Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PeteP » Mon Nov 11, 2013 11:14 pm UTC

steve waterman wrote:
JudeMorrigan wrote:That's not how anyone else on the planet is using it though. If you use it in the manner I've described, you can answer a useful question. (If I know the coordinates of a point in one coordinate system [note, the coordinates are what are in the coordinate system, not the point itself] and the relationship of a second coordinate system to the first, can I find the coordinates of the point in the second coordinate system?) Your definitions simply don't do anything useful.

Sigh...still you insist on points not having equal coordinates in the manifold and elect to ignore equal abscissa. These are not MY definition for abscissa but rather THE accepted mathematical definition(s) for abscissa. Again I agree that point P has a different set of coordinate values regarding S and S'...which is irrelevant to the fact that the abscissa lengths are equal after vt. For 6000 posts I hear that the coordinate values for point P are different in the manifold. For about the 100th time...I totally agree, yes that is 100 percent true!
also have grasped from page one that your points do not move when a system moves.

I am however, not talking about the relationship of the coordinates of point P, rather I am talking about x and x' the abscissa being the distance from their own origin. x is NOT a damn point, x is the abscissa of S. Do you deny that after vt, the abscissa x of S = the abscissa x' of S'????? Taking a posting break until tomorrow.

I repeat the abscissa of which point? Abscissa is sometimes also used for the horizontal axis itself, but then it wouldn't have a value since it's an axis. So it has to be the abscissa of a point. The point doesn't change, x doesn't change therefore x' changes. You agree that the coordinates of S'(x',y',z') change, therefore you agree that x' changes, because it's the abscissa of S'(x',y',z').

Yay steve agrees all further objections from him can only be considered lies!

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Mon Nov 11, 2013 11:38 pm UTC

steve waterman wrote:
JudeMorrigan wrote:That's not how anyone else on the planet is using it though. If you use it in the manner I've described, you can answer a useful question. (If I know the coordinates of a point in one coordinate system [note, the coordinates are what are in the coordinate system, not the point itself] and the relationship of a second coordinate system to the first, can I find the coordinates of the point in the second coordinate system?) Your definitions simply don't do anything useful.

Sigh...still you insist on points not having equal coordinates in the manifold and elect to ignore equal abscissa. These are not MY definition for abscissa but rather THE accepted mathematical definition(s) for abscissa. Again I agree that point P has a different set of coordinate values regarding S and S'...which is irrelevant to the fact that the abscissa lengths are equal after vt. For 6000 posts I hear that the coordinate values for point P are different in the manifold. For about the 100th time...I totally agree, yes that is 100 percent true!
also have grasped from page one that your points do not move when a system moves.

I am however, not talking about the relationship of the coordinates of point P, rather I am talking about x and x' the abscissa being the distance from their own origin. x is NOT a damn point, x is the abscissa of S. Do you deny that after vt, the abscissa x of S = the abscissa x' of S'????? Taking a posting break until tomorrow.

Say, let's look at a couple of those citations you provided.

http://www.mathsisfun.com/definitions/abscissa.html

The horizontal ("x") value in a pair of coordinates. How far along the point is.


http://dictionary.reference.com/browse/abscissa

the x-coordinate of a point


http://www.icoachmath.com/math_dictionary/abscissa.html

In the "Examples of abscissa" section, we have:
The ordered pair of point P is (1, 4). The abscissa of the ordered pair (1, 4) is 1.
The abscissa of the point (2, 5) is 2.

(emphasis added in all cases)

The "of a point" part is *really* important.

Since you agree that "point P has a different set of coordinate values regarding S and S'", surely you can see that the abscissa of the point in each coordinate frame will be different?

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Tue Nov 12, 2013 1:50 am UTC

ahammel wrote:Steve: the discussion behind the spoiler is entirely for my own edification. You may defer opening until after cyanyoshi has responded or, if you prefer, indefinitely.

Spoiler:
Just checking to see if I understand coordinate transforms any better than Steve by now.

steve waterman wrote:Given S(x,y,z)...x = distance from S(0,0,0) to S(x,0,0) and
Given S'(x',y',z')...x' = distance from S'(0,0,0) to S'(x',0,0) and
Wait, didn't he just define x the distance in terms of x element of a coordinate? Is that kosher? It's certainly potentially confusing. I'm going to go ahead and call the distances p and q.

(By the way, if we've got a coordinate, S(x,y), is it correct to say that x is an element of that coordinate, or is there some other term for that relationship?)

Given S and S' coincident with x = x'.

