The rod can only stay horizontal in the hole frame, if in the rod frame, according to LT, the right-hand side of the rod falls forever with a higher velocity than the left-hand side of the rod, because the acceleration has started there earlier. That would not be compatible with "non-destructively".PAllen said:No, I think it would be horizontal in the hole frame, by construction, and slightly distorted and not quite horizontal in the rod frame (assuming we remain talking about the Rindler example from post #2, with a thin right hole edge, rather than a wall). Given the tiny angle of deformation required over a light year, even the most rigid material imaginable would easily accommodate the deformation non-destructively.
True, but I think this would reach a breaking point on a time scale longer than it takes the rod to get through the hole. It's hard to realize how much give even e.g. diamond would have over a 1 ly long rod of e.g. 1 cm thickness. But, yes, eventually the rod must break if the conditions are maintained long enough.Sagittarius A-Star said:The rod can only stay horizontal in the hole frame, if in the rod frame, according to LT, the right-hand side of the rod falls forever with a higher velocity than the left-hand side of the rod, because the acceleration has started there earlier. That would not be compatible with "non-destructively".
Or, alternatively, there would be time for the rod to be pulled horizontal in the rod frame by molecular forces before it broke. I would have to calculate this, at least to order of magnitude, to know which is a more plausible model.PAllen said:True, but I think this would reach a breaking point on a time scale longer than it takes the rod to get through the hole. It's hard to realize how much give even e.g. diamond would have over a 1 ly long rod of e.g. 1 cm thickness. But, yes, eventually the rod must break if the conditions are maintained long enough.
It is removed simultaneously in the ground frame. This leads to the rod being straight in the ground frame, as shown in the diagrams that he draws.pervect said:This is a bit ambiguous due to the relativity of simultaneity, I was assuming the trap door was removed "at the same time" as seen in the rod's frame, which seems simplest to analyze. But it's not entirely clear as which simultaneity convention is implied when Rindler states that the floor is removed.
And it should be obvious that such a rod will not be rigid.PAllen said:You can work around this by e.g. having a rod whose thickness would be less than an atom, if you want to consider that acceptable.
I think I mentioned in a previous thread that I believe it's a bad idea to try to learn SR using these so-called paradoxes. Instead, you should learn SR properly first and then tackle these paradoxes.alan123hk said:At my level, it seems that I can't immediately understand it, but anyway, I would like to thank you for providing precious reference information.
Your suggestion might be a good idea. The problem is that I am reading a book on basic special relativity, and I read each chapter in order. These famous "paradoxes" have been mentioned in the previous chapters of this book. So far, I understand almost everything except for incomplete understanding or not completely satisfied with the explanation of the "paradox" in the book, so I can't restrain myself from trying to find an explanation that I think is satisfactory. I will actively consider your suggestions, but by the way, I think only a small part of these 100 posts are directly related to the basic issues I want to discuss.PeroK said:In any case, I suggest you consider whether you are trying to learn SR in an efficient manner.
Which one? We may have found the problem, as many basic ones are fairly awful (failure to explain what is meant by a reference frame, failure to introduce relativity of simultaneity early and often, confusing time dilation and the twin paradox, overemphasis on length contraction and time dilation are among the more common failings).alan123hk said:I am reading a book on basic special relativity,
Thank you for your reminder, I don't know if the explanation of this book is not detailed and clear enough, but this very old book is written in Chinese, my native language, so when I read this book, it feels more fluent and easier to understand from a language perspective.Nugatory said:Which one? We may have found the problem, as many basic ones are fairly awful (failure to explain what is meant by a reference frame, failure to introduce relativity of simultaneity early and often, confusing time dilation and the twin paradox, overemphasis on length contraction and time dilation are among the more common failings).
