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

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

eran_rathan wrote:spoilered for minor pedantry.
Spoiler:
to be fair, coordinate systems can move in relation to each other - re: GRS80 vs. NAD83 - because they are defined to different things (center of Earth's mass for the GRS80 ellipsoid, center of mass for the North American plate for NAD83 - considering that the North American plate moves on the order of centimeters per year, the two coordinate systems are roughly 4-6 feet divergent from their relative positions when they were defined.

Spoiler:
If you're trying to locate events, then you need to include time in your manifold. Then you'll have a Galilean component to your transformation, but there's no "meta-time" in which the coordinate systems move. Otherwise, the manifold of the earth's surface in 2000 is not the same manifold as the earth's surface in 2010, so we can't compare coordinate systems at those two times.

Of course, when you know what you're talking about, as you do, you can bend the definitions to account for this movement. But I was very strict with my definitions for Steve specifically to rule out this possibility.

steve waterman wrote:Thanks, indeed, that is my bottom line, logic.

So, you believe that given two coincident systems S(x,y,z) and S(x',y',z') that x does not equal x'?
What do you suggest then, that x equals in terms of x' then, when S and S' are coincident?

Of course x = x' when the systems are coincident. But all interesting coordinate transformations, including the Galilean, are between non-coincident systems. (Recall that time is a coordinate in reference frames, and the Galilean is between reference frames. The spatial parts being coincident at t=0 does not make the frames coincident.)
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WibblyWobbly
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
eran_rathan wrote:spoilered for minor pedantry.
Spoiler:
to be fair, coordinate systems can move in relation to each other - re: GRS80 vs. NAD83 - because they are defined to different things (center of Earth's mass for the GRS80 ellipsoid, center of mass for the North American plate for NAD83 - considering that the North American plate moves on the order of centimeters per year, the two coordinate systems are roughly 4-6 feet divergent from their relative positions when they were defined.

Spoiler:
If you're trying to locate events, then you need to include time in your manifold. Then you'll have a Galilean component to your transformation, but there's no "meta-time" in which the coordinate systems move. Otherwise, the manifold of the earth's surface in 2000 is not the same manifold as the earth's surface in 2010, so we can't compare coordinate systems at those two times.

Of course, when you know what you're talking about, as you do, you can bend the definitions to account for this movement. But I was very strict with my definitions for Steve specifically to rule out this possibility.

steve waterman wrote:Thanks, indeed, that is my bottom line, logic.

So, you believe that given two coincident systems S(x,y,z) and S(x',y',z') that x does not equal x'?
What do you suggest then, that x equals in terms of x' then, when S and S' are coincident?

Of course x = x' when the systems are coincident. But all interesting coordinate transformations, including the Galilean, are between non-coincident systems. (Recall that time is a coordinate in reference frames, and the Galilean is between reference frames. The spatial parts being coincident at t=0 does not make the frames coincident.)

Ah, but don't you see? Since x and x' are not points, they are clearly abscissal lengths (and not just, numbers, you know, which have none of this bullshit Steve made up surrounding his magical "abscissa"), which means they are once and forever equal! Once coincident, always coincident, even if they're not coincident, right, Steve?

THIS IS PURE MATH, PEOPLE. GIVEN! LET! ABSCISSA! WRT!

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

Schrollini wrote: Allow me to cut out a bunch of junk that you aren't actually using in this proof:

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'),
where x = x', y = y', z = z', d* = 0
let d* = distance of separation between S and S' along the common x/x' axis

1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ). like 2 = -1.
2 whereas, S(x,y,z) = S'(x',y',z') always properly equates the given and eternally fixed abscissa ( lengths ).
3 x = x', therefore x' ≠ x - d, when d > 0.

Which is obvious.

Schrollini wrote:Of course x = x' when the systems are coincident.

the rest of my proof you had lined out
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WibblyWobbly
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
Schrollini wrote: Allow me to cut out a bunch of junk that you aren't actually using in this proof:

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'),
where x = x', y = y', z = z', d* = 0
let d* = distance of separation between S and S' along the common x/x' axis

1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ). like 2 = -1.
2 whereas, S(x,y,z) = S'(x',y',z') always properly equates the given and eternally fixed abscissa ( lengths ).
3 x = x', therefore x' ≠ x - d, when d > 0.

Which is obvious.

Schrollini wrote:Of course x = x' when the systems are coincident.

the rest of my proof you had lined out

The rest of your proof was superfluous frippery, made to sound like you had an actual proof. Schrollini cut away everything that wasn't useful, leaving you with something trivial instead.

