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

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

I am troubled by the inclusion of your third point with the first two. It's true, but trivial and redundant.

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

1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point...
restricted to the notational form of either S(x,y,z) or S'(x',y',z')
3 the first coordinate, by itself, is not a point.
eSOANEM wrote: x and x' do not exist on the manifold. They exist as part of a co-ordinate system
(technically they specify a subspace of its domain).

edit
4 x and x' are not inherent to the manifold. x and x' only exist as part of a coordinate system.
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ucim
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point...
restricted to the notational form of either S(x,y,z) or S'(x',y',z')
3 the first coordinate, by itself, is not a point.
4 x and x' are not inherent to the manifold. x and x' only exist as part of a coordinate system.

1 and 3 areTRUE. Furthermore, the way you have phrased them indicates to me that you understand these ideas.

The first part of 4 is true. The second part is.. well... sort-of true. x and x' (and anything else you might put into the ordered triple you are using) are just numbers. Numbers have an independent existence. Being numbers, they are amenable to ordinary math.

2 is... well... sloppy. The way you have phrased it tells me that you haven't gotten your head around it completely. It's not the (a,b,c) that identifies a point, it is the entirety of
S(a,b,c) that does so, because you must specify the system.
Spoiler:
Let's look at a concrete example in my library first, where there is a card catalog that allows slots to point to books. We've discussed three different catalog systems: M (the Marian system), J (the Julian system), and G (the Gregorian system) We could have a one-dimensional coordinate system where M(x) means "the book identified by the xth slot in the Marian catalog)", and similarly J(x) means "the book identified by the xth slot in the Julian catalog)", and similarly G(x) means "the book identified by the xth slot in the Gregorian catalog)". As it turns out in my example M(5) identifies An Aardvark's Journey, by Mel Richardson, J(5) identifies Nightspring, by Anthony Asimov, and G(5) identifies the Book of Kells.

The notational form that identifies an actual book is G(x) or M(x) or J(x). Just (x) by itself does not identify anything. We could also identify books using the notational form G(y) or M(y) or J(y). If we let y equal 5 in the notational form G(y) we'd identify the same book as if we let x equal 5 in the notational form G(x). We could use w or u or v or even Greek letters in place of x and it would work the same. We could put primes and double primes on it too.

If we use the notational form G(x'), and let x' equal 5, then we'd identify the very same book. (In all these examples, it would be the Book of Kells

On the other hand, if we used the notation G'(x) instead, we'd be speaking nonsense. There is no G' function defined on the manifold ("cataloging the pile of books") yet.

Now, in the Library of Hades, they only actually use catalog, and therefore have only one mapping function. Those librarians there will sometimes use a shorthand (x) to refer to the xth book in that catalog. If "everyone knows" what catalog that is, everyone can still get the book they want. But anybody who does not know what catalog is in use is toast.

That's where we are now.

Next, looking at coordinate systems that identify real points in a mathematical manifold (such as the 3D Cartesian coordinate system), you might say something like (3,1,7) to identify a point.... but to do so you are acting like the Librarian of Hades. You have to "know" what mapping function is in use. By itself, (3,1,7) or (x,y,z) or even (d,w,i) means nothing.

Take us out of Hades. In order to identify a point, the proper, complete notation looks like
A(x,y,z) where A is an already defined coordinate system. You can substutue anything you want for A, as long as whatever identifier you use is completely defined as a function that maps your ordered pairs (or triples, or whatevers) to points in the manifold.
A much better way of writing your #2 is to say something like:
For any given mapping function S which maps an ordered triple to points in the manifold, the notation S(x,y,z) identifies a point in that manifold. When x, y, and z are given actual values, it identifies a specific point in that manifold.

(The definition can be altered or generalized to deal with ordered pairs, quadruples, or ordered sets of any positive integer number of elements)

An example: Once we decide on an origin and axis orientation, R(x,y) can be defined as the rectangular coordinates of points in a manifold. R(x,y) would identify a point. R(0,0) would identify a point which we call the origin (of this system). R(1,0) would identify a point one unit out on the x axis (of this system).

Another example: Using the same axis orientation but a different origin, B(x,y) can be defined as the rectangular coordinates of points in a manifold. B(x,y) would identify a point. That point will be different from the point identified by R(x,y). B(0,0) would identify a point we call the origin (of this system). It will be a different point from the origin of the R system, above.

