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

Posted: Fri Oct 04, 2013 2:22 pm UTC
xkcd -

The issue...
x' = x-vt refers to the distance the abscissa is from their own origin, not to points, nor to axis, nor to the manifold.

So, in the case of r or radius r' = r-vt refers to distance...again, only one dimension. So, that posted reference does nothing to impact upon my argument, regarding the Galilean restriction to a mere one dimensional lineal movement from coincidence.

Regardless of how one believes the prime notation should be applied to points in the manifold,
they have nothing to do with the distance issue manifested by the abscissa of S and S'.

If xkcders do not wish to focus upon my abscissa issue re x' - x-vt, then so be it. I have had my say, and if xkcders do not grasp my contention, so be it. I really cannot make my contention clearer, and feel more explanation will only serve to repeat itself.

When this all started, i knew not of the manifold, and have learned about it, from xkcd posts. This behavior was not a refusal to learn, as I have often been accused. I now grasp the manifold thing/logic, and it still has no impact upon my abscissa based logic from post one nearly a year ago. That is, the manifold and its point comparison is the straw man, and the abscissa comparison remains my contention why x' = x-vt is mathematically wrong.

Hence, although I never received acknowledgement on any xkcd thread that the x' = x-vt issue was about abscissa, I see no need to keep flogging this over and over and over.

Additionally, be it known, that I also do not feel my contention was defeated in any manner, nor that either side has won this argument. I am pleased to have my position on record and thrilled to have had all the educated feedback that I was given. I have long anticipated to go to my grave with "my awareness" and am now quite comforted that my views have been made public. I had suspected/feared that my site would eventually disappear and a majority of my life's work would be lost forever with its closure. That depressing scenario is no longer a concern at this point in time.

So, I am about out of clever things to say, and I am ready to accept this impasse. There are no hard feelings, just admiration for all those willing to express their views, even those I disagree(d) with. Thanks too to the powers of xkcd for this special opportunity that I was given.

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

Posted: Fri Oct 04, 2013 2:27 pm UTC
Abscissa between two manifolds doesn't seem to exist without defining at least one point on each system. And once you define one point on each system, one point is as good as any other point for discussing the relationship between the two.

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

Posted: Fri Oct 04, 2013 2:54 pm UTC
SecondTalon wrote:Abscissa between two manifolds doesn't seem to exist without defining at least one point on each system. And once you define one point on each system, one point is as good as any other point for discussing the relationship between the two.

No need to involve the manifold or points in the manifold...
this is about comparing the abscissa distances from their origin.

No. You are thinking of abscissa as a point or as points...they are the DISTANCE from their origin.
Points have NO DIMENSION! DISTANCE has dimension.

Abscissa are distances from their own origin.
Pick a distance, any distance, as an example...say, let abscissa x = 2.
That means that ONLY x' = 2 can possibly be equal abscissa distance to the x = 2 abscissa distance.
According to the Galilean textbook x' = x - vt...if vt =3...
then x= 2 when x' = -1, whereas only abscissa x' = 2, is EVER equal to abscissa x = 2.

Now try another example...abscissa x = 257 and vt = 3
therefore, ONLY abscissa x' = 257 wrt S' manifests equality to abscissa x at distance 257,
whereas the Galilean textbook equation have would us believe/accept that 257 = 254!

Note...you are confusing a dimensional-less point ( having all 3 coordinates ) with an abscissa with only the one distance coordinate. This is not about points, in any form what-so-ever.
This is not about say, S(2,0,0). It is rather about say, distance x = 2...from S(0,0,0).

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

Posted: Fri Oct 04, 2013 2:56 pm UTC
0,0,0 is still a point. So anything relating distance from there by precise amounts - that is, not a generic "2 away" but specifically 2 along the X and 0 on the others refers to a specific point as well.

Abscissa are distances from their own origin.
Pick a distance, any distance, as an example...say, let abscissa x = 2.
That means that ONLY x' = 2 can possibly be equal abscissa distance to the x = 2 abscissa distance.
According to the Galilean textbook x' = x - vt...if vt =3...
then x= 2 when x' = -1, whereas only abscissa x' = 2, is EVER equal to abscissa x = 2.

Now try another example...abscissa x = 257 and vt = 3
therefore, ONLY abscissa x' = 257 wrt S' manifests equality to abscissa x at distance 257,
whereas the Galilean textbook equation have would us believe/accept that 257 = 254!

So, x=257, and vt =3

Then x' = 254.

