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

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

Schrollini wrote:
steve waterman wrote:So the Galilean for you, only ever transforms a point within the manifold itself, since points only exist in the manifold.

No. The Galilean transformation is a coordinate transformation. Quoting this post:
A coordinate transformation is a function that takes the coordinates of a point in one coordinate system to the coordinates in another system.
So a coordinate transformation transforms coordinates to coordinates. It does not act on points.

steve waterman wrote:Yes, my statement above was very poorly worded.

Schrollini wrote:So a coordinate transformation transforms the coordinates to coordinates

I need some help with this.
So a "coordinate transformation" equates the coordinates in one system to the coordinates in the other system?
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:So a "coordinate transformation" equates the coordinates in one system to the coordinates in the other system?

A transformation maps a co-ordinate in one co-ordinate system to a co-ordinate in another co-ordinate system, both of which refer to the same point.
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'?

yurell wrote:
steve waterman wrote:So a "coordinate transformation" equates the coordinates in one system to the coordinates in the other system?

A transformation maps a co-ordinate in one co-ordinate system to a co-ordinate in another co-ordinate system, both of which refer to the same point.

the equation is x' = x-vt that I challenge, not x' maps from x-xt or the like.

So a "coordinate transformation" equates the coordinates in one system to the coordinates in the other system?
Are coordinates inherent to a system?
"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

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

steve waterman wrote:So a "coordinate transformation" equates the coordinates in one system to the coordinates in the other system?

No. It equates the point identified by the coordinates given in one system to the point identified by coordinates given in the other system.

First consider the pile of books again.
Three different librarians came up with three different cataloging systems. Each slot (in any given catalog) identifies a specific book in the pile of books on the floor. Conversely, each book is associated with a specific slot in the catalog for any given catalog. Given a different catalog, the same book will typically have a different slot that points to it.

A "coordinate transformation" function for the library would, for any given slot number of one catalog, figure out which slot number in the other catalog it is that points to the same actual book. So, if you wanted "book number seven" (Note - this is sloppy shorthand for "the book that slot number seven identifies) in the Julian system, and you applied the Julian to Gregorian transformation for this library, it would tell you that slot number forty-five is the slot you need to push in order to get that same book. You may be tempted to think of it as "Julian seven" = "Gregorian forty-five", but that would be incorrect.

What is correct is the book pointed to by Julian seven is the same as the book pointed to by Gregorian forty-five.

The thing that is the same is the book.

Something similar can be done with dates and calendar systems.

Now, consider points (in space) and coordinate systems (such as the Cartesian one).

"Space" is the manifold. It is the analog of the pile of books. DO NOT yield to the temptation of thinking of space as having a coordinate system of its own. That is what will lead you astray. Space, being the manifold, is just a pile of points.

If you want to identify these points with coordinates, you need to impose a coordinate system. This is the analog of the card catalog. There are many coordinate systems to choose from - you can choose rectangular or polar for example, you can orient the axes any way you like, and you can put the origin anywhere you like.
Spoiler:
What does this mean? Put the origin anywhere I like? It means that you can define (0,0) in your system as pointing to a specific point in the manifold, or to a specific other point in the manifold. Axis orientation is done the same way.
. I'll call the two systems R and P (rectangular and polar, just to keep things easy to differentiate).

Once you set up two different coordinate systems that each span the space (are capable of identifying any point in the manifold we call space), you will want to be able to find which coordinates of R refer to the same point that a given set of coordinates of P refers to. The function that does this is a coordinate transform.

You may have seen something written that looks like
R(1,1) = P(sqrt(2), pi/4)
This is sloppy notation which is usually harmless, but is what is throwing you. It is not the coordinates that are equal, but the points to which they refer.

The correct (albeit clumsier) way to say it is: The point identified by R(1,1) is the same as (equal to) the point identified by P(sqrt(2), pi/4).

It is the points that are equal, just like in the library, it's the books that were equal.

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

steve waterman wrote:the equation is x' = x-vt that I challenge, not x' maps from x-xt or the like.

First, I purposely used u and s for the coordinates in the second frame, since the primes can be easy to miss in typing or in reading. I'd like to suggest that we stick with them, to avoid these sorts of mistakes.

