# Thread: relativistic puzzle

1. The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.

Imagine in the below diagram that the grey boxes are the solid objects (say glass blocks with a rod embedded in them). Are you saying that the result of length contraction is not as I have shown it?

In the blue frame, let's say the box with the red rod is contracted in the axis of motion to half the length it was at rest (due to having ), and the same is true for the blue box when viewed from the red frame. What am I doing wrong here?

2. Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.

3. Originally Posted by xyzt
Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.
The language of physics is English. Math is just a "tool".

4. Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.
The language of physics is English. Math is just a "tool".
Actually, you are wrong. The presentation can be in any language, the backbone is in the universal language of math, you don't know math, you can't do physics.

5. Originally Posted by xyzt
Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.
The language of physics is English. Math is just a "tool".
Actually, you are wrong. The presentation can be in any language, the backbone is in the universal language of math, you don't know math, you can't do physics.
Yes I know. But English has been adopted as the international language.

6. Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.
The language of physics is English. Math is just a "tool".
Actually, you are wrong. The presentation can be in any language, the backbone is in the universal language of math, you don't know math, you can't do physics.
Yes I know. But English has been adopted as the international language.
French physics journals are published in French. Japanese physics journals are published in Japanese. The only common is the math. Like I said, you don't know math, you aren't doing physics. BTW: what are you doing here , in this thread? I don't see any contribution.

7. Originally Posted by xyzt
Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.
The language of physics is English. Math is just a "tool".
Actually, you are wrong. The presentation can be in any language, the backbone is in the universal language of math, you don't know math, you can't do physics.
Yes I know. But English has been adopted as the international language.
French physics journals are published in French. Japanese physics journals are published in Japanese. The only common is the math. Like I said, you don't know math, you aren't doing physics. BTW: what are you doing here , in this thread? I don't see any contribution.
Threads are open to all. I'm still reading through whatever you guys wrote.

8. Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by ostkef
Originally Posted by xyzt
Originally Posted by SpeedFreek
The very math I linked to looks different to the equation you gave. I don't see a g there for a start. What is g? I am very dumb so you will have to walk me through this, using the equations shown in the link. Assume I am a simpleton.
You need to learn trigonommetry, then.

are two alternate standard notations for the "tangent" function.

We cannot have an intelligent conversation if you do not understand the math.

What am I doing wrong here?
You are trying to substitute pictures for math. The language of physics is math. You need to learn it.
The language of physics is English. Math is just a "tool".
Actually, you are wrong. The presentation can be in any language, the backbone is in the universal language of math, you don't know math, you can't do physics.
Yes I know. But English has been adopted as the international language.
French physics journals are published in French. Japanese physics journals are published in Japanese. The only common is the math. Like I said, you don't know math, you aren't doing physics. BTW: what are you doing here , in this thread? I don't see any contribution.
Threads are open to all. I'm still reading through whatever you guys wrote.
Feel free to read, just do not pollute with frivolous posts.

9. Of course math is a pre-requisite for understanding physics, but are you denying that with a gamma of 2, each object will be length contracted by a factor of 2 in the direction of motion, as shown in my simple picture?

The length of each object is halved from the frame of the other object, yes or no?

If no, then why?

If yes, then what is wrong with my picture?

Oh, and please don't talk down to me.

10. Originally Posted by SpeedFreek
Of course math is a pre-requisite for understanding physics, but are you denying that with a gamma of 2, each object will be length contracted by a factor of 2 in the direction of motion, as shown in my simple picture?
Yes, so what? Have you read the wiki page? it deals with any , there is nothing special about
What did you understand from the wiki page you, yourself linked in?

11. I asked you to stop talking down to me, in the same way you asked me earlier in the thread.

I didn't say there was anything special about . I was asking if the situation was as shown in my picture, which uses 2 for . I was giving you the gamma so you could ascertain as to whether my picture was a good representation, or not. You answered in the affirmative, but then asked so what?

Well, look at the blue and red rods in that picture and tell me, if the blocks are length contracted as shown in my picture, and you confirm that the situation is as I showed it, how the rods can be parallel in one frame but not parallel in the other. Either the length contraction of the blocks they are embedded in is shown correctly, and the rods present different angles to each other, or the length contraction I showed in that picture is wrong, no?

So, which is it? And what happens if we remove the blocks and just leave the rods? Does the situation change?

Oh and by the way, I am completely familiar with the abbreviation of tan for tangent, whereas I can't remember ever seeing it expressed as tg before.

12. Originally Posted by SpeedFreek
I asked you to stop talking down to me, in the same way you asked me earlier in the thread.

I didn't say there was anything special about . I was asking if the situation was as shown in my picture, which uses 2 for . I was giving you the gamma so you could ascertain as to whether my picture was a good representation, or not. You answered in the affirmative, but then asked so what?

Well, look at the blue and red rods in that picture and tell me, if the blocks are length contracted as shown in my picture, and you confirm that the situation is as I showed it, how the rods can be parallel in one frame but not parallel in the other.
easy,

results into independent of

13. But neither rod makes an angle of 0 degrees with the x-axis, which is how is defined for the red rod. The red rod has an angle of 0 degrees in relation to the blue rod whilst in motion, when viewed from the frame of the blue rod, for sure, but neither rod is aligned with the direction of motion (the x-axis).

From that link:

Assuming that motion occurs along the x-axis, suppose the rod has a proper length (rest length) of metres and makes an angle of degrees with the x'-axis in its rest frame.
If motion occurs along the x-axis, then where are you getting your figure of 0 from?

From this picture, it looks like the blue rod makes an angle of 45 degrees in relation to the x-axis in its rest frame, whereas the red rod makes an angle more like 30 degrees.

14. Originally Posted by SpeedFreek
But neither rod makes an angle of 0 degrees with the x-axis,
That is false, in the frame of the train car , the rod falls parallel with the axis. You appear to not understand the problem explained in the OP and you keep substituting your (incorrect) pictures.

15. Oh for heaven's sake!

I made it completely clear in post #74 that I was proposing a different thought experiment, one which I thought might shed light on the problem as to whether a rod meeting a parallel surface in one frame ALWAYS has to meet that surface parallel in another frame! I have repeatedly posted the diagram, asking how this works for MY example, and you kept replying that . Is it any wonder I was questioning your answers when it is now totally obvious that you haven't been addressing the problem I posted in post #74 at all, even though you kept replying as if you were.

You claim to have solved the problem already in your OP, but for the last half of this thread I have been asking for an answer to mine! Your replies have been completely unhelpful and show a complete lack of comprehension.

16. Originally Posted by SpeedFreek
Oh for heaven's sake!

I made it completely clear in post #74 that I was proposing a different thought experiment, one which I thought might shed light on the problem as to whether a rod meeting a parallel surface in one frame ALWAYS has to meet that surface parallel in another frame!
I couldn't care less for your different gedank, try sticking with the subject of this thread.