(0,0) does not identify the origin, because nobody knows which system is in use! The identifier of the system is just as important as the identifiers for the actual coordinates.

Schrollini wrote:Not to get into a fight over notation, but I purposely defined the coordinate systems to be functions mapping coordinates to points, so in this notation R(1,1) really is a point.
Point (koff) taken. I wasn't referring to your notation though, I was merely trying to emphasize that the points are "in" the manifold, and "identified by" the notation. (The person called) Fred is in France, but (the name) Fred is... um.... right on this page.

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

ucim wrote:
steve waterman wrote:1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point...
restricted to the notational form of either S(x,y,z) or S'(x',y',z')
3 the first coordinate, by itself, is not a point.
4 x and x' are not inherent to the manifold. x and x' only exist as part of a coordinate system.

1 and 3 areTRUE. Furthermore, the way you have phrased them indicates to me that you understand these ideas.

I'm not sure that's actually true. I'm worried about the way he emphasizes that the first coordinate by itself is not a point.

To clarify, steve, could I trouble you to explain exactly what the motivation of the "by itself" qualification is in that line? Just to make sure I'm worrying about nothing?

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

JudeMorrigan wrote:
ucim wrote:
steve waterman wrote:1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point...
restricted to the notational form of either S(x,y,z) or S'(x',y',z')
3 the first coordinate, by itself, is not a point.
4 x and x' are not inherent to the manifold. x and x' only exist as part of a coordinate system.

1 and 3 areTRUE. Furthermore, the way you have phrased them indicates to me that you understand these ideas.

I'm not sure that's actually true. I'm worried about the way he emphasizes that the first coordinate by itself is not a point.

To clarify, steve, could I trouble you to explain exactly what the motivation of the "by itself" qualification is in that line? Just to make sure I'm worrying about nothing?

x as a number and not seen as part of S(x,0,0). If that is unclear still, then "3" stays in the debate bucket, as these 4 are all under currently. There was better wording for another, so I will post that new set of 4...next.
Hmmm, I think part of 4 in not needed too...

Thanks all, for the recent feedback.

No, I am not happy with suggested number 2, so,

1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point.
3 the first coordinate, by itself, is not a point.
4 x and x' only exist as part of a coordinate system
5 all coordinates are either S(x,y,z) or S'(x',y',z')
Last edited by steve waterman on Thu Jul 18, 2013 11:43 pm UTC, edited 1 time in total.
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JudeMorrigan
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### Re: Galilean:x' with respect to S'?

Oh, like I said. Your third statement is completely true. (It is, however, still a coordinate and maintains all the properties of coordinates.) There's no need to debate its veracity. My problem is that I think that the way you've phrased it hints at your eventual proof. Based on your previous posts and the way you're writing these, I'm fairly certain it's going to involve claiming (trying to use the phrasing you've used in the past):

1 - The "Gallilean given" is x = x'-vt
2 - It may be true S(x,t) = S'(x-vt,t), but x and x' are not, by by themselves, points
3 - Mathematically speaking, that means the Gallilean given must be referring to their "abscissa lengths"
4 - Those are not the same if t does not equal zero
5- Therefore, the Gallilean given is violated, disproving Gallilean relativity.

I would be genuinely delighted if you told me that you weren't trying to do anything like that and that the proof you say you're working towards looks nothing at all like that. I also think it's why you don't like the suggested revisions to #2 though.

And your new #5 is erroneous.
Last edited by JudeMorrigan on Thu Jul 18, 2013 11:50 pm UTC, edited 1 time in total.

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

JudeMorrigan wrote:Oh, like I said. Your third statement is completely true. (It is, however, still a coordinate and maintains all the properties of coordinates.) There's no need to debate its veracity. My problem is that I think that the way you've phrased it hints at your eventual proof. Based on your previous posts and the way you're writing these, I'm fairly certain it's going to involve claiming (trying to use the phrasing you've used in the past):

1 - The "Gallilean given" is x = x'-vt
2 - It may be true S(x,t) = S'(x-vt,t), but x and x' are not, by by themselves, points
3 - Mathematically speaking, that means the Gallilean given must be referring to their "abscissa lengths"
4 - Those are not the same if t does not equal zero
5- Therefore, the Gallilean given is violated, disproving Gallilean relativity.

I would be genuinely delighted if you told me that you weren't trying to do anything like that and that the proof you say you're working towards looks nothing at all like that.

