Talk:Ladder paradox

From Wikipedia, the free encyclopedia

	Portal This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.
???	This article has not yet received a rating on the assessment scale. [FAQ]
???	This article has not yet received an importance rating within physics.
Help with this template Please rate this article, and then leave comments to explain the ratings and/or to identify its strengths and weaknesses.

[edit] Explanation

[edit] Part 1

At a first read the solution to this paradox seems hard to digest. Is it not that the very hypothesis of fitting the ladder in the garage is based on an observer at rest relative to the garage? If not so, then what does the "fit" mean?

To "fit" means that the entire ladder is contained within the garage with the doors closed. It is easy to see this from the reference frame of the garage. From the reference frame of the ladder itself this seems impossible until you realize that for the garage to contain the ladder it must bring the ladder to a complete halt within itself (make the ladder be at rest relative to the garage). In order for that to happen the ladder must be accelerated into the reference frame of the garage. The acceleration doesn't happen all at once but one part of the ladder at a time (think of a series of beads connected by springs), as is mentioned in the article.

Coincidentally, I've been thinking about including the original man falling into grate version because although the solution is simpler to visualize in the ladder and garage version (i.e. two-door garage) it's easier to visualize the application in the man and grate version (as opposed to the contracting ladder). DonQuixote 03:58, 21 May 2005 (UTC)

I have read the above explanation, and the explanation in the article and it is wrong. The ladder is "contained" in the garage if you are in the rest frame of the garage, but it is not "contained" in the garage when you are in the rest frame of the ladder. "Contained" is a relative term, not an absolute one. Two inertial observers can disagree over whether the ladder was "contained" just as they can disagree over whether two events were simultaneous. Use the idea of two doors which are initially closed, and the ladder is heading towards the garage. The front one opens just when front of the ladder enters, and the back one opens just when the front of the ladder exits. The front door closes just when the back of the ladder enters, and the back door closes just as the back of the ladder exits, leaving both doors closed. The ladder is "contained" if there is a point in time during which both doors are simultaneously closed after the first door has opened and then closed. "Simultaneously" is a relative term. It can very well be that what the person at rest with the garage says is simultaneous is not simultaneous with respect to the person on the ladder. That is in fact the case. Both people agree that each door opened and closed, they disagree on whether they were both closed simultaneously after the first door had opened and closed. The resolution of the paradox is not to show that both see the ladder as contained, but to show that "containment" is a relative term, not absolute.PAR 04:56, 22 Jun 2005 (UTC)

The ladder is contained in both reference frames -- so it's correct. The part that you have to understand is that the ladder has to accelerate into the reference frame of the garage and come to a complete halt with respect to it. Read Relativity: Special, General and Cosmological by Wolfgang Rindler. Come to think of it, I should probably include a link to it in the article (silly me). Also, don't mix up the two-door garage example with the one-door garage example. Yes, in the two-door garage example the ladder cannot be contained within the garage (the two-door example is used to illustrate the concept of relative simultaneity), however in the one-door example the ladder has to be contained within the garage in both reference frames, hence the ladder contracting under its own acceleration. DonQuixote 04:04, 22 Jun 2005 (UTC)

Ok, we agree how things work in the "two door example", the fact that they disagree on whether the ladder is contained, right? But I don't get the "one door example". If you require that the ladder be contained in the garage in both reference frames, then yes, you have to crush the ladder to make it fit, because its rest length is longer than the garage, but its not going to crush itself. If the force of deceleration doesn't happen to crush it, then it won't fit, and there's no special reason that the force of deceleration should crush it, you may have to apply extra force to crush it. And theres no guarantee that once crushed it will spring back to its original rest length. What's the paradox? PAR 04:56, 22 Jun 2005 (UTC) (PS I will try to find Rindler)

Actually, it is going to crush itself. There is no such thing as a rigid rod in relativity. Here's where relative simultaneity comes into play. From the perspective of the garage, every point of the ladder comes to a halt within the garage simultaneously while from the perspective of the ladder, each point of the ladder comes to a halt within the garage sequentially. And it will spring back to its original length because it is not being physically crushed but relativistically crushed (i.e. trash compacter vs. relativistic length contraction). This is because during the acceleration (or deceleration), each point of the ladder occupies a different inertial reference frame as a result of the sequential acceleration. The paradox is that without this contraction there is no way a ladder will fit into a one-door garage from the perspective of the ladder. DonQuixote 14:04, 22 Jun 2005 (UTC)

