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'? AND SPECIAL BONUS x' = x

Postby steve waterman » Wed Jul 24, 2013 2:33 pm UTC

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

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

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


NOT TRUE! if S'(x-d,0,0) = S(x,0,0) does not mathematically extrapolate to x' = x-d.
This is apparently what others may believe as well. It would seem obvious that it should, but it does not.
x is not manifold material, only S(x,0,0) as point P is, and the equation x' = x-d has no mention nor connections to a point.

Someone needs to make this distinction between x the coordinate and (x,0,0) the point. Thanks for dropping the vt requirement. Now a further request, please...to drop anything/logic re manifold, mapping, point or in the point format (a,b,c). The discussion/equation/thread evolution is now about x the coordinate. P had left the building.

added -
So. No point P, an ignored manifold and given two coincident Cartesian systems S(x,y,z) and S'(x',y',z') with x = x', y = y', z= z'. Then, we reposition so that the two origins are separated by d along the common x/x' axis.
x = x'.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Wed Jul 24, 2013 2:43 pm UTC

steve waterman wrote:NOT TRUE! if S'(x-d,0,0) = S(x,0,0) does not mathematically extrapolate to x' = x-d.

Steve, it's simple algebra.

S'(x',0,0) = S(x,0,0)
x' = x - d
S'(x-d,0,0) = S(x,0,0)

Someone needs to make this distinction between x the coordinate and (x,0,0) the point. Thanks for dropping the vt requirement. Now a further request, please...to drop anything/logic re manifold, mapping, point or in the point format (a,b,c). The discussion/equation/thread evolution is now about x the coordinate. P had left the building.

We can't do that and have any of this have any meaning at all. If your goal is still to "disprove" the Galilean, you can't do it by attacking a strawman.

Allow me to recommend you go back and reread Schrollini's posts. You don't seem to have really groked what they were saying.

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

Postby steve waterman » Wed Jul 24, 2013 3:01 pm UTC

JudeMorrigan wrote:
steve waterman wrote:NOT TRUE! if S'(x-d,0,0) = S(x,0,0) does not mathematically extrapolate to x' = x-d.

Steve, it's simple algebra.

S'(x',0,0) = S(x,0,0)
x' = x - d
S'(x-d,0,0) = S(x,0,0)

Someone needs to make this distinction between x the coordinate and (x,0,0) the point. Thanks for dropping the vt requirement. Now a further request, please...to drop anything/logic re manifold, mapping, point or in the point format (a,b,c). The discussion/equation/thread evolution is now about x the coordinate. P had left the building.

We can't do that and have any of this have any meaning at all. If your goal is still to "disprove" the Galilean, you can't do it by attacking a strawman.

Allow me to recommend you go back and reread Schrollini's posts. You don't seem to have really groked what they were saying.


Allow me to not discuss points or manifolds any more. The thread is now about the coordinate x and the coordinate x', and what that means.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Wed Jul 24, 2013 3:05 pm UTC

steve waterman wrote:Allow me to not discuss points or manifolds any more.

Hey, you're allowed to discuss whatever you'd like. Just don't pretend that what you're talking about has a durned thing to do with coordinate transformations in general or the Galilean transformation specifically if you're taking points and manifolds out of the discussion.

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

Postby Schrollini » Wed Jul 24, 2013 3:18 pm UTC

JudeMorrigan wrote:
steve waterman wrote:NOT TRUE! if S'(x-d,0,0) = S(x,0,0) does not mathematically extrapolate to x' = x-d.

Steve, it's simple algebra.

S'(x',0,0) = S(x,0,0)
x' = x - d
S'(x-d,0,0) = S(x,0,0)

Or, reverse the order:
(1) Given S'(x-d,0,0) = S(x,0,0),
(2) Define x' such that S'(x',0,0) = S(x,0,0).
(3) Then x' = x-d, since both S and S' are S' is one-to-one.

Edit: only the one-to-one nature of S' is necessary for this proof.
Last edited by Schrollini on Wed Jul 24, 2013 3:27 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Wed Jul 24, 2013 3:19 pm UTC

JudeMorrigan wrote:
steve waterman wrote:Allow me to not discuss points or manifolds any more.

Hey, you're allowed to discuss whatever you'd like. Just don't pretend that what you're talking about has a durned thing to do with coordinate transformations in general or the Galilean transformation specifically if you're taking points and manifolds out of the discussion.


