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

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

Postby ucim » Wed Jul 17, 2013 3:41 pm UTC

steve waterman wrote:
ucim wrote:What does it mean for a coordinate system to be "coincident"?

Think about the question, and keep in mind Schrollini's first lesson; that is, separating the idea of a manifold from the idea of a coordinate system, and remembering that points live in a manifold, not in a coordinate system,

So, does that make it true? The idea, is that I indeed DISAGREE and I start with a different set of initial premises.

Then you are doing a different kind of mathematics. The stuff in Schrollini's first lesson underlies the abbreviated notation you found and are using for the Galilean. If you reject it, then the notation ceases to actually mean what it is supposed to mean.

steve waterman wrote:given S(x,y,z) and only S(x,y,z)...
Do any points exist "on top of, but not inherent to" ( according to Scrollini ) manifold M?
If we removed manifold M as a thought experiment, what would remain? Would we know exactly where to place S(2,3,4) wrt S(0,0,0)?


I'm not sure what you mean by S(x,y,z), although I know what others often mean by it. But in any case, the answer to your second question ("Do any points..") is no. All the points are in the manifold. The manifold is just a pile of points, with no structure1, no order. It's the only place where the points "are". And if we removed the manifold as a thought experiment, you would have nothing useful... essentially, you'd have the card catalog to a library that does not exist. Your question: Would we know exactly where to place S(2,3,4) wrt S(0,0,0)? is actually backwards, for this reason. S(2,3,4) isn't "placed" anywhere. S(2,3,4) is the card in the card catalog that identifies a specific point in the manifold (book in the library). Without the manifold, there wouldn't be a point for S(2,3,4) to refer to. The same is true of S(0,0,0). No manifold, no point.

Remember also that points are not located "with respect to" other points. They are just piled up in the manifold, jumbled up. A coordinate system gives you the ability to identify them "from outside", so to speak, but if you want to determine distance between points, you need to decide what that means first. That involves a metric. (Key takeaway: a coordinate system, by itself, does not have a metric. You need to explicitly pick one.)

Jose
eta: 1not quite true, but close enough for now
Last edited by ucim on Wed Jul 17, 2013 9:31 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 3:49 pm UTC

beojan wrote:Later addition (Steve: notice that I'm only adding to my post because nobody had posted after me. Addition made 2013-07-17 14:44 UTC)
The problem with the abscissa, ordinate, applicate terminology is that it locks you into using just one coordinate system. When doing geometric work ("here does the point at (0,5) end up when reflected in the x-axis?", or "What does a line from (0,6) to (4,7) look like when rotated through 45° clockwise about the point at (0,4)?"), this terminology may be sufficient, but in physics, coordinates are more than just a way to address points. We are interested in looking at the same 'experiment' from more than one point of view, and coordinate transforms enable us to express this mathematically. When working with more than one coordinate system (as we inherently must, when discussing coordinate transforms), the abscissa, ordinate, applicate terminology becomes cumbersome, or even meaningless (look at the drawing giving an example of rotation in the "Several Examples, Chosen with Malice Aforethought" section by Schrollini. In the u-v system, which is the abscissa, and which is the ordinate?)


Yes, this is a nice kind of possible counter-example to investigate.
Image

LOGIC
Since both systems were coincident, P(2,1) mathematically cannot exist without there also being a corresponding P'(2,1) in the opposite system, at t = 0.

You rotated red and only kept P. You could also have only rotated blue counter-clock, in which case for you
you merely use P' and ignore that P also, mathematically exists.

Now, I am fully aware that you do not wish to rotate P' along with the coordinate system.
I understand that your points are stationary wrt manifold M.

MY approach - what if we did allow points in the manifold to move as the origin of the moving system does?
as shown in the diagram included. So, all one needs to do, to see through my mindset, is to allow this one premise.
This does not mean the premise is right or wrong, just that ...when we do, the red abscissa magically appears and is equal to the blue abscissa.
Last edited by steve waterman on Wed Jul 17, 2013 4:02 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'?

Postby PM 2Ring » Wed Jul 17, 2013 3:59 pm UTC

JudeMorrigan wrote:
beojan wrote:Much of the 'misunderstanding', in my opinion, is because of Steve's refusal to answer the simple question asked in the original thread: what is your native language?

I'm pretty sure he did give a real answer to it, actually. He said he's from Massachusetts, iirc.

ETA: The relevant post:
viewtopic.php?f=2&t=96231&start=720#p3043064


While reading Steve's posts, I've occasionally wondered if the misunderstandings are simply due to language differences and I almost asked him if English is his mother tongue in one of my earlier posts in this thread. So thanks, JudeMorrigan, for clearing that up. I guess it's still possible that he acquired his idiosyncratic style from the people he grew up with. But I suspect that there is another explanation...

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

Postby JudeMorrigan » Wed Jul 17, 2013 4:01 pm UTC

steve waterman wrote:MY approach - what if we did allow points in the manifold to move as the origin of the moving system does?

Then you wouldn't be doing a coordinate transformation and what you would be would have exactly zero applicability to the question at hand. At least, not the way you're doing it. In effect, that would mean that your "moving" system would be stationary, and that it would be the other one that was moving. You could do a coordinate transformation using that set-up, but it's not what you're actually doing. All of your talk of "placing points in the opposite system" and such are, again, non-sense.