I understand that there is no such thing as a rigid rod in relativity. As to the crushing force, there will have to be an external force applied to the ladder to make it come to rest in the garage frame, and this is the crushing force I am referring to. Lets say this crushing force is delivered by the (doorless) back wall of the garage when the front of the ladder hits it. Lets say that when the front of the ladder hits the back wall, it stops rather quickly. Simultaneously with that instant, the garage guy will say that the back end of the ladder is inside the garage, while the ladder guy will say that it is not. This is because they disagree on what is simultaneous. The collision event at the rear of the garage can only propagate its influence at the speed of light, no faster. Therefore, the back end of the ladder will continue on its course, unaffected by what happened at the head of the ladder. Eventually it will be influenced by the collision at the back of the garage, as the crushing effect moves along the ladder at less than or equal to the speed of light, as you say. At the instant the back of the ladder feels the effect of the collision at the front, it will be inside the garage. The point is, the ladder has been trash-compacter crushed by the force of the impact with the rear wall. Whether it now springs back to its original length or not depends on the physical characteristics of the ladder. I expect if it is wood, it would be quite damaged, and not resume its original condition.

To say it another way: If the rest length of the ladder never changes, then it will never fit inside the garage while it is at rest in the garage. Never. If different parts of the ladder are in different inertial frames, this is exactly, identically, the same as saying that it is being trash-compacter crushed. PAR 15:31, 22 Jun 2005 (UTC)

The problem is that you're thinking of the ladder as being in one inertial reference frame. It's not. If you were to stand at the back of the ladder, when the front part of the ladder accelerates into the reference frame of the garage it won't be in the same reference frame that you're in. The front part of the ladder doesn't physically get crushed, rather it undergoes a relativistinc length contraction. It appears to be crushed because you're no longer in the same reference frame as it is. Think of two ladders that are side-by-side. If one of them suddenly accelerates into another reference frame, it won't be trash compactor crushed but will be relativistically length contracted with respect to the other ladder. Instead of having them side-by-side we have them in single file, we'll have the same phenomenon. If we make the individual ladders smaller and smaller and put a large number of them in single file, then we'll have the ladder in a garage situation where the ladder itself isn't being trash compactor crushed but rather each individual part of the ladder is being length contracted at different times. In other words, from the perspective of the garage each part of the ladder is being length contracted simultaneously, however from the perspective of the ladder each part of the ladder is being length contracted sequentially. DonQuixote 16:54, 22 Jun 2005 (UTC)

[edit] Part 2

Ok - this is getting simpler. The definition of physically (trash-compacter) deformed is that at some instant, one part of the ladder is simultaneously moving at a velocity that is different from another part of the ladder according to someone at rest with some part of the ladder. (this is the person who defines "simultaneous"). Unless all parts of the ladder are simultaneously at rest with each other part, (again, according to someone at rest with respect to the ladder), then the ladder is being physically deformed at that time. Can we agree on this definition? PAR 18:19, 22 Jun 2005 (UTC)

PS - this excludes someone who is not in the rest frame of the ladder, who is watching it perhaps accelerate/decelerate, and undergoing the subsequent relativistic "sequential" or "non-simultaneous" length contractions/expansions.PAR 18:19, 22 Jun 2005 (UTC)

No, the definition of physical deformation (trash compacter) is that the rest length is changed. No such thing happens in the acceleration contraction of the ladder. The only time anything remotely like that happens is after the ladder comes to a complete halt in the garage and reverts back to its rest length where it is deformed by hitting the walls of the garage. DonQuixote 04:19, 23 Jun 2005 (UTC)

You could have the ladder's front half being crushed while its back half was stretched. There would be physical deformation, yet the rest length would stay the same. To be precise, there is no physical deformation only if all parts of the ladder have the same velocity, according to someone at rest with the ladder. (That velocity would be zero). It follows that the rest length is unchanging, but an unchanging rest length does not imply the absence of physical deformation. Its a good definition isn't it? PAR 05:01, 23 Jun 2005 (UTC)