So the abscissa of x' is 254, as it's that many units distant from the origin 0,0,0
(Note : I have no idea what " ONLY abscissa x' = 257 wrt S' manifests equality to abscissa x at distance 257," is trying to say. I can hazard a guess that you're saying that x' = 257 within S', except that by the formula, it doesn't. x' on the S' whazitozit is 254 because x' = x - vt, and we defined x as 257 and vt as 3)

I'm just going to point out once again, Steve - you've been arguing with people who have mathematic backgrounds and college level mathematic educations. I lack such education. So, things that are above my head I'm.... not really going to be able to address and more or less gloss over. Me just talk gud, I maff bad.

Dumb it down a couple hairs.

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

Posted: Fri Oct 04, 2013 3:14 pm UTC

You've been talking about the "textbook definition" of the transformation for a little bit now. I'd like to ask you to give us that textbook definition. Not your definition, not what you think the definition is or what it means, give us the actual definition, verbatim from an accepted textbook. Include reference, if possible, so that we can verify. Copy and paste, if you like, but make sure to give the whole thing. I positively GUARANTEE you that no textbook worth the paper it's printed on asserts that the Galilean transformation is only " x' = x - vt ". Include all the assumptions, all the names given, all the logic included. Then, perhaps we can start addressing the definitions of specific words.

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

Posted: Fri Oct 04, 2013 3:19 pm UTC
steve waterman wrote:xkcd -

The issue...
x' = x-vt refers to the distance the abscissa is from their own origin, not to points, nor to axis, nor to the manifold.

x' is the abscissa distance that a point is from the S' origin. x is the abscissa distance of that SAME point from the S origin. We can relate these two distances using x' = x-vt if S' and S are moving with respect to one another at velocity v (in the x direction) and that the x and x' axes are parallel, that t = t' and that at t=0 the origins of both S and S' coincide (technically without specifying anything for y and z, the origins need only have their x coordinates the same at t=0 for the above to hold).

Without all the extra clarification, all you have is an equation without any discernible meaning.

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

Posted: Fri Oct 04, 2013 5:04 pm UTC
steve waterman wrote:The issue...
x' = x-vt refers to the distance the abscissa is from their own origin, not to points, nor to axis, nor to the manifold.

That is not what the people who wrote the equation mean by it. Nor is that what scientists and mathematicians who use the equation mean by it.

The equation, in the context of the Galilean, is very much about "being the same point in the manifold".
It is not about "being the same distance from its own origin".

That is the crux of it all.

Jose

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

Posted: Fri Oct 04, 2013 9:20 pm UTC
Okay, Steve, here is a diagram I have drawn (in Paint! as you can tell from the less-than-marvellous quality) with v=1

You will notice that I have drawn the S' frame in red, and the S in black, and superimposed them. I have only included the axes x and t, which I shall write as (x,t). In the centre I have made sure that their origins are co-incident — S(0,0) = S'(0,0) — and that t'=t (that is, lines of constant time are co-incident).

To aid in interpreting this, I have drawn in three points — S(1,-1), S(1,0) and S(1,1). These represent an object that is stationary in the S frame (not moving in space, but moving in time, as we all are). What are the co-ordinates of each of the notes?

Note 1: S(1,-1) = S'(2,-1)
Note 2: S(1,0) = S'(1,0)
Note 3: S(1,1) = S'(0,1)

Now, I hope you can see that if we have S(x,t) = S'(x',t'), we have in this case S(x,t) = S'(x',t); that is, the only disagreement between our S and S' frames is not in the time dimension.

Now, what formula can we use to relate x and x' when both are referring to the same point? At present we have:
IF x=1 & t=-1 THEN x'=2 — this is the point S(1,-1)
IF x=1 & t=0 THEN x'=1 — this is the point S(1,0)
IF x=1 & t=1 THEN x'=0 — this is the point S(1,1)
IF x=0 & t=0 THEN x'=0 — this is the point S(0,0)

How about the solution x' = x-vt? I've already said that v=1 by design of this situation, so let's see how this holds up.

IF x=1 & t=-1 WHAT WE EXPECT IS x'=2
x' = x - vt
x' = 1 - 1*-1
x' = 2
Success!

IF x=1 & t=0 WHAT WE EXPECT IS x'=1
x' = x - vt
x' = 1 - 1*0
x' = 1
Success!

IF x=1 & t=1 WHAT WE EXPECT IS x'=0
x' = x - vt
x' = 1 - 1*1
x' = 0
Success!

IF x=0 & t=0 WHAT WE EXPECT IS x'=0
x' = x - vt
x' = 0 - 1*0
x' = 0
Success!

As you can see, for every point we've examined (and I can assure you it holds for every other point, too), x' = x-vt is a true relation for describing the position of a single point on a manifold in the frames S(x,t) and S'(x',t') if those frames are related by a Galilean transform. It is not meant to do anything else. That is all the equation is implying, nothing further. Any other meaning is the result of taking the equation out of context.