17. Originally Posted by xyzt
I couldn't care less for your different gedank, try sticking with the subject of this thread.
Well stop replying to my posts about my different gedank with the wrong answers then. Read through my replies in this thread from #74 onwards and see what a mess you made answering them.

My suggestion was that my different gedank might help resolve the question of how two different frames views of events can seemingly have different physical results, but that seems to have gone completely over your head.

Oh, and I would recommend you drop the attitude too.

18. Originally Posted by SpeedFreek
Originally Posted by xyzt
I couldn't care less for your different gedank, try sticking with the subject of this thread.
Well stop replying to my posts about my different gedank with the wrong answers then. Read through my replies in this thread from #74 onwards and see what a mess you made answering them.

My suggestion was that my different gedank might help resolve the question of how two different frames views of events can seemingly have different physical results, but that seems to have gone completely over your head.

A different gedank doesn't help resolving the gedank in the OP. If you put garbage in, you get garbage out.

19. Originally Posted by SpeedFreek

You claim to have solved the problem already in your OP, but for the last half of this thread I have been asking for an answer to mine! Your replies have been completely unhelpful and show a complete lack of comprehension.
Open another thread, don't muddy this one with your own misconceptions.

20. Originally Posted by xyzt
Originally Posted by SpeedFreek
Originally Posted by xyzt
I couldn't care less for your different gedank, try sticking with the subject of this thread.
Well stop replying to my posts about my different gedank with the wrong answers then. Read through my replies in this thread from #74 onwards and see what a mess you made answering them.

My suggestion was that my different gedank might help resolve the question of how two different frames views of events can seemingly have different physical results, but that seems to have gone completely over your head.

A different gedank doesn't help resolving the gedank in the OP. If you put garbage in, you get garbage out.
But my gedank is definitely not garbage, and nor is yours. In mine we have two rods approaching each other such that both ends of the red rod intersect with the blue rod simultaneously in the blue frame, but not simultaneously in the red frame. It is a different gedank, for sure, but it raises EXACTLY the same question as yours - how can they meet parallel in one frame but have one end of the red rod hitting the blue rod before the other end, in the red frame.

If you think it is garbage then that says a lot about you and your understanding of these concepts. I am done here.

Oh, and Janus is correct, by the way. But with your attitude you will be lucky if you find out why.

21. Originally Posted by SpeedFreek
Originally Posted by Markus Hanke
I think the rod cannot hit the cabin floor in a different way in the two frames. It either falls straight in both frames or it doesn't.
I'm not quite so sure of that fundamental principle, Markus.

I believe this might be related to the puzzle at hand and might be worth looking into in order to resolve this puzzle:

Imagine two long rods, at rest. One rod is sitting at an angle in relation to the other, such that the rods are neither parallel or perpendicular. See the diagram below. The blue and red lines represent the rods, whereas the grey boxes are there to aid visualisation of length contraction along the axis of motion.

If the distance between the rods was decreasing, due to there being relative motion between them, then:

From the frame of A (blue), the length contraction of B (red) causes both ends of B to intersect with A simultaneously, as the length contraction of B makes it parallel with A.

But from the frame of B, the length contraction of A causes both ends of B to intersect with A at different times. The two rods are not parallel.

In case anyone is wondering about this, look here:
Special Relativity/Simultaneity, time dilation and length contraction
As the link clearly states, the angle is dependent on the relative speed of the observers:

The above is true with ONE notable exception, when which is PRECISELY the case of the puzzle in the OP.
So, your pictures are utterly irrelevant for solving the puzzle. GiGo.

22. Originally Posted by xyzt
Originally Posted by SpeedFreek

You claim to have solved the problem already in your OP, but for the last half of this thread I have been asking for an answer to mine! Your replies have been completely unhelpful and show a complete lack of comprehension.
Open another thread, don't muddy this one with your own misconceptions.
There are misconceptions in this thread, but they are not mine, my friend. If you had some insight, you might see I was trying to help you towards answering your OP. Once again, good luck with that. If you have any friends, perhaps you could ask them to help you resolve this puzzle.

23. Originally Posted by SpeedFreek
Originally Posted by xyzt
Originally Posted by SpeedFreek
Originally Posted by xyzt
I couldn't care less for your different gedank, try sticking with the subject of this thread.
Well stop replying to my posts about my different gedank with the wrong answers then. Read through my replies in this thread from #74 onwards and see what a mess you made answering them.

My suggestion was that my different gedank might help resolve the question of how two different frames views of events can seemingly have different physical results, but that seems to have gone completely over your head.

A different gedank doesn't help resolving the gedank in the OP. If you put garbage in, you get garbage out.
But my gedank is definitely not garbage, and nor is yours. In mine we have two rods approaching each other such that both ends of the red rod intersect with the blue rod simultaneously in the blue frame, but not simultaneously in the red frame. It is a different gedank, for sure, but it raises EXACTLY the same question as yours - how can they meet parallel in one frame but have one end of the red rod hitting the blue rod before the other end, in the red frame.
Because the angle is zero. The math shows that a zero angle is zero in ALL frames, this is why. If you cannot follow the simple math in your link, butt out.

24. Why do you insist on replying to my descriptions of my gedank with the incorrect answer that the angle is zero? Do you have a comprehension problem?

In my example, the angle is not zero, but for only one of the frames the rods meet in parallel. Go figure...

Who knows, perhaps if you were to solve mine, you might find the correct solution to your own. But I doubt it.

25. Originally Posted by SpeedFreek
Why do you insist on replying to my descriptions of my gedank with the incorrect answer that the angle is zero?
I am trying to explain why "your gedank" is irrelevant in solving ANYTHING. Since you are unable to follow simple math and simple reasoning, I failed. Sorry, my fault.

In my example, the angle is not zero, but for only one of the frames the rods meet in parallel. Go figure...
See , this is the problem of your attempts at "proof by picture", the math says that your pictures are wrong. If the rods meet parallel in one frame, they meet parallel in ALL frames because the math of the very link you cited shows that if the rods make a zero angle in one frame, they make a zero angle in ALL frames. The fact that you are unable to progress past your silly drawings is what is disabling you to understand why is it so. Perhaps when you progress to understand the simple math in the wiki link, you will understand why you are wrong. I am not holding my breath.

26. Originally Posted by xyzt
Originally Posted by SpeedFreek
Why do you insist on replying to my descriptions of my gedank with the incorrect answer that the angle is zero?
I am trying to explain why "your gedank" is irrelevant in solving ANYTHING. Since you are unable to follow simple math and simple reasoning, I failed. Sorry, my fault.
I would be wary of making unsubstantiated claims like that if I were you. Just because I have shown no math doesn't mean I cannot.

