Quick calculation check please

[name123]

In other words, why couldn't the distance between the atoms that are lined up change? We can verify that that doesn't change too, by measurements. In addition to counting the atoms, we also use strain gauges to measure the inter-atomic forces, and verify that they don't change either--all of the strain gauges mounted on the sphere read the same when it is moving at 87% of the speed of light relative to the poles, as when it is at rest relative to the poles. (Or think up any other measure of density you like; the measurement, if made by apparatus at rest relative to the sphere, will be the same regardless of the sphere's speed relative to the poles.)

Well just rewinding to the conveyor belt scenario, where roughly 29,979,245.8 A-Team 1m rulers are laid out between each of the B-Team observers and then the conveyor belt is started. When it gets up to speed and they make their own 1m rulers from the same material and they find that the distance between the B-Team observers is 30,130,275.702 B-Team 1m rulers. Are you saying that the atomic density in those B team rulers are the same but they are just shorter (have less atoms)?

No. The length contraction is only in the $x$ direction. The relative motion in the $y$ direction between the sphere and the poles is too slow to have any significant length contraction effect.

It might have been traveling faster in the y than the x.

This sentence is a bit garbled, but I assume you mean "they're pretty sure something has changed even though neither of them can notice anything changing in their rest frame".

As you state it, this is false; each of them, in their own rest frame, measures the other to be length contracted, whereas if they were both at rest relative to each other they would not. So their relative motion does cause an observable change. The point is simply that there is no observable change in measurements that each one makes in their own rest frame of themselves--no measurement of the sphere in the sphere's rest frame changes, and no measurement of the poles in the poles' rest frame changes. But that doesn't mean there is no observable change anywhere.

Yes it is the observable change with relative motion that we are talking about.
Is it ok for the cartoon characters to conclude that something has changed span?

 

[PeterDonis]

roughly 29,979,245.8 A-Team 1m rulers are laid out between each of the B-Team observers

Ok.

it gets up to speed and they make their own 1m rulers from the same material and they find that the distance between the B-Team observers is 30,130,275.702 B-Team 1m rulers

No, they wouldn't. They would find that the distance between the B-Team observers is 29,979,245.8 B-Team rulers. But now that the conveyor is moving, the distance between B-Team observers, as measured in Team A's frame, is less than 29,979,245.8 Team A rulers.

It might have been traveling faster in the y than the x.

If so, then the analysis of the scenario gets a lot more complicated, because you have length contraction in both the x and the y directions. I was not trying to analyze that case; I was assuming that the relative speed in the y direction is too small to cause any significant relativistic effects, so that we only have length contraction in the x direction.

 

[name123]

No, they wouldn't. They would find that the distance between the B-Team observers is 29,979,245.8 B-Team rulers. But now that the conveyor is moving, the distance between B-Team observers, as measured in Team A's frame, is less than 29,979,245.8 Team A rulers.

Assume the A-Team and the B-Team are both at rest, the conveyor belt hasn't started. The A-Team make their A-Team 1m rulers, and the B-Team make their first batch of 1m rulers, and they both measure the span between the A-Team members standing opposite B-Team members and find the distance between the A-Team members and the distance between the B-Team members to be the equivalent to the span of roughly 29, 979,245.8 1m rulers, each of the rulers that the A-team made being roughly the same size as each of the rulers the B-Team made. The conveyor belt then starts up to 0.1c. Presumably there will still be the same amount of rulers between the B-Team members (both types can be laid out between them). Imagine the B-Team make a new batch of 1m rulers (while going at 0.1c on the conveyor belt). How many of this new batch will the B-Team members find it will take to span the distance between the A-Team and B-Team members?

Regarding the poles and the sphere, you had said:"So their relative motion does cause an observable change."

And I asked: "Is it ok for the cartoon characters to conclude that something has changed span?"

But you didn't reply.

 

[PeterDonis]

the conveyor belt hasn't started. The A-Team make their A-Team 1m rulers, and the B-Team make their first batch of 1m rulers, and they both measure the span between the A-Team members standing opposite B-Team members and find the distance between the A-Team members and the distance between the B-Team members to be the equivalent to the span of roughly 29, 979,245.8 1m rulers

Ok. (I don't know why you use the word "roughly", though--they're both measuring to 9 significant figures.)

The conveyor belt then starts up to 0.1c. Presumably there will still be the same amount of rulers between the B-Team members (both types can be laid out between them).

The same amount of B-Team rulers, yes, because those rulers are at rest relative to the B Team. But not the same amount of A-Team rulers; those rulers are moving relative to the B-Team, so the number of A-Team rulers between two B-Team members will be smaller. The gamma factor for a relative speed of 0.100000000 c (I'm using 9 significant figures since that's your stated accuracy) is 1.00503781; dividing 29,979,245.8 by this gives 29,828,973.1, so that is how many A-Team rulers now fit between two B-Team members.

Imagine the B-Team make a new batch of 1m rulers (while going at 0.1c on the conveyor belt). How many of this new batch will the B-Team members find it will take to span the distance between the A-Team and B-Team members?

There will be 29,979,245.8 of these new B-Team rulers between two B-Team members (the same as the original B-Team rulers)--this is assuming that these new rulers are at rest relative to the B-Team. There will be a smaller number of these new B-Team rulers between two A-Team members, because these rulers are moving relative to the A-Team. This will be the same number we calculated above: 29,828,973.1 new B-Team rulers (and the same number of original B-Team rulers) will fit between two A-Team members.

Regarding the poles and the sphere, you had said:"So their relative motion does cause an observable change."

And I asked: "Is it ok for the cartoon characters to conclude that something has changed span?"

But you didn't reply.

Because that scenario brings in other elements that are irrelevant, like the whole computer simulation thing. The A-Team and B-Team scenario with the rulers is simpler and clearer, so I'm sticking to that one. You should be able to figure out the answer to the cartoon character question from what I'm saying about the rulers.

