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

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

Postby yurell » Wed Jul 31, 2013 10:48 pm UTC

steve waterman wrote:
Spoiler:
Image


Image


No Steve, you are naming two different axes the same thing, and using x' to refer to the same thing.
x' does not refer to 'the vector (x,0) translated to the origin of the S' co-ordinate system'. You're the only one that believes it does. If we're taking S'(x')=P (which is the way everyone else defines it), then x' in the second picture is equal to -1 (otherwise S'(x')=P is false when it's a statement we defined to be true).
You're obfuscating this point by naming two different co-ordinate systems S'
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Pfhorrest » Thu Aug 01, 2013 12:23 am UTC

Steve, you're still talking about Lebanon in a conversation about lesbians. It's ok. You misunderstood what someone was saying. Please make an honest effort to understand that nobody is arguing with what you are trying to say; we are all talking about something else entirely, and the only thing at issue is you think everyone means one thing when we've all said a bajillion times that we mean something else.

Here it is explained on your own picture:

Image

Can anyone here with some kind of Flash or other interactive animation skills make a little app to help here? I'd like it to shows two sets of axes like this, with the equation x' = x - vt (or maybe just x' = x - d) at the top, and those variables fleshed out into concrete numbers at the bottom (e.g. -1 = 2 - 3), with arrows pointing from the numbers and variables to the corresponding marks on the two sets of axes, and allows you to click and drag the two sets of axes and the point P around? Possibly also label the point with its coordinates in the two coordinate systems (which update as it is dragged around)?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby addams » Thu Aug 01, 2013 6:48 am UTC

Pfhorrest wrote:It was my birthday vacation this weekend, so I made a present for Steve that I hope the rest of you can enjoy too.

Steve, please watch it all the way through (until it loops) and tell me what you think about it, I put a lot of work into this gift for you.

Image


This thing is so great.
Is it correct?

If it is correct, then even I can, sort of, understand it.
So, clear.

In the past I have run Head On into a concept that I could not grasp.
I was told I had to let go of One before I could grasp the Other.

It is hard work to let go of an Idea that fits.
Often, it is required to do something else with the mind.

Long walks were prescribed for some.
Painting for others. Just Stop! for others.

I met a Number's Guy one time.
He was Far Far from Home.

He was there because he was changing his life, his goals, his career, all over some Number Thing.
We went for a long walk. He explained a bunch of stuff.

How hard it was to let go of a Dream.
How it was effecting the way he thought of himself.
He had worked a lot of that out. Self Esteem issues.

He had been counting his blessings.
He did not know what he would do to contribute and make money, but, he was not crying, anymore.

As we walked back, he started to tell me about the Problem that was driving him away from his life's work.
It was something I can not really understand.

Dividing something by zero.
or Some impossible thing that Math Theory says can not be done.
In advanced physics, that Can be done.

It was a long walk back.
As he was explaining to me how it Works, it became clear to him.
He was shy and happy! He called Home that night.
His Dream is Back on Track!

His mind needed to rest.
His mind needed to cavort with other kinds of Ideas as it blossomed open.
Once he Had it. It was Obvious.

That was a Big Complicated idea.
It was all about numbers that are either So small or So Large, I can only vaguely get it.

For a while, I understood some.
His mind carried my mind.
His was a strong mind.

He had been using it and using it well for a long time.
It was not failing him. It was tired.
Life is, just, an exchange of electrons; It is up to us to give it meaning.

We are all in The Gutter.
Some of us see The Gutter.
Some of us see The Stars.
by mr. Oscar Wilde.

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Those that do not Know; Don't tell them.
They do terrible things to people that Tell Them.

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

Postby steve waterman » Thu Aug 01, 2013 11:32 am UTC

Given S(x,y,z) , x was given/always assigned to the left/right axis of S.

Yet, you say the x axis is the up/down axis in the manifold!?!

Please explain the mathematical justification for your 90 degree axial shifts from the given x axis, please.

Does your z (up/down) axis rotate 90 degrees too from their given, in your manifold?...
otherwise z would also share your common x/x' manifold axis,
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby yurell » Thu Aug 01, 2013 11:42 am UTC

It doesn't matter. There is no privileged direction; the manifold doesn't have an inherent 'up' or 'down', you're just jumping on semantics.

Edit: Turns out I was mistaken as to the source of the confusion. You can ignore this, but leaving it here so the discussion still makes sense to people outside this timezone.
Last edited by yurell on Thu Aug 01, 2013 1:18 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 » Thu Aug 01, 2013 11:49 am UTC

steve waterman wrote:Given S(x,y,z) , x was given/always assigned to the left/right axis of S.

Yet, you say the x axis is the up/down axis in the manifold!?!

Please explain the mathematical justification for your 90 degree axial shifts from the given x axis, please.

Does your z (up/down) axis rotate 90 degrees too from their given, in your manifold?...
otherwise z would also share your common x/x' manifold axis,


Nothing has been shifted. The names we give things are entirely arbitrary (as Pfhorrest's gif from a while back illustrated). Maths is not about the labels we assign points, distances, operations, numbers etc. it is about the underlying objects. Which axis we call x and which we call y makes no difference to the mathematics.
my pronouns are they

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

Postby beojan » Thu Aug 01, 2013 12:20 pm UTC

Steve as per my previous reply,

beojan wrote:Ok, with those images, can you locate the point (2,3) in the red system (i.e. the point S(2,3)), and post images, with this point marked, with both coincident and non coincident coordinate systems, as before.


