Basic Special Relativity Question

Now that we've slogged through the pages and pages of slow algebra, let's "talk about the rest."

 

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It isn't the algebra that is "slow"

you haven't really admitted to error, so let's not.
Pete is still thinking, so, let's not.
RJBeery and Motor Daddy are STILL polluting the thread with their crank trolling, so let's not.

 

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Tach, I'm laughing at you, do you understand that?

 

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Of course I do, you and RJBeery are equals.

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Good. So we're on the same sheet of music. Drive on, buddy. I'm enjoying the show (train wreck).

 

... I opened myself up to that one.

I admitted that I made an error in my algebra. Your expression from the first page was right.
If you want to put things on hold until Pete verifies that he's on the same page, that's fine. Now that someone has actually specified which step is wrong, I'm sure he'll do so soon.
If we wait for Motor Daddy to stop trolling, we might be waiting a long time, so let's not wait for that

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.

 

Thank you.

Well, we need for BOTH cranks (MD AND RJBeery) to stop polluting the thread.
Let's wait for Pete to admit he was in error.

 

A Crank is useful. A Twit? Not so much.

 

So, you aren't useful

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You're confused. You be the Twit, Yo. LOL

 

Seeing as how he was right, I retract my statement that Tach is an elitist twit, and replace is with the statement that he is just an elitist. As for cranks being useful, no comment.

Edit: Where is your avatar from? Its eyes shine with a delightful and fitting madness.

 

I got it from another site a long time ago, I still haven't figured out what the heck it is.

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Thank you Tach and Fednis, I did make an algebra error when transforming \(y'_B\) to \(y''_B\)

The correction to the OP is as follows:

In \(S''\):

\(\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= ... &= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right) \end{align}\)
\(\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} - \frac{uV}{c^2} \right) \end{align}\)
The angle of the rod with the x-axis at time t'' is:
\(\begin{align} \tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ &= \frac{-uV/c^2}{1 / \gamma'} \\ &= \frac{-uV\gamma'}{c^2} \end{align}\)​

Tach,
You also said that the physics treatment is incorrect.
What do you mean?

 

The sign on your final formula is still wrong, it needs to be a minus:

\(\begin{align} \tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ &=- \frac{uV\gamma'}{c^2} \end{align}\)[/indent]

The negative angle is consistent with endpoint \(A\) dropping first, I hope you agree.
I'll get back to you tomorrow with the physics errors in the above. Judging by how long it took to admit that the math was wrong, I can only shudder to thinking how long admitting that the physics is wrong.

 

I look forward to your clear explanation.

 

The way Pete defined it, \(x''_B-x''_A\) is positive, so if the angle is negative, \(y_B''-y_A''<0\). This rearranges to \(y_B''<y_A''\), which would mean side B drops first. Right? I'm sorry to belabor the algebra further; if there's a mistake here, please point it out to me right away so we can get to the physics asap.

 

\(v' = (V, 0)\) is the velocity of \(S''\) relative to \(S'\)

This means that the platform \(S"\) moves in the posititive x direction (or, otherwise said, the train moves in the negative x direction wrt the platform) , so , the light from the source hits \(B\) FIRST because of the closing speed being \(c+V\) whereas , the light "chases" \(A\) at the closing speed \(c-V\), so , \(B\) drops first.
On the other hand, \(y^"_B<y^"_A\) means that \(B\) drops first.

 

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Positive \(\theta\) means \(A\) meets the floor first. Negative \(\theta''\) means \(B\) meets the floor first.

\(\theta\) if negative if u is positive (ie the train is moving in the positive y direction in the rod frame), and V is positive (the platform is moving in the positive x direction in the train frame).

I see that in the animation I did earlier, I have made V negative, which is why the angle is positive.

 

Good, we agree, B drops first (not that it is very important).
Do you also agree that the time between B dropping and A dropping is dictated by RoS? \(\gamma' \frac{VL}{c^2}\) where \(L=\) rod length?

 

Is this computerized math equations checking program generally available to everyone or only to someone working in the field? Anyone can input whatever math equations and get out a diagnostic of what is wrong or missing or inconsistent and such error message? Then one just attacks each section and makes corrections to typos and missing or wrong signs and such until the error messages stop? Is that what Tach is using? Why don't everyone have this program so we don't have to be hostage to his computer generated messages? Maybe he is using these error messages from the computer to say there is "something wrong there" and wait till you find it for him so he can input it to his computer program and it tells him what was wrong and he comes back and says he found it? Sounds suspicious because he doesn't make clear identification of what is wrong but waits until he has something from you that he can compute right away and pretend he did it all? It would explain his reticence to be straight and clear about where the error is and what it is? He playing for time until one of you finds the error bit for him to input into his math equations analyzer and it spits out the right bit in question? Is this kind of disrespectful exploitative manipulation and sneakiness allowed on this forum by moderators as "fair debate"? I am shocked if that is what goes for fair debate here. It's all too unfriendly and mean, so I will keep out and just watch it happen until moderator does something to make people not so sneaky and rude.