the diamond "paradox"

[Thales et al]

 

Two diamond shaped bodies are moving towards one another. The lower right side of one is aligned with the upper left side of the other.

 

 

In a relativistic world, in the inertial frame of reference of the upper diamond at rest, the lower moving diamond is length contracted in the direction of its motion. And, in the inertial frame of reference of the lower diamond at rest, the upper moving diamond is length contracted in the direction of its motion.

 

 

In the inertial frame of reference of the upper diamond at rest, the leading edge of the upper diamond and the upper edge of the lower diamond collide first. And, in the inertial frame of reference of the lower diamond at rest, the lower edge of the upper diamond and the leading edge of the lower diamond collide first.

 

This means, in the inertial frame of reference of the upper diamond at rest, after the initial collision the lower diamond rotates clockwise, and, in the inertial frame of reference of the lower diamond at rest, after the initial collision the upper diamond rotates clockwise. (Once they begin to rotate they are no longer “inertial frames of reference.”)

 

This is not a paradox. In a relativistic world the sequence of events in one inertial frame of reference can be the inverse of the sequence of those same events in another frame of reference. And here, in both frames of reference, once the length contracted diamond rotates and expands and collides fully with the other diamond, the other diamond will also be set in clockwise motion. And so, from the perspective of both frames of reference, in the end both diamonds will be in clockwise rotation. It’s just the sequence of events that differs.

 

 

There is now a water balloon next to one of the diamonds and along the wall that will collide with the other diamond. It is in the same inertial frame of reference as the diamond it is next to but it is not attached to that diamond. When at rest it is round in shape, and so when in motion it is length contracted into an oval.

 

 

 

The water balloon is large enough so that before the two diamonds collide with each other, it collides with the other diamond first. It is then squeezed in between the walls of the two diamonds as they close in on one another.

 

In a nonrelativistic world the walls of the two non-length contracted diamonds would be parallel to each other and so the balloon would have no pressure to move in one direction or another. However, here in a relativistic world, in the inertial frame of reference of the upper diamond at rest, the water balloon is squeezed down and to the left, and in the inertial frame of reference of the lower diamond at rest, the water balloon is squeezed up and to the right. In one inertial frame of reference the water balloon will move in one direction and in the other inertial frame of reference the water balloon will move in the opposite direction.

 

This is a paradox.

 

What is the solution?

[mathematic]

What happens in the reference frame of the balloon?

[arc]

What happens in the reference frame of the balloon?

You get wet.

[swansont]

The answer may be related to Terrell rotation

https://en.wikipedia.org/wiki/Terrell_rotation

[Sensei]

What is the solution?

 

While calculations of relativistic collisions there is often used center-of-mass/energy. Instead of lab frame.

Like they did it for example here:

http://galileo.phys.virginia.edu/classes/252/particle_creation.html

Try it whether it helps to solve your case.

Edited May 14, 2016 by Sensei

[michel123456]

First of all the effects must be related to the line of sight. In your diagram the line of sight is inclined. It is the red line here below

 

If the objects are aligned there is no paradox (I think). See below

 

 

 

 

 

The answer may be related to Terrell rotation

https://en.wikipedia.org/wiki/Terrell_rotation

 

Do you mean something like this? see below where the objects are skewed. The green lines indicate the parallelism of the sides.(this is wishful thinking in order to eliminate the paradox)

 

[Janus]

Part of the problem is the the OP treats the diamonds as perfectly rigid objects, which can never be the case. even in the the example of with no balloon, the force of impact of the points of the diamonds can't propagate through the diamonds at any speed greater than c. This means that the bulk of both diamonds will not react to the impact until after the impact has occurred and the shape of both diamonds will be distorted. The other point to consider is that once you add the balloon the initial point of impact has shifted. Without the balloon the point of impact occurred along the horizontal axis of one of the diamonds and at the apex of the vertical axis of the other. With the balloon, the impact is shifted to a point somewhere between for both.

[Thales et al]

What happens in the reference frame of the balloon?