Edit: And dibs on the name "Superfluous Frippery" for a trance-hop trio. In case you were thinking about stealing it.
Last edited by WibblyWobbly on Fri Jul 19, 2013 3:50 pm UTC, edited 1 time in total.

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

Schrollini wrote:Of course x = x' when the systems are coincident.

And to try to prevent misunderstandings: Coincident systems implies x = x'. x = x' for some points does not imply that the systems are coincident. The Galilean transformation is not between coincident systems, even though x = x' at t = t' = 0.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:Of course x = x' when the systems are coincident.

And to try to prevent misunderstandings: Coincident systems implies x = x'.

x = x' for some points does not imply that the systems are coincident.

However, We are NOT trying to determine if systems are coincident or not...we know that
we have two cases...one as coincident, one as not coincident..
We are comparing distances from each origin of each system to one another along the common x/x' axis.
Schrollini wrote:x = x' for some points

You said points only exist in the manifold, whereas x and x' are not points as you agreed.

Schrollini wrote:The Galilean transformation is not between coincident systems, even though x = x' at t = t' = 0.

My proof is only about the mathematical validity of x' = x- d. I am only out to prove this. I will not be applying this to Physics myself, personally, nor plan to do so or in the future, here at xkcd or even elsewhere.

So, the proof presented as ..."the short form"
Given x = x', therefore x' ≠ x-d.

Allowing x' = x - d, spawns equations like 2 = -1...x' = x-d is not about points nor about a 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

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

WibblyWobbly wrote:Once coincident, always coincident, even if they're not coincident, right, Steve?

NOOOOO!

Once coincident with x = x', x = x' always, even if they're not coincident.

Two inches on one ruler = two inches on the other ruler no matter where the rulers are,
be they coincident or not.
"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

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

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

steve waterman wrote:
Schrollini wrote:So, the proof presented as ..."the short form"
Given x = x', therefore x' ≠ x-d.

Allowing x' = x - d, spawns equations like 2 = -1...x' = x-d is not about points nor about a manifold.

Allow me to apply your *exact* logic to a completely different equation. Even you'd think I look like a complete fool/tool if I weren't being incredibly tongue-in-cheek. Ready?

c2 = a2 + b2

When b=0, c2 = a2, therefore c = a (or c = -a, but we're not going to deal with negative lengths here)

Let's then decide that c = a is the most important part of the Pythagorean Theorem.

But if b ≠ 0, then c ≠ a, but c must equal a because we said so. So clearly the Pythagorean Theorem is wrong and we can't calculate the length of the hypotenuse of a triangle.

(edited to be slightly more, if undeservedly more, polite about it)
(edited again to be snarky about it since the snarky version was already quoted anyway )
Last edited by Vetala on Fri Jul 19, 2013 5:02 pm UTC, edited 1 time in total.

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

steve waterman wrote:Allowing x' = x - d, spawns equations like 2 = -1...x' = x-d is not about points nor about a manifold.

Wouldn't the 2 and the -1 be x-coordinates in different coordinate systems?

Wow, this thread got me to stop lurking. :/

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

Vetala wrote:
steve waterman wrote:
Schrollini wrote:So, the proof presented as ..."the short form"
Given x = x', therefore x' ≠ x-d.

Allowing x' = x - d, spawns equations like 2 = -1...x' = x-d is not about points nor about a manifold.

Allow me to apply your *exact* logic to a completely different equation. Even you'd think I look like a complete fool/tool if I weren't being incredibly tongue-in-cheek. Ready?

c2 = a2 + b2

When b=0, c2 = a2, therefore c = a (or c = -a, but we're not going to deal with negative lengths here)

Let's then decide that c = a is the most important part of the Pythagorean Theorem.

But if b ≠ 0, then c ≠ a, but c must equal a because we said so. So clearly the Pythagorean Theorem is wrong and we can't calculate the length of the hypotenuse of a triangle.

That's almost nice as some comic humor.
If b = 0, then you ain't got no "right" triangle, more like what is called a line segment. A tad off topic.

let x = x' = 2, do some "Physics" and now bingo, we get to equate x' = x-d.
so, with Physics, we can prove that 2 = 2 - d when d > 0.
Then Physics can launch that "math" into more equations...and generate more equalities/paradoxes.
Last edited by steve waterman on Fri Jul 19, 2013 4:57 pm UTC, edited 1 time in total.
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eran_rathan
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
Vetala wrote:
steve waterman wrote:
Schrollini wrote:So, the proof presented as ..."the short form"
Given x = x', therefore x' ≠ x-d.