You're just making it more complex that it has to be. It's like assuming a cow is spherical and spurts milk isotropically to make the calculations easier. Similarly, assuming a ladder that doesn't change its rest length makes understanding this problem easier. The bottom line is that the rest length of our simple ladder doesn't change and therefore it doesn't get crushed physically; the only thing that happens is that it undergoes a relativistic length contraction (simultaneously with respect to the garage and sequentially with respect to the ladder). DonQuixote 20:47, 23 Jun 2005 (UTC)

When you say "sequentially with respect to the ladder" are you saying that, according to someone sitting on the ladder, there is some point in time when part of the ladder is relativistically contracted, and part of the ladder is not? PAR 20:58, 23 Jun 2005 (UTC)

Yes, exactly. DonQuixote 03:21, 24 Jun 2005 (UTC)

Then (with respect to the ladder person) the part of the ladder that is relativistically contracted is travelling at a different velocity than the part that is not. right? PAR 05:19, 24 Jun 2005 (UTC)

I know what you're trying to get to, and the answer is that it's still not being physically crushed. Look, the thing is that all this takes place in a split second and what we're interested in is the part that happens before anything physical happens to the rest length of the ladder. In that split second it's purely relativistic. It's like assuming a frictionless surface in order to assess the equations of motion of a moving object. Similarly, we're disregarding any crushing effects and are focusing directly on the relativistic effects. Another way of looking at it is that instead of a ladder made out of paper or aluminum, we're considering a non-crushable super-strong material. This material cannot be physically crushed, ever--i.e. it's the extreme case scenario. Even with this indestructible material, the ladder will undergo a relativistic length contraction: simultaneous in the garage rest frame and sequential in the ladder rest frame. DonQuixote 18:33, 24 Jun 2005 (UTC)

I'm just trying to nail down where we disagree. I know that you know what I'm trying to get to, but how do you answer the above question? I think its a reasonable question. Is it right or wrong? PAR 19:34, 24 Jun 2005 (UTC)

It's both right and wrong. From the perspective of the garage, the every part of the ladder is travelling at the same speed and every part of the ladder accelerates simultaneously into the reference frame of the garage. From the perspective of the ladder, every part of the ladder is travelling at the same speed but each part of the ladder accelerates at different times, and therefore have different speeds, when it accelerates into the reference frame of the garage. However, as stated above, this doesn't mean that the ladder gets physically crushed (the non-crushable material). Rather, the ladder undergoes a relativistic length contraction. In other words it's saying that there is no such thing as a rigid object in relativity because even a non-crushable object (something that cannot be trash compacter crushed) will under go length contraction. DonQuixote 20:30, 24 Jun 2005 (UTC)

It's a clear, unambiguous question. It cannot be both right and wrong. From what I understand, you are saying that it's right, different parts of the ladder have different speeds at some instant in time with respect to an observer at rest with some part of the ladder, and that this does not imply physical deformation of the ladder. Is that right? Because if it is, we have nailed down the basic nature of the disagreement. PAR 21:01, 24 Jun 2005 (UTC)

From the reference frame of the ladder it's right; from the reference frame of the garage it's wrong. It can be both right and wrong based on the observer. The problem with common every-day speed and acceleration is that it's not a true relativistic vector and therefore variant based on the observer. That's why we have things like time-dilation and length-contraction. That's why rods appear to bend and and distort. The ladder only appears to deform but it isn't actually deforming. Look, as stated above, if we start out with an invulnerable ladder--one that's not subject to being crushed, even with this miraculous highly indestructible thing, given the sitation, this thing will still appear to be crushed. It's not being physically deformed, since we have specifically stated that that is impossible, so something else is going on. That something else is just relativistic length contraction. Forget speeds, forget accelerations--what we have is an uncrushable object that appears to have been crushed. DonQuixote 04:57, 25 Jun 2005 (UTC)

The question was:

"Then (with respect to the ladder person) the part of the ladder that is relativistically contracted is travelling at a different velocity than the part that is not. right?"

The question has always been framed with the proviso that it was with respect to someone at rest somewhere on the ladder, so I take it you are saying that it is right. This is the crux, the central core of the disagreement. I understand that the relativistic contraction is just an appearance, not a physical distortion. The thing that makes all the difference is the simultaneous presence of different distortions (according to the ladder person). Relativistic distortion is a function of velocity. If the ladder person sees one piece of the ladder contracted and another piece uncontracted, then it is absolutely certain that those pieces are moving at different velocities. This difference in velocity is the very definition of a physical distortion in progress. If all parts were equally contracted, they would all have the same velocity, and there would be no physical distortion although there would be a relativistic contraction.