Originally Posted by xyzt
In my example, the angle is not zero, but for only one of the frames the rods meet in parallel. Go figure...
See , this is the problem of your attempts at "proof by picture", the math says that your pictures are wrong. If the rods meet parallel in one frame, they meet parallel in ALL frames because the math of the very link you cited shows that if the rods make a zero angle in one frame, they make a zero angle in ALL frames. The fact that you are unable to progress past your silly drawings is what is disabling you to understand why is it so. Perhaps when you progress to understand the simple math in the wiki link, you will understand why you are wrong. I am not holding my breath.
No, the math DOES NOT say my pictures are wrong, because in my example there is no frame where the angle between the rod and the x-axis is zero! You seem to have a blind spot where this is concerned.

Once again, in my gedank, the angle between either rod and the axis of motion (the x-axis) is NOT zero. You might learn something by plugging in some numbers and working through it. But I am guessing you won't, and with your attitude I am certainly not going to do it for you.

27. Originally Posted by SpeedFreek
Originally Posted by xyzt
Originally Posted by SpeedFreek
Why do you insist on replying to my descriptions of my gedank with the incorrect answer that the angle is zero?
I am trying to explain why "your gedank" is irrelevant in solving ANYTHING. Since you are unable to follow simple math and simple reasoning, I failed. Sorry, my fault.
I would be wary of making unsubstantiated claims like that if I were you. Just because I have shown no math doesn't mean I cannot.

Originally Posted by xyzt
[
In my example, the angle is not zero, but for only one of the frames the rods meet in parallel. Go figure...
See , this is the problem of your attempts at "proof by picture", the math says that your pictures are wrong. If the rods meet parallel in one frame, they meet parallel in ALL frames because the math of the very link you cited shows that if the rods make a zero angle in one frame, they make a zero angle in ALL frames. The fact that you are unable to progress past your silly drawings is what is disabling you to understand why is it so. Perhaps when you progress to understand the simple math in the wiki link, you will understand why you are wrong. I am not holding my breath.
No, the math DOES NOT say my pictures are wrong, because in my example there is no frame where the angle between the rod and the x-axis is zero! You seem to have a blind spot where this is concerned.
Sigh, according to you, there is a frame where the rods are parallel, i.e. they make a zero angle. Therefore, according to the wiki link that you cited but you are unable to understand, the angle is zero in all frames. There is not much that I can do for you, you are mathematically illiterate. Sorry.

Once again, in my gedank, the angle between either rod and the axis of motion (the x-axis) is NOT zero. You could learn a lot by working through it. But you won't.
Forget about the x-axis. Concentrate on the fact that the angle between rods is zero.

28. is not the angle between the rods, it is the angle between a rod and the axis of motion in the rest frame of the rod. Length contraction does not care what the angle of the other rod is, it is only concerned with the angle relative to the axis of motion.

You seem to be stuck in a universe where all rods are aligned with their axis of motion...

29. Originally Posted by SpeedFreek
is not the angle between the rods, it is the angle between a rod and the axis of motion in the rest frame of the rod. Length contraction does not care what the angle of the other rod is, it is only concerned with the angle relative to the axis of motion.

You seem to be stuck in a universe where all rods are aligned with their axis of motion...
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.

30. Originally Posted by xyzt
Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.
hmmm.....
What's your problem?
nothing, since i had it right.
Good for you, when you posted something totally incoherent you can always claim that you "had it right" afterwards.
afterwards? re-read the thread and tell me where your answer came before mine. in fact you said i'm wrong.

31. Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.
hmmm.....
What's your problem?
nothing, since i had it right.
Good for you, when you posted something totally incoherent you can always claim that you "had it right" afterwards.
afterwards? re-read the thread and tell me where your answer came before mine. in fact you said i'm wrong.
Here, just as incoherent as usual.

32. and? i said they will disagree on when the rods hit the ground, not how. and i don't know any math where you can multiply or divide by 0.

33. Originally Posted by curious mind
and? i said they will disagree on when the rods hit the ground, not how. and i don't know any math where you can multiply or divide by 0.
You can always multiply by 0, you didn't know that? I suggest that you go troll elsewhere.

34. yea with the result being 0. but it's ok.

35. Originally Posted by xyzt
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
You are missing a vital point here. Look at the quick and dirty sketch I made below.

There are three scenarios, seen from both sides shown here.

Look at the 4 lines across the top of the image, 2 blue and 2 red. The direction of motion (x-axis) is horizontal. The blue rod on the left is at 45 degrees in relation to it, whilst the red rod on the right is at 30 degrees in relation to it. At rest, the angle between the rods is 15 degrees. Whilst in motion, the blue rod shrinks to 60 degrees and thus there is an angle of 30 degrees between them according to the rest frame of the red rod, whereas the red rod shrinks to 45 degrees and thus there is 0 degrees between them, according to the rest frame of the blue rod.

Now look at the lines down the left hand side, where the x-axis is now vertical, and the situation is reversed. Here, they still show an angle of 15 degrees at rest, but when in motion it is now the red rod rest frame which finds 0 degrees between them, whilst the blue rod rest frame now finds an angle of 30 degrees.

Lastly we have the diagonal scenario. The direction of motion is now at 45 degrees, which is perpendicular to the blue rod, so the angle it presents to the rest frame of the red rod is the same whether there is motion between them or not. But the same is not true for the red rod when viewed from the rest frame of the blue rod - the initial angle of 15 degrees is changed and gets smaller, because the red rod is not perpendicular to the direction of motion.

The all at rest angle is 15 degrees in all the scenarios. The relative positions are arbitrary of course, but when in motion the direction of motion relative to the angle of the rod is NOT arbitrary.

In the horizontal scenario, the impact is parallel for the blue frame only.

In the vertical scenario, the impact is parallel for the red frame only.

In the diagonal scenario, the impact is not parallel in either frame, and the angle in motion is different to the angle at rest as viewed from the blue frame only.

The direction of motion in relation to the rest angle makes a BIG difference to who sees (or calculates) what.

Apologies for the crude diagram.

36. Originally Posted by xyzt
This is a variant of the Einstein train experiment. There is a human body suspended horizontally by two wires in a train car moving on the tracks. The body is positioned symmetrically wrt the center of the car. Two flashes of light are emitted simultaneously from the car center activating two devices that cut the wires simultaneousy (in the frame of the car). Obviously, the body falls parallel to the car floor hitting the floor simultaneously with all its points (in the car frame).
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor : one end of the body will hit the floor before the other end. This will result into one end (say, the head) absorbing the full impact. This cannot be since the same experiment has differing results in the two frames. What is the resolution of the puzzle?
This proves nothing.

Sure from the view of the rest frame, the body falls simultaneously.

But, in the moving frame, light is not infinite. So, the head falling reaches the moving frame center before the falling feet reaches.