Yes, I am only talking about x' = x-d in a 100 percent all mathematical Cartesian setting.
So, no Galilean, no Physics, no time, no motion, no events, no objects, no observers, no light, no points, no manifold, no transformations.

x' = x-d?

case 1
given S and S' coincident with x = x'
case 2
the above coincident S and S' subsequently repositioned by d along the common x and x' axis...
x = x'.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Wed Jul 24, 2013 3:25 pm UTC

If you have no manifold, and hence no points, you also don't have coordinates or coordinate systems or coordinate transformations, and x' = x - d is just an equation containing three variables, none of which are defined.

Edit: Just noting, S and S' are also now undefined.

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

Postby JudeMorrigan » Wed Jul 24, 2013 3:36 pm UTC

steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:Allow me to not discuss points or manifolds any more.

Hey, you're allowed to discuss whatever you'd like. Just don't pretend that what you're talking about has a durned thing to do with coordinate transformations in general or the Galilean transformation specifically if you're taking points and manifolds out of the discussion.


Yes, I am only talking about x' = x-d in a 100 percent all mathematical Cartesian setting.
So, no Galilean, no Physics, no time, no motion, no events, no objects, no observers, no light, no points, no manifold, no transformations.

Seriously, Steve. Points, manifolds and transformations are all very much mathematical. The others arguably can be as well, but there is *nothing* inherently "physics" about coordinate geometry.

Also, what beojan said.

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

Postby steve waterman » Wed Jul 24, 2013 3:39 pm UTC

beojan wrote:If you have no manifold, and hence no points, you also don't have coordinates or coordinate systems or coordinate transformations, and x' = x - d is just an equation containing three variables, none of which are defined.

Edit: Just noting, S and S' are also now undefined.


Yes, thanks for reminding me. I just got a tad lazy.

case 1
Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',
case 2
the above coincident S and S' subsequently repositioned and now, are no longer coincident, x = x'.


These statements are better as I have dropped my d term completely.
Last edited by steve waterman on Wed Jul 24, 2013 4:02 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Wed Jul 24, 2013 3:54 pm UTC

steve waterman wrote:the above coincident S and S' subsequently repositioned and now, are no longer coincident

Coordinate
systems
do
not
move!


If you're talking about repositioning a coordinate system, you're doing it wrong.

This has been explained and re-explained many times, so I won't bother to repeat it again.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Wed Jul 24, 2013 4:02 pm UTC

steve waterman wrote:
beojan wrote:If you have no manifold, and hence no points, you also don't have coordinates or coordinate systems or coordinate transformations, and x' = x - d is just an equation containing three variables, none of which are defined.

Edit: Just noting, S and S' are also now undefined.


Yes, thanks for reminding me. I just got a tad lazy.

case 1
Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',
case 2
the above coincident S and S' subsequently repositioned and now, are no longer coincident
x = x'.

You're missing the point. (So to speak.) Without a manifold, you can't have your Cartesian coordinate systems at all. If the idea of a manifold bothers you since you feel it's too modern, note that a euclidean plane is a specific type of manifold, and we can work on one of those.

Also, it seems ironic that you insist that you want to do pure mathemetics and no physics and then talk about repositioning coordinate systems. It's an intuitive way to think about it due to obvious real-world analogies, but it's not really what's happening mathematically. See the earlier part of this discussion where I fell into that trap myself.

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

Postby steve waterman » Wed Jul 24, 2013 4:07 pm UTC

Schrollini wrote:
steve waterman wrote:the above coincident S and S' subsequently repositioned and now, are no longer coincident

Coordinate
systems
do
not
move!


If you're talking about repositioning a coordinate system, you're doing it wrong.

This has been explained and re-explained many times, so I won't bother to repeat it again.


So what does vt act upon? ( if not one of the coincident systems along the direction of the common x/x' axis? )

added -
http://en.wikipedia.org/wiki/Galilean_transformation
Please view the depiction...of course, coordinate systems can move!
Last edited by steve waterman on Wed Jul 24, 2013 4:17 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Wed Jul 24, 2013 4:16 pm UTC

steve waterman wrote:Someone needs to make this distinction between x the coordinate and (x,0,0) the point.

(x,0,0) is not a point. (Why? What is missing?)

Why not drop the other two dimensions (with which we aren't doing anything right now) and consider one dimensional space?

Let L be a function that maps coordinates to points in the manifold.
x is a variable which represents a real number.
(x) is the set of coordinates (containing one member, since it's 1D) in an unspecified system (and is thus not useful on its own)
L(x) is a point in the manifold.
L(0) is a point in the manifold we call the origin of system L.
L(4) is a point in the manifold I'm going to call Fred.
L(x) is Fred, when x=4.
L(x) is NOT Fred when x=3.
L(x-1) is Fred when x=5.


