# Relativistic parallel rods

Discussion in 'Physics & Math' started by Pete, Apr 30, 2013.

1. ### brucepValued Senior Member

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I'm not sure what the second point means. "......constructed from pairs of points on bodies in one reference frame .....". That seems to support a common local proper frame exists for that relationship?

3. ### OnlyMeValued Senior Member

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I thought of that too, while away today. I also did a lot of other thinking around the issue.

My concerns originate not from a purely SR perspective and I should not have raised them within the context of a hypothetical constructed purely within the context of SR.

I was commingling material stresses, emerging from some work with inertia and QM that I have been struggling to understand. This was a deviation from the framework of the hypothetical and should have been left, unmentioned.

So.., within the context of the hypothetical and SR there is no issue to resolve. I got side tracked.

5. ### brucepValued Senior Member

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Most of the universe is the weak field. The experiments at CERN are conducted in a laboratory frame where the effects of gravity can be ignored. The reason the effects of gravity can be ignored, in the weak field, is because they generally don't effect empirical measurements in a meaningful way. That's not the case for an experiment such as the GPS or the Gravity Probe B. If the particle experiments at the LHC required the inclusion of the effects of gravity then they would model the experiment to do just that.

7. ### mikelizziRegistered Senior Member

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Thank you for your response. I’m just getting around to a reply. With regards to your above statement/question, I am not sure how to interpret it. But one can always define a reference frame in which a body is at rest, no? And with respect to that reference frame 2 points on a body will always have the same geometric relationship no matter the time. My definition of a pair of points on a body is a pair of x, y, z coordinates with one common time. The body does not have to be at rest to define a pair of points on it like that. Once a pair of points on a body is defined in one reference frame, I can transform them to another reference frame, as long as I have complete information about the body. By complete information I mean the body’s linear/angular velocity, linear/angular acceleration, whether or not it’s exploding, etc. In the parent of this thread, the body, a rod, is declared to have linear velocity in the initial reference frame. Transforming a pair of end points on such a rod to another reference frame is straightforward. But it’s not the same thing as Lorentz transforming 2 events. Because the end result of the transformation must be another pair of x, y, z coordinates with one common time. So characteristics of event transformations may not apply.

If someone is interested in the actual calculations I can run thru an example.

Oops.
I left out something important. (I hope what I am writing isn’t so obvious as to be totally boring) Strictly speaking there are an infinite number of solutions to the problem of transforming 2 points on a body. Any time in the target reference frame will satisfy. Often a specific time can be determined from the context of the problem, like “what is the arrangement of the 2 bodies when they collide”. And often the time in the target reference frame is irrelevant, if I just wanted to know the angle of a translating rod, for instance.

Last edited: May 13, 2013
8. ### TachBannedBanned

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I agree with you. I have long contended that the "solution" presented in the original thread has nothing to do with the physics. It also contradicts the much simpler, more straightforward conclusion of solving the problem in full, in the presence of the gravitational field responsible for the rod motion.

Yes, I would like to see your solution.
Yes, I agree with this statement as well. I tried (in vain) to explain this to Pete over hundreds of posts. No success, maybe you will manage.

9. ### brucepValued Senior Member

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If you can find a rest frame for the relationship you describe then relativity requires measurements made in the local proper frame of the spacetime event transform as invariants [as you know]. Local proper frame measurements are made at a specific time during the event and it's that measurement which is required to transforms as an invariant. That's what I was getting at. I'd like to see how you model the relationship between points.

Last edited: May 13, 2013
10. ### mikelizziRegistered Senior Member

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That's funny because I consider my position on this subject to be the same as Pete's. Could all of this painful arguing be the result of a misunderstanding? Then it wouldn't be funny anymore. I am cleaning up my example/solution and will post it in a couple of hours.

11. ### mikelizziRegistered Senior Member

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I don't know what to say. The resolution to the "Meterstick and Hole Paradox" is that the initially horizontal surface with a hole in it, transforms to a tilted surface with a hole in it. For the problem from which this thread originated, it is the horizontal rod that I (and others) believe transforms tilted. I do 3D transformations of bodies all the time using my computer program. I study the results and believe they are all faithful to SR. And a lot of those transform horizontal bodies to tilted bodies. I am going to post the simplest problem I can think of to demonstrate. There has got to be a misunderstanding someplace.

12. ### rpennerFully WiredRegistered Senior Member

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I tried to present generic calculations for Pete's diagram.
From Post #1 of this thread:

From post #8:

Pete also has $z_A = z_B = Z_{\tiny 0} \neq Z_{\tiny 1} = z'_C = z'_D$ but that does not enter the calculations.

13. ### Fednis48Registered Senior Member

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I'm having a hard time understanding what you mean here. What is a "rest frame for a relationship"?

14. ### TachBannedBanned

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....since it is just a collection of labels attributed by the observer in the platform frame to the collision events in the train car frame....

15. ### mikelizziRegistered Senior Member

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Looks like rpenner posted a solution way back in post #8. Here's mine. It for a specific example.

Given:
Two horizontal rods.
Rod A is at rest with respect to observer S. ------------------------------------(Analogous to the train floor in Pete’s example)
Rod B has constant velocity u (vx, vy) = (0, -.5c) with respect to observer S. --(Analogous to the falling rod in Pete’s example, but there is no gravity in my example)
Observer S’ has velocity v (vx, vy) = (.866c, 0) with respect to S. --------------(Analogous to the observer on the ground in Pete’s example)

Question:
Are the rods parallel with respect to observer S’?

