Benefits of time dilation / length contraction pairing?

In summary, there is often confusion about the use of time dilation and length contraction in regards to frames in motion. These concepts are consequences of the Lorentz transformation and Einstein's SR postulates. While there may be some inconsistency in the use of primes in equations, the frames are actually consistent and explain a variety of empirical observations. However, there may be a more intuitive way to express these concepts that would not lead to confusion. The use of time dilation and length contraction may have a historical and practical significance, but it is important to understand the fundamental theory in order to fully grasp their utility.
  • #421
neopolitan said:
It seems to me that we have two stopping points here.

First, the need to prove that there must be one single factor (as in my derivation shown earlier), rather than possibly two factors.
Yes.
neopolitan said:
Second, generality, in that you think that if I chose two totally arbitrary events in order to to measure the interval between them in two frames, and that you feel that there is a "need for an explanation as to what specific property of the two events (I) chose ensures that the constants in those two equations will be the same".
How is that different from the above? I'm just asking for a demonstration that the specific properties of the events you chose ensure that the factor will be the same in both equations, and pointing out that since you presumably agree the factor wouldn't be the same for an arbitrary pair of events, then your demonstration will have to somehow make use of those specific properties rather than just making use of more general facts like the first postulate of SR.
neopolitan said:
Do you further agree that, if I were to convince you that there being one single factor is the default, that feat would negate the need for a proof?
I would say any totally airtight argument for why they must be the same would constitute a "proof", so I don't really understand this distinction. Unless by "default" you just mean "the assumption that seems most plausible a priori even if we aren't sure it's actually correct", in which case I don't think that would negate the need for a proof.

Keep in mind, also, my later criticisms involving the final equations you derived, which I say have a physical meaning that's totally unlike the Lorentz transformation equations...it seems in a way a bit pointless to spend a lot of time thinking about how to justify this one step in your demonstration (which I agree happens to be true even if we haven't found a way to justify it) if the final endpoint of your demonstration is just an equation that looks superficially like the Lorentz transformation but really has almost nothing to do with it, and has no real utility outside of relating quantities in the specific physical scenario you describe where a light ray crosses path with two observers (so it cannot even really be understood as a special case of the Lorentz transformation, since the variables that appear in the equations aren't all defined in terms of the same single event or pair of events)
 
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  • #422
JesseM said:
How is that different from the above?

I must be getting tired.

My second point was supposed to be that generality seems to be a stopping point, in that you think my selection of event has a significant bearing on the final equation (to the extent that it is not actually the Lorentz Transform or even equivalent to it, but something less general).

I had meant to finish the phrase "you think that if I chose two totally arbitrary events in order to to measure the interval between them in two frames" with "then my derivation would not stand". I did mean to end with the ending I had, for reasons which might become clear when, at a time when I am not so dog tired, I will try to explain that my starting scenario is so general that A should be totally interchangeable with B. Generality in, generality out.

cheers,

neopolitan
 
  • #423
neopolitan said:
I had meant to finish the phrase "you think that if I chose two totally arbitrary events in order to to measure the interval between them in two frames" with "then my derivation would not stand".
OK, yes, I think your derivation seems to depend on the fact that both events lie on the path of a photon. What's more, even with this restriction, your final equations don't even end up measuring "the interval between them" (i.e. the variables in your final equations don't all refer to the space and time intervals between a single pair of events).
neopolitan said:
I did mean to end with the ending I had, for reasons which might become clear when, at a time when I am not so dog tired, I will try to explain that my starting scenario is so general that A should be totally interchangeable with B. Generality in, generality out.
Showing that A could be exchanged with B might show that the constants in those equations would be the same, although it wouldn't address the issue of the final equations you derive. In any case it doesn't seem to me A can be exchanged with B here, since after all both frames agree the photon passes B before it passes A.
 
  • #424
JesseM said:
In any case it doesn't seem to me A can be exchanged with B here, since after all both frames agree the photon passes B before it passes A.

Surely you see that that depends on v and the x values being positive. If we plug in a negative v, the photon passes A before it passes B. Similarly if we start with a negative displacement for the event (while v stays positive), then the photon passes A before it passes B. To complete the picture, if we put in a negative v and a negative starting displacement, then the photon passes B first.

We're not constrained to using positive values.

Can you understand that if I say "t time units before it passes A, the photon passes B" and the scenario makes t negative, then that is the same as saying "|t| time units after it passes A, the photon passes B" where |t| is the magnitude of t.

Does this help you see that A and B are interchangeable?

If so, that will leave us with what the derived equations describe.

I do have one other question, you've not said anything about it but do you have issue with using (0,0) as an agreed location (conceptually the same as a colocation, if not an actual colocation, of A and B)? If you do have a problem with it, I will have to talk exclusively about intervals (even though I think we have agreed that coordinates are implicitly intervals anyway).

cheers,

neopolitan
 
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  • #425
neopolitan said:
We have a couple of options for consistency here: compare "in the rest frame" with "in the moving frame" in each of the pairs, ie priming the "in the moving frame" values; or consider a single rod clock which is in motion relative to a notional rest frame, ie priming the rod clock values.

Comparing these approaches, the rod clock despite having a single construction has two natures for our purposes "rod" and "clock". If we have "rod clock frame" and "observer frame", then the "rod clock frame" is the rest frame for considering the rod clock's rod-nature, but the "observer frame" is the rest frame for considering the rod clock's clock-nature.

I think it is this inconsistency that Rasalhague is getting at (and which I have touched on once or twice).

Anyways, Rasalhague might want to use the rod clock approach to frame "space-like" and "time-like" in an internally consistent way, since using the terminology "moving ends of a journey" is fraught with danger :)

That's true. Partly I wanted to take advantage of the ambiguity in words like length that could refer to a length of time or a length of space. I was wondering if we could come up with a rigorous definition of a length of travel time that would intuitively match our idea of the length of an object, but I don't know how practical this would be, or whether the risk of confusion would outweigh the benefits.

Your "rod clock" is an interesting idea, although it's taken me a while to get my tongue around "rod clock rest frame" - maybe our next example can involve a red lorry and a yellow lorry... I'd been thinking along similar (albeit more primative) lines with candle clocks, or a burning fuse, but I always seemed to come back to this apparent asymmetry, as if even an object that's both ruler and clock, using the same physical markings to measure time along its length as it does to measure distance, leads to this strange feeling that we have to treat time and space as behaving differently from each other with respect to the object, or else switch which frame we take as out starting point.

