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

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steve waterman
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:Imagine having no points and t = 0....

Schrollini wrote:I'm not addressing this, and I think you know why.

Nope, not a clue even. Please explain...hmmm, crappy wording indeed.
I was really only saying...image two coincident systems, so rather then
go back there, instead, I have two more related mapping questions please.

given only S(x,y,z)...does S have an infinite set of coordinates?
given only S(x,y,z)...is manifold M a null set of points?

Getting ready to close up shop for the evening. Thanks, RS, for all the work and thoughts and responses.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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### Re: Galilean:x' with respect to S'?

Schrollini wrote:Steve, if you truly want to understand the Galilean transform, I'm willing to work you through it with you. But, it's not going to be quick and it's going to take quite a bit of work on your behalf. If you're willing to do that work, here's what'll happen.

I'll make a series of posts, each of which will define a few terms, work a few examples, and ask a few questions. You'll need to read each post carefully. Each time you find a term you don't understand or a definition that's unclear, you need to stop and ask for clarification. Do not continue until we're clear. You need to work each example yourself and ensure you get the same answer as I give. You need to solve each problem, using the definitions and notations I've introduced, and post your answers. Once I check your answers, I'll make my next post, and we'll repeat the process with that. I'm not quite sure how many steps this will take, but I'm guessing it'll be 5-10. Only at the end will we get to the Galilean transformation, so you'll need patience and endurance.

This is asking a lot, I know, so let me point out what I'm not asking: I'm not asking you to believe the definitions I give are "right" or the best to represent the world. I'm just asking you to use them as we go through and derive the Galilean transform, since they are the ones we need. Once we've done that, you're free to decide that these definitions are stupid and you can present your new system which is obviously better. But you won't be able to say the Galilean transformation is wrong.

Are you up for it? If so, respond here and I'll write up the first post.

ok. I am so impressed.
You can do that?

Three hours a day for as long as it Takes.
That is how I learned what little I know about Math.

Math is Like Psychothrapy.
In Psychothreapy the Patient does most of The Work.

In Math the Student does most of The Work.
At least in Math the Teacher has also Suffered.

The terms have to be learned by Each and Everyone that wants to speak the Language.
You are Fluent in Math?

You are so fluent that you are able to be Kind, Patient and Forgiving?
As Long as The Student does not decide The Student is The Master before The Master does.

Who ever The Student is; That is a Lucky Student!
I might stop in to see Gooliean Unfolding.

I am not fluent in math.
I may be able to recognize The Joy of discovery on my Screen.

What good friends you two may be by the Time you Numbers Guys are done.
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Schrollini
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### Re: Galilean:x' with respect to S'?

yurell wrote:Schrollini, when you are listing functions and letters by themselves (or even together), it may be a good idea to italicise them; it's hard to distinguish f from a typo (or just skip over it entirely with your eye), but f is somewhat more obvious. It also creates a solid break between mathesese and general text that I find makes it easier to read.

I've been being lazy, since it's really quite a pain to type {i}f{/i}({i}x{/i}) to get f(x). But if it's getting confusing, I suppose it's better to be clear than fast. Or I can give up having non-italic parenthesis and just type f(x), even though that looks ugly. Oh, for proper TeX support.

Out of curiosity, does πππππππ show up for other people? How about π’π£π€π₯π¦π§π¨? These are from the mathematical symbols unicode block, but they may show up for me because I have math fonts that normal people don't have.

steve waterman wrote:
Schrollini wrote:I'm not addressing this, and I think you know why.

Nope, not a clue even. Please explain...hmmm, crappy wording indeed.

As I noted in my edit, I thought you were asking a question about time evolution, and we're not discussing that.

steve waterman wrote:given only S(x,y,z)...does S have an infinite set of coordinates?
given only S(x,y,z)...is manifold M a null set of points?

What is S? A coordinate system? If so, it must be associated with a manifold, since a coordinate system is a mapping to a manifold. The manifold has an uncountable infinitude of points (since I'm not allowing us to consider the null set to be a manifold, since that would be silly). The set of coordinates mapped to M will also be uncountably infinite. (Math details: We should require the domain of β2 mapped to M to be an open, non-empty set, and any such subset of β2 will have an uncountably infinite number of elements.)

Steve, take the evening off and relax. When you come back, please go through this post again and start working the example and exercises. I think you're grasping these ideas, and the exercises will be a little test. I know you want to move on to coordinate transformations and time evolution; I do too. But we're going to do it carefully and a step at a time, so I need to know that we're all set with these definitions before we move on.

Last edited by Schrollini on Thu Jul 11, 2013 11:38 pm UTC, edited 1 time in total.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
Sorry, I think I misunderstood your last point. But please, don't mention time anymore. There is no time (yet).

Good. Agreed.

Schrollini wrote:Anyway, if there are no points, there is no manifold. If there's no manifold, there's no mapping to that manifold. And therefore there's no coordinate system*. So your question really doesn't make sense.

So, are you saying that there is no such thing as an Cartesian system with only coordinates and no points?
Yet Cartesian coordinate systems did exist then between 1650 and the early 1900's and manifold M was yet to be imagined?

How then, can the Galilean, be given coincident S(x,y,z) and S'(x',y',z') WHERE x = x', y = y' and z = z'
as it clearly states,

if there ARE no coordinates should there be no points, hence no manifold AT ALL??

* I made this bold for emphasis
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steve

yurell
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:Out of curiosity, does πππππππ show up for other people? How about π’π£π€π₯π¦π§π¨? These are from the mathematical symbols unicode block, but they may show up for me because I have math fonts that normal people don't have.

Just squares, unfortunately. And proper TeX support would be nice, but unfortunately we have to live in lieu of that; if you consider italicised ugly, then I shall just have to pay more attention
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

Pronouns: Feminine pronouns please!

Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:So, are you saying that there is no such thing as an Cartesian system with only coordinates and no points?
Yet Cartesian coordinate systems did exist then between 1650 and the early 1900's and manifold M was yet to be imagined?

How many times (snerk) must I say it: There is no time! This includes 1650 to the early 1900s.

More to the point, I'm not teaching the history of mathematics, I'm teaching the current conventions. We currently derive the integers from set theory, even though the integers were invented well before set theory.

steve waterman wrote:if there ARE no coordinates should there be no points, hence no manifold AT ALL??