 

[name123]

Assume the A-Team and the B-Team are both at rest, the conveyor belt hasn't started. The A-Team make their A-Team 1m rulers, and the B-Team make their first batch of 1m rulers, and they both measure the span between the A-Team members standing opposite B-Team members and find the distance between the A-Team members and the distance between the B-Team members to be the equivalent to the span of roughly 29, 979,245.8 1m rulers, each of the rulers that the A-team made being roughly the same size as each of the rulers the B-Team made. The conveyor belt then starts up to 0.1c. Presumably there will still be the same amount of rulers between the B-Team members (both types can be laid out between them). Imagine the B-Team make a new batch of 1m rulers (while going at 0.1c on the conveyor belt). How many of this new batch will the B-Team members find it will take to span the distance between the A-Team and B-Team members?

The same amount of B-Team rulers, yes, because those rulers are at rest relative to the B Team. But not the same amount of A-Team rulers; those rulers are moving relative to the B-Team, so the number of A-Team rulers between two B-Team members will be smaller. The gamma factor for a relative speed of 0.100000000 c (I'm using 9 significant figures since that's your stated accuracy) is 1.00503781; dividing 29,979,245.8 by this gives 29,828,973.1, so that is how many A-Team rulers now fit between two B-Team members.

There will be 29,979,245.8 of these new B-Team rulers between two B-Team members (the same as the original B-Team rulers)--this is assuming that these new rulers are at rest relative to the B-Team. There will be a smaller number of these new B-Team rulers between two A-Team members, because these rulers are moving relative to the A-Team. This will be the same number we calculated above: 29,828,973.1 new B-Team rulers (and the same number of original B-Team rulers) will fit between two A-Team members.

So they can observe that the A-Team observers have 29,979,245.8 of the old B-Team rulers between them, and that you wouldn't get as many of the new ones. So an old B-Team ruler according to the B-Team is 1/29,979,245.8th of the span between the A-Team members and according to their observations a new B-Team ruler is 1/ 29,828,973.1th of the span between A-Team members. When they slow back down again the B-Team find that the new B-Team rulers were only 1/ 29,979,245.8th of the span? Would their be any difference in the clock times?

Regarding the poles and the sphere, you had said:"So their relative motion does cause an observable change."

And I asked: "Is it ok for the cartoon characters to conclude that something has changed span?"

But you didn't reply.

Because that scenario brings in other elements that are irrelevant, like the whole computer simulation thing. The A-Team and B-Team scenario with the rulers is simpler and clearer, so I'm sticking to that one. You should be able to figure out the answer to the cartoon character question from what I'm saying about the rulers.

They are two different points. The cartoon character scenario is an analogy where the cartoon characters are analogous to physicists, or philosophers, and the computer simulation is analogous to imagined physical causes. So since it's a different question, and it seems a simple enough question, and was less effort to answer I would have thought, could you just say why the computer simulation could not be thought to be analogous to imagined physical causes (you've already accepted the question has meaning with both) or answer the question “Is it ok for the cartoon characters to conclude that something has changed span?"

 

[PeterDonis]

So they can observe that the A-Team observers have 29,979,245.8 of the old B-Team rulers between them

Not if the old B-Team rulers are moving with the B-Team. I assumed that they were, but now it seems like you're saying they aren't.

If the old B-Team rulers are left behind when the B-Team starts moving, so they stay at rest relative to the A-Team, then they are just the same as the A-Team rulers. Is this what you mean? I'll assume it is for the rest of this post.

So an old B-Team ruler according to the B-Team is 1/29,979,245.8th of the span between the A-Team members and according to their observations a new B-Team ruler is 1/ 29,828,973.1th of the span between A-Team members.

While the B-Team is moving relative to the A-Team, yes. But a new B-Team ruler (which is at rest relative to the B-Team) is 1/29,979,245.8th of the span between B-Team members.

When they slow back down again the B-Team find that the new B-Team rulers were only 1/ 29,979,245.8th of the span?

Meaning, when the conveyor stops again and the B-Team is now at rest again relative to the A-Team, the new B-Team rulers are 1/29,979,245.8th of the span between A-Team members? Yes, that's correct.

Would their be any difference in the clock times?

Yes, the B-Team's clocks would show less elapsed time than the A-Team's clocks.

They are two different points.

No, they're not--at least, not if you intend the computer simulation to be correct. If the computer simulation doesn't have to model what actually happens in the real world, then I have nothing to say about it, because you can make the numbers anything you like. If the computer simulation does model what actually happens in the real world, then, as I said, you can figure out what the cartoon characters can conclude by analogy with the A-Team/B-Team case that we're discussing.

it seems a simple enough question, and was less effort to answer I would have thought

It didn't seem at all simple to me, because if the computer simulation is a correct model of what happens in the real world, why not just talk about what happens in the real world? And if the computer simulation does not have to correctly model what happens in the real world, what's the point of it?

 

[name123]

Not if the old B-Team rulers are moving with the B-Team. I assumed that they were, but now it seems like you're saying they aren't.

I thought I'd made it clear, there are two sets of old B-Team rulers, one set between the A-Team members and one set between the B-Team members.

Would their be any difference in the clock times?

Yes, the B-Team's clocks would show less elapsed time than the A-Team's clocks.

When the conveyor belt stops and they are both in the A-Team's frame of reference?

They are two different points.

No, they're not--at least, not if you intend the computer simulation to be correct. If the computer simulation doesn't have to model what actually happens in the real world, then I have nothing to say about it, because you can make the numbers anything you like. If the computer simulation does model what actually happens in the real world, then, as I said, you can figure out what the cartoon characters can conclude by analogy with the A-Team/B-Team case that we're discussing.

Ok, we can have the cartoon characters watching videos on the A-Team/B-Team case, but we can change it slightly and imagine that each observer is represented by a thin pole. And that the B-Team have a y velocity also. So we can imagine a pair of adjacent B-Team members thinking that they are going to pass two specific adjacent A team members on the y axis. So if you imagine the B-Team member closer to x'=0 to be facing in the positive y direction (I know it is also being thought of as a thin pole, but you can just switch as convenient) can it think that it passed the A-Team pair on the y-axis with both of them to its right, and the other member of the B-Team pair think that it passed the A-Team pair on the y-axis with both of them on its left (imagine it too is facing in positive y)?