 

The way I set up the thought experiment the balloon is in one or the other of the two diamonds inertial frames of reference. It could be either one. In the drawings I have the balloon in the upper diamond’s inertial frame of reference. And so, to answer your question, in the balloon’s initial inertial frame of reference, it ends up moving down and to the left. (Once the balloon squeezed it is then in motion in all different kinds of directions and is no longer “inertial.”) (And if the balloon where initially in the lower diamond’s inertial frame of reference, it would then end up moving up and to the right.)

 

Did I answer your question?

 

The answer may be related to Terrell rotation

https://en.wikipedia.org/wiki/Terrell_rotation

 

 

 

 

In a nonrelativistic world, from the perspective of the lower diamond at rest, the colliding walls of the two diamonds remain parallel, and so the balloon, in the end, has no tendency to move up or down. And, from the perspective of the upper diamond at rest, the colliding walls of the two diamonds also remain parallel, and so the balloon, in the end, has no tendency to move up or down. And so there is no paradox.

 

Here, in a relativistic world, to also end up without a paradox, the balloon, in the end, must do the same thing in both frames of reference. I can’t see a way how it would move up in both or down in both. The only possibility I can see is that the balloon must remain in place in both. And the only way I can see it remaining in place in both is if the walls of the two diamonds are parallel at the time of impact. If the walls are not parallel to one another, then the balloon will be squeezed in one direction or another.

 

 

I thought this must be the solution: in addition to being length contracted, the moving body is also rotated relative to the body at rest. And I found support for this idea in the “Bar and Ring Paradox.” (https://en.wikipedia.org/wiki/Ladder_paradox#Bar_and_ring_paradox)

 

 

The problem with this is what happens if you have two sets of two relatively moving bodies. If I’m right (… if, if, if …) and the only way to get the same result in the end is for the balloon to remain in place in both inertial frames of reference. And if I’m right (… if, if, if …) and the only way for the balloon to remain in place in both inertial frames of reference is for the colliding walls to be parallel to one another. And if this is achieved by the moving bodies not only being length contracted but also tilted, then both of the two bodies in each inertial frame of reference must be tilted when considered to be the bodies in motion.

 

But this leads to a problem. (And I ran into the same problem in the “Bar and Ring Paradox” if its extended to include two bars and two rings.) If the two bodies are connected, then in the inertial frame of reference of them at rest, they remain connected, while in the inertial frame reference of them in motion (where they are tilted) they cannot remain connected. And so, by resolving the initial “diamond paradox” with “relativistic titling,” I found myself ending up in a second and different paradox.

 

Maybe someone else here in this forum can see what I can’t see, and tilting can actually work.

 

Thank you for the Terrell rotation link. I think “relativistic rotation” would end up in this same secondary paradox. No? Something I can’t see? What do you think?

 

 

 

While calculations of relativistic collisions there is often used center-of-mass/energy. Instead of lab frame.

Like they did it for example here:

http://galileo.phys.virginia.edu/classes/252/particle_creation.html

Try it whether it helps to solve your case.

 

 

Thank you for the center-of-mass/energy link.

 

“Recall that this is the speed in the center of mass frame, and for practical purposes, like designing the accelerator, we need to know the energy necessary in the “lab” frame—that in which one of the protons is initially at rest.”

 

I read through it a few times, but I couldn’t find a solution here from that.

 

Maybe there is something that I missed. Let me know.

 

First of all the effects must be related to the line of sight. In your diagram the line of sight is inclined. It is the red line here below

 

I googled “length contraction” and “line of sight” but I couldn’t really understand it.

 

 

 

Are you saying that a body in motion does not length contract in the direction of its motion in the other body’s inertial frame of reference at rest, but rather it length contracts in its direction of motion relative to the other body itself at rest? And so, I was wrong in my thought experiment to suppose that the moving diamond will be length contracted as shown on the left, but rather it will length contract along the x axis relative to the body at rest, as shown on the right, in its horizontal motion relative to the body at rest and not in the direction of its actual motion in the entire inertial frame of reference of the body at rest?

 

Do I understand you correctly? No?