Now, the equation u = x - vt is only half the Galilean transformation. The other half is s = t. Together, these are the shorthand way (discussed in this post) to indicate that the Galilean transformation maps (x,t) to (x-vt,t). Or equivalently, that the Galilean transformation is given by h(x,t) = (x-vt,t).

So exactly what it is that bothers you is rather unclear.

ucim wrote:You may have seen something written that looks like
R(1,1) = P(sqrt(2), pi/4)
This is sloppy notation which is usually harmless, but is what is throwing you. It is not the coordinates that are equal, but the points to which they refer.

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.
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Dark Avorian
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### Re: Galilean:x' with respect to S'?

(Author's note: I am aware that Kentucky is a state, not a city. I would claim it to be a humorous reference to my miscommunication analogy, but to be honest it was the first thing that came to mind)

Steve,

I've noticed you said you were somewhat confused by the idea that a manifold contained an uncountable infinitude of points. In brief, the reasoning is that we have a map from some area around (x,y,z, ...) into the manifold injectively (that means distinct coordinates map to distinct points). This region contains an uncountable infinitude of points (any small ball around the coordinate will), and as each has a distinct point of the manifold it is mapped to, the manifold has at least that many points. If you'd like more rigor, and perhaps to understand some of our more precise terminology, see the spoiler for the quickest primer on topology I could write.

Spoiler:
If you are reading this, understand that these terms are not up for debate. These are the standard mathematical terms. If you know topology, skip to Part II

Part I: Topology

A topological space is composed of two things: a set X, known as the space, and a topology T. We will formally define a topology soon. Topologies generalize and formalize a certain notion of the structure of space, distance, and nearness that we see.

A topology on a set X is a set T of subsets of X. Elements of X are points. Elements of T are called open sets, and are a subset of X, thus a collection of points. To be called a topology, T must have certain properties.

(i) Both the empty set, and the entire set X, are open sets, that is to say, elements of T.
(ii) The union of open sets is open. This allows for arbitrary unions, even unions over uncountably infinitely many open sets.
(iii) The intersection of finitely many open sets is open.

Exercise, if desired: Let X = R, the set of all real numbers. Define T to be the set of all sets S such that for any point x in S, there is some (possibly small) positive number h such that the entire interval [x-h, x+h] lies in S. Show that this is a topology on R. This is the canonical topology of R.

Exercise: Look at my previous post defining metric spaces. Let (X,d) be a metric space. Define a topology T on X as follows: Open sets are sets such that for any point x in X, there is a small positive number h (dependent on both x, S) such that if d(x,y) < h, then y is also an element of S. Show that this a topology. This is the canonical topology on a metric space.

Exercise: Let X be a set. Define a topology T as follows: S is an open set if either T-S is a finite set, or if S is the empty set. Show this is a topology. This is the finite-complement topology.

We say that a set C is closed if its complement ( that is, the set of all elements of X not in C), is open.

We say that an point "x" is a limit point of a set A (a subset of X, obviously), if every open set S containing x also contains points of A.

Exercise: Show that a set is closed if and only if it contains all of its limit points.

Part II: Why Topology Matters Here.

Let X and Y be spaces, with topologies T,V respectively. A function "f" mapping X to Y is said to be continuous if for every open set S that is a subset of Y, the set A of all elements x in X such that f(x) is in Y, is open. More compactly: f^(-1)(S) is open.

A function f is said to be a homeomorphism if it is bijective, thus has an inverse, and both the function and its inverse are continuous. Homeomorphisms preserve all topological properties. A set is open if and only if its image under a homeomorphism is open.

Now we get down to the nitty gritty.

Its been mentioned that manifolds have no structure. That's not quite true. A manifold is not just a set. A manifold is a topological space with a very special property. Every point "x" of a manifold M, has some open set S in n-dimensional space R^n, that can be homeomorphically mapped to an open set U in the manifold, where U contains x (this homeomorphism is the coordinate system). This does not mean that there is a right coordinate system. It simply means there are coordinate systems.

A manifold is defined by being able to have coordinate systems defined on it. This does not mean any particular choice of coordinate systems is inherent to the manifold, thus the notion that they are "on top of it"
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### Re: Galilean:x' with respect to S'?