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

Postby Schrollini » Wed Jul 17, 2013 4:05 pm UTC

steve waterman wrote:You rotated red

No! No, no, no, no, no, a thousand times no.

We have a single manifold M. We place two coordinate systems on M, xy and uv. Essentially, we draw the blue axes to define xy and the red axes to define uv. I did not rotate anything. In the xy system, P has coordinates (2,1). In the uv system it has coordinates (3/sqrt(2), -1/sqrt(2)).

We call the transformtion between the two systems a rotation, which I already pointed out could be misleading. And you've been misled -- nothing has rotated!
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Re: Galilean:x' with respect to S'?

Postby JudeMorrigan » Wed Jul 17, 2013 4:07 pm UTC

Schrollini wrote:And you've been misled -- nothing has rotated!

An important point. I'm letting myself get stuck in the mindset of us talking about cases where things are actually moving. My sitting back and not trying to add substanative posts was the better strategy on my part. I need to go back to it.

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

Postby steve waterman » Wed Jul 17, 2013 4:09 pm UTC

Schrollini wrote:
steve waterman wrote:You rotated red

No! No, no, no, no, no, a thousand times no.

We have a single manifold M. We place two coordinate systems on M, xy and uv. Essentially, we draw the blue axes to define xy and the red axes to define uv. I did not rotate anything. In the xy system, P has coordinates (2,1). In the uv system it has coordinates (3/sqrt(2), -1/sqrt(2)).

We call the transformation between the two systems a rotation, which I already pointed out could be misleading. And you've been misled -- nothing has rotated!

Did you start with our two coordinate systems as coincident with the manifold?

added -
so,
P re xy(2,1) = P re uv (3/sqrt(2), -1/sqrt(2))

So, yes, the points in the manifold have different relative coordination to each other for the same point.
I agree 100 percent. Hear me please, no need to keeping flog this, I agree.

In the context of x' = x-vt, does x represent the abscissa in S?
Last edited by steve waterman on Wed Jul 17, 2013 4:27 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'?

Postby Schrollini » Wed Jul 17, 2013 4:21 pm UTC

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

No.

This question suggests that you're still thinking of coordinate systems as dynamic objects that evolve in some sort of "time". They aren't. There is no "time" in which they can evolve. They simply exist.
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Re: Galilean:x' with respect to S'?

Postby firechicago » Wed Jul 17, 2013 4:30 pm UTC

JudeMorrigan wrote:
steve waterman wrote:MY approach - what if we did allow points in the manifold to move as the origin of the moving system does?

Then you wouldn't be doing a coordinate transformation and what you would be would have exactly zero applicability to the question at hand.


What you're doing is not a coordinate transformation, as the term is defined. Perhaps it will help to think about what the different functions we're talking about do:

A coordinate system maps a point in a manifold to a coordinate (usually defined as a couple of real numbers in an ordered pair or trio). To use the notation you've been using, a coordinate system might look like Point P -> S(x,y,z) (or S(x,y,z) -> Point P since most of the coordinate systems that we work with are one-to-one and therefore we don't make much distinction between the function and its inverse).

A coordinate transformation maps a coordinate to another coordinate, such that the two coordinates refer to the same point in two different coordinate systems. So S(a,b,c) -> S'(d,e,f) or you could think of the function as having an intermediate step, like S(a,b,c) -> Point P -> S'(d,e,f)

What you are proposing is a function that maps a point onto a different point that happens to share the same coordinate in a different coordinate system. So your proposed function would look like Point P -> Point Q, or to show the intermediate steps: Point P -> S(a,b,c) -> S'(a,b,c) -> Point Q.

This is a perfectly good definition of a function, but it is not a coordinate transformation. It's not even in the same category as a coordinate function, because it is mapping from points to points and not coordinates to coordinates, which is why Schrollini spent so much time trying to make the distinction between the points in the manifold and the coordinates that describe them. The galilean transformation is defined as a coordinate transformation, so whatever you may show about your function is irrelevant to the galilean, since they're not in the same class.

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

Postby steve waterman » Wed Jul 17, 2013 4:41 pm UTC

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

No.

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


NO? That was a complete surprise!

MY thinking is that GIVEN S(x,y,z) means
1 no manifold required
2 no mapping required
3 infinite exact coordinate point set including all (x,y,z) within S exists mathematically
4 eternal, fixed, not "transformable abscissa wrt own system"
5 no time

I keep asking how exactly that you achieve coordination in S, given only S(x,y,z)...how do you?
Explain the manifold under those conditions please...
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Re: Galilean:x' with respect to S'?

Postby Schrollini » Wed Jul 17, 2013 4:55 pm UTC

steve waterman wrote:NO? That was a complete surprise!

That's fantastic! This means that this is one of those Eureka moments that Jose was talking about. It represents a crucial difference between the standard conception of coordinate systems and your way of thinking. If you're going to understand what we mean by the Galilean transformation, you must keep this difference in mind!

steve waterman wrote:I keep asking how exactly that you achieve coordination in S, given only S(x,y,z)...how do you?
Explain the manifold under those conditions please...

