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

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

Postby Izawwlgood » Sun Nov 10, 2013 6:14 pm UTC

I love this thread, it's been an interesting insight into a bunch of people, as well as an indication of how stubborn the delusional can be.

The only thing I wanted to add is this new 'appeal to maturity'; Steve, I urge you to consider that it is perhaps the fora's youth and your advanced aged (the last being somewhat glib of course, but you are older than most posters here) as a sign that your thinking is disconnected, stagnant, removed from the education and information and rigorous thinking that is a product of being fresh from academia, or even still in it. It is an INCREDIBLY bad scientist who looks at the progress of younger academics and tells them they're wrong, especially when you yourself seem to have misunderstood the basics. Respectfully, as a 29 year old graduate student, I'm telling you you need to work on your listening and comprehension, because your increased insistence on not doing so will result in an increased intellectual isolation, that seems to have already significantly impaired your ability to participate in and especially contribute to academic discussion.

I'm at a conference now, and something that distinctly marks the good older members is how excitedly they listen to the undergrads and grad students poster sessions. You're not even investigating a new field of mathematics; you're arguing from a position of a lack of understanding, and now asking for the age of posters as some sort of proof that they don't know what's going on. It's pretty appallingly backwards actually.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby cyanyoshi » Sun Nov 10, 2013 7:52 pm UTC

steve waterman wrote:
ucim wrote:Steve, when you say x=x', you must also staple your two rulers together at that point.... x in (ruler) S gets stapled to x' in (ruler) S'. This ensures that x will always equal x'.

Now move the rulers around.

Jose

The Galilean example however would do this...
1 commence with the two rulers imagined to be in the same place ( coincident )
2 moves one ruler
s since x is the distance from the zero inches location and not just the point at x, at that same vector head,
then of course that ensues that x will always equal x'.


The Galilean is not what you think it is, then. I'll explain as clearly as I can what is going on when everyone talks about the Galilean transformation. This is not a matter of opinion. The Galilean transformation is defined a specific way for a specific scenario, and arguing otherwise is a bit like someone insisting that 2+3=6 when he interprets "+" to mean multiplication. It may be consistent and work all fine and dandy in his head, but there will be no hope to communicate with the rest of us.

Imagine there are two rulers: a purple ruler, and a yellow ruler. The purple ruler is lying on the ground, and that the yellow ruler is moving along the length of the purple ruler with a known speed, 2 inches per second. We will call the time that they are on top of teach other "0". So t=0 refers to the time where the two rulers are coincident.

Place a dot on the ground near the purple ruler at the 10 inches mark (let's call this number x). When the rulers are on top of each other, then both rulers will read the location as "10 inches" (i.e. the distance from the left end of each ruler to the dot is 10 inches). I think we all agree here.

However, something interesting happens when you move the yellow ruler to the right a bit. For example, after one second, the yellow ruler measures the point the be 8 inches from its zero mark (we will call the yellow ruler's measurement x'). This is what the Galilean transformation is all about. If we know what the reading on the purple ruler is (x = 10 inches), the speed of the yellow ruler (v = 2 inches per second), and the time since the rulers were on top of each other (t=1 second), then you can calculate the reading on the second ruler (x' = x - vt = 8 inches).

Notice that x' takes on a different value for each value of t you put into the equation. This is because the dot isn't on the yellow ruler; the dot is on the ground. Since you seem to like thinking of distances, the Galilean transformation implies that the distance from the dot to the left end of the yellow ruler (x') changes whenever the yellow ruler moves, even if the distance from the dot to the left end of the purple ruler (x) remains the same. If you want to talk about the Galilean transformation, please understand what the rest of us are talking about, and try to speak in our language. It doesn't do anyone any good to insist that the definition of the Galilean transformation is wrong; it is what it is. If you are talking about something else, then please don't call it the Galilean. It is confusing.

steve waterman wrote:Care to mention you age?

Chances are that if someone is old enough to use these forums, they are experienced enough have a basic understanding of the Galilean transformation. Everything here is simple algebra. I don't see how age is terribly relevant in this situation. But before you ask, I am 21 years old and a tutor for calculus and physics at my university. Please tell me if anything I typed is unclear or does not make any sense.

steve waterman wrote:btw, the red dot on the ruler ( or not on the ruler ) under the mattress is critical, and of infinite interest.