Dark Avorian wrote:(Author's note: I am aware that Kentucky is a state, not a city. I would claim it to be a humorous reference to my miscommunication analogy, but to be honest it was the first thing that came to mind)
(Note my location)
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heuristically_alone wrote:I have been informed that this is called writing a book.

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

Herp derp. Wish I could take credit for cleverly doing that.
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beojan
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### Re: Galilean:x' with respect to S'?

1. Steve, are you actually interested in understanding what coordinate transformations are for / what they do, or are you only interested in arguing and trying to trick people into accepting whatever you think you have discovered?

2. You accepted that mapping was a synonym for function, then started saying things like "S(x,y,z) before mapping". Do you know what a function is? If so, define / explain it.

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

steve waterman wrote:the equation is x' = x-vt that I challenge, not x' maps from x-xt or the like.

Steve, x' = x-vt (or u = x-vt in the notation that we're using) is shorthand. We define the co-ordinate systems f and g such that:
f(s,u) = g(t,x) = P, for some point P in the manifold.
In order for f and g to be Galilean transforms of each other, we require that
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)
This is not a 'conclusion' that can be argued against, this is a definition.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

PM 2Ring
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:So, yes, it only makes absolute total sense that points are inherent to the manifold. When I kept saying on top of, nobody asked me why, in amongst all the other confusion, is most understandable. So, I am glad that little thingie is cleared up.

I did try:
PM 2Ring wrote:Could you clarify this for me, Steve?
steve waterman wrote:1 the point on top of manifold M at (x',0,0) = S(x-vt,0,0)

A manifold is a set of points, so in standard mathematical language we talk about points being in manifold M. So what do you mean by a point on top of manifold M?

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

Schrollini-
g(x') = f(x-vt) = P?

This is all I meant to ask.
"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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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:Schrollini-
I equate a point P in the manifold as (x',0,0) = (x-vt,0,0)?
perhaps a point P in the manifold is g(x',0,0) = f(x-vt,0,0)?

yurell gave you the answer two posts before yours:
yurell wrote:In order for f and g to be Galilean transforms of each other, we require that
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

Note that she's using the (t,x) convention, and associating t,x with g rather than f, but these are cosmetic differences. I can equally well write g(x-vt,t) = f(x,t) (though that would, for the sticklers in the audience, flip the sign of v).

Note that t is a coordinate. That's because the manifold we're considering with Galilean transformations is spacetime. You seem to be consistently neglecting this point.

Edit: Thanks PM 2 Ring
Last edited by Schrollini on Thu Jul 18, 2013 2:57 pm UTC, edited 1 time in total.
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steve waterman
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### Re: Galilean:x' with respect to S'?

yurell gave you the answer two posts before yours:
yurell wrote:In order for f and g to be Galilean transforms of each other, we require that
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

Excellent. I can work with that notation.

I find it all being in italics most confusing. Is this what you mean?
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

or better yet, place t/s and the end...?
g(x,t) = P = f(u,s) = f(u,t) = f(x-vt,t)
"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

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

For what it's worth, I really enjoyed Schrollini's course. I learned what a manifold is, and got a much clearer understanding of coordinate mappings/transformations than the intuitive understanding I had from beforehand. Thank you!

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

steve waterman wrote:I find it all being in italics most confusing. Is this what you mean?
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

or better yet, place t/s and the end...?
g(x,t) = P = f(u,s) = f(u,t) = f(x-vt,t)

Correct! ^.^
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'?

yurell wrote:
steve waterman wrote:I find it all being in italics most confusing. Is this what you mean?
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

or better yet, place t/s and the end...?
g(x,t) = P = f(u,s) = f(u,t) = f(x-vt,t)

Correct! ^.^

Great.

Can we focus only upon this part, please?
g(x,t) = P = f(x-vt,t)

I suspect a typo?...did you mean to say that,
g(x',t) = P = f(u,s) = f(u,t) = f(x-vt,t)
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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brenok
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### Re: Galilean:x' with respect to S'?

No, the x' is represented by the "u"

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

brenok wrote:No, the x' is represented by the "u"

Thanks. So, someone write me the equation where u is represented by x' please.
"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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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
brenok wrote:No, the x' is represented by the "u"

Thanks. So, someone write me the equation where u is represented by x' please.

Again, I recommend that we stick with u and s instead of x' and t', so we can avoid these issues of "Did you really mean....?"