To say it a different way, just because two relativistic distortions are each an "appearance" does not mean that the fact that they are different is also an "appearance".

Sorry, I haven't read through the complete discussion above, but I believe that it is not necessary to invoke the notion of acceleration; the ladder doesn't have to come to a halt in my understanding. Specifically, I disagree with the statement

for the garage to contain the ladder it must bring the ladder to a complete halt within itself (make the ladder be at rest relative to the garage)

In my view, the question is simply which of the following two events happens first:

• event A: the front of the ladder crashes into the end of the garage
• event B: the end of the ladder enters the garage

Both observers agree to say that "the ladder fits" if and only if event A happens after B. The two observers simply disagree which of the two events occurs first (because of relativity of simultaneity), so we conclude that the statement "the ladder fits" is also relative and both observers are correct to reach opposite conclusions. AxelBoldt 20:25, 25 Jun 2005 (UTC)

Hello Axel - Thanks for responding to our discussion - I know we have generated a lot of verbiage but I think I can summarize as follows: (Please correct me if I'm wrong, DonQuixote)

• There are a number of versions of the ladder problem. You have described what I will call "ladder paradox 1". I think we all agree on the nature and solution to this problem.
• What we are discussing is what I will call "ladder paradox 2" and whether it is indeed a paradox. This consists of assuming that once the ladder is contained in the garage, (according to the "garage guy" who is at rest with respect to the garage), it then experiences, at each point, simultaneously, an equal deceleration which brings the ladder to rest inside the garage. Again, this "simultaneous" is with respect to the garage. In the ladder frame of reference, this deceleration will not be simultaneous, but "sequential". The question is, does the ladder undergo a physical deformation during this deceleration, and what happens after the deceleration is complete?

I have put two headers on the discussion (part 1 and part 2). Part I think delineates the initial part of the discussion where I didn't really understand the nature of the disagreement. Part 2 is where we zero in on the exact nature of the disagreement. If you have time to read and comment on part 2, I think we would both appreciate it. PAR 22:54, 25 Jun 2005 (UTC)

"Consider the admittedly unrealistic situation of a man carrying horizontally a 20-ft pole and wanting to get it into a 10-ft garage. He will run at a speed v = 0.866c to make γ = 2, so that the pole contracts 10 ft. It will be well to insist on having a sufficiently massive block of concrete at the back of the garage, so that there is no question of whether the pole finally stops in the inertial frame of the garage, or vice versa [note: in the article I just generalized it to an undefined acceleration]. So the man runs with his (now contracted) pole into the garage and a friend quickly closes the door. In principle we do not doubt the feasibility of this experiment; that is, the reality of length contraction. When the pole stops in the rest-frame of the garage, it will tend to assume, if it can, its original length relative to the garage [emphasis mine]. Thus, if it survived the impact, it must now either bend, or burst the door, or remain compressed [i.e. trash-compacter crushed]."--Wolfgang Rindler, Relativity: Special, General and Cosmological pg.63

Also, the ladder fitting "if and only if event A happens after event B is wrong", unless you change event A to "the front of the ladder exits the other end of the garage". And the two observers don't "disagree which of the two events occurs first" (at least not in the way that its being referred to). The disagreement arises because the garage sees both events occuring simultaneously while the ladder sees event A occuring before event B (relative simultaneity). DonQuixote 08:06, 26 Jun 2005 (UTC)

Hello DonQ - I have rearranged the Ladder paradox page in a way I dont think you will disagree with. I separated into 3 parts, and put in some explicit calculations and a Minkowski diagram for the garage with two doors problem. I have left the second part, that we are discussing, alone. I modified the Minkowski diagram to illustrate the situation we are discussing. These are shown below, with my argument. We have to discuss this more quantitatively, and the Minkowski diagram is the way to do it.

You have introduced above a new twist - the idea that the ladder stops as a result of collision with the rear wall. This is different than saying all of its points stop simultaneously in the garage (in the garage reference frame) This is because the effects of the impact can only travel at the speed of light or less, so the deceleration will be sequential in the garage frame. I'm not sure which we want to discuss so I diagrammed both. Also you introduced events A and B and I don't know what they mean.