Allow S' or S to move by vt along the common x/x' axis[...]
It's a little weird to say that S and S' are coincident and then 'move' one of them, but OK. I guess that's Steve's way of telling us that the two coordinate systems are related by a translation rather than a shear.

Don't know why we we're representing a constant distance by what looks like two previously undefined things multiplied by one another either, but I'll assume that vt is some constant.

in which case, it is absolutely obvious that [p] = [q].
Looks correct to me. A translation between coordinate systems doesn't change the scaling of the axes.

Therefore, one can deduce that x' = x-vt is mathematically incorrect unless vt = 0, since [p] = [q].
I think swapping out the symbols adequately highlights the equivocation. Steve's result is about distances, and coordinate transformations, uh, transform coordinates.

Just because this disagrees with the Galilean transformation equation results does not make my logic invalid. Just because point P in your manifold shares the same spatial location has zero impact upon the absolute fact that [p] = [q].
In other words, this thing that Steve's concluded has nothing much to do with coordinate transformations, except with regards to what happens to the scaling of the axes when you perform a translation (viz. nothing).

I do not comprehend why this simple logic is sooo difficult for everyone here at xkcd to grasp/agree with.
Everybody grasps it just fine, it's just that Steve is convinced he's knocked over special relativity, is wrong, and interprets any attempt at correction as a misunderstanding of his argument.

Prove to me that [p] does not equal [q] after vt gets applied using the definitions above for [p] and [q], or at least explain why you disagree with that definition for [p] and [q].
As above, p and q are distances, x and x' are elements of coordinates. Coordinate transforms are about the relationship between coordinates identifying the same points in different coordinate systems, and results about distances are irrelevant.

Also, SR doesn't even use the Galilean transformation, so why are we even talking about this?

It looks to me that you've got it. We can define coordinates without having a notion of distance, so your separation of the coordinates x and x' from the distances p and q is exactly the right thing to do.

If only Steve would take the time to understand your post....
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Tue Nov 12, 2013 2:05 am UTC

Schrollini wrote:We can define coordinates without having a notion of distance,


I'm a little curious about this statement. I wholeheartedly agree we can define coordinates without a set-in-stone metric, but "notion" seems like a strong statement. Isn't "distance", at the most fundamental level, the notion of "this point is not the same as that point"?

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ahammel » Tue Nov 12, 2013 3:37 am UTC

WibblyWobbly wrote:
Schrollini wrote:We can define coordinates without having a notion of distance,


I'm a little curious about this statement. I wholeheartedly agree we can define coordinates without a set-in-stone metric, but "notion" seems like a strong statement. Isn't "distance", at the most fundamental level, the notion of "this point is not the same as that point"?
I guess (and 'guess' is the word) that you need to have a concept of the difference between zero and nonzero distance (i.e., this point is not that point), but you don't need to distinguish between different nonzero distances (i.e., you don't need to know whether (2,3) is further from (0,-1) than (4,pi). Or whether p=q, for that matter.)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Tue Nov 12, 2013 3:49 am UTC

WibblyWobbly wrote:
Schrollini wrote:We can define coordinates without having a notion of distance,


I'm a little curious about this statement. I wholeheartedly agree we can define coordinates without a set-in-stone metric, but "notion" seems like a strong statement. Isn't "distance", at the most fundamental level, the notion of "this point is not the same as that point"?

I'm using "distance" in the mathematical sense of "the thing that is measured by a metric". From the definition of a metric, you can see it's slightly more than "this point is not the same as that one". Notably, the metric must satisfy the triangle inequality d(x, z) ≤ d(x, y) + d(y, z).

As Dark Avorian explained, a manifold is a topological sp​ace, so it does have a concept of "nearness" built-in. This concept of "nearness" may satisfy your notions of "distance", but it's not enough for the notion defined by a metric. There's no way to turn it into a distance function.

Once you have a coordinate system on your manifold, it's easy to define metric on it. (Well, I'm ignoring issues that may arise if your coordinate system doesn't cover the whole manifold.) Several metrics, in fact, which may disagree on which points are closer to each other. None is preferred based on the manifold or the coordinate system, which is another way of saying that the coordinate system doesn't have an inherent concept of distance.

Disclaimer: This is all based on a physicist's understanding of math. The broad strokes are right, but there may be details I've screwed up, especially about topological spaces.

Edit to clarify, or possibly confuse:
The discrete metric d(x,y) = {0 iff x=y; 1 otherwise} exactly captures "this point is not the same as that point" without introducing anything else, so WibblyWobbly's basic description is in fact enough to define a metric. My statements above about the triangle inequality are irrelevant.