The cartoon character scenario is an analogy where the cartoon characters are analogous to physicists, or philosophers, and the computer simulation is analogous to imagined physical causes. So since it's a different question, and it seems a simple enough question,

It didn't seem at all simple to me, because if the computer simulation is a correct model of what happens in the real world, why not just talk about what happens in the real world? And if the computer simulation does not have to correctly model what happens in the real world, what's the point of it?

The point is that it is clear with the computer model that it could have been modeled using any frame as absolute rest. So it analogous, and yet highlights that jumping to a certain conclusion involves a logical fallacy.

and was less effort to answer I would have thought, could you just say why the computer simulation could not be thought to be analogous to imagined physical causes (you've already accepted the question has meaning with both) or answer the question “Is it ok for the cartoon characters to conclude that something has changed span?"

So could you now say why the computer simulation couldn't be thought of as analogous to imagined physical causes, or answer the question?

 

[PeterDonis]

I thought I'd made it clear, there are two sets of old B-Team rulers, one set between the A-Team members and one set between the B-Team members.

It's clear now, but it wasn't in previous posts.

When the conveyor belt stops and they are both in the A-Team's frame of reference?

Yes. The B-Team's clocks will be running at the same rate as the A-Team's clocks once the conveyor belt stops, but they won't show the same total elapsed time, because they were running slower than the A-Team's clocks during the period when the conveyor belt was running.

we can have the cartoon characters watching videos on the A-Team/B-Team case, but we can change it slightly

Can you draw a diagram? I'm having real trouble following your verbal descriptions. You keep piling on scenario after scenario and I can't keep up. It looks like you're trying to reproduce the sphere and poles scenario with the conveyor belts, but I don't understand how you're doing it.

Or, if you can't draw a diagram, at least pick some reference frame (the A-Team's would be fine), and give coordinates (time and space) of all relevant events in this frame. You should be doing this anyway in order to analyze the scenario; just waving your hands with verbal descriptions without doing the math is not likely to give useful results.

it is clear with the computer model that it could have been modeled using any frame as absolute rest

And doing that means you're not simulating the real world, since in the real world there is no such thing as absolute rest. I get that you believe you are able to draw some kind of useful conclusion about the real world by thinking about this computer model, but I don't see how. If the model is not constrained by the actual laws of physics, then how can its conclusions be relevant?

could you now say why the computer simulation couldn't be thought of as analogous to imagined physical causes

I have said so, repeatedly. I just said it again, above.

 

[name123]

Can you draw a diagram? I'm having real trouble following your verbal descriptions. You keep piling on scenario after scenario and I can't keep up. It looks like you're trying to reproduce the sphere and poles scenario with the conveyor belts, but I don't understand how you're doing it.

Or, if you can't draw a diagram, at least pick some reference frame (the A-Team's would be fine), and give coordinates (time and space) of all relevant events in this frame. You should be doing this anyway in order to analyze the scenario; just waving your hands with verbal descriptions without doing the math is not likely to give useful results.

Is there a way I can attach a diagram?

Imagine the A-Team/B-Team scenario except that the A-Team are on suspended walkway (the z axis is used for the height) which is parallel to the conveyor belt, but having greater Y coordinates. You can imagine poles coming down underneath where each of the A-Team members are standing. The major difference is that the floor the conveyor belt is on moves in the +ve y direction so that the whole conveyor belt passes under the suspended walkway. The poles from the suspended walkway being long enough to touch a standing B-Team observer, but not long enough to hinder the conveyor belt passing underneath. Can you get the idea from that description?.

So from what you’ve said once up at 0.1c the B-Team will think there is less space between the A-Team members than the B-Team members and the A-Team members will think there is less space between the B-Team members than the A-Team members.

So now there is some y momentum so that the two team will pass on the y. And the question is whether a pair of B-Team will be able to pass under the suspended walkway with a pair of A-Team members poles passing in between the B-Team members. Presumably the A-Team members will think the gap between the B-Team members is smaller, and it is that two B-Team members will be able to pass in between 2 of the suspended poles. I find it difficult to see how they could both be right, and so wondered what would happen. Does that help at all?

The point is that it is clear with the computer model that it could have been modeled using any frame as absolute rest. So it analogous, and yet highlights that jumping to a certain conclusion involves a logical fallacy.

And doing that means you're not simulating the real world, since in the real world there is no such thing as absolute rest. I get that you believe you are able to draw some kind of useful conclusion about the real world by thinking about this computer model, but I don't see how. If the model is not constrained by the actual laws of physics, then how can its conclusions be relevant?

So could you now say why the computer simulation couldn't be thought of as analogous to imagined physical causes, or answer the question?

I have said so, repeatedly. I just said it again, above.

I’ve already said that the computer simulation follows the same laws of physics. But you are surely aware that with those laws could have been modeled using any rest frame as absolute rest. It’d just work everything out from that frame’s perspective.Are you denying that this is possible? So everything in the video library would look as you'd expect it to, but the videos themselves wouldn't allow you to tell if it was modeled using a particular frame as the absolute rest, or no frame as absolute rest. So the point is that if it was is a logical fallacy for the computer characters to conclude that there isn’t absolute rest (made clear by us being able to understand that from the equations it can modeled that way) , then why wouldn’t it be a logical fallacy for you to conclude that there isn’t absolute rest? What information do you have about the physical underlying which they don’t have about the computer model?

 

[PeterDonis]

Is there a way I can attach a diagram?

Yes, there is an "Upload a file" button at the bottom right of the editing window.

Can you get the idea from that description?

Yes, thanks.

at 0.1c the B-Team will think there is less space between the A-Team members than the B-Team members and the A-Team members will think there is less space between the B-Team members than the A-Team members.

Yes. However, that doesn't make the situation completely symmetrical, because the poles are at rest relative to the A-Team members, not the B-Team members. See below.

the question is whether a pair of B-Team will be able to pass under the suspended walkway with a pair of A-Team members poles passing in between the B-Team members.

The answer to this is "no". Here's why:

First, if we look at things in the A-Team's rest frame, it is obvious that the B-Team members will pass in between the poles. But, as I noted above, what makes it obvious is that the poles are at rest relative to the A-Team members, so the only things that are moving are the B-Team members. (Note that we are also assuming that the relative velocity in the $y$ direction is slow enough that we can ignore length contraction in the $y$ direction. As I noted a number of posts ago, adding length contraction in the $y$ direction would make this scenario considerably more complicated.)