PAllen said:No, I think it would be horizontal in the hole frame, by construction
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | -1 |
| 0 | 0 | 0 | 0 | 0 | 0 | -1 | -2 |
| 0 | 0 | 0 | 0 | 0 | -1 | -2 | -4 |
| 0 | 0 | 0 | 0 | -1 | -2 | -4 | -8 |
You missed that I said "with a thin right hole edge, rather than a wall", i.e. the rod just continues to fall and move to the right in the hole frame (or the hole continues to move to the left in the rod frame). This is right in the post your replied to. The rest of your post is generally consistent with what I have been saying. Except that I proposed that nearby atoms will apply a counter to free fall only at a delay of molecular distance divided by speed of sound. I think this means that for your light year long rod and ordinary g, there is plenty of time for back reaction before breakage; while for any lab scale case, the only response is for molecules to stabilize to an independent, unbound state. Ultimately, everything depends on the distance between a molecule and its neighbors by the time it can "communicate" with them. If this is too large, "dust" is the result. If not, attractive forces will pull the molecules back together, thus straightening the rod in the rod frame (and curving it in the hole frame).Sagittarius A-Star said:In Rindlers scenario, the right wall of the hole hits the right-hand side of the rod in the rod frame, before the rod is completely in free fall (see figure b). So, Rindlers "by construction" statement is only valid until this event.
The following table shows in the rod-frame the z'-coordinate of 8 atoms on the bottom of the rod (columns = x'-position, rows = instance in time, increasing downwards.
The shown time granularity is ##D * v / c^2##, with D= distance between the atoms.
0 0 0 0 0 0 0 -1 0 0 0 0 0 0 -1 -2 0 0 0 0 0 -1 -2 -4 0 0 0 0 -1 -2 -4 -8
Because the point, where the falling starts (due to missing support of the trap-door), is moving to the left with more than c, for example the atom in the 3rd table row and z'-position "-2" thinks, that its originally right neighbor is also at z'=-2 (instead of z'=-4). So, the right neighbor can freely fall and the rod stays horizontally in the hole-frame.
The news can propagate no faster than the speed of light and cannot reach the adjacent atom before a time interval D/c has passed. Until then the shear forces cannot act.
PAllen said:Except that I proposed that nearby atoms will apply a counter to free fall only at a delay of molecular distance divided by speed of sound.
PAllen said:I think this means that for your light year long rod and ordinary g, there is plenty of time for back reaction before breakage;
I disagree. When 'news' of displacement of neighbor reaches a given molecule, it responds to that. It does not respond to a yet later position of the molecule until later yet. Note, irrespective of the faster than c propagation of where force is applied, the relative speed of molecules is quite small in the light year rod case. So there is plenty of time for restoring forces. In the lab scale model, the only problem is that displacement becomes too large before a response can occur.Sagittarius A-Star said:But after that time, the nearby atom has already itself a new z'-coordinate and sees then not need to react to the (same) old z'-coordinate of it's neighbor.
PAllen said:I disagree. When 'news' of displacement of neighbor reaches a given molecule, it responds to that. It does not respond to a yet later position of the molecule until later yet.
Horizontal versus vertical are irrelevant. For simplicity, imagine a molecule connected to its neighbors by springs. Neighbor undergoes displacement, wave propagates on spring, given molecule responds to wave produced by initial displacement, and only later finds out about any further change of position by neighbor. Note that if there were anything really mysterious about FTL initial displacement, then a rod sliding off a table at 100 mph, with a trapdoor pulled at 2g at the right moment (and no right side of hole at all, which is just a distraction for this discussion) would have all the features of this case. The 1 g force would propagate FTL right to the left in the rod frame, but not quite simultaneous. Do you really think any strange breakage would occur in this case?? On the other hand, there would certainly be molecular restoration forces, and some minute flexing in the rod before equilibrium is reached.Sagittarius A-Star said:I don't understand this argument. I think it is always 'news' of an apparent non-displacement of the neighbor (same z'-coordinate). The horizontal "signal" from the neighbor at z'=-2 reaches the atom, when it has arrived in the meantime itself at z'=-2.
PAllen said:Do you really think any strange breakage would occur in this case??