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

Posted: Sat Oct 05, 2013 2:03 pm UTC
SecondTalon wrote:0,0,0 is still a point. So anything relating distance from there by precise amounts - that is, not a generic "2 away" but specifically 2 along the X and 0 on the others refers to a specific point as well.

You bet. S(0,0,0) is indeed a point! Just as S(2,3,4) is a point. This is denied by xkcders. They insist that points only exist in the manifold. They insist that a coordinate system cannot exist without a manifold too. They insist that x' = x-vt is about points in the manifold. They will not allow me to use the term "coordinate point", nor to differentiate these points from "manifold points".

I posted a bunch of urls supporting the concept that coordinates were points. These were simply ignored. If I recall, the respond was something like, we do not care about any of these urls. I said that a Cartesian systems was the infinite set of points. This was denied. I tried other terms to differentiate these two types of points, like "selected points" or "itemized points", etc, all of which were all simply rejected.

I keep talking about x being a distance, and all I get back is that x represents the first coordinate of a point in the manifold. I mention that for 200 years that Cartesian systems were not simply devoid of points...again simply ignored. Always I hear that your points in the manifold are equal...and I agreed, numerous times...but say/note that is just a straw man.

I see new diagrams about the point relationship and want to tear my hair out. I am tired of having the point thing rammed down my throat. I am also tired of being told I am wrong because Relativity is right. Like a parent that sets the rules - "because I say so, that's why". I am stunned that only you, it would seem, Sexy Talon, have given the distance vs point concept any thought/consideration at all. I am ever so tired of being pontificated to, merely because Relativity cannot be challenged...because it has right for over a hundred years.

I am perplexed that everyone sees/insists that x = 2 refers to the manifold point S(2,0,0) and the not distance from S(0,0,0) to S(2,0,0). I will never ever ever believe that a Cartesian system is not the infinite set of points, nor that those coordinates are not eternally fixed with respect to their origin, once given S(x,y,z). I also shall never accept that given S(x,y,z)...that x is not the abscissa. I shall never accept that one cannot have a Cartesian system and not also require a manifold M.

So, resolution of this thread is certainly going nowhere soon, or ever for that matter. I tire of seeing new diagrams posted that deal with points in the manifold as proof for this straw man. I give up upon explaining that distance from their origin IS the issue and not whether or not a point in the manifold has equality of location. I want to laugh when I hear that the manifold maps coordination yet requires no reference origin, as only some of coordinates need be mapped. This seems downright "fudged" to me...but only me, apparently.

i am tired of supplying urls and diagrams and then hearing about manifold points. As I said recently , if this is not about comparing distances from origins, then I am not going to waste more time and energy simply hearing that same crap about the manifold points relationship justifying the Galilean coordinate transformation equations.

So, I guess I am done with explaining my side. xkcders deny the distance issue...so be it. To recap; xkcders deny that
a Cartesian coordinate system is an infinite set of points and I deny that the manifold relationship has squat to do with the textbook equations for the Galilean; x' = x-vt, y = y', z = z' t = t'. So, I guess i am done now, having said all that. It appears quite evident that neither side is remotely close to accepting what the other side believes - what x means in THE equation. Future discussion is therefore obviously futile.

There seems to be no reason for me to keep posting to this thread, nor to even mention any of this in some other thread down the road. As well, when I showed that with an applet, that three different sets of circles all independently converged upon the transmission site, if I recall, first this was initially agreed to. Then this was denied, and before I could respond, on that "multilateration" thread, that thread was locked. Hence, even mathematical proof of the applet's results is now denied. Any chances of actual words launching the discussing forward re distance vs points is a complete waste of everyone's time, in my opinion.

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

Posted: Sat Oct 05, 2013 3:55 pm UTC
On a phone (and admitting that I don't really know what a manifold is exactly) but if you think the others are denying all that ...

I don't think you read the thread very well. I think you read what you wanted to read, not what was written. I think you may need to take a week or two and re-read the thread.

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

Posted: Sat Oct 05, 2013 4:24 pm UTC
Sexy Talon,

http://forums.xkcd.com/viewtopic.php?f=2&t=96231

Their denial of a given Cartesian coordinate system as an infinite set of points...is sufficient enough for me to stop flogging the distance thingie too or in continuing this thread at all. Re-reading this or the monster thread will do nothing to change this scenario, so be it.

### The wheels on the bus

Posted: Sat Oct 05, 2013 5:27 pm UTC
Hi Steve -- It's good to see that your ticker is tocking correctly. I hope your recovery and rehab continue to go well. But I have to wonder what your purpose in this thread is. Is it...

... to understand the Galilean transformation? If so, you have to accept and work with the definitions we give, even if you don't like them.

... to prove the Galilean transformation is wrong? First you have to understand what the Galilean transformation is saying. No one will believe your claim when it's so obvious that you don't understand the math behind the Galilean transformation.