Maybe. But again, that is so not the purpose of the Galilean transformation. The big thing about coordinate transformations is that there is only one dot and it doesn't move. What "moves" is the ruler that is measuring where the dot is. If you are talking about something else, then you are not referring to a coordinate transformation.

I realize that this post has a 0.01% chance of changing your mind, Steve, but maybe it will be helpful to someone, somewhere.

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

Postby steve waterman » Mon Nov 11, 2013 12:59 am UTC

cyanyoshi wrote: The big thing about coordinate transformations is that there is only one dot and it doesn't move. What "moves" is the ruler that is measuring where the dot is. If you are talking about something else, then you are not referring to a coordinate transformation.

I wish to have an exclusive discussion with cyanyoshi, for a few posts, ( maybe as many as a half dozen, please. ) That is, I am ONLY asking cyanyoshi for a response to the following three statements.

1 Cartesian coordinate systems existed in the 1800's, thus existed before the Galilean transformation had been created/conceived.
Agreed?
2 A Cartesian coordinate system contains an infinite set of points, including the point (0,0,0) called the origin.
Agreed?
3 Given coincident systems S and S',
mathematically allow a point P at (2,0,0) in/wrt S and also allow a point P' at (2,0,0) in/wrt S'.
Allowed?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby yurell » Mon Nov 11, 2013 1:36 am UTC

steve waterman wrote:1 Cartesian coordinate systems existed in the 1800's, thus existed before the Galilean transformation had been created/conceived.
Agreed?


Cartesian co-ordinate systems were invented in the 17th century, but Galilean relativity pre-dates this invention. That said, the formalism of the transform was made for Cartesian co-ordinate systems, yes.

steve waterman wrote:2 A Cartesian coordinate system contains an infinite set of points, including the point (0,0,0) called the origin.
Agreed?


False. Points are contained on a manifold. A co-ordinate system contains an infinite number of co-ordinates. (0,0,0) is a co-ordinate called the origin.

steve waterman wrote:3 Given coincident systems S and S',
mathematically allow a point P at (2,0,0) in/wrt S and also allow a point P' at (2,0,0) in/wrt S'.
Allowed?


Point P and P' are the same point if S is co-incident with S' and each are mapped to by the same co-ordinates.
Last edited by yurell on Mon Nov 11, 2013 1:40 am UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Izawwlgood » Mon Nov 11, 2013 1:39 am UTC

I won't interfere with that, but I would like an answer to why you felt bringing age up was suddenly, somehow, appropriate. I don't really expect a response to my assertion that it was intellectually dishonest and dangerously backwards, but I do think you ought to reconsider your intent with respect to your approach to this discussion.

If you're looking for sympathy because of your age, or some sort of age camaraderie, I think you're somewhat abandoning the intention of intellectual debate, but then, I'm not convinced you're really after such a thing in the first place.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Nov 11, 2013 1:57 am UTC

Izawwlgood wrote:I won't interfere with that, but I would like an answer to why you felt bringing age up was suddenly, somehow, appropriate. I don't really expect a response to my assertion that it was intellectually dishonest and dangerously backwards, but I do think you ought to reconsider your intent with respect to your approach to this discussion.

If you're looking for sympathy because of your age, or some sort of age camaraderie, I think you're somewhat abandoning the intention of intellectual debate, but then, I'm not convinced you're really after such a thing in the first place.


As I said, just curious. I also said that no matter what age, one can still be (100 percent) logical.

Yurell, I asked you to let me have a few posts ONLY with cyanyoshi...so I am not going to respond to you until after that discussion is over. It is a nightmare arguing against more than a couple of people at a time. I did that for thousands of posts and no longer wish to do so.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Izawwlgood » Mon Nov 11, 2013 2:06 am UTC

Then it is curious that you associated surprise at the age thread distribution with the 'unified front of opposition' that you were receiving here.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby cyanyoshi » Mon Nov 11, 2013 2:22 am UTC

Sorry Steve, but I won't be able to get into a serious conversation until probably tomorrow. I am a little busy at the moment.

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

Postby yurell » Mon Nov 11, 2013 2:25 am UTC

Sorry, I didn't notice that; only saw the assertions and so corrected them.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Nov 11, 2013 12:15 pm UTC

cyanyoshi,

In an attempt for you and I to have a valid discussion, I am initially trying to establish a working scenario for the terms "point" and "coordinate system" that we can mutually agree upon.

adding these two statements to the three statements previous posted...