How would the situation change, I wonder, if we replace the photons in the rod clock with a massive particle? Then we could think in terms of the particle's rest frame. I was looking at the way the idea of frame dependent distance is introduced in Fishbane et al.: Physics for Scientists and Engineers with a muon formed in the Earth's atmosphere and traveling from its point of creation to the ground. It lives just long enough to reach the ground. Of all the possible values they could take in all possible reference frames, the following two variables are minimum in the muon's rest frame: the time [tex]t_{\mu}[/tex] that passes between the events of the muon's birth and death, and the distance [tex]x_{\mu}[/tex] that extends between the receding bit of air that marks the spot where the muon was created and the approaching ground that marks the spot where it will be annihilated. So in the ground's rest frame (where the muon is moving), both the t component and the x component of the separation of these events (muon birth and death) will be greater than they were in the muon's rest frame.

But in that example, the muon is the clock and the Earth is the ruler. Although I subscripted the space coordinate with a mu there, this seems at odds with the fact that it's the Earth that's actually doing the measuring of the muon's journey, since in the muon's rest frame, the muon makes no journey. Will this always be the case, due to the way the distortions of time and space complement each other, even when we construct an object that incorporates the roles of clock and ruler: will the part or aspect of it that's used to determine time need to be moving in the rest frame of the part or aspect of it that's used to measure distance?
 
  • #426
neopolitan said:
Surely you see that that depends on v and the x values being positive. If we plug in a negative v, the photon passes A before it passes B. Similarly if we start with a negative displacement for the event (while v stays positive), then the photon passes A before it passes B. To complete the picture, if we put in a negative v and a negative starting displacement, then the photon passes B first.

We're not constrained to using positive values.

Can you understand that if I say "t time units before it passes A, the photon passes B" and the scenario makes t negative, then that is the same as saying "|t| time units after it passes A, the photon passes B" where |t| is the magnitude of t.

Does this help you see that A and B are interchangeable?
I don't understand what you mean by "interchangeable" here, or how it proves the constant should be the same in your equations. It's true that whatever xa and v we pick, then whatever is seen in A's frame vs. B's frame, we could then pick a different xa and v so that B saw what A formerly saw and vice versa. Still, for any particular choice of xa and v, their viewpoints are asymmetrical in that they both agree the photon passed one of them first. And your equations both assume we have picked a particular choice of xa and v, and then assert the constants relating the times will be the same.

As an analogy, suppose that instead of the two events being the photon crossing the x=0 axis of A's frame and the event of the photon crossing the x=0 axis of B's frame, we instead chose Event #1 to be the event of the photon crossing the x=5 axis of A's frame, and Event #2 to be the event of the photon crossing the x=5 axis of B's frame. Now suppose that someone said that since the two frames are "interchangeable", we should expect the constant in the following pair of equations to be the same:

(time of Event #2 in B's frame) = (constant)*(time of Event #1 in A's frame)
(time of Event #2 in A's frame) = (constant)*(time of Event #1 in B's frame)

Without doing any new calculations to check the validity of this, are you confident enough in the meaning of "interchangeability" that you would bet a large sum of money that this guy's argument is correct or incorrect? If you feel any uncertainty here that perhaps shows why the use of "interchangeability" in your own argument is not really rigorous, and is more of what physicists would call a "handwavey" argument. You really need to provide a step-by-step proof where each statement clearly follow from the last, and you show specifically where the two postulates of relativity come into play (since I think you'd agree that if we came up with a coordinate transformation that didn't respect these postulates the constants in your equations would not be the same), and what the relevance is of the fact that the two events in question are the events of a photon's worldline intersecting each frame's x=0 axis.
neopolitan said:
I do have one other question, you've not said anything about it but do you have issue with using (0,0) as an agreed location (conceptually the same as a colocation, if not an actual colocation, of A and B)? If you do have a problem with it, I will have to talk exclusively about intervals (even though I think we have agreed that coordinates are implicitly intervals anyway).
What do you mean "an agreed location"? Are you just referring to the idea that whenever we talk about the space and time coordinates of an event, we're implicitly talking about the space and time intervals between that event at the event at (0,0)? Is so I have no problem with that, again my problem with your final equations is that all the variables don't refer to intervals between a single pair of events.
 
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  • #427
I started to reply to this last night, what I wrote it probably sitting on the home computer in "Preview Post" form rather than being submitted. Oh well.

I'm not going to be able to reply in depth to your first chunk because you seem to have gone off on a tangent. I've highlighted the phrases which are of concern.
JesseM said:
I don't understand what you mean by "interchangeable" here, or how it proves the constant should be the same in your equations. It's true that whatever xa and v we pick, then whatever is seen in A's frame vs. B's frame, we could then pick a different xa and v so that B saw what A formerly saw and vice versa. Still, for any particular choice of xa and v, their viewpoints are asymmetrical in that they both agree the photon passed one of them first. And your equations both assume we have picked a particular choice of xa and v, and then assert the constants relating the times will be the same.

As an analogy, suppose that instead of the two events being the photon crossing the x=0 axis of A's frame and the event of the photon crossing the x=0 axis of B's frame, we instead chose Event #1 to be the event of the photon crossing the x=5 axis of A's frame, and Event #2 to be the event of the photon crossing the x=5 axis of B's frame. Now suppose that someone said that since the two frames are "interchangeable", we should expect the constant in the following pair of equations to be the same:

(time of Event #2 in B's frame) = (constant)*(time of Event #1 in A's frame)
(time of Event #2 in A's frame) = (constant)*(time of Event #1 in B's frame)

Without doing any new calculations to check the validity of this, are you confident enough in the meaning of "interchangeability" that you would bet a large sum of money that this guy's argument is correct or incorrect? If you feel any uncertainty here that perhaps shows why the use of "interchangeability" in your own argument is not really rigorous, and is more of what physicists would call a "handwavey" argument. You really need to provide a step-by-step proof where each statement clearly follow from the last, and you show specifically where the two postulates of relativity come into play (since I think you'd agree that if we came up with a coordinate transformation that didn't respect these postulates the constants in your equations would not be the same), and what the relevance is of the fact that the two events in question are the events of a photon's worldline intersecting each frame's x=0 axis.