Allowing x' = x - d, spawns equations like 2 = -1...x' = x-d is not about points nor about a manifold.

Allow me to apply your *exact* logic to a completely different equation. Even you'd think I look like a complete fool/tool if I weren't being incredibly tongue-in-cheek. Ready?

c2 = a2 + b2

When b=0, c2 = a2, therefore c = a (or c = -a, but we're not going to deal with negative lengths here)

Let's then decide that c = a is the most important part of the Pythagorean Theorem.

But if b ≠ 0, then c ≠ a, but c must equal a because we said so. So clearly the Pythagorean Theorem is wrong and we can't calculate the length of the hypotenuse of a triangle.

That's almost nice as some comic humor.
If b = 0, then you ain't got no "right" triangle, more like what is called a line segment. A tad off topic.

steve - This is PRECISELY what your so-called proof is doing.
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:If b = 0, then you ain't got no "right" triangle, more like what is called a line segment. A tad off topic.

If x = x', you ain't got no non-trivial transformation, more like what is called the identity. A tad off topic for a thread about the Galilean transformation.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
steve waterman wrote:If b = 0, then you ain't got no "right" triangle, more like what is called a line segment. A tad off topic.

If x = x', you ain't got no non-trivial transformation, more like what is called the identity. A tad off topic for a thread about the Galilean transformation.

You are right. Perhaps I should end my posts to this thread, as I have just quite recently re-focused my challenge from x' = x-vt to x' = x-d. Indeed that seems the proper thing to do, as I no longer have anything to post about the Galilean transformation on this thread or any thread any more. Since, x' = x -d is indeed off topic, this will end my posting to this thread. I likely will open a new thread call x' = x-d today, some time. Thanks for pointing this out.
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### Re: Galilean:x' with respect to S'?

I'd keep it right here, steve - just to keep the moderators happy.
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beojan
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:If b = 0, then you ain't got no "right" triangle, more like what is called a line segment. A tad off topic.

If x = x', you ain't got no non-trivial transformation, more like what is called the identity. A tad off topic for a thread about the Galilean transformation.

You are right. Perhaps I should end my posts to this thread, as I have just quite recently re-focused my challenge from x' = x-vt to x' = x-d. Indeed that seems the proper thing to do, as I no longer have anything to post about the Galilean transformation on this thread or any thread any more. Since, x' = x -d is indeed off topic, this will end my posting to this thread. I likely will open a new thread call x' = x-d today, some time. Thanks for pointing this out.

x' = x - d is a translation. There is no need to include time as a coordinate when discussing this transform.
However remember that the primed and unprimed coordinate systems are not, never were, and never will be, coincident (that is, their origins are not coincident)*.

* Again disregarding the trivial case of d = 0

steve waterman
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### x' = x-d?

Does x' = x -d, when restricted to a mathematical Cartesian setting?

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d = 0.
*d = distance of separation between S and S' along the common x/x' axis

Allow one of the two systems to be repositioned by distance d > 0.

1 x' = x-d, if mathematically allowed, would equate unequal coordinate ( lengths ), like 2 = -1.

2 whereas, S(x,y,z) = S'(x',y',z') equates the given coordinate ( lengths ), like 2 = 2.

Observation, given x = x', we cannot do any kind of spiffy math and wind up with x ≠ x'.

( wording of the above is subject to change/feedback, as would be a draft )
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

I'll bite (because few things would please me better than to see you learn what this stuff means and why the questions you've asked don't make sense and/or are trivial).

steve waterman wrote:Does x' = x -d, when restricted to a mathematical Cartesian setting?

Maybe. For a given value of d and a cartesian co-ordiante system, x'=x-d (I'm assuming y'=y, z'=z. w'=w ... for any other co-ordinates) specifies a single cartesian co-ordinate system which satisfies that relationship i.e. the one whose origin sits at (d,0,0,0,...) in and has the same orientation and handedness as the given system. If we let d range over all real values, we get the set of all cartesian co-ordinate systems of the same handedness and orientation which label every point with identical y, z, w, ... co-ordinates.

steve waterman wrote:Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d = 0.
*d = distance of separation between S and S' along the common x/x' axis

Your first line here assumes that d is 0 (by assuming the trivial identity transformation). As such, all you're considering is a special case of x'=x-d where d=0 and pretending you're including the whole set by still using d is either a glaring error or deliberate obfuscation.

steve waterman wrote:Allow one of the two systems to be repositioned by distance d > 0.

