Basic Special Relativity Question

Before you start making ridiculous claims : RJBeery is a troll. There is no "program" that calculates these things, all calculations are done by hand. The errors are quite obvious except for people who have a blind spot. It would be good for you to stay out of this since you obviously understand very little to nothing.

 

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Yes

 

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Ok, I will stay out because it is not very nice atmosphere here so far. But please tell me, there is no such program anywhere in use to check formal equations and such in SR, GR, QM formalisms? Or any combination of such? It would be very useful to weed out trivial errors and garbage in garbage out situations wouldn't it? Why haven't the mathematicians come up with a algorithm for it that computer scientists and programmers can make to ease the load of checking complicated and long equations and calculations? It would be a good idea I think.

 

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I agree. For what it's worth, I like Pete's choice of distance units such that \(L=1\) to reduce the number of letters floating around, but I agree either way.

 

Sorry about the atmosphere; Tach is a little snappy, it seems. I, for one, will keep answering any questions you have. There are programs like Mathematica and MATLAB that solve symbolic equations, and they are indeed useful for avoiding errors in scientific calculations. For specialists in a given field, I'm sure there are packages for those programs that streamline them for use with Lorentz transforms or the Schrodinger equation or whatever. But for the rest of us, and especially in a casual context like a forum, it's actually a lot more effort to translate the problem into computer code than it is to solve it by hand. Remember, Pete essentially solved the problem in about an hour, and Tach caught all the errors within another hour. The only reason the algebra took so long for us is that Tach wasn't pointing out where the errors were, and Pete and I were looking for them in the wrong places. If we'd really been aiming for speed, the three of us could have solved the system in under an hour, which is less time than it would have taken to program the Lorentz formalism into our computers.

 

Ok,

Let's continue:

For \(\gamma' \frac{VL}{c^2}\) seconds, the endpoint \(B"\) is describing a circular arc with the center in \(A"\) in the frame of the platform. \(A"\) is not moving.

 

Correct. The only thing is that I only pointed out the math errors.....

I very much doubt it, I told you guys exactly where the errors were, you were much too blind to your own faults, this is why it took so long.

 

No.
B'' drops first.
A'' and B'' are both moving inertially in that time.

B'' has velocity \((-V, \ 0)\).
A'' has velocity \((-V, \ \frac{-u}{\gamma'})\)

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\(A"\) yes, \(B"\), no.

Yes.

Nope:
\(A"\) has not been released yet, the hanging wire is still intact, you just agreed to that.
\(B"\) has been released, look at your own picture.

 

Edited as you posted. I thought you were talking about the landing, rather than the release.

 

Release, clearly.

 

In \(S''\), in the time period \(\frac{\gamma'VL}{c^2}\) units long when \(B''\) has been released and when \(A''\) is still attached:

A'' has velocity \((-V, \ 0)\).
B'' has velocity \((-V, \ \frac{-u}{\gamma'})\)

B'' is inertial, not describing a circular arc.

 

Correct.

False, \(B"\) is connected to \(A"\) via a rod of length \(L\) , so, the only way it is allowed to move is in a circular arc. You REMOVED your own picture, I see! This one, though much worse, contradicts you just the saame.
Your own picture contradicts your claim. You realize that your analysis is wrong, now the many weeks before admitting that your physics is wrong start.

 

No, the picture shows B moving in a straight line. Put a ruler on the screen and see.

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This is trivial - B is inertial in S. It must be inertial in all inertial frames.

 

In the frame of the platform:
\(A"\) is moving left to right.
\(B"\) is connected to \(A"\) via a rigid rod ,so it is being dragged left to right while the rod describes a circular arc with the center in \(A"\), so \(B"\) is accelerated, not inertial.
Think about it, \(B"\) acts as the end of a moving pendulum while \(A"\) is the anchor point.

True,it should but it isn't, this is how you get contradictions: by generating scenarios that are unphysical in the first place. The scenario that you generated is a prime example of unphysical scenario. As such, it is self contradictory, on one hand \(B"\) moves inertially, on the other hand \(B"\) does NOT move inertially.

 

The length of the rod in S'' is \(L/\gamma'[tex] before it is released, and longer after. [quote]You REMOVED your own picture, I see![/quote] No, I edited it a while ago to make the time scales consistent. See [post=3063241][u]this post[/u][/post]. Seriously, put a ruler on the screen. B moves in a straight line. [quote="Tach, post: 3063398"]In the frame of the platform: [tex]A"\) is moving left to right. [/quote]
Right to left.

A perfectly rigid rod? In a special relativity exercise?

You know the motion of B' in S' is inertial.
Therefore we know the motion of B'' in S'' is inertial. It's easy to do the transform to prove it.

 

I note that the release of the rod isn't well defined in the OP. It wasn't my intention to analyse the release, but rather the landing. In this scenario, the release can't be as simple as cutting the wires as in the other thread.

 

Maths. Do it.

 

Ah, now comes the desperate attempt at escaping, you need to analyze the WHOLE scenario, not only the parts that suit your prejudices.

You are grasping at straws now and you know it.