But it's worth noting that the discrete metric doesn't use a coordinate system, and it doesn't use any properties of the manifold. It can be defined on any set. So while we can define a metric just from a manifold without any additional input, we could do that with any set.

I don't know the proper way to say this in math, but the discrete metric doesn't "respect" the topology of the manifold. The "nearness" of points isn't reflected in the distance function. Or put another way, the discrete metric suggests the discrete topology, not the manifold topology. So there is something that the manifold is telling us about "acceptable" metrics. Hopefully a real mathematician will wander by and tell me what I mean.

In any event, none of this depends on the coordinate system, nor does the coordinate system require us to pick a metric, which is the point I so inelegantly tried to make.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby cyanyoshi » Tue Nov 12, 2013 8:28 am UTC

Wow. Things sure exploded. You'd think I was gone for a week.

It would be nice if you at least acknowledged that you have read my post before wanting to have a discussion, Steve. But sure, I'll bite. Just a fair warning: I may not be able to reply to you right away, and my answers may disappoint you. Also I'm not the kind to beat my head against the wall very much, so don't be surprised if I drop out this thread for good if it becomes clear that you expect me to listen to you without you doing the same for me. In the meantime, however, I'll gladly weigh in on your thoughts.
steve waterman wrote:1 Cartesian coordinate systems existed in the 1800's, thus existed before the Galilean transformation had been created/conceived.
Agreed?

This is probably true, but it is irrelevant.
2 A Cartesian coordinate system contains an infinite set of points, including the point (0,0,0) called the origin.
Agreed?

As you wrote it, no. This is because (0,0,0) is not a point; it is an ordered list of numbers (often called a [url="http://en.wikipedia.org/wiki/Tuple"tuple[/url]). To use the analogy I put forth, you must not confuse the measurement "zero inches" with a dot on the ground. There is a one-to-one correlation, sure, but that does not mean they are the same thing. In most cases this would be me just nitpicking, but it is absolutely crucial to understand.

But you are partially right. A three-dimensional Cartesian coordinate system implies that there are an infinite number of points, and the point corresponding to the tuple (0,0,0) is typically called the origin. That does not mean that the origin is special in any way, though.
3 Given coincident systems S and S',
mathematically allow a point P at (2,0,0) in/wrt S and also allow a point P' at (2,0,0) in/wrt S'.
Allowed?

From here on out, my statement that a tuple and point are two separate things still stands. It is just pointless for me to repeat myself. Back to the question though: If two coordinate systems are coincident, then the same coordinates in each coordinate system would both correspond to the same point. Calling one P and the other P' implies that they are not the same point. You can just call them both P because they are one in the same.

steve waterman wrote:cyanyoshi,

In an attempt for you and I to have a valid discussion, I am initially trying to establish a working scenario for the terms "point" and "coordinate system" that we can mutually agree upon.

That's fine by me. May I then suggest following accepted naming conventions? It's why they are there, after all: to let people communicate their ideas with each other.
4 Given a Cartesian coordinate system with a line drawn between point A(1,1) and point B(-1,-1).
Allowed? ( noting that there is only just the one system in this mathematical statement )

This is precisely why we need to get on the same page notation-wise. Are A and B functions? If so, then how are they defined? Are you simply naming the points that correspond to the tuples (1,1) and (-1,-1)? We are speaking two different languages.

You can certainly draw a line between any two distinct points in a Cartesian coordinate system, if that's what you are asking.
5 IFF number 4 is allowed, we could make a distinction between
the "coordinate point" (1,1) [ btw, I am quite aware that Relativity has no such term ]
and the "named point" A(1,1).
Do you allow these two differentiating terms, cyanyoshi, for the sake of our discussion?

Sorry, but unless you provide a rigorous definition of their meanings, then there is no chance of me using those words. I can't read your mind. It makes much more sense for us to use only words that we can look up the meaning of from an outside source (like Wikipedia, or a dictionary). How else can I we determine the truth of a statement? If however you want to try and give us definitions or suitable metaphors for those terms, then I'm all ears.
I will skip to the bottom line...

Given S(x,y,z)...x = distance from S(0,0,0) to S(x,0,0) and
Given S'(x',y',z')...x' = distance from S'(0,0,0) to S'(x',0,0) and
Given S and S' coincident with x = x'.