 

Part of the problem is the the OP treats the diamonds as perfectly rigid objects

 

The other point to consider is that once you add the balloon the initial point of impact has shifted.

 

Right.

 

I glossed over the non-rigid aspects of the solid diamond shaped bodies.

 

In the first example, no balloon, when the leading edge of the diamond at rest collides with the edge of the length contracted diamond in motion (in both frames of reference) the diamond at rest will contract from the collision and the diamond at rest will begin to accelerate due to the collision. This contraction and acceleration will take some time (at the speed of light or less) to makes its way through the entire body of the diamond. And when the two diamonds collide, the diamond in motion (in both frames of reference) will start to expand (except for at its very tip), and the diamond in motion will start to rotate, and the diamond in motion will start to decelerate. This expansion and rotation and deceleration will take some time (at the speed of light or less) to make its way through the entire body of the diamond.

 

 

In the second example, the one with the balloon, at one point I thought I had found the solution to this “paradox” with this concept of “no body is a perfectly rigid body.” After the balloon collides with the second diamond, and it is in contact with the two closing in diamonds, this impact and pressure not only distorts and squeezes the balloon, it also distorts the more rigid, though not perfectly rigid, diamonds. And, I thought, if this could lead to the walls of the diamonds being distorted into being parallel, just after the impact, then this could possibly lead to the solution. This could mean that the balloon remains in place. And so no paradox between inertial frames reference.

 

There are three problems I came up against in this line of thought. First, when the impact between the balloon and the second diamond distorts the length contracted diamond and when the impact between the balloon and the second diamond distorts the diamond at rest (no matter if the balloon is in one inertial frame of reference or the other), the distortions will likely be in the opposite direction needed for the walls to become parallel. Second, while the distortion from the contraction of the diamond at rest the distortion from the expansion of the diamond in motion do occur in the directions needed to make the walls parallel, by the time this happens (first the ends of the diamonds must collide), the balloon in the two different inertial frames of reference has already been squeezed differently and in two different directions. And, third, while the diamonds will contract, and while this will have an effect on the squeeze of the balloon and therefore its resultant motion, these contractions of the diamonds will then become expansions, as the diamonds reassert their shapes after the impact. And this contraction and expansion will occur at something less than the speed of light across their bodies and then back, but it should occur at a must faster rate than the actual squeezing out of the balloon from between them, and so may be of no impact in the end.

 

Well, that’s where I ended up.

 

Let me know what you think.

[swansont]

And the only way I can see it remaining in place in both is if the walls of the two diamonds are parallel at the time of impact. If the walls are not parallel to one another, then the balloon will be squeezed in one direction or another.

 

 

That assumes they remain straight lines in the other frame.

[Thales et al]

 

 

That assumes they remain straight lines in the other frame.

 

 

 

 

If length contraction of the moving diamond occurs in the direction of its motion, then, yes, its length contracted walls remain straight lines.

 

(If a square is length contracted in the direction of its motion, then it’s walls too remain straight lines. If a circle is length contracted in the direction of its motion, then, somewhat differently, its circular shape becomes an oval.)

 

And so, yes, I am assuming straight lines. If I’m wrong in this, please let me know.

 

 

Is there a way for the moving diamond to length contract so that its walls are not straight but curved?

 

If so, then if curved in just the right way, then this could lead to the colliding walls of the two diamonds being parallel to one another at the point of impact with the balloon. And so, this could then lead to a solution to this “paradox.”

 

But I can’t see it.

 

 

If a moving body is not only length contracted in the direction of its motion but it is also tilted, and if a moving body is not only titled but curved, then if curved in just the right way, then this could lead to the colliding walls of the two diamonds being parallel to one another at the point of impact with the balloon. And so, again, this could then possibly lead to a solution to this “paradox.”

 

However, I have not found any support for this in either the “Bar and Ring Paradox” or in “Terrell Rotation.” If my understanding of these is correct, there is a “tilting” or a “rotating” but not a “bending.” And, further, while the Bar and Ring Paradox suggests an actual physical tilting, if I’m reading the Terrell Rotation Wikipedia link correctly, this refers to appearances and not the actual physical contracted configurations. And of course to resolve the “paradox” here with the balloon, the moving diamond would actually have to be physically tilted and/or bent for the walls of the two colliding diamonds to be parallel at the point of impact with the balloon.