Haha!! Nicely done Jude. That's almost verbatim what I was tempted to post that I thought he was trying when I first saw that third statement (I just didn't want to risk giving him ideas in case I was wrong )

But yeah, I'm worried, but really really hoping that's not where he's going either. Would be a shame after Schrollini's well done lesson.

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

JudeMorrigan hopes I am not headed here...
JudeMorrigan wrote:1 - The "Gallilean given" is x = x'-vt
2 - It may be true S(x,t) = S'(x-vt,t), but x and x' are not, by by themselves, points
3 - Mathematically speaking, that means the Gallilean given must be referring to their "abscissa lengths"
4 - Those are not the same if t does not equal zero
5- Therefore, the Gallilean given is violated, disproving Gallilean relativity.

Yes, very close indeed. I will use your skeleton and amend it...

1 - The Galilean equation is x = x'-vt.
2 - It is true that g(x,t) = f(x-vt,t), noting that x and x' are not, by themselves, points.
3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa (lengths) and not to points in the manifold.
4 - Sx = S'x' regardless of t, same as x = x'.
5- Therefore, the Galilean equation is violated, disproving x' = x-vt.

These 5 statements of proof too are now open for debate, and not yet written in stone at all, a first draft, if you will.
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yurell
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point.
3 the first coordinate, by itself, is not a point.
4 x and x' only exist as part of a coordinate system
5 all coordinates are either S(x,y,z) or S'(x',y',z')

1. Is true, although could be stated more clearly.
2. Is false. What you have listed is a co-ordinate. It takes a co-ordinate and a co-ordinate system to identify a point. So if we have a co-ordinate system ℥, then ℥(a,b,c) identifies a point.
3. Is true. No co-ordinate is a point.
4. True enough for the purpose of this discussion.
5. Not strictly true; I can create any number of co-ordinate systems I like. If you're saying "let there be two co-ordinate systems, S(x,y,z,t) and S'(x',y',z',t')", then you should be explicit

Also note that to do this, you need a 't' dimension in there, otherwise you're probably not doing what you think you're doing
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yurell
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:1 - The Galilean equation is x = x'-vt.
2 - It is true that g(x,t) = f(x-vt,t), noting that x and x' are not, by themselves, points.
3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa (lengths) and not to points in the manifold.
4 - Sx = S'x' regardless of t, same as x = x'.
5- Therefore, the Galilean equation is violated, disproving x' = x-vt.

1. We've been saying x' = x - vt, not what you wrote. To be more precise, if g(x,t) = P = f(x',t'), then f and g are related by a Galilean transform iff f(x',t') = f(x-vt,t)
2. True
3. We're not talking about lengths, we're talking about how the co-ordinates in one co-ordinate system relate to the co-ordinates in another co-ordinate system if those two co-ordinate systems are related by a Galilean transform.
4. This is where we stop — use the same notation as us, please, don't go inventing your own. If you don't know how to express something, ask us in words.

Edit: Sorry for the double post, Steve's post appeared after mine, and I thought it important to address this separately.
Last edited by yurell on Fri Jul 19, 2013 12:20 am UTC, edited 1 time in total.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

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

let vt = d thus, removing time from my ALL mathematical proof...

1 - The Galilean equation is x = x'-d.
2 - It is true that g(x,y,z) = f(x-d,y,z), noting that x and x' are not, by themselves, points.
3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa (lengths) and not to points in the manifold.
4 - x wrt S(0,0,0) = x' wrt S'(0,0,0), same as x = x'.
5- Therefore, the Galilean equation is violated, disproving x' = x-d, hence disproving x' = x-vt, too.
Last edited by steve waterman on Fri Jul 19, 2013 1:19 am UTC, edited 1 time in total.
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yurell
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:let vt = d thus, removing time from my ALL mathematical proof...

No, that's not a Galilean transform, that's a translation; the Galilean transform requires that one ordinate be dependent upon the other — I can write if g(x,y) = P = f(x',y'), then f and g are related by a Galilean transform iff f(x',y') = f(x-vy,y) , if you prefer.

Edit: I should also add that if I do this, I am now dealing with maths and not physics at all, which will make things more confusing, not less.

steve waterman wrote:4 - x wrt S(0,0,0) = x' wrt S'(0,0,0) regardless of t, same as x = x'.