... to prove that the Galilean transformation isn't an isometry of the world we live it? That's a physics issue, not a math issue, so you'll never do this through pure reason. That said, no one will disagree with you on this point, because we know that physics is not invariant under Galilean transformations!

... to prove that if you give different meanings to symbols, you don't get the same equations? Again, none of us will argue this point. None of us is particularly impressed by it, either.

... to offer an alternative model for coordinate transformations? Ok, but stop claiming that you've proven the Galilean "wrong". Furthermore, you'll need to lay out precise definitions of all the terms you're using, since they seem to be subtly different from the usual definitions. Also, don't be surprised if no one cares.

To everyone else: Keep up the good fight. And remember, Coordinate systems do not move. This is a bit of sophistry that Steve likes to use to try to obscure the distinction between the coordinates and the underlying manifold. There are (at least) two valid ways to look at the Galilean transformation. I've been promoting the view that it is a coordinate transformation of the manifold of spacetime. This makes it obvious that we're not dealing with coincident systems (another of Steve's favorite go-tos) and provides the frame of mind necessary to move on to special relativity. The other is to treat the Galilean transformation as a family of coordinate transformations on space, parameterized by time. That is, the Galilean transformation is a function that takes a number (the time) and returns a function (the coordinate transformation at that time). I've been avoiding this approach, but given Steve's unwillingness to include a time coordinate, perhaps someone else will have success with it.

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

Posted: Sat Oct 05, 2013 5:53 pm UTC
Glad you're feeling better Steve - but of course take it easy, and don't overdo it yet.

That being said... I wanted to let some of this slide.. but I'm apparently not that generous. A couple of these points *really* needed to be addressed to set the record straight.

steve waterman wrote:I posted a bunch of urls supporting the concept that coordinates were points. These were simply ignored. If I recall, the respond was something like, we do not care about any of these urls.

I *might* be remembering wrong, but I believe URLs were dismissed by the majority of the xkcd forum posters from non-reputable sources only (and there weren't that many that were dismissed). The reputable ones that you posted, the common response was to point out that they didn't actually support your claim, and to try to explain how you were misreading them. But since you know how you want to read them, that was never effective (and so you may "remember" them being ignored).

steve waterman wrote:I keep talking about x being a distance, and all I get back is that x represents the first coordinate of a point in the manifold. I mention that for 200 years that Cartesian systems were not simply devoid of points...again simply ignored. Always I hear that your points in the manifold are equal...and I agreed, numerous times...but say/note that is just a straw man.

The thing about the points being equal isn't a straw-man because that's the way the equation is used. To use it differently, claim it doesn't make sense, and then try to argue that means the equation is wrong is the straw man.

steve waterman wrote:I am also tired of being told I am wrong because Relativity is right.

And this one *really* fires me up, because as far as I can tell, NOBODY is saying that you are wrong *because* relativity is right. They're saying you're wrong because you're using the equation wrong, for the wrong reasons, with the wrong definitions - best demonstrated by the fact that it *works* for what we say it's used for, and doesn't work for what you claim it's trying to do. In fact, I believe, you've even been told that the Galilean is particularly for non-relativistic environments/speeds, because it doesn't exhibit length contraction for example (in fact Schrollini already reiterated that before I got this posted, re: non-invariance of physics under Galilean transformations).

People have been saying that you won't be able to disprove Relativity without very good evidence/arguments, because it's held up so well for so long as a workable model. But that's completely independent of your misunderstanding of the Galilean.

So no, "you're wrong because, well... Relativity" is not what you've been told, or at least not by the majority. And if that's what you think you've been told, then SecondTalon is right - you need to reread the thread and understand it.

And definitely read Schollini's post in case you missed it. I do like his question about your purpose here. If we knew that, maybe we could all tailor the discussion better to it.

### Re: The wheels on the bus

Posted: Sat Oct 05, 2013 7:44 pm UTC
Schrollini wrote:Coordinate systems do not move.

i will respond to the two recent posts later...likely tomorrow.
I cannot believe that you would say the above, and actually mean it.

I also cannot believe that xkcders still will not acknowledge that points exist in a coordinate system. So be it.

On the plus side, I do believe that Schrollini asks some really good and pertinent questions. I am already taking up too much energy with this thread again and so will only want to answer Schrollini properly...but only tomorrow.

This thread is about x' = x-d..as mentioned in its title. However, when I do that, xkcders insist that this has squat to do with x' = x-vt...and that I basically cannot say it involves the Galilean. More of a semantic rebuttal, imo.
Please note that iff x' = x-d, can x' possibly = x-vt. (rhetorical)

i am actually finally getting progressively more upset by all this denial and surely this cannot be advantageous to my health. I am no longer willing to buckle under to crap like Cartesian systems do not contain points nor are they unable to move. These are both deal breakers for my continuance in this thread.