The problem is that I don't understand the question. If S is a coordinate system (I'm assuming), then it is a mapping between the coordinates (sets of real numbers) and points in a manifold. This mapping is (fairly) arbitrary. There's no specific coordinate system that's "right" or "wrong". You can do it any way you want.
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 5:03 pm UTC

Schrollini wrote:
steve waterman wrote:NO? That was a complete surprise!

That's fantastic! This means that this is one of those Eureka moments that Jose was talking about. It represents a crucial difference between the standard conception of coordinate systems and your way of thinking. If you're going to understand what we mean by the Galilean transformation, you must keep this difference in mind!

steve waterman wrote:I keep asking how exactly that you achieve coordination in S, given only S(x,y,z)...how do you?
Explain the manifold under those conditions please...

The problem is that I don't understand the question. If S is a coordinate system (I'm assuming), then it is a mapping between the coordinates (sets of real numbers) and points in a manifold. This mapping is (fairly) arbitrary. There's no specific coordinate system that's "right" or "wrong". You can do it any way you want.


Perfect. Yes, this is a good thing. ( sorry, Martha)
so, you map from the GIVEN coordination wrt S?
in some manner onto the manifold...remember, our condition, is only given S(x,y,z)...

Also please, in our case, are we restricted to only one mapping per transformation ?

added -
Schrollini wrote:The problem is that I don't understand the question.

I so much appreciate hearing this rather than just having some question ignored.
if you can, also explain what parts you cannot parse, whenever pertinent.
Last edited by steve waterman on Wed Jul 17, 2013 5:21 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'?

Postby Schrollini » Wed Jul 17, 2013 5:20 pm UTC

steve waterman wrote:so, you map from the GIVEN coordination wrt S?

What do you mean by "coordination"? We have "coordinates" (a set of real numbers) and "coordinate systems" (a mapping from coordinates to points). You need to specify the coordinate system before you know how to map coordinates to points.

To take Jose's example: The manifold is like a bunch of books. The coordinate system is like an ordering that maps integers to books. You can have a coordinate system that orders books by title, or you can have one that orders books by length. It's meaningless to ask what the 27th book is until you specify the ordering. Similarly, it's meaningless to ask what point has coordinates (2,7) until you specify the coordinate system.
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 5:58 pm UTC

Schrollini wrote:
steve waterman wrote:so, you map from the GIVEN coordination wrt S?

What do you mean by "coordination"? We have "coordinates" (a set of real numbers) and "coordinate systems" (a mapping from coordinates to points). You need to specify the coordinate system before you know how to map coordinates to points.

To take Jose's example: The manifold is like a bunch of books. The coordinate system is like an ordering that maps integers to books. You can have a coordinate system that orders books by title, or you can have one that orders books by length. It's meaningless to ask what the 27th book is until you specify the ordering. Similarly, it's meaningless to ask what point has coordinates (2,7) until you specify the coordinate system.


The issue is really, what is meant by x in the equation x' = x-vt??? ?
x = abscissa wrt S
x = (x,0,0) wrt S
x = (x,0,0) wrt the manifold
x = something else altogether

added -
Imagine a blue ruler with inches called x instead, and a red ruler with inches called x'.
No matter the location of the two rulers, x = x'.

Imagine a blue ruler with inches called (x,0,0) instead, and a red ruler with inches called (x',0.0).
if the rulers are separated by three inches along the x/x' direction,
(x,0,0) = (x',0,0) +/- (3,0,0)

added again -
So, for you, it would appear, it is impossible to have S(x,y,z) and not also have an optional manifold M?

and again -. Terms. I really am having problems not being able to say "coordinate points". This should not be in debate at all. Making the term points "exclusive" to the manifold is not fair play.
I will be using these two terms to clarify the difference. "manifold points" and "coordinate points".

You can insist that points mean only the manifold, and I can premise that there are two kinds of points, involved.
Last edited by steve waterman on Wed Jul 17, 2013 6:36 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'?

Postby beojan » Wed Jul 17, 2013 6:24 pm UTC

Steve:
It appears Schrollini's attempt to define all terms used rigorously has just left us with a group of terms, all of which have definitions, all of which you are unable to use correctly (this, strangely enough, includes terms you have introduced (e.g. "abscissa of x")).

Steve, in your opinion, what is mathematics? That is, what does a mathematician do? And what is physics? What does a physicist do?

A few more questions for Steve (added later):
What is the purpose of coordinates / coordinate systems? What do you do with them?

What is the purpose of the Galilean transform?
Where did the equation x' = x - vt come from? That is, was it written down by a person, was it found carved in a stone, was it divinely inspired in your mind or the mind of another person?

Note to others: the purpose of these questions is to try and find out what Steve thinks, not what the right answers actually are.

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

Postby Schrollini » Wed Jul 17, 2013 6:44 pm UTC

steve waterman wrote:The issue is really, what is meant by x in the equation x' = x-vt??? ?

x is the first coordinate, in the xt reference frame, of some event in spacetime. To answer the follow-up question, x' is the first coordinate, in the x't' reference from, of that same event.

To understand this answer, you must understand what we mean by "reference frame" and "spacetime", which in turn means you must understand "coordinate system" and "manifold". If you do not understand these basic building blocks, you will never understand this equation.

I have done my best to provide lessons to provide this understanding. But if you're unwilling to work through them, I will never be able to explain this equation to you.
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 6:49 pm UTC

beojan wrote:Steve:
It appears Schrollini's attempt to define all terms used rigorously has just left us with a group of terms, all of which have definitions, all of which you are unable to use correctly (this, strangely enough, includes terms you have introduced (e.g. "abscissa of x").