Can you post the images I asked for?

Also, the manifold has no axes. It is the coordinate system that has axes. The manifold just has points.

For example, we could take the surface of the Earth as our manifold. Places on Earth are our points. The surface of the Earth has no inherent axes, no inherent coordinate system.
Customarily, we use latitude and longitude, measuring latitude from the equator, and longitude from the prime (Greenwich) meridian. However, there is nothing fundamentally privileged about this coordinate system. We could just as easily decide to measure longitude from the Boston meridian, in which case the longitude of any given place (say New York City, or Montreal, or Mt Everest) would be different to what it was when we were measuring from the Greenwich meridian. The axes (in this case the prime meridian and the equator) are defined by the coordinate system, not the manifold.

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

Postby steve waterman » Thu Aug 01, 2013 12:30 pm UTC

yurell wrote:It doesn't matter. There is no privileged direction; the manifold doesn't have an inherent 'up' or 'down', you're just jumping on semantics.


yurell wrote:the manifold doesn't have an inherent 'up' or 'down',

Yet the statiionary manifold is mapped DIRECTLY from S(x,y,z) itself, which DOES ONLY have an x axis that goes left/right.



My gripe is that mathematically one cannot call P as S(x,y,z) if there is no S(0,0,0) in the manifold
When you call P both S(x-vt,y,z) and S'(x',y',z') , that mathematically impacts the given, whilst also describing the manifold relationship.

In truth, in the manifold, P = (x-vt,y,z) = (x'y'z)
since the manifold has no origin...we mathematically CANNOT use that S notation..as the S portion means wrt the origin at S(0,0,0). Since the manifold has no origins, then M(x,y,z), as notation, is also wrong.

In truth, we were given
S(x,y,z) coincident S'(x',y',z')
and neither S(x,y,z) wrt S(0,0,0)
nor S'(x',y',z') wrt S'(0,0,0)
can be mathematically changed.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Thu Aug 01, 2013 12:41 pm UTC

Steve,
1) Please see my last post and post the images I requested.
2) I think you may be misunderstanding the function notation Schrollini introduced.

(2,0,5), for example, is a set of coordinates.
S, for example, is a function that relates a set of three coordinates to a point in the manifold. (For the purposes of this thread, we are assuming, unless stated otherwise, that all spatial coordinate systems are cartesian.)

S(x,y,z) is the point returned when S is give the coordinates (x,y,z).
S(2,0,5), hence, is a point.

To reiterate, (2,0,5) is a set of coordinates, while S(2,0,5) is a point.

If you say S(x,y,z), this is, in itself, the point x units along the x axis, y units along the y axis, and z units along the z axis from the origin of S.
As a result, it doesn't make sense to say "S(x,y,z) with regard to S(0,0,0)".

As a special case, if we set x = 0, y = 0, and z = 0, S(0,0,0) is 0 units along the x axis, 0 units along the y axis, and 0 units along the z axis from the origin of S. Hence, S(0,0,0) is the origin of S. This is simply a special case of the above.

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

Postby yurell » Thu Aug 01, 2013 1:06 pm UTC

steve waterman wrote:Yet the statiionary manifold is mapped DIRECTLY from S(x,y,z) itself, which DOES ONLY have an x axis that goes left/right.


EDIT: See ucim's post, below.

steve waterman wrote:My gripe is that mathematically one cannot call P as S(x,y,z) if there is no S(0,0,0) in the manifold


a) Yes we can, the co-ordinate system doesn't have to be onto;
b) In the case we're discussing, S(0,0,0) is a point in the mannifold, no one has said it's not.

steve waterman wrote:When you call P both S(x-vt,y,z) and S'(x',y',z') , that mathematically impacts the given, whilst also describing the manifold relationship.


Lucky we don't say that. We said P = S(x,y,z,t) = S'(x',y',z',t')=S'(x-vt,y,z,t). Primed notation is hard to read and easy to get confused, but your'e the one who insisted upon it.

steve waterman wrote:In truth, in the manifold, P = (x-vt,y,z) = (x'y'z)


Blatantly wrong; co-ordinates are not points.

steve waterman wrote:since the manifold has no origin...we mathematically CANNOT use that S notation..as the S portion means wrt the origin at S(0,0,0). Since the manifold has no origins, then M(x,y,z), as notation, is also wrong.


What the hell are you going on about? The manifold has no origin just like the surface of the Earth has no inherent origin. We choose a point and label it the origin for a particular co-ordinate system. M(x,y,z) as notation is wrong because the manifold is not a function and you're writing it as a function. It's an abuse of notation and one that only you are using.

steve waterman wrote:In truth, we were given S(x,y,z) coincident S'(x',y',z')


That's not what we're given. Try again.

steve waterman wrote:and neither S(x,y,z) wrt S(0,0,0) nor S'(x',y',z') wrt S'(0,0,0) can be mathematically changed.