Here's what you are referring back to:
neopolitan said:
If we make an alternative hypothesis that:

(time of colocation of B and photon in the A frame) = (some factor) * (time of colocation of B and photon in the B frame)

and

(time of colocation of A and photon in the B frame) = (some factor) * (time of colocation of A and photon in the A frame)


I never said it was a constant. You are correct in that the factor is dependent on a particular value of v, and therefore not a constant. I don't call it a "function" because within the scenario v is fixed (if we changed v, then we would be talking about a different relationship between a different pair of frames a different relative speed with respect to each other). It would be a function if it varied as a result of changing our input values from within the pair of frames to which v applies (ie x and t values).

The selection of a different xa in the scenario does not change the factor and the factor does not vary with time.

By interchangeable, I mean that I could assign either of two observers, one in each frame, with the label A, and the other B. It won't affect the end result in any meaningful way (we might as a result prime A values).

I'm going to redo the equations with v as a velocity, rather than a speed. This may help.

JesseM said:
What do you mean "an agreed location"? Are you just referring to the idea that whenever we talk about the space and time coordinates of an event, we're implicitly talking about the space and time intervals between that event AND the event at (0,0)? Is so I have no problem with that, again my problem with your final equations is that all the variables don't refer to intervals between a single pair of events.


I made a correction, I think that is what you meant.

You are part of the way there. I mean what you wrote in the corrected sentence, plus the fact that both A and B see that (0,0) event as (0,0), ie x=0,t=0 is the same event as x'=0,t'=0. There's agreement between A and B as to that event.

They can agree on another event rather than (0,0) if it is convenient to them as well, it will just make the coordinate transformation a little different.

When I redo the equations with v as a velocity rather than a speed, I will do them all with only intervals involved, rather than aiming for a coordinate transformation. I will also choose an event which is not necessarily simultaneous with a colocation of A and B.

At the same time, I will attempt to maintain enough rigour to satisfy you that "some factor" is in fact the same in both frames. If I can't manage that, then I might have to try a proof by elimination, eliminating the possibility of two different factors. I've already eliminated a single factor which is equal to 1.

I'll give it some thought and get back to you.

cheers,

neopolitan
 
  • #428
neopolitan said:
I'm not going to be able to reply in depth to your first chunk because you seem to have gone off on a tangent. I've highlighted the phrases which are of concern.
Here's what you are referring back to:
I never said it was a constant.
All I meant was that for any given choice of xa and v it's just a number, but I probably shouldn't have used the word "constant" since it could vary as you vary xa and v (although in fact it turns out that it doesn't). However, I wasn't actually assuming it remained constant as you varied these, so my argument in no way depends on this assumption, nor is it a "tangent". Please just mentally replace "constant" with "factor" in the block of text you quoted and reread it.
neopolitan said:
By interchangeable, I mean that I could assign either of two observers, one in each frame, with the label A, and the other B. It won't affect the end result in any meaningful way (we might as a result prime A values).
What "end result" are you referring to? The end result that the same factor appears in both equations? In this case you are simply asserting what you are supposed to prove, but perhaps you mean something else. In any case, my point is that you haven't made any coherent argument as to why your ill-defined notion of "interchangeability" proves that for a specific choice of xa and v, the factor in your two equations must be the same. The actual values of the variables that appear in the equations are certainly not invariant under a switching of labels--for example, the value of (the time that the photon is colocated with B, as measured in the A frame) is not equal to (the time that the photon is colocated with A, as measured in the B frame) for most specific choices of xa and v.

Note that since A and B's worldlines are assumed to lie along the x=0 axis of their respective frames, the event of the photon being colocated with A is equivalent to the event of the photon crossing A's x=0 axis, and likewise for B...this means your two equations can be written as:

(time of photon crossing B's x=0 axis in A's frame) = (factor)*(time of photon crossing B's x=0 axis in B's frame)

(time of photon crossing A's x=0 axis in B's frame) = (factor)*(time of photon crossing A's x=0 axis in A's frame)

So maybe now you can see why it wasn't a tangent when I asked in the previous post if you'd agree with a hypothetical guy making the argument that because of the "interchangeability" of the two frames, we could also assume the same factor will appear in the following two equations (same as one another, not same as in the previous two equations):

(time of photon crossing B's x=5 axis in A's frame) = (factor)*(time of photon crossing B's x=5 axis in B's frame)

(time of photon crossing A's x=5 axis in B's frame) = (factor)*(time of photon crossing A's x=5 axis in A's frame)

All that's changed in these two equations is that x=0 has been replaced with x=5. Please tell me whether you'd be confident about whether this guy's argument is right or wrong, without doing any SR calculations to check explicitly what factor appears in each. If you're unsure, that suggests that you don't really have good justification for being confident that "interchangeability" means the factor should be the same in your own two equations.
neopolitan said:
You are part of the way there. I mean what you wrote in the corrected sentence, plus the fact that both A and B see that (0,0) event as (0,0), ie x=0,t=0 is the same event as x'=0,t'=0. There's agreement between A and B as to that event.

They can agree on another event rather than (0,0) if it is convenient to them as well, it will just make the coordinate transformation a little different.
Agree on another event for what purposes? Do you mean some of the variables in equations in your derivation would no longer represent the interval between 0,0 and some specific event like the photon crossing B's worldline (i.e. the coordinate of the latter event), but would instead represent the interval between a different event other than 0,0 on one side of the interval, but the same event on the other side? I think I'd need to see you go through at least some of the same steps of the derivation as in post 397, but with whatever altered assumptions about "another event" you want to make, in order to follow what you're talking about here.
 
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  • #429
JesseM said:
I just meant that for any given choice of xa and v it's just a number, but I probably shouldn't have used the word "constant" since it could vary as you vary xa and v (although in fact it turns out that it doesn't). However, I wasn't actually assuming it remained constant as you varied these, so my argument in no way depends on this assumption, nor is it a "tangent". Please just mentally replace "constant" with "factor" in the block of text you quoted and reread it.

It will vary with v, but not xa as long as you don't do something weird.

JesseM said:
What "end result" are you referring to? The end result that the same factor appears in both equations?

The end result is the the final pair of equations. The same factor in both equations is a middle stage in the process. But I am going to have a go at redoing it. We can continue to go over what you don't understand here though, until you are satisfied.