Steve, in your opinion, what is mathematics? That is, what does a mathematician do? And what is physics? What does a physicist do?

A few more questions for Steve (added later):
What is the purpose of coordinates / coordinate systems? What do you do with them?

What is the purpose of the Galilean transform?
Where did the equation x' = x - vt come from? That is, was it written down by a person, was it found carved in a stone, was it divinely inspired in your mind or the mind of another person?

Note to other: the purpose of these questions is to try and find out what Steve thinks, not what the right answers actually are.

beojan wrote:What is the purpose of coordinates / coordinate systems? What do you do with them?

Given S(x,y,z) Let POINT P in/wrt S at S(2,3,4) = "coordinate point P" in S
Galilean -
Given S(,x,y,z) map POINT P on top of manifold M from/wrt S(2,3,4) = "manifold point P" on top of, but not inherent to, manifold M
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Re: Galilean:x' with respect to S'?

Postby eSOANEM » Wed Jul 17, 2013 6:59 pm UTC

steve waterman wrote:
beojan wrote:Steve:
It appears Schrollini's attempt to define all terms used rigorously has just left us with a group of terms, all of which have definitions, all of which you are unable to use correctly (this, strangely enough, includes terms you have introduced (e.g. "abscissa of x").

Steve, in your opinion, what is mathematics? That is, what does a mathematician do? And what is physics? What does a physicist do?

A few more questions for Steve (added later):
What is the purpose of coordinates / coordinate systems? What do you do with them?

What is the purpose of the Galilean transform?
Where did the equation x' = x - vt come from? That is, was it written down by a person, was it found carved in a stone, was it divinely inspired in your mind or the mind of another person?

Note to other: the purpose of these questions is to try and find out what Steve thinks, not what the right answers actually are.

beojan wrote:What is the purpose of coordinates / coordinate systems? What do you do with them?

Given S(x,y,z) Let POINT P in/wrt S at S(2,3,4) = "coordinate point P" in S
Galilean -
Given S(,x,y,z) map POINT P on top of manifold M from/wrt S(2,3,4) = "manifold point P" on top of, but not inherent to, manifold M


This means nothing.

S(x,y,z,t) refers to the same point as S'(x',y',z',t).

(x,y,z,t) are the co-ordinates of that point in the S co-ordinate system and (x',y',z',t) are the co-ordinates in the S' co-ordinate system. The galilean gives you a way to find the co-ordinates in one system in terms of those in another (up to certain assumptions about the two co-ordinate systems).

The x and x' are not really distances in any useful sense. They are simply co-ordinates in their respective co-ordinate systems.
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Re: Galilean:x' with respect to S'?

Postby beojan » Wed Jul 17, 2013 7:00 pm UTC

steve waterman wrote:
beojan wrote:What is the purpose of coordinates / coordinate systems? What do you do with them?

Given S(x,y,z) Let POINT P in/wrt S at S(2,3,4) = "coordinate point" in S
Galilean -
Given S(,x,y,z) map POINT P on top of manifold M from/wrt S(2,3,4) = "manifold point" on top of, but not inherent to, manifold M


You haven't actually answered this or any other question. However, I'd like to post a simple demonstration that might help you understand the concepts everyone else is using in a more intuitive manner.

You don't have to carry out the demonstration, imagining it will be fine (assuming you can do so accurately), but it may help to actually carry it out.
You will need:
  • Two equal sized square sheets of perspex (plexiglass)
  • Red and blue dry-erase markers
  • A square, equal in size to the perspex sheets, marked out on the floor (should be flat)
  • Two small toy cars, with a small round yellow dot on top of one, and a small round green dot on top of the other. The one with the green dot should be taller than the one with the yellow dot.
Of course, it doesn't actually need to be toy cars, just some small object that can be easily moved on the floor.

Once you post that you are ready to carry out the demonstration, or to imagine it, I will post the demonstration.

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

Postby steve waterman » Wed Jul 17, 2013 7:17 pm UTC

Schrollini wrote:
steve waterman wrote:The issue is really, what is meant by x in the equation x' = x-vt??? ?

x is the first coordinate, in the xt reference frame, of some event in spacetime.

Great.
Would the first coordinate, by itself, represent a manifold point after mapping?
Would the first coordinate, by itself, represent the abscissa before mapping?


Still working to get some notation straightened out...
Would S(x,y,z) represent the manifold point P after mapping?
Would (x,y,z) represent the manifold point P after mapping?
Would M(x,y,z) represent the manifold point P after mapping?
What notation can we use to represent the manifold point P after mapping, if not one of those?
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Re: Galilean:x' with respect to S'?

Postby WibblyWobbly » Wed Jul 17, 2013 7:48 pm UTC

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:The issue is really, what is meant by x in the equation x' = x-vt??? ?

x is the first coordinate, in the xt reference frame, of some event in spacetime.

Great.
Would the first coordinate, by itself, represent a manifold point after mapping?
Would the first coordinate, by itself, represent the abscissa before mapping?