I still see thread this going absolutely nowhere soon and will forever believe that x = 2 is about distance and not exclusively about some dimensional-less point. I will never accept that a point has mapped/assigned coordination without also having a related origin. There is just so much that I disagree with that this becomes just more and more a waste of my time and this going round and round again and again is just not worth it.

I grasp that xkcders will NEVER agree that x refers to distance...so be it. I surely will lose much sleep tonight because I will not be able to stop myself from running all this silliness through my thoughts.

I am actually finally getting angry, which is not my style at all. I only see this getting worse for me, post by post.
Recap - way too many deal breakers at present, with no hope in hell of any one of them being mutually resolved.

Eleven months and some 6000 total responses is more than enough for me. If this discussion lasts more than a few more days I will be amazed. As I said, I will answer Schrollini's questions. I will shortly stop making any assertions. As well, I feel that my normally pleasant mood is currently strained to its limits and I do not need to blow a gasket on any of this Galilean challenge. It now has become stupid for me to keep flogging any of this...so, I have decided to stop all the flogging, nor in trying to convince others here at xkcd, that the Galilean generates a mathematical inequality. So be it. I will not be making any more posts today.

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

Posted: Sat Oct 05, 2013 11:12 pm UTC
steve waterman wrote:I am perplexed that everyone sees/insists that x = 2 refers to the manifold point S(2,0,0) and the not distance from S(0,0,0) to S(2,0,0).

x=2 refers to the infinite plane S(2,y,z). And keep in mind that 'distance from the origin' is a very vague term; in a Cartesian system, it's the square root of the sum of the squares of all the co-ordinates, but in other systems it's not — when we use the Galilean transform, we are no longer dealing with a Cartesian system because not all the axes are parallel. If you have a look at my diagram, you will see the t' axis is oblique with respect to the x' axis — would you say that a point at S'(1,1) is the 'same distance' from the origin S'(0,0) as S(1,1)?

You could say that S'(1,1) is one unit parallel to the x'-axis and one unit parallel to the t'-axis from the origin S'(0,0), and by the same token S(1,1) is one unit parallel to the x-axis and one unit parallel to the t-axis from the origin S(0,0), but if you look on the actual graph the 'distance' is different despite us using the same number of units! The only thing we could do is define a Cartesian system and use the length along this system to represent distance, and define all other systems in such a way that we can translate between the systems. What would such a function look like?

Well, in the case of the Galilean, x = x' + vt will give your 'distance from the origin in the x-axis', that is to say that it will project the point of interest onto the x-axis; isn't that what you want?

### Re: The wheels on the bus

Posted: Sun Oct 06, 2013 5:53 am UTC
steve waterman wrote: I grasp that xkcders will NEVER agree that x refers to distance

Of course not. x is an axis. Once you assign it a value, then it has some measurable worth.... but even then it might not be distance. Might be time. Or number of countries with broadband internet access. Or whatever.

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

Posted: Sun Oct 06, 2013 6:06 am UTC
steve waterman wrote:S(0,0,0) is indeed a point! Just as S(2,3,4) is a point. This is denied by xkcders. They insist that points only exist in the manifold. They insist that a coordinate system cannot exist without a manifold too.
(Emphasis mine)

Actually that first part is not denied by xkcders.

S(0,0,0) and S(2,3,4) are indeed both points. The points exist in the manifold, but are named in a coordinate system. This is why those same points can have different names in a different coordinate system. For example, there could be a coordinate system R in which those two points are (respectively) named R(5,0,0) and R(7,3,4). With those two coordinate systems (R and S)
S(0,0,0) = R(5,0,0) (which means that S(0,0,0) and R(5,0,0) are the same point)
S(2,3,4) = R(7,3,4) (which means that S(2,3,4) and R(7,3,4) are the same point)

A coordinate system is really a naming system for points (in the manifold).

That is the key idea.

Jose

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

Posted: Sun Oct 06, 2013 10:45 am UTC
steve waterman wrote:You bet. S(0,0,0) is indeed a point! Just as S(2,3,4) is a point. This is denied by xkcders. They insist that points only exist in the manifold. They insist that a coordinate system cannot exist without a manifold too. They insist that x' = x-vt is about points in the manifold. They will not allow me to use the term "coordinate point", nor to differentiate these points from "manifold points".

Steve, surely you can accept that we can talk about a collection of points without us being forced to introduce a coordinate system. After all, people were doing Euclidean-style geometry many centuries before Descartes developed coordinate geometry. The term "manifold" is a relatively recent invention (compared to Cartesian coordinates), but the concept is not. A manifold is basically a collection of points. And a manifold doesn't automatically include a coordinate system or require one. But we are free to add various coordinate systems to a manifold if we so desire.