Therefore, the head appears to fall before the feet strictly based on the finite speed of light.

No paradox and no problem.

37. Originally Posted by SpeedFreek
Originally Posted by xyzt
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
You are missing a vital point here. Look at the quick and dirty sketch I made below.
I am not missing anything, you, on the other hand keep pushing silly sketches as "proofs". You really need to learn to express your thought in math. I am done with your nonsense.

38. Originally Posted by xyzt
I am not missing anything, you, on the other hand keep pushing silly sketches as "proofs". You really need to learn to express your thought in math. I am done with your nonsense.
What you say is true. But I don't agree. In order to teach the science, one must use the language- Math.
This is not a teaching environment, however. The ability to express the hard science to members who do not have the math skills is very important.
So, yes, what you say is true. But you're being hard nosed about it when you need to be more realistic about delivery in this setting.
As such, both the mathematician and the "sketchatician" are invaluable against cranks like JohnWhisp.

Now... it's time to piss you off.
You are in error. The mathematics was already sufficiently presented by mod Janus ( a well done job) and this debacle afterward is you squabbling with another member who is trying to discuss the nuances of simultaneity (Which should be interesting since JohnWill-o-th'-whisp entered the thread...) while you pound your chest and bleat.

So here's my post- You're not fooling anyone. Maybe yourself and if you insist on continuing- that is fine. But you're not fooling anyone else.
Asking to re-hash the mathematics and calling other posts nonsense is pointless. If you didn't get it the first time and if you didn't get it after someone tried to break it down for you, it is only you that can re-examine the position and reassess. No one can do that for you.

Considering that you have the intellect, I would like to believe that you will read this without letting your annoyance at being called out on it guide your next step. I would like to believe that you'll correct the error- But I won't hold my breath. You will probably just complain about what I said, too.
Feel free. I'm fighting in enough threads right now that I'm content to leave you to SpeedFreek to deal with.

39. Originally Posted by Neverfly
You are in error. The mathematics was already sufficiently presented by mod Janus ( a well done job) and this debacle afterward is you squabbling with another member who is trying to discuss the nuances of simultaneity (Which should be interesting since JohnWill-o-th'-whisp entered the thread...) while you pound your chest and bleat.
I think the bleating sheep is you. First off, I presented two very detailed solutions. Second off, you contributed nothing to the solution (well, you have one outright laughable post), so you have no say. Third off, Janus presented no math to support his point, despite my helping him in formulating the framework. So, bleat off.

40. Originally Posted by xyzt
So, bleat off.
You got it.

41. I see, this is your "contribution" to the thread. Totally laughable, shows you understand...absolutely nothing about the subject being discussed.

42. Originally Posted by xyzt
I see, this is your "contribution" to the thread. Totally laughable, shows you understand...absolutely nothing about the subject being discussed.
Aww... I really did piss you off, didn't I? Sweet. It's nice to see that you don't disappoint. Rather than re-assess, you actually went to all the trouble of digging up a one liner I threw in in response to something particular someone said and say, "Laughable."
Well, that's fine- I certainly agree it's laughable.
So is your behavior.

43. Originally Posted by xyzt
Originally Posted by SpeedFreek
Originally Posted by xyzt
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
You are missing a vital point here. Look at the quick and dirty sketch I made below.
I am not missing anything, you, on the other hand keep pushing silly sketches as "proofs". You really need to learn to express your thought in math. I am done with your nonsense.
It's your loss.

If you care to read Relativity - Special and General theory, by Einstein, and look at his chapter on the relativity of simultaneity, you will find no math presented, just a simple sketch and a body of text describing the implications of that sketch. From that chapter alone, one can gain insight into this issue.

I cannot understand why you find the concept that an objects angle respective to its direction of motion in its rest frame makes a difference to the outcome seen by the other frame so hard to grasp. It is trivially obvious.

Listen. Actually read what I am saying, rather than dismissing it without comprehending it.

Imagine a vertical rod A, and another rod B that is sitting at 45 degrees.

The angle that rod A presents to rod B will not change if rod A moves sideways, in a direction perpendicular to its orientation. This is trivially obvious. If a rod is perpendicular to its direction of motion, length contraction will NOT change the angle it presents to anyone. Its angle will not vary with

But if rod B is sitting at an angle that is neither aligned with or perpendicular to its direction of motion, the angle it presents to anyone will be changed by length contraction.

If the two rods are in relative motion, Rod A presents the same angle to rod B regardless of whether it is length contracted or not. But the same is not true for the angle Rod B presents to Rod A, due to the angle Rod B makes in relation to its directon of motion. The angle it presents will vary with

It really is that simple, and your total refusal to accept this trivial fact is inexplicably stupid.

44. Originally Posted by SpeedFreek

If the two rods are in relative motion, Rod A presents the same angle to rod B regardless of whether it is length contracted or not. But the same is not true for the angle Rod B presents to Rod A, due to the angle Rod B makes in relation to its directon of motion
This is patently FALSE, the angle A makes with B is equal to the angle B makes with A.

The angle it presents will vary with

Good, you finally learned this basic fact. The only (very important) exception is when the angle is ZERO in ONE frame . In this particular case, the angle is ZERO in ALL frames, INDEPENDENT of . Ain't math a bitch?

It really is that simple, and your total refusal to accept this trivial fact is inexplicably stupid.
It is pretty evident who the one posting stupid stuff in this thread is.

45. Originally Posted by xyzt
This is a variant of the Einstein train experiment. There is a human body suspended horizontally by two wires in a train car moving on the tracks. The body is positioned symmetrically wrt the center of the car. Two flashes of light are emitted simultaneously from the car center activating two devices that cut the wires simultaneousy (in the frame of the car). Obviously, the body falls parallel to the car floor hitting the floor simultaneously with all its points (in the car frame).
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor : one end of the body will hit the floor before the other end. This will result into one end (say, the head) absorbing the full impact. This cannot be since the same experiment has differing results in the two frames. What is the resolution of the puzzle?
The release events are separated by the rod length, and therefore will be perceived differently depending on viewer speed.
G (ground viewer) sees back end of rod fall first, and contact floor first. Momentum rotates rod clockwise until front end contacts floor. (possible bounce, depending on floor and rod composition)
There is no expectation of G seeing the same motions as P (passenger), since the experiment is done in the P frame. SR only predicts an simultaneous (flat) contact per the initial conditions, in the absolute rest frame, or an equivalent pseudo rest frame, aka the P frame.
If G performed the same experiment in HIS frame, he would see the rod fall in the horizontal position.

46. Originally Posted by xyzt
Originally Posted by Strange
Does the length of the wires matter?
If you make them to be zero length then the transmission of the "cut" signal becomes virtually instantaneous.
I am confused by this. I thought the problem was the non-simultaneous cutting as seen from a different frame of reference. There is a (possibly irrelevant) complication of the time it takes the information that the wire has been cut at one end to propagate along the rod as it rotates.