4 Given a Cartesian coordinate system with a line drawn between point A(1,1) and point B(-1,-1).
Allowed? ( noting that there is only just the one system in this mathematical statement )

5 IFF number 4 is allowed, we could make a distinction between
the "coordinate point" (1,1) [ btw, I am quite aware that Relativity has no such term ]
and the "named point" A(1,1).
Do you allow these two differentiating terms, cyanyoshi, for the sake of our discussion?
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Re: Carry on my wayward son

Postby eran_rathan » Mon Nov 11, 2013 12:36 pm UTC

Schrollini wrote:What devilry is this? Steve Waterman, late of these parts, vowed ne'er to return, doth appear. Speak, O specter -- what tidings doth ye bring? What witchcraft hath doom'd thy soul to haunt this message board?

My friends, my compatriots, who I hold to my breast, remember ye this: For as the sun need not a cloud to shine, neither doth a coordinate system need a metric. Verily I tell you, thou mustn't consider distances whilst transforming thy coordinates. And while thy coordinate systems and metric spaces are oft together, as the sun and the cloud doth share the heavenly sphere, thou must keep them separate in thy thoughts, lest foul tidings befall you!

Good day, and thus, good bye.


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

Postby PolakoVoador » Mon Nov 11, 2013 1:13 pm UTC

Steve, please be honest here: do you really want to learn something, or do you just want to preach your ideas until someone, anyone, agrees with you?


For the rest of the people here: guys/gals, we all know the answer to the above question. Steve already stated he does not wish for a honest intellectual argument. There is no point trying to teach him anymore. If all that was amazingly done, explained, and said for the last threads and thousands of posts was not enough to make a dent in Steve's position, I don't believe there's anyway we could change his views. For the last dozen (more, maybe?) of pages, it's just a matter of how much we like listening to our own voices (ok, reading our own words), because I really doubt anyone here still thinks Steve can be persuaded out of his beliefs.

Or maybe it just a severe case of this:
Spoiler:
Image


Anyway, it should be clear it is a pointless persuit by now.

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

Postby ucim » Mon Nov 11, 2013 3:41 pm UTC

steve waterman wrote:
ucim wrote:Steve, when you say x=x', you must also staple your two rulers together at that point.... x in (ruler) S gets stapled to x' in (ruler) S'. This ensures that x will always equal x'.

Now move the rulers around.

Jose

The Galilean example however would do this...
1 commence with the two rulers imagined to be in the same place ( coincident )
2 moves one ruler
s since x is the distance from the zero inches location and not just the point at x, at that same vector head,
then of course that ensues that x will always equal x'.
No, that's not what the actual Galilean does. I agree, as does pretty much everyone here, that the distance (appropriately defined*) between two coordinates in a coordinate system does not change just because the coordinate system is "moved around". However, I do not see the utility of this observation.

Distance is measured between two points. If there are two distances involved, there are four points (two pairs). If we are talking about the kinds of things the Galilean talks about, one point is shared between the two pairs. We ask "how far is this from here" and "how far is this from there", where "this" is the shared point, and where "here" and "there" are different. We transform between the "here" system and the "there" system.

You are perhaps conflating this with the relativistic idea that the rulers themselves change size. However that is not the case here.

(*)
Spoiler:
I am gliding over the fact that "coordinate systems do not move" and am using what I think you are imagining when you use that phrase, despite its lack of mathematical rigor)
steve waterman wrote:Care to mention you age?
I do not post my age on public media because of the large number of abusive robotic data scavengers on modern networked computer systems. However, I will say that I was three or four years older than four years old when the Beatles first came to New York. I was already out of college when John Lennon was shot.

steve waterman wrote:btw, the red dot on the ruler ( or not on the ruler ) under the mattress is critical, and of infinite interest.
Why?

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

Postby steve waterman » Mon Nov 11, 2013 5:10 pm UTC

cyanyoshi, [ others are asked to please hold off comments until I get a few responses from cyanyoshi first, thanks ]

I will skip to the bottom line...

Given S(x,y,z)...x = distance from S(0,0,0) to S(x,0,0) and
Given S'(x',y',z')...x' = distance from S'(0,0,0) to S'(x',0,0) and
Given S and S' coincident with x = x'.

Allow S' or S to move by vt along the common x/x' axis, in which case, it is absolutely obvious that x = x'.
Therefore, one can deduce that x' = x-vt is mathematically incorrect unless vt = 0, since x = x'.

Just because this disagrees with the Galilean transformation equation results does not make my logic invalid. Just because point P in your manifold shares the same spatial location has zero impact upon the absolute fact that x = x'.