This does not make sense. Do you mean d(P,S(0,0,0)) and d(P,S'(0,0,0)) do not change? If so, what do you mean by that? Do not change for different choices of x'? Do not change for different choices of S'? Your problem still seems to be that you're labelling two different co-ordinate systems S' and then claiming they're not the same co-ordinate system.

Between Pfhorrest's images, Schrollini's maths and everyone else pointing this out to you, you should really consider that you are referring to something else that no one else is with the notation we're using, and attacking straw shadows of your own creation.
Last edited by yurell on Thu Aug 01, 2013 1:17 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 » Thu Aug 01, 2013 1:14 pm UTC

steve waterman wrote:Given S(x,y,z) , x was given/always assigned to the [b]left/right axis of S.

Yet, you say the x axis is the up/down axis in the manifold!?!
[/b]
Please explain the mathematical justification for your 90 degree axial shifts from the given x axis, please.

Does your z (up/down) axis rotate 90 degrees too from their given, in your manifold?...
otherwise z would also share your common x/x' manifold axis,
You are misunderstanding x.

In a 2D Cartesian plane, think about what the axis is.

The first-coordinate axis is a line where the first coordinate takes on all possible values while the second coordinate always has the constant value of zero. If we identify a point on the first-coordinate axis, we will do so by giving the value of the first coordinate at that point. (Since it's on the first-coordinate axis, we already know the value of the second-coordinate is zero). If we are calling the first coordinate x, and the second coordinate y, we can place the letter x on (or near) the first-coordinate axis to indicate the value of the first coordinate at that point. So far I think you're with me, because you do the same thing.

Now, a line of constant first-coordinate will be parallel to the second coordinate axis, because we are allowing the second coordinate to take on all values while holding the first coordinate constant. When we need to refer to this line, we will typically label it x because that is the value of the first-coordinate of every point on the line. It is a line of constant first-coordinate (whose value is x), where the second coordinate takes on all possible values.

This is why we keep distinguishing between x (the value) and the x-axis. They are different.

I see how it can be confusing if you are not careful. It is second nature to those who use the notation a lot however, because they already have in their minds the difference between a value and an axis. But in any case, that is why it looks like we are moving things 90 degrees. In actuality, we are not. We are referring to different things, things that you have conflated in your mind.

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

Postby yurell » Thu Aug 01, 2013 1:17 pm UTC

Oh, that's what he was referring to? I couldn't figure it out or find the problem, so I just presumed someone had worded something badly. Going to edit my post, then, to correct myself.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Thu Aug 01, 2013 1:39 pm UTC

beojan wrote:Steve,
1) Please see my last post and post the images I requested.
2) I think you may be misunderstanding the function notation Schrollini introduced.

(2,0,5), for example, is a set of coordinates.
S, for example, is a function that relates a set of three coordinates to a point in the manifold. (For the purposes of this thread, we are assuming, unless stated otherwise, that all spatial coordinate systems are cartesian.)

S(x,y,z) is the point returned when S is give the coordinates (x,y,z).
S(2,0,5), hence, is a point.

To reiterate, (2,0,5) is a set of coordinates, while S(2,0,5) is a point.

If you say S(x,y,z), this is, in itself, the point x units along the x axis, y units along the y axis, and z units along the z axis from the origin of S.
As a result, it doesn't make sense to say "S(x,y,z) with regard to S(0,0,0)".

As a special case, if we set x = 0, y = 0, and z = 0, S(0,0,0) is 0 units along the x axis, 0 units along the y axis, and 0 units along the z axis from the origin of S. Hence, S(0,0,0) is the origin of S. This is simply a special case of the above.


S is not a function of the given, S IS the given...I have all the S(x,y,z) coordinates, given S(x,y,z)

P in the manifold is a function of (x,y,z).
P in the manifold is NOT a function of the given and unchangeable S(x,y,z) wrt S(0.0.0).

given J = 2, is 2 a function of J, in your understanding?

I am working upon the requested P (2,3) coincident/non-coincident depictions...

Seems we still have not agreed upon what x means yet. Jude posted a def and I agreed, but that has been lost
in the current discussion. I would really like to have a common definition for x that we all agree upon.

You all are comparing line segments and I am comparing distances from the respective origin.

All abscissa are line segments, not all line segments are abscissa. In this case, your x' line segment in the manifold
is not longer equal to your given x abscissa in S. Mathematically, this should be the abscissa equality x = x', not the origin-less line segment equality of x' = x-vt of the manifold.

So, I will cease posting until we all agree upon what x means, including any new depictions. I too, am quite tired of having this discussion go round in circles.

x = The first coordinate (in S) of a point, is the distance along the first axis (of S) from the origin (of S) to that point. correct?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Thu Aug 01, 2013 1:50 pm UTC

steve waterman wrote:S is not a function of the given, S IS the given...I have all the S(x,y,z) coordinates, given S(x,y,z)

To be clear, people are not saying that "S is a function of the given". People are saying that S is a mathematical function.

Seems we still have not agreed upon what x means yet. Jude posted a def and I agreed, but that has been lost

I suspect you're either misremembering an exchange we had or misremembering an attribution. The only thing that I recall posting that you agreed to was that the proper expression was "S'(x-d,0,0) = S(x,0,0)" vice "S'(x',0,0) = P = point S(x-d,0,0)". (Although you recently made the exact same error in another post.) I may, of course, be forgetting something myself though.