There is another problem with this. I claim the news travels at the speed of sound, not light, for mechanical purposes. Then there is definitely news of a different z.Sagittarius A-Star said:I don't understand this argument. I think it is always 'news' of an apparent non-displacement of the neighbor (same z'-coordinate). The horizontal "signal" from the neighbor at z'=-2 reaches the atom, when it has arrived in the meantime itself at z'=-2.
No, he is wrong on glass at lab scale example where accelerations are centimeter per nanosecond squared. Gron was talking about a case where a glass rod went through a small hole at lab dimensions.Sagittarius A-Star said:In this 100 mph example obviously no breakage occurs, according to our experience. That would also be valid for glass. But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with ##z'=z##, without being slowed down by internal forces. Maybe, Gron is therefore correct with the glass?
I completely disagree. The separation between molecules in their local rest frames would increase without bound. Either there are restoring forces, or dust is the result. In cases where the growth of displacement is less than the speed of sound for the material, restoring forces would produce a modified equilibrium shape in the rod frame, such that further free fall involves no further displacement from each element’s rest frame.Sagittarius A-Star said:And what about my argument "But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with ##z'=z## , without being slowed down by internal forces."?
Let me be clear about something. If the trap door were pulled in rod frame, then rod would simply free fall with no stresses. If moving trap door is pulled in its own frame, then there would be small stesses and restoration forces leading to delayed equlibrium for the rod in its own frame (for ordinary g forces; for extreme cases, there would transition to dust state) There cannot be absence of stress in both cases.Sagittarius A-Star said:In this 100 mph example obviously no breakage occurs, according to our experience. That would also be valid for glass. But LT from frame A to frame B suggests, that in it's frame B, the rod's slices must fall freely and sequentially with ##z'=z##, without being slowed down by internal forces. Maybe, Gron is therefore correct with the glass?
PAllen said:then there would be small stesses and restoration forces leading to delayed equlibrium for the rod in its own frame.
Of course, so what?Sagittarius A-Star said:After this delay, the falling rod must be then distorted in the hole's frame, according to LT.
I thought of another fundamental problem with this table. Interactions are two way communications. Even in this table, if you model each element doing a round trip communication with its neighbor, the round trip time increases without bound. Meanwhile, the rod free falling in its own rest frame horizontally, has constant round trip times. It is the round trip time that matters - interactions are two way exchanges. And this is why the frame invariant kinematic decomposition is what really determines whether stresses are present to respond to (at whatever delay). And this clearly shows the hole frame horizontal fall has ever growing stress, while the rod frame horizontal fall has none. And thus, if the acceleration in the hole frame scenario for initial state is not too extreme, an equilibrium will be reached that is Born rigid in the rod frame and bent in the hole frame.Sagittarius A-Star said:The following table shows in the rod-frame the z'-coordinate of 8 atoms on the bottom of the rod (columns = x'-position, rows = instance in time, increasing downwards.
The shown time granularity is ##D * v / c^2##, with D= distance between the atoms.
0 0 0 0 0 0 0 -1 0 0 0 0 0 0 -1 -2 0 0 0 0 0 -1 -2 -4 0 0 0 0 -1 -2 -4 -8
Because the point, where the falling starts (due to missing support of the trap-door), is moving to the left with more than c, for example the atom in the 3rd table row and z'-position "-2" thinks, that its originally right neighbor is also at z'=-2 (instead of z'=-4). So, the right neighbor can freely fall and the rod stays horizontally in the hole-frame.
The news can propagate no faster than the speed of light and cannot reach the adjacent atom before a time interval D/c has passed. Until then the shear forces cannot act.
Each segment of the rod is in free fall and never "knows" that the adjacent segments are at different heights. The shape of the rod in frame B is determined solely by how long each segment has been falling and has nothing to do with the elastic properties of the material of which it is composed. A rubber rod and a steel rod would have precisely the same shape.