I do not comprehend why this simple logic is sooo difficult for everyone here at xkcd to grasp/agree with.
Prove to me that x does not equal x' after vt gets applied using the definitions above for x and x', or at least explain why you disagree with that definition for x and x'.

btw, combining these two threads on this topic, has now reached an unbelievable 6000 posts.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PeteP » Mon Nov 11, 2013 5:35 pm UTC

steve waterman wrote:Given S(x,y,z)...x = distance from S(0,0,0) to S(x,0,0) and
Given S'(x',y',z')...x' = distance from S'(0,0,0) to S'(x',0,0) and
Given S and S' coincident with x = x'.

Allow S' or S to move by vt along the common x/x' axis, in which case, it is absolutely obvious that x = x'.
Therefore, one can deduce that x' = x-vt is mathematically incorrect unless vt = 0, since x = x'.

Because x and x' are chosen so that S'(x',y',z')=S(x,y,z) is fulfilled, therefore x doesn't equal x' when vt isn't zero.

(Yes you asked others not to answer, but it's not like you actually seriously consider objections. So in effect I may have as well written nothing.)

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

Postby ahammel » Mon Nov 11, 2013 5:58 pm UTC

Steve: the discussion behind the spoiler is entirely for my own edification. You may defer opening until after cyanyoshi has responded or, if you prefer, indefinitely.

Spoiler:
Just checking to see if I understand coordinate transforms any better than Steve by now.

steve waterman wrote:Given S(x,y,z)...x = distance from S(0,0,0) to S(x,0,0) and
Given S'(x',y',z')...x' = distance from S'(0,0,0) to S'(x',0,0) and
Wait, didn't he just define x the distance in terms of x element of a coordinate? Is that kosher? It's certainly potentially confusing. I'm going to go ahead and call the distances p and q.

(By the way, if we've got a coordinate, S(x,y), is it correct to say that x is an element of that coordinate, or is there some other term for that relationship?)

Given S and S' coincident with x = x'.

Allow S' or S to move by vt along the common x/x' axis[...]
It's a little weird to say that S and S' are coincident and then 'move' one of them, but OK. I guess that's Steve's way of telling us that the two coordinate systems are related by a translation rather than a shear.

Don't know why we we're representing a constant distance by what looks like two previously undefined things multiplied by one another either, but I'll assume that vt is some constant.

in which case, it is absolutely obvious that [p] = [q].
Looks correct to me. A translation between coordinate systems doesn't change the scaling of the axes.

Therefore, one can deduce that x' = x-vt is mathematically incorrect unless vt = 0, since [p] = [q].
I think swapping out the symbols adequately highlights the equivocation. Steve's result is about distances, and coordinate transformations, uh, transform coordinates.

Just because this disagrees with the Galilean transformation equation results does not make my logic invalid. Just because point P in your manifold shares the same spatial location has zero impact upon the absolute fact that [p] = [q].
In other words, this thing that Steve's concluded has nothing much to do with coordinate transformations, except with regards to what happens to the scaling of the axes when you perform a translation (viz. nothing).

I do not comprehend why this simple logic is sooo difficult for everyone here at xkcd to grasp/agree with.
Everybody grasps it just fine, it's just that Steve is convinced he's knocked over special relativity, is wrong, and interprets any attempt at correction as a misunderstanding of his argument.

Prove to me that [p] does not equal [q] after vt gets applied using the definitions above for [p] and [q], or at least explain why you disagree with that definition for [p] and [q].
As above, p and q are distances, x and x' are elements of coordinates. Coordinate transforms are about the relationship between coordinates identifying the same points in different coordinate systems, and results about distances are irrelevant.

Also, SR doesn't even use the Galilean transformation, so why are we even talking about this?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Mon Nov 11, 2013 6:14 pm UTC

steve waterman wrote:Allow S' or S to move by vt along the common x/x' axis, in which case, it is absolutely obvious that x = x'.
Therefore, one can deduce that x' = x-vt is mathematically incorrect unless vt = 0, since x = x'.

Just because this disagrees with the Galilean transformation equation results does not make my logic invalid. Just because point P in your manifold shares the same spatial location has zero impact upon the absolute fact that x = x'.

I do not comprehend why this simple logic is sooo difficult for everyone here at xkcd to grasp/agree with.