Allow S' or S to move by vt along the common x/x' axis, in which case, it is absolutely obvious that x = x'.
Therefore, one can deduce that x' = x-vt is mathematically incorrect unless vt = 0, since x = x'.

Just because this disagrees with the Galilean transformation equation results does not make my logic invalid. Just because point P in your manifold shares the same spatial location has zero impact upon the absolute fact that x = x'.

I do not comprehend why this simple logic is sooo difficult for everyone here at xkcd to grasp/agree with.
Prove to me that x does not equal x' after vt gets applied using the definitions above for x and x', or at least explain why you disagree with that definition for x and x'.

Okay, straight to the good stuff (for some definition of "good"). Read my original post. Seriously, read it and try to visualize what is going on with the dot and the rulers. Have you read it yet? No? What a shame.

I will still address your post, even though you might not like it. You start off by defining coincident coordinate systems where x=x'. Then you say that since x=x' that the Galilean transformation must be wrong? That's not how that works, no matter how much you want to believe it. More to the point, the Galilean transformation is applicable to certain pairs of coordinate systems that are not coincident. Your analysis is not. Have you seriously considered that people might have issue with your logic not because they are close minded, but rather because your conclusion does not logically follow from your premise?

Now your turn. What part of my logic do you find fault with in my other post, the one with the colored rulers?
btw, combining these two threads on this topic, has now reached an unbelievable 6000 posts.

unbelievable indeed
steve waterman wrote:
JudeMorrigan wrote:Groovy. Now why should the distance from S(0,0,0) to S(x,0,0) be the same as the distance from S'(0,0,0) to S'(x',0,0) if the axes are not coincident?
Simply because is was the Galilean given that at t = 0, x = x' so AFTER vt gets applied, x still equals x'.

You know what, I think we're done here. It is abundantly clear that you have no interest in understanding what the Galilean transformation is referring to, or basic logic for that matter. Dozens have come before me and dozens have failed, and the time has come for me to join their likes. Take care of yourself, Steve.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Carlington » Tue Nov 12, 2013 11:58 am UTC

At Steve's nearest convenience, and provided that nobody in this thread has a problem with it, I'd like to try explaining this with an analogy that may be more relevant to Steve's field of expertise, and thus Steve might hopefully find it a little easier to digest.
Steve, everyone else, does this sound acceptable?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PeteP » Tue Nov 12, 2013 12:05 pm UTC

Carlington wrote:At Steve's nearest convenience, and provided that nobody in this thread has a problem with it, I'd like to try explaining this with an analogy that may be more relevant to Steve's field of expertise, and thus Steve might hopefully find it a little easier to digest.
Steve, everyone else, does this sound acceptable?

Just as a warning I believe something like that has already been done. Well have fun.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Chen » Tue Nov 12, 2013 1:21 pm UTC

steve waterman wrote:Drawing 1
Spoiler:
Image


Drawing 2
Spoiler:
Image



Lets go with the definitions you have here. Lets put a point down, in space, called P with coordinates called S(a,0), S'(a',0). In the first picture lets call a = 2, a' = 2. So my point is defined as S(2,0)=S'(2,0).

Now lets move on to the second diagram. The generalization of the point remains the same S(a,0), S'(a',0), however since the point has not moved (only the S' frame moved) we can agree that S(a,0) = S' (a',0). Since S did not move, we still have a = 2. However to retain the equality S(a,0) = S'(a',0), a' MUST change. In this case a' = -1.

If we were to continue with another drawing where S' (0,0,0) was 2 more units over, a' = -3 to retain the above equality. As such we can construct a relationship here. It appears that a' = a - d, where d is the distance moved by the frame a'. That is all the Galilean is trying to say.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Tue Nov 12, 2013 1:54 pm UTC

cyantoshi,
Yes, I agree with your statement about our discussion being over.

Jude,
I am the one stating that a coordinate system contains points. Relativity says that points only exist in the manifold.
So, your highlighting that the term abscissa refers to some point that represents the distance from (0,0,0) agrees with me and disagrees with Relativity.

All,
I believe I am done explaining and justifying and asking for agreement and now I elect to resort to pontificating.

"Given Cartesian systems S(x,y,z) coincident with S'(x',y',z') where the abscissa x = the abscissa x', then regardless of the manner to manifest a condition wherein S and S' are no longer coincident, it remains true that the abscissa x = the abscissa x'." steve waterman November 2013

There is no more logic for me to discuss as I will never be changing my mind about the inherent logic of this pontification. I see no reason for me to plow through more analogies, or depictions, or try to agree upon what "point" means or what a "coordinate system" is. I am done dickering about the point relationship in the manifold as it has absolutely no impact upon the above pontification. I certainly can assume that not one poster here is in agreement with this pontification, and that posters will continue to insist that the point relationship in the manifold gets represented by x' = x-vt.