No manifold => no coordinate system. But you can't turn that arrow around. You can perfectly well have a manifold without a coordinate system. Review the definition of the manifold. It says nothing about coordinates. Review the definition of the coordinate system. It requires a manifold.

yurell wrote:Just squares, unfortunately. And proper TeX support would be nice, but unfortunately we have to live in lieu of that; if you consider italicised ugly, then I shall just have to pay more attention

Drat. I do like italic symbols; it's the italic parenthesis that look bad. But turning italics on and off for each character is a drag.

I'll start italicizing symbols with my next post. Maybe.

Edit: I can haz grammar?
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beojan
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### Re: Galilean:x' with respect to S'?

I can see the symbols, but I also have a lot of mathematical fonts installed.
You could use mathurl.com. You could also ask people to install the necessary mathematical fonts, although I'm not sure if all browsers / environments will use symbols from fonts other than the selected text font.

steve waterman
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### Re: Galilean:x' with respect to S'?

re - using appropriate notation

You have strayed away from the Galilean given.

WE are given coincident Cartesian SYSTEMS S(x,y,z) and S'(x',y',z') where x = x', y = y' and z = z'.

The Galilean has NO f(x,y,z). Using little f, usually means a function. So, for each set of values chosen, one point is determined/gets mapped.
GIVEN or LET Cartesian coordinate system S(x,y,z) means the INFINITE set of values, and ALL coordinates point locations determined/assigned/known/present/, no mapping required.

So, f is denied as pertinent/appropriate notation as the Galilean has only S and S'. f is your own concoction/device/logic crutch to explain things.
.......................................................................................................................................

Imagine we are given Cartesian coordinate system S(x,y,z) and point P at S(2,0,0).

So, manifold M would possess only the one coordinate-free point P as the total point set of elements of M?

How does system S obtain its infinite coordination? what gets mapped to what?

Does point P in manifold M eventually get assigned coordination? If so, how?
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yurell
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### Re: Galilean:x' with respect to S'?

He hasn't strayed away from anything, Steve; he's trying to work from the ground up, establishing each and every term to be used explicitly so that you're using the same terminology (speaking the same language, as it were) as everyone else. That's why he said it would take a while, and you just need a bit of patience to stick with it. You've already admitted that one of the terms you used must be confusing here, because it was oxymoronic in the language we use. A lot of the communication issues between you and the rest of us are probably due to more mistaken terminology (as is demonstrated by your latest post), but we won't know unless you're willing to humour us.

And if you're not willing to work with Schrollini, who is taking the time to carefully prepare this information in a form that is both easy to understand and interactive (insofar as you can ask him questions), then what do you hope to gain by posting here?
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

Pronouns: Feminine pronouns please!

steve waterman
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### Re: Galilean:x' with respect to S'?

yurell wrote:He hasn't strayed away from anything, Steve;

Schrollini is not using the Galilean given...there is no f(x,y,z). Your comments are not helping things.

x' = x-vt x transforms to x to transforms to x'?

Still no thread readers/posters at xkcd have an opinion/side/stance that they are willing to actually express here.

added - to Schrollini
Given S(x,y,z) means?
a) the infinite set of all coordinates is contained in S
b) a null of all of coordinates in contained in S
c) or if you offer another choice of phrasing/wording/logic/definition

Also, how does your S(x,y,z) obtain/manifest its coordinates? What is the exact procedure please.
Last edited by steve waterman on Fri Jul 12, 2013 1:38 pm UTC, edited 1 time in total.
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JudeMorrigan
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### Re: Galilean:x' with respect to S'?

No, really, Steve. He'll get to it. If you have patience, work through schrollini's exercises step by step, they'll eventually get there. (By "exercises", I mean the entire conversation Schrollini is attempting to have with you, not simply the example problems.) He's trying to build a foundation. If you don't do that first, the walls will collapse. He is *not* straying from the overall topic. Have faith.

steve waterman
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### Re: Galilean:x' with respect to S'?

JudeMorrigan wrote:No, really, Steve. He'll get to it. If you have patience, work through schrollini's exercises step by step, they'll eventually get there. (By "exercises", I mean the entire conversation Schrollini is attempting to have with you, not simply the example problems.) He's trying to build a foundation. If you don't do that first, the walls will collapse. He is *not* straying from the overall topic. Have faith.

I repeat. There is no f(x,y,z) mentioned in the Galilean process ever.
There are only Cartesian coordinate SYSTEMS S(x,y,z) and S'(x',y',z').

Logically we need to commence with the Galilean given and not the Schrollini given.

x' = x-vt; x transforms to x or does x transform to x'?
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
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JudeMorrigan
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### Re: Galilean:x' with respect to S'?

And I'll repeat. You need to be patient. He will get there. You're trying to write a haiku without even having decided on what language you're going to use yet.

Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:Using little f, usually means a function. So, for each set of values chosen, one point is determined/gets mapped.

Yes. A coordinate system is a function. It's a function that maps coordinates to points in a manifold. This is why I've purposely been using functional notation.

steve waterman wrote:I repeat. There is no f(x,y,z) mentioned in the Galilean process ever.
There are only Cartesian coordinate SYSTEMS S(x,y,z) and S'(x',y',z').

Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold. To understand the Galilean transformation, you must understand this point.

If it helps, here's a rough outline of how we get to the Galilean transformation:
1. Manifolds and Coordinate Systems
2. Coordinate Transformations
3. A Note on Notation, and Several Examples
4. Two More Examples, Chosen with Malice Aforethought
5. Spacetime and Reference Frames
6. The Galilean Transformation

We're still on step 1 at the moment, and it's a long way to the Galilean. If you truly want to understand it, you have to buckle down, accept the definitions I give, and work your way through. It'll take some time, but it'll take exponentially longer if you keep trying to find excuses not to continue.

I think you do understand my definition of a manifold and a coordinate system, from the questions you've been asking. So the way forward is for you to prove this knowledge by doing the exercises so that I can post part 2.
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steve waterman
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### Re: Galilean:x' with respect to S'?

JudeMorrigan wrote:And I'll repeat. You need to be patient. He will get there. You're trying to write a haiku without even having decided on what language you're going to use yet.

