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If moving object decelerated into your frame, will it lengthen?

  1. Sep 15, 2012 #1
    Let's say something is moving very fast past you. You see the thing to be length contracted. But let's say that thing suddenly slows down very fast and enters your inertial reference frame. Will it suddenly appear to you to have lengthened? My tutor said that this is not the case because of general relativity which I dont yet understand perfectly, but here is my argument as to why he is wrong, using just special relativity:

    Suppose something like a stick in the shape of a "T" (such that it has horizontal arms on either sides) falls vertically into a hole (at very high speed) whose rest length is equal to the rest length of the vertical length of the stick. The hole is only wide enough to let the vertical part of the T in. Now if the stick's bottom touches the bottom of the hole, there will be an explosion instantly. If however the arm of the T touches the top of the hole, a signal will be sent such as to deactivate the bomb.

    Ok so causality is relative and so according to the T stick, the bottom of the stick will touch the bottom of the hole before the arm of the stick touches the top of the hole, and so there will be explosion.

    And according to the hole, the arm of the T will touch the top of the hole before the bottom of the T touches the bottom of the hole. ie, the order of the events is reversed as happens with spacelike separated events.

    But if an explosion occurs in one frame it must occur in all others. So here is the meat of my argument: according to the hole's ref frame, the T is length contracted, such that once the arms of the T hit the sides of the hole, the T shouldn't be able to continue falling into the hole such as to touch the bottom...unless the deceleration that the T experiences results in it becoming lengthened. Only then will the bottom of the T be able to touch the bottom of the hole and explode.

    (And just to tie loose ends, the fact that according to the hole, the signal to deactivate the bomb was sent before the bottom of the T touched the bottom of the hole doesn't matter because the signal would not have made it in time anyway)

    Am i right? does this argument make sense? I want to know before i go yell at my tutor haha:D
     
  2. jcsd
  3. Sep 16, 2012 #2
    A deceleration is an acceleration in the opposite direction. No difference.
     
  4. Sep 16, 2012 #3
    um yea i kno but thats not my question at all..this has absolutely nothing to do with my question..
     
  5. Sep 16, 2012 #4

    Nugatory

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    This question is a variant of the old pole-barn paradox.

    General relativity has nothing to do with it; if your tutor says it does, he's wrong.

    As with so many SR paradoxes, the problem has been carefully phrased to sneak in a bogus "at the same time" assumption, which leads to the contradictory results. In this case, the hidden bogus assumption is that the bottom of the T and the crossbars of the T will both stop moving at the same time in all reference frames. They don't, because of the relativity of simultaneity.
     
  6. Sep 16, 2012 #5

    ghwellsjr

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    Ignoring your post but just answering the question on your title: if moving object decelerated into your frame, will it lengthen?

    The answer is yes, if I can rephrase your scenario:

    If an object is moving in my inertial rest frame, it will be length contracted along the direction of its motion and if it then comes to rest in my inertial rest frame, it's previously contracted length will lengthen.

    Another way of saying this is that the length of the object in its rest frame while it is in motion in my rest frame is the length that it will become when it comes to rest in my rest frame.

    You don't need to complicate the situation with the contents of your post, it's a simple issue.
     
  7. Sep 16, 2012 #6
    Yes you only need SR for the above discussion. Note that in SR everything is in every frame. And perhaps useful to clarify that "suddenly" can normally not exceed the speed of sound (less than the speed of light), and that for gravitation things (as you do next) one sometimes needs GR.
    Apart of the complication from gravitation (which doesn't seem to matter here) I think that it's quite correct, but the dynamics from inertia will be much greater than effects of length de-contraction. Pertinent is that the elastic forces to keep the material together cannot exceed the speed of sound, so that there is nothing to stop the bottom of the T to touch the bottom of the hole! Thus I'm afraid that it's not very convincing for what you want to prove.

    However, the simple answer given by ghwellsjr suffices.

    If you like such riddles about things falling in holes, there are several papers about them, all slightly different (e.g. http://iopscience.iop.org/0143-0807/26/1/003/)
     
  8. Sep 16, 2012 #7
    thanks! nice paper but im kind of confused about what you mean by "Pertinent is that the elastic forces to keep the material together cannot exceed the speed of sound, so that there is nothing to stop the bottom of the T to touch the bottom of the hole! Thus I'm afraid that it's not very convincing for what you want to prove."...do you mean that the T cannot lengthen "instantly" in the rest frame of the hole? I know that...but the point is that it will lengthen fast enough so as to hit the bottom before the signal from the top of the hole reaches the bottom..
     
  9. Sep 17, 2012 #8
    As I think is also discussed in that paper, those atoms and the ones near them are not affected by the collision on the other end (they don't "know" that something happened there) - therefore they are still falling at full speed! The length of an object is the result of force equilibrium, and your example doesn't permit any such equilibrium to install.