So to be clear, I will continue to repetitively spout this singular pontification as mathematically true. I no longer am willing to banter about what any these following terms/phrases mean; Cartesian system, coordinates, tuples, points, manifold, transformation, equality, shared spatial location , distance, abscissa, ordinate, applicate, distance from their own origin, point P, point P', (x-vt,y,z), Galilean coordinate transformation, Relativity, time dilation, length contraction, Lorentz transformation, Albert Einstein, Woldemar Voigt nor anything else even remotely connected with Relativity.

Indeed, the above singular pontification is my bottom line. As well, I am not going to be available to supply any future logic regarding the singular pontification. I am simply placing my singular pontification on record. All that i am prepared to do on this thread going forward, is to repeat the pontification, when their is objection to it.

I repeat this above statement for emphasis...
"Given Cartesian systems S(x,y,z) coincident with S'(x',y',z') where the abscissa x = the abscissa x', then regardless of the manner to manifest a condition wherein S and S' are no longer coincident, it remains true that the abscissa x = the abscissa x'."
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Chen » Tue Nov 12, 2013 2:06 pm UTC

steve waterman wrote:I repeat this above statement for emphasis...
"Given Cartesian systems S(x,y,z) coincident with S'(x',y',z') where the abscissa x = the abscissa x', then regardless of the manner to manifest a condition wherein S and S' are no longer coincident, it remains true that the abscissa x = the abscissa x'."


Despite any possible subtle problems with that sentence, I don't think anyone is arguing that the point you're attempting to make is wrong with respect to distances and moving rulers and such. We ARE arguing that it has absolutely nothing to do with the Galilean though.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PeteP » Tue Nov 12, 2013 2:13 pm UTC

I would say he misunderstands what abscissa means. Or at least pretends to.

"The 12 November was a historical day. On this day steve waterman finally realized that since his viewpoint wasn't based on logic it was useless to defend it with logic. Continuing to pretend that he wasn't just intellectually dishonest and never had the intention to actually think about his beloved pet theory, would just serve to further frustrate him and those who had the vain hope that given enough different viewpoints he would actually be forced to actually think instead of willfully misunderstanding and twisting terms.
And so he decided to do openly, what he had already done, under the guise of discussion, for hundreds of posts. To just spam his beliefs without justification since there wasn't one. Nor had there ever been one.
At last it had ended."
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Tue Nov 12, 2013 2:22 pm UTC

steve waterman wrote:Jude,
I am the one stating that a coordinate system contains points. Relativity says that points only exist in the manifold.
So, your highlighting that the term abscissa refers to some point that represents the distance from (0,0,0) agrees with me and disagrees with Relativity.

No, it doesn't. You're wildly misinterpreting those quotes.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Tue Nov 12, 2013 3:21 pm UTC

There are two reasons steve waterman is having trouble following this.
1. We keep using symbols, and steve has trouble keeping track of what the symbols stand for. We need to use as many concrete numbers as possible FIRST, and come up with several specific examples that we can agree on FIRST.

2. We keep using only spacial coordinates. This is a bad idea because then we start talking about one coordinate system "moving" on top of another, which gets confusing for steve.

Steve, this is an important point that you MUST address for us to move forward. In your diagram with t and t' equal and non-zero
Spoiler:
Image

do you agree that the S(2,0,0,t) and S'(2,0,0,t) are NOT mapping to the same point?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Tue Nov 12, 2013 4:23 pm UTC

All,
If this does not become too tedious, I will PM responses directly....from some of your posts that are made to the thread.
However, I do not desire to have those PM messages posted to the thread.
If that happens, then I will simply stop writing any Private Message's.
btw, the two threads combined just passed 170,000 total views.

"Given Cartesian systems S(x,y,z) coincident with S'(x',y',z') where the abscissa x = the abscissa x',
then regardless of the manner to manifest a condition wherein S and S' are no longer coincident,
it remains true that the abscissa x = the abscissa x'."
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Tue Nov 12, 2013 4:23 pm UTC

Роберт wrote:There are two reasons steve waterman is having trouble following this.

You're forgetting the third reason.

steve waterman wrote:I only got into Physics following the path of the ccp, as I continue to do. I kept getting confronted with
"you cannot say that because of relativity", in numerous meetings at mcGill. Finally I thought, well I have no oyher choice do I, let's check it out.