In any event, with the functional notation, you don't need either. The Galilean is, as you previously noted:
steve waterman wrote:g(x,t) = P = f(x-vt,t)

Or, taking f-1 of each side:
f-1(g(x,t)) = (x-vt,t)
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
steve waterman wrote:
brenok wrote:No, the x' is represented by the "u"

Thanks. So, someone write me the equation where u is represented by x' please.

Again, I recommend that we stick with u and s instead of x' and t', so we can avoid these issues of "Did you really mean....?"

In any event, with the functional notation, you don't need either. The Galilean is, as you previously noted:
steve waterman wrote:g(x,t) = P = f(x-vt,t)

Or, taking f-1 of each side:
f-1(g(x,t)) = (x-vt,t)

This is getting silly again.
How is x related to x' in the manifold? Show me that equation! Please use x', not u, this is hard enough.

here is my confusion - I think that the equation is g(x',t) = P = f(x-vt,t), and so I want to know
why not? We are after x' - x-vt, hence the problem, for me. I am still working on understanding your notation and meanings, Just about ready to try and explain my side, once I get your formula you are using that supports x = x-vt in the manifold. If you write it as functions, then so be it. So, your equation, therefore x = x-vt, would be ever so helpful. You needed to derive x' = x-vt. So fine, if your function equation does that, then I wish to see it.
If your function equation does not, in turn, derive x' = x -vt, I would see that as a problem.
Last edited by steve waterman on Thu Jul 18, 2013 3:47 pm UTC, edited 1 time in total.
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PM 2Ring
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### Re: Galilean:x' with respect to S'?

eran_rathan wrote:I think steve is a living example of the "Hungarian Phrasebook" sketch from Monty Python.

I will not buy this Galilean transformation, it is scratched.

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

Schrollini

P = g(first coordinate, fourth coordinate) mapped from S'
= f(first coordinate -vt, fourth coordinate) mapped from S?
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Schrollini
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:here is my confusion - I think that the equation is g(x',t) = P = f(x-vt,t), and so I want to know
why not?

Because this equation says: g maps the coordinates (x',t) to the same point that f maps the coordinates (x-vt,t). This isn't what that Galilean says.

What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t).

Or in other words: g(x,t) = f(x-vt,t).

Or in other other words: (u,s) = f-1(g(x,t)) = (x-vt,t).

Or in other other other words: u = x-vt, s = t.

Or in other other other other words, using the notation that I'm trying to avoid: x' = x-vt, t' = t.

All of these statements are equivalent.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t) ...Or x' = x-vt, t' = t.

Great. I will just check first to be sure that is what you are saying, even after my removal of some of it.

If you stand by that statement, which surely you do, I can begin my challenge using our current notation.

btw, I hate the white background for posting and find proof-reading it hideous, so I need to read the stuff in the posted blue background to find the errors, as I do not handle the white to black contrast well at all. Please let a post sit for like two minutes to fix the typos.
"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'?

Aleksander wrote:For what it's worth, I really enjoyed Schrollini's course. I learned what a manifold is, and got a much clearer understanding of coordinate mappings/transformations than the intuitive understanding I had from beforehand. Thank you!

Thank you for saying so! It's nice to know these masses of text I've fired off into the ether haven't been in vain.

If you're interested in understanding manifolds beyond "something that looks like Cartesian space", check out Dark Avorian's post.

steve waterman wrote:btw, I hate the white background for posting and find proof-reading it hideous, so I need to read the stuff in the posted blue background to find the errors, as I do not handle the white to black contrast well at all. Please let a post sit for like two minutes to fix the typos.

The preview button is your friend.
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steve waterman
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:
Aleksander wrote:For what it's worth, I really enjoyed Schrollini's course. I learned what a manifold is, and got a much clearer understanding of coordinate mappings/transformations than the intuitive understanding I had from beforehand. Thank you!

Thank you for saying so! It's nice to know these masses of text I've fired off into the ether haven't been in vain.

If you're interested in understanding manifolds beyond "something that looks like Cartesian space", check out Dark Avorian's post.

steve waterman wrote:btw, I hate the white background for posting and find proof-reading it hideous, so I need to read the stuff in the posted blue background to find the errors, as I do not handle the white to black contrast well at all. Please let a post sit for like two minutes to fix the typos.