It's easy to grasp what's going on, and every regular here (except possibly use) has grasped it. You are insisting that the galilean is saying something it's not. Then, you are trying to explain how your version of the Gallilean is clearly not true.

We all agree that your version of the Gallilean is not true, but we don't care. Now, I know you want only cyanyoshi to respond, but your actual intent of the question is unclear, so I'm going to prod you for more specifics so that cyanyoshi can know what you are asking.

Let me quote the entry that you quoted from wikipedia.
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 ,


You are wanting to think in terms of distances, but you are taking a short cut and just using the distance from the origin. The important thing here is that the spatial origins coinciding at time t=t'=0. This means that S(0,0,0,0) and S'(0,0,0,0) refer to the same point. And if I pick any arbitrary values v1, v2, and v3, I know that S(v1,v2,v3,0) refers to the exact same point at S'(v1,v2,v3,0).

Got it? At t=0, they are spatially coincident.
But for coordinate systems where the spacial relationship is changing as time changes (thinking of "rulers moving along an x axis), at t=t' is not equal to zero, this will not hold true necessarily hold true. So S(0,0,0,t) is not referring to the same point as S'(0,0,0,t'). Also important, and the point you are missing, is that S'(0,0,0,0) is not referring to the same point as S'(0,0,0,1). They are different points.

So, quoting the gallilean again:
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 ,

You are trying to make x and x' refer to distances. What distance exactly? Please answer this question for cyanyoshi, so they can respond.
Let's use the example where, like the simplified galilean situation, we the only relative spatial motion in the coordinate systems is in the x portion, not the y or z.

At t=t'=0, we can look at the distance from S'(0,0,0,0) to S'(1,0,0,0), which is 1 (measurement a), and we can look at the distance from S(0,0,0,0) to S(1,0,0,0). (Measurement b.) This distance is also 1. Since at t=t'=0, we are looking at the same spatial coordinates, we are measuring the distance between the exact same two points.

Let say S' is moving in the x/x' diraction at a speed of 1. This means that At t=t'=1, S'(0,0,0,1) and S(1,0,0,1) refer to the same point. I could measure a few things here.
The distance between S'(0,0,0,1) and S'(1,0,0,1) is 1 (measurment c). The distance between S(1,0,0,1) and S(2,0,0,1) is also 1 (measurment d), and actually is referring to the same points we were look at in S'.

I could also look at the distance between S(0,0,0,1) and S(1,0,0,1), and this would also be 1 (measurement e), looking at the ruler, but this would be a different measurement than the one I made using the S1 ruler.

Please clarify what distances you are actually referring to so that cyanyoshi can appropriately answer. Draw diagram for every measurement. Use specific numbers, please.

In my diagram, there are five points
p1, at both S(0,0,0,0) and S'(0,0,0,0),
p2, at S(1,0,0,0) and S'(1,0,0,0)
p3, at S(0,0,0,1) and S'(-1,0,0,1)
p4, at S(1,0,0,1) and S'(0,0,0,1)
and
p5 at S(2,0,0,1) and S'(1,0,0,1)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Nov 11, 2013 8:27 pm UTC

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

Postby JudeMorrigan » Mon Nov 11, 2013 9:02 pm UTC

steve waterman wrote:
Po6ept wrote:You are trying to make x and x' refer to distances. What distance exactly? Please answer this question for cyanyoshi, so they can respond.

Given S(x,y,z) x is the abscissa where abscissa is the the value of the coordinate.
x is the distance from S(0,0,0) to S (x,0,0)

Groovy. Now why should the distance from S(0,0,0) to S(x,0,0) be the same as the distance from S'(0,0,0) to S'(x',0,0) if the axes are not coincdent?

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

Postby steve waterman » Mon Nov 11, 2013 9:13 pm UTC

JudeMorrigan wrote:
steve waterman wrote:
Po6ept wrote:You are trying to make x and x' refer to distances. What distance exactly? Please answer this question for cyanyoshi, so they can respond.

Given S(x,y,z) x is the abscissa where abscissa is the the value of the coordinate.
x is the distance from S(0,0,0) to S (x,0,0)

Groovy. Now why should the distance from S(0,0,0) to S(x,0,0) be the same as the distance from S'(0,0,0) to S'(x',0,0) if the axes are not coincident?

Simply because is was the Galilean given that at t = 0, x = x' so AFTER vt gets applied, x still equals x'.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PeteP » Mon Nov 11, 2013 9:30 pm UTC

No it doesn't because it's conditional on t=0 and thus doesn't hold if t changes? Or rather it's always x'=x-vt. So the question remains.

