated by a direct impact with the floor, not by the indirect impact transmitted from the head.
In fact, by the time the head's acceleration is felt by any part of the rod, the part of the rod has already met the floor and is accelerating with the head, so the impact transmitted from the head has no effect.

16. ### TachBannedBanned

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Not, it can't be "safely modeled in SR" both the car frame and the platform frame are accelerated frames and you are being asked to compare the results between the two frames. The only inertial frame in the OP is the rod frame. I pointed all this out early on in the thread.

Yes, I pointed all this earlier in the thread.

But you are not being asked to judge the results in frame $S'$. The introduction of frame $S'$ is just a red herring. You are being asked to compare the results in the two accelerated frames, the platform and the car. So, you aren't solving the OP problem.

Nonsense, the bending has nothing to do with any propagation of the floor resistance force from the head (the point of first impact) to the tail (the point of last impact). You copied the same line of argument from Janus. The rod bends precisely because some points are subject to a (resistance) force while the others are not.

The tail end of the rod is accelerated by a direct impact with the floor, not by the indirect impact transmitted from the head.

Think some more. BTW, my contention to your "solution" is valid not only for the accelerated example (the OP) but also for the dumbed down case you concocted. It is interesting to note that the bogus frame you introduced, $S'$ plays no role in elucidating the puzzle. Nice try, though.

17. ### TachBannedBanned

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I don't misrepresent you, I simple pointed out your hilarious errors. Both endpoints $A",B"$ have the same y coordinate , $-ut'$ <shrug>

18. ### PeteIt's not rocket surgeryModerator

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Come and chat in the other thread, Tach. This one's done.

19. ### TachBannedBanned

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No, it isn't , look up. But thank you for taking the garbage out of this thread and genuinely trying to engage on the OP problem. Try to stick with the OP problem, you haven't solved anything.

20. ### TachBannedBanned

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While the above is certainly true, one of the hallmarks of rookie attempts at solving relativity problems is resorting to using RoS as the means of solving SR problems (the other two rookie mistakes are using length contraction or time dilation). To boot, the OP is not even an SR problem, as pointed out to you repeatedly.

21. ### Neddy BateValued Senior Member

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I agree that typical rookie mistakes are to use only time dilation, or only length contraction. However, I do not agree that rookies tend to use RoS. Usually the problem with using only time dilation or length contraction is that RoS is overlooked entirely. Consider the barn-pole "paradox". Using length contraction alone, we have an apparent paradox -- the pole fits inside the barn in one frame, but it does not fit in the other frame. Once RoS is included, the apparent paradox goes away.

22. ### Fednis48Registered Senior Member

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You seem to be of the opinion that SR is unusable in any experiment with any gravitational field. In the limit of a weak field, GR reduces to SR, so SR can be a good approximation. I don't know mathematically what the condition is for a "weak field"; if you have a formula, I'd appreciate seeing it. But intuitively, a rod falling under earth's gravity for a few feet seems pretty weak compared to the near-c velocity of the train that makes the problem relativistic in the first place.

Of course, if you actually had a good solution using GR that didn't match the SR solution, that would be a compelling argument. But the only GR solution proposed so far (Thomas precession) was shot down in post 151. I've been trying to get you to take a break from attacking my position and defend your own (see post 231 most recently, in case you missed it), but you seem intent on not doing so. Tellingly, with this post, you ignored my latest request for a defense of the Thomas precession picture in favor of answering something that wasn't even addressed to you. Please, please, please just give me one good reason post 151 is wrong. (Again, post 231 clarifies why your two attempts so far to do so are non-responsive, so don't go saying "I already did.")

23. ### TachBannedBanned

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Well, this rookies did. Otherwise they would not be so obsessed with "proofs" based on the rod hitting the floor with all components simultaneously in the car train frame.