I had a book with in the transformations with lorentz too, so I started with x' = x-vt, as it does...and worked out ifthe math was okay in each equation...it was. So i concluded that the only chance of it being wrong would have to be that equation. So, my search, quite luckily, started there. Whitle, whittle, whitle,...brings me here to xkcd, now.

viewtopic.php?f=2&t=96231&p=3039321#p3039321

steve waterman wrote:Here is the deal.

My sole goal is to mathematically not allow the equation x' = x-vt to represent equivalence

viewtopic.php?f=2&t=96231&start=720#p3042221

That is, that Steve doesn't want to follow it.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Tue Nov 12, 2013 4:32 pm UTC

steve waterman wrote:All,
If this does not become too tedious, I will PM responses directly....from some of your posts that are made to the thread.
However, I do not desire to have those PM messages posted to the thread.
If that happens, then I will simply stop writing any Private Message's.
btw, the two threads combined just passed 170,000 total views.

"Given Cartesian systems S(x,y,z) coincident with S'(x',y',z') where the abscissa x = the abscissa x',
then regardless of the manner to manifest a condition wherein S and S' are no longer coincident,
it remains true that the abscissa x = the abscissa x'."

Steve, you need a new diagram for us to be discussing, since you agreed to me that your diagram doesn't apply to the Galilean. Can you please select and post a new diagram for us to discuss that does apply to the Galilean?

And please, stop doing PMs. It's nonsense and will make everything way more inefficient. I let your recent PM slide, but any further PMs I will post in this thread. (And I suggest others do likewise so that we can understand what has been discussed.)

We are waiting for you to post a new diagram to tell use what you are talking about. It needs real values so we can discuss. If you insist on using x and x', make sure that the x and x' meet the situation that galilean is talking about. That is, your given values for your coordinates need to be such that S(x,y,z,t) and S'(x',y',z',t') map to the same event.

Thank you. I look forward to seeing your diagram.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Tue Nov 12, 2013 4:48 pm UTC

Роберт wrote:
steve waterman wrote:All,
If this does not become too tedious, I will PM responses directly....from some of your posts that are made to the thread.
However, I do not desire to have those PM messages posted to the thread.
If that happens, then I will simply stop writing any Private Message's.
btw, the two threads combined just passed 170,000 total views.

"Given Cartesian systems S(x,y,z) coincident with S'(x',y',z') where the abscissa x = the abscissa x',
then regardless of the manner to manifest a condition wherein S and S' are no longer coincident,
it remains true that the abscissa x = the abscissa x'."

Steve, you need a new diagram for us to be discussing, since you agreed to me that your diagram doesn't apply to the Galilean. Can you please select and post a new diagram for us to discuss that does apply to the Galilean?

And please, stop doing PMs. It's nonsense and will make everything way more inefficient. I let your recent PM slide, but any further PMs I will post in this thread. (And I suggest others do likewise so that we can understand what has been discussed.)

We are waiting for you to post a new diagram to tell use what you are talking about. It needs real values so we can discuss. If you insist on using x and x', make sure that the x and x' meet the situation that galilean is talking about. That is, your given values for your coordinates need to be such that S(x,y,z,t) and S'(x',y',z',t') map to the same event.

Thank you. I look forward to seeing your diagram.


Sorry, no more diagrams and I have posted my last PM to you. Since I only have read this above message after I made a second Private message to you, then I would see posting it here as not playing fair. Perhaps I should not offer any PM's at all, I will see how it goes. I suspect it may just as futile as the thread has been.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Tue Nov 12, 2013 5:13 pm UTC

JudeMorrigan wrote:
Роберт wrote:There are two reasons steve waterman is having trouble following this.

You're forgetting the third reason.

steve waterman wrote:I only got into Physics following the path of the ccp, as I continue to do. I kept getting confronted with
"you cannot say that because of relativity", in numerous meetings at mcGill. Finally I thought, well I have no oyher choice do I, let's check it out.

I had a book with in the transformations with lorentz too, so I started with x' = x-vt, as it does...and worked out ifthe math was okay in each equation...it was. So i concluded that the only chance of it being wrong would have to be that equation. So, my search, quite luckily, started there. Whitle, whittle, whitle,...brings me here to xkcd, now.

viewtopic.php?f=2&t=96231&p=3039321#p3039321

steve waterman wrote:Here is the deal.