Schrollini just tried using f(x,y,,z). I do indeed grasp, that if that is what I believed was meant by
S(x,y,z), then with that basis, I too would think that Relativity concepts were golden, as Relativistic concepts would all be quite logical to extrapolate, with that foundation. I will also accept that Schrollini believes that given S(x,y,z) means f(x,y,z). However, I cannot assume that, so I asked him what he takes given S(x,y,z) to mean to him.

I am challenging logic based solely upon the Galilean given.
I am logically not allowing any given that the Galilean does not ever mention/use/employ/require.

Galilean: x' = x-vt; does x transform to x or does x transform to x'?
Galilean inverse: x = x' + vt; does x' transform to x' or does x' transform to x?
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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steve waterman
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### Re: Galilean:x' with respect to S'?

I repeat. There is no f(x,y,z) mentioned in the Galilean process ever.
There are only Cartesian coordinate SYSTEMS S(x,y,z) and S'(x',y',z').

Schrollini wrote:Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold.
I made your statement bold and green for emphasis/easier to find if I want to come back to it with with a visual search.

Perfectly stated, and at the very heart of all this.
I will accept that that represents how the Galilean interprets S(x,y,z).

My stance.
Given S(x,y,z) means the infinite point set (x,y,z) is contained by system S.

I just agreed in the previous post that if S(x,y,z) = f(x,y,z), that I will accept Relativity logic as valid.
With the understanding/caveat that S(x,y,z) refers to all points and f(x,y,z) refers to one point.

So, let's see if we look at my "what if" premise/stance/thought experiment:
Given S(x,y,z) instead means the infinite point set (x,y,z) is contained by/within system S?

added - I replace my one personal unanswered question with these 4 equations.
the Galilean equations...
x' = x -vt means stationary x transforms to x' in the moved system
x = x' +vt means stationary x' transforms to x in the moved system

not Galilean equations...
x = x -vt means stationary x transforms to x in the moved system
x' = x' +vt means stationary x' transforms to x' in the moved system

btw, I had already posted to Jude, and then saw Schrollini's post, which got posted AFTER
my post to Jude. So, I posted to Schrollini, only to see I had double posted on the thread
but not in real time. Please excuse the double post. I am trying to be very careful/diligent to not do that.
hmmm, I just noticed they both messages got posted at the exact same time. So, since I write them one at a time,
the initial message was obviously a delayed posting, which caused my double post.
also need to add -
Nope...I was looking in wrong place for post time. Sorry about the double post.

added - I just fixed the typo in final equation of these 4.
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:My stance.

... is irrelevant.

I've given you the definitions I'm prepared to work with to derive the Galilean transformation. I use them or nothing at all.

Remember, I'm not asking you to believe these definitions; I'm just asking you to conditionally accept them for the time being.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
steve waterman wrote:My stance.

... is irrelevant.

I've given you the definitions I'm prepared to work with to derive the Galilean transformation. I use them or nothing at all.

Remember, I'm not asking you to believe these definitions; I'm just asking you to conditionally accept them for the time being.

Schrollini wrote: Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold.

Okay. I will observe your derivation with your definitions.
Please do not attempt to force ME to jump through hoops to see YOUR presentation.
Use your defined terms and walk us all through, please.
I will avoid answering your/all questions/posts until you post your stand-alone derivation, okay?
The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (xβ²,yβ²,zβ²,tβ²) of a single arbitrary event, as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and xβ directions, with their spatial origins coinciding at time t=t'=0:

x'=x-vt
y'=y
z'=z
t'=t

Are you commencing with the same conditions as the Galilean states above?

Thank you so much, SexyTalon and all other moderators for allowing this thread to continue in the General section.
After just 48 hours it got 1000 views with over 50 posts. I posted my video to YouTube over a week ago,
roughly, and it now has about 30* views with 0 likes and 0 dislikes and 0 comments. The exposure here at xkcd is just stunning. My math says YouTube exposure to xkcd exposure is like ,well, silly. I so appreciate the opportunity to get this feedback and exchange.

*Most, if not all of those TouTube views were due to posting the video url on xckd in the, um, wrong section/thread.

Schrollini wrote: Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold.

The manifold is mapped from where?...where does the manifold obtain its mapping coordinates from?
Does the manifold indeed assign the mapping coordinates in accordance with the moved system's coordination?
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:Okay. I will observe your derivation with your definitions.
Please do not attempt to force ME to jump through hoops to see YOUR presentation.
Use your defined terms and walk us all through, please.
I will avoid answering your/all questions/posts until you post your stand-alone derivation, okay?

Of what, the Galilean transform? We've tried big blocks of text before, and they didn't work. Heck, we've had several dozen posts just about the few paragraphs I've already posted. I'm not posting pages of text, only to have confusion set in two paragraphs in. If you want me to do this, I'm going to do it step by step.

Before moving on to the next step, I need to know that you've understood the current one. The best way to do that is for me to pose problems, you to post solutions, and the two of us to discuss them. Now if you're too shy to post your solutions, just say so. I'll post the solutions and let you check yourself. But this would put the onus on you to figure out if you're understanding something or not.

steve waterman wrote:
Schrollini wrote: Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold.

The manifold is mapped from where?...where does the manifold obtain its mapping coordinates from?
Does the manifold indeed assign the mapping coordinates in accordance with the moved system's coordination?

You're attributing entirely too much structure to the manifold. The manifold has points, nothing else. No mapping, no coordinates, no coordinate system. All of these things exist on top of the manifold, but they aren't inherent to the manifold.

Perhaps you're asking how you know which is the right coordinate system for a given manifold. The answer is that there isn't a "right" coordinate system. You can create whatever coordinate system you want to go on a particular manifold. The next step is to discuss what happens when you have two coordinate systems for the same manifold. But first I need to make sure we're okay with a single coordinate system. Let me know if you're going to post answers to the exercises, or if you want me to so you can check yourself.
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steve waterman
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### Re: Galilean:x' with respect to S'?

re - trying to learn how your mapping procedure works....
Schrollini wrote: Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold.

The manifold is mapped from where?...where does the manifold obtain its mapping coordinates from?
Does the manifold indeed assign the mapping coordinates in accordance with the moved system's coordination?
Schrollini wrote: You're attributing entirely too much structure to the manifold. The manifold has points, nothing else. No mapping, no coordinates, no coordinate system. All of these things exist on top of the manifold, but they aren't inherent to the manifold.