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

Postby ucim » Thu Aug 01, 2013 1:56 pm UTC

steve waterman wrote:x = The first coordinate (in S) of a point, is the distance along the first axis (of S) from the origin (of S) to that point. correct?
Not quite. The following is more correct:

IF x is the first coordinate (in S) of a point,
AND a metric is defined (I'll assume normal Cartesian distance to be defined here)
THEN the value of x is the distance along the first axis (of S) from the origin (of S) to that point.

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

Postby steve waterman » Thu Aug 01, 2013 2:01 pm UTC

JudeMorrigan wrote:
steve waterman wrote:S is not a function of the given, S IS the given...I have all the S(x,y,z) coordinates, given S(x,y,z)

To be clear, people are not saying that "S is a function of the given". People are saying that S is a mathematical function.

Seems we still have not agreed upon what x means yet. Jude posted a def and I agreed, but that has been lost

I suspect you're either misremembering an exchange we had or misremembering an attribution. The only thing that I recall posting that you agreed to was that the proper expression was "S'(x-d,0,0) = S(x,0,0)" vice "S'(x',0,0) = P = point S(x-d,0,0)". (Although you recently made the exact same error in another post.) I may, of course, be forgetting something myself though.

x = The first coordinate (in S) of a point, is the distance along the first axis (of S) from the origin (of S) to that point. Do you agree?

Added - perhaps this was someone else...I cannot find the actual quote yet. Regardless, we need a definition for x to proceed, this discussion. If you disagree with the above definition, then please write what you believe is correct.

Since we are discussing x = x', I think we need to agree upon what x means first. Unless, there is dissent, I will assume you are all 100 percent fine with that definition of x, under the conditions that S(x,y,z) is given.
Last edited by steve waterman on Thu Aug 01, 2013 2:15 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Vetala » Thu Aug 01, 2013 2:15 pm UTC

steve waterman wrote:I would really like to have a common definition for x that we all agree upon.

[...snip...]
So, I will cease posting until we all agree upon what x means, including any new depictions. I too, am quite tired of having this discussion go round in circles.


You and I (and probably every other person who has read/contributed to this thread) know why that will never happen. Half of your approach to this "debate" is to misunderstand what's going on, and once something has been explained perfectly clearly you then turn around and say "well... but but but we never agreed on the definitions."

There's nothing to AGREE on, you've been told how the terms and variables are being used and what they mean in context. You keep insisting on "agreement" because you only want to define them in ways that contradict their use. You will NEVER find agreement because you refuse to define them as everybody else does, and nobody else is going to agree to define them in ways that deliberately contradict their use because that would be the most brilliant idea ever (if by "brilliant" we can all agree we mean "absolutely moronic". Or perhaps we can all mean "kumquat".. does that work in context? :roll: ).

Math doesn't care if you agree with it. It works. If you want to set up a system based off of bad definitions and misunderstandings and then claim math doesn't work because of it, you're not disproving anything. Really, everyone here is trying to help you - when even a self-professed math-challenged moderator can make sense of this stuff, how do you think it looks when a self-proclaimed mathematician can't? Either this really isn't making sense to you - in which case many people here are willing to help you understand it, or you're deliberately misunderstanding it - in which case you're just sabotaging yourself ever being taken seriously as a mathematician. Anyone who reads any of your arguments over this topic will have to doubt anything you say because they won't know if you're using the same math as everyone else, or something different and calling it the same. Give up on the whole "we need to agree" kick, use the definitions that are obviously intended, and see if there's still a problem.

Spoiler:
There won't be.

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

Postby oauitam » Thu Aug 01, 2013 2:22 pm UTC

Steve,

You are standing in the middle of the road.
10 feet further up the middle of the road is an orange.
10 feet further along past the orange is a car, heading towards the orange, and you, at a constant speed of 5 feet per second.

You can work out when the orange will get squashed by the car.


Congratulations, you have just used the "Galilean equation", and probably because it was blindingly obvious to you.

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

Postby ElWanderer » Thu Aug 01, 2013 2:29 pm UTC

oauitam wrote:Steve,

You are standing in the middle of the road.
10 feet further up the middle of the road is an orange.
10 feet further along past the orange is a car, heading towards the orange, and you, at a constant speed of 5 feet per second.

You can work out when the orange will get squashed by the car.


Congratulations, you have just used the "Galilean equation", and probably because it was blindingly obvious to you.

Someone tried something very similar to this during the Pressures thread. Steve's response was that the car's windscreen was always an arm's length in front of the car driver, or words to that effect...
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Thu Aug 01, 2013 2:54 pm UTC

steve waterman wrote:Added - perhaps this was someone else...I cannot find the actual quote yet. Regardless, we need a definition for x to proceed, this discussion. If you disagree with the above definition, then please write what you believe is correct.

Since we are discussing x = x', I think we need to agree upon what x means first. Unless, there is dissent, I will assume you are all 100 percent fine with that definition of x, under the conditions that S(x,y,z) is given.


It is the term S(x,y,z) that defines x.
x is the first coordinate (in the coordinate system S) of the point S(x,y,z).