 

?

[michel123456]

Motion is relative. IOW we must always talk in terms of relative motion. In all cases, the length contraction happen in the direction of apparent motion, which means along the line of sight.

 

 

 

I googled “length contraction” and “line of sight” but I couldn’t really understand it.

 

 

dp 09.jpg

 

Are you saying that a body in motion does not length contract in the direction of its motion in the other body’s inertial frame of reference at rest, but rather it length contracts in its direction of motion relative to the other body itself at rest? And so, I was wrong in my thought experiment to suppose that the moving diamond will be length contracted as shown on the left, but rather it will length contract along the x axis relative to the body at rest, as shown on the right, in its horizontal motion relative to the body at rest and not in the direction of its actual motion in the entire inertial frame of reference of the body at rest?

 

Do I understand you correctly? No?

 

 

The object is observed as "length contracted" along the direction it is observed as approaching. So I guess you understand me, yes.

[Mike-from-the-Bronx]

“Terrell Rotation” is not physical. It is a change in the appearance of an object as a consequence of viewing it from a reference frame from which is moving. The issue here is has nothing to do with appearances.

 

Please disregard this comment ----- I thought the paradox was about a different arrangement.-------

The “Bar and Ring Paradox” involves a rotation that is a consequence of transforming the geometry of a moving object to a reference frame which is moving in a direction NOT parallel to the objects original velocity. In this problem the diamonds and reference frames only have relative velocities in the X-direction. Hence the phenomenon has nothing to do with the issue described here.

 

The resolution to this apparent paradox lies in the phenomenon of rigidity, the rigidity of the balloon. When you fall into a body of water at high speed, that water will act like a solid. It will bring you to a stop before getting out of your way, killing you. (That’s not scientifically precise, but you get the idea.) When the diamond which is moving at high speed relative to the balloon hits it, the diamond will be brought to a stop before the balloon breaks. By stop, I mean it will be accelerated/decelerated approximately to the velocity of the other diamond. Once both diamonds are at the same (or close to the same) velocity, they will have the same geometry. The balloon will break and the water will flow away in a symmetric fashion.

Edited May 16, 2016 by Mike-from-the-Bronx

[michel123456]

And in your diagrams what is interesting is the length contraction at the moment of impact. After some thinking, I wonder maybe at impact the 2 objects share the same FOR and thus there should be no length contraction at all. Maybe I am wrong on this...

[Janus]

 

dp 01.jpg

 

Two diamond shaped bodies are moving towards one another. The lower right side of one is aligned with the upper left side of the other.

 

dp 02.jpg

 

In a relativistic world, in the inertial frame of reference of the upper diamond at rest, the lower moving diamond is length contracted in the direction of its motion. And, in the inertial frame of reference of the lower diamond at rest, the upper moving diamond is length contracted in the direction of its motion.

 

dp 03.jpg

 

In the inertial frame of reference of the upper diamond at rest, the leading edge of the upper diamond and the upper edge of the lower diamond collide first. And, in the inertial frame of reference of the lower diamond at rest, the lower edge of the upper diamond and the leading edge of the lower diamond collide first.

 

This means, in the inertial frame of reference of the upper diamond at rest, after the initial collision the lower diamond rotates clockwise, and, in the inertial frame of reference of the lower diamond at rest, after the initial collision the upper diamond rotates clockwise. (Once they begin to rotate they are no longer “inertial frames of reference.”)

 

No, this is what I meant about improperly treating the diamond as perfectly rigid. At the point when the top vertex of the lower diamond hits the right vertex of the upper diamond. the result of this impact propagates through the lower diamond. This can not happen faster than the speed of light. This means that any part of the lower diamond that is further away from the impact point than the distance light could have traveled since the impact has no way of reacting to the impact. It will just keep moving right to left at its present speed until that information reaches it.