Still working to get some notation straightened out...
Would S(x,y,z) represent the manifold point P after mapping?
Would (x,y,z) represent the manifold point P after mapping?
Would M(x,y,z) represent the manifold point P after mapping?
What notation can we use to represent the manifold point P after mapping, if not one of those?


What is a "manifold point", Steve? And how do you imagine this "mapping"? Why would the coordinates given to a point change after this "mapping", as you see it? Did you actually read and attempt to understand anything Schrollini wrote, or were you simply paying lip service to him?

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

Postby Dark Avorian » Wed Jul 17, 2013 7:54 pm UTC

Steve,

It was previously mentioned that you are like a man standing in the middle of an American city (from now on, I will just refer to Kentucky), wildly speaking Vietnamese. I would beg to disagree. If only that were the case. This is much worse.

What you insist upon doing is not speaking Vietnamese in Kentucky, if that was the case, there would be no shared ambiguous terms, or at least far fewer, and they would be pretty damn obvious.

Consider an analogy. One man is speaking Vietnamese in Kentucky, and wants to find the nearest public library. At first we would not be able to understand, but there would be an obviously unbridgeable gap between our languages of choice. We might eventually, through frantic gesticulation, teach each other the basics of each other's languages, communicate in brutal terms, and maybe one day you'd get to the public library.


The other man is sitting there, asking "How good is the food at McDonalds"? But he also wants to find the public library. By "How good" he means "Where is", by "the food", and by "at Mcdonalds" he means the nearest. Or worse, maybe he isn't even using the same types of sentence structure, but still uses a rich English vocabulary (just large swathes are being totally misused). We may never know if we've gotten through to him, although every once and a while when he says "Is red fish a deep fried warzone?" at which point we know we're in deep shit. The problem is that we never know when a word we're saying is being interpreted as in proper English, or the strange version.

That's the problem we have here. Mathematics is very precise, and if you misuse language, and choose your own definitions, no one will ever know what you're doing. The famous mathematician Grothendieck, while in graduate school (I think), derived results that were basically equivalent to Lebesgue's theory of measure and integration. Maybe you've done something new, but maybe you're just talking in different terms about something we already know.

Oh, and just as an aside, "metric" is a damn precise word mathematically, and if you are not even comfortable with that, you might want to look into it.

Spoiler:
A metric, rigorously speaking is a function that takes ordered pairs of elements (x,y) from some underlying space X, and maps these pairs to real numbers in a very specific way that accords with our notion of "distance between x and y".

A metric space (X, d) is composed of a set X, and a function "d" mapping pairs of elements of X to real numbers with the following properties

(i) Positive-Definite : d(x,y) >= 0 for all x, y in X. d(x,y) = 0 if and only if x=y
(ii) Symmetric : for any x,y in X, d(x,y) = d(y,x)
(iii) Triangle Inequality: for any x,y,z in X , d(x,y) + d(y,z) >= d(x,z)

(">=" is being used to mean greater than or equal to)

This is precise. Perhaps you might want to know the basic elements of mathematics before trying to attack something as subtle as a coordinate transform
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 7:56 pm UTC

WibblyWobbly wrote:What is a "manifold point", Steve?

a point on top of, but not inherent to, the manifold.

added -
"coordinate point"
one of the infinite point set inherent to every given Cartesian system
Last edited by steve waterman on Wed Jul 17, 2013 8:02 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'?

Postby eran_rathan » Wed Jul 17, 2013 8:00 pm UTC

I think steve is a living example of the "Hungarian Phrasebook" sketch from Monty Python.
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Re: Galilean:x' with respect to S'?

Postby Dark Avorian » Wed Jul 17, 2013 8:03 pm UTC

steve waterman wrote:
WibblyWobbly wrote:What is a "manifold point", Steve?

a point on top of, but not inherent to, the manifold.


NO! NO! NO!

That is not how manifolds work, if you were paying any attention to Schrollini.

A manifold is a set, this set has elements (strictly speaking it also has a topology, but I'm not sure oyu're ready to handle that yet). We call elements of a manifold "points". Points lie in a manifold. What is not inherent to a manifold is the coordinate systems that allow use to locally associate coordinates with the points of the manifold, there are many such coordinate systems for any manifold. The coordinates of a point, and the coordinate structure are not inherent to a manifold and in some way lie atop it, but the points are inherent to the manifold, as the manifold is nothing more and nothing less than the set of those points.
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Re: Galilean:x' with respect to S'?

Postby ucim » Wed Jul 17, 2013 8:03 pm UTC

steve waterman wrote:
WibblyWobbly wrote:What is a "manifold point", Steve?

a point on top of, but not inherent to, the manifold.


There is no such thing in the mathematics whence comes the Galilean.

Either you are doing it wrong, or you are doing a new kind of math.

If you are doing it wrong, stop doing it and let Schrollini show you the right way.

If you are doing a new kind of math, then what we call the Galilean will not result from it.

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

Postby steve waterman » Wed Jul 17, 2013 8:05 pm UTC

for Schrollini -
Would the first coordinate, by itself, represent a point in the manifold after mapping?
Would the first coordinate, by itself, represent the abscissa before mapping?
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 8:11 pm UTC

Dark Avorian wrote:
steve waterman wrote:
WibblyWobbly wrote:What is a "manifold point", Steve?

a point on top of, but not inherent to, the manifold.


NO! NO! NO!

That is not how manifolds work, if you were paying any attention to Schrollini.