With regard to the quote by Rindler, read the sentence after the highlighted one - "Thus, if it survived the impact, it must now either bend, or burst the door, or remain compressed [i.e. trash-compacter crushed]". remain compressed - this means it is compressed, i.e. physically deformed. This is my point - Rindler says it is compressed, I say it is compressed, the Minkowski diagrams say it is compressed (in the green regions). The compression is real, not apparent. There is no way that a 12 foot ladder can be at rest inside a 10 foot garage without being physically trash-compacter compressed! Not even for an instant! If you still disagree, please, lets get quantitative and tell me where the Minkowski diagram(s) are wrong.

Figure 1 - This is the case where the ladder is stopped by impact with the back wall of the garage. The ladder is in red, the garage in blue. Everything simultaneous to the garage guy is parallel to the x axis. Everything simultaneous to the ladder guy is parallel to the x' axis. The impact is event A. At impact, the garage guy sees the ladder as AB, the ladder guy sees it as AC. The ladder does not move out of the garage, so its front end now goes directly upward, through point E. The back of the ladder will not change its trajectory in space-time until it feels the effects of the impact. The effect of the impact can propagate outward from A no faster than the speed of light, so the back of the ladder will never feel the effects of the impact until point F or later. When the back of the ladder enters the garage at point D, the ladder guy will see the ladder as DE. It is seen that this length is not the same as CA. The ladder is being crushed physically everywhere in the green region.

Figure 1 - This is the case where the ladder is stopped all along its length, simultaneously in the garage frame. Again, when back of the ladder enters the garage at point D, it has not yet felt the effects of the deceleration. According to someone at rest with respect to the back of the ladder, the front of the ladder will be at point E. Again, when the back of the ladder enters the garage at point D, the ladder guy will see the ladder as DE. It is seen that this length is again not the same as CA. The ladder is being crushed physically everywhere in the green region.

"Remain compressed" means that the length contraction compressed ladder becomes a trash compacter compressed ladder as the walls of the garage physically deform it when it returns to its rest length. The ladder is still not being crushed physically until after the event. In the Minkowski diagrams, CA is the rest length of the ladder but DE is not the rest length just like BA is not the rest length. This is because, just like the object described by BA is in a different reference frame from the garage, each part of the object described by DE are in different reference frames. That is, the green area is relativistically length contracted and if you mathematically uncontract it you'll get a length (rest length of that section) which you add to the length of the non-contracted part of the ladder to get the rest length. I'll sit down and work on the quantative description during any free time I have this week. DonQuixote 28

June 2005 14:05 (UTC)

Yes, thats an excellent idea, we have to get quantitative, because this qualitative discussion is not doing it. Just to respond to the above:

• You say: Remain compressed means that the length contraction compressed ladder becomes a trash compacter compressed ladder as the walls of the garage physically deform it when it returns to its rest length. I cannot make sense of this, we need a diagram. Especially since you have changed the definition from what you wrote previously: Thus, if it survived the impact, it must now either bend, or burst the door, or remain compressed [i.e. trash-compacter crushed].

Does it "remain" or "become" trash compacter compressed?

• You say: The ladder is still not being crushed physically until after the event. I agree totally. If you are saying that yes, the ladder is crushed after it hits the back wall, but is purely relativistically contrancted beforehand, then I fully agree and we have no argument.

• You say: In the Minkowski diagrams, CA is the rest length of the ladder but DE is not the rest length just like BA is not the rest length. This is because, just like the object described by BA is in a different reference frame from the garage, each part of the object described by DE are in different reference frames. Yes, CA is the rest length of the ladder because it is parallel to the x' axis. BA is not the rest length, because it is parallel to the x axis. It is the length according to the garage. DE is the length of the ladder accoding to someone at rest with the back of the ladder. (i.e. not in the green region). In the green region only, different parts of the ladder are moving at different velocities.

• You say: That is, the green area is relativistically length contracted and if you mathematically uncontract it you'll get a length (rest length of that section) which you add to the length of the non-contracted part of the ladder to get the rest length.. The rest length of the ladder is the length of the ladder when all of its parts are at rest with respect to each other, according to an observer that is also at rest with those parts. If different parts of the ladder are moving at different velocities then
  • The ladder is not at rest
  • Its rest length is undefined
  • It is being physically deformed

This is an important point, please let me know if you agree or disagree with all or part of this last point. PAR 28 June 2005 15:30 (UTC)

Does it "remain" or "become" trash compacter compressed?