Perhaps you're asking how you know which is the right coordinate system for a given manifold. The answer is that there isn't a "right" coordinate system. You can create whatever coordinate system you want to go on a particular manifold. The next step is to discuss what happens when you have two coordinate systems for the same manifold. But first I need to make sure we're okay with a single coordinate system. Let me know if you're going to post answers to the exercises, or if you want me to so you can check yourself.

Yes, post them please...hopefully leading to you posing your derivation of x' =x-vt using your definitions, your terms, and your understanding, your diagrams, and your notation you believe is appropriate..

I am surprised to hear you say,
Schrollini wrote: Both S and S' are coordinate systems, which means they are both functions mapping coordinates to a manifold.

So, coordinate system map coordinates to the manifold?
and also say that,
Schrollini wrote:The manifold has points, nothing else. No mapping, no coordinates, no coordinate system.

These seems unclear to me still, taken together. Perhaps answering the following will let me better grasp where coordinates get mapped from as I now know where coordinates get mapped to..."on top of the manifold".

So, how does the given S(x,y,z) itself achieve coordination for its system?
Do coordinates get mapped from the moved system on top of ( but not inherent to ) the stationary manifold as vt occurs?

Schrollini wrote:Perhaps you're asking how you know which is the right coordinate system for a given manifold. The answer is that there isn't a "right" coordinate system. You can create whatever coordinate system you want to go on a particular manifold.

Did you mean to say...
"You can create whatever coordinate system you want to go on top of, but not inherent to, a particular manifold?"
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
Schrollini wrote:Perhaps you're asking how you know which is the right coordinate system for a given manifold. The answer is that there isn't a "right" coordinate system. You can create whatever coordinate system you want to go on a particular manifold.

Did you mean to say...
"You can create whatever coordinate system you want to go on top of, but not inherent to, a particular manifold?"

Yes, that's a fine way to put it.

I've been trying to avoid analogies, because analogies are always false if you take them too far. Maybe this will help, but please don't take it too seriously:

Consider a text. Moby Dick, for example. The text is just a series of words. There are no pages inherent to the text. But then someone typesets it. They choose where the page breaks go, and therefore each word gets a page number. The structure of text + pages is sort of like a coordinate system, while the text alone is sort of like a manifold. It's perfect fine to talk about the text without page numbers, just like it's perfect fine to talk about the manifold without a coordinate system. There are many ways to put the text on pages, just like there are many ways to put coordinate systems on the manifold. (There are a lot of ways in which this analogy falls apart, so don't take it seriously.)
Edit: See Jose's following post for a better analogy.

steve waterman wrote:Yes, post them please...

It's against my better judgement, but okay. Let me reiterate: If you truly want to understand the Galilean transformation, you should work them yourself first, and then check them against mine.

Excercise 1: Find f-1(f(x,y)).
Spoiler:
Define point P such that f(x,y) = P. Thus, f-1(f(x,y)) = f-1(P). By the definition of f-1, f-1(P) = (a,b), where f(a,b) = P = f(x,y). Since f is one-to-one, (a,b) = (x,y) and therefore f-1(f(x,y)) = (x,y).

Excercise 2: Find f(f-1(P)).
Spoiler:
Similar to the above, we find f(f-1(P)) = P.

Exercise 3: Draw the point f(-1, 2) = Q on the manifold above.
Spoiler:

Exercise 4: Find the coordinates of the point R marked above in the coordinate system f.
Spoiler:
R = f(1,-1).

Let me know when you get all the same answers that I have, or if you have trouble working some of them.
Last edited by Schrollini on Sat Jul 13, 2013 2:46 am UTC, edited 1 time in total.
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ucim
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### Re: Galilean:x' with respect to S'?

Moth...flame - what have I done!

Steve, perhaps the following would be helpful. If not, feel free to ignore. If so, keep it in mind.
Spoiler:
The first key idea is that a set of points is not the same as a set of coordinates. Shrollini (bless him!) is carefully setting up this distinction using a manifold ("set of points") and β2 (set of pairs of real numbers). They are different, but can be linked using a mapping (called a coordinate system). (I'm using β2 for convenience, the idea generalizes to any number of dimensions)

You are used to considering them the same. Sometimes it is a useful abbreviation to do so, but sometimes it leads to problems in understanding. Your understanding of the coordinate transforms is such a case. It has worked for your maps and for your sphere packing because the differences were not critical to the results you got. But the difference does exist, and this is one of those times where that difference is critical.

You need to completely separate those two concepts in your mind, and you cannot do that if you bring up your prior understanding of coordinate transforms, because that understanding will obscure the distinction that needs to be made.

Think of it this way. Consider a pile of (nonidentical) books. There is no inherent order to that pile, and if twice you are shown a book from the pile, the only thing you could say is that they are the same book, or they are not the same book.

It would be nice if the books could be put in some sort of order. So Marian (the librarian) creates a high tech card catalog. Each card has the name of a book and an RFID tag. Each book has a corresponding RFID tag. Then she arranges those cards in alphabetical order by the first word in the title. VoilΓ‘ - a mapping (or coordinate system) that lets you find a book easily! Just go down the alphabetical list, find the card, and sqeeze it. Beep beep beep! Pick up the beeping book.

Julian (the head honcho) doesn't like Marian to get all the credit, so he takes all the cards and rearranges them alphabetically by last name of author. VoilΓ‘ - a different mapping (or coordinate system) that lets you find a book a different way. Whereas in Marian's (the original) system, the fifth card "points to" An Aardvark's Journey, by Mel Richardson, instead in Julian's system, the fifth card "points to" Nightspring, by Anthony Asimov.

The pile of books doesn't care. It just sits there, fat, dumb, and happy. It is the card catalog that lets you find the books. But the card catalog isn't really much of anything if there is no pile of books to begin with. Nothing is going to go "beep beep beep".

Not to be outdone, Gregory proposes yet a third system, arranging the cards by publication date and time. The Gregorian system's fifth card "points to" the Book of Kells.

No book "belongs to" the fifth slot, in and of itself. There is no equivalence of slot to book. Only the card catalog system in use establishes the mapping.