But the time for the cut information to travel down the wires would be them same at each end for both frames. And I assumed they were cut at the point of attachment anyway.

But, as you say, it has not effect so ...

47. One thing that we have been ignoring is that neither end of the falling rod is in either the ground's or the train's inertial frame. Both ends, once they are falling, are in frames moving at non-horizontal directions as seen from the ground and train frames. Moreover, we would have to integrate over the motion for each end, since neither end is in an inertial frame at all. Simultaneity could be quite complicated in this situation.

Is there a non-obvious simplification going on here, so that the tilted, non-inertial frames don't matter, or does the complication move the calculation toward a more intuitive result?

With the moving, but not accelerating rods, do we need some kind of similar correction? All the standard derivations put the two observers at the same point at one time, usually t=0, and here we are analyzing objects whose positions never coincide. My intuition is that the never-coincidence ( ) wouldn't matter, but I am a little worried that I should in principle verify that. Another way of putting my concern is that light expands out in a circle in each frame, but a sound wave does not, and I haven't convinced myself that this fact doesn't matter.

48. Originally Posted by mvb
One thing that we have been ignoring is that neither end of the falling rod is in either the ground's or the train's inertial frame. Both ends, once they are falling, are in frames moving at non-horizontal directions as seen from the ground and train frames. Moreover, we would have to integrate over the motion for each end, since neither end is in an inertial frame at all. Simultaneity could be quite complicated in this situation.
This is a valid point.

49. Originally Posted by xyzt
Originally Posted by mvb
One thing that we have been ignoring is that neither end of the falling rod is in either the ground's or the train's inertial frame. Both ends, once they are falling, are in frames moving at non-horizontal directions as seen from the ground and train frames. Moreover, we would have to integrate over the motion for each end, since neither end is in an inertial frame at all. Simultaneity could be quite complicated in this situation.
This is a valid point.
Exactly how does acceleration change the problem since it is absolute under SR.

Also, I tried to bring it to your level in my post that under SR events under the relativity of simultaneity occur when the light strikes the observer for the events.

You claim an absoluteness that the body must drop in parallel in both frames which is not required under SR.

50. Originally Posted by JohnWisp
Originally Posted by xyzt
Originally Posted by mvb
One thing that we have been ignoring is that neither end of the falling rod is in either the ground's or the train's inertial frame. Both ends, once they are falling, are in frames moving at non-horizontal directions as seen from the ground and train frames. Moreover, we would have to integrate over the motion for each end, since neither end is in an inertial frame at all. Simultaneity could be quite complicated in this situation.
This is a valid point.
Exactly how does acceleration change the problem since it is absolute under SR.
By disallowing application of the SR formalism to this particular puzzle. I would prefer if you kept your crackpot posts out of this thread.

51. Originally Posted by xyzt
I would prefer if you kept your crackpot posts out of this thread.
Seconded. Some of us are ignorant, some of us even have misconceptions or more learning to do.
But crackpot is the arrogance of believing you're absolutely correct in spite of being shown to be in error, JohnWisp. Your input will only be a disruption. You already have a thread on the topic which you started and you cannot even support what you've said there while desperately closing your eyes and ears to anything said to show your errors there. Try not to thread-\$h!t all over the board. One cesspool is enough.

52. Originally Posted by JohnWisp

Exactly how does acceleration change the problem since it is absolute under SR.

Also, I tried to bring it to your level in my post that under SR events under the relativity of simultaneity occur when the light strikes the observer for the events.

You claim an absoluteness that the body must drop in parallel in both frames which is not required under SR.
As an end falls, its velocity relative to the ground observer has a constant horizontal component and an increasing vertical component. Its rest frame is moving at a changing velocity relative to the ground, hence the frame it is in changes. This can be handled on the reasonable assumption that the conditions are chosen so that General Relativity doesn't have a significant effect, but it complicates the determination of simultaneity.

The simultaneity of two events in a frame is not governed by the simultaneity of the arrival of photons at the origin. It is governed by the time variable in that frame at the points of the two events. These two need not agree if one event takes place further from the origin of the coordinates. [Convention, by the way, puts the observer at the origin]. I do agree, however, that it would be intriguing if the collisions of the two ends of the rod were nonsimultaneous but that the photons carrying the information arrived at the middle of the rod at the same time.

The complications of the two ends of the rod not being released at the same time in the ground frame and the varying rest frames of the ends of the rod simply mean that the discussion of the relative nature of the collision of two rods in different frames would be better judged from rods that are moving with respect to each other at a constant speed. As we have seen earlier in this thread, this situation is complicated enough and also gives unintuitive results.

53. Originally Posted by xyzt
Originally Posted by JohnWisp
Originally Posted by xyzt
Originally Posted by mvb
One thing that we have been ignoring is that neither end of the falling rod is in either the ground's or the train's inertial frame. Both ends, once they are falling, are in frames moving at non-horizontal directions as seen from the ground and train frames. Moreover, we would have to integrate over the motion for each end, since neither end is in an inertial frame at all. Simultaneity could be quite complicated in this situation.
This is a valid point.
Exactly how does acceleration change the problem since it is absolute under SR.
By disallowing application of the SR formalism to this particular puzzle. I would prefer if you kept your crackpot posts out of this thread.

Perhaps you are confused, but constant acceleration is still the domain of SR.

This would make you and your followers crackpots for not knowing this.

This link will teach you constant/uniform acceleration is in the domain of SR.

The Relativistic Rocket

54. Originally Posted by mvb
Originally Posted by JohnWisp

Exactly how does acceleration change the problem since it is absolute under SR.

Also, I tried to bring it to your level in my post that under SR events under the relativity of simultaneity occur when the light strikes the observer for the events.

You claim an absoluteness that the body must drop in parallel in both frames which is not required under SR.
As an end falls, its velocity relative to the ground observer has a constant horizontal component and an increasing vertical component. Its rest frame is moving at a changing velocity relative to the ground, hence the frame it is in changes. This can be handled on the reasonable assumption that the conditions are chosen so that General Relativity doesn't have a significant effect, but it complicates the determination of simultaneity.

The simultaneity of two events in a frame is not governed by the simultaneity of the arrival of photons at the origin. It is governed by the time variable in that frame at the points of the two events. These two need not agree if one event takes place further from the origin of the coordinates. [Convention, by the way, puts the observer at the origin]. I do agree, however, that it would be intriguing if the collisions of the two ends of the rod were nonsimultaneous but that the photons carrying the information arrived at the middle of the rod at the same time.