In other words, the notion of a collection of points is more primitive (both logically and chronologically) than the notion of a collection of points with a coordinate system imposed on it. A coordinate system gives us a handy way of naming individual points and describing relationships between them algebraically, but the points and their relationships exist independently of any given coordinate system and they certainly do not require a coordinate system.

Does that make sense?

(To yurell, ucim, Schrollini, et al: please feel free to correct any blunders I've inadvertently made in this post!)

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

Posted: Sun Oct 06, 2013 1:46 pm UTC
My personal understanding/take/conjecture/awareness/speculation/premises/ of Galilean transformations...

Given Cartesian coordinate S(x,y)
x is the abscissa; the distance from S(0,0) to S(x,0).
(x,y) is a point.
The x axis is called the x axis, and not simply x, which is the abscissa.
S(x,y) is an infinite set of points.
S(x,y) is eternally fixed.
S(x,y) can be moved.

The Galilean equation says x' = x-vt,
it does not say that the x' axis = x axis -vt,
nor does it say (x',y') = (x,y) -vt
nor does it say S'(x',y') = S(x,y) -vt
nor does it say S'(x',y') = S(x-vt,y)

if/since x = abscissa, then when S' is moved from coincidence, x' = x, and therefore x' = x-vt is wrong since vt > 0.
A Cartesian coordinate system can exist without the need for a manifold.
Manifold points having mapped coordinates, do so wrt an origin, whether the origin itself is mapped/included in or not.
if x' ≠ x-d, then x' ≠ x-vt

Then there are these...I am NOT looking for agreement as they as purely stated as a premise. These are not up for discussion any more, either.

"Coordinate points" can be specified, for example, point P = S(2,3)...and differentiated from that coordinate point S(2,3) as a "specified point"; that itself IS also IN that coordinate system.

Specified points move as the coordinate system does.

Coincident coordinate systems cannot possess a selected point in only one system.

Recap - all these are simply for the ( public ) record, my record. I fully comprehend that no-one at all actually agrees with any of them. So be it. The reasons for your disagreement are also well documented. In the end, this all comes down to whether - given S(x,y)...does x exclusively mean the abscissa; the DISTANCE from S(0,0) to S(x,0)? Your responses to that are also well documented. I am unwilling to engage in further dickering over any of the above as neither my understanding nor the currently accepted understanding by all xkcd posters for all this is going to change. So be it.

This is all about the/my math. This is not about me personally, nor my character, i believe that covers it, and I have nothing else left to say/defend/discuss/dicker/bicker/depict/justify/pontificate/premise/state/challenge.

Many thanks to all the thread posters ( and thread viewers ) for ALL your
input/feedback/knowledge/efforts/time/energy/interest...
it has been ever so appreciated and borders upon being treasured by me.

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

Posted: Sun Oct 06, 2013 1:52 pm UTC
Dude, Even I know at this point that coordinate systems don't move.

A premise you are unwilling to budge on is flawed.

Ergo, all conclusions you draw with that premise will be incorrect.

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

Posted: Sun Oct 06, 2013 1:58 pm UTC
steve waterman wrote:My personal understanding/take/conjecture/awareness/speculation/premises/ of Galilean transformations...
[...]
S(x,y) is eternally fixed.
S(x,y) can be moved.

You know, there's that Principle of explosion...

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

Posted: Sun Oct 06, 2013 1:59 pm UTC
steve waterman wrote:My personal understanding/take/conjecture/awareness/speculation/premises/ of Galilean transformations...
S(x,y) is eternally fixed.
S(x,y) can be moved.

Seems to be a bit of a contradiction in these two points?

Edit: beaten to it by a minute...

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

Posted: Sun Oct 06, 2013 2:05 pm UTC
SecondTalon wrote:Dude, Even I know at this point that coordinate systems don't move.

A premise you are unwilling to budge on is flawed.

i appreciate your brevity in expressing this objection to this one stated concept.

The Galilean commences with S and S' being coincident.

Then, one system is moved from coincidence by vt along the common x/x' axis...
as shown here in their depiction...http://en.wikipedia.org/wiki/Galilean_transformation

Do you agree that applying vt moves an entire coordinate system from once being coincident?

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

Posted: Sun Oct 06, 2013 2:15 pm UTC
ElWanderer wrote:
steve waterman wrote:My personal understanding/take/conjecture/awareness/speculation/premises/ of Galilean transformations...
S(x,y) is eternally fixed.
S(x,y) can be moved.

Seems to be a bit of a contradiction in these two points?

Edit: beaten to it by a minute...

Yes, these need a bit more clarity...thanks for pointing this confusing set of statements out.

Given S(x,y), therefore -

S(x,y) is eternally fixed...wrt S(0,0).