Except you previously assumed that d=0. As such, you've assumed two contradictory statements and so the principle of explosion applies and no conclusion you reach has any relevance to anything. It is worth noting that, the statement 2 = -1 is true in the logical system as your assumptions define it. This is in fact true of any proposition. If we work in this world where d = 0 and d != 0, it is true that "badgers and blue than has".

steve waterman wrote:1 x' = x-d, if mathematically allowed, would equate unequal coordinate ( lengths ), like 2 = -1.

Only because you assumed d = 0 in your first line and are now assuming that d != 0. In the system as you have defined it, it is indeed correct that 2 = -1 and no contradiction lies in this. The contradiction lies in the assumption that d = 0 on the first line.

steve waterman wrote:2 whereas, S(x,y,z) = S'(x',y',z') equates the given coordinate ( lengths ), like 2 = 2.

It does no such thing. Go back and read Schrollini's posts about co-ordinate transformations in the other thread. S(x,y,z)=S'(x',y',z') says that the points S(x,y,z) and S'(x',y',z') are the same. It says absolutely nothing about any lengths (because no such concept need be introduced in order to produce such statements as demonstrated by Schrollini in the other thread) and it says nothing about the relations between the co-ordinates themselves. Without knowing the co-ordinate systems that S and S' refer to, the only thing this tells us is that S(x,y,z) is the same point as S'(x',y',z').

steve waterman wrote:Observation, given x = x', we cannot do any kind of spiffy math and wind up with x ≠ x'.

No, because you assumed it at the beginning.

If you had managed to derive it, your logic would by definition be circular and not useful to anything.

It's clear that you have not understood what Schrollini seemed to have taught you. Go back and read the post when he first introduced co-ordinate transformations.

Edit: wrote this reply when this was another thread so a few things might be worded slightly confusingly.
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WibblyWobbly
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### Re: x' = x-d?

steve waterman wrote:Does x' = x -d, when restricted to a mathematical Cartesian setting?

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d = 0.
*d = distance of separation between S and S' along the common x/x' axis

Allow one of the two systems to be repositioned by distance d > 0.

1 x' = x-d, if mathematically allowed, would equate unequal coordinate ( lengths ), like 2 = -1.

2 whereas, S(x,y,z) = S'(x',y',z') equates the given coordinate ( lengths ), like 2 = 2.

Observation, given x = x', we cannot do any kind of spiffy math and wind up with x ≠ x'.

( wording of the above is subject to change/feedback, as would be a draft )

Let me try this (I may be a bit sloppy; calling all pedants for clarification):

I have two friends, Dave and Kristin. My house is five miles due east of Dave's house. Kristin's house is three miles due east of Dave's house. Kristin comes over for dinner one night, and we play our favorite game: "Let's Be Cartesian Coordinate Systems!" We consider the surface of the earth to be our manifold, and identify Dave's house as a point of interest. Then I say "I'm a Cartesian coordinate system origin! My axes are defined in these directions!" <waves hands like dying water buffalo> Kristin then says "I want to do that too! Let's be coincident coordinate systems!" <She, too, does the dying water buffalo dance>

So, we are now coincident. As far as I can tell, your assertion boils down to this: When Kristin goes home for the night, she is still five miles from Dave's house. Because when we were coincident, five miles is five miles, and that can never be changed, right?

Coordinate transformations don't care about your red and blue rulers. Yes, when two rulers start at the same point and are oriented in the same direction, two inches on blue is the same as two inches on red. (Coincidentally, that would be true if the rulers were never coincident, but that's because that's how rulers are supposed to be constructed, not some vague "magic abscissa" notion of Cartesian coordinates). What coordinate transformations are interested in is that object two inches away from the origin of the blue ruler. When the blue and red rulers are in the same place, the red ruler also shows that object as being two inches away. When we move the red ruler an inch to the right, however, that object we were interested in isn't two inches away, it's one inch away. And that point that is now two inches away from the red ruler origin? Who cares? It might not even exist, for all I give a shit. Wasn't interested, doesn't mean anything for actual, mathematical coordinate transformations. Because coordinate transformations are about getting two sets of coordinates that represent the same point, not two points that have the same coordinates. And that's where your mistake seems to be...

... among others.

Addition (with respect to nothing in particular):

beojan wrote:You should try using a text editor with the Solarized colour scheme to compose your posts.

I know this quote was directed at Steve, but I took a look at this color scheme, and it is quite nice! I use vim quite a lot, but never got into plugins or color schemes. So, many thanks for that, and if you've got any other tips or tricks like that, I'd be interested.
Last edited by WibblyWobbly on Fri Jul 19, 2013 7:43 pm UTC, edited 1 time in total.