JesseM said:
Note that since A and B's worldlines are assumed to lie along the x=0 axis of their respective frames, the event of the photon being colocated with A is equivalent to the event of the photon crossing A's x=0 axis, and likewise for B...this means your two equations can be written as:

(time of photon crossing B's x=0 axis in A's frame) = (factor)*(time of photon crossing B's x=0 axis in B's frame)

(time of photon crossing A's x=0 axis in B's frame) = (factor)*(time of photon crossing A's x=0 axis in A's frame)

So maybe now you can see why it wasn't a tangent when I asked in the previous post if you'd agree with a hypothetical guy making the argument that because of the "interchangeability" of the two frames, we could also assume the same factor will appear in the following two equations:

(time of photon crossing B's x=5 axis in A's frame) = (factor)*(time of photon crossing B's x=5 axis in B's frame)

(time of photon crossing A's x=5 axis in B's frame) = (factor)*(time of photon crossing A's x=5 axis in A's frame)

All that's changed in these two equations is that x=0 has been replaced with x=5. Please tell me whether you'd be confident about whether this guy's argument is right or wrong, without doing any SR calculations to check explicitly what factor appears in each. If you're unsure, that suggests that you don't really have good justification for being confident that "interchangeability" means the factor should be the same in your own two equations.

Actually no, it convinces me more that you have gone off on a tangent.

A and B are colocated at (0,0) ie (x=0,t=0) and (x'=0,t'=0) is the same event, so there is agreement about that event. There is no agreement about (5,0).

So I am pretty confident that "this guy's argument is wrong". (It might be right if the scenario is changed enough, but then it would be another scenario.)

JesseM said:
Agree on another event for what purposes? Do you mean some of the variables in equations in your derivation would no longer represent the interval between 0,0 and some specific event like the photon crossing B's worldline (i.e. the coordinate of the latter event), but would instead represent the interval between a different event other than 0,0 on one side of the interval, but the same event on the other side? I think I'd need to see you go through at least some of the same steps of the derivation as in post 397, but with whatever altered assumptions about "another event" you want to make, in order to follow what you're talking about here.

Well, agree on another event for the purposes that your friend "this guy" could make the claim he did above about (5,0). Not that I know why they would do that.

However, as I said before: "When I redo the equations with v as a velocity rather than a speed, I will do them all with only intervals involved, rather than aiming for a coordinate transformation."

cheers,

neopolitan
 
  • #430
neopolitan said:
The end result is the the final pair of equations. The same factor in both equations is a middle stage in the process. But I am going to have a go at redoing it. We can continue to go over what you don't understand here though, until you are satisfied.
We seem to be going in circles here. You've been saying that the magic word "interchangeable" somehow justifies the claim that the factors in your two equations should be the same, but then you said "By interchangeable, I mean that I could assign either of two observers, one in each frame, with the label A, and the other B. It won't affect the end result in any meaningful way (we might as a result prime A values)." So here again it sounds like you're asserting what you're supposed to be proving, that "interchangeability" somehow will lead to the same "end result" (keeping in mind that what you got for the 'end result' itself depended on the earlier step where you were supposed to show that the factors would be the same in the two equations you wrote). Can you define "interchangeability" in a way that doesn't make reference to any steps in your derivation that happened after the step where you assume the factors are in fact the same in both equations?
neopolitan said:
Actually no, it convinces me more that you have gone off on a tangent.

A and B are colocated at (0,0) ie (x=0,t=0) and (x'=0,t'=0) is the same event, so there is agreement about that event. There is no agreement about (5,0).
Huh? The event (0,0) is not referred to in the two equations where you assert the factors will be the same. The two events referred to in those equations of yours are the colocation of A and the photon (at x=0,t=ta in A's frame) and the colocation of B and the photon (at x'=0, t=t'b in B's frame). Of course the time coordinate of either event in a given frame can be understood as the time interval between that event and (0,0), but then exactly is true about the time coordinate of either event of the photon crossing the x=5 axis of one of the frames. Perhaps there is something in what you mean by "interchangeable" that equations involving the time coordinates of a pair of events in both frames can only be considered interchangeable if the two events happened on the worldlines of observers at rest in each frame who crossed paths at (0,0), but if so nothing you have written about interchangeability so far even hinted at such a requirement. And what about observers who crossed paths at (0,0) but who are not at rest? What if we considered two observers who did cross paths there, with the first observer traveling at velocity V in the A frame and the second traveling at the same velocity V in the B frame? Then if we defined our two events in terms of where the light crossed each of these observer's paths, then without doing any numerical calculations do you think "interchangeability" means the factors in the equations relating the time coordinates of these two events in each frame would be the same?

Even if your answer is once again no, I say that these questions are not "tangential" because they help pin down exactly what you do or do not mean by the slippery word "interchangeable" and how it's supposed to be relevant to showing the factors are the same in those equations--this would be unnecessary if you would be willing to actually spell this stuff out, but so far you've been exceedingly vague. Remember my other request from the "tangent" post which you just ignored:
You really need to provide a step-by-step proof where each statement clearly follow from the last, and you show specifically where the two postulates of relativity come into play (since I think you'd agree that if we came up with a coordinate transformation that didn't respect these postulates the factors in your equations would not be the same), and what the relevance is of the fact that the two events in question are the events of a photon's worldline intersecting each frame's x=0 axis.
Here's another tack: if physicists say that certain things are "interchangeability", they are usually referring to some sort of symmetry where we see that something (like a dynamical equation or the values of some variables) remains the same under a certain transformation of labels (like which charge you call positive and which you call negative, or which of two frames you label A and B, etc.) Is this the sort of thing you're referring to here? If so, what is the specific thing that remains unchanged under a switch of which frame we call A and B, and how does this relate to proving that the factors will be the same in both equations?
neopolitan said:
You are part of the way there. I mean what you wrote in the corrected sentence, plus the fact that both A and B see that (0,0) event as (0,0), ie x=0,t=0 is the same event as x'=0,t'=0. There's agreement between A and B as to that event.

