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

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

Schrollini wrote: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.

The top half is meant more as a "this is the way we were talking about it at the start of Pressures" - and I suppose as how someone unfamiliar with relativity would say they perceive the scenario. And as a more practical matter, I couldn't figure out how to represent an event in a way that made sense without a time axis being shown. I could have the dot at P suddenly appear at some time, with the appearance being the event, but it seemed odd to me to be measuring spatial distance to something that hasn't occurred yet for the first however many frames.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

@ivnja: Looks good to me, given Schrollini's comment. (Picturing a "point in space" tied to one reference frame privileges that reference frame.) It is equally valid to tie your "point in space" to the other frame; it would look weird though because we are used to privileging the frame that stands still on our monitor. What we are used to, however, is wrong.

As a "what we were using" and "what we should be using" comparison, it's pretty good.

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

The thing is, Steve has made it clear that he is trying to present a mathematical challenge and not a physical one.

As such, we should not be bringing in any physical interpretation and so including things about reference frames, moving observers and only considering the spatial co-ordinates (when looking at the Galilean, obviously we don't need to do this for the shear x'=x-d) only obfuscates the real point we need to drive home which is what a co-ordinate transformation actually is.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

I honestly don't understand what Steve is contending any more. Steve, can you state cogently and strictly the statement that everyone else believes is true, what you believe is true, and each logical step used to disprove what everyone else thinks is true?
Because from what I can see it looks like you're striking at strawmen and shadows.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

yurell wrote:I honestly don't understand what Steve is contending any more. Steve, can you state cogently and strictly the statement that everyone else believes is true, what you believe is true, and each logical step used to disprove what everyone else thinks is true?
Because from what I can see it looks like you're striking at strawmen and shadows.

I got really ill this weekend from what seems to a freon leak in my fridge. Just getting back on my feet. I will write a post tomorrow.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Sorry to hear that, Steve. Best wishes on a speedy recovery.

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

Ditto. Hope you're feeling better.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:I got really ill this weekend from what seems to a freon leak in my fridge.

You know, Einstein designed a refrigerator specifically to avoid this problem:
wikipedia wrote:The Einstein-Szilard or Einstein refrigerator is an absorption refrigerator which has no moving parts, operates at constant pressure, and requires only a heat source to operate. ... The two were motivated by contemporary newspaper reports of a Berlin family who had been killed when a seal in their refrigerator broke and leaked toxic fumes into their home.

Hope you feel better!
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

yurell wrote: Steve, can you state cogently and strictly the statement that everyone else believes is true.

Pardon please, it is likely some sloppy wording that follows...please check first, as perhaps I have used one of your terms incorrectly, so, I am throwing this out there for mutual discussion.... let vt = d,

everyone else believes x' = x - d equates the relationship of points in the manifold between the two mappings.
yurell wrote:what you believe is true,

I would equate point (x',0,0) = point (x-d,0,0) in the manifold

and coordinate x = coordinate x'.
Allowing x' = x -d manifests two variant solutions for x'....
Given x = x', x' is also ALWAYS equal to the abscissa of S', as HIS coordinate x' now becomes doubly equated, this time as x -d!, therefore generating the mathematical inequality ...x' = x = x - d.
yurell wrote:and each logical step used to disprove what everyone else thinks is true?

x is a coordinate ( length ) not a point
(x,0,0) is a point, and as such, has no associated length, indeed, a point has no dimensions.

Thanks for the kind words for good health and thanks too for all your collective extended patience.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:
yurell wrote:everyone else believes x' = x - d equates the relationship of points in the manifold between the two mappings.

This is not a relationship between points. It is a relationship between coordinates in different systems for a single point.

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

steve waterman wrote:everyone else believes x' = x - d equates the relationship of points in the manifold between the two mappings.

Do you have any quote of someone actually saying that?

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

steve waterman wrote:
yurell wrote: Steve, can you state cogently and strictly the statement that everyone else believes is true.

Pardon please, it is likely some sloppy wording that follows...please check first, as perhaps I have used one of your terms incorrectly, so, I am throwing this out there for mutual discussion.... let vt = d,

everyone else believes x' = x - d equates the relationship of points in the manifold between the two mappings.