I will make the analysis as easy as possible. I will use 1 light second for the length of both rods and place the left end of both at the origin of S. Now the rods are parallel, of equal length and coincident in S. Will they transform parallel?

You’re probably thinking to yourself “Both rods may be described by exactly the same pair of space time coordinates in S. How can I get different results when I transform them to S’?”

Transforming rod A is trivial. I’ll skip the math but I do want to write down the process because I will use the exact same process to transform rod B.

Let the left end of rod A be identified as AL = (x, y) = (0, 0) and the right end be identified as AR = (1, 0). The Lorentz transformation transforms events, not positions, so I have to convert the positions to events. Since the rod is at rest in S the two ends will always be at the given positions so I can construct a pair of events with any time. I pick t = 0.

So now I have, in S,
Event AL = (t, x, y) = (0, 0, 0)
and
Event AR = (0, 1, 0).

Transforming to reference frame S’ using gamma = 2, I get
Event AL’ = (t’, x’, y’) = (0, 0, 0)
and
Event AR’ = (1.73, 2, 0).

Let me interpret what I got. The Lorentz Transformation tells me the left end of rod A is at (x’, y’) = (0, 0) at time t’= 0 in S’ and the right end is at (2, 0) at time t’ = 1.73 sec. The vector between those two x/y coordinates does not define the rod in S’. What I need is a pair of x/y coordinates with a common time. Any time will do because I’m not concerned with the position of the rod, just the orientation. I’ll pick t’ = 0 since I already know the location of the left end for that time. Where is the right end at t’ = 0? I just back up the position given to me using Newtonian Physics:

(Right end at t=0) = (right end at t=1.73) – (velocity of rod times the time difference)
x = 2 – .866 x 1.73 = .5
y = 0
Now I can construct a vector between the transformed left end and the transformed (and repositioned right end) to get a definition of the rod in S’.
In S the rod is the vector <1, 0>
In S’ the rod is the vector <.5, 0>

No surprises. The rod transformed parallel.

Transforming Rod B:
Let me label the left end of the rod “BL” and the right end “BR”. Then
BL (x, y,) = (0, 0)
and
BR = (1, 0).
Again, that’s two positions on a body. Again I need to convert the x/y coordinates to events with a common time. Again I pick t = 0. And I get the same two events.
BL (t, x, y) = (0, 0, 0)
and
BR = (0, 1, 0).
There is one small difference. Rod B is moving in S. What I have done is declare that the left end of the moving rod passed thru point (0, 0) at time t = 0 and the right end of the rod passed thru (1, 0) at time t = 0. Again transforming each event to S’ using a gamma factor of 2 results in;
BL’ (t’, x’, y’) = (0, 0, 0)
and
BR’ = (1.73, 2, 0).
Same results as earlier. Again I need to convert the two events to two pairs of x/y coordinates with a common time. And again I pick t’ = 0 so I only have to adjust the right end. And this is the part that’s really different because rod B has a velocity component in the y-direction. I am going to move the right end of a rod that is presently on the X-Axis using a velocity that has both x and y components and a non-zero time. Do I need to even do the calculation? I hope it is obvious that the right end of the rod will not end up on the X-Axis.

So,
Both pairs of events transformed parallel.
One rod transformed parallel. One did not.

To anyone interested in the actual position of the right end of rod B in S’ I will post that part of the solution later. It requires relativistic velocity composition. Not difficult. It just takes space.

16. ### TachBannedBanned

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No one is arguing the above, what you have described is the Thomas precession exhibited by the rod moving in the $y$ direction.
The debate is not about that, it is about what happens at the collision between the two rods.

17. ### TrooperSecular SanityValued Senior Member

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Are you essentially saying that all three of Pete’s examples are correct?

Any chance that you'd make it a little bit easier by considering the two rockets in problem 12? Both observers agree that the missile misses the rocket and both are correct. If the rockets have equal and opposite velocity, how is this any different from Pete’s examples? I'm way out of my element here but I’m starting to think that Tach may be right.

http://oyc.yale.edu/sites/default/files/problem_set_7_solutions_6.pdf

18. ### TachBannedBanned

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Pete is trying to use a trick, his rockets have different lengths, one is $L$, the other one is $\gamma L$. If both rockets have the same length, as described in the above exercise, his scenario falls apart immediately. By using different lengths, it takes a little longer for the scenario to fall apart.

19. ### PeteIt's not rocket surgeryRegistered Senior Member

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Wrong thread, Tach. In this thread, the boxes have the same length.
The different length scenario is closer to Einstein's original train-platform-lightning thought experiment.

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi Trooper,
Tach's arguments are confusing, because he's arguing issues from other threads in this one.
I believe he agrees that the animations in the opening post of this thread are valid.

21. ### TachBannedBanned

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You need to pay attention to what Trooper asked. You are too focused on what you wrote.

22. ### TachBannedBanned

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Yes, they are valid . Unfortunately, the animations are totally irrelevant in solving the puzzle, like most of the posts spread over three threads. The irony is that this problem requires only a few lines if you don't dumb it down by removing the gravitational effects.

23. ### PeteIt's not rocket surgeryRegistered Senior Member

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You need to pay attention to what thread you're in.