 

Here is an image showing the result of this impact after five equal time periods. the yellow circle expand at the speed of light. Only parts of the lower diamond that are inside of this circle at any given time period can be effected by the collision of the vertexes. (no part of the lower diamond feels the effect of the vertex collision until it is inside the yellow circle. This includes any torque trying to rotate the diamond)

 

 

The green outline shows the how the lower diamond reacts to the impact. The entire shape is what would happen if the upper vertex were to just suddenly stop and this was the only point at which an outside force was applied to the diamond. Any part of the diamond outside of the expanding yellow circle continues to move right to left at 0.866 c. You will notice that this shape overlaps the upper diamond. Since in reality, no part of the lower diamond can go past the face of the upper diamond, when any part of the lower diamond strikes the upper diamond, it will stop due to the collision (we will assume that it doesn't bounce off.) You will note that every part of the upper left face of the lower diamond strikes the lower right face of the upper diamond before the effect of the initial collision between the diamonds can reach that part of the face of the lower diamond. In other words, every part of the face of the lower diamond just continues on until it hits the upper diamond face without having any reaction to the vertexes of the diamonds impacting.

 

Please note, this is not a matter of the "stiffness" of the diamond's material, this will come into play, but not until after the force from the impact has propagated from the point of impact.

 

Another thing to consider is what things look like from the frame where neither diamond is at rest but they are measured as coming at each other at equal speeds from opposite directions. In this frame both diamonds are equally length contracted and the two faces are at equal angles and thus all parts of both faces collide instantaneously. The point is that once you take the manner in the forces propagate through the diamond, this scenario produces exactly the same result as the one above. Adding the balloon will not change this, it just makes it harder to see.

 

[Thales et al]

 

No, this is what I meant about improperly treating the diamond as perfectly rigid.

 

 

 

Yep. I get it. Thank you.

 

I felt comfortable that I had resolved the two colliding diamonds, and there was no paradox, but you resolved it better. My analysis was not right.

 

 

The problem I’m having is how to resolve this when a third body (such as a water balloon) is added.

 

Whether the water balloon is large and it collides with the second diamond first, or whether the water balloon is small and the diamonds collide first, doesn’t matter. I’m having the same problem in either case.

 

 

Starting with your analysis of the diamonds colliding first, it still appears to me, that if there is a third body (such as a water balloon) between the two colliding walls, that in the inertial frame of reference of the upper diamond at rest the balloon will be pushed down and to the left.

 

 

While in the inertial frame of reference of the lower diamond at rest, the water balloon will be pushed up and to the right.

 

?

 

 

Motion is relative. IOW we must always talk in terms of relative motion. In all cases, the length contraction happen in the direction of apparent motion, which means along the line of sight.

The object is observed as "length contracted" along the direction it is observed as approaching. So I guess you understand me, yes.

 

 

Say there is an X moving in inertial frame of reference xy in direction 1. And say there are two people (observer 1 and observer 2) at rest in xy.

 

From the perspective of observer 1, arm ab is perpendicular to the direction of the motion and is not length contracted, and arm cd is in line with the direction of the motion and so is length contracted.

 

Observer 2 is in the same position as the diamond at rest in thought experiment presented above.

 

Are you saying that for observer 2, who is in the same inertial frame of reference as observer 1, the X is somehow length contracted differently than it is for observer 1?

 

(And, if so, are you saying this is physically true and not just a matter of appearances?)

 

 

And in your diagrams what is interesting is the length contraction at the moment of impact. After some thinking, I wonder maybe at impact the 2 objects share the same FOR and thus there should be no length contraction at all. Maybe I am wrong on this...

 

I think in the end, when the diamonds are spinning around clockwise with one another, they are in the same frame of reference. But I think at the moment of impact they are necessarily in two different inertial frames of reference.

 

No?