No. u = x-d (to borrow Schrollini's terminology) is the relationship between two coordinate systems of the same manifold.

steve waterman wrote:
yurell wrote:what you believe is true,

I would equate point (x',0,0) = point (x-d,0,0) in the manifold

If you change that to S'(x',0,0) = P = S(x,0,0) you would be correct.

steve waterman wrote:and coordinate x = coordinate x'.

this is meaningless without more information.

steve waterman wrote:Allowing x' = x -d manifests two variant solutions for x'....
Given x = x', x' is also ALWAYS equal to the abscissa of S', as HIS coordinate x' now becomes doubly equated, this time as x -d!, therefore generating the mathematical inequality ...x' = x = x - d.

this is occurring only because you are equating 3 different coordinate systems.
S(x,0,0)
S'(x',0,0)
S''(x-d,0,0)

S does not equal S'' (except if and only if d=0).

steve waterman wrote:
yurell wrote:and each logical step used to disprove what everyone else thinks is true?

x is a coordinate ( length ) not a point
(x,0,0) is a point, and as such, has no associated length, indeed, a point has no dimensions.

Thanks for the kind words for good health and thanks too for all your collective extended patience.

No. x is a number. It is not a length, a coordinate, or anything else. Its just a number.

EDIT for quote-tag wierdness.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Steve, I too hope you are feeling better and continue to do so.
steve waterman wrote:let vt = d,

everyone else believes x' = x - d equates the relationship of points in the manifold between the two mappings.
No, x' = x - d is simply an ordinary algebraic equation using real numbers. Real numbers are used as coordinates. Coordinates, in the context of a mapping, identify points in the manifold.

math:
Given a 2 dimensional manifold M and a mapping R...
R(3,6) identifies a point in M.
R(3,6) is the name of that point in M.
It is often said that R(3,6) is that point in M. This is true to the extent that steve waterman is you, and you are steve waterman. To nitpick, steve waterman is a name, and it identifies a real person who answers to that name. Normally we don't make that distinction except to quote Monty Python, but is important to recognize the teeny shortcut we take when we say:
R(3,6) "is" a point in M.

example:
Given ordinary 2 dimensional space (a flat plane), a cartesian coordinate system R whose origin and axis alignment is defined to everyone's satisfaction, R(3,6) uniquely identifies a point in that plane.

If you also consider a different cartesian coordinate system, even one that is similar to the one above except for a translation along the x-axis, it is still a different system, and requires a different name. You've used primes, I'll use different letters. Either way works.

I'll call this "different" system D. D(3,6) also uniquely identifies a point in the plane. It is (of necessity) a different point than the one identified by R(3,6).

Nothing at all is identified by (3,6) by itself. This is true even if you say "the point at (3,6)". You must say something like "the point at (3,6) in system R", or "R(3,6)".

application:
R(3,6) identifies a different point in the plane from D(3,6)
in shorthand, R(3,6) =/= D(3,6)
This is true even though 3=3 and 6=6. This is because R is not the same system as D.

Now, the 3 that is used as a coordinate does not care how it was created. It is still three. It could have been created by adding seventeen to negative fourteen. It could have been created by adding two to one. It could have been created by starting with three and doing nothing. It could have been created by an algebraic expression like x-d where x=4 and d=1. But, no matter how that three was "created", it is still three, and each of the functions (R or D) will use that three at face value.

steve waterman wrote:I would equate point (x',0,0) = point (x-d,0,0) in the manifold
If you are good to here, I'll explain in my next post the reason this does not fly. In short, it is that you have not specified any system to use in your mapping, and absent a specified system, coordinates do not mean anything.

Let me know.

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

steve waterman wrote:I would equate point (x',0,0) = point (x-d,0,0) in the manifold

and coordinate x = coordinate x'.

This first line is correct but only for a specific set of transformations. The second is true only for a single transformation (which is also an element of the set that satisfies the first). If you assume that all transformations that satisfy the first line also satisfy the second, you're obviously going to run into trouble. You have taken two contradictory (except in the trivial case) axioms; of course you get contradictions.

ucim wrote:math:
Given a 2 dimensional manifold M and a mapping R...
R(3,6) identifies a point in M.
R(3,6) is the name of that point in M.
It is often said that R(3,6) is that point in M. This is true to the extent that steve waterman is you, and you are steve waterman. To nitpick, steve waterman is a name, and it identifies a real person who answers to that name. Normally we don't make that distinction except to quote Monty Python, but is important to recognize the teeny shortcut we take when we say:
R(3,6) "is" a point in M.

No, R(3,6) definitely is a point. R is the co-ordinate system and acts on the tuple (the co-ordiantes which is analogous to the name) returning a point. As such, R(3,6) very definitely is a point and in no way is the name.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

everyone else believes x' = x - d equates the relationship of points in the manifold between the two mappings.[/quote]
yurell wrote:This is not a relationship between points. It is a relationship between coordinates in different systems for a single point.

Yes, thanks. That's sounds much better, I am okay with that, so...
Everyone else believes x' = x - d equates the relationship between coordinates in different systems for a single point
How does that sound for wording regarding what everyone else believes?
This is a very logical place to start.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:Everyone else believes x' = x - d equates the relationship between coordinates in different systems for a single point
How does that sound for wording regarding what everyone else believes?
This is a very logical place to start.

That's true, but it's not really a belief. There's no meaning inherent to the symbols "x' = x - d"; we have to imbue them with meaning by deciding what they mean. And the meaning we choose is that this gives a relationship between coordinates in different systems for a single point.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Steve.

You seem to be getting the gist of things. Sort of.

I am going to ask a few questions.

Consider the function f(x) = x^2

f(1) = 1 = f(-1)

Do you disagree?

Does 1 = -1 ? No.

Let f(x) = x + 3, Let g(x) = x^3

g(2) = 8 = f(5)

does 2 =5 ? No.

Do these seem like stupid questions? If so, then understand that this is basically what you are doing. You are taking our claim that two points on the manifold, with different coordinates, are equal, to mean that the coordinates are equal. That is not true. A coordinate system is a function. We put in the coordinates, and it outputs a point on the manifold. If two different coordinate systems output the same point, we have no idea about the inputs and how they are related. (Well, we can define a coordinate transform, but that's the whole point).

Now, onto one of your language choices.

Abscissa: abscissa is, as you know, the x-coordinate of a point. Here's the problem, saying two points are the same doesn't mean they have the same abscissa, there is no privileged axis that we can measure from, no privileged point is the origin. Worse yet, the units of distance in the coordinate systems do not relate in a privileged way to distances in the manifold.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Schrollini wrote:
steve waterman wrote:Everyone else believes x' = x - d equates the relationship between coordinates in different systems for a single point
How does that sound for wording regarding what everyone else believes?
This is a very logical place to start.

That's true, but it's not really a belief. There's no meaning inherent to the symbols "x' = x - d"; we have to imbue them with meaning by deciding what they mean. And the meaning we choose is that this gives a relationship between coordinates in different systems for a single point.

Or in other words, everyone else believes that:

IF
by x' we mean the coordinate of a point in one coordinate system
AND
by x we mean the coordinate of that same point in a different coordinate system
AND
by d we mean the offset between those two coordinate systems
THEN
x' = x - d

Without that antecedent (the "if"), we have no idea what x', x, or d mean, and thus we have no more idea whether x' = x - d than we do whether q = j - w. It's just an equation between meaningless variables, until we give them meanings. And the meanings we give them make that equation trivially true. If you give them different meanings, it may make it just as trivially false, but then you've proven something different false than the thing we've proven true.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Everyone else believes x' = x - d equates the relationship between coordinates in different systems for a single point
How does that sound for wording regarding what everyone else believes?
This is a very logical place to start.
Schrollini wrote:That's true, but it's not really a belief. There's no meaning inherent to the symbols "x' = x - d"; we have to imbue them with meaning by deciding what they mean. And the meaning we choose is that this give a relationship between coordinates in different systems for a single point.

Schrollini wrote:There's no meaning inherent to the symbols "x' = x - d";

That is getting closer to the essence of things.
Schrollini wrote:we have to imbue them with meaning by deciding what they mean.

True, Keeping in mind, please, "mathematics in not subjective". So, from my
point of view, given S(x,y,z) defines x, as the first coordinate (distance).
For me, there is no "choice", up for grabs.
x, for me, is inherent to S, given Cartesian coordinate system S(x,y,z).
Just as x', is mathematically defined, given S(x',y',z').

Physicists cannot decide what they apply to! They are coordinates.
Physicists can decide that POINT P = (x-d,0,0) = (x',y',z')....but that has nothing to do with x the abscissa/distance.
The manifold, for me, is all smoke and mirrors, and x' = x, is incredibly obvious, and ONLY has to do with
coordinates/abscissa/distances/lengths/numbers/quanta. So, I hear you loud and clear.

As a mathematician, given S(x,y,z), imo, Physicists do not get to "choose"/imbue. Physicists needs to accept the mathematical reality, that x IS the first coordinate of S, Given S(x,y,z).

btw, they just now removed my dead fridge after I found out that unplugging it may not stop the leak at all.
So, I am running on fumes...and finished posting for tonight. I will try and read all the posts, I think I might have missed a couple recently. Need a break, I'll fix typos tomorrow.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

No Steve, you are not a mathematician. You have displayed that time and time again. There is fucking choice in mathematics. x can be a variable, a point, a constant, a vector, an element of a group a ring or a field, it can be an arbitrary element of a topological set, a multiplication symbol, a cross product etc. Furthermore, you claim to be a mathematician, and yet you dismiss manifolds as smoke and mirrors.

Manifolds are a mathematical construct, the source of a plethora of deep and rich fields of study. To act as if that is the brainchild of physicists is stupid beyond all belief. mathematicians have all sorts of structures, rings, groups, fields vector spaces, Banach spaces, hilbert spaces, non-hausdorff topological spaces. Why in the name of fuck does your insistence on the existence of "distances", "abscissa" and what-not have any more meaning or "pure mathematical sense" than any other choice of structure: HINT: It doesn't.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:True, Keeping in mind, please, "mathematics in not subjective". So, from my
point of view, given S(x,y,z) defines x, as the first coordinate (distance).
For me, there is no "choice", up for grabs.
x, for me, is inherent to S, given Cartesian coordinate system S(x,y,z).
Just as x', is mathematically defined, given S(x',y',z').

This is where Schrollini's objection to using primes makes sense, because I can't tell if you just left out a prime in S'(x',y',z') or you're still unclear on the concept that a coordinate transformation involves two different coordinate systems which need not produce the same abcissa. In fact, the second coordinate system need not even have an abcissa. To drag up an image Schrollini used a couple pages back: Do you agree that point P has coordinates of (2,1) in the blue coordinate system? And do you agree that point P has coordinates of (.5, 1.5) in the red coordinate system? Do you see how the abcissa is a feature of the coordinate system, and not of the point? Do you see how saying that B(2,1) = R(.5,1.5) does not mean that 2 = .5?

Edited to fix the error that was pointed out in my labeling of the coordinates.
Last edited by firechicago on Wed Jul 24, 2013 1:59 am UTC, edited 1 time in total.

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

I will never be able to do the Math.
I can see that 2 does not equal 0.5.
Thank you, for the picture and a question even I can understand.

Back to Real Math with all of you Numbers Guys.
Math is a lovely language some people share.

We never know when the next arch will be invented in the minds of a group of Numbers Guys.
The arch is a big deal.

Some ding-dong may have come up with one all by him or her self.
It took Math people to make it so useful. Arches are strong and last a long time.

Go! Think of some other wonderful thing!
An imaginary donut in Time and Space and Giggles.

Please, Remember to laugh. Numbers can be funny, too.
Not to me. They frighten me.

You can laugh at them.
You know what they mean.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

And in this moment, I suddenly decided addams is wonderful.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

firechicago wrote: Do you agree that point P has coordinates of (2,1) in the blue coordinate system? And do you agree that point P has coordinates of (.5, 1.5) in the red coordinate system? Do you see how the abcissa is a feature of the coordinate system, and not of the point? Do you see how saying that R(0.5,1.5) = B(2,1) does not mean that 2 = .5? [Edited in quote, see footnote]

And to carry this on: if we don't assume that we can eyeball the coordinates of P here, so although they're close to what firechicago says, they might be slightly different numbers we're not sure of, do you (Steve) agree that we could call the unknown coordinates of P in the red system R(x,y), and the unknown coordinates of P in the blue system B(u,v), since that's how we've labelled the axes of those two systems?

And that, if we call the unknown horizontal offset between them d, that u (whatever it is) is equal to x (whatever that is) minus d (whatever that is)? So that if, as it appears to the naked eye, these systems are really horizontally offset by -1.5 units (so d = -1.5), and P really is aligned to the 0.5 value of the x axis as it appears (so x = 0.5) and the 2 value of u (so u = 2), then 2 = 0.5 - (-1.5)? That's true, right? No matter what, 2 = 0.5 - (-1.5), right? Because that's all that "u = x - d" means in this context.

Now what if we labelled the blue axes with primed x and y instead of u and v, so instead of u we write x' and instead of v we write y'. That changes the previous equation from u = x - d to x' = x - d, but it still just means the same thing, which in this case still fleshes out to 2 = 0.5 - (-1.5). And that's still true, isn't it?

If x' always had to equal x, as you insist, then in this case 0.5 would have to equal 2, which would be absurd, wouldn't it?

Edited because I switched blue and red locations in my mind while writing.

And again because I think firechicago made the same mistake when writing those R() and B() functions, assuming R is supposed to be the function of the red-colored coordinate system, and B is supposed to be the function of the blue-colored coordinate system
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:True, Keeping in mind, please, "mathematics in not subjective". So, from my
point of view,

Snerk.

Dark Avorian wrote:And in this moment, I suddenly decided addams is wonderful.

Hear, hear. (Or is that, "here, here"?)

I'll own up to skimming over addams' posts in the past, but now that I read it, it's quite delightful. I wonder what I missed before.
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Dark Avorian
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Schrollini wrote:
steve waterman wrote:True, Keeping in mind, please, "mathematics in not subjective". So, from my
point of view,

Snerk.

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

Schrollini wrote:I'll own up to skimming over addams' posts in the past, but now that I read it, it's quite delightful. I wonder what I missed before.
The beauty inherent in something, which becomes apparent when looked at through an unfamiliar filter.

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

Schrollini wrote:
steve waterman wrote:True, Keeping in mind, please, "mathematics in not subjective". So, from my
point of view,

Snerk.

I'm glad someone else caught that. I was going to say something but didn't want to sound... snerky.

I'll own up to skimming over addams' posts in the past, but now that I read it, it's quite delightful. I wonder what I missed before.

I still don't see what everyone else sees in them. No offense intended to to addams, just... people talk like it's poetry, but I don't see it, even when I look for it. It's just kind of hard-to-read prose for me.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:Everyone else believes x' = x - d equates the relationship between coordinates in different systems for a single point
How does that sound for wording regarding what everyone else believes?
This is a very logical place to start.
Schrollini wrote:That's true, but it's not really a belief. There's no meaning inherent to the symbols "x' = x - d"; we have to imbue them with meaning by deciding what they mean. And the meaning we choose is that this give a relationship between coordinates in different systems for a single point.

Schrollini wrote:There's no meaning inherent to the symbols "x' = x - d";

That is getting closer to the essence of things.
Schrollini wrote:we have to imbue them with meaning by deciding what they mean.

True, Keeping in mind, please, "mathematics in not subjective". So, from my
point of view, given S(x,y,z) defines x, as the first coordinate (distance).
For me, there is no "choice", up for grabs.
x, for me, is inherent to S, given Cartesian coordinate system S(x,y,z).
Just as x', is mathematically defined, given S(x',y',z').

Physicists cannot decide what they apply to! They are coordinates.
Physicists can decide that POINT P = (x-d,0,0) = (x',y',z')....but that has nothing to do with x the abscissa/distance.
The manifold, for me, is all smoke and mirrors, and x' = x, is incredibly obvious, and ONLY has to do with
coordinates/abscissa/distances/lengths/numbers/quanta. So, I hear you loud and clear.

As a mathematician, given S(x,y,z), imo, Physicists do not get to "choose"/imbue. Physicists needs to accept the mathematical reality, that x IS the first coordinate of S, Given S(x,y,z).

btw, they just now removed my dead fridge after I found out that unplugging it may not stop the leak at all.
So, I am running on fumes...and finished posting for tonight. I will try and read all the posts, I think I might have missed a couple recently. Need a break, I'll fix typos tomorrow.

Saying S(x,y,z) is a point and that S'(x',y',z') is also a point does part of the job of imbuing the symbols x and x' with meaning. Saying that they're the same point does the rest.

Also, whilst maths is objective, the labels we give to objects are not. The maths doesn't care what we call our variables in exactly the same way the manifold doesn't care what we call our points (i.e. what co-ordinate system we use). It is completely acceptable to relabel anything in a proof as anything else provided you don't end up with two objects with the same label. This can be done completely arbitrarily as well. The reason people don't do this is that there are certain subjective standards which people have adopted so that they can actually communicate about stuff without having several pages of definitions before each and every two line proof.

Note, this is all mathematical here, I have made no resort to physics.

The manifold too is entirely mathematical. Physically, I am restricted in the co-ordinates I can actually observe things in. All I really get to use is some set of spherical co-ordinates centred with the origin between my eyes. I can't actually observe points with a cartesian co-ordinate system, I can't use cylindrical co-ordinates. I can't observe the origin to be over there. All I can do is see things in spherical polar co-ordinates centred between my eyes. Any other co-ordinate system is therefore a purely mathematical entity which I can relate (mathematically) to my spherical co-ordinates in order to gain a physical interpretation of my observations.

The co-ordinates are entirely mathematical. This is important. The manifold is an abstraction but is certainly closer to reality than the co-ordinates are.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

addams wrote:Go! Think of some other wonderful thing!
An imaginary donut in Time and Space and Giggles.

Mathematician & sci-fi author, Rudy Rucker, wrote a story called Spacetime Donuts.

addams wrote:Please, Remember to laugh. Numbers can be funny, too.
Not to me. They frighten me.

Numbers are wonderful, but numbers can make you numb, to paraphrase Rudy.

ucim wrote:
Schrollini wrote:I'll own up to skimming over addams' posts in the past, but now that I read it, it's quite delightful. I wonder what I missed before.
The beauty inherent in something, which becomes apparent when looked at through an unfamiliar filter.

Jose

Indeed.

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

Have some more reading to do, as I just briefly read through the recent posts.

Meanwhile, perhaps it is best to go one question at at time, for a bit.

When we are given S(x,y,z) coincident S'(x',y',z') with x = x', y = y', z = z' and d = 0, that mathematically means
the first coordinate in S equals the first coordinate in S', as would be say, 2 = 2. Correct?

added -

I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

Since x = x' was given at coincidence/d = 0, I equate coordinate x' = coordinate x as remaining true, regardless of any
math exercise being foisted upon that given x = x'. We agree in principle here, "given equalities do not change".
It is now a matter of figuring out if x is mathematically defined/inherent to a Cartesian system, or has a choice for meanings, for Physicists, as Schrollini has suggested.
Last edited by steve waterman on Wed Jul 24, 2013 12:39 pm UTC, edited 3 times in total.
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beojan
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

If S(x,y,z) and S'(x',y',z') are coincident, with x = x', y = y', z=z' for all points, then, making the usual assumptions (same unit of measurement, both are Cartesian, etc), S and S' are two names for the same coordinate system.

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

beojan wrote:If S(x,y,z) and S'(x',y',z') are coincident, with x = x', y = y', z=z' for all points, then, making the usual assumptions (same unit of measurement, both are Cartesian, etc), S and S' are two names for the same coordinate system.

Nope. S has only (x,y,z) coordinates and S' has only (x',y',z') coordinates. Given two systems.
After one system gets repositioned, it should be obvious that there are two UNIQUE origins/systems.
As a mathematician, I could simply say, okay, now remove the S' system leaving the S system.

Simply being coincident does not make either system temporarily disappear, the two systems merely share/equate the same spatial locations.
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JudeMorrigan
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

I'll let others deal with the nature of coincident coordinate systems, but real quick - this is backwards, isn't it? (Demonstrating the dangers of those primes.) Shouldn't it be S'(x-d,0,0) = S(x,0,0)?

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

JudeMorrigan wrote:
steve waterman wrote:I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

I'll let others deal with the nature of coincident coordinate systems, but real quick - this is backwards, isn't it? (Demonstrating the dangers of those primes.) Shouldn't it be S'(x-d,0,0) = S(x,0,0)?

Indeed. S'(x-d,0,0) = S(x,0,0), That math would be fine. x in one system transformed to x in the other.

However, the equation IS x' = x-d, which means, mathematically, x in one system transformed to x' in the other.
Their equation is not x = x- d. The lack of ANY system notation in that equation is cause for great concern.
The fact that their equation is NOT written as point S(x,0,0) yada yada and only as coordinate x, should be a huge clue.
If this were about the manifold, then why x, not S(x,0,0), shows up in THEIR transformation equation.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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Chaoszerom
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

I'll let others deal with the nature of coincident coordinate systems, but real quick - this is backwards, isn't it? (Demonstrating the dangers of those primes.) Shouldn't it be S'(x-d,0,0) = S(x,0,0)?

Indeed. S'(x-d,0,0) = S(x,0,0), That math would be fine. x in one system transformed to x in the other.

However, the equation IS x' = x-d, which means, mathematically, x in one system transformed to x' in the other.
Their equation is not x = x- d. The lack of ANY system notation in that equation is cause for great concern.
The fact that their equation is NOT written as point S(x,0,0) yada yada and only as coordinate x, should be a huge clue.
If this were about the manifold, then why x, not S(x,0,0), shows up in THEIR transformation equation.
Jude said that S'(x-d,0,0) = P = S(x,0,0) (well, they forgot the P, but that's excusable, given the context of the correction)
We know that x' = x-d
Thus, S'(x-d,0,0) = P = S(x,0,0) is equivalent to S'(x',0,0) = P = S(x,0,0) (the one you used was S'(x',0,0) = P = S(x', 0, 0), which doesn't really work unless d=0)
Does that work for you?

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

steve waterman wrote:I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

No. That is precisely what no one here believes. (Also, we don't believe things, we choose to notate them in given ways, and this is not how we are notating things).

WE EQUATE S'(x',0,0) = P = S(x,0,0)

x and x' are the abscissa of the point P in the coordinate systems S and S' respectively. We want to find a relationship between them.

steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

I'll let others deal with the nature of coincident coordinate systems, but real quick - this is backwards, isn't it? (Demonstrating the dangers of those primes.) Shouldn't it be S'(x-d,0,0) = S(x,0,0)?

Indeed. S'(x-d,0,0) = S(x,0,0), That math would be fine. x in one system transformed to x in the other.

However, the equation IS x' = x-d, which means, mathematically, x in one system transformed to x' in the other.
Their equation is not x = x- d. The lack of ANY system notation in that equation is cause for great concern.
The fact that their equation is NOT written as point S(x,0,0) yada yada and only as coordinate x, should be a huge clue.
If this were about the manifold, then why x, not S(x,0,0), shows up in THEIR transformation equation.

So you agree that S'(x-d, 0 , 0) = S(x,0,0)?

Our entire goal is to find x' such that S'(x',0,0) = S(x,0,0). Therefore, x' = x-d.

Also, x = x-d IS INCORRECT. THIS IS NOT TRUE. WE ARE NOT EQUATING THE COORDINATES OF TWO DIFFERENT SYSTEMS. WE ARE EQUATING TWO REPRESENTATIONS OF THE COORDINATE OF ONE POINT IN ONE SYSTEM.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Chaoszerom wrote:
steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:I equate point S'(x',0,0) = P = point S(x-d,0,0) in the manifold.
Is that what you believe too?

I'll let others deal with the nature of coincident coordinate systems, but real quick - this is backwards, isn't it? (Demonstrating the dangers of those primes.) Shouldn't it be S'(x-d,0,0) = S(x,0,0)?

Indeed. S'(x-d,0,0) = S(x,0,0), That math would be fine. x in one system transformed to x in the other.

However, the equation IS x' = x-d, which means, mathematically, x in one system transformed to x' in the other.
Their equation is not x = x- d. The lack of ANY system notation in that equation is cause for great concern.
The fact that their equation is NOT written as point S(x,0,0) yada yada and only as coordinate x, should be a huge clue.
If this were about the manifold, then why x, not S(x,0,0), shows up in THEIR transformation equation.
Jude said that S'(x-d,0,0) = P = S(x,0,0) (well, they forgot the P, but that's excusable, given the context of the correction)
We know that x' = x-d
Thus, S'(x-d,0,0) = P = S(x,0,0) is equivalent to S'(x',0,0) = P = S(x,0,0) (the one you used was S'(x',0,0) = P = S(x', 0, 0), which doesn't really work unless d=0)
Does that work for you?

Chaoszerom wrote:We know that x' = x-d
Thus, S'(x-d,0,0) = P = S(x,0,0) is equivalent to S'(x',0,0) = P = S(x,0,0) (the one you used was S'(x',0,0) = P = S(x', 0, 0), which doesn't really work unless d=0)
Does that work for you?

Not at all. You are flip-flopping x coordinates with the first coordinate of a point S(x,0,0).
These can not be substituted for one another. Also, we do not know x' = x- d, so that cannot be part of the given in your counter-proof. You are trapped in the manifold head space still. The equations says x, the equation does not say S(x,0,0) = yada yada. Why are we talking about ANYTHING S(x,0,0) or P? I want to talk about x, not about P,

x' = x-d has ZERO to do about the math of the manifold. x' = x-d is only about the math of the coordinates.

added -
the equation says x' = x -d
does not say S'(x',0,0) = S(x-d,0,0) nor even S'(x,0,0) = S(x-d,0,0) nor equate anything at all about a point P,
it says x' = x - d. I am NOT challenging your manifold equation re point P, whatever that may be, notation-wise.
Given x = x', yada yada, therefore x = x'.
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beojan
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

The labels x, y, and z aren't special. They do not inherently mean horizontal, vertical, and into plane. We simply usually define them that way. When we have more than one coordinate system, we use different labels for the second coordinate system, sometimes x', y', and z', sometimes u, v, w, sometimes others.

steve waterman wrote:Indeed. S'(x-d,0,0) = S(x,0,0), That math would be fine.

Apparently we all agree. This is what we were all trying to show is true (for translations).
It's simply that, for the sake of ease of communication, we use x' to refer to the first argument of S', y' for the second, and z' for the third. Hence, if S'(x-d,0,0) = S(x,0,0), x' = the first argument of S' = x - d.

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

steve waterman wrote:
Chaoszerom wrote:We know that x' = x-d
Thus, S'(x-d,0,0) = P = S(x,0,0) is equivalent to S'(x',0,0) = P = S(x,0,0) (the one you used was S'(x',0,0) = P = S(x', 0, 0), which doesn't really work unless d=0)
Does that work for you?

Not at all. You are flip-flopping x coordinates with the first coordinate of a point S(x,0,0).
These can not be substituted for one another. Also, we do not know x' = x- d, so that cannot be part of the given in your counter-proof. You are trapped in the manifold head space still. The equations says x, the equation does not say S(x,0,0) = yada yada. Why are we talking about ANYTHING S(x,0,0) or P? I want to talk about x, not about P,

x' = x-d has ZERO to do about the math of the manifold. x' = x-d is only about the math of the coordinates.

Apologies, I might need to step back a bit. You have said "S'(x',0,0) = P = S(x-d,0,0)". This equates to "In some system S, we have a point S(x-d,0,0). We do a transformation on this system, and the transformed system is S'. The point S(x-d,0,0) is equivalent to the point S'(x',0,0)". Does that sound right?
If it is, for some transformation, and some value of d, this may be correct. My question is, so what? What does it show?

Also (seperating this as I believe it is mildly distinct from above):
steve waterman wrote:Since x = x' was given at coincidence/d = 0, I equate coordinate x' = coordinate x as remaining true, regardless of any
math exercise being foisted upon that given x = x'

Will this remain true if d does not equal 0?

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