The preview button is your friend.

Hmm, yes, I do use that. I should just stop trying to fix them in the white at all, as I type, and make ALL the corrections in preview.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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steve

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

steve waterman wrote:Hmm, yes, I do use that. I should just stop trying to fix them in the white at all, as I type, and make ALL the corrections in preview.

You should try using a text editor with the Solarized colour scheme to compose your posts.
Preview is most useful for checking things like quote nesting, url tags, and math tags back when those worked.

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

Schrollini wrote: What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t) ...Or x' = x-vt, t' = t.

Are you saying that
since the mapped coordinates (x,t) = the mapped coordinates (x-vt,t), therefore x' = x-xt?
"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'?

Schrollini wrote: What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t) ...Or x' = x-vt, t' = t.

Are you saying that
since the mapped coordinates (x,t) = the mapped coordinates (x-vt,t), therefore x' = x-xt?

added - hoping for some agreement on some of these
1 mapping coordinates manifests a point in the manifold.
2 a point can be identified as they come in the mathematical form of (a,b) or say (a,b,c).
3 the first coordinate, by itself, is not a point.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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

steve waterman wrote:
Schrollini wrote: What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t) ...Or x' = x-vt, t' = t.

Are you saying that
since the mapped coordinates (x,t) = the mapped coordinates (x-vt,t), therefore x' = x-xt?

We're saying g(x,t) = f(x-vt,t).

If you typically call the arguments of f u and s, then you might write u = x-vt, s = t. If you typically call the arguments of f ⅊ and ℥, then you might write ⅊ = x-vt, ℥ = t. And if, despite my warnings, you typically call the arguments of f x' and t', then you might write x' = x-vt, t' = t.

There is no math in any of this -- it's just an issue of naming.
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:yurell gave you the answer two posts before yours:
yurell wrote:In order for f and g to be Galilean transforms of each other, we require that
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

Excellent. I can work with that notation.

I find it all being in italics most confusing. Is this what you mean?
g(t,x) = P = f(s,u) = f(t,u) = f(t,x-vt)

or better yet, place t/s and the end...?
g(x,t) = P = f(u,s) = f(u,t) = f(x-vt,t)

The italics do not mean anything at all. Feel free to ignore them.

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:
brenok wrote:No, the x' is represented by the "u"

Thanks. So, someone write me the equation where u is represented by x' please.

Again, I recommend that we stick with u and s instead of x' and t', so we can avoid these issues of "Did you really mean....?"

In any event, with the functional notation, you don't need either. The Galilean is, as you previously noted:
steve waterman wrote:g(x,t) = P = f(x-vt,t)

Or, taking f-1 of each side:
f-1(g(x,t)) = (x-vt,t)

This is getting silly again.
How is x related to x' in the manifold? Show me that equation! Please use x', not u, this is hard enough.

here is my confusion - I think that the equation is g(x',t) = P = f(x-vt,t), and so I want to know
why not? We are after x' - x-vt, hence the problem, for me. I am still working on understanding your notation and meanings, Just about ready to try and explain my side, once I get your formula you are using that supports x = x-vt in the manifold. If you write it as functions, then so be it. So, your equation, therefore x = x-vt, would be ever so helpful. You needed to derive x' = x-vt. So fine, if your function equation does that, then I wish to see it.
If your function equation does not, in turn, derive x' = x -vt, I would see that as a problem.

No, 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).

And that assumption would be incorrect.

The Galilean is that f(x,t)=g(x',t')=g(x-vt,t). Yes I am certain that I have got my f's and g's correct (note, if you can replace all the g's with f's and get, up to the sign of v, an equivalent result, what I mean is I am certain that the two systems I've labelled as g here are the same and that the other system is the other one).

(Sorry Schrollini about not using the s and u convention, given various other common uses of those symbols in contexts where the Galilean is also commonly used, I think it will actually lead to more confusion in the long-run)

steve waterman wrote:
Schrollini wrote: What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t) ...Or x' = x-vt, t' = t.

Are you saying that
since the mapped coordinates (x,t) = the mapped coordinates (x-vt,t), therefore x' = x-xt?