It remains compressed by becoming trash compacter compressed. I.e. it was length contraction compressed but has become trash compacter compressed. And don't forget that the other two possibilites were that the ladder buckles or garage door bursts. So becoming trash compacter compressed is only one possibility. If the ladder were truly physically deformed during the event, then the other two possibilites would be impossible as it's rest length would already have been changed because of the compression.

If different parts of the ladder are moving at different velocities then

• The ladder is not at rest
• Its rest length is undefined
• It is being physically deformed

Yes, the first two points are correct. But if we examine each section of the ladder at the ladder's instantaneous inertial reference frame (i.e. references frames corresponding to the instantaneous velocity), then each part of the ladder has its own rest length. These rest lengths do not change. Being physically deformed means that these rest lengths are changed. That's why there isn't any physical deformation during this event. That's the point, you're confusing relativistic compression (where the lengths only appear to change while the rest lengths remain constant) with physical compression (where the rest lengths are changed).

DonQuixote 29 June 2005 16:00 (UTC)

I agree that a ladder could have different parts moving at different velocities, and still not be physically deformed. My point is that this can only occur for an instant. If the ladder is undeformed at time t and different parts have different velocities, then at time t+dt, it will be physically deformed. Do you agree with that? PAR 29 June 2005 18:04 (UTC)

I think this page needs some revising there are several resolution of the paradox each with their own section.--SurrealWarrior 2 July 2005 04:25 (UTC)

Sorry it's taking me so long to work it out quantitatively, but I'm a little swamped with teaching summer classes right now and my free time is rather low. DonQuixote 18:53, 24 July 2005 (UTC)

[edit] Minor change, Major change.

Removed the word "clearly" from the phrase the situation is clearly illustrated. For most people the Minkowski diagram is clear, and telling them it's clear is pointless. I'm not saying anything against the diagram by removing the word. But for some people (a small minority I'm sure) it will not be clear. They will have to work hard to understand it, and telling them it's clear will be unhelpful, making them think that they should be able to understand it without effort.

I hope this is okay. Today is the first day I've looked at Wikipedia properly (I haven't got an account yet) and on checking out SR discussion page it says you need help with the ladder paradox.

I'm only a relativity student, so I'm not going to make the substantive change myself. As I see it, the premise that the ladder can be brought to a simultaneous halt in the frame of the garage is where the problem starts. Arranging for every part of the ladder to be "simultaneously" decelerated is conceptually complex and should be way beyond the scope of this page. It would have to be an effectively infinite deceleration with the origin of the garage frame clearly defined. I think (although I haven't worked it out) that DonQuixote is correct that in this case the physical crushing occurs after the deceleration, as the deceleration itself causes comparable length contraction to the original velocity.

However for people interested in SR paradox resolution, the effect they want to understand is the ladder hitting the wall and that causing the deceleration, which leads to a physical crushing effect and the ladder breaking. The ladder never comes to a simultaneous halt in the frame of the garage as it is broken long before then.

If you want a simple explanation, consider the back end of the ladder at the time the front end hits the wall (in the frame of the garage). The information that the front end has hit takes a time to reach the back end (as the force's effect can't travel faster than c) and so it will just keep going at speed v until the force reaches it. I suspect that the force will actually travel at the speed of sound in the ladder. The front end stopped, the back end moving, something's got to give. If it isn't the wall, it's the ladder.

Hope this helps. Jon Hurwitz

Hi Jon - With regard to the statement that "simultaneous" deceleration (simultaneous in the garage frame, that is) is complex and infinite: Yes, it would be a complicated matter to make this happen, but the deceleration would not have to be infinite, just quick enough to keep the ladder in the garage. The physical crushing does not occur after the deceleration, the deceleration and the crushing are one and the same. Since the deceleration IS simultaneous in the garage frame, it therefore IS NOT simultaneous in the ladder frame. The deceleration in the ladder frame works its way along the length of the ladder. It takes a finite time for the deceleration to complete in the ladder frame. You simply cannot look at a ladder at rest, then watch different parts of it accelerate (change velocities) at different times, and deny that it is being physically deformed. The very definition of physical deformation is that different parts move at different velocities simultaneously in the rest frame. Right? PAR 13:22, 22 August 2005 (UTC)

Hi, PAR. Yes, I agree that the acceleration would not have to be instantaneous. I was thinking it would happen as the ladder touched the far wall, but if (in the garage frame) it was after B and before A, the deceleration would not have to be infinite. That it is not simultaneous in the ladder frame is beside my point. And so what if we disagree? This too is beside my point. You may well be right. You and DonQuixote can argue it out.