My sole goal is to mathematically not allow the equation x' = x-vt to represent equivalence

viewtopic.php?f=2&t=96231&start=720#p3042221

That is, that Steve doesn't want to follow it.

Also relevant is this post, where Steve says, (1) the Galilean equation works for coordinates, but (2) the Galilean isn't supposed to be about coordinates, it's supposed to be about distances, but (3) it doesn't work for distances. So Steve understands and can come up with no objection to the Galilean-for-coordinates argument. Instead he invents a straw-man Galilean-for-distances argument, disproves it, and then tries to convince all of us that the Galilean-for-distances argument is the one we're supposed to be defending!

So there's really no use in trying to come up with a clever argument or diagram to explain the Galilean-for-coordinates -- Steve already gets it. Likewise, there's no use in disproving the Galilean-for-distances argument, since Steve's already done that. What we have to do is convince him that when we talk about the Galilean, we're always talking about the Galilean-for-coordinates. I fear this is an impossible task -- Steve has so much ego invested in his sphere packing model, which (in his mind) conflicts with relativity, which (in his mind) follows from the equation x' = x - vt, that he must prove this equation wrong, regardless of how. I suspect that this is an example of cognitive dissonance.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Tue Nov 12, 2013 5:15 pm UTC

steve waterman wrote:Sorry, no more diagrams and I have posted my last PM to you. Since I only have read this above message after I made a second Private message to you, then I would see posting it here as not playing fair.

"Playing fair?" What does that even mean? That's silly.

I had planned to go ahead and let my response PM be only a PM and not post it on this thread, but since you posted here to say you will no longer PM me, I'm going to go ahead and post my response here so it can be discussed:
Роберт wrote:
steve waterman wrote:That is a problem...what does x mean for you in respect to x' = x-vt?

You seem to disagree...
that given S(x,y,z), that x is the abscissa of S; aka the distance from S(0,0,0) to S(x,0,0,0),
that given S'(x',y',z'), that x' is the abscissa of S'; aka the distance from S'(0,0,0) to S(x',0,0,0).

This is exactly what I mean by you tripping up in the notation.

S(0,0,0,0) has a spacial distance distance of 0 from S(0,0,0,0)
S(1,0,0,0) has a spacial distance distance of 1 from S(0,0,0,0)
S(2,0,0,0) has a spacial distance distance of 2 from S(0,0,0,0)
Generalizing this, for a given value of x
S(x,0,0,t) has a spacial distance distance of x from S(0,0,0,t)
We can say the same for S' and x', for a given value of x'
S'(x',0,0,t') has a spacial distance distance of x' from S'(0,0,0,t')

How does this apply to the galilean? For this specific example, let v=2 and t=t'=3. Well, I know a point S(10,0,0,3) has a spacial distance of 10 from S(0,0,0,3). If I want to know how far that point is from the point S'(0,0,0,3), I the easiest way to do so is to S(10,0,0,3) into the S' coordinate system. I do so using the galilean. S(10,0,0,3) maps to the same event as S'(10-vt,0,0,3), which for this example is S'(10-2*3,0,0,3), which I can simplify to S'(4,0,0,3). Now we can find the spacial distance to S'(0,0,0,3).
S'(4,0,0,3) has a spacial distance distance of 4 from S'(0,0,0,3)

So here, in this specific example, x is 10, v is 2, and z is 3. We used that knowledge to find x'.
x'=10-2*3=10-6=4

So the spacial distance of the event we are talking about to S(0,0,0,3) is 10.
So the spacial distance of the event we are talking about to S'(0,0,0,3) is 4.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Tue Nov 12, 2013 5:38 pm UTC

I will respond to this; Schrollini's misunderstanding...

Schrollini wrote:[ I fear this is an impossible task -- Steve has so much ego invested in his sphere packing model, which (in his mind) conflicts with relativity,

Never said that, never will.
Sphere packing has nothing to do Relativity. You have wildly misrepresented what I said.
Schrollini wrote:which (in his mind) follows from the equation x' = x - vt,

Sphere packing has absolutely nothing to do with x' = x-vt as I have said numerous times before.
Schrollini wrote:I suspect that this is an example of cognitive dissonance.

This is more akin, to you putting words in my mouth that do not now, nor have ever represented my stance.

added -
I see now that offering to PM people is likely pure folly. I have changed my mind and will not be using the PM in the future. I am not discussing my pontification either, as I have said. That leaves me only to correct statements that have put words in my mouth that I have not said. I assume that this thread will quickly degenerate into mostly negative comments about me and my stance regarding my pontification and soon will be locked too.