What the hell is this two-faced crap? You demanded we remove time, and then you go and use it as an integral part of your 'proof'? Also, I asked you to use the same terminology we use; this 'proof' is gibberish.
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:let vt = d thus, removing time from my ALL mathematical proof...

1 - The Galilean equation is x = x'-d.
2 - It is true that g(x,y,z) = f(x-d,y,z), noting that x and x' are not, by themselves, points.
3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa (lengths) and not to points in the manifold.
4 - x wrt S(0,0,0) = x' wrt S'(0,0,0), same as x = x'.
5- Therefore, the Galilean equation is violated, disproving x' = x-d, hence disproving x' = x-vt, too.

Your proof says that it is impossible to have two coordinate systems separated by a translation. If this is true, it would be impossible to draw this picture:

Since it is possible to draw this picture, your proof must be flawed.
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Dark Avorian
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### Re: Galilean:x' with respect to S'?

Steve,

steve waterman wrote:let vt = d thus, removing time from my ALL mathematical proof...

What the hell? Why is time non-mathematical? Have you ever met the wave equation (doubtful), or any of the other plethora of functions that use time to represent time dependent processes. Did you discover that time is logically inconsistent. If so, stop the presses, all of our proofs now work only when we're graphing things wrt distance, not time.
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yurell
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### Re: Galilean:x' with respect to S'?

Okay, Steve, we can work in three spatial dimensions if you want, using your S and S' notation.

First, we define a Galilean transform:
If S(x, y, z, t) = P = S'(x', y', z', t'), then S and S' are related by a Galilean transform iff S'(x', y', z', t') = S'(x-vxt, y-vyt, z-vzt, t), where vn is the component of v in the n direction.

For the sake of simplicity, let us align just one axis along v, so we have:
If S(x, y, z, t) = P = S'(x', y', z', t'), then S and S' are related by a Galilean transform iff S'(x', y', z', t') = S'(x-vt, y, z, t)

Given this (it's a definition, and so is not subject to debate), we have
S'(x', y', z', t') = S'(x-vt, y, z, t)
=> S'-1∘S'(x', y', z', t') = (x-vt, y, z, t)
You'll notice that we now have a function composed with its inverse; this essentially cancels out the two functions (and remember, co-ordinate systems are functions. S': ℝ4↦M), leaving us with:
(x', y', z', t') = (x-vt, y, z, t)
Or component wise:
x' = x-vt
y' = y
z' = z
t' = t
You'll probably find the dashes hard to read, which is why we've been insisting using f, g, s, u.

We can show that the origins are coincident quite easily. We have:
S(x, y, z, t) = S'(x'=x-vt, y'=y, z'=z, t'=t)
At the origin, x=y=z=t=0
=> S(0,0,0,0) = S'(x'=0-v*0, y'=0, z'=0, t'=0)
Now, we know that v*0=0, giving us:
=> S(0,0,0,0) = S'(x'=0, y'=0, z'=0, t'=0) or simply S(0,0,0,0)=S'(0,0,0,0)

If we want to find what happens as we move along the t-axis, keeping x,y,z fixed at 0, we can do that too:
S(0,0,0,t) = S'(0-vt,0,0,t)
Now since you want pure maths, that's all there is here, pure maths. No physical explanation is needed: our co-ordinate systems are co-incident only when t=0 or v=0, and are not co-incident otherwise. However, that's the only time the Galilean transform requires them to be co-incident, so that's not a problem.

What were you hoping this would disprove?
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:I am modifying these based upon various posts from others...

1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b,c) identifies a point...
restricted to the notational form of either S(x,y,z) or S'(x',y',z')
3 the first coordinate, by itself, is not a point.

2. No, this is now wrong. S(x,y,z) or S'(x',y',z') identify points, but (a,b,c) does not. Ucim posted a very good expansion on this.

steve waterman wrote:JudeMorrigan hopes I am not headed here...
JudeMorrigan wrote:1 - The "Gallilean given" is x = x'-vt
2 - It may be true S(x,t) = S'(x-vt,t), but x and x' are not, by by themselves, points
3 - Mathematically speaking, that means the Gallilean given must be referring to their "abscissa lengths"
4 - Those are not the same if t does not equal zero
5- Therefore, the Gallilean given is violated, disproving Gallilean relativity.

Yes, very close indeed. I will use your skeleton and amend it...