A manifold is a set, this set has elements (strictly speaking it also has a topology, but I'm not sure oyu're ready to handle that yet). We call elements of a manifold "points". Points lie in a manifold. What is not inherent to a manifold is the coordinate systems that allow use to locally associate coordinates with the points of the manifold, there are many such coordinate systems for any manifold. The coordinates of a point, and the coordinate structure are not inherent to a manifold and in some way lie atop it, but the points are inherent to the manifold, as the manifold is nothing more and nothing less than the set of those points.


Dark Avorian wrote:Points lie in a manifold.

Schrollini is the one who educated me in that points are not inherent to the manifold, but lie on top of the manifold.
So, take up your beef with him please. I was merely repeating what I was being "taught" by him.

I really do not care, as I have no manifold requirement given S(x,y,z)....my entire coordinate system is complete without it.

added -
I will use the phrase "points inherent to the manifold" since that is what seems appropriate.
I will also switch to the long phrase
"premised points inherent to the given coordinate system"
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Re: Galilean:x' with respect to S'?

Postby JudeMorrigan » Wed Jul 17, 2013 8:15 pm UTC

steve waterman wrote:Schrollini is the one who educated me in that points are not inherent to the manifold, but lie on top of the manifold.

Citation needed, please.

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

Postby induction » Wed Jul 17, 2013 8:26 pm UTC

This is what Schrollini actually said before steve ran it through the transmogrifier:

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

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

Postby SecondTalon » Wed Jul 17, 2013 8:36 pm UTC

Dark Avorian wrote:... in the middle of an American city (from now on, I will just refer to Kentucky),...


Bravo. 5 points to Gryffindor.
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Re: Galilean:x' with respect to S'?

Postby eSOANEM » Wed Jul 17, 2013 8:37 pm UTC

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:The issue is really, what is meant by x in the equation x' = x-vt??? ?

x is the first coordinate, in the xt reference frame, of some event in spacetime.

Great.
Would the first coordinate, by itself, represent a manifold point after mapping?
Would the first coordinate, by itself, represent the abscissa before mapping?


Still working to get some notation straightened out...
Would S(x,y,z) represent the manifold point P after mapping?
Would (x,y,z) represent the manifold point P after mapping?
Would M(x,y,z) represent the manifold point P after mapping?
What notation can we use to represent the manifold point P after mapping, if not one of those?


Unnecessary and confusing (as pointed out by schrollini earlier) words struck through by me (although in one of your later posts you did correct the first line).

No, you need a full set of co-ordinates in order to reach a specific point after mapping using your co-ordinate system. What you get (if anything at all) if you give your co-ordinate system a single co-ordinate depends on many factors including exactly what you mean by "give it one co-ordinate".

Not necessarily. Abscissa only makes sense in certain co-ordinate systems (in a polar co-ordinate system for instance, none of your co-ordinates are the abscissa. Also, you seem to be misinterpreting what abscissa means for the reasons I posted earlier.

If S and M are co-ordinate systems then S(x,y,z) and M(x,y,z) represent points as it is, the "S" or "M" is the mapping between co-ordinates and points. They already have been mapped. (x,y,z) might represent a point after mapping but that would depend on the map you choose (if I pick a polar co-ordinate system, the co-ordinates (-10,17,4) do not map to any point because they have a negative radius and angles greater than 2pi and pi respectively).

steve waterman wrote:
WibblyWobbly wrote:What is a "manifold point", Steve?

a point on top of, but not inherent to, the manifold.

added -
"coordinate point"
one of the infinite point set inherent to every given Cartesian system


There is no such thing. The manifold is made of points. It is the set of all points. The only points that are relevant to this discussion are those in the manifold because if we postulate a point not part of the manifold, we can simply change our precise definition of the manifold to include it. We lose no generality whatsoever by assuming that all points in the system are part of the manifold.

Your term co-ordinate point is similar to the image of a co-ordinate system (at least in intent). It describes all the points (which, by definition, exist in our manifold) that can be mapped to by our co-ordinate system.

steve waterman wrote:
Dark Avorian wrote:
steve waterman wrote:
WibblyWobbly wrote:What is a "manifold point", Steve?

a point on top of, but not inherent to, the manifold.


NO! NO! NO!

That is not how manifolds work, if you were paying any attention to Schrollini.

A manifold is a set, this set has elements (strictly speaking it also has a topology, but I'm not sure oyu're ready to handle that yet). We call elements of a manifold "points". Points lie in a manifold. What is not inherent to a manifold is the coordinate systems that allow use to locally associate coordinates with the points of the manifold, there are many such coordinate systems for any manifold. The coordinates of a point, and the coordinate structure are not inherent to a manifold and in some way lie atop it, but the points are inherent to the manifold, as the manifold is nothing more and nothing less than the set of those points.


Dark Avorian wrote:Points lie in a manifold.

Schrollini is the one who educated me in that points are not inherent to the manifold, but lie on top of the manifold.
So, take up your beef with him please. I was merely repeating what I was being "taught" by him.

I really do not care, as I have no manifold requirement given S(x,y,z)....my entire coordinate system is complete without it.

added -
I will use the phrase "points inherent to the manifold" since that is what seems appropriate.
I will also switch to the long phrase
"premised points inherent to the given coordinate system"


No.