Let K be another function that maps coordinates to points in the manifold.
--> We choose K in such a manner that K(0) is Fred, and the scale and orientation of the axis matches L.
--> This is like saying K is "offset from L" by 4 units.
x is still a variable which represents a real number
(x) is the set of coordinates (containing one member, since it's 1D) in an unspecified system (and is thus not useful on its own)
K(x) is a point in the manifold.
K(0) is a point in the manifold we call the origin in system K.
K(0) is Fred
K(4) is not Fred
K(x) is Fred when x=0
K(x) is not Fred when x = 4.
K(x+12) is Fred when x=-12


K(0) is not L(0)
K(0) is L(4). We call that point Fred. We also call it the origin in system K (but not the origin in system L)

K(x) is not L(x) (presuming the numerical value of x is the same in each expression, which is what makes the notation useful in the first place)

Now, lets introduce u.
u is a variable which represents a real number.
u can represent a different real number from x.


L(u) is Fred ONLY WHEN u=4.
K(u) is Fred ONLY WHEN u=0.
L(x) is Fred ONLY WHEN x=4.

K(u) = L(x) means that K(u) is the same point in the manifold as L(x).
--> This puts a constraint on the relationship between u and x.
K(u) = L(x) ONLY WHEN u = x-4.
--> Note specifically that K(u) does not equal L(x) when u=x.
K(u) = L(x) = Fred ONLY WHEN u = x-4 AND x=4. This forces u to be 0, so that x-4 would equal zero.

Remember that:
x and u are JUST NUMBERS.
(x) and (u) have NO MEANING by themselves.
K(x) and K(u) and L(x) and L(u) are POINTS IN THE MANIFOLD. They may well be different points, depending on the actual values of the NUMBERS x and u.
An algebraic equation like u=x-4 puts a constraint on the relationship.

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

Postby JudeMorrigan » Wed Jul 24, 2013 4:25 pm UTC

steve waterman wrote:So what does vt act upon? ( if not one of the coincident systems along the direction of the common x/x' axis? )

added -
http://en.wikipedia.org/wiki/Galilean_transformation
Please view the depiction...of course, coordinate systems can move!

I don't think it would be fair to say that vt "acts upon" anything. If one insists on keeping space and time separate, I think you'd have to say that S' at t=3 and S' at t=5 are simply different coordinate systems. If you're willing and able to think in four-space, time is simply an additional dimension.

As for the wiki diagram, remember when Schrollini mentioned that people can sometimes be sloppy in how they describe things?

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

Postby steve waterman » Wed Jul 24, 2013 4:38 pm UTC

Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',

regardless of either system rotating or if any system is in motion or not,

it is still true that x' = x
.

added - Which as the first coordinate means that S(0,0,0) to S(x,0,0) = S'(0,0,0) to S'(x',0,0).

Jude wrote:As for the wiki diagram, remember when Schrollini mentioned that people can sometimes be sloppy in how they describe things?

One system moves, at least...look anyplace on the net. You will see they all say the same. Who else thinks that coordinate system do not move?? This is a very odd and silly stance/logic/understanding to have, imo.
Last edited by steve waterman on Wed Jul 24, 2013 4:54 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Wed Jul 24, 2013 4:46 pm UTC

steve waterman wrote:Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',

regardless of either system rotating or if any system is in motion or not,

it is still true that x' = x
.

Well, yes. That is the defintion of coincident systems. It will not, of course, be true for systems that are not coincident.

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

Postby Dark Avorian » Wed Jul 24, 2013 4:54 pm UTC

STEVE!


Image

Look at this image. Let us say that S is the x,y system, S' the u,s system (We could freely relabel x' =u, y' = s)

x,y are the coordinates of P in the first system. u,s are the coordinates in the other

Do you understand?


S(x,y) = P = S'(x- y/2 , y)
S(x,y) = P = S'(u,s)

Do you agree?

u = x -y/2

Do you agree?

Notice that this is not a simple offset, the offset DEPENDS ON Y.

But the coordinate systems are fixed.

Do you understand?

The coordinate systems in the Galilean are 4-dimensional, notated S(x,y,z,t) and S'(x',y',z',t') and the offset in x depends on the value of t. But it is still two FIXED coordinate systems on space time. The confusion is that the restriction of this coordinate system to space at any particular time sort of appears to be moving( relative to the other system, in a sense)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PM 2Ring » Wed Jul 24, 2013 5:13 pm UTC

Dark Avorian,
There appears to be some confusion re: s & v in your last post.

