## Search found 1602 matches

- Sat Jul 20, 2013 4:05 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Steve , I really wish you'd remember this surprise. Your intuition says that coordinate systems start aligned, and then get moved to separate positions. But that's not how they work with the conventional definitions. Instead, they come into being separately, and then we have to figure out how to co...

- Fri Jul 19, 2013 5:52 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### x' = x-d?

Does x' = x -d, when restricted to a mathematical Cartesian setting? Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d = 0. *d = distance of separation between S and S' along the common x/x' axis Allow one of the two systems to be repositioned...

- Fri Jul 19, 2013 5:14 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

If b = 0, then you ain't got no "right" triangle, more like what is called a line segment. A tad off topic. If x = x', you ain't got no non-trivial transformation, more like what is called the identity. A tad off topic for a thread about the Galilean transformation. You are right. Perhaps...

- Fri Jul 19, 2013 4:51 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

So, the proof presented as ..."the short form" Given x = x', therefore x' ≠ x-d. Allowing x' = x - d, spawns equations like 2 = -1...x' = x-d is not about points nor about a manifold. Allow me to apply your *exact* logic to a completely different equation. Even you'd think I look like a c...

- Fri Jul 19, 2013 4:37 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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

NOOOOO!

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

Two inches on one ruler = two inches on the other ruler no matter where the rulers are,

be they coincident or not.

- Fri Jul 19, 2013 4:19 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Of course x = x' when the systems are coincident. And to try to prevent misunderstandings: Coincident systems implies x = x'. x = x' for some points does not imply that the systems are coincident. However, We are NOT trying to determine if systems are coincident or not...we know that we have two ca...

- Fri Jul 19, 2013 3:46 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Allow me to cut out a bunch of junk that you aren't actually using in this proof: Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x' , y = y', z = z', d* = 0 let d* = distance of separation between S and S' along the common x/x' axis 1 x = x'-d, if mathematically...

- Fri Jul 19, 2013 3:30 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Allow me to cut out a bunch of junk that you aren't acutally using in this proof: Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d* = 0 let d* = distance of separation between S and S' along the com...

- Fri Jul 19, 2013 3:08 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d* = 0 let d* = distance of separation between S and S' along the common x/x' axis 1 x = x'-d , if mathematically allowed, would equate unequal abscissa ( lengths ). 2 whereas, S(x,y,z) = S'(x',y...

- Fri Jul 19, 2013 2:54 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z', d* = 0 let d* = distance of separation between S and S' along the common x/x' axis 1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ). like 2 = -1. 2 whereas, S(x,y,z...

- Fri Jul 19, 2013 2:37 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

In my proof, there is no t, there is no Galilean. There is only what is shown in the proof. Please discuss the proof and not anything Galilean transformation. There is no Physics anything here. There is math and only math. Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), whe...

- Fri Jul 19, 2013 2:20 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

Given coincident Cartesian coordinates systems S(x,y,z) and S'(x',y',z'), where x = x', y = y', z = z' let d = distance along the common x/x' axis 1 x = x'-d, if mathematically allowed, would equate unequal abscissa ( lengths ). No it doesn't. It equates two real numbers. so you are okay with 2 = -1...

- Fri Jul 19, 2013 1:58 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

4 S(x,y,z) = S'(x',y',z') also equates abscissa ( lengths ). What are S and S'? Coordinate systems? You've already used f and g for your coordinate systems. I was trying to merely copy your equation regarding points in the manifold. I was using your function notation, but it is too much a hassle an...

- Fri Jul 19, 2013 1:13 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

second draft ( after reading yesterday's comments ) x = x'-vt, let vt = d The validity of the math of the following equation is challenged; x = x'-d. g (x,y,z) = f ( x-d,y,z), noting that x and x' are not points. 1 x = x'-d equates abscissa ( lengths ). 2 S(x,y,z) = S'(x',y',z') also equates abscis...

- Fri Jul 19, 2013 12:20 am UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

let vt = d thus, removing time from my ALL mathematical proof... 1 - The Galilean equation is x = x'-d. 2 - It is true that g (x,y,z) = f( x-d,y,z), noting that x and x' are not, by themselves, points. 3 - Mathematically speaking, that means the Galilean equation must be referring to their abscissa ...

- Fri Jul 19, 2013 12:05 am UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

JudeMorrigan hopes I am not headed here... 1 - The "Gallilean given" is x = x'-vt 2 - It may be true S(x,t) = S'(x-vt,t), but x and x' are not, by by themselves, points 3 - Mathematically speaking, that means the Gallilean given must be referring to their "abscissa lengths" 4 - T...

- Thu Jul 18, 2013 11:31 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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

- Thu Jul 18, 2013 9:48 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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

- Thu Jul 18, 2013 9:08 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### 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 ...

- Thu Jul 18, 2013 8:45 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### 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

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

- Thu Jul 18, 2013 8:33 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### 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. And for your recent objection, a point need not have coordinates in a given coordinate system. Indeed, as I post...

- Thu Jul 18, 2013 8:10 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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 ...

- Thu Jul 18, 2013 7:53 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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 po...

- Thu Jul 18, 2013 7:33 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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...

- Thu Jul 18, 2013 6:57 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### 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?

- Thu Jul 18, 2013 4:38 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### 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! Thank you for saying so! It's nice to know these masses of text I'v...

- Thu Jul 18, 2013 4:16 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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, whic...

- Thu Jul 18, 2013 3:57 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

### 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?

P = g(first coordinate, fourth coordinate) mapped from S'

= f(first coordinate -vt, fourth coordinate) mapped from S?

- Thu Jul 18, 2013 3:13 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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 n...

- Thu Jul 18, 2013 2:51 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
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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.

- Thu Jul 18, 2013 2:25 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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) added - I s...

- Thu Jul 18, 2013 2:14 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

yurell gave you the answer two posts before yours: 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 (...

- Thu Jul 18, 2013 1:24 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
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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.

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

This is all I meant to ask.

- Thu Jul 18, 2013 12:15 am UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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 tha...

- Wed Jul 17, 2013 11:55 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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 coord...

- Wed Jul 17, 2013 10:52 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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...

- Wed Jul 17, 2013 9:07 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
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### Re: Galilean:x' with respect to S'?

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 ...

- Wed Jul 17, 2013 8:47 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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

- Wed Jul 17, 2013 8:11 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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'r...

- Wed Jul 17, 2013 8:05 pm UTC
- Forum: General
- Topic: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x-d?
- Replies:
**908** - Views:
**172532**

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

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?

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?