Schrollini did the exact opposite and has said so explicitly in response to you claiming this before.

It is co-ordinates, co-ordinate systems and metrics (measures of "distance" in some sense) which are external to the manifold. Points are what is used to define the manifold in the first place. Here's a quote from his first post explaining that points are elements of (exist in) the manifold if you don't believe me:

Schrollini wrote:Definition 2: A point is an element in a manifold. If I want to say that the point P is an element of manifold M, I could write P ∊ M. (But I'll try to avoid doing that and just put things in English.)
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Re: Galilean:x' with respect to S'?

Postby Vetala » Wed Jul 17, 2013 8:38 pm UTC

induction wrote:This is what Schrollini actually said before steve ran it through the transmogrifier:

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


I've been staying out of the reprise of the carousel so far, but I'll take that one step further, just to drive the point home:

Schrollini wrote:Definition 2: A point is an element in a manifold. If I want to say that the point P is an element of manifold M, I could write P ∊ M. (But I'll try to avoid doing that and just put things in English.)

Schrollini wrote:If P and Q each label a point in the manifold [...]

Schrollini wrote:[...] may not necessarily map (0,0) to a point in the manifold

Schrollini wrote:The manifold contains an uncountable infinity of points

Schrollini wrote:[...] maps coordinates to points in a manifold.

Schrollini wrote:Points are elements of a manifold

Schrollini wrote:[...] a manifold is a set, and the points of the manifold are the elements of the set.

Schrollini wrote:[...] given a specific x and y, f(x,y) is a point in the manifold.

Schrollini wrote:[...] you may mentally replace point with "element of the manifold".

Schrollini wrote:Points live in the manifold.

Schrollini wrote:[...] the manifold M (whose elements are the points).

Schrollini wrote:The manifold has an uncountable infinitude of points [...]

Schrollini wrote:[...] function that maps coordinates to points in a manifold.

Schrollini wrote:The manifold has points, nothing else.

Schrollini wrote:the manifold only has points

Schrollini wrote:[...] f(x,y) and g(u,v) are both points in manifold

Schrollini wrote:If S is a coordinate system (I'm assuming), then it is a mapping between the coordinates (sets of real numbers) and points in a manifold.


Nowhere are points claimed to be anything but part of the manifold. Even if "All of these things exist on top [...]" could be misinterpreted as including the points in Schrollini's "attributing entirely too much" post, *every* other mention of manifolds and points makes it very clear that the points are elements of the manifold - not something separate from it.

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

Postby steve waterman » Wed Jul 17, 2013 8:47 pm UTC

induction wrote:This is what Schrollini actually said before steve ran it through the transmogrifier:

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


Thanks for finding and posting the original post, induction !

Wow. I took "they" to mean points, as I kinda read "all the other things" and thought he meant points aren't inherent.
"All of these things" I took to mean, well ALL of the things just itemized, meaning points too!

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.

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

Postby steve waterman » Wed Jul 17, 2013 9:07 pm UTC

Schrollini wrote:The manifold contains an uncountable infinity of points


Thanks for isolating this, Vetala, it got lost in the shuffle of reading and posting and preparing stuff.

This is really what is confusing to me about this statement...

I thought the infinite coordinate point set is on top of the manifold
and that the manifold point set would only have selected elements like point P.

For me, the coordinate systems contains all the elements of any given system.
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Re: Galilean:x' with respect to S'?

Postby eSOANEM » Wed Jul 17, 2013 9:29 pm UTC

This is what Schrollini was saying.

A co-ordinate system does not contain any points. What it does is take a set of co-ordinates and return a point in the manifold. All it does is refer to points which exist as part of that manifold.

The set of all points that the co-ordinate system can refer to is its image. This need not be all the points in the manifold. Neither must the set of all co-ordinates which map to points (the co-ordinate system's domain) include all possible co-ordinates.
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Re: Galilean:x' with respect to S'?

Postby steve waterman » Wed Jul 17, 2013 10:52 pm UTC

eSOANEM wrote:This is what Schrollini was saying.

A co-ordinate system does not contain any points. What it does is take a set of co-ordinates and return a point in the manifold. All it does is refer to points which exist as part of that manifold.

The set of all points that the co-ordinate system can refer to is its image. This need not be all the points in the manifold. Neither must the set of all co-ordinates which map to points (the co-ordinate system's domain) include all possible co-ordinates.


Very nice explanation. That clears up some things for me.
So the Galilean for you, only ever transforms a point within the manifold itself, since points only exist in the manifold.

If we ignore the manifold for a second, and only consider the co-ordinate system's domain,
does the the first coordinate of one system equal the first coordinate of the other system?

If you cannot parse that question, please let me know.

added =
1 so, in the manifold, Point P maps from S as (x,0,0) and point P also maps from S' as (x-vt,0,0)?

2 so, in the once coincident coordinate systems, now separated by vt along the common x/x' direction,
the first coordinate in the domain of S = the first coordinate in the domain of S' regardless of the manner of their placement on top of the manifold?

added again,
Since, in the manifold itself, (x,0,0) = (x-vt,0,0), do you believe that this means we can mathematically, logically, extrapolate that x = x'-vt from that equation?
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Re: Galilean:x' with respect to S'?

