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

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

Postby beojan » Sun Jul 28, 2013 4:44 pm UTC

steve waterman wrote:equating x with x',
vt = 3
S(2,0) = P
S'(-1,0) = P
therefore,
2 = -1?

NB: I'm going to ignore the error in the number of dimensions in this post, and simply add an extra ",0" to all 2d coordinates in grey.

If x' = x - vt
and vt ≠ 0
then x' ≠ x.
The mistake is in assuming that S(x,y,z) = S'(x',y',z') implies x = x'.
steve waterman wrote:equating point P relative coordinate sets
vt = 3
S(2,0) = P
S'(-1,0) = P
therefore, in the manifold
S(2,0) = S'(-1,0) = P
However, since S(x,y,z) = S'(x',y',z') was given, it is also quite true that S(2,0) = S'(2,0).

No... How? How does S(x,y,z) = S'(x',y',z') mean S(2,0,0) = S'(2,0,0)?
steve waterman wrote:So, S(2,0) = S'(2,0) = S'(-1,0)?

edit - took out the quote, which I had misread. The above is stand-alone.

added -
ucim -
The Galilean has no double prime notation/logic, nor do I.
Hence, why I backed down on any possible M(x,y,z)...no such thing.

The Galilean has no inherent labels. The primes and the double primes and the 'x'es and the 'y's and the 't's and the 'z's have no inherent meaning beyond simply being placeholders.

When we say something like

"Given two systems S(x,y,z) and S'(x',y',z') on a manifold M"

It is the term "S(x,y,z)" that defines x, y, and z, and the term "S'(x',y',z')" that defines x', y', and z'.
It is the statement "Given two systems S(x,y,z) and S'(x',y',z') on a manifold M" as a whole that defines S, S', and M.
The assignment of labels can be done however we wish, although consistency is useful. The labels are not magical.
Statements like "x' = x - v*t" were not discovered written on a cave wall or handed down from above to be decoded. The meaning was discovered first, and then expressed algebraically.

Edit: If two systems are spacially coincident, they are by definition not separated, aren't they?

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

Postby ucim » Sun Jul 28, 2013 4:59 pm UTC

But you contradict your givens right off the bat.

Given spatially coincident cartesian systems S and S', it IS true that
S(x,y,z) = S'(x',y',z')

but the only way
P = S'(x',y',z') = S(x-vt,y,z) = P
is true is if
x'=x-vt
and the only way that is true is if vt=0.

If vt is not zero, then the systems are not coincident, and the premise
S(x,y,z) = S'(x',y',z')
no longer applies.

Case 1 works, because it does not contradict its own premise.

Case 2 does not work because it contradicts the premise: if
S and S' are separated by vt=3, then it is NOT TRUE that
S(x,y,z) = S'(x',y',z')

You are asking that systems be coincident and not coincident at the same time.

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

Postby beojan » Sun Jul 28, 2013 5:18 pm UTC

ucim wrote:But you contradict your givens right off the bat.

Given spatially coincident cartesian systems S and S', it IS true that
S(x,y,z) = S'(x',y',z')

Can't that be true with non coincident S and S'? Spatially coincident S and S' means
S(x,y,z) = S'(x,y,z)
ucim wrote:but the only way
P = S'(x',y',z') = S(x-vt,y,z) = P
is true is if
x'=x-vt
and the only way that is true is if vt=0.

Again, I think you meant
P = S'(x',y',z') = S'(x-vt,y,z) = P

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

Postby ucim » Sun Jul 28, 2013 5:53 pm UTC

beojan wrote:Can't that be true with non coincident S and S'? Spatially coincident S and S' means
S(x,y,z) = S'(x,y,z)
Yes, you are correct, but Steve has another given, that x=x', y=y', and z=z', and I'm trying to stick with his notation so much as possible, to avoid introducing another issue. But maybe that's not the best way.

edit to add:
steve waterman wrote:added -
ucim -
The Galilean has no double prime notation/logic, nor do I.

The Galilean is an idea. The notation is a way to communicate the idea.

The logic in using double prime notation is to communicate the idea that when you "move" a coordinate system, it no longer is the same coordinate system, and therefore, the prior givens do not apply. You keep trying to apply the prior givens to the new situation. It doesn't lead to sense.

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

Postby Schrollini » Sun Jul 28, 2013 9:13 pm UTC

steve waterman wrote:The Galilean has no double prime notation/logic, nor do I.
Hence, why I backed down on any possible M(x,y,z)...no such thing.


Steve, when I was three, I wanted to write music.

Stay with me here -- I swear that it's relevant.

When I was three, I wanted to write music. And I knew what music looked like. There were five lines, 𝄚, and at the beginning of the five lines was this squiggly thing, 𝄞, and then there were all these notes that looked like 𝅝 and 𝅗𝅥 and 𝅘𝅥 and 𝅘𝅥𝅮 and 𝅘𝅥𝅯. So I drew five lines across a sheet of paper. I drew that squiggly thing at the beginning. (I'm sure it took a bunch of practice for me to manage something even remotely like 𝄞.) I put notes in and around the five lines. And they I gave the paper to my mother for her to play on the piano, so I could find out what my song sounded like.

I hope you see the problem here: No one writes music like that. (Well, aside from certain 20th century composers that we'll ignore for the time being.) Composers don't start with the symbols; they start with a musical idea, a melody. They play with that idea and twerk it and rearrange it and twist it until they're happy with it, and only then do they write it down in those symbols I was so proud I could draw. The symbols are just a way to convey that idea to others in a compact form. There's nothing special about them; you can use other symbols if you like (and people that read tabulature do) or you can use no symbols and communicate the idea directly (and many great musicians don't read notation). But three-year-old-me confused the symbols for the idea they represented and tried to create music with just the symbols and no understanding of the ideas behind them.

Do you see where I'm going?

You are making the same mistake here. You are focused on the symbols, the x and the x', the M and the S and the S'. You are hoping that if you just get the right combinations of symbols into the right order, you'll have done math. But math isn't about manipulating symbols. It's about ideas. The symbols let us convey ideas to others in a compact form. There's nothing special about them; you could use other symbols, or you could use no symbols at all, and you're still doing math. But without an understanding of the ideas behind them symbols, you can't possibly do math.

What happened to me? I went to school and took music class. And in music class they taught us the ideas of music. I learned about scales and chords and rhythms and time signatures and accidentals. And they made me do exercises, writing out scales and rhythms and practicing turning those symbols into musical ideas. And these exercises were stupid things: write this scale, fill in the missing note. They had nothing to do, it seemed, with getting to that melody that I wanted to write at three. Had I been more belligerent, I would have complained. "I don't want to learn about measures or keys. I want to write music. I can't hear a bar line or a key change." Oh, but I can. Now that I've learned this ideas, I can hear measures and keys and accidentals and all sorts of things that aren't the melody that I wanted to write at three.

Do you see where I'm going?

This is what we've tried to do with you. We've tried to teach you the basic ideas in this field of math. The manifolds and the coordinate systems and the coordinate transformations. And we've tried to make you do exercises. Stupid things, it must seem. Nothing to do with the x' and the vt and the Galilean. "There's no manifold in x' = x - vt." Oh, but there is. And everyone who knows the idea of a manifold can see it as clearly in that equation as I can hear the beats of a measure or the chord structure of the blues.

What happened to me? Well, I never became that great composer that three-year-old-me wanted to be, and I probably never will. But I've played music and transcribed music and arranged music, transmuting symbols to ideas and ideas to symbols. And I'm glad I did all those stupid exercises that gave the understanding of these ideas, so I can do more than just draw symbols and wait for someone to recognize it as music.

Do you see where you're going?

You've resisted learning these new ideas, and you haven't practiced converting between these ideas and symbols. You've put too much stock in the importance of symbols and not enough into the ideas behind them. And you keep throwing symbols on the page and hoping that someone will recognize it as math. Is there hope? Certainly. But only if you take the time to learn these ideas and stop this attachment to symbols. Until then, you have no more hope of writing a proof than three-year-old-me had of writing a symphony.
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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 » Sun Jul 28, 2013 10:00 pm UTC

ucim wrote:If vt is not zero, then the systems are not coincident,

Correct.
ucim wrote:and the premise
S(x,y,z) = S'(x',y',z')
no longer applies.

CORRECT!
You are thinking about the manifold, where NOW P = S(x,y,z) = S'(x',y',z') does no longer apply!
THAT IS THE EXACTLY THE PROBLEM! Since S(x,y,z) = S'(x',y',z') was given.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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

Postby WibblyWobbly » Sun Jul 28, 2013 10:14 pm UTC

Schrollini wrote:
steve waterman wrote:The Galilean has no double prime notation/logic, nor do I.
Hence, why I backed down on any possible M(x,y,z)...no such thing.


Steve, when I was three, I wanted to write music.

Stay with me here -- I swear that it's relevant.

When I was three, I wanted to write music. And I knew what music looked like. There were five lines, 𝄚, and at the beginning of the five lines was this squiggly thing, 𝄞, and then there were all these notes that looked like 𝅝 and 𝅗𝅥 and 𝅘𝅥 and 𝅘𝅥𝅮 and 𝅘𝅥𝅯. So I drew five lines across a sheet of paper. I drew that squiggly thing at the beginning. (I'm sure it took a bunch of practice for me to manage something even remotely like 𝄞.) I put notes in and around the five lines. And they I gave the paper to my mother for her to play on the piano, so I could find out what my song sounded like.

I hope you see the problem here: No one writes music like that. (Well, aside from certain 20th century composers that we'll ignore for the time being.) Composers don't start with the symbols; they start with a musical idea, a melody. They play with that idea and twerk it and rearrange it and twist it until they're happy with it, and only then do they write it down in those symbols I was so proud I could draw. The symbols are just a way to convey that idea to others in a compact form. There's nothing special about them; you can use other symbols if you like (and people that read tabulature do) or you can use no symbols and communicate the idea directly (and many great musicians don't read notation). But three-year-old-me confused the symbols for the idea they represented and tried to create music with just the symbols and no understanding of the ideas behind them.

Do you see where I'm going?

You are making the same mistake here. You are focused on the symbols, the x and the x', the M and the S and the S'. You are hoping that if you just get the right combinations of symbols into the right order, you'll have done math. But math isn't about manipulating symbols. It's about ideas. The symbols let us convey ideas to others in a compact form. There's nothing special about them; you could use other symbols, or you could use no symbols at all, and you're still doing math. But without an understanding of the ideas behind them symbols, you can't possibly do math.

What happened to me? I went to school and took music class. And in music class they taught us the ideas of music. I learned about scales and chords and rhythms and time signatures and accidentals. And they made me do exercises, writing out scales and rhythms and practicing turning those symbols into musical ideas. And these exercises were stupid things: write this scale, fill in the missing note. They had nothing to do, it seemed, with getting to that melody that I wanted to write at three. Had I been more belligerent, I would have complained. "I don't want to learn about measures or keys. I want to write music. I can't hear a bar line or a key change." Oh, but I can. Now that I've learned this ideas, I can hear measures and keys and accidentals and all sorts of things that aren't the melody that I wanted to write at three.

Do you see where I'm going?

This is what we've tried to do with you. We've tried to teach you the basic ideas in this field of math. The manifolds and the coordinate systems and the coordinate transformations. And we've tried to make you do exercises. Stupid things, it must seem. Nothing to do with the x' and the vt and the Galilean. "There's no manifold in x' = x - vt." Oh, but there is. And everyone who knows the idea of a manifold can see it as clearly in that equation as I can hear the beats of a measure or the chord structure of the blues.

What happened to me? Well, I never became that great composer that three-year-old-me wanted to be, and I probably never will. But I've played music and transcribed music and arranged music, transmuting symbols to ideas and ideas to symbols. And I'm glad I did all those stupid exercises that gave the understanding of these ideas, so I can do more than just draw symbols and wait for someone to recognize it as music.

Do you see where you're going?

You've resisted learning these new ideas, and you haven't practiced converting between these ideas and symbols. You've put too much stock in the importance of symbols and not enough into the ideas behind them. And you keep throwing symbols on the page and hoping that someone will recognize it as math. Is there hope? Certainly. But only if you take the time to learn these ideas and stop this attachment to symbols. Until then, you have no more hope of writing a proof than three-year-old-me had of writing a symphony.

I want you to know, even if Steve doesn't reply or even acknowledge this post, I found it illuminating. I'm not much of a mathematician, and barely a semi-pro physicist, but I keep hearing people say that one shouldn't be ashamed not to know something, only in refusing to learn. I'd like to think that you and several others here have taught me at least a few things I didn't know.

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

Postby eSOANEM » Sun Jul 28, 2013 11:04 pm UTC

steve waterman wrote:
ucim wrote:If vt is not zero, then the systems are not coincident,

Correct.
ucim wrote:and the premise
S(x,y,z) = S'(x',y',z')
no longer applies.

CORRECT!
You are thinking about the manifold, where NOW P = S(x,y,z) = S'(x',y',z') does no longer apply!
THAT IS THE EXACTLY THE PROBLEM! Since S(x,y,z) = S'(x',y',z') was given.


This is all true.

It is however completely irrelevant to the Galilean (or other non-trivial co-ordinate transforms) because with those, it is not given that S(x,y,z)=S'(x',y',z').
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Sun Jul 28, 2013 11:17 pm UTC

steve waterman wrote:You are thinking about the manifold, where NOW P = S(x,y,z) = S'(x',y',z') does no longer apply!
THAT IS THE EXACTLY THE PROBLEM! Since S(x,y,z) = S'(x',y',z') was given.
I'm going to draw you an analogy.

My friend and I went to Paris. We didn't take many pictures, but we did want a picture of the Eiffel tower. In fact, we each wanted identical pictures from our identical cameras. So, we found a field from which we could see the Eiffel tower, and we stood in exactly the same spot. Well, as close as we could anyway (we're very close friends, and it is, after all, France. :) ) We each pointed our cameras in the same direction, used the same lens settings, and took the picture, fully expecting to get the same image on the film.

Having done this, we did it again just to be sure. But this time I sidled away. She was so focused on the pictures we were taking that she didn't notice, but I moved to the left about a hundred yards. We then pointed our cameras in the same direction (our cameras have little compasses on them; great feature for experiments like this!) and took the second picture.

When the film were developed, the first picture on my roll matched the first picture on her roll. However, the second picture on my roll did not match anything. Looking more closely, I could see part of the Eiffel Tower on the right side of the frame of my second picture, but her second picture showed the full Eiffel Tower right in the center of the frame, just like her first picture.

She accused me of leaving her, but I insisted that it was already established that we were in the same position, so I couldn't have left her.

Well, she smashed my camera and left in a huff. I didn't even know you could find a huff in Paris, but live and learn.

Aside from any lessons in romance that this story provides, my question for you, Steve, is: What are the components of the analogy I am making here?

I'll spoiler a few structural hints, but try to answer first without them.
Spoiler:
I am, of course, drawing an analogy between these events and the S, S', S" examples under discussion. So, more specifically,

1: What is taking the place of the manifold here?
2: What is taking the place of S?
3: What is taking the place of S' and S"?
4: What is taking the place of point P?

5: My girlfriend accused me of leaving her, but I insisted that it was already established that we were in the same position. Who's right? What is this an analogy of?
Jose
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Sun Jul 28, 2013 11:40 pm UTC

WibblyWobbly wrote:I want you to know, even if Steve doesn't reply or even acknowledge this post, I found it illuminating. I'm not much of a mathematician, and barely a semi-pro physicist, but I keep hearing people say that one shouldn't be ashamed not to know something, only in refusing to learn. I'd like to think that you and several others here have taught me at least a few things I didn't know.

Thanks for saying so. I've known there's only a slim chance of convincing Steve of anything, but I've gone through with it anyway in the hope that it'll help someone who wants to learn. I'm sure those several others have the same thing in mind. So it's nice to know that this was successful, even if the apparent goal wasn't.

ucim wrote:Well, she smashed my camera and left in a huff.

Rufus T. Firefly wrote:If that's too soon, I'll leave in a minute and a huff!
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Jul 29, 2013 1:01 am UTC

eSOANEM wrote:
steve waterman wrote:
ucim wrote:If vt is not zero, then the systems are not coincident,

Correct.
ucim wrote:and the premise
S(x,y,z) = S'(x',y',z')
no longer applies.

CORRECT!
You are thinking about the manifold, where NOW P = S(x,y,z) = S'(x',y',z') does no longer apply!
THAT IS THE EXACTLY THE PROBLEM! Since S(x,y,z) = S'(x',y',z') was given.


This is all true.

It is however completely irrelevant to the Galilean (or other non-trivial co-ordinate transforms) because with those, it is not given that S(x,y,z)=S'(x',y',z').


Of course it is the given, I have posted the wiki URL so many times already.
At t = t' = 0, S(x,y,z) = S'(x',y',z') because they are coincident.
Schrollini wrote:Of course x = x' when the systems are coincident.


wiki wrote:The notation below [ my case 2] describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) of a single arbitrary event, as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and x’ directions, with their spatial origins coinciding at time t=t'=0 [my given case1]

x'=x-vt
y'=y
z'=z
t'=t

You will note that at t = 0, then vt = 0, and x = x '.
Last edited by steve waterman on Mon Jul 29, 2013 1:17 am UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Mon Jul 29, 2013 1:14 am UTC

steve waterman wrote:Of course it is the given, I have posted the wiki URL so many times already.
At t = t' = 0, S(x,y,z) = S'(x',y',z') because they are coincident.


Given: I'm standing in front of the Empire State Building.

I then go to France, stand in front of the Eiffel Tower and take a picture of what's in front of me.

Question: When I get my film back, what image do I see?

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

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

Postby eSOANEM » Mon Jul 29, 2013 1:26 am UTC

steve waterman wrote:
eSOANEM wrote:
steve waterman wrote:
ucim wrote:If vt is not zero, then the systems are not coincident,

Correct.
ucim wrote:and the premise
S(x,y,z) = S'(x',y',z')
no longer applies.

CORRECT!
You are thinking about the manifold, where NOW P = S(x,y,z) = S'(x',y',z') does no longer apply!
THAT IS THE EXACTLY THE PROBLEM! Since S(x,y,z) = S'(x',y',z') was given.


This is all true.

It is however completely irrelevant to the Galilean (or other non-trivial co-ordinate transforms) because with those, it is not given that S(x,y,z)=S'(x',y',z').


Of course it is the given, I have posted the wiki URL so many times already.
At t = t' = 0, S(x,y,z) = S'(x',y',z') because they are coincident.
Schrollini wrote:Of course x = x' when the systems are coincident.


Do you even read the things you quote?

I've bolded and made enormous the bit of text you somehow missed.

That the co-ordinates at a particular time (i.e. in a certain constant-t-section of the manifold/spacetime) are equal says nothing about their relation when t != 0.

Furthermore, what you've quoted is not what you mean to have said, the statement you've given is simply defining the usual names of the co-ordinates in that system, what you mean to have said was that S(x,y,z)=S'(x,y,z) which implies x=x', y=y', z=z'.

Anyway, you claim to have looked at how SR follows from the lorentz tranform from a logical pov and found no flaw there. Let us look at this statement too from a logical point of view.

Let O be the statement that t=t'=0 and let C be the statement that s=s', y=y' and z=z'.

What you quoted (or meant to quote) is the statement that O => C.

Now, let's draw the truth table of that statement:

O C O => C
F F T
F T T
T F F
T T T

Now, we have been told that O => C is true meaning the third case is irrelevant. We are now interested in cases when t != t' though so we don't need to consider the bottom row.

Now, we know that O=F and we know that (O => C) = T so what does this tell us about C?

The answer is absolutely nothing.

If you look at the truth table, you'll see that both the first and second cases satisfy those conditions. This means there is no inconsistency through having the co-ordiantes remain coincident (i.e. they are related by the trivial transformation) nor through them no longer being coincident (them being related by something more complicated such as the Galilean).

Now. This is propositional logic. There are no words here to argue about. There is no interpretation here. This is simply indisputable logic. It says that, the text you quoted (once corrected to make it a non-trivial statement and into one which you could possibly read as supporting your case) has no inconsistency with the Galilean.