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

Postby Роберт » Mon Nov 11, 2013 9:41 pm UTC

steve waterman wrote:
Po6ept wrote:You are trying to make x and x' refer to distances. What distance exactly? Please answer this question for cyanyoshi , so they can respond.

Given S(x,y,z) x is the abscissa where abscissa is the the value of the coordinate.
x is the distance from S(0,0,0) to S (x,0,0)

So, for example, using the numbers I did in my diagram, let's look at point p4. S(1,0,0,1) and S'(0,0,0,1).
The distance from S(1,0,0,1) to the origin S(0,0,0,1) is 1. This is x.
The distance from S'(0,0,0,1) to the origin S'(0,0,0,1) is 0. This is x'.
Remember v was 1.
So here, the gallilean says
x'=x-vt
0 = 1 - (1*1)
0 = 0

What is your problem with this? Again, I'm not trying to make any assertions to you, I'm trying to get you to phrase what you are saying clearly, using diagrams and examples with real numbers, so that cyanyoshi can give a useful response when they get time.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Nov 11, 2013 9:43 pm UTC

PeteP wrote:No it doesn't because it's conditional on t=0 and thus doesn't hold if t changes? Or rather it's always x'=x-vt. So the question remains.

Obvious too, is that when t = 0, then x = x' for x' = x-vt. All that I am saying is that x = x' when t > 0, and that x' = x-vt is not true when t > 0. This really is not complicated at all once one accepts that x means the abscissa.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Mon Nov 11, 2013 9:48 pm UTC

steve waterman wrote:
JudeMorrigan wrote:
steve waterman wrote:
Po6ept wrote:You are trying to make x and x' refer to distances. What distance exactly? Please answer this question for cyanyoshi, so they can respond.

Given S(x,y,z) x is the abscissa where abscissa is the the value of the coordinate.
x is the distance from S(0,0,0) to S (x,0,0)

Groovy. Now why should the distance from S(0,0,0) to S(x,0,0) be the same as the distance from S'(0,0,0) to S'(x',0,0) if the axes are not coincident?

Simply because is was the Galilean given that at t = 0, x = x' so AFTER vt gets applied, x still equals x'.

That doesn't follow. Something being true at t = 0 does not necessarily mean that it is also true when t does not equal 0.

The issue here is still that S(x) and S'(x') refer to points that exist in the ... let's talk about euclidean planes, since the more generalized term bothers you. Euclidean planes are made up of an infinite set of points, right? They're a concept that has existed for several thousand years. If you lay two coordinate systems on the euclidean plane and then pick a specific point on it, why should the distance from the origin of one coordinate system to the point be the same as the origin of the other coordinate system to that point unless the coordinate systems are coincident?

It's like ... if I met up with you somewhere to buy you a tasty beverage to celebrate your mostly cheerful persistence in these threads and, while we were there, we looked up our distance to the city of London. While we were in the same pub, the distance would be pretty much the same for both of us. And then I went home to Alabama and then looked up my new distance to London. It would be different than what it was when we were both in the same pub, right? And why shouldn't it be?

Fundamentally, the Galilean (and coordinate transformations in general) are trying to ask an entirely different question than the one you're attempting to answer. Hence the disconnect.

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

Postby eran_rathan » Mon Nov 11, 2013 9:50 pm UTC

steve waterman wrote:
PeteP wrote:No it doesn't because it's conditional on t=0 and thus doesn't hold if t changes? Or rather it's always x'=x-vt. So the question remains.

Obvious too, is that when t = 0, then x = x' for x' = x-vt. All that I am saying is that x = x' when t > 0, and that x' = x-vt is not true when t > 0. This really is not complicated at all once one accepts that x means the abscissa.


Cart before the horse, mate.

x' = x -vt (for all t)
at t=0, x' = x.


not the other way around.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PeteP » Mon Nov 11, 2013 9:59 pm UTC

steve waterman wrote:
PeteP wrote:No it doesn't because it's conditional on t=0 and thus doesn't hold if t changes? Or rather it's always x'=x-vt. So the question remains.

Obvious too, is that when t = 0, then x = x' for x' = x-vt. All that I am saying is that x = x' when t > 0, and that x' = x-vt is not true when t > 0. This really is not complicated at all once one accepts that x means the abscissa.