Equivalently, no point in the manifold "belongs to" any given real number pair. It is only through the use of a mapping (coordinate system) that you impose on the manifold that you can assign number pairs to points in the manifold. Change the mapping and you break the connection (while establishing a new one).

Completely separating the concept of a set of points, a mapping (coordinate system), and a set of pairs of real numbers (the resulting coordinates), and keeping them separate, will allow you to see in stereo, as it were, and give you the depth of understanding you need to ultimately see the answer to your original question. But give it time. You need to see in stereo in order to "get it".
Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

beojan
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### Re: Galilean:x' with respect to S'?

Since it appears steve is utterly unwilling to follow the Schrollini course, here's something that he may just understand

Imagine you and a group of friends are standing in a field, which also contains trees. You see a particular pine tree, and you say that the tree is 3 meters north of you, and 2 meters east. Your friend, who is standing 1 meter due east of you says the same pine tree is 3 meters north, and 1 meter east of him.
You then see an elm tree that is 4 meters north of you, and 5 meters east. What does your friend say about the same tree?

He is standing due east of you, so, since the tree is 4 meters north of you, it is also 4 meters north of him.
He is standing 1 meter east of you, so, since the tree is 5 meters east of you, it is 5 - 1 = 4 meters east of him.

Now, you and your friend each decide to write down where the other 20 trees in the field are. Since it is a lot of work to write down "so many meters north of me, and so many meters east of me", you invent a shortcut:
You will write down (x,y) for each tree, where x is the distance to the east of you, and y is the distance to the north of you.
You friend will write down (x',y'), where x' is the distance to the east of him, and y' is the distance to the north of him.

Given the piece of paper on which you have written twenty lines of the form

Code: Select all

`tree name --- (x,y)`

What is the corresponding line on the piece of paper your friend has written on, in terms of x and y?
In other words, express (x',y') in terms of x and y.

yurell
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### Re: Galilean:x' with respect to S'?

For reference, the thread that spawned this discussion originally (afaik) is here.
It's over a hundred pages, though, so if you hadn't kept up with it when it was being made, chances are you're not going to read it now (although there are some really nice maths explanations in there, in a multitude of different ways).
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

Pronouns: Feminine pronouns please!

steve waterman
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### Re: Galilean:x' with respect to S'?

Okay, I have just read the posts since my last post. I wish to know the following before I respond to any of them.

Given S(x,y,z) - how does Cartesian coordinate system S obtain it coordinates?

Given a blue and a red 12 inch ruler, no matter where they are, even both in motion,...blue inches (x') = red inches (x)
Once given, the inch coordination:( abscissa,ordinate,applicate) is FIXED wrt their own ruler:(system)
[ the same coordination as it was given ]

I am not asking, rather I am stating the following as mathematical certainties!
Given S(x,y,z)...if S moves, then S(x,y,z) wrt S(0,0,0) is FIXED/eternal.
likewise,
Given S'(x',y',z')...if S' moves, then S'(x',y',z') wrt S'(0,0,0) is FIXED/eternal.

Are you possibly suggesting that x' = f(x-vt)?

added again - still working on finding out what the manifold M parameters are...are these all true?

1 manifold M is coincident with S and S', when the given systems are coincident themselves.
2 manifold M never moves.
3 coordinates from the moved system are mapped to points on top of, but not inherent to, manifold M.
4 manifold M itself has no coordinates, however, the points on top of manifold M do, since they were mapped.
5 mapping is never done from the stationary system to on top of manifold M
6 given only S(x,y,z) requires mapping to manifest the infinite set of coordinates (x,y,z) on top of manifold M.
Last edited by steve waterman on Sat Jul 13, 2013 1:42 pm UTC, edited 1 time in total.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

yurell
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### Re: Galilean:x' with respect to S'?

Steve, this is what we're trying to explain to you β you are not speaking the same language as us. That is why Schrollini is trying to teach you the maths as we use it, so you can understand what we mean when we talk about a Galilean transform, and you can communicate your ideas in notation we can understand.

As for your last question, Schrollini is not 'suggesting' anything of the kind β in fact, he's not suggesting anything whatsoever, merely stating what he has stated. He hasn't touched on the Galilean transform yet because you haven't demonstrated that you understand his first lesson. If you just answer his questions, we will very quickly get to Galilean relativity. If you don't, we'll be stuck on the carousel of the damned again. We're not refusing to answer your questions out of malice, we're not answering them because we're not sure what you're trying to express with your notation, nor are we sure you understand what we mean when we use standard notation.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

Pronouns: Feminine pronouns please!

steve waterman
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### Re: Galilean:x' with respect to S'?

yurell wrote:Steve, this is what we're trying to explain to you β you are not speaking the same language as us. That is why Schrollini is trying to teach you the maths as we use it, so you can understand what we mean when we talk about a Galilean transform, and you can communicate your ideas in notation we can understand.

As for your last question, Schrollini is not 'suggesting' anything of the kind β in fact, he's not suggesting anything whatsoever, merely stating what he has stated. He hasn't touched on the Galilean transform yet because you haven't demonstrated that you understand his first lesson. If you just answer his questions, we will very quickly get to Galilean relativity. If you don't, we'll be stuck on the carousel of the damned again. We're not refusing to answer your questions out of malice, we're not answering them because we're not sure what you're trying to express with your notation, nor are we sure you understand what we mean when we use standard notation.

Is this how coordinate mapping functions?

IMAGINE please; S ruler with x markings, S' ruler with x' markings, M ruler with no markings

let S and S' and M be coincident,
moving S', S' is mapped on top of manifold M
moving S, S is mapped on top of manifold M

yurell wrote:We're not refusing to answer your questions out of malice, we're not answering them because we're not sure what you're trying to express with your notation,

Thank you, I have been wondering why, since I did not even get a - "cannot parse your equation/concept" or nothing.
Zero feedback to stuff is really never helpful.

The Galilean uses the notation S(x,y,z) and S'(x'y'z').

How would YOU notate the concept that red ruler inches = blue ruler inches no matter where our rulers be?
using the Galilean notation that they use .......other than ...........S(x,y,z) = S'(x'y'z')?
Last edited by steve waterman on Sat Jul 13, 2013 2:20 pm UTC, edited 2 times in total.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

beojan
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### Re: Galilean:x' with respect to S'?