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

Postby WibblyWobbly » Wed Jul 24, 2013 5:22 pm UTC

I posted this once before and then deleted it because I thought it wasn't a very good exercise, but all the mathematical rigor in the world doesn't seem to convince Steve that a manifold isn't something physicists invented to pull the wool over his eyes. So here goes; I'll be using some of Steve's terminology; apologies in advance.

steve waterman wrote:Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',

regardless of either system rotating or if any system is in motion or not,

it is still true that x' = x
.


OK, Steve, now take a slightly different example:

Given: two Cartesian coordinate systems, S(x,y,z) and S'(x',y',z'), which have the same scale, orientation, etc. ... S and S' are identical, but they are not coincident; their origins are separated by a distance d along a shared axis (make it the x-axis/x'-axis).

Question: Does x' = x?

Now, reposition S' so that it shares the same origin as S. That implies they now have the same scale, orientation, and position.

Question: Are these coordinate systems now coincident? If not, why not? If so, is x' = x?

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

Postby steve waterman » Wed Jul 24, 2013 5:38 pm UTC

WibblyWobbly wrote:I posted this once before and then deleted it because I thought it wasn't a very good exercise, but all the mathematical rigor in the world doesn't seem to convince Steve that a manifold isn't something physicists invented to pull the wool over his eyes. So here goes; I'll be using some of Steve's terminology; apologies in advance.

Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',
regardless of either system rotating or if any system is in motion or not,
it is still true that x' = x
.
WibblyWobbly wrote:OK, Steve, now take a slightly different example:

Given: two Cartesian coordinate systems, S(x,y,z) and S'(x',y',z'), which have the same scale, orientation, etc. ... S and S' are identical, but they are not coincident; their origins are separated by a distance d along a shared axis (make it the x-axis/x'-axis).
Question: Does x' = x?

Yes. By your wording, I believe we can conclude that x would equal x' under the conditions stated.
WibblyWobbly wrote:Now, reposition S' so that it shares the same origin as S. That implies they now have the same scale, orientation, and position.

Yes., under those conditions.
WibblyWobbly wrote:Question: Are these coordinate systems now coincident?

Yes.
WibblyWobbly wrote:If so, is x' = x?

Of course. Once x = x' is GIVEN someplace, it stays x = x' no matter what moves where or when or how.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby eSOANEM » Wed Jul 24, 2013 5:44 pm UTC

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

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

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


NOT TRUE! if S'(x-d,0,0) = S(x,0,0) does not mathematically extrapolate to x' = x-d.
This is apparently what others may believe as well. It would seem obvious that it should, but it does not.
x is not manifold material, only S(x,0,0) as point P is, and the equation x' = x-d has no mention nor connections to a point.


The ability to relabel expressions however we like without affecting what's going on is a very important principle in all of mathematics and that's all that's going on here. All we're saying is "let's call the first co-ordinate in the system, S', "x' "". This is completely legitimate and has no effect on anything.

I could just as easily call it "u" or "λ" or "Ϙ" or "@" or "€" or indeed any squiggle which can be completely arbitrary.

steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:Someone needs to make this distinction between x the coordinate and (x,0,0) the point. Thanks for dropping the vt requirement. Now a further request, please...to drop anything/logic re manifold, mapping, point or in the point format (a,b,c). The discussion/equation/thread evolution is now about x the coordinate. P had left the building.

We can't do that and have any of this have any meaning at all. If your goal is still to "disprove" the Galilean, you can't do it by attacking a strawman.

Allow me to recommend you go back and reread Schrollini's posts. You don't seem to have really groked what they were saying.


Allow me to not discuss points or manifolds any more. The thread is now about the coordinate x and the coordinate x', and what that means.


If we do that, the co-ordinates have no meaning whatsoever and you're trying to disprove an arbitrary equation between real numbers (exactly as if you were trying to disprove f(x)=x-d ).

steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:Allow me to not discuss points or manifolds any more.

Hey, you're allowed to discuss whatever you'd like. Just don't pretend that what you're talking about has a durned thing to do with coordinate transformations in general or the Galilean transformation specifically if you're taking points and manifolds out of the discussion.


Yes, I am only talking about x' = x-d in a 100 percent all mathematical Cartesian setting.
So, no Galilean, no Physics, no time, no motion, no events, no objects, no observers, no light, no points, no manifold, no transformations.


Cartesian co-ordinates (what you mean by a cartesian setting) imply the existence of a manifold (certainly if two systems can have any relation whatsoever between them).

Furthermore, a relationship between co-ordinates (what you're talking about) is by definition a transformation.

You can't remove these things if you want your discussion to make any sense whatsoever because they are in a very real sense the basis of the terms you're using.