1 - The Galilean equation is x = x'-vt.
2 - It is true that g(x,t) = f(x-vt,t), noting that x and x' are not, by themselves, points.
3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa (lengths) and not to points in the manifold.
4 - Sx = S'x' regardless of t, same as x = x'.
5- Therefore, the Galilean equation is violated, disproving x' = x-vt.

These 5 statements of proof too are now open for debate, and not yet written in stone at all, a first draft, if you will.

This doesn't follow. As I said in my first post, whilst both x and x' (and therefore x-vt because x' is simply a relabelling of x-vt) are abscissae, they are not both measured from the same 0. Look at the co-ordinate system Schrollini gave a diagram of in his post where he gave the Galilean transformation, notice how the points S(0,t) and S'(0,t) are not the same and so, the distance from that point to the point S(x,t) which is the same as the point S'(x',t) will also be different.

Also, Sx and S'x' have no meaning. S and S' act only on ordered tuples of a specific length (strictly for the Galilean this ought to be 3+1 but problems often ignore 1 or 2 of the spatial dimensions by orienting their co-ordinates cannily). x is not a 3+1-tuple and so cannot be acted on by S. It's like trying to take sin(x,y,z,t). It makes no sense. Just as sin takes a single argument, S takes 3+1.

steve waterman wrote:let vt = d thus, removing time from my ALL mathematical proof...

1 - The Galilean equation is x = x'-d.
2 - It is true that g(x,y,z) = f(x-d,y,z), noting that x and x' are not, by themselves, points.
3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa (lengths) and not to points in the manifold.
4 - x wrt S(0,0,0) = x' wrt S'(0,0,0), same as x = x'.
5- Therefore, the Galilean equation is violated, disproving x' = x-d, hence disproving x' = x-vt, too.

No, don't get into that wrt stuff. That's a short hand and it doesn't mean what you think it means. You seem to understand the rigorous notation Schrollini introduced and we've tried to stick with that for that reason.

Anyway, Yurell's latest post is an elegant illustration of why what you're arguing doesn't really make sense and implies a fundamental misunderstanding of what the Galilean is. I suggest you read it.
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:1 mapping coordinates identifies a point in the manifold.

This does not, as phrased, really make sense. However,
Applying a mapping (namely an coordinate system) to a tuple (ordered pair / triple / etc. according to the number of dimensions) returns a point in the (relevant) manifold

steve waterman wrote:2 - It is true that g(x,t) = f(x-vt,t), noting that x and x' are not, by themselves, points.

Let's rephrase this a little to make things clearer.
• x is a (real) number (like 1, 2, 3.4565, ⅔, √2, π, or e)
• x' is a number
• t is a number
• t' is a number
f(x',t') = f(x-vt,t) = g(x,t)
where f and g are coordinate systems related by the Galilean transform.

In making the statement f(x',t') = f(x-vt,t) we use the two substitutions:
• x' = x - vt
• t' = t

steve waterman wrote:4 - Sx = S'x' regardless of t, same as x = x'.

This makes no sense. I'll assume you meant
S(x,t) = S'(x',t) regardless of t, therefore x = x'

This is incorrect. S and S' are different mappings. To give a simple example.

sin(x) = cos(y) does not imply that x = y, because if x = π/6 (30°) and y = π/3 (60°)
sin(x) = sin(π/6) = ½
cos(y) = cos(π/3) = ½
Hence sin(x) = cos(y) but x ≠ y.

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

x = x'-vt, let vt = d

The validity of the math of the following equation is challenged; x = x'-d.

g(x,y,z) = f(x-d,y,z), noting that x and x' are not points.

1 x = x'-d equates abscissa ( lengths ).

2 S(x,y,z) = S'(x',y',z') also equates abscissa ( lengths ).

3 x = x', therefore x' ≠ x - d
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:4 S(x,y,z) = S'(x',y',z') also equates abscissa ( lengths ).

What are S and S'? Coordinate systems? You've already used f and g for your coordinate systems.
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yurell
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### Re: Galilean:x' with respect to S'?

1) You're the one who made up that equation, so challenge it all you like. The rest of us have said x' = x-vt, which is the definition of a Galilean transform.
2) You keep missing the t dimension
3) I have no idea what you're trying to say here, please elaborate
4) I have no idea what you're trying to say here, please elaborate
5) x=x' only on the trajectory t=0 or for v=0, which does not contradict x' = x-vt

Since you're talking about 'lengths', I'd like to point out that we haven't defined length at all. If you want to discuss length, we need a metric space and units, and you need at least two points.