Now look at things in the B-Team's rest frame. In this frame, the poles are closer together, in the $x$ direction, than the B-Team members are. However, and this is the key point, in this frame, the poles are offset in the y direction, i.e., they are at different $y$ coordinates at any given instant of time in the B-Team frame (in the A-Team frame, they are at the same $y$ coordinate). This is because of relativity of simultaneity: if the poles are at the same $y$ coordinate at the same time in the A-Team frame, they cannot be at the same $y$ coordinate at the same time in the B-Team frame, because "at the same time" means different things in the two frames.

So in the B-Team frame, what happens is that the two B-team members slip between the poles because the combination of the poles' separation in the $x$ direction and their separation in the $y$ direction creates an opening that is large enough for them to pass through. (Note that the poles are moving "diagonally" in the B-team frame.)

I’ve already said that the computer simulation follows the same laws of physics.

Yes, and I've already said that that is not consistent with this:

you are surely aware that with those laws could have been modeled using any rest frame as absolute rest.

No, I am not aware of that. See below.

It’d just work everything out from that frame’s perspective.Are you denying that this is possible?

You can run the model in any frame you like, yes. That is not the same as defining that frame as "absolute rest". The existence of an "absolute rest" frame would have physical consequences--there would be experimental results that would be different from the ones we actually find in the real world. (For example, the Michelson-Morley experiment would have different results.) Otherwise "absolute rest" is just a meaningless label.

why wouldn’t it be a logical fallacy for you to conclude that there isn’t absolute rest?

Because we've done experiments that test whether there is absolute rest, and those experiments said there isn't. See above.

 

[name123]

The answer to this is "no". Here's why:

First, if we look at things in the A-Team's rest frame, it is obvious that the B-Team members will pass in between the poles. But, as I noted above, what makes it obvious is that the poles are at rest relative to the A-Team members, so the only things that are moving are the B-Team members. (Note that we are also assuming that the relative velocity in the $y$ direction is slow enough that we can ignore length contraction in the $y$ direction. As I noted a number of posts ago, adding length contraction in the $y$ direction would make this scenario considerably more complicated.)

Now look at things in the B-Team's rest frame. In this frame, the poles are closer together, in the $x$ direction, than the B-Team members are. However, and this is the key point, in this frame, the poles are offset in the y direction, i.e., they are at different $y$ coordinates at any given instant of time in the B-Team frame (in the A-Team frame, they are at the same $y$ coordinate). This is because of relativity of simultaneity: if the poles are at the same $y$ coordinate at the same time in the A-Team frame, they cannot be at the same $y$ coordinate at the same time in the B-Team frame, because "at the same time" means different things in the two frames.

So in the B-Team frame, what happens is that the two B-team members slip between the poles because the combination of the poles' separation in the $x$ direction and their separation in the $y$ direction creates an opening that is large enough for them to pass through. (Note that the poles are moving "diagonally" in the B-team frame.)

Consider the B-Team pair. One is closer to x' = 0. You seem to be saying that when it passed the A-Team pair member closest to x = 0 with the A-Team member being to its left (imagine the B-Team member in question was looking in the +ve y direction), is that correct?

You can run the model in any frame you like, yes. That is not the same as defining that frame as "absolute rest". The existence of an "absolute rest" frame would have physical consequences--there would be experimental results that would be different from the ones we actually find in the real world. (For example, the Michelson-Morley experiment would have different results.) Otherwise "absolute rest" is just a meaningless label.

I'm not sure what an experiment to do with whether anything could be considered to physically support light waves has anything to do with it. Perhaps you'd like to mention what properties you were thinking absolute rest would need to have. I was simply thinking that it would have to be that one was absolute rest either by virtue of the frame of reference the computer simulation used, in which velocity = 0. Or that there was perhaps some imagined physical reason why something was not at rest. I thought that space wasn't exactly empty, so doesn't motion in relation to that background have experimental consequences? Or that God maintains the model, and like the computer simulation any frame could have been chosen as absolute rest. All three make sense to me, so I don't see how you can say they have no meaning, as opposed to perhaps you not understanding the meaning. Perhaps when you say what properties you thought absolute rest implied I'll understand.

 

[PeterDonis]

Consider the B-Team pair. One is closer to x' = 0. You seem to be saying that when it passed the A-Team pair member closest to x = 0 with the A-Team member being to its left (imagine the B-Team member in question was looking in the +ve y direction), is that correct?

"To its left" is ambiguous, and also doesn't fully capture the relative motion. Also, more importantly, "when it passed" is ambiguous, and in fact is not even well-defined in the B-Team Frame. That is to say, in the A-Team frame, there is a single instant of time (which we can call $t = 0$) at which all four objects--the two poles, and the two B-Team members passing between them--have the same $y$ coordinate (which we can designate as $y = 0$). But in the B-Team frame, there is no single instant of time at which this is true. That is, there is no instant of time, in the B-Team frame, at which all four objects have the same $y'$ coordinate. That is because of relativity of simultaneity (as I have already explained), and is part of the key to understanding this scenario.

To answer the question as best I can, given the limitations I've just mentioned, if we designate the A-Team pair (with the poles) as A1 (at $x = 0$) and A2 (at $x = L$, where $L$ is the separation between the poles in the A-Team frame), and if we designate the B-Team pair as B1 (the one that passes closest to A1) and B2 (the one that passes closest to A2), then when B1 passes A1 (meaning, when both of them have the same $y$ or $y'$ coordinate, depending on which frame we are using), B1's $x$ (or $x'$) coordinate is larger than A1's; and when B2 passes A2, B2's $x$ (or $x'$) coordinate is smaller than A2's.

I have suggested a couple of times now that, instead of waving your hands with ordinary language descriptions, you either draw a spacetime diagram or write down explicitly the math involved--the coordinates of all important events, and how they transform between the two frames. Doing that will make the answers to this and a lot of other questions obvious.

Perhaps you'd like to mention what properties you were thinking absolute rest would need to have.

It depends on whether you think "absolute rest" has physical consequences. If it doesn't, then it's not a physical property or a physical thing, and talking about it is off topic in this forum. That's why I haven't bothered addressing that possibility.

If "absolute rest" does have physical consequences (which is how I've been using the term), then, as I've said several times, now, there will be experiments that will give different results depending on whether there is "absolute rest" or not. (One famous one is the Michelson-Morley experiment.)

As far as your computer simulation is concerned, once again, if "absolute rest" has physical consequences, and if the simulation is correct, then the simulation will simulate different experimental results depending on whether "absolute rest" exists or not. So it's easy to tell whether the simulation is using "absolute rest" by just looking at what experimental results appear in it.

I thought that space wasn't exactly empty, so doesn't motion in relation to that background have experimental consequences?

What do you mean by "background"? There are certainly other objects in the universe, and those objects have particular states of motion, so one can measure whether one is at rest or moving relative to those objects. But there is no "background" independent of the objects; there is no way to measure "motion" relative to some "background" that is different from motion relative to any of the objects.

All three make sense to me, so I don't see how you can say they have no meaning

I didn't say they have "no meaning" period. I said they have no physical meaning--that is, if we have two different simulations, and they both make the same predictions for all experimental results, then there is no physical meaning to saying that one uses the "absolute rest" frame but the other uses some other frame that is not at "absolute rest". There might well be non-physical meaning to that statement--after all, you can just look at the numbers in the two computers and see that they're different. But the label "absolute rest" for one set of numbers is not a physical label; it doesn't correspond to any physical difference, because all the experimental results are the same in both simulations.

 

[name123]

To answer the question as best I can, given the limitations I've just mentioned, if we designate the A-Team pair (with the poles) as A1 (at $x = 0$) and A2 (at $x = L$, where $L$ is the separation between the poles in the A-Team frame), and if we designate the B-Team pair as B1 (the one that passes closest to A1) and B2 (the one that passes closest to A2), then when B1 passes A1 (meaning, when both of them have the same $y$ or $y'$ coordinate, depending on which frame we are using), B1's $x$ (or $x'$) coordinate is larger than A1's; and when B2 passes A2, B2's $x$ (or $x'$) coordinate is smaller than A2's.

Two questions here. Firstly, how do the B-Team explain that they passed through the A-Team poles as you described when they are saying the gap between those two poles is narrower than the gap between the B-Team pair?

Secondly can both teams conclude that something has changed span?

It depends on whether you think "absolute rest" has physical consequences. If it doesn't, then it's not a physical property or a physical thing, and talking about it is off topic in this forum. That's why I haven't bothered addressing that possibility.

The whole point was that it wouldn't have any physical consequences, that is why they could no more tell in the simulation which way it was done, than you could tell how it was done in an imagined physical universe. There could be an imagined physical reason why it should be considered that the spaceship changed size when it applied its thrusters and not the universe. If it was imagined that way round, then it would clearly be a physical property that was being imagined. Understanding that several physical realities with different physical properties could be measured to be the same, is like understanding that the cartoon characters could understand that although they might not be able to detect which way it was modeled, because different models could be measured the same, it doesn't follow that how it was modeled wasn't a property of the model.

What do you mean by "background"? There are certainly other objects in the universe, and those objects have particular states of motion, so one can measure whether one is at rest or moving relative to those objects. But there is no "background" independent of the objects; there is no way to measure "motion" relative to some "background" that is different from motion relative to any of the objects.

Virtual particles appearing in a vacuum for example (as the background). Presumably there is some effect that is measured for the theory that it happens.

 

[PeterDonis]

Firstly, how do the B-Team explain that they passed through the A-Team poles as you described when they are saying the gap between those two poles is narrower than the gap between the B-Team pair?

Um, because they are not saying the gap between the poles is narrower? They are saying it is shorter in the x direction; but the gap in the B-Team frame is not just in the x direction. It is also in the y direction. The full gap, combining both the x and y components, is large enough for the two B-Team members to fit through.

can both teams conclude that something has changed span?

If "changed span" means "shows relativistic length contraction", then yes. Otherwise I don't know what you mean by "changed span". You keep talking as if this has some other meaning, but I don't know what it is. The only "change in length" that occurs in this scenario is relativistic length contraction.

The whole point was that it wouldn't have any physical consequences

Then any questions about it are not physics questions and are off topic in this forum.

Virtual particles appearing in a vacuum

The quantum vacuum is Lorentz invariant; it does not define an "absolute rest" frame. It looks the same in every frame.

 

[name123]

Um, because they are not saying the gap between the poles is narrower? They are saying it is shorter in the x direction; but the gap in the B-Team frame is not just in the x direction. It is also in the y direction. The full gap, combining both the x and y components, is large enough for the two B-Team members to fit through.

Oh, so they are saying the gap between the A-Team members spans more rulers that the gap between the B-Team members?

If "changed span" means "shows relativistic length contraction", then yes. Otherwise I don't know what you mean by "changed span". You keep talking as if this has some other meaning, but I don't know what it is. The only "change in length" that occurs in this scenario is relativistic length contraction.

I find it hard to believe that you couldn't understand any other meaning, such as certain physical lengths changing, because I had explained it, and showed why it makes sense as a physical property and thus it would be a logical fallacy to conclude there was no rest frame. Remember you effectively said that Ok, you could understand it as a physical property but because it didn't make any difference to the physics it was off topic on this forum. Which is fine, but if you assume there is no absolute rest frame it is still a logical fallacy even if you don't wish it discussed. Another sense of the absolute rest frame would be that its clocks are in synch with the "now" (so presentism). I'm not sure of the motivation for promoting the fallacy.

 

[PeterDonis]

Oh, so they are saying the gap between the A-Team members spans more rulers that the gap between the B-Team members?

No. They are saying that the gap between the B-Team members is purely in the $x'$ direction; but the gap between the A-Team members is diagonal in the B-Team frame (it has components in both the $x'$ and the $y'$ directions). So the component of the B-Team gap that is in the same direction as the A-Team gap is shorter than the A-Team gap; but that doesn't mean the full B-Team gap is shorter than the A-Team gap.

Note also that length contraction of the A-Team rulers in the B-Team frame only applies in the $x$ direction, so you can't just plug numbers into the length contraction formula and expect it to give correct answers for the comparison between the rulers. This is why I have said (repeatedly) that you need to actually assign coordinates to all the events of interest and actually do the math of transforming them from one frame to the other.

you effectively said that Ok, you could understand it as a physical property but because it didn't make any difference to the physics it was off topic on this forum.

No, that is not what I said. I said that since it is not a physical property (because it doesn't make any difference to any physical measurements), it is off topic on this forum. Which means that discussion of it is closed.

 

[name123]

No. They are saying that the gap between the B-Team members is purely in the $x'$ direction; but the gap between the A-Team members is diagonal in the B-Team frame (it has components in both the $x'$ and the $y'$ directions). So the component of the B-Team gap that is in the same direction as the A-Team gap is shorter than the A-Team gap; but that doesn't mean the full B-Team gap is shorter than the A-Team gap.

But remember we can imagine the rulers laid out between the A-Team pair, and rulers laid out between the B-Team pair before they start the conveyor belt. So there is a set number of rulers between them. But the B-Team think that the rulers in the A-Teams frame of reference have shrunk, and that fewer rulers of the span they make now would fit. Presumably if we imagine the A-Team poles and the B-Team members to be replaced by very thin filaments, the thinner they are imagined to be the better, then the y momentum could be relatively quite slow say 1/10,000,000th m/s. Do you think it shows that the 29,979,245.8 old B-Team rulers between the A-Team pair span a greater distance than the 29,979,245.8 old B-Team rulers between the B-Team pair, or what type of angle between them were you thinking that y-momentum would make?

I find it hard to believe that you couldn't understand any other meaning, such as certain physical lengths changing, because I had explained it, and showed why it makes sense as a physical property and thus it would be a logical fallacy to conclude there was no rest frame. Remember you effectively said that Ok, you could understand it as a physical property but because it didn't make any difference to the physics it was off topic on this forum.

No, that is not what I said. I said that since it is not a physical property (because it doesn't make any difference to any physical measurements), it is off topic on this forum. Which means that discussion of it is closed.

Which is fine, but if you assume there is no absolute rest frame it is still a logical fallacy even if you don't wish it discussed. Another sense of the absolute rest frame would be that its clocks are in synch with the "now" (so presentism). I'm not sure of the motivation for promoting the fallacy.

Regarding my understanding that you were effectively saying that you understood it as a physical property but because it didn't make any difference to the physics it was off topic in this forum, I got the impression when I said:

The whole point was that it wouldn't have any physical consequences, that is why they could no more tell in the simulation which way it was done, than you could tell how it was done in an imagined physical universe. There could be an imagined physical reason why it should be considered that the spaceship changed size when it applied its thrusters and not the universe. If it was imagined that way round, then it would clearly be a physical property that was being imagined. Understanding that several physical realities with different physical properties could be measured to be the same, is like understanding that the cartoon characters could understand that although they might not be able to detect which way it was modeled, because different models could be measured the same, it doesn't follow that how it was modeled wasn't a property of the model.

And you replied to the part "The whole point was that it wouldn't have any physical consequences"

Then any questions about it are not physics questions and are off topic in this forum.

I had assumed you were saying that if it hadn't any physical consequences that it wasn't a topic for the forum, and were accepting the point that it could have no physical consequences but still be a physical property, by virtue of you not disputing that the assumption that there is no rest frame is a logical fallacy.

 

[PeterDonis]

I had assumed you were saying that if it hadn't any physical consequences that it wasn't a topic for the forum, and were accepting the point that it could have no physical consequences but still be a physical property, by virtue of you not disputing that the assumption that there is no rest frame is a logical fallacy.

You assumed incorrectly. The fact that I did not bother to dispute something you said does not imply that I agree with it. I may just have not bothered because it's off topic for this forum.

The topic is closed. If you bring it up again I will issue a warning.

 

[name123]

You assumed incorrectly. The fact that I did not bother to dispute something you said does not imply that I agree with it. The topic is closed. If you bring it up again I will issue a warning.

So regarding the other point we can imagine the rulers laid out between the A-Team pair, and rulers laid out between the B-Team pair before they start the conveyor belt. So there is a set number of rulers between them. But the B-Team think that the rulers in the A-Teams frame of reference have shrunk, and that fewer rulers of the span they make now would fit. Presumably if we imagine the A-Team poles and the B-Team members to be replaced by very thin filaments, the thinner they are imagined to be the better, then the y momentum could be relatively quite slow say 1/10,000,000th m/s. Do you think it shows that it can't be as the B-Team thought which is that 29,979,245.8 new B-Team rulers would fit between the B-Team pair but only 29,828,973.1 between the A-Team pair they wouldn't have fitted through, or were you thinking a 1/10,000,000th m/s momentum would create a larger enough angle that it'd fit through?

 

[PeterDonis]

we can imagine the rulers laid out between the A-Team pair, and rulers laid out between the B-Team pair before they start the conveyor belt.

Yes. But also, before the belt is started, both sets of rulers are pointing solely in the $x$ direction. After the belt is started, that is not the case.

the B-Team think that the rulers in the A-Teams frame of reference have shrunk, and that fewer rulers of the span they make now would fit

Yes.

if we imagine the A-Team poles and the B-Team members to be replaced by very thin filaments, the thinner they are imagined to be the better, then the y momentum could be relatively quite slow say 1/10,000,000th m/s.

Why don't you work out the actual numbers, as I have repeatedly suggested, instead of guessing?

Do you think it shows that the 29,979,245.8 old B-Team rulers between the A-Team pair span a greater distance than the 29,979,245.8 old B-Team rulers between the B-Team pair

Obviously the same set of rulers, pointed in the same direction, at rest relative to each other, will span the same distance. However, once the conveyor belt starts, the B-Team pair is not pointed in the same direction as the old B-Team rulers. I have repeatedly pointed this out. Why don't you work out the actual numbers?

what type of angle between them were you thinking that y-momentum would make?

Why don't you work out the actual numbers, as I have repeatedly suggested? I have repeatedly given you the solution of the problem. All you have to do is work out the numbers.

 

[name123]

Why don't you work out the actual numbers, as I have repeatedly suggested? I have repeatedly given you the solution of the problem. All you have to do is work out the numbers.

Well, well I've laid out the numbers in the scenario, but I've not done the calculation with x and y before. And if others are reading it, maybe it'd be the same for them. Could you do it please?

 

[PeterDonis]

Could you do it please?

No, but I'll give the Lorentz transformation equations that apply. If the relative velocity between the frames is $v$ in the $x$ direction and $w$ in the $y$ direction, then the transformation from the unprimed (A-Team) frame to the primed (B-Team) frame is

$$
t' = \gamma \left( t - v x \right)
$$
$$
x' = \gamma \left( x - v t \right)
$$
$$
y' = y - w t
$$

where $\gamma = 1 / \sqrt{1 - v^2}$ (we assume that $w$ is too small to affect length contraction or time dilation), and we leave out the $z$ coordinate since there is no relative motion in that direction.

(Note that this transformation is approximate; it is not valid if $w$ is not small. That case is considerably more complicated mathematically, and you don't need that extra complication to solve this problem.)

So if you define coordinates for all the relevant events in the A-Team frame, you can use the above transformation to get the coordinates of those events in the B-Team frame. Answers to all the questions you want to ask can then be read off of those coordinate values.

 

[name123]

No, but I'll give the Lorentz transformation equations that apply. If the relative velocity between the frames is $v$ in the $x$ direction and $w$ in the $y$ direction, then the transformation from the unprimed (A-Team) frame to the primed (B-Team) frame is

$$
t' = \gamma \left( t - v x \right)
$$
$$
x' = \gamma \left( x - v t \right)
$$
$$
y' = y - w t
$$

where $\gamma = 1 / \sqrt{1 - v^2}$ (we assume that $w$ is too small to affect length contraction or time dilation), and we leave out the $z$ coordinate since there is no relative motion in that direction.

(Note that this transformation is approximate; it is not valid if $w$ is not small. That case is considerably more complicated mathematically, and you don't need that extra complication to solve this problem.)

So if you define coordinates for all the relevant events in the A-Team frame, you can use the above transformation to get the coordinates of those events in the B-Team frame. Answers to all the questions you want to ask can then be read off of those coordinate values.

I thought the gamma equation was $\gamma = 1 / \sqrt{1 - (v^2/c^2)}$ was that a mistake?

 

[PeterDonis]

I thought the gamma equation was $\gamma = 1 / \sqrt{1 - (v^2/c^2)}$ was that a mistake?

I am using units in which $c = 1$. So $v = 0.1$ in these units (given your specification of the problem). The easiest units of time and distance, given your numbers, are probably seconds and light-seconds, so 29,979,245.8 meters is a distance of 0.1 (i.e., 0.1 light-seconds).

 

[name123]

Ok, so I had a try at doing the calculations.

So with v = 29,979,245.8 and w = 0.00000001 gamma is roughly 1.0050378.

And at t = 0 the x coordinates are as follows:
A1: x = 0
A2: x = 29,979,245.8

B1: x = 50,000
B2: x = 29878973.1

According to the B-Team the time and coordinates are roughly as follows:
A1: t' = 0 x' = 0
A2: t' = -0.01005038 x' = 30,130,275.70

B1: t' = -0.00001676 x' = 50,251.89
B2: t' = -0.01001676 x' = 30,029,497.85

It seemed to me that from the B-Team's perspective at t' = -.006
A1: x' = 179,875.47
A2: x' = 30,008,848.42

B1: x' = 50,251.89
B2: x' = 30,029,497.85

So A-Team is within the B-Team bounds, and it seems to me that there is no time according to the B-Team when: B1.x' > A1.x' AND B2.x' < A2.x' is true from their perspective. I presume I've done it wrong. Taking into account the Y distance didn't seem to make any significant difference (as it was so small).

 

[PeterDonis]

I had a try at doing the calculations.

On re-checking the approximation I gave in my previous post, I realized that it's not good enough; there are additional terms that have to be taken into account. (Briefly, I was assuming that all of the corrections were second order in $w$, but I was wrong; there are correction terms that are first order in $w$ and therefore have to be included.)

A better approximation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = \gamma \left( x - v t \right) + \left( \gamma - 1 \right) \frac{w}{v} y
$$

$$
y' = y - \gamma w t + \left( \gamma - 1 \right) \frac{w}{v} x
$$

Note that even this might not be good enough, depending on how small you pick $w$ and how many significant figures you look at. The fully correct transformation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = - \gamma v t + \left[ 1 + \left( \gamma - 1 \right) \frac{v^2}{v^2 + w^2} \right] x + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} y
$$

$$
y' = - \gamma w t + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} x + \left[ 1 + \left( \gamma - 1 \right) \frac{w^2}{v^2 + w^2} \right] y
$$

where now we are using $\gamma = 1 / \sqrt{1 - v^2 - w^2}$ so we take both speeds into account (and remember units are always such that $c = 1$).

Sorry for the mixup on my part.

Taking into account the Y distance didn't seem to make any significant difference (as it was so small).

If you leave out the motion in the y direction, you leave out the critical feature of the problem, because if you only look at the "fit" in the x direction, the A-Team will not fit. You have to look at the y coordinates. What you should be checking for is the path in the x'-y' plane of the two A-Team poles, and where each pole passes relative to the two B-Team members (who are at rest in the B-Team frame). You can't figure that out just from looking at x coordinates.

I also suggest picking a value for w that doesn't require you to do calculations accurate to 30 or more significant figures in order to see actual variation in the y coordinates. Try, for example, a value of w that is 1/100 or 1/1000 the value of v, so you only need 7 or 8 figures at most.

 

[name123]

On re-checking the approximation I gave in my previous post, I realized that it's not good enough; there are additional terms that have to be taken into account. (Briefly, I was assuming that all of the corrections were second order in $w$, but I was wrong; there are correction terms that are first order in $w$ and therefore have to be included.)