I am suggesting you abandon the idea of having the ladder decelerate simultaneously in the garage frame. Then the problem goes away (to return should you knowledgeable people ever agree). I'm saying it's not what people want to know from this page. It is a complicated, counterintuitive case.

If you let the ladder crash into the far end, there is no argument; the ladder breaks, you wrap up the paradox nicely and everyone goes home happy.

Cheers, Jon

Ok, I see what you are saying. I'm still arguing with DonQ I guess. But I think you have a good idea there. What changes would you make to the article, in a bit more detail? PAR 18:16, 22 August 2005 (UTC)

I think the bulk of the article is fine as far as I’ve read it. It’s just the resolution I think needs work. I’m not sure I’m the right person to write it, but here’s a first stab at an update. It'll need someone else to do the spacetime diagram. I'm pretty so-so on those.

"When the stationary garage traps the moving ladder, what happens after the event is either 1) the ladder continues out the other side of the garage (the above two-door garage example) or 2) the ladder is prevented from leaving the garage by the back wall. Considering just the latter: in the frame of the garage, the front of the ladder hitting the back wall, event A, occurs after the back of the ladder enters the garage, event B. The effect of the impact on the front of the ladder is not transmitted to the rear instaneously; it takes time, as nothing travels faster than the speed of light. During this time the rear of the ladder travels forward, compressing the ladder causing it to break.

From the frame of the ladder, the garage rushes forward, the back wall hitting the front of the ladder, event A, before the entrance of the garage passes the rear of the ladder, event B. But once again the effect of the impact does not reach the whole of the ladder immediately. It is impossible to accelerate the rear of the ladder by pushing on the front in less time than it takes light to travel the length of the ladder. By that time the entrance of the garage, travelling at v, has already passed the rear of the ladder, event B, and the front of the ladder has been compressed by the wall of the garage. The compression will cause the ladder to break at the front end before the rear of the ladder has had chance to move.

So in both frames the ladder and the wall collide and the rear of the ladder enters the garage. In both frames the ladder breaks. Only the order of events differs due the relativity of simultaneity."

--JonHurwitz 20:56, 22 August 2005 (UTC)

[edit] On stopping the ladder

I have been asked to put in my "two cents" on this, so here it is.

The real issue is this: If there is only one door to the garage and an impenetrable wall on the other side, can you be guarenteed a time when you can shut the door? The answer is "yes". Never mind whether the rod is crushed or not. As has been noted above, there is no such thing as a classically rigid rod in relativity. The issue now is why that is the case.

Look at it this way: When the front of the rod hits the far wall of the garage, how does the back of the rod know that this has happenned? The answer is that it does not! The back of the rod will continue to move forward until it receives a signal that the front has hit something and been accelerated by it. Such a signal will travel at the speed of sound through the rod. However, in relativity no signal can travel faster than lightspeed (c). So look at where back of the rod would receive the signal that the front has hit if it was traveling at c in both frames of reference. You will find that the rod must end up fitting within the garage in either case. At that point, my advice is to close the door quickly and hope that this is not a perfectly elastic rod such that it may spring back and ruin your nice door. --EMS | Talk 03:23, 23 August 2005 (UTC)

[edit] How to make a 20' ladder fit into a 10' garage

1. Simultaneously (from the point of reference of the ladder) accelerate all points of the ladder to a speed v = 0.866c relative to the garage.
2. Simultaneously (from the point of reference of the garage) decelerate all points of the ladder to a stop.