Postby eSOANEM » Wed Jul 17, 2013 11:28 pm UTC

steve waterman wrote:
eSOANEM wrote:This is what Schrollini was saying.

A co-ordinate system does not contain any points. What it does is take a set of co-ordinates and return a point in the manifold. All it does is refer to points which exist as part of that manifold.

The set of all points that the co-ordinate system can refer to is its image. This need not be all the points in the manifold. Neither must the set of all co-ordinates which map to points (the co-ordinate system's domain) include all possible co-ordinates.


Very nice explanation. That clears up some things for me.
So the Galilean for you, only ever transforms a point within the manifold itself, since points only exist in the manifold.

If we ignore the manifold for a second, and only consider the co-ordinate system's domain,
does the the first coordinate of one system equal the first coordinate of the other system?

If you cannot parse that question, please let me know.


A few things.

There is no "for me" involved in the Galilean. It is phrased in terms defined as I am using them. If you are using terms differently then the Galilean is completely irrelevant and obviously does not apply to anything.

The galilean does not transform points, it transforms the co-ordinates of a point in one co-ordinate system (x,y,z,t) to the co-ordiantes of that same point in another co-ordinate system (x',y',z',t).

The first co-ordinates are not equal. As was shown in Schrollini's examples with two co-ordinate systems, the co-ordinates of a given point in two different systems will, generally, be different. This is true in the case of the systems assumed in the Galilean (unless t=0).

Anyway, you're right to think about the domains of the co-ordinate systems because that's what the Galilean is interested in. It looks at a set of co-ordinates in the domain of one system and returns the set of co-ordinates in the domain of another co-ordinate system which refer to the same point.

steve waterman wrote:added =
1 so, in the manifold, Point P maps from S as (x,0,0) and point P also maps from S' as (x-vt,0,0)?

2 so, in the once coincident coordinate systems, now separated by vt along the common x/x' direction,
the first coordinate in the domain of S = the first coordinate in the domain of S' regardless of the manner of their placement on top of the manifold?


1. Yes. Although this is phrased a little oddly but I think you have the right idea. You've dropped your fourth co-ordinate (the t co-ordinate, you have to keep this, it is very important), but otherwise that is correct. The Galilean tells us that S(x,y,z,t)=S'(x-vt,y,z,t)=S'(x',y',z',t) for all values of x, y, z and t (provided those values are the same on both sides of the equation). v is some constant which tells us in some sense how slanted the t-axis of one co-ordinate system looks in the other one.

The galilean assumes a few things about our co-ordinate systems (at least, when it's written in this form). These are that they have three mutually perpendicular axes (the x, y and z axes) and one other axis. We assume that both systems are aligned with their axes x, y and z axes parallel to each other. We assume that the t axis is only ever slanted towards or away from the x axis and never towards or away from the y or z axis. We also assume that any point has the same t-co-ordinate in both systems and that both systems map the co-ordinates (0,0,0,0) to the same point. If you want to break any of these assumptions, you end up with slightly different forms of the Galilean (or possibly the Lorentz transformation of special relativity if you break the right one in the right way) that all express the same basic idea.

2. No. The co-ordinate systems are not really moving so they cannot "now" be separated. When we say the co-ordinate systems were co-incident at t=0 we mean that they label the same point as (0,0,0,0) i.e. S(0,0,0,0)=S'(0,0,0,0). Given the types of co-ordinate systems the Galilean applies to, we find that S(0,0,0,t) is not the same point as S'(0,0,0,t) any more than S(2,17,8,1) has to be the same point as S'(2,17,8,1).

You seem to be confusing the common interpretation of the Galilean frequently employed in physics (where it's used to transform between reference frames) with the mathematical concept where time is just another co-ordinate. This a very important and deep concept and you really do need to devote some time to reading Schrollini's posts about spacetime and adding a time co-ordinate.
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Re: Galilean:x' with respect to S'?

Postby Schrollini » Wed Jul 17, 2013 11:32 pm UTC

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:
Schrollini wrote: 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:If we ignore the manifold for a second, and only consider the co-ordinate system's domain,
does the the first coordinate of one system equal the first coordinate of the other system?

You can't ignore the manifold. That's the only thing that relates the two coordinate systems. The only meaningful question you can ask about a set of coordinates in one system and a set in the other system is, do these coordinates map to the same point or not? I say as much in this very same post; maybe you should go back and reread it now that you've got a better grip on what a coordinate system is.

I'm not going to discuss the Galilean transformation now; we need to get the basic concept shored up before we start discussing an example fraught with additional baggage.
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Re: Galilean:x' with respect to S'?

Postby yurell » Wed Jul 17, 2013 11:38 pm UTC

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, a Galilean transform is a transformation between two co-ordinate systems. Remember a co-ordinate system maps to a point; it's the map that's being transformed, not the point.

steve waterman wrote:If we ignore the manifold for a second, and only consider the co-ordinate system's domain,
does the the first coordinate of one system equal the first coordinate of the other system?

If you cannot parse that question, please let me know.


The problem is that the question doesn't make sense. If what you're asking is 'for the Galilean transform, does the origin of the (s,u) co-ordinate system coincide with the origin of the (t,x) co-ordinate system (that is, does (s=0,u=0) map to the same point as (t=0, x=0))', then the answer is 'yes', although this question doesn't make sense without a manifold (I italicised the bit that shows why).

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