Also, just in case it is the term "transformation" that is confusing you.
There are two types of transformation. To demonstrate, take a piece of paper, and cut out a large [, and place it in front of you and look at it.
Now move it to the right 10 cm. This is an active transformation.

Now put it back where it was before, and step to the left, by 10 cm. Relative to you, it will look like the [ has moved to the right 10 cm, just as before.
This is a passive transformation, and the Galilean transformation is of this sort. Note, when you step 10 cm to the left, the [ does not magically move with you.
As I said before, the purpose of the Galilean transformation is to describe how the same object or experiment looks different depending on the state of motion of the observer.

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

Postby Schrollini » Thu Aug 01, 2013 3:00 pm UTC

ElWanderer wrote:
oauitam wrote:Steve,

You are standing in the middle of the road.
10 feet further up the middle of the road is an orange.
10 feet further along past the orange is a car, heading towards the orange, and you, at a constant speed of 5 feet per second.

You can work out when the orange will get squashed by the car.


Congratulations, you have just used the "Galilean equation", and probably because it was blindingly obvious to you.

Someone tried something very similar to this during the Pressures thread. Steve's response was that the car's windscreen was always an arm's length in front of the car driver, or words to that effect...

Here is the post in question. Note that Steve actually does the Galilean transformation correctly; he just denies that this is what he's done. His problem isn't conceptual, it's notational. He's convinced himself that the symbols x and x' mean something different than what everyone else takes them to mean. Until he reconsiders this or lets us use different symbols, we can't make any progress. The most brilliant analogy won't do us any good, because Steve already has the idea. He just won't admit it, to us or to himself.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Thu Aug 01, 2013 3:24 pm UTC

Vetala wrote:
steve waterman wrote:I would really like to have a common definition for x that we all agree upon.

[...snip...]
So, I will cease posting until we all agree upon what x means, including any new depictions. I too, am quite tired of having this discussion go round in circles.


You and I (and probably every other person who has read/contributed to this thread) know why that will never happen. Half of your approach to this "debate" is to misunderstand what's going on, and once something has been explained perfectly clearly you then turn around and say "well... but but but we never agreed on the definitions."

There's nothing to AGREE on, you've been told how the terms and variables are being used and what they mean in context. You keep insisting on "agreement" because you only want to define them in ways that contradict their use. You will NEVER find agreement because you refuse to define them as everybody else does, and nobody else is going to agree to define them in ways that deliberately contradict their use because that would be the most brilliant idea ever (if by "brilliant" we can all agree we mean "absolutely moronic". Or perhaps we can all mean "kumquat".. does that work in context? :roll: ).

Math doesn't care if you agree with it. It works. If you want to set up a system based off of bad definitions and misunderstandings and then claim math doesn't work because of it, you're not disproving anything. Really, everyone here is trying to help you - when even a self-professed math-challenged moderator can make sense of this stuff, how do you think it looks when a self-proclaimed mathematician can't? Either this really isn't making sense to you - in which case many people here are willing to help you understand it, or you're deliberately misunderstanding it - in which case you're just sabotaging yourself ever being taken seriously as a mathematician. Anyone who reads any of your arguments over this topic will have to doubt anything you say because they won't know if you're using the same math as everyone else, or something different and calling it the same. Give up on the whole "we need to agree" kick, use the definitions that are obviously intended, and see if there's still a problem.

Spoiler:
There won't be.


So, it appears that you all refuse to define x in words, and assert x is not the abscissa of S, given S(x,y,z).
You assert P has mapped coordinates S(x,y,z), yet the manifold no origin.
You assert x axis = x' axis in the manifold, yet x' = x-vt.
You will not tell me if YOUR x is the abscissa or the axis, or why x is not the given axis, and the given abscissa of S.

So be it. Then this discussion/thread is over for me.

I leave the thread with this, ( unless there is a willingness to mutually define x in words )
Given S(x,y,x) coincident S'(x',y',z'),
x = the distance along the first axis (of S) from S(0,0,0).
x' = the distance along the first axis (of S') from S'(0,0,0).
the x axis is of infinite length and ALWAYS intersects S(0,0,0).
x axis and x' axis and y axis and y' axis and z axis and z' axis ALL intersect in the manifold at one point,
which is not an origin. I personally view this denial of these intersections as the origin as pure mathematical bullshit!

PM me directly, if so inclined. I do not even want to look at this thread to see what parting shots might be taken.
However, I probably will, after a while, out of sheer curiosity.
Time to reflect what I have learned recently about your manifold, and manifest that understanding onto my site.

Sorry, but without a working definition for x, the idea of continuing, is totally useless, and will only bring out
more confusion, frustration, anxiety, anger, and negative thoughts from all sides and angles.

added -
Schrollini wrote: His problem isn't conceptual, it's notational.

Indeed, 100 percent correct!

Given S(x,y,z) What does x mean mathematically to YOU! Please just tell me, without me having to keep guessing at what YOUR x means. You can disagree with my defined x, fine...but PLEASE PLEASE PLEASE, tell what YOUR x is, given S(x,y,z).
In the manifold x is...??????????
in S, x is...??????????