They can agree on another event rather than (0,0) if it is convenient to them as well, it will just make the coordinate transformation a little different.
JesseM said:
Agree on another event for what purposes? Do you mean some of the variables in equations in your derivation would no longer represent the interval between 0,0 and some specific event like the photon crossing B's worldline (i.e. the coordinate of the latter event), but would instead represent the interval between a different event other than 0,0 on one side of the interval, but the same event on the other side? I think I'd need to see you go through at least some of the same steps of the derivation as in post 397, but with whatever altered assumptions about "another event" you want to make, in order to follow what you're talking about here.
neopolitan said:
Well, agree on another event for the purposes that your friend "this guy" could make the claim he did above about (5,0). Not that I know why they would do that.
I have no idea what this comment has to do with the questions in my post above--what "purposes" are you talking about? It's difficult to communicate with you when you a) often seem to conceptualize things in very idiosyncratic ways, and likewise seem to have your own idiosyncratic vocabulary for describing your own concepts, and b) don't seem to be aware of the fact that these ways of thinking/talking are idiosyncratic, or just not make allowances for the fact that they are, and thus you often act as though you expect others to pick up on your meaning from a few enigmatic keywords, that somehow you think the meaning you infer from these words is the "natural" one that everyone else should infer by default too (much like with your use of the word 'interchangeable' and how you immediately dismissed my questions as a 'tangent' when I didn't read your mind and detect all the subtle nuances of what you meant by that, or like when you expected me to immediately pick up on exactly what you meant by the phrase 'number of ticks' in that other recent post...this seems to happen time and time again in my discussions with you).
 
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  • #431
JesseM said:
We seem to be going in circles here. You've been saying that the magic word "interchangeable" somehow justifies the claim that the factors in your two equations should be the same, but then you said "By interchangeable, I mean that I could assign either of two observers, one in each frame, with the label A, and the other B. It won't affect the end result in any meaningful way (we might as a result prime A values)." So here again it sounds like you're asserting what you're supposed to be proving, that "interchangeability" somehow will lead to the same "end result" (keeping in mind that what you got for the 'end result' itself depended on the earlier step where you were supposed to show that the factors would be the same in the two equations you wrote). Can you define "interchangeability" in a way that doesn't make reference to any steps in your derivation that happened after the step where you assume the factors are in fact the same in both equations?

Ok, you'll probably not like it, but I am going to have to jump past the factors stage to the end result which, for me, is:

[tex]x_b'=\gamma . (x_a - vt_a)[/tex]

and

[tex]x_a=\gamma . (x_b' + vt_b')[/tex]

(There is a similar pair for time, but the same argument will apply to that pair as to this pair.)

In this pair, can you see that all b terms are also primed and all a terms are unprimed. So we could, if we wanted to, drop the subscripts leaving us with:

[tex]x'=\gamma . (x - vt)[/tex]

and

[tex]x=\gamma . (x' + vt')[/tex]


Alternatively, we could swap the A and B terms (what we called A before is now B and what we called B before is now A). Our v was (implicitly) defined as positive in the direction that B moves away from A, and negative in the direction that A moves away from B. We keep the same priming notation, since we have shifted focus from the erstwhile A to the new A.

This gives us:

[tex]x_a=\gamma . (x_b' - vt_b')[/tex]

and

[tex]x_b'=\gamma . (x_a + vt_a)[/tex]

Again all our b terms are primed and all our a are unprimed. So we could drop our subscripts, giving:

[tex]x'=\gamma . (x - vt)[/tex]

and

[tex]x=\gamma . (x' + vt')[/tex]


A and B are interchangeable labels. The possible confusion here is that in my idiosyncratic way, I have considered A and B to be labels. You seem to want to have more concrete (and thus less general) determinations of what A and B are.


JesseM said:
Huh? The event (0,0) is not referred to in the two equations where you assert the factors will be the same. The two events referred to in those equations of yours are the colocation of A and the photon (at x=0,t=ta in A's frame) and the colocation of B and the photon (at x'=0, t=t'b in B's frame). Of course the time coordinate of either event in a given frame can be understood as the time interval between that event and (0,0), but then exactly is true about the time coordinate of either event of the photon crossing the x=5 axis of one of the frames. Perhaps there is something in what you mean by "interchangeable" that equations involving the time coordinates of a pair of events in both frames can only be considered interchangeable if the two events happened on the worldlines of observers at rest in each frame who crossed paths at (0,0), but if so nothing you have written about interchangeability so far even hinted at such a requirement. And what about observers who crossed paths at (0,0) but who are not at rest? What if we considered two observers who did cross paths there, with the first observer traveling at velocity V in the A frame and the second traveling at the same velocity V in the B frame? Then if we defined our two events in terms of where the light crossed each of these observer's paths, then without doing any numerical calculations do you think "interchangeability" means the factors in the equations relating the time coordinates of these two events in each frame would be the same?

I've mentioned a few times that there are only three colocation events. I honestly thought that I didn't need to make explicit that the colocation of A and B was at (0,0). I do think I have done it anyway - posts https://www.physicsforums.com/showpost.php?p=2232756&postcount=397" - sadly I don't have time to go on but I think there are more references where I have stated that A and B are colocated at t=0,t'=0. In at least one of those I have even made explicit that that event is (0,0).

As for the tangent, your asking questings for clarifications doesn't constitute a tangent, but your assertion that events associated with photons crossing the x=5 axis can be profitably used in my scenario indicates that you have gone off on a tangent.

The difference is like between these:

"I don't understand what you are saying, what are you saying"

and

"You are claiming this wrong thing, you are wrong"

when I never actually claimed the wrong thing you asserted that I claimed. In the real life example where I said it seems you were going off on a tangent, it might not have been so clear that you were asserting that I claimed something that I had not claimed. That's why I said you were going off on a tangent rather than saying you were making unfounded assertions.

But yes, we are going around in circles.

cheers,

neopolitan
 
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  • #432
neopolitan said:
Ok, you'll probably not like it, but I am going to have to jump past the factors stage to the end result which, for me, is:

[tex]x_b'=\gamma . (x_a - vt_a)[/tex]

and

[tex]x_a=\gamma . (x_b' + vt_b')[/tex]

(There is a similar pair for time, but the same argument will apply to that pair as to this pair.)

In this pair, can you see that all b terms are also primed and all a terms are unprimed. So we could, if we wanted to, drop the subscripts leaving us with:

[tex]x'=\gamma . (x - vt)[/tex]

and

[tex]x=\gamma . (x' + vt')[/tex]Alternatively, we could swap the A and B terms (what we called A before is now B and what we called B before is now A). Our v was (implicitly) defined as positive in the direction that B moves away from A, and negative in the direction that A moves away from B. We keep the same priming notation, since we have shifted focus from the erstwhile A to the new A.

This gives us:

[tex]x_a=\gamma . (x_b' - vt_b')[/tex]

and

[tex]x_b'=\gamma . (x_a + vt_a)[/tex]

Again all our b terms are primed and all our a are unprimed. So we could drop our subscripts, giving:

[tex]x'=\gamma . (x - vt)[/tex]

and

[tex]x=\gamma . (x' + vt')[/tex]A and B are interchangeable labels. The possible confusion here is that in my idiosyncratic way, I have considered A and B to be labels. You seem to want to have more concrete (and thus less general) determinations of what A and B are.
Sigh. Of course I understand that once you have the final Lorentz transformation (which is not what you actually derive, but never mind that for now), A and B are interchangeable labels, "interchangeable" in the sense that the transformation equations still work fine if you switch the subscripts for A and B and also switch which frame v is defined in terms of. How does this have anything whatsoever to do with the question of how you intend to show that the factor relating the two specific events you chose--and only those two events, not two other events like the ones I mentioned--should be the same in the two equations you wrote down, without assuming the Lorentz transformations to begin with? Do you really not understand what it means to "show" something in a derivation without engaging in circular reasoning?
neopolitan said:
I've mentioned a few times that there are only three colocation events. I honestly thought that I didn't need to make explicit that the colocation of A and B was at (0,0).
I'm not talking about how many colocation events appear in your overall derivation, I'm just talking about the two events that appear in these two equations you wrote earlier:

(time of colocation of B and photon in the A frame) = (some factor) * (time of colocation of B and photon in the B frame)

and

(time of colocation of A and photon in the B frame) = (some factor) * (time of colocation of A and photon in the A frame)

Can you please stop your mind from wandering all over the place and imagining I am talking about all these other aspects of your derivation, when I think I made it pretty clear I'm just talking about the specific question of how in these two equations you intend to show that the factor must be the same? If you don't in fact believe that your notion of "interchangeability" is sufficient to show this specific thing, and you haven't actually thought of a good way to show this yet, then that's fine, just say so. But if you do, then please stick to this specific topic for now.
neopolitan said:
As for the tangent, your asking questings for clarifications doesn't constitute a tangent, but your assertion that events associated with photons crossing the x=5 axis can be profitably used in my scenario indicates that you have gone off on a tangent.
I didn't say it "can be profitably used" in your scenario. I was saying that your argument about "interchangeability" is so vague that there's no apparent reason it shouldn't apply to those two events just as much as the two events you chose, and thus it "proves too much". You could respond by actually elaborating on what you meant by "interchangeability" (as I keep asking you to do but you keep not doing), giving an expanded definition which would show why interchangeability demands the factor be the same in your two equations but not in my closely analogous equations. Likewise for my question about another fairly analogous pair of events in the last post:
And what about observers who crossed paths at (0,0) but who are not at rest? What if we considered two observers who did cross paths there, with the first observer traveling at velocity V in the A frame and the second traveling at the same velocity V in the B frame? Then if we defined our two events in terms of where the light crossed each of these observer's paths, then without doing any numerical calculations do you think "interchangeability" means the factors in the equations relating the time coordinates of these two events in each frame would be the same?
Again, if your "proof by interchangeability" is supposed to make any sense, you should have some clear explanation as to why "interchangeability" does or doesn't imply that the factors in equations relating these two events would be the same, and if it doesn't what the crucial difference is that makes it imply that for your pair of events but not this pair of events.
neopolitan said:
The difference is like between these:

"I don't understand what you are saying, what are you saying"

and

"You are claiming this wrong thing, you are wrong"

when I never actually claimed the wrong thing you asserted that I claimed.
I never asserted that you claimed anything about the x=5 example, my point was that if you didn't have a clear idea of why interchangeability is supposed to imply one thing about your equations and something different about the equations I wrote, then the concept is too vague to make for a convincing demonstration that the factors should in fact be the same in your own pair of equations.
 
  • #433
Going back to this point:

neopolitan said:
It seems to me that we have two stopping points here.

First, the need to prove that there must be one single factor (as in my derivation shown earlier), rather than possibly two factors.

Second, generality, in that you think that if I chose two totally arbitrary events in order to to measure the interval between them in two frames, and that you feel that there is a "need for an explanation as to what specific property of the two events (I) chose ensures that the constants in those two equations will be the same".

Is this correct?

Do you further agree that, if I were to convince you that there being one single factor is the default, that feat would negate the need for a proof? I'm going to think about the proof angle anyway, but I am not abandoning my argument that what you are asking me to prove is actually the default.

I am going to assume that I have not convinced you that there being one single factor is the default.

That means we still have two stopping points, and I have to address both of them.

From here I suggest that I provide a general scenario, rather than one which is limited by having an event which is simultaneous with colocation of A and B in one frame, and work from there, with as much rigour as I can muster.

Included in that rigour is a requirement to prove, by elimination of other options if necessary, that a single factor applies in the intermediate step which we have been discussing.

Is that right?

cheers,

neopolitan
 
  • #434
neopolitan said:
I am going to assume that I have not convinced you that there being one single factor is the default.
Not sure what you mean by "default". Certainly if someone had no familiarity with SR and saw your two equations, if forced to guess, they might by default choose the same factor in each. But if this is supposed to be a derivation you need to establish that it must be the same factor.
neopolitan said:
From here I suggest that I provide a general scenario, rather than one which is limited by having an event which is simultaneous with colocation of A and B in one frame, and work from there, with as much rigour as I can muster.
Neither of the events in your two equations is simultaneous with colocation of A and B. They're just two events on the path of a totally arbitrary light beam that crosses paths with A and B.
neopolitan said:
Included in that rigour is a requirement to prove, by elimination of other options if necessary, that a single factor applies in the intermediate step which we have been discussing.
Yes, that's what must be proved.
 
  • #435
JesseM said:
Neither of the events in your two equations is simultaneous with colocation of A and B. They're just two events on the path of a totally arbitrary light beam that crosses paths with A and B.

This statement means that you clearly have not grasped something. It's good to know this now, rather than later.

The light beam is totally arbitrary, because I intended a general proof.

Take a totally arbitrary light beam, let it pass A and B (order not really important), then select an event on that light beam (I didn't make it totally arbitrary, I restricted it to being simultaneous with colocation of A and B in one frame).

You then have four events:

colocation of A and B, at (0,0) - ie x=0,t=0 and x'=0,t'=0
the event I just talked about selecting
colocation of the light beam and A
colocation of the light beam and B

With these four events (plus the laws of physics) you have enough information to derive the Lorentz Transforms (or analogues thereof).