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:the above coincident S and S' subsequently repositioned and now, are no longer coincident

Coordinate
systems
do
not
move!


If you're talking about repositioning a coordinate system, you're doing it wrong.

This has been explained and re-explained many times, so I won't bother to repeat it again.


So what does vt act upon? ( if not one of the coincident systems along the direction of the common x/x' axis? )

added -
http://en.wikipedia.org/wiki/Galilean_transformation
Please view the depiction...of course, coordinate systems can move!


t is just another co-ordinate. Neither system moves unless you look at a constant-time subspace (which is the physics). You only see co-ordinate systems moving when you introduce a physical interpretation.

You, Steve, are the only one doing any physics here (and you're doing it wrong whilst insisting it's maths as well).
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Wed Jul 24, 2013 5:54 pm UTC

steve waterman wrote:
WibblyWobbly wrote:I posted this once before and then deleted it because I thought it wasn't a very good exercise, but all the mathematical rigor in the world doesn't seem to convince Steve that a manifold isn't something physicists invented to pull the wool over his eyes. So here goes; I'll be using some of Steve's terminology; apologies in advance.

Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',
regardless of either system rotating or if any system is in motion or not,
it is still true that x' = x
.
WibblyWobbly wrote:OK, Steve, now take a slightly different example:

Given: two Cartesian coordinate systems, S(x,y,z) and S'(x',y',z'), which have the same scale, orientation, etc. ... S and S' are identical, but they are not coincident; their origins are separated by a distance d along a shared axis (make it the x-axis/x'-axis).
Question: Does x' = x?

Yes. By your wording, I believe we can conclude that x would equal x' under the conditions stated.


Then if x' = x when the systems are explicitly given as coincident, and x' = x when they're explicitly given as not coincident, in your statement

steve waterman wrote:Given two coincident Cartesian coordinate systems, S(x,y,z) and S'(x',y',z') with x = x' and y = y' and z = z',
regardless of either system rotating or if any system is in motion or not,
it is still true that x' = x
.

coincidence actually makes no difference, does it? Whether you're given coincident S and S' or non-coincident S and S', x' = x anyway? If we start with these systems:
firechicago wrote:Image
as "GIVEN", you're saying 2 = 0.5?

Basically, you're saying that x' = x no matter what the relationship between the two states. So if coincidence doesn't matter, what is the point of your "given", with coincident S and S' and x = x', etc.? If you define all systems as having x' = x, what's actually special about this? And if your "GIVEN" is just a definition you made up (because it's not something actual mathematicians use), why should your conclusion be any less made-up?

steve waterman wrote:
WibblyWobbly wrote:Now, reposition S' so that it shares the same origin as S. That implies they now have the same scale, orientation, and position.

Yes., under those conditions.
WibblyWobbly wrote:Question: Are these coordinate systems now coincident?

Yes.
WibblyWobbly wrote:If so, is x' = x?

Of course. Once x = x' is GIVEN someplace, it stays x = x' no matter what moves where or when or how.

Emphasis mine. So, what you're saying is, even if we have one initial set of 'givens' at the beginning of a proof, we can change the 'givens' whenever we want? Because here I thought you were arguing that once an initial condition is "given", it cannot be changed. So does changing the given retroactively change everything else about the proof?

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

Postby beojan » Wed Jul 24, 2013 6:09 pm UTC

steve waterman wrote:
Schrollini wrote:
steve waterman wrote:the above coincident S and S' subsequently repositioned and now, are no longer coincident

Coordinate
systems
do
not
move!


If you're talking about repositioning a coordinate system, you're doing it wrong.

This has been explained and re-explained many times, so I won't bother to repeat it again.


So what does vt act upon? ( if not one of the coincident systems along the direction of the common x/x' axis? )

added -
http://en.wikipedia.org/wiki/Galilean_transformation
Please view the depiction...of course, coordinate systems can move!

To put it simply, the depiction in that article is (a 2 dimensional projection of) a 3 dimensional slice (at t'=t=0) of a 4 dimensional diagram.

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

Postby Dark Avorian » Wed Jul 24, 2013 7:50 pm UTC

Honestly, there are so many misconceptions floating around that I'm not sure we'll ever get to the bottom of this (one thought, if someone is willing to animate a video with a voiceover, lucid commentary, and some nice animations, that might help).


That said. I'll bite.


Steve,

You are doing yourself a disservice by only thinking about the Galilean transform (I know you say this is pure math, but it's very obvious that your misconceptions are rooted in your conception of the Galilean). Coordinate systems that deal with spacetime, like the Lorentz transform and the Galilean transform (and yes, these are physics, but they are also purely mathematical objects) assign coordinates to a 4-dimensional space please take note of that.