I am looking for the words to replace ????????, and NOT the notation!!!
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby addams » Thu Aug 01, 2013 3:45 pm UTC

Sorry, but without a working definition for x, the idea of continuing, is totally useless, and will only bring out
more confusion, frustration, anxiety, anger, and negative thoughts from all sides and angles.

That is funny.
Geometry? All sides and Angles?
Funny. very funny.

How many sides and angles are there?
Geometry Always sets limits. Right?

I know nothing about Geometry.
I know All sides and angles is Doable when we set limits.

All sides and angles without limits is Funny.
I listened to a Number's Guy. Or; He was a comedian.

He said there is an imaginary shape that exists in 165 different dimensions.
His numbers. His mind. Other people were going agone with it. (shrug)

I handle 3D poorly at best. 162 more would be a bit much for me.
All Sides and Angles! That is so funny. It is a good thing this is 2D.

One more D and someone would find that Program a new Hobby.
How about Pick Up Teeth?

I am sure you are learning a ton. I can't do the Math.
Why does Waterman? Is it a Program to ask question that challenge fundamental understanding?

Good Program?
Have the Regular Posters Reinforced complex ideas within their own knowledge base?

Did you learn by attempting to Teach?
What fun! Good Math People. Good.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Thu Aug 01, 2013 3:51 pm UTC

steve waterman wrote:Given S(x,y,z) What does x mean mathematically to YOU!

You're asking a circular question. To answer it, we have to know what S(x,y,z) is, and to know that, we have to know what S, x, y, and z are, which is the question your asking.

In conventional notation, S is a coordinate system, and x, y, and z are real numbers. Therefore, S(x,y,z) is a point in a manifold.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Thu Aug 01, 2013 4:23 pm UTC

steve waterman wrote:So, it appears that you all refuse to define x in words, and assert x is not the abscissa of S, given S(x,y,z).
I did exactly that, here, pretty much right after you asked the question.

I had also elaborated on it here, with special emphasis on the notational confusion you seem to be having.

Does this not make sense to you?

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

Postby eSOANEM » Thu Aug 01, 2013 4:25 pm UTC

steve waterman wrote:So be it. Then this discussion/thread is over for me.


I wondered how long that would take. Anyway, we all know it's not true. How many times did Steve say that in the pressures thread?

steve waterman wrote:I leave the thread with this, ( unless there is a willingness to mutually define x in words )
Given S(x,y,x) coincident S'(x',y',z'),
x = the distance along the first axis (of S) from S(0,0,0).
x' = the distance along the first axis (of S') from S'(0,0,0).
the x axis is of infinite length and ALWAYS intersects S(0,0,0).
x axis and x' axis and y axis and y' axis and z axis and z' axis ALL intersect in the manifold at one point,
which is not an origin. I personally view this denial of these intersections as the origin as pure mathematical bullshit!


No. Everyone agrees that this is an origin.

Emphasis on "an". Where we disagree is that we know that this point is not inherent to the manifold, just the co-ordinate system we choose.

steve waterman wrote:In the manifold x is...??????????


Nothing.

x has no meaning to the manifold, only to the co-ordinate system (and only because we defined it as S(x,y,z) and not A(flibberty-gibbet,gobbledy-gook,hufflepuff)).
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Thu Aug 01, 2013 4:27 pm UTC

Schrollini wrote:
steve waterman wrote:Given S(x,y,z) What does x mean mathematically to YOU!

You're asking a circular question. To answer it, we have to know what S(x,y,z) is, and to know that, we have to know what S, x, y, and z are, which is the question your asking.

In conventional notation, S is a coordinate system, and x, y, and z are real numbers. Therefore, S(x,y,z) is a point in a manifold.

Many thanks R. Now I wish to get a clearer understanding of what you are taking x to mean...okay?
That is, I wish to discern the parameters for x, and these further questions should help me.
If, they seem circular, then please do say so, as you did above.

1 Given S(x,y, z)...are x and y and z coordinates wrt S?
2 Given S(x,y, z) and given P = S(2,0,0)...are x and y and z coordinates wrt S or wrt P or wrt to both S and P?
3 Given S(x,y, z), does the x coordinate always represent the distance from its own origin S(0,0,0), along its own x axis?
4 Given S(2,0,0) is at point P in a manifold, could we also have a point O at S(0,0,0) in the manifold, or does the manifold only ever have just one point/event??
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Schrollini » Thu Aug 01, 2013 4:39 pm UTC

steve waterman wrote:1 Given S(x,y, z)...

I'm going to stop you right there, because I suspect what you wrote isn't what you mean. What you wrote is essentially, "Given a point". I think what you mean to say is, "Given a coordinate system and some coordinates." Or, if you want to introduce some symbols, "Given a coordinate system S and coordinates x, y, and z." But again, this is assuming that you're using conventional notation. To avoid confusion, you should specifically say what each symbol you introduce means.

(Note that sometimes we refer to a function f, and sometimes we refer to a function f(x). In the latter case, the x is just acting as a dummy, indicating that f takes a single argument. It has no meaning. I will try to avoid this notation, and I encourage everyone else to avoid it as well.)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby beojan » Thu Aug 01, 2013 5:01 pm UTC

Steve, here is a link to Khan Academy's explanation of functions, and function notation. I would like you to watch the first two videos in that playlist (you can watch the rest too, but the first two are important here).

