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 steve waterman » Tue Oct 01, 2013 12:25 pm UTC

SecondTalon wrote:Steve,

I don't know you personally. This is no big shocker. I also don't know your doctor nor the medical advice you've been given. I also don't know how worked up you get while reading and debating in this thread.

I do know that a triple bypass and valve replacement with a pace maker (and some complications on the side) is not something you just bounce right back from, even after a month stay in the Hospital.

So, with that being said and with your remark about jumping the gun - we aren't going anywhere anytime soon. Can this wait another couple of weeks?

In the past, I would find myself actually dreaming of responses while I slept. I will no longer allow myself to be going to such depths for this thread. Much of the confusion over various terms has been addressed at this point in time. I am hoping for some feedback upon my previous post. I do so also hope, that any conversation does not explode as it has in the past. For me, my contention has been boiled down to differentiating between x as a coordinate or as x as an axis.

In conclusion, I am seeking any others that would acknowledge that the equation x' = x-vt is correct iff we are comparing the x axis to the x' axis. Likewise, that the equation x' = x -vt is not correct iff we are comparing the x distance from its origin to the x' distance from its origin.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby firechicago » Tue Oct 01, 2013 3:08 pm UTC

It's not at all clear what you mean by "the x distance from its origin" and "the x' distance from its origin." How are you defining x and x'? How are you defining their origins? The distance from the origin to what? Are they the distance to the same point or to different points? (And by the same point, I mean a single location which will have different coordinates in any non-coincident coordinate system.)

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

Postby steve waterman » Tue Oct 01, 2013 3:33 pm UTC

firechicago wrote:It's not at all clear what you mean by "the x distance from its origin" and "the x' distance from its origin." How are you defining x and x'? How are you defining their origins? The distance from the origin to what? Are they the distance to the same point or to different points? (And by the same point, I mean a single location which will have different coordinates in any non-coincident coordinate system.)

Thanks for the feedback and your questions.

Given coincident Cartesian coordinate systems S(x,y,z) and S'(x',y',z')...where at t = 0, let x = 2 and x' = 2.
I define x, not as the point S(2,0,0) along the x axis, rather as the lineal distance of 2 from S(0,0,0) to S(2,0,0).
I define x', not as the point S'(2,0,0) along the x' axis, rather as the lineal distance of 2 from S'(0,0,0) to S'(2,0,0).

The point at S(2,0,0) has no dimensions, that is, a point has no associated distance.
Letting x = 2, mandates a dimension...one; the lineal distance along the x axis.

Hence, my concern is NOT about a point(s), at all. My concern is ONLY about x being a coordinate; hence the distance from S(0,0,0) - aka the abscissa distance from S(0,0,0).

Thus, I am comparing abscissa distances from EACH origin to one another...so x = x'
Einstein in essence, is comparing distances between each axis to one another in the manifold...so x' = x-vt.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby firechicago » Tue Oct 01, 2013 4:04 pm UTC

We certainly agree that if we have two coordinate systems, S(x,y,z) and S'(x',y',z') that map to points in the same manifold, and if we define these coordinate systems as being coincident, then the coordinates x=x', y=y', z=z', for any given point in the manifold. This is the definition of coincident coordinate systems. But now I'm confused by this part:
steve waterman wrote:Einstein in essence, is comparing distances between each axis to one another in the manifold...so x' = x-vt.

We've defined two coordinate systems with the coordinates (x,y,z) and (x',y',z'), so where are v and t coming from? I don't see them in any of the definitions we've laid out just now. What do they mean?

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

Postby steve waterman » Tue Oct 01, 2013 4:49 pm UTC

firechicago wrote:We certainly agree that if we have two coordinate systems, S(x,y,z) and S'(x',y',z') that map to points in the same manifold, and if we define these coordinate systems as being coincident, then the coordinates x=x', y=y', z=z', for any given point in the manifold. This is the definition of coincident coordinate systems. But now I'm confused by this part:
steve waterman wrote:Einstein in essence, is comparing distances between each axis to one another in the manifold...so x' = x-vt.

We've defined two coordinate systems with the coordinates (x,y,z) and (x',y',z'), so where are v and t coming from? I don't see them in any of the definitions we've laid out just now. What do they mean?

At t = 0, the two systems are coincident - for one thing, that means that both origins share the same location in 3 space. Then we allow the S' system to be relocated to the right along the common x/x' axis by a distance of vt = 3.

So, NOW the x distance from the S origin = 2 ...(since we let x = 2 when both systems were coincident) and
the x' distance from the S' origin is also = 2....(since we let x' = 2 when both systems were coincident).
x coordinate = x' coordinate

As well, now the two systems are no longer coincident,
as the x axis is separated from the x' axis...(by a length of 3).
x' axis = x axis -vt
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby firechicago » Tue Oct 01, 2013 5:18 pm UTC

Wait, so the systems were coincident but now they're not? How did that happen? I thought we had defined them as being coincident. I don't see anything like that in the definitions we've laid out.

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

Postby steve waterman » Tue Oct 01, 2013 5:35 pm UTC

firechicago wrote:Wait, so the systems were coincident but now they're not? How did that happen? I thought we had defined them as being coincident. I don't see anything like that in the definitions we've laid out.

CASE 1 - S and S' begin as coincident at t = 0.
CASE 2 - Then at t > 0, we move S' to the right along the common x'/x axis...in this case by vt = 3.

edit - added
given coincident S(x,y,z) with S'(x',y',z')...