Steve, since you seem to be having some trouble, can you read my last post again, and answer the question I set at the end. If you do so, my next post will be about the Galilean transformation.

Evidently steve is online, but using the 'appear offline' feature all the time. To answer a part of the edit to your last post that you made after I made this post:
steve waterman wrote:How would YOU notate the concept that red ruler inches = blue ruler inches no matter where our rulers be?

This is the invariance of the distance between two points under the Galilean transform. That is, given any two cartesian coordinate systems, and two points, while the coordinates of each of the points may be different in the two coordinate systems, the distance ((x1-x2)2 + (y1-y2)2 + (z1-z2)2) is the same for both systems (so long as (x1,y1,z1) and (x2,y2,z2) are in the same coordinate system).

Now, please answer my last post so that I can get on to explaining what the Galilean transform means.

steve waterman
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### Re: Galilean:x' with respect to S'?

beojan wrote:Steve, since you seem to be having some trouble, can you read my last post again, and answer the question I set at the end. If you do so, my next post will be about the Galilean transformation.

Sorry but no. I am waaay past the new examples phase. I am just about ready to have to put on hold any of the non-Schrollini posts. Too many to answer every one already. So sorry, most responses will have to wait. I must indeed, cut down on time spent typing and reading as it causes migraines after a while; last night.

Schrollini and I are still working on the parameters of his defined terms; "mapping' is still under discussion.
For one, I was not aware that they have no inherent points wrt manifold M, and that the points are on top of
manifold M. So, I have asked some questions about manifold M, so that we can eventually see his derivation
using f(x,y,x) instead of using the Galilean given S(x,y,z).

SO, please...if you absolutely must post to this thread now, make it ever so brief and I will respond.
I am sorry to have to do this, but there is no way I can possibly keep up. Thank you for your understanding and patience.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

beojan
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
beojan wrote:Steve, since you seem to be having some trouble, can you read my last post again, and answer the question I set at the end. If you do so, my next post will be about the Galilean transformation.

Sorry but no. I am waaay past the new examples phase. I am just about ready to have to put on hold any of the non-Schrollini posts. Too many to answer every one already. So sorry, most responses will have to wait. I must indeed, cut down on time spent typing and reading as it causes migraines after a while; last night.

Schrollini and I are still working on the parameters of his defined terms; "mapping' is still under discussion.
For one, I was not aware that they have no inherent points wrt manifold M, and that the points are on top of
manifold M. So, I have asked some questions about manifold M, so that we can eventually see his derivation
using f(x,y,x) instead of using the Galilean given S(x,y,z).

SO, please...if you absolutely must post to this thread now, make it ever so brief and I will respond.
I am sorry to have to do this, but there is no way I can possibly keep up. Thank you for your understanding and patience.

For the purposes of this thread, mapping is a synonym for function. I'm sure schrollini has made this clear before.

steve waterman
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### Re: Galilean:x' with respect to S'?

beojan wrote:For the purposes of this thread, mapping is a synonym for function. I'm sure schrollini has made this clear before.

Yes, that makes complete sense. Thanks for the comment/input.
I suspect (t)his connection was always implied, but I do not recall it being stated.

"function points" exist on top of manifold M, and get mapped wrt the coordinates of the moved system.
TRUE?

I want to know if a given coordinate system needs to be mapped in order to manifest coordination,
or if it does not need to be mapped if all we are given is S(x,y,z)...as per the Galilean.
Pick one, I do not care. I will not argue with what ever that is, Just as I have said squat so far about say,
"on top of".

added - the Galilean WITH system notation...
S'(x') = S(x-vt)
S'(y') = S(y)
S'(z') = S(z)
S'(t') = S(t)

All I did was to add the system notation to their existing posted 4 equations.

Deny it as much as you wish, regardless of the value of t, S'(x') = S(x).
Last edited by steve waterman on Sat Jul 13, 2013 4:40 pm UTC, edited 3 times in total.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

beojan
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
beojan wrote:For the purposes of this thread, mapping is a synonym for function. I'm sure schrollini has made this clear before.

Yes, that makes complete sense. Thanks for the comment/input.
I suspect (t)his connection was always implied, but I do not recall it being stated.

http://fora.xkcd.com/viewtopic.php?f=2&t=103552#p3405998
This was explicitly stated by Schrollini.
steve waterman wrote:"function points" exist on top of manifold M, and get mapped wrt the coordinates of the moved system.
TRUE?

This make not sense. What are "function points"? That term was never used by Schrollini.
What do you mean "get mapped wrt the coordinates of the moved system"? Schrollini has stated many times that we are not close to involving any motion. Motion, and time, has not yet been introduced in the Schrollini course.
What he stated was that a coordinate system maps coordinates (pairs or triplets of the form (x,y) or (x,y,z)) to points.

steve waterman wrote:I want to know if a given coordinate system needs to be mapped in order to manifest coordination,
or if it does not need to be mapped if all we are given is S(x,y,z)...as per the Galilean.
Pick one, I do not care. I will not argue with what ever that is, Just as I have said squat so far about say,
"on top of".

This makes no sense. What does it mean for a coordinate system to "be mapped" or "manifest coordination"?

May I once again recommend you read and reply to the post I made about the trees in a field. I think you may find it easier to communicate if we use a 'real-world' scenario. It really shouldn't take much time or give you a migrane.

steve waterman
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### Re: Galilean:x' with respect to S'?

I want to know if a given coordinate system needs to be mapped in order to manifest coordination,
or if it does not need to be mapped if all we are given is S(x,y,z)...as per the Galilean.
Pick one, I do not care. I will not argue with what ever that is, Just as I have said squat so far about say,
"on top of".
beojan wrote:This makes no sense. What does it mean for a coordinate system to "be mapped" or "manifest coordination"?]

Try this please. given S(x,y,z)...how do we know that S(2,0,0) is exactly 2 units from S(0,0,0)?
Some explanation would be welcomed too, on this question, from you. My thoughts are posted here
http://www.watermanpolyhedron.com/MATHVS.html

I believe I should be able to follow your notation. If not, I will you ask you to please clarity. I will
allow your terms and their definitions too. Thanks for offering to walk us through your derivation.
If you wish diagrams, that will also be logical. I will, for the time being, suspend all requests for
answers to whatever is currently unanswered at this point in time, from you.