Coordinate systems related by a Galilean will both associate the same time coordinate with a given point. A coordinate system needn't do this. The points we talk about in spacetime have 4 coordinates, the last one is time. This is of course, a purely mathematical four dimensional space, no time or space needed. Now, this coincidence wrt the 4th axis allows us to do something unfortunate, we can trick ourselves into thinking we're talking about two relatively moving coordinate systems, because we can drop the time factor, and then consider the transform between (x,y,z) and (x'y',z') [where S(x,y,z,t) = S'(x',y',z',t')]. This transform varies with time.

But here's the thing, we actually are talking about two coordinate systems, already fixed, just with skewed axes.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Wed Jul 24, 2013 8:43 pm UTC

WibblyWobbly wrote:Given: two Cartesian coordinate systems, S(x,y,z) and S'(x',y',z'), which have the same scale, orientation, etc. ... S and S' are identical, but they are not coincident; their origins are separated by a distance d along a shared axis (make it the x-axis/x'-axis).

Question: Does x' = x?


x and x' are both variables. It is meaningless to speak of equality between them in isolation. It is only when we say something like
In the case where S(x,y,z,) and S'(x',y',z') refer to the same point...
does x' = x ?


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

Postby beojan » Wed Jul 24, 2013 9:14 pm UTC

Dark Avorian wrote:Honestly, there are so many misconceptions floating around that I'm not sure we'll ever get to the bottom of this (one thought, if someone is willing to animate a video with a voiceover, lucid commentary, and some nice animations, that might help).

I considered doing this, partly as a way to learn blender, but decided it would be a lot of work, and might not come out that well anyway (because I would be learning blender on the way).

Hence, I first posted the trees in a field scenario, and then offered to post a demonstration with toy cars and clear acrylic sheets that Steve could carry out, but he ignored / refused both.

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

Postby WibblyWobbly » Wed Jul 24, 2013 9:24 pm UTC

ucim wrote:
WibblyWobbly wrote:Given: two Cartesian coordinate systems, S(x,y,z) and S'(x',y',z'), which have the same scale, orientation, etc. ... S and S' are identical, but they are not coincident; their origins are separated by a distance d along a shared axis (make it the x-axis/x'-axis).

Question: Does x' = x?


x and x' are both variables. It is meaningless to speak of equality between them in isolation. It is only when we say something like
In the case where S(x,y,z,) and S'(x',y',z') refer to the same point...
does x' = x ?


Jose


Yeah, I realize that. See the disclaimer:

WibblyWobbly wrote:... all the mathematical rigor in the world doesn't seem to convince Steve that a manifold isn't something physicists invented to pull the wool over his eyes. So here goes; I'll be using some of Steve's terminology; apologies in advance.


Granted, my Stevean math is rusty, so I'm not sure it was a faithful translation.

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

Postby eSOANEM » Wed Jul 24, 2013 10:24 pm UTC

beojan wrote:
steve waterman wrote:http://en.wikipedia.org/wiki/Galilean_transformation
Please view the depiction...of course, coordinate systems can move!

To put it simply, the depiction in that article is (a 2 dimensional projection of) a 3 dimensional slice (at t'=t=0) of a 4 dimensional diagram.


Technically it's a section (a constant-t section) not a projection because events at different times don't appear on it.
Last edited by eSOANEM on Thu Jul 25, 2013 6:16 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Pfhorrest » Wed Jul 24, 2013 11:48 pm UTC

I'm keep getting reminded of an anecdote I'd like to share.

When I was in late elementary school, I was in an independent study program and working on math at a higher level than the kids my age who were in public schools.

When I mentioned to some of them at a park playground one day that I was learning algebra, they said something like "Oh yeah, well, if you're so smart, what's A plus B times C?"

I don't remember what retort I had for them, but the point of the anecdote, and this is for you here Steve: these kids thought that A and B and C somehow had intrinsic mathematical meaning in algebra, and that with no further context I could do some kind of arithmetic on them like they were used to doing on numbers. They didn't understand that the letters used in algebra were just placeholders for different numbers of unknown value, and the answer to their question depended entirely on the meaning of "A", "B", and "C", which had not been given.

Steve, in all these coordinate transformations, x and y and z and x' and y' and z' and u and v and w and d and vt and so on and so forth are all just placeholders for unknown numbers. They have no intrinsic meaning whatsoever. We could use x and x' as you do to mean the distances between the origins of two coordinate systems and points some unknown common distance away, e.g. we could use x to mean the distance between S(0) and S(c), and x' to mean the distance between S'(0) and S'(c), in which case, presuming the two coordinate systems have the same scale, then yes it would be trivially true that x = x' no matter what.