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

Steve,

This has been asked before. What is your mathematical background? I'm honestly appalled at your seeming complete lack of understanding of how to prove things, define things, or even talk about mathematics.

Mathematics requires clarity and precision. You use nebulous terms that you refuse to define, and refuse to look at the language that every other mathematician uses. Mathematics is also a community. We have an agreed upon language that allows us to proceed, your obvious flouting of that impresses no one.

You seem to have this notion that these small, snappy lines of overly abbreviated statements are what constitute a proof. They are not. Go read a goddamn book or paper. There is exposition and motivation in any good paper. Especially one that treads new ground. You must explain yourself in full sentences, using English, without abbreviations, and only invoking results, definitions, and concepts that you fully understand.
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### Re: x' = x-d?

WibblyWobbly wrote:I have two friends, Dave and Kristin. My house is five miles due east of Dave's house. Kristin's house is three miles due east of Dave's house. Kristin comes over for dinner one night, and we play our favorite game: "Let's Be Cartesian Coordinate Systems!" We consider the surface of the earth to be our manifold, and identify Dave's house as a point of interest. Then I say "I'm a Cartesian coordinate system origin! My axes are defined in these directions!" <waves hands like dying water buffalo> Kristin then says "I want to do that too! Let's be coincident coordinate systems!" <She, too, does the dying water buffalo dance>

So, we are now coincident.

Kinky!
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beojan
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### Re: x' = x-d?

WibblyWobbly wrote:I have two friends, Dave and Kristin. My house is five miles due east of Dave's house. Kristin's house is three miles due east of Dave's house. Kristin comes over for dinner one night, and we play our favorite game: "Let's Be Cartesian Coordinate Systems!" We consider the surface of the earth to be our manifold, and identify Dave's house as a point of interest. Then I say "I'm a Cartesian coordinate system origin! My axes are defined in these directions!" <waves hands like dying water buffalo> Kristin then says "I want to do that too! Let's be coincident coordinate systems!" <She, too, does the dying water buffalo dance>

So, we are now coincident. As far as I can tell, your assertion boils down to this: When Kristin goes home for the night, she is still five miles from Dave's house. Because when we were coincident, five miles is five miles, and that can never be changed, right?

This is essentially what my post with the trees in the field was about.
Of course, I doubt Steve will read yours either (or perhaps he'll read it then fail to respond, or respond with a statement that has nothing whatsoever to do with this post).

WibblyWobbly wrote:Addition (with respect to nothing in particular):

beojan wrote:You should try using a text editor with the Solarized colour scheme to compose your posts.

I know this quote was directed at Steve, but I took a look at this color scheme, and it is quite nice! I use vim quite a lot, but never got into plugins or color schemes. So, many thanks for that, and if you've got any other tips or trickses like that, I'd be interested.

Glad to be of help. You should also try Powerline.

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

Steve, we have told you this already — no one disagrees with you that if the systems are co-incident, then x=x', and x'!=x-d (for non-trivial d). The thing you don't seem to realise is that you don't need a proof for this — it is the definition of 'co-incident'. We all agree.

With that said, what's your point?
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ucim
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d = 0.
*d = distance of separation between S and S' along the common x/x' axis
Allow one of the two systems to be repositioned by distance d > 0.
We will need some way to tell whether we are looking at the repositioned system before "repositioning", or afterwards. Usually this is understood, but in this case let's be explicit. Let's let the system

S"(x",y",z") represent the "after repositioning" version. There are now three systems to consider.

We'll also need a way to identify which value of d we are talking about, since repositioning (as you like to call it) actually changes d. Let's use d' for the original value of d, and use d" for the new value of d when the "repositioning" occurs. Look closely, the primes are hard to see.

When coincident, x=x'. Also, x'=x-d' This implies that d' = 0. Under the circumstances, this means that the systems are coincident.

So far, so good.

Now reposition, and use the correct system. We need the double-prime system, with d" being nonzero.

x"=x-d" is the correct way to express the situation.

x'=x-d' is still true, but does not apply to this (moved) system. d' is zero, if you recall, while d" is not.

x'=x-d" is however false. The error is using the "new" d" with the "old" x'
x"=x-d' is also false. The error is using the "old" d' with the "new" x"

This is the crux of it. You are making one of the errors above, but your notation and exposition hide it from your own brain. See if you can follow it with the notation above, which will help you keep in mind which is the "about to be moved" system, and which is the "already moved" system.