Meanwhile, I will respond to short posts. Longer posts, I will read and then put on the back-burner,
most likely, and unfortunately, not have the capacity/strength and energy to get to them, to eventually,
write back.
Last edited by steve waterman on Sat Jul 13, 2013 5:16 pm UTC, edited 1 time in total.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

beojan
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
beojan wrote:
steve waterman wrote:I want to know if a given coordinate system needs to be mapped in order to manifest coordination,
or if it does not need to be mapped if all we are given is S(x,y,z)...as per the Galilean.
Pick one, I do not care. I will not argue with what ever that is, Just as I have said squat so far about say,
"on top of"

This makes no sense. What does it mean for a coordinate system to "be mapped" or "manifest coordination"?]

Try this please. given S(x,y,z)...how do we know that S(2,0,0) is exactly 2 units from S(0,0,0)?
Some explanation would be welcomed too, on this question, from you. My thoughts are posted here
http://www.watermanpolyhedron.com/MATHVS.html

NB: Quotes can be nested.

Code: Select all

`[quote="steve waterman"][quote="beojan"][quote="steve waterman"]I want to know if a given coordinate system needs to be mapped in order to manifest coordination,or if it does not need to be mapped if all we are given is S(x,y,z)...as per the Galilean.  Pick one, I do not care. I will not argue with what ever that is,  Just as I have said squat so far about say,"on top of"[/quote]This makes no sense. What does it mean for a coordinate system to "be mapped" or "manifest coordination"?][/quote]Try this please. given S(x,y,z)...[color=#40BF00][b]how do we know that S(2,0,0) is exactly 2 units from S(0,0,0)[/b][/color]?Some explanation would be welcomed too, on this question, from you.  My thoughts are posted herehttp://www.watermanpolyhedron.com/MATHVS.html[/quote]`

Please use these nested quotes in future, it would greatly improve the readability of your posts.
On the matter of how we know that S(2,0,0) is exactly 2 units from S(0,0,0), see my previous post where I talked about the distance between two points. As I said in that post, this distance is the same no matter which coordinate system we use, so long as we use the same coordinate system for both points. (I am assuming the notation S(2,0,0) means "the point at coordinates (2,0,0) as measured in system S", that is, treating the coordinate system S as a function that takes a triplet of numbers and gives a point, as in Schrollini's definition)

I reiterate, I would like you to read my post about trees in a field, and give a reply to it.

Edit at 2013-07-13 18:29 UTC to fix missing quote mark
Last edited by beojan on Sat Jul 13, 2013 6:30 pm UTC, edited 1 time in total.

steve waterman
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### Re: Galilean:x' with respect to S'?

beojan wrote:(I am assuming the notation S(2,0,0) means "the point at coordinates (2,0,0) as measured in system S

Yes, exactly.
S'(2,0,0) means the point at coordinates (2,0,0) as measured in system S' from S'(0,0,0)

Given S(2,0,0) and given S'(2,0,0) are coincident,
if we then elect to move either system, does S(2,0,0) = S'(2,0,0)?

added - in case of possible notational confusion...
Given S(2,0,0) and given S'(2,0,0) are coincident,
does S(0,0,0) to S(2,0,0) = S'(0,0,0) to S'(2,0,0)?
SOOO...I are talking about lineal distances being equal, NOT points being nor even coordinates being equal nor locations being equal!
The lineal abscissa distance is the DISTANCE +/- FROM the system origin to (abscissa,0,0).

Seems I am comparing absolute lengths from the origin and you are comparing points, not lengths!!!
Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a vector from (0,0,0) 3d comparing dude and the Galilean is a location/point 0d comparing dude.
That might do well explain a portion of our verbal mental conceptual stances.
Last edited by steve waterman on Sat Jul 13, 2013 6:34 pm UTC, edited 3 times in total.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

beojan
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
beojan wrote:(I am assuming the notation S(2,0,0) means "the point at coordinates (2,0,0) as measured in system S [

Yes, exactly.
S'(2,0,0) means the point at coordinates (2,0,0) as measured in system S'

Given S(2,0,0) and given S'(2,0,0) are coincident,
if we then elect to move either system, does S(2,0,0) = S'(2,0,0)?

It's important to start with two systems that are stationary relative to each other, but not coincident.
Take two coordinate systems, S and S', such that S'(0,0,0) = S(1,0,0)*. Does S(2,0,0) = S'(2,0,0)? Obviously not, S(2,0,0) = S'(1,0,0).

Please read my post about the trees in the field and reply to it, this is essentially what I am trying to explain in that post.

*also assuming the x, y and z axes of the two coordinate systems are each parallel, i.e. there is no rotation.

And... he's decided to add something to his previous post instead of creating a new reply.
steve waterman wrote:SOOO...I are talking about lineal distances being equal, NOT points being nor even coordinates being equal nor locations being equal!
The lineal abscissa distance is the DISTANCE +/- FROM the system origin to (abscissa,0,0).

Seems I am comparing absolute lengths from the origin and you are comparing points, not lengths!!!
Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a distance comparing dude and the Galilean is a point comparing dude.

1. I think you mean linear, not lineal.
2.
The lineal abscissa distance is the DISTANCE +/- FROM the system origin to (abscissa,0,0).

So you've now defined "lineal abscissa distance" to be the absolute value of the x coordinate. If we are talking about an individual, fixed point, and multiple coordinate systems, this is not the same in all coordinate systems, for the simple reason that the origins of the multiple coordinate systems are not all (necessarily) in the same place.

3.
Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a distance comparing dude and the Galilean is a point comparing dude.

I have no idea what you are talking about here.

And he's edited his post again
steve waterman wrote:Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a vector from (0,0,0) 2d comparing dude and the Galilean is a location/point 1d comparing dude.
That might do well explain a portion of our verbal mental conceptual stances.

The second part ("the Galilean is a location/point 1d comparing dude") is wrong. The Galilean transform can be used with any number of dimensions, although, due to living in 3 spatial dimensions, its 1, 2, and 3d versions are the ones used more often.