Sexy Talon - anytime you wish to lock this thread, starting as of right now, please feel free to do so, as I would appreciate completely shutting down this thread now. I envision nothing productive going forward and likely just anger and frustration from both sides.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Tue Nov 12, 2013 6:56 pm UTC

I suggest we take his advice before he changes his mind ... again.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Tue Nov 12, 2013 7:14 pm UTC

steve waterman wrote:I only got into Physics following the path of the ccp, as I continue to do. I kept getting confronted with
"you cannot say that because of relativity", in numerous meetings at mcGill. Finally I thought, well I have no oyher choice do I, let's check it out.

["ccp" being an abbreviation for cubic close packed, another name for the face-centered cubic sphere packing.]

steve waterman wrote:Sphere packing has nothing to do Relativity.


steve waterman wrote:I had a book with in the transformations with lorentz too, so I started with x' = x-vt, as it does...and worked out ifthe math was okay in each equation...it was. So i concluded that the only chance of it being wrong would have to be that equation.


steve waterman wrote:Sphere packing has absolutely nothing to do with x' = x-vt as I have said numerous times before.


steve waterman wrote:It kinda logically all goes together for me.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Tue Nov 12, 2013 7:28 pm UTC

Yeah, it started with sphere packing, and then "relativity must be wrong" and the Pressures thread. Then, after realizing he couldn't assault that, it became the Galilean. Then, in the midst of realizing he couldn't assault that, it became "I AM NOT TALKING ABOUT THE GALILIEAN, I AM ONLY TALKING ABOUT x' = x - vt", and then "the Galilean says this" and "look at all these definitions of abscissa that I don't understand" and yadda yadda yadda. If I go to the top of the new 1 WTC1, I can probably see my house in Ohio, but I won't be able to see where Steve left the goalposts.


1 By the way, that spire on 1 WTC counting as an integral part of the building for purposes of height measurement? Give me a break.

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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Tue Nov 12, 2013 7:39 pm UTC

Steve, I'm glad you now agree with x'=x-vt as defined in the Galilean. For good closure, could you post it more clearly in this thread?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Tue Nov 12, 2013 7:44 pm UTC

Schrollini wrote:
steve waterman wrote:I only got into Physics following the path of the ccp, as I continue to do. I kept getting confronted with "you cannot say that because of relativity", in numerous meetings at McGill. Finally I thought, well I have no other choice do I, let's check it out.

["ccp" being an abbreviation for cubic close packed, another name for the face-centered cubic sphere packing.]


I did not say that the ccp was connected to Relativity. I said that I continue to follow the path of the ccp. You mistook that by saying that I was getting into Physics came from following the path of the ccp. I can see how that wording could be unclear and one might think that I was saying that there was some sort of connection between the ccp and Relativity.
"Let's check it out", only meant that that I would read up on Relativity to try and discern its parameters. So, I looked into Relativity because Mcgill professor's said that my ccp mathematics could not be correct due to the realities of Relativity. As I had had the math proven for ccp by various others, I felt forced into Physics and into checking out what Relativity was saying.

So, I hope this post straightens out why I see no connection of any sort between the mathematics of sphere packing and the Physics of Relativity...as there is none, and both concerns are completely separate entities/concepts.

added -
[quote="Po6ept"][Steve, I'm glad you now agree with x'=x-vt as defined in the Galilean. For good closure, could you post it more clearly in this thread?/quote]
I have not said that ever. I disagree/challenge/do not accept with the equation x' = x-vt now and as I have for 6000 posts. I hope that is clear enough for you to properly understand my stance.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Tue Nov 12, 2013 8:44 pm UTC

steve waterman wrote:I have not said that ever. I disagree/challenge/do not accept with the equation x' = x-vt now and as I have for 6000 posts. I hope that is clear enough for you to properly understand my stance.

I see. I misunderstood you I guess. I can't understand what you actually think is the truth, since you agreed that your diagram was not applicable, and other diagrams have shown it nicely.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby SecondTalon » Tue Nov 12, 2013 8:45 pm UTC

steve waterman wrote:Sexy Talon - anytime you wish to lock this thread, starting as of right now, please feel free to do so, as I would appreciate completely shutting down this thread now. I envision nothing productive going forward and likely just anger and frustration from both sides.


Please, no. Stop twisting my arm. It hurts. Ow. Ow. Ow. Okay, okay, ow. I'll do it. I'll lock it. Ow.

(Note : If anything, negative arm twisting just took place. For the record.)
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