Perhaps then you will be able to ask questions that make a little more sense.

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

Postby ucim » Thu Aug 01, 2013 5:29 pm UTC

steve waterman wrote:1 Given S(x,y, z)...are x and y and z coordinates wrt S?
2 Given S(x,y, z) and given P = S(2,0,0)...are x and y and z coordinates wrt S or wrt P or wrt to both S and P?
3 Given S(x,y, z), does the x coordinate always represent the distance from its own origin S(0,0,0), along its own x axis?
4 Given S(2,0,0) is at point P in a manifold, could we also have a point O at S(0,0,0) in the manifold, or does the manifold only ever have just one point/event??

Schrollini is right. "Given S(x,y,z)" does not actually mean what you think it does. But if rephrased as "Given a mapping S to the three dimensional manifold M, whose coordinates are called x, y, and z respectively" (which is what I think you have in your head),

then

1: yes.
x, y, and z are coordinates of points in M wrt the mapping S.
Spoiler:
This is only because we said so in the definition. If we had said: "Given a mapping S to the three dimensional manifold M, whose coordinates are called moose, squirrel, and natasha respectively", then moose, squirrel, and natasha would be coordinates of points in M wrt the mapping S. In that case, x, y, and z would have no meaning.

2: neither.
x, y, and z are coordinates of point P (which is in M), wrt the mapping S
Spoiler:
P is the specific point being referred to, S is the mapping function that allows us to refer to specific points using coordinates.

3: sort of, given a suitable metric. Also, distance is from something, to something else. You need to specify what it is to.
Assuming the normal pythagorean metric, the numeric value of x would be equal to the distance in question. It's not however what x "means".
Spoiler:
The distance would be from the origin, to the point S(x,0,0), which is the projection of S(x,y,z) onto the x axis. Since we defined x as part of the definition of S, I'll accept "along its own x axis" as a valid utterance, even though it's not the way mathematicians or physicists would say it. They would just say "along the x axis", since there is only one.

4: this is confuddled.
S(2,0,0) is the name of a specific point (in the manifold) which we've chosen to call point P. It is said that S(2,0,0) is point P. Think of P as a nickname for S(2,0,0). There are an infinite number of points in the manifold. One of them is the point S(0,0,0), which we call the origin of S. We can call it point O if we like. That point is in the manifold M along with all the other points, but is not special to M, any more than point P is special to M.
Spoiler:
Point O is neat and useful if we are dealing with mapping S. However, point O is probably of little interest if we are using a different mapping, such as S' or T (neither of which is defined yet, but could be). The mapping S' and the mapping T, when suitably defined, will also have an origin; that origin is likely to be a different point (in the mainfold) than point O, so will need a different nickname if we want to refer to it as anything other than S'(0,0,0) or T(0,0,0).
Jose
edit: formatting, clarifying #3
Last edited by ucim on Fri Aug 02, 2013 7:51 pm UTC, edited 5 times in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Pfhorrest » Thu Aug 01, 2013 6:59 pm UTC

Steve, did you miss this? This explains exactly what we mean by x in your own setup, and how it differs from what you mean by x. Also x' and vt (or d if you like).

Since you asked for it "in words" specifically today, I'll put it in text here too:

When we are considering a point P and its coordinates in two coordinate systems S and S'

x' = the place on the first axis of S' (the x' axis) which lines up with P
x = the place on the first axis of S (the x axis) which lines up with P
vt = d = the place on the first axis of S (the x axis) which lines up with the origin of S'

In your first setup depicted below (the non-coincident one), that means:

-1 = x' = the place on the first axis of S' (the x' axis) which lines up with P
2 = x = the place on the first axis of S (the x axis) which lines up with P
3 = vt = d = the place on the first axis of S (the x axis) which lines up with the origin of S'

In your second setup depicted below (the coincident one), that means:

2 = x' = the place on the first axis of S' (the x' axis) which lines up with P
2 = x = the place on the first axis of S (the x axis) which lines up with P
0 = vt = d = the place on the first axis of S (the x axis) which lines up with the origin of S'

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

Postby SecondTalon » Thu Aug 01, 2013 8:25 pm UTC

Schrollini wrote:Here is the post in question. Note that Steve actually does the Galilean transformation correctly; he just denies that this is what he's done. His problem isn't conceptual, it's notational. He's convinced himself that the symbols x and x' mean something different than what everyone else takes them to mean. Until he reconsiders this or lets us use different symbols, we can't make any progress. The most brilliant analogy won't do us any good, because Steve already has the idea. He just won't admit it, to us or to himself.

Do you realize how hard it is to not insist that all axis notation now be named after the Three Tenors?

vt = 0
S(Luciano, Placido, Jose) = 'S(Pavarotti, Domingo, Carreras)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Dark Avorian » Fri Aug 02, 2013 12:43 am UTC

SecondTalon wrote:
Schrollini wrote:Here is the post in question. Note that Steve actually does the Galilean transformation correctly; he just denies that this is what he's done. His problem isn't conceptual, it's notational. He's convinced himself that the symbols x and x' mean something different than what everyone else takes them to mean. Until he reconsiders this or lets us use different symbols, we can't make any progress. The most brilliant analogy won't do us any good, because Steve already has the idea. He just won't admit it, to us or to himself.