The complications of the two ends of the rod not being released at the same time in the ground frame and the varying rest frames of the ends of the rod simply mean that the discussion of the relative nature of the collision of two rods in different frames would be better judged from rods that are moving with respect to each other at a constant speed. As we have seen earlier in this thread, this situation is complicated enough and also gives unintuitive results.
How exactly does constant acceleration change the conclusions of the rest frame?

55. Originally Posted by JohnWisp
How exactly does constant acceleration change the conclusions of the rest frame?
As indicated in the discussion by Baez that you referenced in your previous post, the frame of the falling rod is only instantaneously a rest frame. You have to add to the discussion of the rod an integration over a series of frames that are moving in different directions. Granted the resulting velocities could be treated as being small compared to c and ignored, but I would be worried about doing simultaneity arguments that way. The lack of simultaneity is also going to be small, and there would be a danger of getting it wrong by a significant amount. I would at least want to investigate the size of the first neglected terms. Since we are investigating a question of principle, it seems to me to be best to start with a problem that has no accelerations in it.

Another point is the internal response of the rod to its release from the supports. You can't even estimate this without looking at the frame of the rod.

In any event, analyzing a dropping a rod is bringing accelerations into a special relativity problem, where they can be handled only approximately. I would rather tackle issues of principle with an example that has no approximations.

56. Originally Posted by mvb
Originally Posted by JohnWisp
How exactly does constant acceleration change the conclusions of the rest frame?
As indicated in the discussion by Baez that you referenced in your previous post, the frame of the falling rod is only instantaneously a rest frame. You have to add to the discussion of the rod an integration over a series of frames that are moving in different directions. Granted the resulting velocities could be treated as being small compared to c and ignored, but I would be worried about doing simultaneity arguments that way. The lack of simultaneity is also going to be small, and there would be a danger of getting it wrong by a significant amount. I would at least want to investigate the size of the first neglected terms. Since we are investigating a question of principle, it seems to me to be best to start with a problem that has no accelerations in it.

Another point is the internal response of the rod to its release from the supports. You can't even estimate this without looking at the frame of the rod.

In any event, analyzing a dropping a rod is bringing accelerations into a special relativity problem, where they can be handled only approximately. I would rather tackle issues of principle with an example that has no approximations.

I will answer the question for you. The acceleration of the body in the rest frame does not change the conclusions of the rest frame.

Therefore, the body will appear to drop simultaneously to an observer centered with the body in the rest frame.

In the rest frame, Newtonian mechanics hold true.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the “stationary system.”
On the Electrodynamics of Moving Bodies

57. Originally Posted by JohnWisp
Originally Posted by mvb
Originally Posted by JohnWisp
How exactly does constant acceleration change the conclusions of the rest frame?
As indicated in the discussion by Baez that you referenced in your previous post, the frame of the falling rod is only instantaneously a rest frame. You have to add to the discussion of the rod an integration over a series of frames that are moving in different directions. Granted the resulting velocities could be treated as being small compared to c and ignored, but I would be worried about doing simultaneity arguments that way. The lack of simultaneity is also going to be small, and there would be a danger of getting it wrong by a significant amount. I would at least want to investigate the size of the first neglected terms. Since we are investigating a question of principle, it seems to me to be best to start with a problem that has no accelerations in it.

Another point is the internal response of the rod to its release from the supports. You can't even estimate this without looking at the frame of the rod.

In any event, analyzing a dropping a rod is bringing accelerations into a special relativity problem, where they can be handled only approximately. I would rather tackle issues of principle with an example that has no approximations.

I will answer the question for you. The acceleration of the body in the rest frame does not change the conclusions of the rest frame.

Therefore, the body will appear to drop simultaneously to an observer centered with the body in the rest frame.

In the rest frame, Newtonian mechanics hold true.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the “stationary system.”
On the Electrodynamics of Moving Bodies
The rest frame of one object may contain other objects moving at speeds close to c. In that event it is not true that Newtonian mechanics holds good. Moreover, in this problem, the crucial question is simultaneity of the collisions of ends of a body as viewed in a frame in which the body is moving (horizontally) at relativistic velocities. By definition we are not doing Newtonian mechanics, even if we eventually get the same answer as Newtonian mechanics. The point is to test the behavior of an object in special relativity; if we manage to remove relativity from the problem in principle, we have failed in what we started out to do.

58. Originally Posted by mvb
Originally Posted by JohnWisp
Originally Posted by mvb
Originally Posted by JohnWisp
How exactly does constant acceleration change the conclusions of the rest frame?
As indicated in the discussion by Baez that you referenced in your previous post, the frame of the falling rod is only instantaneously a rest frame. You have to add to the discussion of the rod an integration over a series of frames that are moving in different directions. Granted the resulting velocities could be treated as being small compared to c and ignored, but I would be worried about doing simultaneity arguments that way. The lack of simultaneity is also going to be small, and there would be a danger of getting it wrong by a significant amount. I would at least want to investigate the size of the first neglected terms. Since we are investigating a question of principle, it seems to me to be best to start with a problem that has no accelerations in it.

Another point is the internal response of the rod to its release from the supports. You can't even estimate this without looking at the frame of the rod.

In any event, analyzing a dropping a rod is bringing accelerations into a special relativity problem, where they can be handled only approximately. I would rather tackle issues of principle with an example that has no approximations.

I will answer the question for you. The acceleration of the body in the rest frame does not change the conclusions of the rest frame.

Therefore, the body will appear to drop simultaneously to an observer centered with the body in the rest frame.

In the rest frame, Newtonian mechanics hold true.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the “stationary system.”
On the Electrodynamics of Moving Bodies
The rest frame of one object may contain other objects moving at speeds close to c. In that event it is not true that Newtonian mechanics holds good. Moreover, in this problem, the crucial question is simultaneity of the collisions of ends of a body as viewed in a frame in which the body is moving (horizontally) at relativistic velocities. By definition we are not doing Newtonian mechanics, even if we eventually get the same answer as Newtonian mechanics. The point is to test the behavior of an object in special relativity; if we manage to remove relativity from the problem in principle, we have failed in what we started out to do.
OK I have brought you to the point that you agree the uniform acceleration does not change the conclusions of the rest frame. As Einstein said, those calculations are Newtonian. Didn't you know this?

Now, I will ask you how this changes the conclusions of the moving frame since uniform acceleration is the domain of SR.

59. Originally Posted by JohnWisp
OK I have brought you to the point that you agree the uniform acceleration does not change the conclusions of the rest frame. As Einstein said, those calculations are Newtonian. Didn't you know this?

Now, I will ask you how this changes the conclusions of the moving frame since uniform acceleration is the domain of SR.
No, I do not agree that uniform acceleration does not change the conclusions of the rest frame. By Mach's Principle, uniform acceleration is locally equivalent to a uniform gravitational field, and gravitational fields are the province of General Relativity. If the acceleration is small enough the differences may not matter for a particular calculation, but this always has to be checked. An accelerated rest frame is not an inertial rest frame, and special relativity requires inertial frames.