 

 

 

The resolution to this apparent paradox lies in the phenomenon of rigidity, the rigidity of the balloon. When you fall into a body of water at high speed, that water will act like a solid. It will bring you to a stop before getting out of your way, killing you. (That’s not scientifically precise, but you get the idea.) When the diamond which is moving at high speed relative to the balloon hits it, the diamond will be brought to a stop before the balloon breaks. By stop, I mean it will be accelerated/decelerated approximately to the velocity of the other diamond. Once both diamonds are at the same (or close to the same) velocity, they will have the same geometry. The balloon will break and the water will flow away in a symmetric fashion.

 

The water balloon, like and even more so than the diamonds, is not a perfectly rigid body. When you fall fast into a pool of water, it hurts like hitting a solid, but the water does move.

 

When the water balloon collides with the second diamond, and is then squeezed between them, it will not act like a rigid body. It will be squeezed and distorted and move.

 

Or, so I think.

 

(For the purposes of this thought experiment, we can stipulate an unbreakable rubber balloon.)

 

 

[Janus]

 

impact02.jpg

 

Starting with your analysis of the diamonds colliding first, it still appears to me, that if there is a third body (such as a water balloon) between the two colliding walls, that in the inertial frame of reference of the upper diamond at rest the balloon will be pushed down and to the left.

 

impact03.jpg

 

While in the inertial frame of reference of the lower diamond at rest, the water balloon will be pushed up and to the right.

 

?

 

 

Here's the thing, in my images, for simplicity's sake I ignored the effect of the collision on the upper diamond and showed it as if it would maintain its own shape (or put another way, The mass of the lower diamond was insignificant compared to the mass of the Upper diamond. In such a scenario, you cannot just flip the images as you did to show how the same collision would look like from the frame of the Lower diamond. In fact, after the initial impact, the lower diamond would not have a single rest frame that would work for all its parts. Parts of it would be moving to the right with respect to to the initial rest frame they had before the collision before the collision.

 

If you want to model it as a collision between two equally massive diamonds, then you have to take into account how the effect of the collision travels through both cubes. In which case part s of both diamonds will be moving at different speed with respect to other parts of the same diamond at moments after the collision. If you add the balloon, you then also have to model how different parts of the balloon react to the collision ( the force of collision has to propagate through it too.) and this becomes quite a complex problem and It will not match what your intuition tells you (that the balloon will be forced up in one frame and down in the other.) That "scissoring" effect you show won't happen.

 

Consider what happens when the first part of the lower diamond hits the balloon. it will take time, even at the speed of light for the upper part of the balloon to react to the impact (and even longer if the balloon is less rigid than the diamond since the speed at which the force can propagate through it will be less than the speed of light). Mean while, the upper parts of the lower diamond's face are traveling unimpeded at 0.886 c to the left. Those parts of the face will hit the upper part of the balloon before it can start moving. In effect, the face of the lower diamond will "cup" the balloon, preventing it from being forced upward or downward. I know this seems strange, but you have to remember that our normal experience with colliding objects just doesn't apply when you are dealing with relativistic speeds.

[michel123456]

 

dp 15.jpg

Say there is an X moving in inertial frame of reference xy in direction 1. And say there are two people (observer 1 and observer 2) at rest in xy.

 

From the perspective of observer 1, arm ab is perpendicular to the direction of the motion and is not length contracted, and arm cd is in line with the direction of the motion and so is length contracted.

 

Observer 2 is in the same position as the diamond at rest in thought experiment presented above.

 

Are you saying that for observer 2, who is in the same inertial frame of reference as observer 1, the X is somehow length contracted differently than it is for observer 1?

 

(And, if so, are you saying this is physically true and not just a matter of appearances?)

 

Yes I say that the X is somehow contracted differently. And it will get worse if 1 and 2 are in a different state of motion.

And I say it is a matter of appearance (a matter of measurement). It is indeed "physically true" because it is the result of a measurement (it is not what one would call a optical illusion), however what one measures is different from what another measures.

 

V1 is the approach velocity as observed by observer 1

 

V2 is approach velocity as observed by 2 because the light we measure is the one coming from the source and hitting our eye. All other light going in all other directions are not directly observable. If there is a lateral velocity it is not directly observed by transmission of light and there is no observable length contraction on something that you cannot observe directly.

Edited May 18, 2016 by michel123456