Next thing is to realize that nothing actually moved. We merely considered a new system. You can call it a moved system if you want, and if you understand what is really happening. But in fact, it's just another system.

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

I'm posting mostly so this thread starts showing up in my View New Posts, but I have to say I'm not surprised to see it's turning out this way and I'm kind of glad I didn't bother trying to participate earlier.

To Steve, as I've said many times before (and others have repeated again in this thread): the things you are trying to prove are obvious things that nobody questions. The equation you think you are trying to disprove does not actually dispute them either. The reason you think it does is because you don't understand what that equation is really saying. Everyone here is trying to explain to you what it is really saying, and why it doesn't say anything contradictory to the things you are trying to prove.

It doesn't say that two different numbers are equal.

It doesn't say that two different lengths are equal.

It only says that two different labels apply to the same place, and that there is a pattern in how two different labeling schemes label the same places.
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:Did you start with our two coordinate systems as coincident with the manifold?

No.

This question suggests that you're still thinking of coordinate systems as dynamic objects that evolve in some sort of "time". They aren't. There is no "time" in which they can evolve. They simply exist.

NO? That was a complete surprise!

Steve, I really wish you'd remember this surprise. Your intuition says that coordinate systems start aligned, and then get moved to separate positions. But that's not how they work with the conventional definitions. Instead, they come into being separately, and then we have to figure out how to convert between them. This is true for rotations, and it's also true translations (and it's also true for the Galilean transformation). Until you grasp this point, I don't think you have any chance of understanding what everyone else is trying to tell you.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
Steve, I really wish you'd remember this surprise. Your intuition says that coordinate systems start aligned, and then get moved to separate positions. But that's not how they work with the conventional definitions. Instead, they come into being separately, and then we have to figure out how to convert between them. This is true for rotations, and it's also true translations (and it's also true for the Galilean transformation). Until you grasp this point, I don't think you have any chance of understanding what everyone else is trying to tell you.

wiki -
The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) 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 x’ directions, with their spatial origins coinciding at time t=t'=0

I simply start with the same coincident systems scenario as does the above.
However, my proof only deals with the mathematical challenge to x' = x-d.
Last edited by steve waterman on Sat Jul 20, 2013 4:17 pm UTC, edited 1 time in total.
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Dark Avorian
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Steve,

There is a distinction that needs to be made here. Time has been used in two senses in this discussion.

First, we have considered time as one of the dimensions over which our coordinates run when we are trying to identify points in the manifold of interest (that is, spacetime). This is perfectly mathematical, but it should be noted that we have little freedom in this case, as the time is part of the coordinate system.

We have also implicitly mentioned some type of time when you talk about moving around coordinate systems. We cannot do this. There is no notion of time EXTERNAL to the coordinate systems. Our identification of time is a facet of the coordinate system itself. Once the coordinate system has been fixed, we are locked into it.
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ucim
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote: as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and x’ directions, with their spatial origins coinciding at time t=t'=0

1: The "moving" system did not start of stationary and then get repositioned. It was always "moving".
2: Time is a component of each of the systems. As such, "movement" is illusory. It's a metaphor that helps some people picture it. But if misunderstood, it can lead to nonsense.

There are many ways to notate something. Nonsense results if you mix notation though.

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.

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

Dark Avorian wrote:Steve,

There is a distinction that needs to be made here. Time has been used in two senses in this discussion.

First, we have considered time as one of the dimensions over which our coordinates run when we are trying to identify points in the manifold of interest (that is, spacetime). This is perfectly mathematical, but it should be noted that we have little freedom in this case, as the time is part of the coordinate system.

We have also implicitly mentioned some type of time when you talk about moving around coordinate systems. We cannot do this. There is no notion of time EXTERNAL to the coordinate systems. Our identification of time is a facet of the coordinate system itself. Once the coordinate system has been fixed, we are locked into it.

There is no time element involved in my mathematical proof. It is only about x' = x-d. There is is only about the coordinates relationship between S to S'.
This EQUATION, x' = x-d, is not about mapping or images, or anything manifold what-so-ever!
Just about coordinates; x is the first coordinate, x is not a point.
Given x = x', therefore always x = x'.
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steve

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

steve waterman wrote:wiki -
The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) 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 x’ directions, with their spatial origins coinciding at time t=t'=0

Note what I've bolded. Time is a coordinate of these systems. The fact that the spatial parts are coinciding at a bit in time does not mean that the full coordinate systems are coincident.
I simply start with the same coincident systems scenario as does the above.
However, my proof only deals with the mathematical challenge to x' = x-d.

If that were true, you wouldn't be speaking about "moving" a coordinate system. See Jose's post for more details.
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Dark Avorian
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:
Dark Avorian wrote:Steve,

There is a distinction that needs to be made here. Time has been used in two senses in this discussion.

First, we have considered time as one of the dimensions over which our coordinates run when we are trying to identify points in the manifold of interest (that is, spacetime). This is perfectly mathematical, but it should be noted that we have little freedom in this case, as the time is part of the coordinate system.

We have also implicitly mentioned some type of time when you talk about moving around coordinate systems. We cannot do this. There is no notion of time EXTERNAL to the coordinate systems. Our identification of time is a facet of the coordinate system itself. Once the coordinate system has been fixed, we are locked into it.

There is no time element involved in my mathematical proof. It is only about x' = x-d. There is is only about the coordinates relationship between S to S'.
This EQUATION, x' = x-d, is not about mapping or images, or anything manifold what-so-ever!
Just about coordinates; x is the first coordinate, x is not a point.
Given x = x', therefore always x = x'.

I grow tired of your disingenuous approach to this conversation and your constant repetition of utter babble.

We do not speak of coordinates without speaking of points. The very definition of coordinate entails identifying a point.

You keep referring to this equation x' = x-d. You do realize this is nearly meaningless right. If youd aim to relate this in any way to space, then you need to identify the numbers with points in some way: f(x') = g(x-d) [this means that in the system "f" x' refers to a point P, and the system "g", x-d refers to the same point] in which case we have no way to algebraically manipulate the numbers with relation to each other.

We cannot say that x' = x-d in this situation. The coordinates aren't really a distance, but even insofar as the are, they are measured from f(0) and g(0) in their appropriate systems. If the systems have origins coincident, then we don't necessarily have the that x' = x-d even then.

If you refuse to refer to manifolds, points, or anything else, then you are also not allowed to talk about coordinates. (Coordinate, to repeat for the umpteenth time, means that you are mapping elements of our nicest space (R^n) to elements of another space so as to identify that it has locally nice structure.

If you refuse to refer to these, then good job: x' = x-d if and only if d = 0. This is a fucking trivial result of the definition of zero.
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steve waterman
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### Re: Galilean:x' with respect to S'?

wiki -
The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) 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 x’ directions, with their spatial origins coinciding at time t=t'=0[/quote]
Schrollini wrote:Note what I've bolded. Time is a coordinate of these systems. The fact that the spatial parts are coinciding at a bit in time does not mean that the full coordinate systems are coincident.

S and S are GIVEN as coincident when d = 0, as does Galilean at t = 0. .
Remember, I am discussing only x' = x-d, and the new thread for x' = x-d, was piggy-backed here.

I simply start with the same coincident systems scenario as does the above.
However, my proof only deals with the mathematical challenge to x' = x-d.
Schrollini wrote:If that were true, you wouldn't be speaking about "moving" a coordinate system. See Jose's post for more details.

I tried the concept/phrasing of snapshot 1 and snapshot 2, and I tried repositioned. There is no velocity nor time involved.

Given S(x,y,z) coincident S'(x',y',z') with x = x', y = y', z= z'
since the systems are given as coincident, then obviously, d = 0, whether directly stated or not.
d = distance between S(0,0,0) and S'(0,0,0) along the common x/x' axis
WE have
case 1 d = 0, S and S are coincident
case 2 d > 0, S and S are not coincident

case 1 x' = x
case 2 x' = x
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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steve

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

steve waterman wrote:S and S [sic - Perhaps you mean S'?] are GIVEN as coincident when d = 0, as does Galilean at t = 0. .

No. Not at all. Because the Galilean includes time, and you must consider all space and all time when deciding whether the systems are coincident.

Oh, and remember how I said we should avoid using primes....
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WibblyWobbly
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

blanked
Last edited by WibblyWobbly on Sat Jul 20, 2013 5:43 pm UTC, edited 1 time in total.

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

Steve,

A coordinate system is. It does not change. If you take a snapshot of space, with no time allowed, and the coordinates changed, then that doesn't mean that there is one coordinate system and it changed, it means that you switched to A DIFFERENT COORDINATE SYSTEM

In fact, let me make something clear. We've been using this term "Coincident"

Suppose we have two coordinate systems, S and S'. S and S' are coincident just means they are equal, it means they are the same coordinate system, we cannot change either of them, they are the same object.