A better approximation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = \gamma \left( x - v t \right) + \left( \gamma - 1 \right) \frac{w}{v} y
$$

$$
y' = y - \gamma w t + \left( \gamma - 1 \right) \frac{w}{v} x
$$

Note that even this might not be good enough, depending on how small you pick $w$ and how many significant figures you look at. The fully correct transformation is:

$$
t' = \gamma \left( t - v x - w y \right)
$$

$$
x' = - \gamma v t + \left[ 1 + \left( \gamma - 1 \right) \frac{v^2}{v^2 + w^2} \right] x + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} y
$$

$$
y' = - \gamma w t + \left( \gamma - 1 \right) \frac{vw}{v^2 + w^2} x + \left[ 1 + \left( \gamma - 1 \right) \frac{w^2}{v^2 + w^2} \right] y
$$

where now we are using $\gamma = 1 / \sqrt{1 - v^2 - w^2}$ so we take both speeds into account (and remember units are always such that $c = 1$).

Sorry for the mixup on my part.
If you leave out the motion in the y direction, you leave out the critical feature of the problem, because if you only look at the "fit" in the x direction, the A-Team will not fit. You have to look at the y coordinates. What you should be checking for is the path in the x'-y' plane of the two A-Team poles, and where each pole passes relative to the two B-Team members (who are at rest in the B-Team frame). You can't figure that out just from looking at x coordinates.

I also suggest picking a value for w that doesn't require you to do calculations accurate to 30 or more significant figures in order to see actual variation in the y coordinates. Try, for example, a value of w that is 1/100 or 1/1000 the value of v, so you only need 7 or 8 figures at most.

Have you the equations in the normal form where the unit of c isn't 1? (a link maybe, I'm having trouble finding it)

 

[PeterDonis]

Have you the equations in the normal form where the unit of c isn't 1?

It's easy to modify them. The general rule is to put $ct$, $ct'$ in place of $t$, $t'$ and $v/c$, $w/c$ in place of $v$, $w$. If you work it out, though, you will see that the only equation that actually changes is the $t'$ equation; if you put $v/c^2$, $w/c^2$ in place of $v$, $w$ in that equation, that will do it.

 

[name123]

It's easy to modify them. The general rule is to put $ct$, $ct'$ in place of $t$, $t'$ and $v/c$, $w/c$ in place of $v$, $w$. If you work it out, though, you will see that the only equation that actually changes is the $t'$ equation; if you put $v/c^2$, $w/c^2$ in place of $v$, $w$ in that equation, that will do it.

Not clear if that is $v/c$, $w/c$ in place of $v$, $w$ or $v/c^2$, $w/c^2$ in place of $v$, $w$. Have you a link to the equations in normal form? Also regarding the calculations I'd done they'd be correct for x right? So could the gap between the A-Team members if there was a y-direction also be worked out using gap = SQRT(x-length ^2 + y-length ^2) so that I could work out the minimum y-length that would be required to allow the B-Team members through, and just use the normal equation to see if that y-length is big enough?

 

[PeterDonis]

Not clear if that is $v/c$, $w/c$ in place of $v$, $w$ or $v/c^2$, $w/c^2$ in place of $v$, $w$.

Modify just the $t'$ equation using the latter.

Have you a link to the equations in normal form?

Have you tried googling "Lorentz transformation"?

regarding the calculations I'd done they'd be correct for x right?

Not if you used the formulas I gave before. Note that the new formulas I gave add terms in all the equations ($t'$, $x'$, and $y'$).

could the gap between the A-Team members if there was a y-direction also be worked out using gap = SQRT(x-length ^2 + y-length ^2) so that I could work out the minimum y-length that would be required to allow the B-Team members through, and just use the normal equation to see if that y-length is big enough?

No. You are still missing a key aspect of the situation: in the B-Team frame, the line connecting the two A-Team members is at an angle from the line connecting the two B-Team members. You need to look at the x-y plane; there is no shortcut.

Here is an outline of the steps I recommend:

(1) We have four observers, two A-Team (call them A1 and A2) and two B-Team (call them B1 and B2). The key condition of the problem is that there is an instant of time in the A-Team frame (call it $t = 0$) at which these four observers are lined up along the $x$ axis (i.e., they all have $y = 0$) in the following order (going from smaller to larger $x$ coordinates): A1, B1, B2, A2. This is what it means to say that the two B-Team members pass between the two A-Team members.

(2) The above condition gives us $( t, x, y )$ coordinates for four events, in the A-Team frame.

(3) Use the Lorentz transformation to obtain the $( t', x', y' )$ coordinates for these four events in the B-Team frame.

(4) In the B-Team frame, B1 and B2 are at rest, so the $( x', y' )$ coordinates obtained for them are valid at any time $t'$. So to find out whether they still pass between A1 and A2 in the B-Team frame, compute the worldlines of A1 and A2 in that frame, using the coordinates obtained above and the fact that A1 and A2 both move with speed $( v_x, v_y ) = ( - v, - w )$ in this frame, and check to see where they are in relation to the fixed coordinates of B1 and B2.

Note that you can actually do all of this using general formulas; you don't need to plug in numbers. However, picking specific numbers and then graphing the results in the $x', y'$ plane may help to visualize what is going on.

 

[name123]

Modify just the $t'$ equation using the latter.

With the gamma equation you gave wouldn't it be wrong if $w$ = 0, and I divided $v^2$ by $c$ instead of $c^2$? By wrong I mean different to the normal equation for a single boost.

 

[PeterDonis]

With the gamma equation you gave wouldn't it be wrong if ##w = 0##, and I divided ##v^2$by$c$instead of$c^2##?

I don't think I ever said to divide $v^2$ by $c$ anywhere, but you are correct that you do have to include $c$ in the $\gamma$ equation, and I should have mentioned that. The correct equation with $c$ included is

$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2} - \frac{w^2}{c^2}}}
$$

 

[name123]

I don't think I ever said to divide $v^2$ by $c$ anywhere, but you are correct that you do have to include $c$ in the $\gamma$ equation, and I should have mentioned that. The correct equation with $c$ included is

$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2} - \frac{w^2}{c^2}}}
$$

Could you just include c with the others, I'm sure I'm not the only one that would find it useful?

 

[PeterDonis]

Could you just include c with the others

The $x'$ and $y'$ equations are unchanged. The $t'$ equation becomes

$$
t' = \gamma \left( t - \frac{v}{c^2} x - \frac{w}{c^2} y \right)
$$

 

[name123]

Thanks, and when looking at it from the B-Team's perspective are they considering the A-Team to be going at $-v$,$-w$?