You're using confusing terminology. What the galilean says is simply that S(x,t)=S'(x-vt,t)=S'(x',t').

x' and t' are just notational conveniences, the important bit that is actually useful is the first equality. What this tells you is what the co-ordinate system S' calls the point the co-ordinate system S calls (x,t). What makes this useful is that it does so whilst only referring to the co-ordinates in the system S.

steve waterman wrote:
Schrollini wrote: What the Galilean transformation says is that g maps the coordinates (x,t) to the same point that f maps the coordinates (x-vt,t) ...Or x' = x-vt, t' = t.

Are you saying that
since the mapped coordinates (x,t) = the mapped coordinates (x-vt,t), therefore x' = x-xt?

added - hoping for some agreement on some of these
1 mapping coordinates manifests a point in the manifold.
2 a point can be identified as they come in the mathematical form of (a,b) or say (a,b,c).
3 the first coordinate, by itself, is not a point.

Yes to 1. This follows directly from definition of a co-ordinate system as something that maps from co-ordinates to points in the manifold (in a one-to-one manner).

No to 2. A point is not it's co-ordinates. Often people are sloppy and assume a particular co-ordinate system but all this means is that there's an "S" in front of every set of co-ordinates they've written which they can't be bothered to write.

Yes to 3. The first co-ordinate is not a point. Neither is the second. Or any one co-ordinate. Nor any set of co-ordinates. The only way to get a point from a set of co-ordinates is to put it through a co-ordinate system.
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### Re: Galilean:x' with respect to S'?

Schrollini wrote:We're saying g(x,t) = f(x-vt,t).
And if, despite my warnings, you typically call the arguments of
f x' and t', then you might write x' = x-vt, t' = t.

So, to best be sure of exactly what you are saying/meaning,
SINCE g(x,t) = f(x-vt,t). THEREFORE x' = x-vt, t' = t?

I just started my proof by posting the first three statements on the previous post.
They are up for debate/correction/clearer terms, before moving ahead...
posting them again, so that you do not need to scroll back to find the material...

1 mapping coordinates manifests a point in the manifold.
2 a point can be identified as they come in the mathematical form of (a,b) or say (a,b,c).
3 the first coordinate, by itself, is not a point.
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### Re: Galilean:x' with respect to S'?

eSOANEM wrote:No, 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).

What the galilean says is simply that S(x,t)=S'(x-vt,t)=S'(x',t').

x' and t' are just notational conveniences, the important bit that is actually useful is the first equality. What this tells you is what the co-ordinate system S' calls the point the co-ordinate system S calls (x,t). What makes this useful is that it does so whilst only referring to the co-ordinates in the system S.

I will get back to other things in your post later...thanks for giving your 3 responses. As well as things I clipped above.

1 mapping coordinates manifests a point in the manifold.
2 a point can be identified as they come in the mathematical form of (a,b) or say (a,b,c).
3 the first coordinate, by itself, is not a point.

eSOANEM wrote:No to 2. A point is not it's co-ordinates.

This statement is about identifying the correct mathematical format that is used to represent a point.

eSOANEM wrote:Often people are sloppy and assume a particular co-ordinate system but all this means is that there's an "S" in front of every set of co-ordinates they've written which they can't be bothered to write.
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
Schrollini wrote:We're saying g(x,t) = f(x-vt,t).
And if, despite my warnings, you typically call the arguments of
f x' and t', then you might write x' = x-vt, t' = t.

So, to best be sure of exactly what you are saying/meaning,
SINCE g(x,t) = f(x-vt,t). THEREFORE x' = x-vt, t' = t?

No.

I'm saying these two statements are two ways of expressing the same concept. I wrote a whole post about translating between these two ways of speaking. Perhaps you could read it?

steve waterman wrote:I just started my proof by posting the first three statements on the previous post.
They are up for debate/correction/clearer terms, before moving ahead...
posting them again, so that you do not need to scroll back to find the material...
1 mapping coordinates manifests a point in the manifold.

I don't know what you mean by "manifests". If you mean "identifies", then yes. If you mean "creates", then no.
steve waterman wrote:2 a point can be identified as they come in the mathematical form of (a,b) or say (a,b,c).