    What don't you like about the much simpler proof (post #5)?
     
  10. Sep 21, 2012 #9
    I like the simpler proof but I just want to understand what you are saying as well. Ok so I understand that the atoms at one end don't know what happened at the other end as so are therefore still falling at full speed, but can you please explain why my example doesn't permit such equilibrium?

    Sorry for being so nit-picky but I really want to understand because after I asked this question to the tutor (actually a TA for my class) he apparently suggested a problem similar to my example to the professor who assigned it as hw...basically the question was, given a scenario very similar to the one I described in my first post, would there even be an explosion? You can see in my first post why I said there would be an explosion: the events are spacelike separated, so the signal cannot make it in time....apparently though the correct answer is that there would not be an explosion! This caused huge controversy in my class and even the TA is confused and the professor was out today so we couldnt ask him why....would anyone know why?
    When I have some time later I will take photo of original question. maybe it is the wording that threw us all off. But I really don't think that that's the case....this is really bothering me so please help!
     
  11. Sep 22, 2012 #10
    Aziza,

    You are actually asking a questions that highlights one of the many fundamental paradoxes of SR (also called the 'ladder paradox' or 'pole-barn paradox' as Nugatory stated). Another often encountered paradox is the 'twin paradox'.

    Such paradoxes in SR are explained in different ways by different people, but none are every convincing, and multiple mutually non-compatible explantions exist. The issue has to do with the 'reciprocality' of all observations between the two objects in question in SR (the stick and the hole in your example).

    If you look carefully at the more general GR framework, you will find that this reciprocality because of relative velocity between two objects is no longer present. In the easiest to understand of GR solutions, the Schwarzschild metric, you will notice that the 'relative velocity' is always (somewhat implicitly!) ascribed to a small test mass moving w.r.t. another (again implicitly assumed much larger) stationary spherical mass. This is the principle used to explain the observed time dilation between Earth clocks and GPS satellite clocks (with the GPS satellites having velocity in this context, not Earth!).

    So in that sense, your teacher is right. The absence of a 'preferred frame' in SR has been proved repeatedly wrong (though many still don't accept that fact). GR is the broader framework to look at relativity, as it (implicitly) drops the demand on reciprocality of the relative velocity between objects.
     
  12. Sep 22, 2012 #11
    BTW, causality is never stated to be relative, be it in SR or GR. Only simultaneity is considered relative.
     
  13. Sep 22, 2012 #12
    Good! :smile:
    Of course such an equilibrium will occur, but it's useless for your purpose (that is, if I correctly understood the problem - see next!). You will quickly understand this if you do a little dynamical simulation in your head. Here's mine (from the "ground frame" perspective):

    - First the mechanically stress wave from the destructive shock at the top needs time to reach the bottom end of the T.
    - Next the atoms at the bottom end will slow down due to this pulling force from above. But wait, those atoms already collided with the bottom of the hole! The force equilibrium process with the bottom of the hole starts to happen first, together with the explosion there.
    - etc.

    Overlooking the fact that realistically you may end up with a gas equilibrium above a crater, all the important events take place before your T object has time to "de-contract". So, you end up with a piece of matter (or even a gas) that has been ripped apart at one end, and smashed together at the other end. That does not serve your purpose to demonstrate that is "longer" in a certain way.
    If I correctly understood your verbal problem sketch then I think that there will be two explosions, as elaborated. So yes, it will be useful if you can post a sketch or photo to make sure that the problem is clear. :smile:

    Note: for the usual homework problems we have the homework forum. However, it seems that this is really a discussion about a controversy.
     
    Last edited: Sep 22, 2012
  14. Sep 22, 2012 #13

    PAllen

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    I disagree with many points in the above. GR doesn't have preferred frames any more than SR. The real change is that frames become local, and velocity comparison is only well defined between nearby objects. Within the local frame (the only kind that exists in GR) the laws of SR hold. Within a local region, there is no preferred frame for local physics. Globally, you have coordinate systems, and there are certainly no preferred coordinates system. At least 6 are commonly used for a spherically symmetric gravitational source. As for preferred observers, you can say there are observers that more readily detect global physical symmetries arising from the initial conditions than other observers. However, this is not much of a preference. For example, gravitational time dilation is defined in terms of hovering observers in a sufficiently static field. It is actually not definable at all as a general feature of GR solutions. Note, however, that most bodies in the solar system are near inertial observers (i.e. orbiting in complex ways) not hovering observers. They do not measure simple gravitational time dilation.

    As for SR paradoxes, the pure SR explanations emphasize different useful points to consider. It seems that different ways of explaining so called paradoxes fall into the following useful categories:

    1) It is best to make sure to define everything as measured in one inertial frame. Then Lorentz transform to see what how a different inertial frame would describe the situation. This has a direct analog in GR only for local physics.