CASE 1 - coincident systems
x abscissa = x' abscissa
x axis = x' axis

CASE 2 - once coincident systems are no longer coincident ( where vt was applied to S' along common x'/x axis)
x abscissa = x' abscissa
x' axis = x' axis - vt

abscissa = the coordinate distance from the origin along the x axis for S
and the coordinate distance from the origin along the x' axis for S'.
Last edited by steve waterman on Tue Oct 01, 2013 5:54 pm UTC, edited 1 time in total.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby eSOANEM » Tue Oct 01, 2013 5:47 pm UTC

Steve, all this wonkiness comes because you're trying to picture this as an evolving system by considering time to be external. Whilst this is possible to do reasonably sensibly in classical mechanics, it leads to major issues if you trying doing it with anything even approaching SR. You do not have two systems which start coincident and then diverge, what you have is two 4D systems with coincident origins and all but one of the axes aligned. The axis which is not aligned appears in the orthogonal system to be skewed in the x direction. This gives the skewed co-ordinates Schrollini described at the beginning of the thread.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby firechicago » Tue Oct 01, 2013 5:48 pm UTC

So we're dealing with two separate cases, one in which there are two coincident coordinate systems and one in which there are two non-coincident coordinate systems?

If that's the case, we've proved that x=x' in case 1. But that proof relied upon the definition of coincidence. We can't generalize that proof to cases where the coordinate systems are not coincident.

And, trying to restrict myself to just the mathematical definitions that you've laid down today, I still do not understand what you mean by "at t=0" or "at t>0". If we had a t-dimension in our coordinate systems I would understand those statements to mean "the set of points where t=0, and the other coordinates can equal any real number" but we defined our coordinate systems as having only x-, y- and z-dimensions. Where is t coming from? What relevance does it have to the coordinate systems?

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

Postby steve waterman » Tue Oct 01, 2013 7:12 pm UTC

firechicago wrote:So we're dealing with two separate cases, one in which there are two coincident coordinate systems and one in which there are two non-coincident coordinate systems?

If that's the case, we've proved that x=x' in case 1. But that proof relied upon the definition of coincidence. We can't generalize that proof to cases where the coordinate systems are not coincident.

And, trying to restrict myself to just the mathematical definitions that you've laid down today, I still do not understand what you mean by "at t=0" or "at t>0". If we had a t-dimension in our coordinate systems I would understand those statements to mean "the set of points where t=0, and the other coordinates can equal any real number" but we defined our coordinate systems as having only x-, y- and z-dimensions. Where is t coming from? What relevance does it have to the coordinate systems?

Okay, let's get rid of time as a component please...making this strictly a mathematical problem...
so, as if we have two "frozen/static" snapshots -

given coincident S(x,y,z) = S'(x',y',z') where x axis = x' axis
x abscissa = x' abscissa

given S(x-d*,y,z) = S'(x',y',z') where x' axis = x axis-d* [/b]
where d* = the distance from the S origin to the S' origin along the common x/x' axis.
x abscissa = x' abscissa
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby PolakoVoador » Tue Oct 01, 2013 8:26 pm UTC

First things firts: glad to see you up and running again Steve. Get all the rest you need, and don't overdo things around here. Your health is your priority :)

Second: I think you are made everything right in your last post, at least, as far as I understand your non-standard notation. Yes, the x-coordinate of a point in S will differ by d* from the x'-coordinate of the same point in S'. Yes, the distances between S(x,0,0) and S(0,0,0); and between S'(x',0,0) and S'(0,0,0), will be equal for equal values of x and x'.

Is this what you're saying?

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

Postby SecondTalon » Tue Oct 01, 2013 8:34 pm UTC

steve waterman wrote:Okay, let's get rid of time as a component please...making this strictly a mathematical problem...
so, as if we have two "frozen/static" snapshots -

given coincident S(x,y,z) = S'(x',y',z') where x axis = x' axis
x abscissa = x' abscissa

given S(x-d*,y,z) = S'(x',y',z') where x' axis = x axis-d* [/b]
where d* = the distance from the S origin to the S' origin along the common x/x' axis.
x abscissa = x' abscissa

If we're getting rid of time, what's moving S'?

But yeah, as PolakoVoador said - it looks like you're describing two identical pair of X/Y axes that happen to exist precisely on top of each other. Which is fine, 'cause it's math.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Tue Oct 01, 2013 9:23 pm UTC

PolakoVoador wrote:First things first: glad to see you up and running again Steve. Get all the rest you need, and don't overdo things around here. Your health is your priority :)

Second: I think you are made everything right in your last post, at least, as far as I understand your non-standard notation. Yes, the x-coordinate of a point in S will differ by d* from the x'-coordinate of the same point in S'. Yes, the distances between S(x,0,0) and S(0,0,0); and between S'(x',0,0) and S'(0,0,0), will be equal for equal values of x and x'.

Is this what you're saying?

Your first yes statement above is right on in accordance to what I am trying to express.
Your second yes statement above is very very close...to accordance in what I am trying to mathematically express.
Noting that - these distances are actually vectors...
so, each distance possesses both magnitude as well as +/- direction from their own origin.

" Yes, the distances FROM S(0,0,0) TO S(x,0,0); and FROM S'(0,0,0) TO S'(x',0,0), will be equal for equal values of x and x' ".

Indeed, that is basically 99 percent of what I have tried to express from day 1 ( now, over 5000 responses later). I will prepare and then post a small set of corresponding/appropriate visuals for tomorrow and feel that I should call it quits for posting today. Thanks so much for these uncomplicated responses. Thanks too for everyone waiting for me to get back on my feet a bit before letting me approach this old topic/issue..
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Pfhorrest » Tue Oct 01, 2013 11:22 pm UTC

steve waterman wrote:" Yes, the distances FROM S(0,0,0) TO S(x,0,0); and FROM S'(0,0,0) TO S'(x',0,0), will be equal for equal values of x and x' ".

Indeed, that is basically 99 percent of what I have tried to express from day 1 ( now, over 5000 responses later).

Hi Steve, glad you're feeling better, and glad there's some good communication happening here.

I want to add to this: most of us have been aware from shortly after day 1 that that is what you were trying to express.

What we have been trying to express in return is that, the way it's understood by everyone who uses it, the Galilean transformation doesn't say anything otherwise. It agrees with this too. The only reason you've thought it didn't agree with it was you misunderstood what it was saying.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Wed Oct 02, 2013 1:10 pm UTC

The distances FROM S(0,0,0) TO S(x,0,0); and FROM S'(0,0,0) TO S'(x',0,0), will be equal for equal values of x and x'.
Image



The distances FROM S(0,0,0) TO S(x,0,0); and FROM S'(0,0,0) TO S'(x',0,0), will be equal for equal values of x and x'.
The x-AXIS of a coordinate in S will differ by d* from the x'-AXIS of the same coordinate in S'.
Image
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby SecondTalon » Wed Oct 02, 2013 1:41 pm UTC

Right, but whatever function was applied to S' to make it move has to also be applied to x' (and every other point in S').

D is that function

So with S+D = S'

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

Postby steve waterman » Wed Oct 02, 2013 2:50 pm UTC

SecondTalon wrote:Right, but whatever function was applied to S' to make it move has to also be applied to x' (and every other point in S').

D is that function

So with S+D = S'

Then x+D = x'

Thank you for your input and feedback.

Try to imagine that no movement occurs. That is, both depictions are mathematically static. That they are simply given as two different mathematical scenarios.

The problem is in deciding what x means in the equation x' = x-vt or more specifically, x' = x-D.
Does x refer to the x-AXIS..so, that x' axis = x axis - D, or as I as suggesting,
that x mathematically refers to the x coordinate...because S(x,y,z) was given
- and hence the x coordinate was defined as the abscissa wrt the S system?

One can observe that the x abscissa lengths and the x' abscissa lengths are not altered when the S and S' systems appear in the second depiction wrt their abscissa lengths in the first depiction.

In general, "given coordinates remain fixed FOREVER wrt their own system's origin".
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby SecondTalon » Wed Oct 02, 2013 5:13 pm UTC

Try to imagine that no movement occurs. That is, both depictions are mathematically static. That they are simply given as two different mathematical scenarios.


Then neither one has anything to do with the other. If S = S' here, and S = S' - D here, and they are two different scenarios..


Then x = x' in the first scenario, x = x' - D in the second scenario, and neither one have anything to do with each other. Scenario 1 x (hereafter x1) and Scenario 2 x (hereafter x 2) are not the same

x1 = x'1 and x2 = x'2 - D

x1 =/= x2 because.. as you said, they are two different scenarios and as such, have no relation to each other whatsoever.

The problem is in deciding what x means in the equation x' = x-vt or more specifically, x' = x-D.


Not really? x means any point on the x axis. Since we're talking about two different x axes, we need a way to distinguish them and you're using the ' mark, so that's what I'm using. x is any point on the x axis, and x' is any point on the x' axis.

In general, "given coordinates remain fixed FOREVER wrt their own system's origin".
I.. think I agree with that, as that's the whole point. With an origin of 0,0 (or 0,0,0, or 0,0,0,0.. you get the idea) the entire existence of something like 2,1 depends utterly on 0,0 existing 2 and 1 point away on the respective X and Y axis. So no, 2,1 cannot move with respect to 0,0

Now... that being said...

If 0,0 and 0,0 are the same on two coordinate systems, and 1,1 and 1,1 are the same, and -1,-1 and -1',-1' are the same and both coordinate systems are described as planes... then they're coincident...

I would argue that we cannot directly say that 50,587 = 50',587' because they are on two different systems. Rather, we say that 50,587 is 50 and 587 units away from 0,0 and that 50', 587' is 50 and 587 units away from 0',0', and that 0,0 = 0',0' | 1,1 = 1'1', | -1',-1' = -1,-1, so we can safely operate as though 50','587' and 50,587 occupy the same space. We.. uh, just shorthand that as a given, but if we had to prove it, that's how we'd prove it. That the three points given on the first system (1,1 , 0,0 and -1,-1) are described as coincident with the defined points on the second (1',1', 0',0', and -1',-1') and the two coordinate systems are defined as planes.

Now once 0,0 and 0',0' no longer occupy the same space, any existing relationship we assumed between the two systems goes completely out the window and we have to rebuild our system of understanding. If at one point 0,0 and 0',0' were the same, our rule has to allow for that... but it also has to allow for 15,68 = 0',0' or whatever. So we come up with some formula and can draw relationships between various points of x,y and x',y' based on that formula.

Which, if I'm not mistaken, is the vd thing that keeps being mentioned. As I said, I've got high school level math skills that have been rusting like hell over the last decade, so I doubt I'm using anything close to the right language and hopefully didn't screw up too many things.

One can observe that the x abscissa lengths and the x' abscissa lengths are not altered when the S and S' systems appear in the second depiction wrt their abscissa lengths in the first depiction.


That.. sounds like a really fancy way of saying "On S, X is equal to X in both depictions such that this particular point is still 3 units away from X, and on S', X' is equal to X' in both depictions such that this particular point is still 3 units away from X' " to which I can only say... "... yes? That's the entire point of having an axis? That points defined on it remain a constant distance from their 0 point?"

The only difference between the two is that S and S' were on the same point in space (or as close as we can physically get with the model, but was defined as being overlapping and Coincident) and now S' is no longer sharing space with S but is now instead... if I had to guess, I'd say that S' is now on 3. So 3,0 = 0',0'

Which, of course, now means that 3,0 and 3',0' are no longer in the same space. Rather, x,y = x'-3,y'

Because S(0,0) = S'(0-3,0)

or something. Not sure what the proper way to write that would be. See above re: High School Level Math Skill (Rusty)
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Wed Oct 02, 2013 11:50 pm UTC

Sexy Talon,

Nice feedback, as I do suspect that I grasp what you are stating, "high school notation aside", conceptually,

You certainly seem to have gotten the main idea that I am going for. That is, once we are given a coordinate set like S(x,y,z), even if that set is moved/relocated/displaced, the relationships of (x,y,z) to S(0,0,0) remains the same.

All that I am suggesting is that the equation x' = x - D references
1 the x coordinate, and NOT the x axis.
2 the x' coordinate, and NOT the x' axis.

I suggest that allowing x' = x - D, or as the Galilean x' = x - vt, indeed sets equally numbered x coordinates and x' coordinates to a mathematical inequality...hence generating the numerous paradoxes of Relativity. Relativity people do not believe that any coordinates get changed due to allowing x' = x-vt...and that coordinates are only associated to one another. That IS precisely what I am mathematically challenging.

Thus, this is about, and only about what x means...given S(x,y,z). I say, x refers to the given coordinates.
btw, it is not permitted to define/characterize a coordinate set as a point, (according to the theory of Relativity,
where points only exist in their manifold M.)

I would believe that in the 1700's/1800's that "points" were manifested in/wrt a Cartesian coordinate system...before manifold M was ever dreamed-up, but that is just me, (so far). It is too late to ask Rene Descartes about that now.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby firechicago » Thu Oct 03, 2013 1:36 am UTC

steve waterman wrote:All that I am suggesting is that the equation x' = x - D references
1 the x coordinate, and NOT the x axis.
2 the x' coordinate, and NOT the x' axis.

I suggest that allowing x' = x - D, or as the Galilean x' = x - vt, indeed sets equally numbered x coordinates and x' coordinates to a mathematical inequality...hence generating the numerous paradoxes of Relativity. Relativity people do not believe that any coordinates get changed due to allowing x' = x-vt...and that coordinates are only associated to one another. That IS precisely what I am mathematically challenging.


I have no idea what any of this means. Specifically, I don't know:

What it means to "reference the x coordinate versus the x axis." A coordinate in a cartesian system references the distance between a point and an axis. How can you reference the coordinate without referencing the axis? How can you talk about the abcissa distance (to use your term) of a point without defining what you're measuring from?
What it means that "coordinates get changed due to allowing x' = x-vt." If x'=x-vt is used to define a coordinate transformation, then of course it changes the coordinates of some points. That is what a coordinate transformation is supposed to do. It's right there in the name.
What it means for coordinates to be "only associated to one another" (and what would the alternative be?)

steve waterman wrote:I would believe that in the 1700's/1800's that "points" were manifested in/wrt a Cartesian coordinate system...before manifold M was ever dreamed-up, but that is just me, (so far). It is too late to ask Rene Descartes about that now.

Euclid was talking about points two thousand years before Descartes came up with his way of labeling points with coordinates.

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

Postby SecondTalon » Thu Oct 03, 2013 2:02 am UTC

steve waterman wrote:All that I am suggesting is that the equation x' = x - D references
1 the x coordinate, and NOT the x axis.
2 the x' coordinate, and NOT the x' axis.


I don't understand the difference between

x = a random but specific point along the X axis. Maybe 12. Maybe 185,793,779.28697. Maybe 0. Whatever! In our examples above, apparently 3.

and

x = in potential the entire axis as it has yet to be defined as a single point, like 12 or 185,793,779.28697 or 0. Or, as in our examples above, 3.

Because.. if you're talking about a specific point on the x axis, like 3... then you just say 3.

You only say x when you're talking about no point in particular but all points in potential. In other words, the entire axis at once. 2x+5=y describing a relationship between X and Y such that we can select any point along X (you know the three I'm going to mention) and we can get a result for Y out of it. Or vice versa. While we can talk about a specific point of X, we have to define it along the "Let x=25" sort of way so we know we're talking about a specific point. Otherwise, we mean the entire thing at once.

So.. no, saying that x represents a single point is completely contrary to standard notation. x is all the points on the x axis at once, because we haven't limited ourselves to a single point yet.

If you want to talk about a single point, at least use some semi-standard notation for them, like Point Alpha.

So in our examples, where x=3 y=0 lies Point Alpha and x'=3 y'=0 lies Point Beta, if S(0,0) and S'(0,0) are coincident, the Point Alpha and Point Beta are also coincident.

And when we have x = x' - vt, then Point Alpha and Point Beta are no longer coincident, excepting when x and x' are the same number and vt = 0.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby ucim » Thu Oct 03, 2013 4:33 am UTC

steve waterman wrote:I suggest that allowing x' = x - D, or as the Galilean x' = x - vt, indeed sets equally numbered x coordinates and x' coordinates to a mathematical inequality...hence generating the numerous paradoxes of Relativity.
The equation does not set 3 = 7 or anything like that. Rather, what it says is that when you are referring to the same point, but "addressing it" in two different systems, it can be referred to as 3 in one system, and as 7 in the other. Assuming the axis is aligned properly, this implies that the systems are offset by four units from each other, and that every point will have a value (in one system) that is off by four units (in the other system).

Think of it as translating French to German.
French = German - 4
means that the French will give a name to a point that is four units less than the name the Germans give the same point.

Relativity has nothing to do with any of this. This is just straight math.

As for the point vs axis thing

Consider a single x-y coordinate plane, S, oriented in the usual fashion (the x axis is horizontal, the y axis is vertical).

Consider the coordinates of points in this plane. The point three units to the right and two units up from the origin is the point S(3,2).

A diagnoal line going up at 45 degrees (bottom left to top right), and passing through the origin, will be the set of points S(q,q) where q can take on all real values. Examples are S(3,3), S(72, 72), and S(-7, -7)

A horizontal line passing through the origin would be the set of points S(q,0), where q can take on all real values. It is the x axis. It is the line y=0. Note which is x and which is y.

The line y=4 is a horizontal line which passes through the point S(0,4) It is the set of points S(q,4) where q can take on all real values. Note that the key element of a horizontal line (a line that is parallel to the x axis) is the y value!

When we draw the x axis, we draw a horizontal line. But when we say x=7 (which in this case is the set of all points where the first coordinate is equal to seven, or the set S(7,q) where q takes on all real values, we draw a vertical line! It is a line that crosses the x axis at the point S(7,0). It is parallel to the y axis.

This can be a source of confusion, especially when looking at graphs, or when you are not sure whether x refers to a value or an axis.

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

Postby steve waterman » Thu Oct 03, 2013 1:33 pm UTC

Sexy Talon and ucim.

I appreciate your recent comments and will take this opportunity to do my best to address them.

First, please understand that the Galilean coordinate transformation equations are ALWAYS stated as x' = x-vt.
see - http://en.wikipedia.org/wiki/Galilean_transformation
As shown on that page, x' = x-vt, y = y', z= z' and t = t'. Hence when t = 0, x = x'.
NEVER EVER is the Galilean coordinate transformation stated as S'(x'-vt,y',z') = S(x,y,z).

I totally agree that coordinate location S'(x'-vt,y',z') = the coordinate location S(x,y,z).
This is however, NOT the actual Galilean coordinate transformation equation of x' = x-vt, is it?

I would like to focus upon the actual equation of x' = x-vt, and throw some numbers in to see what happens. Let the x coordinate = a distance of 2 and the x' coordinate = a distance of 2 and vt = a distance of 0.
So, as given by the article x = x' when vt = 0.

We seem to all agree that that when vt = 3,
that the distance from S(0,0,0) to S(2,0,0) = the distance from S'(0,0,0) to S'(2,0,0).
The Galilean says/allows/purports, when vt = 3 and x = distance 2, that S(2,0,0) = S'(-1,0,0)...since x' = x-vt. Instead, the Galilean is cloaked behind this equation that merely deals with location S'(x'-vt,y',z') = S(x,y,z), and NOT with the abscissa distance from an origin, x' = x-vt.

[b]Now, if S'(x'-vt,y',z') = S(x,y,z) IS the equation that supports the Galilean transformation equations...then that should be the equation that we see in all the textbooks. I only see x' = x-vt in textbooks and NEVER EVER see S'(x'-vt,y',z') = S(x,y,z) written as the Galilean coordinate transformation!! This issue is quite key/critical to getting a handle upon my logic.

Logically, when x = distance 2, then x' must also be equal to distance 2 and ONLY 2.
When x = 2 and vt = 3, the actual Galilean equation printed in textbooks suggests/declares x' = x-vt and would have us accept that x is equal to distance -1, while x = 2.

Again, I totally agree that when x = 2 and vt = 3...that S'(x'-vt,y',z') = S(x,y,z) , so
S'(2-3,0,0') = S(2,0,0) as this merely equates location in 3 space.

recap -
the textbook Galilean equation is x' = x-vt...which I suggest is mathematically false.
the textbook Galilean equation is never written/expressed as S'(x'-vt,y',z') = S(x,y,z) ...
which I accept as mathematically true and has acted as a straw man since 1887.
see - http://en.wikipedia.org/wiki/Woldemar_Voigt
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby SecondTalon » Thu Oct 03, 2013 1:55 pm UTC

I suppose I'm confused as to how the underlined parts of S'(x'-vt,y',z') = S(x,y,z) are relevant when we're looking at x and x'

Wouldn't there also be something along the lines of y = y' - vt? Or is it that that whole bit is taken care of as the x,y,z and x',y',z' are represented as forumulas in and of themselves, letting the y and z be defined by the value of x, (and y', z' by x') making the rest of it redundant and irrelevant.

So in entirety the problem is represented as

2x-5=y , (3-x)/y = z | 2x'-5=y' , (3-x')/y' = z' | x = x' -vt

which would allow us to only care about the relationship between x and x' as the rest become self-evident once that relationship is established.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Chen » Thu Oct 03, 2013 2:33 pm UTC

steve waterman wrote:recap -
the textbook Galilean equation is x' = x-vt...which I suggest is mathematically false.
the textbook Galilean equation is never written/expressed as S'(x'-vt,y',z') = S(x,y,z) ...


These two are the same. The textbook definition of the Galilean is not just x' = x - vt. It includes definitions of S and S' and mentions single arbitrary events (or points). From your wikipedia article link:

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: [2] [3] [4] [5]
x'=x-vt,
y'=y ,
z'=z ,
t'=t


That whole paragraph AND the math define the Galilean transformation. Which is pretty much the same as the second line in your recap but with more words and less pure math definition. That's the only difference, which is not a mathematical (or physical) difference but merely one of language.

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

Postby ucim » Thu Oct 03, 2013 2:44 pm UTC

steve waterman wrote:First, please understand that the Galilean coordinate transformation equations are ALWAYS stated as x' = x-vt.
see - http://en.wikipedia.org/wiki/Galilean_transformation
As shown on that page, x' = x-vt, y = y', z= z' and t = t'. Hence when t = 0, x = x'.
NEVER EVER is the Galilean coordinate transformation stated as S'(x'-vt,y',z') = S(x,y,z).
That's because it isn't.

S(x,y,z) = S'(x', y',z') is the correct statement, saying that the two notations refer to the same point.
The "vt" part is how you find the value of the first coordinate in S' if you know the first coordinate in S.

Another way of writing it is (for a horizontal translation of vt units, and no translation in the y or z direction)
S'(x',y',z') = S'(x-vt,y,z) = S(x,y,z)
Note the placement of the primes.
In this case y=y' and z=z', since the only difference is in the horizontal direction.
x'=x-vt indicates the difference in the horizontal direction.

What the notation S(2,3,4) = S'(1,7,3) means is that S(2,3,4) and S'(1,7,3) are two different names for the same point. One in French, and the other in German, to be metaphorical. One in the S system and the other in the S' system, to be mathematical.

Similarly, S(x,y,z) = S'(x',y',z') does not mean that x=x', y=y' and z=z', but rather, that S(x,y,z) and S'(x',y',z') refer to the same point. (In the special case where they are equal for all x, y, and z, the systems are coincident) I think we're in agreement here.

If the coordinate systems are not coincident, then the primed values will not (necessarily) be equal to the nonprimed values, if they refer to the same point in the manifold (which is the whole point).

In the case where we just displace S and S' along the x axis by two units... thus (ignore the vertical components which are just a function of doing this in text form)

----4----5----6----7----8----9 (S)
--------------*--------------- (manifold)
----2----3----4----5----6----7 (S')

the point "*" in the manifold (there's only one under discussion!) aligns with 6 in system S, and with 4 in system S'

The x coordinate of "*" is 6. (x is measured in system S)
The x' coordinate of "*" is 4. (x' is measured in system S')

The distance from the origin of S to "*" is 6.
The distance from the origin of S' to "*" is 4.

Notice that I am measuring always the distance to "*", which means I'm measuring a different distance in each of the systems (because I'm measuring from a different origin, but to the same point in the manifold).

x = x' - d
In this case, d = -2.

You like to think of "moving" coordinate systems. That's not a good way to think of it, but if you insist, think of it this way... when you "move" a coordinate system, you are not moving the points being measured!. It's like moving a ruler along a wall... you are not moving the nail in the wall. So, the ruler is moving with respect to the nail! Therefore, the distance from the origin of the ruler to the nail changes when you move the system.

It is this which is throwing you.

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

Postby eSOANEM » Thu Oct 03, 2013 2:46 pm UTC

steve waterman wrote:Sexy Talon and ucim.

I appreciate your recent comments and will take this opportunity to do my best to address them.

First, please understand that the Galilean coordinate transformation equations are ALWAYS stated as x' = x-vt.
see - http://en.wikipedia.org/wiki/Galilean_transformation
As shown on that page, x' = x-vt, y = y', z= z' and t = t'. Hence when t = 0, x = x'.
NEVER EVER is the Galilean coordinate transformation stated as S'(x'-vt,y',z') = S(x,y,z).


Of course it's never stated as S'(x'-vt,y',z')=S(x,y,z) because the definition of S', x', y' and z' requires that S'(x',y',z')=S(x,y,z). When we substitute in the equations you point out are shown on that page into this expression e.g. S'(x-vt,y,z)=S(x,y,z)

Basically, be very careful with your primes and make 100% sure you know what they actually mean.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby steve waterman » Thu Oct 03, 2013 3:01 pm UTC

SecondTalon wrote:I suppose I'm confused as to how the underlined parts of S'(x'-vt,y',z') = S(x,y,z) are relevant when we're looking at x and x'

Wouldn't there also be something along the lines of y = y' - vt? Or is it that that whole bit is taken care of as the x,y,z and x',y',z' are represented as forumulas in and of themselves, letting the y and z be defined by the value of x, (and y', z' by x') making the rest of it redundant and irrelevant.

So in entirety the problem is represented as

2x-5=y , (3-x)/y = z | 2x'-5=y' , (3-x')/y' = z' | x = x' -vt

which would allow us to only care about the relationship between x and x' as the rest become self-evident once that relationship is established.

Good question, indeed. Why indeed is the ONLY Galilean movement allowed to be along x??
Certainly, mathematically, either system S or S' can be moved by say, (2,-3,4) from coincidence.

For me, this IS/should be about y and z too! The Galilean restricts movement to only one direction. Indeed, what if we move S' by say (2,-3,4) from coincidence?

Then for me...this results with S'(x'-2,y'+3,z'-4) = S(x,y,z)...[ when S' is relocated ]
and in general....S'(x',y',z') = S(x,y,z) +/- delta origins from their given positions of coincident.
[ btw, delta mathematically means...the change between]

Noting too, Sexy Talon, that distance-wise...it remains true that x = x', y = y', and z = z'...wrt to their OWN origin...should either system be moved from coincidence by (a,b,c)
(regardless of what values are selected for a or b or c).

The focus should not be upon the relationship between two SETS of three coordinates...but rather as the actual Galilean states.. the purported equality of the distance x' with the distance x-vt.

Recap...this is about the distance x' vs the distance x. The TEXTBOOK Galilean equation x' = x-vt is NOT about points nor is it about the equality between sets of coordinates either. Do you see why we need to only be discussing the actual written TEXTBOOK equation? Do you see that my argument is about equality of given abscissa distances and not about their/the Galilean's/Relativity's straw man equality of equal point locations?
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Thu Oct 03, 2013 3:15 pm UTC

steve waterman wrote:
SexyTalon wrote:I suppose I'm confused as to how the underlined parts of S'(x'-vt,y',z') = S(x,y,z) are relevant when we're looking at x and x'

Wouldn't there also be something along the lines of y = y' - vt? Or is it that that whole bit is taken care of as the x,y,z and x',y',z' are represented as forumulas in and of themselves, letting the y and z be defined by the value of x, (and y', z' by x') making the rest of it redundant and irrelevant.

So in entirety the problem is represented as

2x-5=y , (3-x)/y = z | 2x'-5=y' , (3-x')/y' = z' | x = x' -vt

which would allow us to only care about the relationship between x and x' as the rest become self-evident once that relationship is established.

Good question, indeed. Why indeed is the ONLY Galilean movement allowed to be along x??
Certainly, mathematically, either system S or S' can be moved by say, (2,-3,4) from coincidence.

For me, this IS/should be about y and z too! The Galilean restricts movement to only one direction. Indeed, what if we move S' by say (2,-3,4) from coincidence?

Then for me...this results with S'(x'-2,y'+3,z'-4) = S(x,y,z)...[ when S' is relocated ]
and in general....S'(x',y',z') = S(x,y,z) +/- delta origins from their given positions of coincident.
[ btw, delta mathematically means...the change between]

Noting too, Sexy Talon, that distance-wise...it remains true that x = x', y = y', and z = z'...wrt to their OWN origin...should either system be moved from coincidence by (a,b,c)
(regardless of what values are selected for a or b or c).

The focus should not be upon the relationship between two SETS of three coordinates...but rather as the actual Galilean states.. the purported equality of the distance x' with the distance x-vt.

Recap...this is about the distance x' vs the distance x. The TEXTBOOK Galilean equation x' = x-vt is NOT about points nor is it about the equality between sets of coordinates either. Do you see why we need to only be discussing the actual written TEXTBOOK equation? Do you see that my argument is about equality of given abscissa distances and not about their/the Galilean's/Relativity's straw man equality of equal point locations?

No, the Galilean does not restrict us to motion along one axis; the example most frequently given for the transformation suggests three spatial dimensions because it's what we're used to on an everyday basis. It talks about motion along one axis only because the transformation for x and x' can be done exactly the same way for y and y' or z and z' (or any combination of x/x', y/y' and z/z') - it generalizes to the other dimensions.

This sounds like cargo-cult math, steve; you're parsing the words and the symbols and everything, but not getting the concept. When you recombine the words and symbols without the concept, you get a basically random rearrangement that misses the point, doesn't give any useful new information (or is trivial), and therefore when you look at it, it looks like a mistake. The mistake isn't in trusting the textbook to help you understand the theory, the mistake is in thinking the textbook is the theory.

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

Postby SecondTalon » Thu Oct 03, 2013 3:22 pm UTC

steve waterman wrote:Recap...this is about the distance x' vs the distance x. The TEXTBOOK Galilean equation x' = x-vt is NOT about points nor is it about the equality between sets of coordinates either. Do you see why we need to only be discussing the actual written TEXTBOOK equation? Do you see that my argument is about equality of given abscissa distances and not about their/the Galilean's/Relativity's straw man equality of equal point locations?
Not entirely, no.

Mostly because - as I know I've said elsewhere and may have said in this or another thread of yours and I'm sure someone else has said... we are long past the armchair days of discovery.

My point there being that while I won't at all doubt that an amateur in a field can make a groundbreaking discovery as that does still happen from time to time, if an amateur in a field can point out a glaring flaw in an accepted notion, that amateur needs to be able to supply lots of evidence to the contrary.

The problem here is that multiple people more educated in the field than I have pointed out where you are making fundamental flaws. Flaws that I'll admit to not fully understanding.

But in order to prove your take is correct, you have to address those flaws and say why they aren't flaws due to something else or revisit your work and correct the flaws.

I'm not sure that you've done that. Either one of them.

Also, between me starting this reply, WibblyWobbly has pointed out that the Galilean does apply to y|y' and z|z', and if I'm reading what WW's saying correctly, it's in the way I describe earlier - that you only need to discover the relationship between one of them (x | x') and the relationships between y|y' and z|z' fall in to place due to the relationships between x, y and z | z', y' and z'.


So the only reason x|x' gets all the love is because x is commonly the first axis listed, so why not start at the beginning rather than a random axis along the way or at the end?

But that right there - not knowing that the Galilean applies to y, y' and z, z' even though it's commonly written as x = x' -vt.... I can excuse myself for not knowing that as - again, last time I did anything serious with math education was 2001.

If you are trying to overthrow the Galilean, you have to know the Galilean inside and out so you can poke holes in it's flaws and break it apart. If you don't understand it completely, you can't overthrow it.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Thu Oct 03, 2013 3:33 pm UTC

What would be most general to say would probably be this: Given coordinates (x,y,z,t) and (x',y',z',t') of a common event in coordinate systems S and S', respectively, with one coordinate system moving relative to the other with a constant velocity v (technically, constant speed), the following holds:

x' = x - vxt
y' = y - vyt
z' = z - vzt
t = t' (This is the assumption of universal time)

with vx2 + vy2 + vz2 = v2

In the normal illustrative case, we just set vy = vz = 0, and everything collapses to the familiar version.

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

Postby steve waterman » Thu Oct 03, 2013 3:44 pm UTC

WibblyWobbly wrote: It talks about motion along one axis only because the transformation for x and x' can be done exactly the same way for y and y' or z and z' (or any combination of x/x', y/y' and z/z') - it generalizes to the other dimensions.

WibblyWobbly wrote:It talks about motion along one axis only because the transformation for x and x' can be done exactly the same way for y and y' or z and z'

Agreed..hence still restricted to ONLY one dimension change between origins.
WibblyWobbly wrote:or any combination of x/x', y/y' and z/z')

No way...show me a url ( or two ) or as written in some/any textbook please, that supports YOUR claim.
Let S'(2,0,0) be transformed by (2,-3,4)...[from being coincident with S].
What is the Galilean transformation?...what equation are you going to use, please, to perform YOUR transformation?

Show me also, a reference that will allow more that one simultaneous point in your manifold, for example...
LET Pa = S(2,0,0) and Pb = S(-3,0,0) and Pc = S(4,0,0)
What are those three transformations please, when vt = 3? Specifically, what does x' equal?

edit - added
Please excuse me from further posting today. I really cannot allow this thread to consume my life, and cause stress due to my own over-indulgence. I will try to keep my responses to one objection at a time. If others have different objections, so be it, however, I will try to handle/respond to these, but only as one at a time. If anyone else has a textbook or url to support either multiple points and multiple dimensions..please do post them. Already, this thread is creeping into my sleep and I should not allow myself to dedicate multiple hours towards crafting my daily posts. Thanks for your feedback so far and for your understanding/acceptance of my current situation.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby WibblyWobbly » Thu Oct 03, 2013 3:56 pm UTC

steve waterman wrote:
WibblyWobbly wrote: It talks about motion along one axis only because the transformation for x and x' can be done exactly the same way for y and y' or z and z' (or any combination of x/x', y/y' and z/z') - it generalizes to the other dimensions.

WibblyWobbly wrote:It talks about motion along one axis only because the transformation for x and x' can be done exactly the same way for y and y' or z and z'

Agreed..hence still restricted to ONLY one dimension change between origins.
WibblyWobbly wrote:or any combination of x/x', y/y' and z/z')

No way...show me a url ( or two ) or as written in some/any textbook please, that supports YOUR claim.
Let S'(2,0,0) be transformed by (2,-3,4)...[from being coincident with S].
What is the Galilean transformation?...what equation are you going to use, please, to perform YOUR transformation?

Show me also, a reference that will allow more that one simultaneous point in your manifold, for example...
LET Pa = S(2,0,0) and Pb = S(-3,0,0) and Pc = S(4,0,0)
What are their transformations please, when vt = 3? Specifically, what does x' equal?


Not sure what you're trying to get at with the rest of the post, but with respect to

steve waterman wrote:No way...show me a url ( or two ) or as written in some/any textbook please, that supports YOUR claim.

I give you:

http://www.encyclopediaofmath.org/index.php/Galilean_transformation
Equation 15.2 gives the general form, using vectors r and V

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

Postby PolakoVoador » Thu Oct 03, 2013 4:03 pm UTC

The carousel of the damned is going 'round'n'round... Once again, when Steve is almost looking like he will understand, he veers wildly into crazyland and we're back at square one.

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

Postby ucim » Thu Oct 03, 2013 4:51 pm UTC

Steve - distance is always between two points. Sometimes one of the points is understood; for example if you are dealing with one coordinate system S, you could treat x as a distance (the distance from the origin to the point S(x,0,0).

However, sometimes one of the points is misunderstood. This is what we have here, because you are using two different coordinate systems (S and S') and switching between them. If you take every time where you say "distance" and rewrite it as "distance between {fill in} and {fill in}", you will find that at some point you switch origins where you shouldn't, or you fail to do so when you should.

That's your sticking point.

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

Postby steve waterman » Thu Oct 03, 2013 4:54 pm UTC



This reference is still only one dimensional ( change along a single vector direction )..."r" or x or y or z, and NOT a 3 dimensional change as is delta (a,b,c). We are avoiding THE major issue...
the Galilean equation is x' = x-vt and not S'(x'-vt,y,z) = S(x,y,z).

These are just minor side issues, and will only serve to dilute the major issue. I do not care about dickering over either multiple points or multiple dimensions any more...as this is about the Galilean equation being x' = x-vt and not
S'(x'-vt,y,z) = S(x,y,z). I have been down these side issues for almost a year now and wish to stay on topic this time.

Because S'(x'-vt,y,z) = S(x,y,z) is true,
this does not mathematically extrapolate to x' = x-vt being true.


So, I am gonna quit for today, and I guess I should not read any new posts as I may get trapped into answering them today too.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby Chen » Thu Oct 03, 2013 5:12 pm UTC

steve waterman wrote:


This reference is still only one dimensional ( change along a single vector direction )..."r" or x or y or z, and NOT a 3 dimensional change as is delta (a,b,c). We are avoiding THE major issue...
the Galilean equation is x' = x-vt and not S'(x'-vt,y,z) = S(x,y,z).


The first link shows the velocity components on X, Y and Z directions so I have no clue what you're on about here.

More importantly, your green text is WRONG. The Galilean is NOT merely x' = x -vt as those links and your own wikipedia link showed. Further, I'm pretty sure your primes are messed up in that last statement since, without simplification its S(x,y,z) =S'(x',y',z'). After replacement this can be written as S'(x-vt,y,z) with no prime on the x (you could have primes on the y and z if you wanted since y'=y and z'=z, but that would just be confusing terminology wise).

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

Postby JudeMorrigan » Thu Oct 03, 2013 5:16 pm UTC

steve waterman wrote:


This reference is still only one dimensional ( change along a single vector direction )..."r" or x or y or z, and NOT a 3 dimensional change as is delta (a,b,c).

You are right that this is an ultimately minor side-issue, but two notes: first, the links there are to separate citations. The encyclopediaofmath one is quite explicit. In the latter link, r is defined by the cite as r = (x,y,z). Eq. 15.2 does, in fact, involve a three dimensional transformation.

Chen wrote:Further, I'm pretty sure your primes are messed up in that last statement since, without simplification its S(x,y,z) =S'(x',y',z'). After replacement this can be written as S'(x-vt,y,z) with no prime on the x (you could have primes on the y and z if you wanted since y'=y and z'=z, but that would just be confusing terminology wise).

Indeed. This is really important, steve. I know I and others have pointed this out to you previously. If the difference is not clear, please do not hesitate to ask why we keep harping on this point.

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

Postby eSOANEM » Thu Oct 03, 2013 6:24 pm UTC

steve waterman wrote:


This reference is still only one dimensional ( change along a single vector direction )..."r" or x or y or z, and NOT a 3 dimensional change as is delta (a,b,c). We are avoiding THE major issue...
the Galilean equation is x' = x-vt and not S'(x'-vt,y,z) = S(x,y,z).


You're messing up your primes again.

x'=x-vt is equivalent to the statement S'(x-vt,y,z) = S(x,y,z).

As for the whole one-dimensional thing, that's a convenience. In almost all cases, it is convenient to align our co-ordinates such at the displacement is parallel to one of the co-ordinates (it means all but one of our equations can be solved trivially). This is what is understood to have been done in the conventional statements of the Galilean.
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Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Postby yurell » Thu Oct 03, 2013 9:01 pm UTC

steve waterman wrote:[color=#00BF00][b]
Because S'(x'-vt,y,z) = S(x,y,z) is true,


No it's not true. This is why we didn't want to use primes — s'(x',y',z') = S(x,y,z) is true, and S'(x',y',z') = S'(x-vxt, y-vyt, z-vzt), as per page 6 of this thread. And, as per usual, it goes against the grain to implicitly include a 't' axis without explicitly stating it.

Edit: and if we orient our axes such that vy=vz=0, then we have S'(x',y',z') = S'(x-vxt,y,z)=S(x,y,z), which is still not what you wrote
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