However, on the first part ("I am a vector from (0,0,0) 2d comparing dude"), ignoring the fact that (0,0,0) is 3 dimensional, such vectors, from a defined origin to a point, are called position vectors. It is these position vectors that the Galilean transform works with, converting a position vector with regard to one origin to a position vector with regard to a different origin, which may be displaced from the first, and may be moving relative to the first. Referring back to the field with trees, if a person, who we shall define as the origin, moves, the position vector of some particular tree changes as the origin (person) moves, even though the location of the tree relative to the center of the earth, and the other trees, does not change. (In fact, it is impossible to give an absolute location for an object or point. All that can be given is a position vector relative to some origin).

steve waterman
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### Re: Galilean:x' with respect to S'?

beojan wrote:
steve waterman wrote:
beojan wrote:(I am assuming the notation S(2,0,0) means "the point at coordinates (2,0,0) as measured in system S [

Yes, exactly.
S'(2,0,0) means the point at coordinates (2,0,0) as measured in system S'

Given S(2,0,0) and given S'(2,0,0) are coincident,
if we then elect to move either system, does S(2,0,0) = S'(2,0,0)?

It's important to start with two systems that are stationary relative to each other, but not coincident.
Take two coordinate systems, S and S', such that S'(0,0,0) = S(1,0,0)*. Does S(2,0,0) = S'(2,0,0)? Obviously not, S(2,0,0) = S'(1,0,0).

Please read my post about the trees in the field and reply to it, this is essentially what I am trying to explain in that post.

*also assuming the x, y and z axes of the two coordinate systems are each parallel, i.e. there is no rotation.

And... he's decided to add something to his previous post instead of creating a new reply.
steve waterman wrote:SOOO...I are talking about lineal distances being equal, NOT points being nor even coordinates being equal nor locations being equal!
The lineal abscissa distance is the DISTANCE +/- FROM the system origin to (abscissa,0,0).

Seems I am comparing absolute lengths from the origin and you are comparing points, not lengths!!!
Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a distance comparing dude and the Galilean is a point comparing dude.

1. I think you mean linear, not lineal.
2.
The lineal abscissa distance is the DISTANCE +/- FROM the system origin to (abscissa,0,0).

So you've now defined "lineal abscissa distance" to be the absolute value of the x coordinate. If we are talking about an individual, fixed point, and multiple coordinate systems, this is not the same in all coordinate systems, for the simple reason that the origins of the multiple coordinate systems are not all (necessarily) in the same place.

3.
Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a distance comparing dude and the Galilean is a point comparing dude.

I have no idea what you are talking about here.

And he's edited his post again
steve waterman wrote:Very very interesting. I am going to take a break and try to reflect upon this new "revelation" that perhaps I am a vector from (0,0,0) 2d comparing dude and the Galilean is a location/point 1d comparing dude.
That might do well explain a portion of our verbal mental conceptual stances.

The second part ("the Galilean is a location/point 1d comparing dude") is wrong. The Galilean transform can be used with any number of dimensions, although, due to living in 3 spatial dimensions, its 1, 2, and 3d versions are the ones used more often.

However, on the first part ("I am a vector from (0,0,0) 2d comparing dude"), ignoring the fact that (0,0,0) is 3 dimensional, such vectors, from a defined origin to a point, are called position vectors. It is these position vectors that the Galilean transform works with, converting a position vector with regard to one origin to a position vector with regard to a different origin, which may be displaced from the first, and may be moving relative to the first. Referring back to the field with trees, if a person, who we shall define as the origin, moves, the position vector of some particular tree changes as the origin (person) moves, even though the location of the tree relative to the center of the earth, and the other trees, does not change. (In fact, it is impossible to give an absolute location for an object or point. All that can be given is a position vector relative to some origin).

Thanks for the input.
me - equality means having equal vector distances from their own (0,0,0) as the one another system.
as would be S'(2,0,0) = S(2,0,0) hence being a coordinate means being a vector from its own (0,0,0).

Galilean - equality means shared location
as would be S'(-1,0,0,) = S(2,0,0) - vt hence being a coordinate means being a point to you, NEVER A DISTANCE.

S(2,0,0) is a point in the Galilean mind-set and whereas for me,
S(2,0,0) is that vector of length of 2, starting from S(0,0,0) and going to S(2,0,0).
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

ivnja
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:me - equality means having equal vector distances from their own (0,0,0) as the one another system.
as would be S'(2,0,0) = S(2,0,0) hence being a coordinate means being a vector from its own (0,0,0).

Galilean - equality means shared location
as would be S'(-1,0,0,) = S(2,0,0) - vt hence being a coordinate means being a point to you, NEVER A DISTANCE.

S(2,0,0) is a point in the Galilean mind-set and whereas for me,
S(2,0,0) is that vector of length of 2, starting from S(0,0,0) and going to S(2,0,0).

I think we're on the right track here, at least as far as sorting out the differences in mindset.
I look forward to Schrollini's next post.
Hi you.
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### Re: Galilean:x' with respect to S'?

If you are not Steve or Schrollini, keep your mouth shut for a while, 'k?
heuristically_alone wrote:I want to write a DnD campaign and play it by myself and DM it myself.
heuristically_alone wrote:I have been informed that this is called writing a book.

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### Re: Galilean:x' with respect to S'?

ivnja wrote:
steve waterman wrote:me - equality means having equal vector distances from their own (0,0,0) as the one another system.
as would be S'(2,0,0) = S(2,0,0) hence being a coordinate means being a vector from its own (0,0,0).

Galilean - equality means shared location
as would be S'(-1,0,0,) = S(2,0,0) - vt hence being a coordinate means being a point to you, NEVER A DISTANCE.

S(2,0,0) is a point in the Galilean mind-set and whereas for me,
S(2,0,0) is that vector of length of 2, starting from S(0,0,0) and going to S(2,0,0).

I think we're on the right track here, at least as far as sorting out the differences in mindset.
I look forward to Schrollini's next post.

This has been the catalyst for all the confusion since my first post 8 month ago. It explains to me why all the xkcd logic sounded silly that was being spouted. If with my mindset, you were to re-read any section of the epic, I suspect my silly logic would make sense to you. I also had thought that the Galilean transformation meant placing a new point into the opposite system, whereas that is not true, which I later learned in the epic thread.

I am not asking anyone to actually go back and try to parse what was done/written/posted/stated.

imo. we are here, finally, with a chance at possibly resolving my 25 years running x' = x-vt challenge:
be that determination as a yes or of a no, so be it.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve