You are not serious are you? What is $$-ut'-(-ut')$$? This is getting ridiculous.
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Relativity paradox
- Thread starter renislaj
- Start date
You are not serious are you? What is $$-ut'-(-ut')$$? This is getting ridiculous.
No, it wouldn't, please stop hacking at the problem. There are enough hacks in the thread already. See what Neddy Bate tried, see what Pete is trying.
In order to improve my naive understandings, I need more than just "no". I need the "why" too. Please can I have the why? Do you mean we can't simplify by placing it all in a very weak, even vanishingly small GR field? Can't we ignore a very very weak GR field this way and go on with Pete's exercise using just SR? Can you say why not?
In order to improve my naive understandings, I need more than just "no". I need the "why" too. Please can I have the why? Do you mean we can't simplify by placing it all in a very weak, even vanishingly small GR field? Can't we ignore a very very weak GR field this way and go on with Pete's exercise using just SR? Can you say why not?
What is "weak"? How weak is "weak"? How do you solve the problem when it isn't 'weak"? Why always hack at problems instead of solving them? Ponder on this.
But you agree that in either formalism the floor of the train, like the rod, will always appear crooked to the platform observer? If yes then the train floor and the platform surface can never appear level parallel to the platform observer unless you combine frames like I think you did there? I am still confused.
Edit/ It is also my naive understanding that in SR problems any reasonably weak Gravitational Field can be sliced up into small distances and any "GR curvature" may be effectively ignored (even if initial motion of the rod is due to that field, it need not be a strong acceleration but a weak one)? Wouldn't the same "simplification" eliminating GR from consideration apply to each subsequent small slice of distances to each subsequent position traced along the path line to the floor (or platform or whatever)? I can't see why we can't do this problem like Pete wants and see what we get so I can improve my naive understandings from it.
Basically, you're right. I don't know enough about GR to mathematically express the conditions under which it's equivalent to SR, but qualitatively, it's in the small gravity/acceleration limit. Like you implied in parenthesis, if gravity is the source of the initial motion of the rod, but its effect is weak compared to whatever large velocity is making you use relativity in the first place, special relativity is a good approximation. Tach's objections are getting more and more obstructionist, so I'd suggest you just follow Pete's arguments if you want to improve your understanding. In fact, if you really want to hone your understanding of relativity, go through his post 182 and try to derive it for yourself. It's not too advanced an exercise, and you might even be able to tell Pete where his mistakes are (if any), since Tach is refusing to point them out unless Pete starts a new thread.
I see you snipped the 'missing' lines from your post.
$$y_A''(t'')=-ut'$$ and $$y_B''(t'')=-ut'$$ have to be transformed to $$S''$$ coordinates.
t' at $$A''$$ does not equal t' at $$B''$$ for fixed t''.
$$y_A''(t'')=-ut'$$ and $$y_B''(t'')=-ut'$$ have to be transformed to $$S''$$ coordinates.
t' at $$A''$$ does not equal t' at $$B''$$ for fixed t''.
Pete said:In $$S''$$:
$$\begin{align}
A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\
&= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\
&= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\
&= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right)
\end{align}$$
$$\begin{align}
B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\
&= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\
&= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\
&= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right)
\end{align}$$
The angle of the rod with the x-axis at time t'' is:
$$\begin{align}
\tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\
&= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\
&= \frac{uV\gamma'^2}{\gamma c^2}
\end{align}$$
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Why always hack at problems instead of solving them? Ponder on this.
Because immediately jumping into the most complex version of a problem, without building up from simplified versions, makes it hard to catch errors and alienates non-experts.
I see you snipped the 'missing' lines from your post.
In $$S''$$, t' at $$A$$ is not the same as t' at $$B$$. This is the result of the Lorentz transform.
No, your stuff is as ridiculous as before, let me ask you again, what is $$-ut'-(-ut')$$. Hint : $$0$$
Even better, what is $$y"_A(t")-y"_A(t")$$?
You understand why I can't take your math seriously, do you?
What is "weak"? How weak is "weak"? How do you solve the problem when it isn't 'weak"? Why always hack at problems instead of solving them? Ponder on this.
As "weak" a curvature as can be effectively ignored. Why not wait to solve the "non weak" formalism after we do the weak case? Is there something stopping us from doing the weak case first and then continuing with the significant GR case afterwards?
As "weak" a curvature as can be effectively ignored.