Do you realize how hard it is to not insist that all axis notation now be named after the Three Tenors?

vt = 0
S(Luciano, Placido, Jose) = 'S(Pavarotti, Domingo, Carreras)


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

Postby WibblyWobbly » Fri Aug 02, 2013 1:53 am UTC

SecondTalon wrote:
Schrollini wrote:Here is the post in question. Note that Steve actually does the Galilean transformation correctly; he just denies that this is what he's done. His problem isn't conceptual, it's notational. He's convinced himself that the symbols x and x' mean something different than what everyone else takes them to mean. Until he reconsiders this or lets us use different symbols, we can't make any progress. The most brilliant analogy won't do us any good, because Steve already has the idea. He just won't admit it, to us or to himself.

Do you realize how hard it is to not insist that all axis notation now be named after the Three Tenors?

vt = 0
S(Luciano, Placido, Jose) = 'S(Pavarotti, Domingo, Carreras)

What, then, is the Luciano-coordinate of Andrea Bocelli when the Tenors play Florence?

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

Postby ucim » Fri Aug 02, 2013 2:15 am UTC

WibblyWobbly wrote:What, then, is the Luciano-coordinate of Andrea Bocelli when the Tenors play Florence?
B flat. Which is the same as A sharp. So, now that we know B=A, we can rename the system Lucibno(Aocelli). So if I move Florence to New York, does that make all tenors into altos? (or would that be bltos?)

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

Postby WibblyWobbly » Fri Aug 02, 2013 2:39 am UTC

ucim wrote:
WibblyWobbly wrote:What, then, is the Luciano-coordinate of Andrea Bocelli when the Tenors play Florence?
B flat. Which is the same as A sharp. So, now that we know B=A, we can rename the system Lucibno(Aocelli). So if I move Florence to New York, does that make all tenors into altos? (or would that be bltos?)

Jose

I AM NOT TALKING ABOUT PHYSICS. ALL I WANT IS PURE MATH.

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

Postby ucim » Fri Aug 02, 2013 2:47 am UTC

WibblyWobbly wrote:
ucim wrote:
WibblyWobbly wrote:...So, now that we know B=A, we can rename the system...

I AM NOT TALKING ABOUT PHYSICS. ALL I WANT IS PURE MATH.

Sorry. I'll change it to pure math.

...So, now that we know y=x, we can rename the system...

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

Postby Dark Avorian » Fri Aug 02, 2013 2:52 am UTC

ucim wrote:
WibblyWobbly wrote:
ucim wrote:
WibblyWobbly wrote:...So, now that we know B=A, we can rename the system...

I AM NOT TALKING ABOUT PHYSICS. ALL I WANT IS PURE MATH.

Sorry. I'll change it to pure math.

...So, now that we know y=x, we can rename the system...

Jose


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

Postby eran_rathan » Fri Aug 02, 2013 11:23 am UTC

SecondTalon wrote:
Schrollini wrote:Here is the post in question. Note that Steve actually does the Galilean transformation correctly; he just denies that this is what he's done. His problem isn't conceptual, it's notational. He's convinced himself that the symbols x and x' mean something different than what everyone else takes them to mean. Until he reconsiders this or lets us use different symbols, we can't make any progress. The most brilliant analogy won't do us any good, because Steve already has the idea. He just won't admit it, to us or to himself.

Do you realize how hard it is to not insist that all axis notation now be named after the Three Tenors?

vt = 0
S(Luciano, Placido, Jose) = 'S(Pavarotti, Domingo, Carreras)


Shouldn't the Axis notation be:

S(Germany, Japan, Italy) = S'(DPRK, Iran, Iraq)?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby oauitam » Fri Aug 02, 2013 11:39 am UTC

Defining x in words...

Here is what we know;
On the 1st of August 2000, at exactly midday, you are standing in the middle of the road.
On the 1st of August 2000, at exactly midday, 10 feet further up the middle of the road is an orange.
On the 1st of August 2000, at exactly midday, 10 feet further along past the orange is a car, heading towards the orange, and you, at a constant speed of 5 feet per second.

Here's what we'd like to know;
When does the car reach the orange?

In other words, the time that interests us is the time when the nose of the car is in the same place as the centre of the orange.


This is so simple we don't even bother imagining an equation in day-to-day life.
If you want to describe an equation, and use the labels that you used above when saying x' = x-vt then, in words;

x = how many feet forward the car would have to travel so that the nose of the car would be in the same place as the centre of the orange, at exactly midday on the 1st of August 2000.
v = the speed of the car, in feet per second, measured in a straight line down the road in the car's forward direction.
t = the time that interests us, measured in seconds after exactly midday on the 1st of August 2000.
x' = how many feet forward the car would have to travel so that the nose of the car would be in the same place as the centre of the orange, at the time that interests us.



I can't believe I'm typing this for grown-ups but;
Notice that it is a time that interests us and that the car is moving, so it's at different places at different times.

Talking about the position of the car is as half-baked as me asking you, "How old were you?"
You would instinctively reply, "When?".
Just as you should when talking about the position of moving things.


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