Edit: Ah, the ninja edits. He just renumbered everything two less than what I've responded here.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

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

Why, oh why do you think x = x'.
The abscissa in the S coordinate system does not need to be equal to the abscissa in the S' coordinate system.
Just because they are both called the abscissa does not mean they are equal.

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

Schrollini wrote:
steve waterman wrote:4 S(x,y,z) = S'(x',y',z') also equates abscissa ( lengths ).

What are S and S'? Coordinate systems?

I was trying to merely copy your equation regarding points in the manifold.
I was using your function notation, but it is too much a hassle and not needed in the proof.

I now post this third draft...

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

1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ).
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
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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

steve waterman wrote:Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z'
let d = distance along the common x/x' axis

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

No it doesn't. It equates two real numbers.

Edit to add: Anyway, it contradicts your original definition x = x'. We can stop here and ignore the rest.
steve waterman wrote:2 whereas, S(x,y,z) = S'(x',y',z') always properly equates the given and eternally fixed abscissa ( lengths ).

No it doesn't. It equates two points.
steve waterman wrote:3 x = x', therefore x' ≠ x - d

Does not follow.
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z'

Then we aren't talking about a Galilean transform. For starters, you've left out the t axis. Secondly, if S and S' are Cartesian, there's not been a Galilean transform because a Cartesian co-ordinate system requires all axes to be perpendicular; this is not the case for the Galilean transform. Thirdly, if they're coincident then there has been no transform at all.

steve waterman wrote:let d = distance along the common x/x' axis

You haven't defined a metric to give us a definition of 'distance'. Also, those axes are infinitely long, so it only makes sense to talk about distances between two points (which you haven't defined).

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

I don't know why you want to rearrange x' and x constantly, but it is incredibly irritating when we've been using x' = x-vt this entire time. Also, (1) is false if the co-ordinate systems are coincident, which you've stated they are. Basically, what you're saying is 'if we assume there's no transform, and then apply the maths for a transform, we end up with a contradiction, thus the transform can't exist!'

steve waterman wrote:2 whereas, S(x,y,z) = S'(x',y',z') always properly equates the given and eternally fixed abscissa ( lengths ).

I have no idea what you are trying to say. Can you use words?

steve waterman wrote:3 x = x', therefore x' ≠ x - d

You derived a false conclusion from false premises, using bad logic. Such a result is not surprising.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

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

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

1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ).
Schrollini wrote:No it doesn't. It equates two real numbers.

so you are okay with 2 = -1?

Yes, that is exactly the problem!

steve waterman wrote:2 whereas, S(x,y,z) = S'(x',y',z') always properly equates the given and eternally fixed abscissa ( lengths ).

Schrollini wrote:No it doesn't. It equates two points.

You said points only exist in the manifold!

S(x,y,z) = S'(x',y',z') equates coordinates, not points.
Last edited by steve waterman on Fri Jul 19, 2013 2:24 pm UTC, edited 1 time in total.
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yurell
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### Re: Galilean:x' with respect to S'?

Steve, the contradiction that you've 'discovered' (namely, if x=x', then x'=x-vt is false) is not due to the fact that the Galilean transform is wrong, but that the two co-ordinate systems you're using are not related by a Galilean transform.

Edit:
steve waterman wrote: S(x,y,z) = S'(x',y',z') equates coordinates, not points.

No. (x,y,z)=(x',y',z') equates co-ordinates. By applying the co-ordinate systems to them, you are mapping them onto the manifold and thus describing points. Remember how we've always said S(x,y,z,t) = P, with P being an element of (a point in) the manifold? Well, it still does.
Last edited by yurell on Fri Jul 19, 2013 2:31 pm UTC, edited 1 time in total.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

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

Equating
S(x,y,z,t) = S'(x',y',z',t')
is not the same as
x = x', y = y', z = z', t = t' (shortened to (x,y,z,t) = (x',y',z',t'))
for the same reason that sin(x) = cos(y) is not the same as x = y.

You keep using the function notation to denote points (as opposed to coordinates) yet when you compare them, you insist on simply comparing the coordinates (the function arguments), even if they are in different coordinate systems (even if the function is different).

Also, it probably is better that you start with a translation, and understand that the transformation for a translated coordinate system is valid before we even begin to discuss the Galilean transform.