But the problem exists even in the absence of curvature, in a uniform gravitational field. The point is that in the presence of any field, the rod is accelerated. The platform and the train frame are non-inertial so you can't use Lorentz transforms. I pointed that out multiple times.
Why not wait to solve the "non weak" formalism after we do the weak case? Is there something stopping us from doing the weak case first and then continuing with the significant GR case afterwards?
You haven't answered any of the questions.
Basically, you're right. I don't know enough about GR to mathematically express the conditions under which it's equivalent to SR, but qualitatively, it's in the small gravity/acceleration limit. Like you implied in parenthesis, if gravity is the source of the initial motion of the rod, but its effect is weak compared to whatever large velocity is making you use relativity in the first place, special relativity is a good approximation. Tach's objections are getting more and more obstructionist, so I'd suggest you just follow Pete's arguments if you want to improve your understanding. In fact, if you really want to hone your understanding of relativity, go through his post 182 and try to derive it for yourself. It's not too advanced an exercise, and you might even be able to tell Pete where his mistakes are (if any), since Tach is refusing to point them out unless Pete starts a new thread.
Thankyou. I need to understand a little more of what is involved logically in either formalism case before I can improve my naive understandings by reading back with what I will have learned following these latter stages of this discussion to the end, for both cases I hope. Very confusing so far. I just don't understand why we can't take it one step at a time and do the weak GR case first. But I will persevere and come back to read again over the next week.
I did, you just seem to refuse to accept the answer. As an aside, you and Pete should jointly open a separate thread dealing with the dumbed down scenario that you seem so intent on analyzing. I would like to keep this thread on the OP. Deal?
Ok. I promise to make no more posts on this thread about the dumbed-down scenario, although I can't speak for Pete.
But as for your objections to post 151, where are they? You say you made them, but at this point, I just don't believe you. Point me to the post(s). Or repeat the arguments. Something. Because right now, I can only see two replies you've made:
1. I don't specify my reference frames, which is objectively false.
2. The Thomas precession solution is right, and my post conflicts with it, therefore my post is wrong.
Neither of these are remotely responsive. If I missed any other replies, I apologize; please point me to them.
You haven't answered any of the questions.
I have not the level of understanding to answer questions at this stage. I am here to improve on my naive understandings. That is why I asked you the "why" of it when I outlined my confusion and my naive understandings around the GR objection. Why not do it with ignorably insignificant weak GR field first and then get on to the GR included solution?
Irrelevant. We want the angle of the rod in $$S''$$ at fixed $$t''$$, not $$t'$$.No, your stuff is as ridiculous as before, let me ask you again, what is $$-ut'-(-ut')$$.
Irrelevant.Even better, what is $$y"_A(t")-y"_A(t")$$?
What is $$y"_B(t")-y"_A(t")$$?
Note that they are functions of $$t''$$, not $$t'$$. If you start from functions of t', remember to transform to functions of $$t''$$ before you answer.
I have not the level of understanding to answer questions at this stage. I am here to improve on my naive understandings. That is why I asked you the "why" of it when I outlined ny confusion and my naive understandings around the GR objection. Why not do it with weak GR field first and then get on to the GR included solution?
Because the "weak" field is still GR, not SR. That's why.
Irrelevant is what it is.
Even more irrelevant.
What is $$y"_B(t")-y"_A(t")$$?
Not irrelevant at all , look at what you wrote. Do I need to keep rubbing your nose in your errors?
Pete said:In $$S''$$:
$$\begin{align}
A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\
&= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right)
\end{align}$$
$$\begin{align}
B'' &= \left(x_B''(t''), \ y_A''(t'') \right) \\
&= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\
\end{align}$$
How many mistakes do you see? I know : none.
Because the "weak" field is still GR, not SR. That's why.
And...?
And...?
and you shouldn't try using SR, didn't you get it?
Sorry, Tach, but I just think you're mistaken.Not irrelevant at all , look at what you wrote. Do I need to keep rubbing your nose in your errors?
The error you point out is not an error at all.
Work through the transformation 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?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}$$
$$\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}$$
Do you agree with this result?B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\
&= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\
&= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\
&= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right)
\end{align}$$
If not, where is the mistake?
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Going around and around. The whole reason for using the weak insignificant GR field is so that we can do it in effectively SR. You can get to the GR inclusive case after can't you?
Yes you can. If a problem is too much to grasp in its entirety, the best approach is often to take a simplifying approximation and work up from there.