To make my proof more general, I need to select from the events along the light beam an event which is not necessarily simultaneous with (0,0) in either frame.

You don't seem to like me using the intervals (0,0)-event in one frame and (0,0)-event in the other frame - which are the coordinates of the event in one frame and the coordinates of the event in the other frame (perhaps I haven't managed to make it abundantly clear that that is what I am doing, despite having done a bit of prepartory work on the issue of coordinates basically being intervals - latest attempt was here). Therefore, I will select from the events along another arbitrary light beam another event which is also not necessarily simultaneous with (0,0) in either frame. Then I will do the derivation based on a totally arbitrary spacetime interval.

cheers,

neopolitan
 
  • #436
neopolitan said:
JesseM said:
Neither of the events in your two equations is simultaneous with colocation of A and B. They're just two events on the path of a totally arbitrary light beam that crosses paths with A and B.
This statement means that you clearly have not grasped something. It's good to know this now, rather than later.

The light beam is totally arbitrary, because I intended a general proof.
What do you think I didn't grasp? Isn't that what I just said, "a totally arbitrary light beam"?
neopolitan said:
You then have four events:
Yes, in your overall derivation you have four events. But in those two equations where you say the factor is the same, only two events appear. Can we please try to stick to the subject of those two equations and the factor in them, unless you are willing to actually acknowledge that your previous "interchangeability" argument isn't sufficient to establish that the factor should be the same?
neopolitan said:
You don't seem to like me using the intervals (0,0)-event in one frame and (0,0)-event in the other frame
Where did you like the idea that I don't like that? In post 430 I wrote:
The event (0,0) is not referred to in the two equations where you assert the factors will be the same. The two events referred to in those equations of yours are the colocation of A and the photon (at x=0,t=ta in A's frame) and the colocation of B and the photon (at x'=0, t=t'b in B's frame). Of course the time coordinate of either event in a given frame can be understood as the time interval between that event and (0,0), but then exactly is true about the time coordinate of either event of the photon crossing the x=5 axis of one of the frames. Perhaps there is something in what you mean by "interchangeable" that equations involving the time coordinates of a pair of events in both frames can only be considered interchangeable if the two events happened on the worldlines of observers at rest in each frame who crossed paths at (0,0), but if so nothing you have written about interchangeability so far even hinted at such a requirement.
If you're talking more generally about the final equations you derive, my objection is not that you're talking about intervals from (0,0) to some other event, it's that for different variables in those equations the "other event" is different, whereas in the Lorentz transformation equations every variable is supposed to refer to the space and time intervals between the same pair of events in different frames (or the space and time coordinates of the same single event in different frames).
 
  • #437
JesseM said:
What do you think I didn't grasp? Isn't that what I just said, "a totally arbitrary light beam"?

I'm sorry, I was taking your comment to have been meant dismissively. If that was not your intention then ok.

Look back at the whole post (I know you have read it all before responding). I said:

neopolitan said:
Take a totally arbitrary light beam, let it pass A and B (order not really important), then select an event on that light beam (I didn't make it totally arbitrary, I restricted it to being simultaneous with colocation of A and B in one frame).

You then have four events:

colocation of A and B, at (0,0) - ie x=0,t=0 and x'=0,t'=0
the event I just talked about selecting
colocation of the light beam and A
colocation of the light beam and B

With these four events (plus the laws of physics) you have enough information to derive the Lorentz Transforms (or analogues thereof).

To make my proof more general, I need to select from the events along the light beam an event which is not necessarily simultaneous with (0,0) in either frame.

I start with "the event", the other (implied) event which allows me to obtain the coordinates of this event is both frames is what I call an "agreed" event, ie the colocation of A and B at (0,0).

The other two events (colocation of A and the photon) and (colocation of B and the photon) are used for the information we can glean from them, namely the time elapsed when they occur since A and B were colocated.

I'm not probably going to get this across properly right now, but I am glad to have a little better insight into what has not been conveyed correctly.

cheers,

neopolitan
 
  • #438
neopolitan said:
Look back at the whole post (I know you have read it all before responding).
Ugh, please don't make comments like this, it comes across as patronizing. Especially when you have a history of imagining I didn't get some point of yours because I "didn't read the whole post" or "broke it up into pieces" when in fact the issue was your idiosyncratic communication style where you seem to have expected others to interpret some random phrase in exactly the way you were thinking of without your having to actually explain your meaning (again think of the 'number of ticks' thing I responded to at the end of post 413)
neopolitan said:
I start with "the event", the other (implied) event which allows me to obtain the coordinates of this event is both frames is what I call an "agreed" event, ie the colocation of A and B at (0,0).

The other two events (colocation of A and the photon) and (colocation of B and the photon) are used for the information we can glean from them, namely the time elapsed when they occur since A and B were colocated.
Here you just seem to be throwing out random statements about your derivation (statements which I think I already understood, unless I am misinterpreting your words in some way), without explaining their relevance to what we are actually discussing. Is this supposed to somehow address either my point that only two events appear in the pair of equations where you said the factor is the same, or my point that the variables in your final equations don't all refer to a consistent pair of events?
 
  • #439
JesseM said:
Here you just seem to be throwing out random statements about your derivation (statements which I think I already understood, unless I am misinterpreting your words in some way), without explaining their relevance to what we are actually discussing. Is this supposed to somehow address either my point that only two events appear in the pair of equations where you said the factor is the same, or my point that the variables in your final equations don't all refer to a consistent pair of events?

I took one section of your post #434 and addressed it in post #435. Here:

neopolitan said:
JesseM said:
Neither of the events in your two equations is simultaneous with colocation of A and B. They're just two events on the path of a totally arbitrary light beam that crosses paths with A and B.

In that section, you specifically refer to two events twice (I accept that I may be reading "neither" a little too strictly).

In second part of post #437, I go over what two events I am talking about, which is (one event on the path of a totally arbitrary light beam) and (the colocation of A and B). I also point out that I use information associated with two more events which are also on the path of the totally arbitrary light beam.

All up that makes three events on the path of the totally arbitrary light beam.

Since I am not talking about two events on the path of the light beam, it is clear to me that something is amiss if you think I am talking about two events on the light beam. I could agree with one or three and I might even understand what you were talking about if you said four.