They are the abscissa of S(x,0,0) and S'(x',0,0). S(x,0,0) = S'(x',0,0) since the point stays the same and the origin is different that means the distance between S'(0,0,0) and S'(x',0,0) is different when t changes. That means x' is different while x doesn't change, which means x doesn't equal x' if t isn't 0.

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

Postby Роберт » Mon Nov 11, 2013 10:04 pm UTC

steve waterman wrote:
PeteP wrote:No it doesn't because it's conditional on t=0 and thus doesn't hold if t changes? Or rather it's always x'=x-vt. So the question remains.

Obvious too, is that when t = 0, then x = x' for x' = x-vt. All that I am saying is that x = x' when t > 0, and that x' = x-vt is not true when t > 0. This really is not complicated at all once one accepts that x means the abscissa.

The question is, why are you asserting x always equals x'? That is obviously not the case when you look at my diagram. If the spatial coordinates are coincident, this will be true. So at t=t'=0, using arbitrary values v1, v2, v3, S(v1, v2, v3, 0) refers to the same point that S'(v1, v2, v3, 0) refers to. But look at my diagram, at point p4 in spacetime. S(1,0,0,1) and S'(0,0,0,1). The first coordinates, x and x', are clearly not equal in this case.

Can you explain to cyanyoshi why on earth you are asserting that x=x' when t>0?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Nov 11, 2013 10:04 pm UTC

eran_rathan wrote:
steve waterman wrote:
PeteP wrote:No it doesn't because it's conditional on t=0 and thus doesn't hold if t changes? Or rather it's always x'=x-vt. So the question remains.

Obvious too, is that when t = 0, then x = x' for x' = x-vt. All that I am saying is that x = x' when t > 0, and that x' = x-vt is not true when t > 0. This really is not complicated at all once one accepts that x means the abscissa.


Cart before the horse, mate.

x' = x -vt (for all t)
at t=0, x' = x.


not the other way around.

Are you unable to see that the abscissa of S = the abscissa of S' at both t = 0 and t > 0?
Why are you trying to deny this? Your only course of denial can be that you disagree with x being the abscissa of S or that x' is not the abscissa of S'. It is not debatable that x = x' when t > 0, as this is obviously true...
That would mean you need to refute the recently dozen posted definitions for abscissa!

x, aka the abscissa of S, the distance from S(0,0,0) to S(x,0,0)
EQUALS
x', aka the abscissa of S', the distance from S'(0,0,0) to S'(x',0,0)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Mon Nov 11, 2013 10:06 pm UTC

steve waterman wrote: It is not debatable that x = x' when t > 0, as this is obviously true...
It's actually obviously false. Look at my diagram.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Mon Nov 11, 2013 10:21 pm UTC

This is nothing new, but steve, what is the distance between the x coordinate of the origin of coordinate system S and x coordinate of point P in this diagram? Now what is the distance between x coordinate of the origin of coordinate system S' and x coordinate of point P in this diagram? Are they the same value? The former value is x and the latter is x' after S' has been "moved" to the right and down.

Image

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

Postby steve waterman » Mon Nov 11, 2013 10:30 pm UTC

Image

Image

x = the abscissa of S and x' = the abscissa of S'
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby JudeMorrigan » Mon Nov 11, 2013 10:33 pm UTC

That's not how anyone else on the planet is using it though. If you use it in the manner I've described, you can answer a useful question. (If I know the coordinates of a point in one coordinate system [note, the coordinates are what are in the coordinate system, not the point itself] and the relationship of a second coordinate system to the first, can I find the coordinates of the point in the second coordinate system?) Your definitions simply don't do anything useful.

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

Postby PeteP » Mon Nov 11, 2013 10:37 pm UTC

x' is the abscissa of which point steve?

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

Postby firechicago » Mon Nov 11, 2013 10:37 pm UTC

steve waterman wrote:Are you unable to see that the abscissa of S = the abscissa of S' at both t = 0 and t > 0?

There is no such thing as "the abscissa of S" or "the abscissa of S' ".

There is only the abscissa in S of a given point. And if a coordinate system moves then the abscissa of a given point changes.

More generally, the way you're using S' here, it isn't a coordinate system, it's a whole family of coordinate systems, governed by a parameter t. S' at t=0 is a different coordinate system than S' at t=1. This is conceptually different (though it will produce the same numeric results) to the way that we have previously defined the galilean transformation, where S' is a four dimensional coordinate system, with a t dimension.