Coordinate system S(x,y) can be moved, ...wrt S'(0,0).

Does that make mathematical sense now?
( note, I am NOT asking you to agree...only asking if that these two statements together are not self- conflicting. )

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

Posted: Sun Oct 06, 2013 2:18 pm UTC
steve waterman wrote:Do you agree that applying vt moves an entire coordinate system from once being coincident?

Of course not. Because the whole coordinate system includes time.

And I ask again, what is your purpose in posting at this point?

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

Posted: Sun Oct 06, 2013 2:37 pm UTC
Schrollini wrote:
steve waterman wrote:Do you agree that applying vt moves an entire coordinate system from once being coincident?

Of course not. Because the whole coordinate system includes time.

And I ask again, what is your purpose in posting at this point?

To try and recap my final stance on numerous issues as clearly as possible.
Schrollini wrote:Because the whole coordinate system includes time

I totally disagree, and this is my opinion...vt = d which represents the distance moved from coincidence between S and S' origins. i accept that that is your opinion and will not be trying to change it. So be it.

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

Posted: Sun Oct 06, 2013 3:12 pm UTC
steve waterman wrote:That is your opinion/conclusion.

No, it's my premise. And it's not just mine; it's the premise that everyone1 who uses the Galilean transformation uses. And it's a premise that is necessary to come up with the equation x' = x - vt.

steve waterman wrote:I totally disagree, and this is my opinion...vt = d which represents the distance moved from coincidence between S and S' origins.

You can work from different premises if you like. But stop acting surprised that you come up with different conclusions.

Ron Awning wrote:I've been told that Othello is a Moorish captain, but in my mind he's a magical, talking cello with a lisp. Who is right? Is there a right?

For that matter, is there a left?

1 As I mentioned before, there are other premises on which you may base the Galilean transformation. But I guarantee that those who prefer another set of premises would grant that mine are also valid.

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

Posted: Sun Oct 06, 2013 10:00 pm UTC
steve waterman wrote:My personal understanding/take/conjecture/awareness/speculation/premises/ of Galilean transformations...

Keep in mind, that these can be correct or incorrect; these are mathematical terms that whose use is as we define them.

steve waterman wrote:Given Cartesian coordinate S(x,y) [that maps to manifold M]
x is the abscissa; the distance from S(0,0) to S(x,0).

Here you're defining terms. We now know S is a Cartesian co-ordinate system (importantly S(x,y) is not a Cartesian co-ordinate system, it's a point defined by the system S), and we've defined distance to be in 'x-axis of Cartesian system S units'. Just a quick note here — you aren't summarising that this last point is true, you are defining it as true. I added in that S maps to M because it's important, if implicit at this point.

steve waterman wrote:(x,y) is a point.

Incorrect. (x,y) is a co-ordinate (this thing lives in your system S). S(x,y) is a (arbitrary) point.

steve waterman wrote:The x axis is called the x axis, and not simply x, which is the abscissa.

Importantly x is an arbitrary value. If I say S(x,0) I am referring to arbitrary 'x', not a specific x, which would otherwise be denoted with a value (e.g. 0) a number (e.g. x0).

steve waterman wrote:S(x,y) is an infinite set of points.

This is, unfortunately, a result of ambiguity in notation. S(x,y) refers to an arbitrary point, or all points on the plane (x,y) depending on context. If we mean the latter, we'll explicitly state it. This ambiguity is, unfortunately, pretty endemic and definitely not your fault.

steve waterman wrote:S(x,y) is eternally fixed.

Correct. Points in our manifold M do not move.

steve waterman wrote:S(x,y) can be moved.

Incorrect. Even were it not false by definition of points on a manifold, it would be defined to be false by the statement you made just above this one.

steve waterman wrote:The Galilean equation says x' = x-vt,

Equation taken out of context. It says S'(x',t')=S'(x-vt,t)=S(x,t); that is, to find x' such that S'(x',t) = S(x,t) one uses the equation x' = x-vt.

steve waterman wrote:it does not say that the x' axis = x axis -vt,

Correct. It states that this is true for all x', not just those along the axis centred at (0,0).

steve waterman wrote:nor does it say (x',y') = (x,y) -vt

Correct. Such a statement doesn't even really make sense without stating what one means when subtracting a scalar from a vector.

steve waterman wrote:nor does it say S'(x',y') = S(x,y) -vt

Correct. Such a statement doesn't even really make sense without stating what one means when subtracting a scalar from a point.

steve waterman wrote:nor does it say S'(x',y') = S(x-vt,y)

Correct.

steve waterman wrote:if/since x = abscissa, then when S' is moved from coincidence, x' = x, and therefore x' = x-vt is wrong since vt > 0.