I will do my best to provide a derivation in which the consistency of the pair of events is very clear. Are we ready to proceed to that?

cheers,

neopolitan
 
  • #440
neopolitan said:
I took one section of your post #434 and addressed it in post #435. Here:
JesseM said:
Neither of the events in your two equations is simultaneous with colocation of A and B. They're just two events on the path of a totally arbitrary light beam that crosses paths with A and B.
Question: what specific equations do you imagine I was referring to when I said "in your two equations" in that post? (see below if it wasn't clear)
neopolitan said:
In that section, you specifically refer to two events twice (I accept that I may be reading "neither" a little too strictly).

In second part of post #437, I go over what two events I am talking about, which is (one event on the path of a totally arbitrary light beam) and (the colocation of A and B). I also point out that I use information associated with two more events which are also on the path of the totally arbitrary light beam.

All up that makes three events on the path of the totally arbitrary light beam.

Since I am not talking about two events on the path of the light beam, it is clear to me that something is amiss if you think I am talking about two events on the light beam. I could agree with one or three and I might even understand what you were talking about if you said four.
Let me just repeat my comment from post 432 again:
I'm not talking about how many colocation events appear in your overall derivation, I'm just talking about the two events that appear in these two equations you wrote earlier:

(time of colocation of B and photon in the A frame) = (some factor) * (time of colocation of B and photon in the B frame)

and

(time of colocation of A and photon in the B frame) = (some factor) * (time of colocation of A and photon in the A frame)

Can you please stop your mind from wandering all over the place and imagining I am talking about all these other aspects of your derivation, when I think I made it pretty clear I'm just talking about the specific question of how in these two equations you intend to show that the factor must be the same? If you don't in fact believe that your notion of "interchangeability" is sufficient to show this specific thing, and you haven't actually thought of a good way to show this yet, then that's fine, just say so. But if you do, then please stick to this specific topic for now.
Are you claiming that more than two events appear in those two specific equations? I only see "colocation of B and photon" on both sides of the first equation, and "colocation of A and photon" on both sides of the second (and these equations are quoted from your own post 427).

Note that I had said something similar in the earlier post 430:
Huh? The event (0,0) is not referred to in the two equations where you assert the factors will be the same. The two events referred to in those equations of yours are the colocation of A and the photon (at x=0,t=ta in A's frame) and the colocation of B and the photon (at x'=0, t=t'b in B's frame).
...but you didn't seem to get it in your response which was why I elaborated in post 432. Then in post 436 I again repeated the fact that I was just talking about those two specific equations where you said the factor was supposed to be the same:
Yes, in your overall derivation you have four events. But in those two equations where you say the factor is the same, only two events appear. Can we please try to stick to the subject of those two equations and the factor in them, unless you are willing to actually acknowledge that your previous "interchangeability" argument isn't sufficient to establish that the factor should be the same?
And then in post 438 I said:
Is this supposed to somehow address either my point that only two events appear in the pair of equations where you said the factor is the same, or my point that the variables in your final equations don't all refer to a consistent pair of events?
...so I don't think the problem here is that I've failed to make clear I'm just talking about the events that appear in those specific equations.
neopolitan said:
I will do my best to provide a derivation in which the consistency of the pair of events is very clear. Are we ready to proceed to that?
Does "the consistency of the pair of events" mean the same thing as your earlier talk of "interchangeability"? If so please address my questions from post 430. If not, please tell me whether you are abandoning your earlier claims that "interchangeability" was sufficient to rigorously establish that the factor should be the same, or whether you stick by that claim.
 
  • #441
Realisation dawns. I think I am talking at cross purposes.

I think that you are talking about these equations:

neopolitan said:
(time of colocation of B and photon in the A frame) = (some factor) * (time of colocation of B and photon in the B frame)

and

(time of colocation of A and photon in the B frame) = (some factor) * (time of colocation of A and photon in the A frame)

In terms of the scenario, this does work. If you change the scenario, it won't work.

It requires that there is a colocation of A and B at (x,t)=(0,0), (x',t')=(0,0).

In my recent emails, I wasn't talking about those pair of events or those equations, I was talking about the pair of events that the final equations refer to (the event whose coordinates we have in one frame and want to find in the other frame, and the colocation of A and B). This is entirely my fault.

I really hope this is the source of confusion, because otherwise I am completely stumped as to what the issue is.

"Interchangeability" is not clearing things up. I know what I mean, but I seem unable to convey that understanding to anyone else. Therefore it is not a useful concept. Therefore while I don't abandon the claim, I don't intend to use the concept.

cheers,

neopolitan
 
  • #442
neopolitan said:
In terms of the scenario, this does work. If you change the scenario, it won't work.

It requires that there is a colocation of A and B at (x,t)=(0,0), (x',t')=(0,0).

In my recent emails, I wasn't talking about those pair of events or those equations, I was talking about the pair of events that the final equations refer to (the event whose coordinates we have in one frame and want to find in the other frame, and the colocation of A and B). This is entirely my fault.

I really hope this is the source of confusion, because otherwise I am completely stumped as to what the issue is.
Yes, that's it, I think we've finally cleared up the miscommunication here (whew!)
neopolitan said:
"Interchangeability" is not clearing things up. I know what I mean, but I seem unable to convey that understanding to anyone else. Therefore it is not a useful concept. Therefore while I don't abandon the claim, I don't intend to use the concept.
OK, fair enough.
 
  • #443
I've been rather busy, but trying to look at this during the quiet moments.

http://www.geocities.com/neopolitonian/checking.html"is an attempt at a proof for one of the stopping points.

cheers,

neopolitan

PS I previously only posted the proof which attempts to justify my "some factor" assumption (where "some factor" is G1 = G2). It was part of a longer thing which I put together in word and tried to save to pdf. The distiller I have here resulted in a 8M pdf file which I was not happy with. I'll have a go later with another distiller I have at home.

Zipped, the size of that file becomes 2M. If you can handle a file that large, it is http://www.geocities.com/neopolitonian/benefits.part1.v1.zip".
 
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  • #444
Referring to last post and the http://www.geocities.com/neopolitonian/checking.html" therein, I noted that a line was missing from one of the text images (I had fixed it in a later variant, but it is on a work computer). Until I can fix that, I have added a line in very ordinary hypertext script.

For JesseM, the proof which may be of more interest is at the end, following the title "Testing hypothesis that A and B measure space differently".

cheers,

neopolitan
 
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  • #445
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