Now, shut the up with this drivel about how the fact that the co-ordinate systems were once coincident means they must always be so we can actually get on and teach you something more interesting.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Jul 29, 2013 1:41 am UTC

ucim wrote:
steve waterman wrote:Of course it is the given, I have posted the wiki URL so many times already.
At t = t' = 0, S(x,y,z) = S'(x',y',z') because they are coincident.


Given: I'm standing in front of the Empire State Building.

I then go to France, stand in front of the Eiffel Tower and take a picture of what's in front of me.

Question: When I get my film back, what image do I see?

Jose

The photograph is like a mapping of the real thing. You can take pictures at different angles of the same object. No matter what travels you make, the ESB and the ET are the given, and they do not change, they are fixed in New York and Paris. Take your picture of the Tower, and place it in front of the ESB, and take a picture...does the ESB = ET because they share the same space in your mapped photo of the two of them? [rhetorical]

These attempted analogies do not help, I know, I have tried many too. They always fail.
WE are now discussing ...
does the Galilean actually say "given coincident S(x,y,z) and S'(x',y',z')" in one form or another?
Image
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Mon Jul 29, 2013 2:19 am UTC

steve waterman wrote:The photograph is like a mapping of the real thing.
This completely misses the point. When I get my film back, what image do I get?

Do I get the image of the Eiffel Tower? (not possible, since the GIVEN is that I was standing in front of the Empire State Building)

Do I get an image of the Empire State Building (not possible, since I moved and was standing in front of the Eiffel Tower at the time I took the picture).

One of these statements is mustard. Which one, and why?

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

Postby addams » Mon Jul 29, 2013 2:33 am UTC

ucim wrote:
steve waterman wrote:The photograph is like a mapping of the real thing.
This completely misses the point. When I get my film back, what image do I get?

Do I get the image of the Eiffel Tower? (not possible, since the GIVEN is that I was standing in front of the Empire State Building)

Do I get an image of the Empire State Building (not possible, since I moved and was standing in front of the Eiffel Tower at the time I took the picture).

One of these statements is mustard. Which one, and why?

Jose

Hey! That is not nice.

I was actually thinking about thinking about this thing and You Did That!
pfft. No wonder Math has such a bad reputation.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby brenok » Mon Jul 29, 2013 2:35 am UTC

addams wrote:Hey! That is not nice.

I was actually thinking about thinking about this thing and You Did That!
pfft. No wonder Math has such a bad reputation.


What do you mean?

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

Postby ucim » Mon Jul 29, 2013 2:37 am UTC

addams wrote:Hey! That is not nice.
What's not nice about it? It illustrates the part where Steven is having trouble. He is hanging on to the given after he has changed it (by "moving" the coordinate systems). I'm translating his statements into a form where the error might be apparent. It's how we learn.

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

Postby SecondTalon » Mon Jul 29, 2013 3:15 am UTC

steve waterman wrote:You are thinking about the manifold, where NOW P = S(x,y,z) = S'(x',y',z') does no longer apply!
THAT IS THE EXACTLY THE PROBLEM! Since S(x,y,z) = S'(x',y',z') was given.


..then it's not the same manifold, is it?

You are saying the following
HOME = Where I keep my shit
HOME = 123 Fake Street
So 123 Fake Street = Where I keep my shit

But I moved to 321 Notreal Ave.

Home = Where I Keep My Shit
Home = 321 Notreal Ave
123 Fake Street =/= Where I keep my shit
HOW CAN THIS BEEEE!?!?!

... it can't. Shit changed. S(x,y,z) =/= S'(x',y',z') anymore, 'cause you changed just what you meant by S'

Because you can't fucking move coordinate systems.

Jesus, I'm starting to get this shit.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby addams » Mon Jul 29, 2013 3:32 am UTC

ucim wrote:
addams wrote:Hey! That is not nice.
What's not nice about it? It illustrates the part where Steven is having trouble. He is hanging on to the given after he has changed it (by "moving" the coordinate systems). I'm translating his statements into a form where the error might be apparent. It's how we learn.

Jose

yes. That may very well be true.
I can't do the math.

Some of the imagery is Fun.
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see? I was thinking about light and you were leading to Location.
but; You led to a Location outside my experience.

I know it is Real! It is also Math; And; That makes it Imaginary!
Throwing the Sear's Tower in with the Eiffel Tower seemed unfair.

If I could do the Math I would have been careful with each and every word.
Math sentences do not have useless redundancies.
They get stashed out and simplified away.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby yurell » Mon Jul 29, 2013 5:06 am UTC

Wait, what just happened? I started reading this page, and suddenly people were saying:
S(x,y,z) ≠ S'(x',y',z')
This confuses the hell out of me, because we defined the systems such that S(x,y,z) = S'(x',y',z'). Have we redefined these systems?
What we've established, from what I can tell is that:
S'(x,y,z) ≠ S'(x',y',z')
for non-zero vt (note the prime above both 'S').
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby eSOANEM » Mon Jul 29, 2013 10:46 am UTC

I think Steve misquoted something and people didn't quite notice and are now accidentally carrying on that mistake.

I tried to address it in my post but no-one (particularly not Steve who I'd really like to see argue his way past a truth table) seems to have paid any attention to.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Jul 29, 2013 11:37 am UTC

Image
added -
"Given" in a physical sense is not at all the same as a "given" in a math setting.

math mindset... given x = 4, x is forever fixed at 4.

physical mindset ...given 4 apples, one apple could be eaten, one could be sold, one lost and the other juiced. so given x apples are NOT fixed forever.

My challenge is restricted to the arena of the mathematically given coincident S(x,y,z) and S'(x',y',z'). and the mathematically given non-coincident S(x,y,z) and S'(x',y',z'). In both mathematical scenarios, x = x'.
Last edited by steve waterman on Mon Jul 29, 2013 12:35 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 » Mon Jul 29, 2013 12:09 pm UTC

yurell wrote:This confuses the hell out of me, because we defined the systems such that S(x,y,z) = S'(x',y',z'). Have we redefined these systems?
This is another way of illustrating the confusion that comes from saying "GIVEN they are coincident, we now move one of the systems". This, of course, breaks the given.

What makes them coincident is the additional conditions: x=x', etc.

steve waterman wrote:Does the Galilean actually say "given coincident S(x,y,z) and S'(x',y',z')" in one form or another?

Image
The depiction commencws with the same conditions as the Galilean given and show the Galilean results from the manifold.

given coincident S(x,y,z) and S'(x',y',z'), the Galilean states that S(x,y,z) S'(x',y,',z') if vt >0.


Steve - I hate to be harsh but the entire thing you've presented as an image (why as an image? It's hard to quote that way) is a bunch of math-words that do not express a coherent thought. I tried to explain why. It's not coming across.

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

Postby beojan » Mon Jul 29, 2013 12:11 pm UTC

steve waterman wrote:Does the Galilean actually say "given coincident S(x,y,z) and S'(x',y',z')" in one form or another?

No.

The Galilean transformation is not like a book of the bible. There is no canonical text. Despite this, it is not specific to coincident reference frames.
It is simply the function that converts coordinates in one reference frame, to coordinates in another reference frame which is moving (at constant velocity [much less than c]) relative to the first.
For example, if we each define the origin from which we make measurements as ourselves, and I am moving with constant velocity relative to you, it will convert coordinates as I measure them to coordinates as you measure them.

ucim wrote:Steve - I hate to be harsh but the entire thing you've presented as an image (why as an image? It's hard to quote that way) is a bunch of math-words that do not express a coherent thought. I tried to explain why. It's not coming across.

I wouldn't be surprised if that (the difficulty of quoting it) were the very reason it is an image.

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

Postby JudeMorrigan » Mon Jul 29, 2013 12:40 pm UTC

yurell wrote:Wait, what just happened? I started reading this page, and suddenly people were saying:
S(x,y,z) ≠ S'(x',y',z')
This confuses the hell out of me, because we defined the systems such that S(x,y,z) = S'(x',y',z'). Have we redefined these systems?
What we've established, from what I can tell is that:
S'(x,y,z) ≠ S'(x',y',z')
for non-zero vt (note the prime above both 'S').

Exactly this. I'm finding myself very confused as well. Shouldn't S(x,y,z) = S'(x',y',z') be true by definition regardless of whether or not the systems are coincident?

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

Postby steve waterman » Mon Jul 29, 2013 1:01 pm UTC

Does the Galilean actually say "given coincident S(x,y,z) and S'(x',y',z')" in one form or another?
beojan wrote:No.

wiki wrote:The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) of a single arbitrary event, as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and x’ directions,
with their spatial origins coinciding at time t=t'=0

x'=x-vt
y'=y
z'=z
t'=t

with their spatial origins coinciding at time t=t'=0
means COINCIDENT systems.
You are certainly not the only one that is unaware, that S and S' start off as coincident in the Galilean given!

Then the Galilean moves one system by vt as depicted here...on this page
http://en.wikipedia.org/wiki/Galilean_transformation

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

Postby ucim » Mon Jul 29, 2013 1:17 pm UTC

steve waterman wrote:You are certainly not the only one that is unaware, that S and S' start off as coincident in the Galilean given!
Having some of the coordinates coincident is not the same as having all of the coordinates coincident. In a properly constructed Galilean transformation, time is a coordinate. When you say "at t=0, this, and at t=5, that", you are not, in fact, "moving" a coordinate system. You are merely looking at another slice in the same coordinate system.

But this isn't even your fundamental problem, because it extends to ordinary translation. Please go back and tell me what picture will appear when I develop the film.

Jose
Last edited by ucim on Mon Jul 29, 2013 1:19 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 » Mon Jul 29, 2013 1:18 pm UTC

JudeMorrigan wrote: Shouldn't S(x,y,z) = S'(x',y',z') be true by definition regardless of whether or not the systems are coincident?

BINGO. Nicely worded question too. Just as two rulers are. Welcome aboard my train of thought.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Mon Jul 29, 2013 1:20 pm UTC

steve waterman wrote:Does the Galilean actually say "given coincident S(x,y,z) and S'(x',y',z')" in one form or another?
beojan wrote:No.

wiki wrote:The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) of a single arbitrary event, as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and x’ directions,
with their spatial origins coinciding at time t=t'=0

x'=x-vt
y'=y
z'=z
t'=t

with their spatial origins coinciding at time t=t'=0
means COINCIDENT systems.
You are certainly not the only one that is unaware, that S and S' start off as coincident in the Galilean given!

Then the Galilean moves one system by vt as depicted here...on this page
http://en.wikipedia.org/wiki/Galilean_transformation

Image

Note: The primes in the above image need re-checking.

Having the two coordinates systems be spatially coincident at time t = t' = 0 ONLY (note italics. This is not the same as just being coincident which would make the two coordinate systems the same), much like rotating the axes such that the motion is along the x axis, is a simplification we often make in order to make it easier to express and understand the Galilean transformation. It is, however, not essential.

In addition, when, as you say, "Then the Galilean moves one system by vt as depicted here...on this page" (this is not correct. It would be more accurate to say "the spatial origin of one coordinate system is moving at a velocity v"), when t ≠ 0, the two systems are not spatially coincident.

Addition:
steve waterman wrote:
JudeMorrigan wrote: Shouldn't S(x,y,z) = S'(x',y',z') be true by definition regardless of whether or not the systems are coincident?

BINGO. Nicely worded question too. Just as two rulers are. Welcome aboard my train of thought.

JudeMorrigan, you are correct.
The problem is Steve believes S(x,y,z) = S'(x',y',z') implies x = x', y = y', z = z'.
Using single variable functions whose domain and range are both the real numbers,
lets define f(x) = x² and g(x') = x'.

According to Stevean logic, f(x) = g(x') would imply x = x'.
However if x = 5, x' = 25
f(x) = 25, and g(x') = 25 so f(x) = g(x') in this case. However, x ≠ x'.
Hence Stevean logic is incorrect.
Last edited by beojan on Mon Jul 29, 2013 1:33 pm UTC, edited 1 time in total.

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

Postby eSOANEM » Mon Jul 29, 2013 1:32 pm UTC

steve waterman wrote:
JudeMorrigan wrote: Shouldn't S(x,y,z) = S'(x',y',z') be true by definition regardless of whether or not the systems are coincident?

BINGO. Nicely worded question too. Just as two rulers are. Welcome aboard my train of thought.


Steve, this is irrelevant to everything.

S(x,y,z)=S'(x',y',z') is simply saying the following "we will use the variable x for the first co-ordinate of a point in the S system and we will use the variable x' for the first co-ordinate of that same point in the S' system (and so on for y & y' and z& z')". It says absolutely nothing about the relation between those co-ordiantes.

It is true of any two sets of co-ordinates whether they are of the same type (by which I mean, cartesian, skew-cartesian, cylindrical, spherical, curvilinear etc.), have the same orientation or the same origin.

It is just a definition of terms. As such, you cannot argue against it and attempting to do so shows that you have a profound misunderstanding.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Jul 29, 2013 1:36 pm UTC

beojan wrote: when, as you say, "Then the Galilean moves one system by vt as depicted here...on this page" (this is not correct. It would be more accurate to say

"the spatial origin of one coordinate system is moving at a velocity v"), when t ≠ 0, the two systems are not spatially coincident.

That uses a double negative, but otherwise I do the inclusion of "spatial origin"...how about?

1 the two systems are spatially coincident, iff t = 0.
2 the spatial origin of one coordinate system is moving at a velocity v, away from mutual coincidence,
while the other system remains stationary.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Mon Jul 29, 2013 1:43 pm UTC

steve waterman wrote:
beojan wrote: when, as you say, "Then the Galilean moves one system by vt as depicted here...on this page" (this is not correct. It would be more accurate to say

"the spatial origin of one coordinate system is moving at a velocity v"), when t ≠ 0, the two systems are not spatially coincident.

That uses a double negative, but otherwise I do the inclusion of "spatial origin"...how about?

1 the two systems are spatially coincident, iff t = 0.
2 the spatial origin of one coordinate system is moving at a velocity v, away from mutual coincidence,
while the other system remains stationary.

"Away from mutual coincidence" isn't strictly neccessary (and is incorrect, if 'coincidence' is anything but a shorthand for statement 1), but other than that, so long as you fully understand both of your statements here, I see no problem.
Last edited by beojan on Mon Jul 29, 2013 1:44 pm UTC, edited 1 time in total.

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

Postby ivnja » Mon Jul 29, 2013 1:44 pm UTC

steve waterman wrote:You are certainly not the only one that is unaware, that S and S' start off as coincident in the Galilean given!
Start as coincident? Why would the spatial origins have to start as coincident (or more accurately, why would you have to start looking at the 3D cross sections of the 3+1D systems at the spot where the time axes intersect)? There's a whole range of negative t values where the spatial origin of S' is off to the "left" of that of S. t'=t=0 isn't special because it's the starting point, it's just that the spot where the time axes intersect is defined as time zero for convenience's sake.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Jul 29, 2013 1:54 pm UTC

beojan wrote:The problem is Steve believes S(x,y,z) = S'(x',y',z') implies x = x', y = y', z = z'..

That is what the Galilean states, whose url I just posted twice already today.
Given S(x,y,z) coincident S'(x',y',z') having x = x', y = y', z = z'...WHEN T = 0.