The idea of coincidence is only useful as something more that equality when we speak about restricted cases. For example

S(x,y,z,t) is coincident with S'(x',y',z',t') when t=0 is something meaningful.

Edit:

So, just to reiterate, so that it maybe gets through your skull:

If your coordinate system changes, it is no longer the same coordinate system.
COORDINATE SYSTEMS CANNOT CHANGE.

If your coordinate system changes, it is no longer the same coordinate system.
COORDINATE SYSTEMS CANNOT CHANGE.

If your coordinate system changes, it is no longer the same coordinate system.
COORDINATE SYSTEMS CANNOT CHANGE.

If your coordinate system changes, it is no longer the same coordinate system.
COORDINATE SYSTEMS CANNOT CHANGE.

If your coordinate system changes, it is no longer the same coordinate system.
COORDINATE SYSTEMS CANNOT CHANGE.

If two coordinate systems are coincident in all coordinates, the are the SAME DAMN COORDINATE SYSTEM. You cannot change one away from the other, YOU CANNOT CHANGE THEM

If two coordinate systems are coincident in all coordinates, the are the SAME DAMN COORDINATE SYSTEM. You cannot change one away from the other, YOU CANNOT CHANGE THEM

If two coordinate systems are coincident in all coordinates, the are the SAME DAMN COORDINATE SYSTEM. You cannot change one away from the other, YOU CANNOT CHANGE THEM

If two coordinate systems are coincident in all coordinates, the are the SAME DAMN COORDINATE SYSTEM. You cannot change one away from the other, YOU CANNOT CHANGE THEM

If two coordinate systems are coincident in all coordinates, the are the SAME DAMN COORDINATE SYSTEM. You cannot change one away from the other, YOU CANNOT CHANGE THEM

Hopefully that'll get through.

Best,
Dark Avorian
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Dark Avorian
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Apologies to all for double posting, but I feel this analogy deserves the attention.

Steve,

Let us say x is some number, not a function, not a moving thing, just a number.

Let us say that 3 = x.

Put x in a box

Go watch a movie.

Open the box

Take another snapshot.

You see four.

You might claim x=4.

You would be wrong.

x is no longer in the box. x was defined to be a fixed number. We can't just ignore our definition now for convenience.
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JudeMorrigan
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:This EQUATION, x' = x-d, is not about mapping or images, or anything manifold what-so-ever!

[yoda]
That is why you fail.
[/yoda]

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

As far as the concept of coordinate systems not moving, that threw me a little at first, too, especially since so much of the Pressures thread was spent talking about "moving" systems - the reference frame of one spaceship passing another, one car passing another, etc, where the second ship/car/person is at rest compared to some spot we called point P. So mostly as a test to see if I got it right, I pulled together the .gif below to show the reconciliation between the way things were being discussed in the original thread (mostly before Schrollini arrived, I think) and coordinate systems fixed in spacetime. The top half shows the three spatial dimensions; passing time is noted in the upper left corner. In the bottom half, the spatial dimensions are collectively represented by x and x'. The vertical red line is the world line of point f(x=7,y=0,z=0). I have the t' axis intersecting that line at t'=t=3.5, since that is when the g frame's spatial origin (would that be the right term?) is at P.
Schrollini, ucim, yurell, et al, does this look correct?

Please ignore the typo in the bottom right at t= -1.0
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Schrollini
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

ivnja wrote:As far as the concept of coordinate systems not moving, that threw me a little at first, too, especially since so much of the Pressures thread was spent talking about "moving" systems - the reference frame of one spaceship passing another, one car passing another, etc, where the second ship/car/person is at rest compared to some spot we called point P.

And this was part of the problem -- by privileging one coordinate system as being "fixed" on the manifold, we obscure the difference between the manifold and the coordinate system. For people who understand the difference, this isn't a problem. For people who don't....

There are at least two rigorous ways around this that I could think of. The one I chose was to insist on talking about spacetime whenever time is involved. Although the spatial part of the coordinate systems are moving, there's no "meta-time" in which the full coordinate systems move. They both just sit there on the manifold of spacetime, neither of them "moving". This has the advantage of providing an easy segue to Special Relativity.

ivnja wrote:Schrollini, ucim, yurell, et al, does this look correct?

The one correction I would make is that we're considering spacetime, so the idea of a "point in space" isn't well defined. You should consider either an event (a point in the manifold of spacetime), which would have both space and time coordinates, or a trajectory. As you've drawn it now, P is actually the trajectory x = 7, or equivalently x' = 7 - vt.
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