But the triviality of that result is exactly why nobody but you uses x and x' to mean that. And if we use those placeholders to mean different things, as we do, then the equations that use them have different truth values, because the equations, although they look the same, mean something completely different.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Maelstrom. » Thu Jul 25, 2013 4:57 am UTC

I dont know if it will help, but I made a small visualisation of some different coordinate systems mapping on to a manifold, and then graphed the result using a Cartesian coordinate system: http://timheap.github.io/coord-vis/

I've not done this kind of manifold/coordinate system math before, so bear that in mind, but the visualisation shows a few different coordinate systems:
  • A plain Cartesian coordinate system, S(x, y)
  • A polar coordinate system, T(θ, r)
  • Another Cartesian coordinate system, Y(x, y), which is offset, rotated, and scaled compared to S
  • A weird coordinate system, G(x, y, s), which skews the x axis based on s

The code is up on Github if anyone would like to take it and make modifications to suit their uses.

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

Postby beojan » Thu Jul 25, 2013 8:54 am UTC

eSOANEM wrote:
beojan wrote:
steve waterman wrote:
Schrollini wrote:http://en.wikipedia.org/wiki/Galilean_transformation
Please view the depiction...of course, coordinate systems can move!

To put it simply, the depiction in that article is (a 2 dimensional projection of) a 3 dimensional slice (at t'=t=0) of a 4 dimensional diagram.


Technically it's a section (a constant-t section) not a projection because events at different times don't appear on it.


To put it simply, the depiction in that article is [[a 2 dimensional projection of] a [3 dimensional slice (at t'=t=0)]] of a 4 dimensional diagram.


The nested square brackets should make it clearer. The 2 dimensional projection is of the constant-t section, not the 4 dimensional full diagram.

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

Postby Schrollini » Thu Jul 25, 2013 1:35 pm UTC

Maelstrom. wrote:I've not done this kind of manifold/coordinate system math before, so bear that in mind, but the visualisation shows a few different coordinate systems:
  • A plain Cartesian coordinate system, S(x, y)
  • A polar coordinate system, T(θ, r)
  • Another Cartesian coordinate system, Y(x, y), which is offset, rotated, and scaled compared to S
  • A weird coordinate system, G(x, y, s), which skews the x axis based on s

One thing to keep in mind is that a manifold has dimension. Therefore, all coordinate systems will have the same number of coordinates. Thus, the G you've defined isn't really a coordinate system; it's really a family of coordinate systems parametrized by s: Gs(x,y). This is a perfectly fine thing to have. It's one way to define the Galilean transformation, in fact. But it requires you to clearly recognize that the coordinates of a point depend on s, since each s is its own coordinate system.

Instead, the approach I've taken here is to insist on including time as a coordinate, making our manifold spacetime. You could do that with your visualization, giving S, T, and Y time coordinates but leaving them fixed relative to each other. But you also must note that this means points are events -- they have a time coordinate and therefore exist only for an instant in that direction. The difficulty of visualizing this is what caused me to restrict myself to 2D manifolds, even when considering spacetime.

I don't mean to bring you down with this critique -- it's a cool tool. Steve has shown a fondness for computer visualizations in the past, so there's a (small) chance that this will get through to him. I just wanted to get these points out right away.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Thu Jul 25, 2013 5:25 pm UTC

I find it quite a surprise, that only very very recently, did I find out that Schrollini was unaware that
the Galilean transformations equations given states
1 S and S' are coincident at t = 0
2 x = x' at t = 0
3 vt is applied to one of the coincident systems

Given two coincident Cartesian systems S(x,y,z) and S'(x',y',z') with x = x', y = y', z = z'
where x = S(0,0,0) to S(x,0,0) and where x' = S'(0,0,0) to S'(x',0,0)
regardless of any motion/movement to S(x,y,z) or S'(x',y',z'), x = x'.

When the Galilean allows x' = x-vt, that generates a mathematical inequality,
like 2 = -1.

It is quite true, that the manifold properly equates point P as S'(x,0,0) = point P = S(x-vt,0,0),
like S'(2,0,0) = point P = S(-1,0,0)

However, the Galilean equation IS x' = x-vt
The Galilean equation IS NOT S'(x,0,0) = point P = S(x-vt,0,0).

Given x = x', no matter what silly math one does, therefore x = x', and hence x' = x-vt is only true when vt = 0.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby vbkid » Thu Jul 25, 2013 5:35 pm UTC

Just waiting for someone else to explain this better thna I can...