    2) Emphasize invariant metric quantities. Compute observables in terms of the SR metric. This generalizes locally or globally to GR.

    3) Emphasize that time rate comparisons at a distance are conventions not physical observables. The only thing you can observe are the behavior of exchanged signals. These can be described in any frame (SR) or coordinates GR. Such observations fall into the general category of Doppler. Interpreting them in reference to an underlying 'time dilation' results from conventions leading to a particular coordinate system, and is not physically observable, nor necessary to understand the physics or compute anything. In SR paradoxes, this leads to Doppler oriented explanations.
     
    Last edited: Sep 22, 2012
  15. Sep 22, 2012 #14

    Fredrik

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    It doesn't have anything to do with general relativity. If you slow down a solid object gently, then every part of it will expand according to the Lorentz contraction formula. This type of motion is called Born rigid motion.

    If you want to talk about an object that slows down "very fast", you need to be more specific. It matters how it slows down. Is it being pushed to a stop from its front? (This would probably crush the object). Is it being pulled to a stop from its rear? (This would probably tear it to pieces). Here we already have two very different answers.

    You were probably thinking that these details don't matter, if we just assume that the object is rigid. The problem is, there's no such thing as a rigid object in relativity. If every part of an object begins to change its velocity at the same time in one frame, they begin to change their velocities at different times in other frames. We can consider an an accelerating object whose motion is "rigid" in a specific frame, but such an object is always being forcefully stretched or compressed.

    You will either have to provide a lot of information about the external forces acting on each component part of the object, and the internal forces in the object, or you will have to specify the motion of every component part. If you choose to specify the motion, you can of course choose to ensure that the object has the desired length after it's been slowed down, but most such specifications involve a lot of stretching and compressing of component parts. Born rigid motion is the motion that minimizes the stretching/compression of the material. (If matter had been a continuum rather than consisting of atoms, there would be no stretching/compression at all when the object is doing Born rigid motion).
     
  16. Sep 22, 2012 #15
    Yea I know causality is not relative...unfortunately I have no clue what you said in your previous post. Is it possible to explain more simply why my teacher is right?
     
  17. Sep 22, 2012 #16

    Fredrik

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    It means that when (in the hole's frame) the top of the T comes to a stop, the bottom is still going at full speed. The sudden stop of the top will send a longitudinal displacement wave from the top of the T to its bottom. What's another word for a longitudinal displacement wave? "Sound". So this wave is traveling (by definition of "sound") at the speed of sound. The bottom can't even begin to slow down until the wave reaches it. So unless the speed of sound in the T is faster than the original speed of the T in the hole's frame, the wave will never reach the bottom of the T. It will crash into the bottom of the hole, unaffected by what happened at the top. Actually, this will happen regardless of how deep the hole is (since the bottom of the T never gets the "message" to slow down). The T is simply being torn apart.

    If you assume Born rigid motion (this would require a gentle slowdown, or that something is magically pulling on every layer of atoms separately by precisely the right amount) instead of assuming that the object is "pulled to a stop from behind", the result will be a simultaneous touchdown at the top and bottom. Edit: Of course, it would also mean that the speed of the T relative to the hole reaches 0 at that moment, so it's a different situation than the one you have in mind.
     
    Last edited: Sep 22, 2012
  18. Sep 22, 2012 #17
    Sorry it took so long but I fell asleep. Here is a picture of the actual problem:


    https://apf.mail.ru/cgi-bin/readmsg/mms_picture2.jpg?id=13483304830000000589%3B0%3B1&exif=1&bs=1737&bl=271692&ct=image%2Fjpeg&cn=mms_picture2.jpg
     
  19. Sep 22, 2012 #18

    Ohh thank you and thanks harrylin for the explanations, I understand this part of my question now!
     
  20. Sep 22, 2012 #19
    Two explosions? I was aiming for just one at the bottom of the hole; the collision at the top doesnt cause explosion, it causes a signal to be sent to the bottom which, upon reaching the bottom, is meant to disarm the bomb. In any case I already posted the picture so hopefully that will make things clear:)

    And I didnt want to post in the hw section because I figured since some people already responded to my initial post here, and no one criticized my proof of why the explosion occurs...ie no one said that in fact the explosion would not occur, then i figured that they would be interested in seeing that my professor is saying we are all wrong!
     
  21. Sep 22, 2012 #20
    Ah yes I forgot that detail - regretfully I can't see the picture. Anyway, based on what you told us I suppose at everyone here will agree that the bomb at the bottom will explode as it cannot be disarmed in time. :smile:
    And this is immediately evident from the perspective of a co-moving frame with the T at the time the bottoms touch.
     
    Last edited: Sep 22, 2012
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