From the point of reference of the garage the acceleration is not simultaneous and manifests as a crushing force that shrinks the ladder to 10' in step 1, and from the point of reference of the ladder the deceleration is not simultaneous and manifests as a crushing force that shrinks it to 10' in step 2, and there is no paradox. --80.175.250.218 13:26, 9 November 2005 (UTC)

[edit] More on the "Resolution of the paradox if the ladder stopsesolution of the paradox if the ladder stops"

I think this section needs to be rewritten as the way it is now conflates the two reference frames. For example, it starts talking of the "stationary garage" implying the garage's reference frame (RF_g), but when it switches to the ladder's reference frame (RF_l) in the 4th sentence, it claims that "what the ladder experiences is one end decelerating..." But note that in RF_l, the experience is of a stationary ladder with the garage moving towards it, so this is incoherent. This confusion continues throughout the rest of the paragraph (with phrases like "catches up to"). I'm not sure the best way to reword things, but the discussion should probably consistently be from the point of view of RF_g. I may give it a shot though... —Preceding unsigned comment added by 74.98.19.154 (talk • contribs)

Re: changing reference frames in the 4th sentence: "Considering just the latter, we can say that every point of the ladder is simultaneously at rest from the perspective of the garage. From the perspective of the ladder this cannot be true." So it's not an incoherent shift. Read that last sentence carefully.

As to the ladder "catching up to" the garage, from the perspective of the garage the ladder stops within the garage. From the perspective of the ladder, each part of the ladder (starting from the right) independently accelerates to "catch up to" the garage. We are still in the reference frame of the ladder--more specifically the leftmost part of the ladder which hasn't accelerated yet.

DonQuixote 19:43, 23 June 2007 (UTC)

Thanks. I think your changing it from "decelerate" to "accelerate" made it a lot clearer too. (I just tried to help a little more with the ending too.)

[edit] Small things

Open before and after

Closed before

1. Is it a barn or a garage? Barns are usually big and can have two doors on opposite sides. Garages are usually small things with one door for a car.
2. Why are the doors open before the ladder approaches? I think this is more clear. Correct me if I am misunderstanding. — Omegatron 17:35, 23 June 2007 (UTC)

Response:

1. It could be either one. It could be a shed, for all we care. Rindler prefers using a garage. Some authors prefer using a barn. All we need is some building and some long pole-like thing. In fact, the original version involved a man falling into a grate. No ladder or garage there. However the original version is a little harder to visualize, so it's been turned into a ladder/garage (or ladder/barn) problem.
2. Having the garage doors open or closed before and after doesn't make much of a difference. However it might be easier to understand the diagrams if the doors started out as open. If you look at the other diagram (from the garage's perspective), you'll see that both doors start out as open, both close at the same time, and finally both open at the same time. Also, it reduces the number of drawings required. In your "closed before" diagram, you have to include an additional panel before the first where both doors are closed and an additional panel after the last where both doors are closed.

Hope this helps. DonQuixote 19:34, 23 June 2007 (UTC)

1. Yeah, but both are used. Pick one and stick to it.
2. Both doors are not closed at the beginning; just the opposite one. I think it makes the diagram easier to follow. Doesn't really matter, though. — Omegatron 01:38, 24 June 2007 (UTC)

Response:

1. Er...yeah, if you read carefully, only the garage is used. The barn version is mentioned within parentheses.
2. Again, that just creates complexity. The point is that the other diagram (from the garage's perspective) both doors close simultaneously and open simultaneously, and they do that only once. Can't get simpler than that. DonQuixote 07:15, 24 June 2007 (UTC)

[edit] Wrong?

I've got a question:

The article says that the garage would get smaller from the point of view of a person sitting on that ladder. But since there is length contraction happening to that person and since the eyes of that person shrink, shouldn't the garage get larger for that person? Because if it got, I wouldn't see any problem for that ladder to fit into the garage (supposed the ladder moves fast enough). —Preceding unsigned comment added by 217.227.123.77 (talk) 02:24, 26 March 2008 (UTC)

The point is that it's not happening all at once. If you sit at the back of the ladder, as the front of the ladder accelerates into the reference frame of the garage the front of the ladder will appear to shrink into an already small garage. That's the simplest case to imagine. If you're sitting at the front of the ladder, you accelerate into the reference frame of the garage first, that is, to you the garage has expanded to its full size. This also means that, to you, the back of the ladder appears shorter to you now and it is already within the garage. DonQuixote (talk) 14:13, 26 March 2008 (UTC)

Theres no such thing as a "Lorentz expansion". In other words, two people moving at different speeds will each see the other contracted. As mentioned above, the contraction is accompanied by a loss of simultaneity as well, and if you forget about this, there will be an apparent paradox. PAR (talk) 17:12, 26 March 2008 (UTC)