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

Postby brenok » Thu Jul 25, 2013 6:03 pm UTC

steve waterman wrote:Given x = x', no matter what silly math one does, therefore x = x', and hence x' = x-vt is only true when vt = 0.


And do you see a problem with this?

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

Postby bouer » Thu Jul 25, 2013 6:09 pm UTC

vbkid wrote:Just waiting for someone else to explain this better thna[sic] I can...


It has been explained excellently hundreds of times already, over the course of several threads and far too many months.

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

Postby beojan » Thu Jul 25, 2013 6:12 pm UTC

The Galilean transformation states that
  • Given two coordinate systems (on the same manifold), S(x,y,z,t) and S'(x',y',z',t')
  • Also given that the x, y, and z axes are parallel to the x', y', and z' axes respectively
  • Also given that the t and t' axes are the same. That is, t is identically equal to t'. Henceforth I will only use t.
  • Also given that S(a,b,c,0) = S'(a,b,c,0) for all (real) a,b,c
  • And finally, given that S'(0,0,0,t) = S(vt,0,0,t).

Then, if and only if
  • x' = x - vt
  • y' = y
  • z' = z
  • t' = t

S'(x',y',z',t') = S(x,y,z,t) holds for all (real) x,y,z,t (hence also for all real x',y',z',t' since these are functions of x,y,z,t respectively)

As you can see, this is an utterly trivial substitution. It's about as trivial as x³ = 5³ implies x = 5.

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

Postby Schrollini » Thu Jul 25, 2013 6:15 pm UTC

steve waterman wrote:I find it quite a surprise, that only very very recently, did I find out that Schrollini was unaware that
the Galilean transformations equations given states
1 S and S' are coincident at t = 0
2 x = x' at t = 0
3 vt is applied to one of the coincident systems

If you had bothered to read what I've written, you'd know that the Galilean transformation is between reference frames; that is, coordinate systems in which time is a coordinate. You cannot say that the frames are coincident only at t=0 any more than you can say two systems related by a skew are coincident only at y=0. What you can say is that the spatial parts are coincident at t=0, just like you can say that the x-part of the skewed systems are coincident at y=0. That of course has no bearing on y≠0, and neither does your statement on t≠0.

If I believed that you want to understand this, I'd explain it again. But at this point, I get the feeling that you don't want to understand this; you're much happier with your conviction that you're a genius and the rest of us are blithering idiots for not recognizing this. If this makes you happy, fine. But that means that there's no point in me continuing to explain the same points again and again.

If at some point in the future you decide that you'd actually like to know what the rest of us mean when we say "x' = x - vt", let me know. But this time, you'll actually have to do the exercises. Until then, ciao.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Thu Jul 25, 2013 6:18 pm UTC

steve waterman wrote:I find it quite a surprise, that only very very recently, did I find out that Schrollini was unaware that
the Galilean transformations equations given states
1 S and S' are coincident at t = 0
2 x = x' at t = 0
3 vt is applied to one of the coincident systems

Given two coincident Cartesian systems S(x,y,z) and S'(x',y',z') with x = x', y = y', z = z'
where x = S(0,0,0) to S(x,0,0) and where x' = S'(0,0,0) to S'(x',0,0)
regardless of any motion/movement to S(x,y,z) or S'(x',y',z'), x = x'.

When the Galilean allows x' = x-vt, that generates a mathematical inequality,
like 2 = -1.

It is quite true, that the manifold properly equates point P as S'(x,0,0) = point P = S(x-vt,0,0),
like S'(2,0,0) = point P = S(-1,0,0)

However, the Galilean equation IS x' = x-vt
The Galilean equation IS NOT S'(x,0,0) = point P = S(x-vt,0,0).

Given x = x', no matter what silly math one does, therefore x = x', and hence x' = x-vt is only true when vt = 0.

bouer wrote:
vbkid wrote:Just waiting for someone else to explain this better thna[sic] I can...


It has been explained excellently hundreds of times already, over the course of several threads and far too many months.

So let's go around again ...

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

Postby steve waterman » Thu Jul 25, 2013 6:40 pm UTC

the Galilean transformations equations given states
S(x,y,z) = S'(x',y',z') where the spatial parts are coincident at t = 0, and where x = x', y = y', z = z'

Schrollini wrote:If at some point in the future you decide that you'd actually like to know what the rest of us mean when we say "x' = x - vt", let me know.


What does x' mean to the rest of you?

What does x mean to the rest of you?

What does given S(x,0,0) mean to the rest of you?

What does x = x' mean to the rest of you?
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