# Basic Special Relativity Question

Discussion in 'Physics & Math' started by Fednis48, Apr 22, 2013.

1. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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I don't see why there is a need to bother bringing SR into such a situation at all except to make it just a little more complicated - nor why this is a confusing issue. Signal delay causes observations of far away events to happen after nearby events even though to an observer in between they happen at the same time. So what? That's a feature of Galilean Relativity, not SR and not even a little bit profound/confusing. Bringing SR into it just changes the timing calculations a little, it doesn't change the qualitative result.

What am I missing about why this is at all interesting?

3. ### TachBannedBanned

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Post 8. You made some basic algebraic mistakes. Once you figure what the mistakes are, we can continue.

5. ### PeteIt's not rocket surgeryModerator

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Please point out the mistakes you found, so I can correct them.

7. ### TachBannedBanned

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Basic algebra would give the result $tan(\theta)=-\frac{uV\gamma'}{c^2}$.
You got the wrong sign and you got the wrong fraction. I do not know why you are so astute in pointing out other's mistakes while you seem totally incapable of detecting your own. This is a recurring problem with your derivation.

8. ### PeteIt's not rocket surgeryModerator

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Sorry, I don't see how you get that result.

9. ### TachBannedBanned

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One way would be by doing fraction simplification correctly. Try calculating $\frac{y^"_B-y^"_A}{x^"_B-x^"_A}$ from your definitions.
Fednis48 managed to figure out the fraction correctly but he still didn't get the sign right.
Neddy Bate managed nothing.

10. ### Fednis48Registered Senior Member

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You misrepresent me. After seeing your criticism, I thought I agreed with your fraction, but not your sign. After closer inspection, I realized I didn't agree with any part of your answer, and I think Pete is entirely right. We started this whole thread just so you could point out errors. Could you please be more specific than "the algebra is wrong somewhere"?

Edit: Since you seem averse to precision, let's break it down with line numbers:

\begin{align} 1.&y_A''(t'') &= -u\gamma'(t'' + \frac{Vx_A''}{c^2})\\ 2.& &= \frac{-ut''}{\gamma'} \end{align}
\begin{align} 3.&y_B''(t'') &= -u\gamma'(t'' + \frac{Vx_B''}{c^2})\\ 4.& &= \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \end{align}
\begin{align} 5.&\tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ 6.& &=(\frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2}- \frac{-ut''}{\gamma'})/(\frac{1}{\gamma'} - Vt''-(-Vt'')) 7.& &= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\ 8.& &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}​

By number, which step(s) are wrong? (Sorry about the ugly alignment; I'm not super-great at Tex.)

11. ### TachBannedBanned

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LOL

So, you can't calculate the fraction either. Let's try to further dumb it down for you: what is $y^"_B-y^"_A$?

12. ### PeteIt's not rocket surgeryModerator

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\begin{align} y_B''(t'') &= \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \\ y_A''(t'') &= \frac{-ut''}{\gamma'} \\ y_B''(t'') - y_A''(t'') &= \frac{uV\gamma'}{\gamma c^2} \end{align}​
Looks OK...

\begin{align} x_B''(t'') &= \frac{1}{\gamma'} - Vt'' \\ x_A''(t'') &= -Vt'' \\ x_B''(t'') - x_A''(t'') &= \frac{1}{\gamma'} \end{align}​
Looks OK...

\begin{align} \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} &= \frac{uV\gamma'}{\gamma c^2}.\gamma' \\ &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}​
I don't see a mistake.

13. ### TachBannedBanned

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Do you think you can calculate $y^"_B-y^"_A$ without making basic algebra mistakes?

14. ### brucepValued Senior Member

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Last edited: Apr 23, 2013
15. ### Fednis48Registered Senior Member

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$y^"_B-y^"_A=\frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2}- \frac{-ut''}{\gamma'}=uV\gamma' / \gamma c^2$

16. ### TachBannedBanned

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I am not done yet with the "three musketeers" (Pete, Fednis48 and Neddy Bate). Once we get past their basic algebraic mistakes, the fun starts

17. ### TachBannedBanned

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Nope. Try again. I really narrowed down the expressions such that you cannot continue to make errors. It is fascinating to see you and Pete making the same basic error (and stubbornly repeating it).

18. ### PeteIt's not rocket surgeryModerator

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Post 29, Tach.
I don't see the error you describe. I'm beginning to suspect you may be mistaken.

19. ### Fednis48Registered Senior Member

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Nope. The only thing I predict to be frame-dependent is the question: do both ends of the rod hit the ground at the same time? Simultaneity is relative, so this is fine. The results of any well-posed experiment should agree in both reference frames.

Our thought experiment is that a metal rod is moving straight down, with no acceleration, in a fast-moving train. If you think this is "contrived", I don't know what to tell you.

20. ### Motor Daddy☼☼☼☼☼☼☼☼☼☼☼Valued Senior Member

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I'm gonna have the last laugh.

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Seriously?

22. ### TachBannedBanned

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seriously, try again:

In $S''$:
\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \end{align}
\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \end{align}​

23. ### Fednis48Registered Senior Member

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...you still think that expression is wrong?! Now I really want you to point out the exact line-number of the error as per post 27, because the expression you disagree with is of the form $a+b-a=b$. It doesn't get more simple than that.