Basic Special Relativity Question

Shh, adults are talking and you are clearly unable to follow the description of what is going on. Your insistence on the naive application of RoS demonstrates that you are unable to follow the exercise. The math is not very complicated, it all fits in half a page in post 651. If you want to do something meaningful try understanding the math. lt is a better use of your time then trolling.

 

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Hi tashja,
Alpha hasn't been around a lot lately, and he's had fruitless arguments with Tach before, so I'm guessing he doesn't want to get involved.
Tach has had corrections from rpenner and przyk.

 

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Come on , Pete
Your gig is up, you have no place to wiggle from the correction in post 651 and you know it, time to stop playing your game. You are grasping at straws.

 

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Not everything, since you haven't transformed the time coordinate of the events.
What you have said agrees with my mathematical description (except for a change of origin).
It does not say that the rod breaks.

You said that if the rod is brittle and bends in \(S''\), then it breaks in \(S''\). (Try to prove it - you'll need a relativistic materials theory.)

This doesn't follow, because the bending in \(S''\) has no more physical effect on the rod than length contraction.
Length contraction of the train in \(S''\) doesn't crush the train.
The bending of the rod in \(S''\) doesn't break the rod.

 

Hi, Neddy!

Maybe Tach will never admit he is wrong, but it would help you guys to have Alpha's backing on this. If Alpha says Tach is wrong, you guys can move on.

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The coordinate time is just a label, labels don't change the physics of an experiment. surely you knew that, Pete.

Of course it doesn't "say that the rod breaks", since the math shows that it doesn't.

You tried this diversion a few posts ago, I already answered.

 

I see you've placed t'=0 when the rod is at \(y'=H/\gamma'\) (why?). The equations are neater if the rod is at y'=0 at t'=0, as in the original equations.
Let's keep the floor at y'=0, and place the collision events at \(t'=0\).
You also haven't included the \(t''\) coordinate of the events, so let's fix that.
Finally, let's describe the whole rod in one set of equations as \(R(k)\), where \(k\) is a parameter that varies from \(0\) to \(L\) along the rod, such that \(R(0) = A\) and \(R(L) = B\).

In \(S'(t',x',y')\):

\( \begin{align} \large R'(k) &= (t', \ k, \ -ut') \ \normalsize \left{0<k<L \\ -\frac{H}{u}<t'<0\right} \\ \large R'(k) &= (t', \ k, \ 0) \ \normalsize \left{0<k<L \\ t'>=0\right} \end{align} \)​

Transforming to \(S''(t'', x'', y'')\):

\( \begin{align} R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ -ut' \right) \ \normalsize \left{0<k<L \\ -\frac{H}{u}<t'<0\right} \\ R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ 0 \right) \ \normalsize \left{0<k<L \\ t'>=0\right} \end{align} \)​

Since we're now in \(S''\) it is useful (but not strictly necessary - the result is the same) to express R''(k) in terms of \(t''\) instead of \(t'\):

\( \begin{align} \large R''(k) &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ -u(\frac{t''}{\gamma'} + \frac{Vk}{c^2}) \right) \ \normalsize \left{0<k<L \\ -\gamma'(\frac{H}{u} - \frac{Vk}{c^2}) < t'' < -Vk\gamma'\right} \\ \large R''(k) &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ 0) \right) \ \normalsize \left{0<k<L \\ t'' >= -Vk\gamma'\right} \\ \end{align} \)​

Either way, we get the following result in \(S''\):

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(There's probably a mistake or two in the equations for this post, which was compiled on the run and not well checked. The calculations for the animation were done separately, and checked much more thoroughly.)

 

The maths shows that the rod bends in \(S''\).

You claim that if the rod bends in \(S''\), it must break. That claim is wrong.

 

In \(S''\), \(t'\) is coordinate time and \(t''\) is proper time.
The \(S''\) observer needs to use \(S''\) clocks to describe the physics of the experiment.
That means that in \(S''\), the train is length contracted, and the rod bends.

 

Err, correction

\( \begin{align} \large R''(k) &= \left(t'', \frac{k}{\gamma'} - Vt'', \ 0 \right) \ \normalsize \left{0<k<L \\ t'' >= -Vk\gamma'\right} \\ \end{align} \)​

You need to stop the wishful thinking, it took me only two minutes to find your rookie error. Or was it just another sleigh of hand?

 

Haha.. yeah. Well, I guess is okay as long as you guys are having fun arguing with Tach..

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Though I've got to say, Tach sure can take on all comers lol.

 

Yep, spotted that too, fixed before you posted. There's probably more a couple more transcription errors in there, it's easy to miss things when copy-paste editing complex tex.

Have you noticed yet that your equations are consistent with this animation?

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They couldn't possibly be since you have the \(y"\) component wrong.

 

Nope. The y'' component matches your equations.

See the red dots? They're the transformed events of the rod at t'=0. They happened parallel to the x' axis, and they happen parallel to the x'' axis.

 

How could it, I just corrected it by pointing out that \(y"=0\)

You are persisting in transforming the rod as if it continued its motion past the time \(t'=\frac{H}{u\gamma'}\) (in my notation), \(t' \ge 0\) (in your notation). The motion stops because of the application of the floor reaction to the rod. You keep trying to get around this fact.

 

Another correction:

\(\large R''(k) = \left(t'', \frac{k}{\gamma'} - Vt'', \ 0 \right)\) \(\color{red} \normalsize \left{0<k<L \\ t'' >= \frac{-Vk\gamma'}{c^2}\right}\)​

 

The animation was made separately, using much more thoroughly checked maths. You can see in the animation that the rod elements remain at y''=0 after the collision.
And I fixed that copy-paste error before you posted your correction.

Nope. Look again:

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So, what is the rod position in the platform frame after collision?

Of course you did

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I looked, especially at the fact that you think that the coordinate time have any physical meaning in "bending" the rod, when , in reality they are just labels. Of course, you knew that as well

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In \(S''\), \(t''\) is proper time.

The bending of the rod in \(S''\) is exactly as physically meaningful as length contraction.

Your equations are a subset of my equations:

\( \begin{align} \large R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ -ut' \right) \ \normalsize \left{0<k<L \\ -\frac{H}{u}<t'<0\right} \\ &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ -u(\frac{t''}{\gamma'} + \frac{Vk}{c^2}) \right) \ \normalsize \left{0<k<L \\ -\gamma'(\frac{H}{u} - \frac{Vk}{c^2}) < t'' < -Vk\gamma'\right} \\ \large R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ 0 \right) \ \normalsize \left{0<k<L \\ t'>=0\right} &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ 0) \right) \ \normalsize \left{0<k<L \\ t'' >= \frac{-Vk\gamma'}{c^2}\right} \\ \end{align} \)​

They predict that the rod bends in \(S''\) and not in \(S'\), and that's OK - it does not imply any absolute physical contradiction.

 

You have made this false statement several times already. Let me show you why yo are wrong:

1. Your rod "bending" is a function of coordinate time (the time labels attributed by \(S"\) to events happening in \(S'\)).
2. Length contraction is not time dependent (you knew that).
3. Labels have no physical meaning, so the bending is NOT "as physically meaningful as length contraction", actually is not physically meaningful at all.