Since we are now arguing rather than discussing and I am satisfied that by now I will have convinced anyone else who is following the thread, I am bowing out. I see no advantage to further posts.

60. Originally Posted by mvb
Originally Posted by JohnWisp
OK I have brought you to the point that you agree the uniform acceleration does not change the conclusions of the rest frame. As Einstein said, those calculations are Newtonian. Didn't you know this?

Now, I will ask you how this changes the conclusions of the moving frame since uniform acceleration is the domain of SR.
No, I do not agree that uniform acceleration does not change the conclusions of the rest frame. By Mach's Principle, uniform acceleration is locally equivalent to a uniform gravitational field, and gravitational fields are the province of General Relativity. If the acceleration is small enough the differences may not matter for a particular calculation, but this always has to be checked. An accelerated rest frame is not an inertial rest frame, and special relativity requires inertial frames.

Since we are now arguing rather than discussing and I am satisfied that by now I will have convinced anyone else who is following the thread, I am bowing out. I see no advantage to further posts.
Your conclusions are false.

Otherwise, prove with math that uniform acceleration in the rest frame of the dropping body changes whether the body drops simultaneously.

That is your assertion now prove your case.

61. Otherwise, prove with math that uniform acceleration in the rest frame of the dropping body changes whether the body drops simultaneously.
This sentence is without any meaning.

62. Originally Posted by xyzt
Originally Posted by SpeedFreek
is not the angle between the rods, it is the angle between a rod and the axis of motion in the rest frame of the rod. Length contraction does not care what the angle of the other rod is, it is only concerned with the angle relative to the axis of motion.
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
I have been contacted by a member here who doesn't post in the forum any more, but who has read through this thread and he has confirmed to me that my diagrams with the blue and red rods are in fact correct for my gedank. This is a variation of the "meter stick and the hole" paradox, and the principle does apply to your OP, just as I thought.

Your rod hits the floor of the train at an angle, as viewed from the embankment, just as Janus said.

The member who contacted me is the author of a website that hosts a java applet that can illustrate all this.

You can find the java applet here (I have tested it and it is safe (on my system at least!) but it is quite large and takes a while to load) Newtonian and Relativistic Simulations

Or you can just read through the tutorial for the app here - http://www.relativitysimulation.com/...ckAndHole.html which illustrates the meter stick and the hole paradox and shows how the angles are different from each frame of reference.

The only difference between the "meter stick and the hole" paradox and your gedank is that in yours there is no hole. And if you don't believe the author of that website, google for the meter stick and the hole paradox to find other equivalent treatments.

The meter stick and the hole

We have a meter stick aligned with the x-axis, moving at 0.866c. We have a panel moving in the y-axis, sitting perpendicular to the y-axis, with a hole in it that is just larger than half a meter. Because the panel is perpendicular to its axis of motion, there is no contraction in the length of the hole. From the frame of the panel, the meter stick is parallel, but approaching at an angle. When they intersect, the meter stick passes through the hole because, in the frame of the panel, the meter stick is length contracted to half its rest length.

But what happens from the view of the meter stick? In its own frame it is a meter long, and we know the hole in the panel is only just over half a meter long in its rest frame. The answer is that they don't meet parallel. From the frame of the meter stick, the panel makes an angle, allowing the meter long stick to pass through the hole.

63. Originally Posted by SpeedFreek
Originally Posted by xyzt
Originally Posted by SpeedFreek
is not the angle between the rods, it is the angle between a rod and the axis of motion in the rest frame of the rod. Length contraction does not care what the angle of the other rod is, it is only concerned with the angle relative to the axis of motion.
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
I have been contacted by a member here who doesn't post in the forum any more, but who has read through this thread and he has confirmed to me that my diagrams with the blue and red rods are in fact correct for my gedank. This is a variation of the "meter stick and the hole" paradox, and the principle does apply to your OP, just as I thought.

Your rod hits the floor of the train at an angle, as viewed from the embankment, just as Janus said.

The member who contacted me is the author of a website that hosts a java applet that can illustrate all this.

You can find the java applet here (I have tested it and it is safe (on my system at least!) but it is quite large and takes a while to load) Newtonian and Relativistic Simulations

Or you can just read through the tutorial for the app here - http://www.relativitysimulation.com/...ckAndHole.html which illustrates the meter stick and the hole paradox and shows how the angles are different from each frame of reference.

The only difference between the "meter stick and the hole" paradox and your gedank is that in yours there is no hole. And if you don't believe the author of that website, google for the meter stick and the hole paradox to find other equivalent treatments.

The meter stick and the hole

We have a meter stick aligned with the x-axis, moving at 0.866c. We have a panel moving in the y-axis, sitting perpendicular to the y-axis, with a hole in it that is just larger than half a meter. Because the panel is perpendicular to its axis of motion, there is no contraction in the length of the hole. From the frame of the panel, the meter stick is parallel, but approaching at an angle. When they intersect, the meter stick passes through the hole because, in the frame of the panel, the meter stick is length contracted to half its rest length.

But what happens from the view of the meter stick? In its own frame it is a meter long, and we know the hole in the panel is only just over half a meter long in its rest frame. The answer is that they don't meet parallel. From the frame of the meter stick, the panel makes an angle, allowing the meter long stick to pass through the hole.
Here's another take on this problem. It looks at the collision between two rods, each at an angle to their relative motion. In the frame of one rod the upper ends collide first, and in the frame of the other the lower ends collide first. They go to show why this is not a paradox because the reaction of the collision cannot propagate faster than c in either frame and thus the collisions of the ends have no causal link between them. It doesn't matter which pair of ends meet first because the collision can have no effect on events further along the length of the rod.

http://arxiv.org/ftp/arxiv/papers/0809/0809.1721.pdf

64. Originally Posted by Janus
Originally Posted by SpeedFreek
Originally Posted by xyzt
Originally Posted by SpeedFreek
is not the angle between the rods, it is the angle between a rod and the axis of motion in the rest frame of the rod. Length contraction does not care what the angle of the other rod is, it is only concerned with the angle relative to the axis of motion.
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
I have been contacted by a member here who doesn't post in the forum any more, but who has read through this thread and he has confirmed to me that my diagrams with the blue and red rods are in fact correct for my gedank. This is a variation of the "meter stick and the hole" paradox, and the principle does apply to your OP, just as I thought.

Your rod hits the floor of the train at an angle, as viewed from the embankment, just as Janus said.

The member who contacted me is the author of a website that hosts a java applet that can illustrate all this.