These two ways of defining the galilean transformation are more or less equivalent, but as usual you're trying to exploit the confusion between the two of them to argue for some fundamental contradiction which doesn't actually exist.

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

Postby JudeMorrigan » Mon Nov 11, 2013 10:42 pm UTC

Image

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

Postby Роберт » Mon Nov 11, 2013 10:50 pm UTC

steve waterman wrote:
Spoiler:
Image

Image

x = the abscissa of S and x' = the abscissa of S'

This diagram doesn't make any sense, because S'(x',y',z',t') doesn't refer to the same point that S(x,y,z,t) does in the lower part of the diagram. This means it is completely unrelated to the Galilean.

The Galilean says specifically:
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


If you look at your diagram, S(2,0,0,t) refers to a totally different point than S'(2,0,0,t) does in the lower picture (where t is not equal to zero). They are not both at a single arbitrary event. So this is clearly not the x and x' that the Galilean refers to.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Mon Nov 11, 2013 10:57 pm UTC

JudeMorrigan wrote:That's not how anyone else on the planet is using it though. If you use it in the manner I've described, you can answer a useful question. (If I know the coordinates of a point in one coordinate system [note, the coordinates are what are in the coordinate system, not the point itself] and the relationship of a second coordinate system to the first, can I find the coordinates of the point in the second coordinate system?) Your definitions simply don't do anything useful.

Sigh...still you insist on points not having equal coordinates in the manifold and elect to ignore equal abscissa. These are not MY definition for abscissa but rather THE accepted mathematical definition(s) for abscissa. Again I agree that point P has a different set of coordinate values regarding S and S'...which is irrelevant to the fact that the abscissa lengths are equal after vt. For 6000 posts I hear that the coordinate values for point P are different in the manifold. For about the 100th time...I totally agree, yes that is 100 percent true!
also have grasped from page one that your points do not move when a system moves.

I am however, not talking about the relationship of the coordinates of point P, rather I am talking about x and x' the abscissa being the distance from their own origin. x is NOT a damn point, x is the abscissa of S. Do you deny that after vt, the abscissa x of S = the abscissa x' of S'????? Taking a posting break until tomorrow.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Роберт » Mon Nov 11, 2013 11:04 pm UTC

steve waterman wrote:
Spoiler:
JudeMorrigan wrote:That's not how anyone else on the planet is using it though. If you use it in the manner I've described, you can answer a useful question. (If I know the coordinates of a point in one coordinate system [note, the coordinates are what are in the coordinate system, not the point itself] and the relationship of a second coordinate system to the first, can I find the coordinates of the point in the second coordinate system?) Your definitions simply don't do anything useful.

Sigh...still you insist on points not having equal coordinates in the manifold and elect to ignore equal abscissa. These are not MY definition for abscissa but rather THE accepted mathematical definition(s) for abscissa. Again I agree that point P has a different set of coordinate values regarding S and S'...which is irrelevant to the fact that the abscissa lengths are equal after vt. For 6000 posts I hear that the coordinate values for point P are different in the manifold. For about the 100th time...I totally agree, yes that is 100 percent true!
also have grasped from page one that your points do not move when a system moves.

I am however, not talking about the relationship of the coordinates of point P, rather I am talking about x and x' the abscissa being the distance from their own origin. x is NOT a damn point, x is the abscissa of S. Do you deny that after vt, the abscissa x of S = the abscissa x' of S'????? Taking a posting break until tomorrow
.

Do you agree that, in the lower picture in your diagram, S(2,0,0,t) clearly does not refer to the same event that S'(2,0,0,t') refers to?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PolakoVoador » Mon Nov 11, 2013 11:10 pm UTC

steve waterman wrote:
Spoiler:
Image

Image


x = the abscissa of S and x' = the abscissa of S'


I think I'm having a deja vu right now, like, something we saw 5800 posts ago.

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

Postby lokar » Mon Nov 11, 2013 11:14 pm UTC

steve waterman wrote:These are not MY definition for abscissa but rather THE accepted mathematical definition(s) for abscissa


No, they're not. Not for the Galilean transformation, at least.

There was an old man, who was driving his car home on the highway. And suddenly the phone in his car rang. It was his wife calling him, and she said, "Hey Steve, I'm sorry to disturb your driving, but you know what? A report on the radio said there was a crazy man driving the wrong way on the highway.So please be careful, okay?" And the husband said, "Oh, yeah! But it's not just one person; there are hundreds of them!"

You're the only one misusing the terminology, here.
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