Oh, and yet another thing about that table: if you consider a signal from one element to its right neighbor (rather than to its left), the right neighbor has the information on the displacement as soon as it gets this signal. But really, changing timings in a continuing sequence of two way probes is a better model of how molecular bonds would detect and respond to changes in their neighbors; and exactly this is equivalent to the kinematic decomposition.PAllen said:I thought of another fundamental problem with this table. Interactions are two way communications. Even in this table, if you model each element doing a round trip communication with its neighbor, the round trip time increases without bound. Meanwhile, the rod free falling in its own rest frame horizontally, has constant round trip times. It is the round trip time that matters - interactions are two way exchanges. And this is why the frame invariant kinematic decomposition is what really determines whether stresses are present to respond to (at whatever delay). And this clearly shows the hole frame horizontal fall has ever growing stress, while the rod frame horizontal fall has none. And thus, if the acceleration in the hole frame scenario for initial state is not too extreme, an equilibrium will be reached that is Born rigid in the rod frame and bent in the hole frame.
And I repeat how amazing it is to me that none of these sources even consider the kinematic decomposition, when this was first published in 1909!
The new thread with the kinematic decomposition calculations is here:PeterDonis said:Ok, when I get a chance I'll start a new thread and post a link to it here.
Note that if you require the two rod ends to enter the hole simultaneously in the rod frame, a hole whose contracted width is greater than the rod length is required. If, instead, you require the the two ends to enter the hole simultaneously in the hole frame, then a big (proper length) rod can get through a small hole. Both of these statements are true, and they are not in conflict because they are different physical setups..RW137 said:In this kind of problem, you will not come to a conclusion with length contraction alone. You have to apply space-time physics: you have to consider events.
Call the end points of the rod, A and B (from left to right) and those of the hole, C and D. The rod moves to the right in the reference frame of the hole, whereas the hole moves to the left in the reference frame of the rod.
Let ‘B at C’ be the reference event. From there, B moves from C to D and C moves from B to A. In space-time physics, the question is not about the rod matching the hole but about the simultaneity of the events ‘B at D’ and ‘C at A’.
Compute the (x, t) coordinates of these events, in each frame. Then choose one of the frames (for instance the rod frame) and compute the space interval as well as the time interval of the events. You will find out, that the space interval dominates: the events are space-like. This means that there exists a third reference frame in which the events are simultaneous. In this reference frame both lengths L0 are contracted to the proper distance of the events. Now, you can say, that in this reference frame the rod matches the hole. However, in all other reference frames the two events will not happen simultaneously.
Note, that length contraction is not a compression of some rigid structure. You cannot establish a connection between space-like events.
PAllen said:To sidestep this largely separate feature of SR (that the Herglotz-Noether theorem places very tight constraints on what motions can be stress free in SR), and focus on understanding how length, simultaneity, and orientation are all frame dependent, we should simply posit a pure inertial version of this problem.
Source:Iyer said:Differing perceptions on the landing of the rod into the slot
...
In the conventional rod and slot paradox, the rod, if it falls, was expected to fall into the
slot due to gravity.
...
Marx [6] observed that as the velocity of the rod increased, the observation from the co-moving
frame of the slot would be rod rotated and contracted but the line of motion of the end points of
the rod would remain the same. This indicated that the rod’s passing or not passing through the
slot was not affected by the magnitude of the approach velocity.
...
In this paper we present a variant of the rod and slot paradox with motion in two directions but
only with constant velocity. The pedagogical contribution of our paper shows how the noninvariance
of angles and proper lengths comes into play.
PAllen said:Note that if you require the two rod ends to enter the hole simultaneously in the rod frame, a hole whose contracted width is greater than the rod length is required. If, instead, you require the the two ends to enter the hole simultaneously in the hole frame, then a big (proper length) rod can get through a small hole. Both of these statements are true, and they are not in conflict because they are different physical setups..
Independent of ##v##?RW137 said:The rod as well as the hole have the same length L0. I said, that there exists a third reference frame (different from the rod and hole frames) in which the two events are simultaneous. If you do the math, you will find out, that in this reference frame, the rod and the hole still have the same length, but are contracted by the factor 0.816.
It's actually not the same.alan123hk said:I found the following article about the rod and hole paradox.