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

In my proof, there is no t, there is no Galilean. There is only what is shown in the proof. Please discuss the proof and not anything Galilean transformation. There is no Physics anything here. There is math and only math.

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

1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ).
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
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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

steve waterman wrote:Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z'
let d = distance along the common x/x' axis

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

[...]

Yes, that is exactly the problem!

Well, there you have it. If you make an assumption that's inconsistent with the Galilean transformation (or any non-trivial transformation), you get a result that's inconsistent with the Galilean transformation (or any non-trivial transformation).

In other news, the Pope is Catholic. And Francisco Franco is still dead.

And posting the same thing again doesn't change any of this.
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### Re: Galilean:x' with respect to S'?

x = x' equates two numbers.

x=x'-d also equates two numbers: x'-d is a number, and x is a number, the two are equal.

The only way for both statements above to be true at the same time is for d to equal zero. This is part of what makes your two systems coincident.

If d is no longer equal to zero, then if the second equation is true, the first no longer is. That is part of what makes the two systems no longer coincident.

In this case, their origins will also no longer be coincident.

When you talk about distance, you need to define a metric and you need to have two points identified using the same system to measure the distance between. You have not done this, but I think you expect us to infer that you mean the origin to be "the other" point, and are using ordinary Pythagorean distance. A math proof should not require inference, but in any case, that's not your problem. It merely hides your problem, which is that once the two systems are no longer coincident, "the origin" of one of the systems is a different point in the manifold from "the origin" of the other system.

To measure distance in a system, both points must be referenced from the same system. This means that if you know the "name" (coordinates) of one point using one system, and the name (coordinate) of a different point using a different system, (Remember, the points "live" in the same manifold; they simply have different names (coordinates) in different coordinate systems) you must apply some sort of transformation to at least one of the "names" (coordinates) in order to have all the coordinates be from the same system! Only then can you use the numbers that the coordinates are made from in order to calculate a distance.

This is your problem. You haven't properly done this transformation so that all your coordinates are expressed in the same system.

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

Steve, if all you're trying to say is:
If x' = x, then x' ≠ x - d
Then no one on the planet will disagree with you*.

It also has nothing to do with the Galilean transform, which makes me confused as to why you would bring it up in a thread titled 'Galilean:x' with respect to S'?'

Edit:
*For non-trivial d, obviously.
Last edited by yurell on Fri Jul 19, 2013 2:50 pm UTC, edited 1 time in total.
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z'

Steve, do you accept the possibility that there might exist coordinate systems which are neither Cartesian nor coincident?

Do you accept that the rules and equations which are true in the case of two coincident Cartesian coordinate systems might not apply to systems which are not either coincident or Cartesian?

Do you accept that mathematical definitions of coordinate systems which are not both coincident and Cartesian might be valid even if they give different coordinates from a coincident, Cartesian system?

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

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

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
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 ).
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.

If d* is now the separation between the origins of similarly oriented Cartesian coordinate systems, what is d?

Edit: If you try pulling this, steve (and it's a terrible trick to try to turn, to tell the truth), your d in Steps 1 and 3 just becomes a number, with no meaning to any sort of transformation. And that makes step 3 trivial, not a breakthrough.

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

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

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
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. And irrelevant for any coordinate transformation where x ≠ x'. Like, for example, the Galilean.
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steve waterman
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### Re: Galilean:x' with respect to S'?

WibblyWobbly wrote:
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
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 ).

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.

If d* is now the separation between the origins of similarly oriented coordinate systems, what is d?

This seems to be hard to get people to understand.

We have S and S" coincident. Take snapshot 1.
Separate [ from once being coincident] the S and S origins by d. Take snapshot 2.
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### Re: Galilean:x' with respect to S'?

GIVEN: steve = steve waterman
steve = steve + waterman, for all waterman > 0

THEREFOR steve ≠ steve waterman. QED.

EDIT:
AND IN SNAPSHOT 2, THEY ARE NO LONGER COINCIDENT. How is this difficult to understand?
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:We have S and S" coincident. Take snapshot 1.
Separate [ from once being coincident] the S and S origins by d. Take snapshot 2.

Coordinate systems do not move!

They simply are.
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
steve waterman wrote:We have S and S" coincident. Take snapshot 1.
Separate [ from once being coincident] the S and S origins by d. Take snapshot 2.

Coordinate systems do not move!

They simply are.

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.
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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 acutally using in this proof:

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
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.

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', when S and S' are coincident?
"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