No. See eSOANEM's answer. And for your recent objection, a point need not have coordinates in a given coordinate system.
steve waterman wrote:3 the first coordinate, by itself, is not a point.

eSOANEM wrote:(Sorry Schrollini about not using the s and u convention, given various other common uses of those symbols in contexts where the Galilean is also commonly used, I think it will actually lead to more confusion in the long-run)

Confusion is better than certainty about a falsehood. I've been purposefully using notation different from what Steve uses to try to make him think about what the symbols mean, rather than just assuming they mean what he's taken them to mean in the past. Whether this has worked or not is another issue.
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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) or say (a,b,c) identifies a point.
3 the first coordinate, by itself, is not a point.
Schrollini wrote:And for your recent objection, a point need not have coordinates in a given coordinate system.

Indeed, as I posted in the statement, without a system letter... as (a,b) or say (a,b,c).
So are you okay with statement 2?
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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) or say (a,b,c) identifies a point.
3 the first coordinate, by itself, is not a point.

1. YES!
2. YES (with the caveat that there is an implied coordinate system of some form).
3. YES!

By george I think he's got it (or at the very least, getting a lot closer)!

steve wrote:
Schrollini wrote:And for your recent objection, a point need not have coordinates in a given coordinate system.

Indeed, as I posted in the statement, without a system letter... as (a,b) or say (a,b,c).
So are you okay with statement 2?

See the caveat above.
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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) or say (a,b,c) identifies a point...
(with the caveat that there is an implied coordinate system: either S or S')
3 the first coordinate, by itself, is not a point.

edit
Last edited by steve waterman on Thu Jul 18, 2013 8:50 pm UTC, edited 1 time in total.
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### Re: Galilean:x' with respect to S'?

steve waterman wrote:
eSOANEM wrote:No, 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).

What the galilean says is simply that S(x,t)=S'(x-vt,t)=S'(x',t').

x' and t' are just notational conveniences, the important bit that is actually useful is the first equality. What this tells you is what the co-ordinate system S' calls the point the co-ordinate system S calls (x,t). What makes this useful is that it does so whilst only referring to the co-ordinates in the system S.

I will get back to other things in your post later...thanks for giving your 3 responses. As well as things I clipped above.

1 mapping coordinates manifests a point in the manifold.
2 a point can be identified as they come in the mathematical form of (a,b) or say (a,b,c).
3 the first coordinate, by itself, is not a point.

eSOANEM wrote:No to 2. A point is not it's co-ordinates.

This statement is about identifying the correct mathematical format that is used to represent a point.

eSOANEM wrote:Often people are sloppy and assume a particular co-ordinate system but all this means is that there's an "S" in front of every set of co-ordinates they've written which they can't be bothered to write.

I stand by what I said. The point cannot be identified by a set of co-ordinates alone. You need to have a co-ordinate system in order to translate (in the colloquial sense) from the co-ordinates to the point. In the sloppy case, it's usually obvious what system they're using because the names they give their co-ordinates are conventionally used almost exclusively in certain co-ordinate systems (e.g. having the variables x and y as your two co-ordinates usually implies you have a cartesian co-ordinate system with one of the axes (the x-axis) being horizontal).

In that sloppy case, there is still a co-ordinate system, it just may not have been explicitly listed.

Schrollini wrote:
eSOANEM wrote:(Sorry Schrollini about not using the s and u convention, given various other common uses of those symbols in contexts where the Galilean is also commonly used, I think it will actually lead to more confusion in the long-run)

Confusion is better than certainty about a falsehood. I've been purposefully using notation different from what Steve uses to try to make him think about what the symbols mean, rather than just assuming they mean what he's taken them to mean in the past. Whether this has worked or not is another issue.

I agree entirely with your aims in using that alternative notation (and it's also useful for illustrating that relabelling variables is completely fine and doesn't imply anything in and of itself); I just don't agree that this implementation is helpful and, rather than go down the xkcd/927 thought it best if I stick with the x' convention.

steve waterman wrote:1 mapping coordinates identifies a point in the manifold.
2 the mathematical notational form of (a,b) or say (a,b,c) identifies a point...
(with the caveat that there is an implied coordinate system of some form)
3 the first coordinate, by itself, is not a point.

Yes although the second is not a good way to think about it.

It is technically true, but if at this stage you think only about it rigorously and require that all co-ordinate systems are stated explicitly it will help you get a proper understanding of this. As with all things, you need to learn the rules and understand them before you can bend them without breaking them.
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### Re: Galilean:x' with respect to S'?

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.
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