You can find the java applet here (I have tested it and it is safe (on my system at least!) but it is quite large and takes a while to load) Newtonian and Relativistic Simulations

Or you can just read through the tutorial for the app here - http://www.relativitysimulation.com/...ckAndHole.html which illustrates the meter stick and the hole paradox and shows how the angles are different from each frame of reference.

The only difference between the "meter stick and the hole" paradox and your gedank is that in yours there is no hole. And if you don't believe the author of that website, google for the meter stick and the hole paradox to find other equivalent treatments.

The meter stick and the hole

We have a meter stick aligned with the x-axis, moving at 0.866c. We have a panel moving in the y-axis, sitting perpendicular to the y-axis, with a hole in it that is just larger than half a meter. Because the panel is perpendicular to its axis of motion, there is no contraction in the length of the hole. From the frame of the panel, the meter stick is parallel, but approaching at an angle. When they intersect, the meter stick passes through the hole because, in the frame of the panel, the meter stick is length contracted to half its rest length.

But what happens from the view of the meter stick? In its own frame it is a meter long, and we know the hole in the panel is only just over half a meter long in its rest frame. The answer is that they don't meet parallel. From the frame of the meter stick, the panel makes an angle, allowing the meter long stick to pass through the hole.
Here's another take on this problem. It looks at the collision between two rods, each at an angle to their relative motion. In the frame of one rod the upper ends collide first, and in the frame of the other the lower ends collide first. They go to show why this is not a paradox because the reaction of the collision cannot propagate faster than c in either frame and thus the collisions of the ends have no causal link between them. It doesn't matter which pair of ends meet first because the collision can have no effect on events further along the length of the rod.

http://arxiv.org/ftp/arxiv/papers/0809/0809.1721.pdf
Thank you, out of principle I do not trust anything published in the European Journal of Physics, it is a junk journal. Out of respect for you, I will read the paper, I am almost certain that the paper is not worth the paper it was published on. I will get back to you with the errors I found, this should be a fun discussion.

65. Originally Posted by xyzt
Originally Posted by Janus
Originally Posted by SpeedFreek
Originally Posted by xyzt
Originally Posted by SpeedFreek
is not the angle between the rods, it is the angle between a rod and the axis of motion in the rest frame of the rod. Length contraction does not care what the angle of the other rod is, it is only concerned with the angle relative to the axis of motion.
You don't get it, do you? The transformation holds for any angle , between any two arbitrary directions. I have shown this early in the thread. The wiki link is just a particular case of my general proof.
I have been contacted by a member here who doesn't post in the forum any more, but who has read through this thread and he has confirmed to me that my diagrams with the blue and red rods are in fact correct for my gedank. This is a variation of the "meter stick and the hole" paradox, and the principle does apply to your OP, just as I thought.

Your rod hits the floor of the train at an angle, as viewed from the embankment, just as Janus said.

The member who contacted me is the author of a website that hosts a java applet that can illustrate all this.

You can find the java applet here (I have tested it and it is safe (on my system at least!) but it is quite large and takes a while to load) Newtonian and Relativistic Simulations

Or you can just read through the tutorial for the app here - http://www.relativitysimulation.com/...ckAndHole.html which illustrates the meter stick and the hole paradox and shows how the angles are different from each frame of reference.

The only difference between the "meter stick and the hole" paradox and your gedank is that in yours there is no hole. And if you don't believe the author of that website, google for the meter stick and the hole paradox to find other equivalent treatments.

The meter stick and the hole

We have a meter stick aligned with the x-axis, moving at 0.866c. We have a panel moving in the y-axis, sitting perpendicular to the y-axis, with a hole in it that is just larger than half a meter. Because the panel is perpendicular to its axis of motion, there is no contraction in the length of the hole. From the frame of the panel, the meter stick is parallel, but approaching at an angle. When they intersect, the meter stick passes through the hole because, in the frame of the panel, the meter stick is length contracted to half its rest length.

But what happens from the view of the meter stick? In its own frame it is a meter long, and we know the hole in the panel is only just over half a meter long in its rest frame. The answer is that they don't meet parallel. From the frame of the meter stick, the panel makes an angle, allowing the meter long stick to pass through the hole.
Here's another take on this problem. It looks at the collision between two rods, each at an angle to their relative motion. In the frame of one rod the upper ends collide first, and in the frame of the other the lower ends collide first. They go to show why this is not a paradox because the reaction of the collision cannot propagate faster than c in either frame and thus the collisions of the ends have no causal link between them. It doesn't matter which pair of ends meet first because the collision can have no effect on events further along the length of the rod.

http://arxiv.org/ftp/arxiv/papers/0809/0809.1721.pdf
Thank you, out of principle I do not trust anything published in the European Journal of Physics, it is a junk journal. Out of respect for you, I will read the paper, I am almost certain that the paper is not worth the paper it was published on. I will get back to you with the errors I found, this should be a fun discussion.
Hi Janus,

It took me just 5 minutes to read through the paper to conclude that, as expected, it is a very bad one. You are correct, the paper would be relevant to our discussion IF it were correct. Unfortunately, it isn't. Assume, using the author's notation that there are two frames, M and K boosted wrt each other with speed along the x axis.

What the authors are saying is this: the order of two spatially separated events, and is such that (and ). Depending on how the frame M is moving along the x axis with respect to K, the order of the events, as perceived in S' can be the same as in S or can be opposite. Indeed:

for M moving in the positive x direction

for M moving in the negative x direction

Since for all it follows that:

for M moving in the negative x direction
for M moving in the positive x direction

Make and one gets our OP as a particular case of the paper. The problem is that coordinate time is devoid of any physical content, the fact that the clocks run in different order depending on the sense of relative motion between frames M and K does not have any bearing on the physical phenomena, it is just the collisions have been labelled differently, not that the endpoints suffered a reversal of the order of collision. The physics of the problem is invariant to the sense of motion between frames M and K, the two rods suffer no distortion (bending) whatsoever, all points along the two rods align perfectly.

66. Is Lorentz Contraction uniform? Or do objects toward the front of the moving train appear to be contracted more than objects toward the rear of the train?

Intuitively I want to say it is uniform, but non-uniform contracting would solve a lot of issues.

67. Originally Posted by kojax
Is Lorentz Contraction uniform? Or do objects toward the front of the moving train appear to be contracted more than objects toward the rear of the train?

Intuitively I want to say it is uniform, but non-uniform contracting would solve a lot of issues.
In SR, there is no such thing as a perfectly rigid object, so, to answer your question: a pushed rod contracts more at the pushing end, a dragged rod contracts more at the pulling end. If you drag both for long enough, the contraction equalizes. Unfortunately, there is no relativistic flavor of theory of elasticity.

68. But is length contraction at all analogous to pushing or pulling at a rod? I was under the impression that length contraction was due to the constancy of the speed of light - which leads us to the conclusion that neither space nor time are absolute.

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