I believe x = x' is given because that is what their quote states as x' = x-vt, when t = 0,
then x' = x-vt mathematically means x' = x.
CORRECT?
[btw, regardless if you think x means "x axis" or x means "x abscissa". ]


You are certainly not the only one that is unaware, that S and S' start off as coincident in the Galilean given!

added -

ivnja wrote: Start as coincident? Why would the spatial origins have to start as coincident

Because that IS what the Galilean does. This is HIS premise, and why I use/copy it. This is kinda key.
Not my whimsy, I am just following along with the conditions that the Galilean uses!!! Granted, the way it is worded in wiki is at best, confusing. They certainly obscure that x = x' at t = 0 is correct for one thing. Logically too, one would start at t = 0, and then apply vt.
Last edited by steve waterman on Mon Jul 29, 2013 2:07 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Mon Jul 29, 2013 2:03 pm UTC

steve waterman wrote:
beojan wrote:The problem is Steve believes S(x,y,z) = S'(x',y',z') implies x = x', y = y', z = z'..

That is what the Galilean states, whose url I just posted twice already today.
Given S(x,y,z) coincident S'(x',y',z') having x = x', y = y', z = z'...WHEN T = 0.

I believe x = x' is given because that is what their quote states as x' = x-vt, when t = 0,
then x' = x-vt mathematically means x' = x.
CORRECT?
[btw, regardless if you think x means "x axis" or x means "x abscissa". ]


The Wikipedia article on the Galilean transformation is in no way canonical. The Galilean transformation is not interchangeable with its Wikipedia article.

I really don't understand why you think x' = x is a given at any time other than t = 0.

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

Postby steve waterman » Mon Jul 29, 2013 2:36 pm UTC

beojan wrote:I really don't understand why you think x' = x is a given at any time other than t = 0.

x = x' is the GIVEN at t = 0.
So, I am not saying at t > 0 anything at all is given, that is all done at t = 0.

Then we take what was given at t = 0, and apply vt to one system, where t > 0.

The given at t = 0, x = x', stay their same respective values at t > 0...., because they WERE given at t = 0, not because they ARE given in any vt applied setting at t > 0. Sorry, hard concept to find the right words.

Try a recap...x = x' is NOT a given when t > 0, however, x = x' remains equal to that given when t > 0.
Do you see the concept beneath these words, that I am going for?

added -
Thanks for keeping the posts short and straight to the mathematical objection. I will try and keep up with them, and with them being kept shortish and right on topic, I hopefully will be able to keep up relative pace. I hope for no mare analogies, please, nor new things not mentioned in the Galilean given. I will say that, the internet is a poor place to find the Galilean basically commencing with x' = x-vt, and the whole starting from coincidence thing, is even missing in some, and hard to decode from others, and most only show the vt applied diagram if they show one at all.
This is kinda glossed over, and almost 'too trivial" to be bothered to be mentioned by some. So, I could see why this "starting from coincidence" issue seems superfluous to you.
Last edited by steve waterman on Mon Jul 29, 2013 2:55 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby vbkid » Mon Jul 29, 2013 2:48 pm UTC

steve waterman wrote:x = x' is NOT a given when t > 0, however, x = x' remains equal to that given when t > 0.
Do you see the concept beneath these words, that I am going for?

We all see "what you are going for". Unfortunately, math disagrees.
Quick question.
Are you open to the idea that you are wrong and need to understand why you are wrong? Or, are you simply frustrated we can't understand what you've discovered?

beojan
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Joined: Mon May 23, 2011 12:11 pm UTC
Location: Oxford / London, United Kingdom, Europe

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

Postby beojan » Mon Jul 29, 2013 2:51 pm UTC

steve waterman wrote:
beojan wrote:I really don't understand why you think x' = x is a given at any time other than t = 0.

x = x' is the GIVEN at t = 0.
So, I am not saying at t > 0 anything at all is given, that is all done at t = 0.

Then we take what was given at t = 0, and apply vt to one system, where t > 0.

The given at t = 0, x = x', stay their same respective values at t > 0...., because they WERE given at t = 0, not because they ARE given in any vt applied setting at t > 0. Sorry, hard concept to find the right words.

Try a recap...x = x' is NOT a given when t > 0, however, x = x' remains equal to that given when t > 0.
Do you see the concept beneath these words, that I am going for?

No, that makes no sense.
"x = x' is NOT a given when t > 0, however, x = x' remains equal to that given when t > 0."
This is nonsense.

Lets try again,
x' = x - vt is true always (given our simplifications and assumptions).
Substituting in t = 0 gives x' = x - v*0 = x - 0 = x.
However, when t ≠ 0, vt ≠ 0, so x - vt ≠ x so x' ≠ x (assuming v ≠ 0 as always).

To make the situation clearer, the S' spatial origin does not start moving at t = 0.
At t < 0, the S' spatial origin is moving at velocity v, approaching the S spatial origin.
At t > 0, the S' spatial origin is moving at velocity v, receding from the S spatial origin.
At t = 0, the S' spatial origin is moving at velocity v, and is instantaneously at the same position as the S spatial origin.

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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 » Mon Jul 29, 2013 3:05 pm UTC

vbkid
2 Quick questions.
Are you open to the idea that you are wrong and need to understand why you are wrong?

Are you simply frustrated because you know you are right, and you know that I am wrong?

These are rhetorical questions of course,
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve


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