S' can't move, so the rest of your statement is irrelevant. It would be like me beginning a proof with 'if Sqrt(2)=1'; it may lead to some interesting conclusions, but it wouldn't be arguing about anything remotely resembling reality.

steve waterman wrote:A Cartesian coordinate system can exist without the need for a manifold.

Incorrect. Cartesian co-ordinate systems map co-ordinates to points. Without manifolds, there are no points, and so by definition a Cartesian co-ordinate system can't exist without a manifold.

steve waterman wrote:Manifold points having mapped coordinates, do so wrt an origin, whether the origin itself is mapped/included in or not.

I believe what you're saying is 'a co-ordinate system has an origin', which is neither here nor there. I'm uncertain if one can be defined without an origin, but the ones we're dealing with definitely have an origin, so I'll just accept this point.

steve waterman wrote:if x' ≠ x-d, then x' ≠ x-vt

Without defining terms, this is symbol-vomit.

steve waterman wrote:Then there are these...I am NOT looking for agreement as they as purely stated as a premise. These are not up for discussion any more, either.

No, they're not up for discussion. If you don't want people to disagree, you'll stop changing the definition of words and symbols, stop using false premises and stop using contradictory premises. We have carefully defined these terms for you. If you want us to, we can carefully define them again. We are using them as they are used in mathematics and physics. If you want to have a discussion about mathematics and physics, you are the one obliged to learn the language. We are happy to teach you, to an extent, but when you blatantly ignore everything we've told you, don't be surprised when people start getting miffed at you.

steve waterman wrote:"Coordinate points" can be specified, for example, point P = S(2,3)...and differentiated from that coordinate point S(2,3) as a "specified point"; that itself IS also IN that coordinate system.

Incorrect. Points live on a manifold, co-ordinate systems map co-ordinates to that manifold. P = S(2,3) is just giving a name to the point S(2,3). S(2,3) is not a co-ordinate, it's a point; (2,3) is a co-ordinate.

steve waterman wrote:Specified points move as the coordinate system does.

Incorrect. Points do not move. Co-ordinate systems do not move.
Most people implicitly have time as part of their co-ordinate system and take 'snapshots' along this axis when discussing the Galilean transform, because high levels of formality are generally not required when discussing this with high school students. We've explicitly described time being part of our co-ordinate system so that there is no ambiguity for the purposes of discussing the transform in detail with you, because you weren't satisfied with lower-level explanations.

steve waterman wrote:Coincident coordinate systems cannot possess a selected point in only one system.

Presuming you mean 'If two systems, S and W are co-incident, and if S maps to a point P ∈ M, then so to does W', correct.

steve waterman wrote:I fully comprehend that no-one at all actually agrees with any of them.

Then why are you trying to use them? It's like having a discussion but using words in a completely different way to everyone else and then complaining when you're misunderstood.

steve waterman wrote:In the end, this all comes down to whether - given S(x,y)...does x exclusively mean the abscissa; the DISTANCE from S(0,0) to S(x,0)?

No, no it doesn't. At the very start of this post, and I believe in my last post as well, I said we could define distance such that S(x,t) is |x| units away from the point S(0,t). This, however, does not imply that the point S'(x',t) is |x'| units away from S'(0,t).

steve waterman wrote:it has been ever so appreciated and borders upon being treasured by me.

I'm glad you enjoy it, although disappointed that we couldn't teach you how the Galilean transform works.

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

Posted: Sun Oct 06, 2013 10:33 pm UTC
yurell wrote:
steve waterman wrote:Manifold points having mapped coordinates, do so wrt an origin, whether the origin itself is mapped/included in or not.

I believe what you're saying is 'a co-ordinate system has an origin', which is neither here nor there. I'm uncertain if one can be defined without an origin, but the ones we're dealing with definitely have an origin, so I'll just accept this point.

There's no need for the coordinates (0,0) to map to a point in the manifold. As an example, consider a coordinate system for the surface of the earth consisting of latitude, as conventionally defined, and schrollinitude, which is the longitude + 360. The domain of this coordinate system is ]-90,90[ x ]180,540[. Clearly the origin isn't in this domain, so there's no point on the manifold with coordinates (0,0). It's still a perfectly well-defined (if slightly silly) coordinate system.

(At least according to the usual definitions. Apparently, the latitude-schrollinitude system isn't a coordinate system under Steve's definitions. This is one reason (among many) that no one uses Steve's definitions.)

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

Posted: Mon Oct 07, 2013 1:05 pm UTC
This applet written by Schrollini that was crafted in a collaboration using my "3 different intersection math methods" works properly to manifest the transmission site and time.
http://rschroll.github.io/relativity/detectors-h.html

How to use this applet properly...
1 input a value for the x coordinate = from -4 to 4.
2 input a value for the y coordinate = from -4 to 4.
3 input a value for the t coordinate = from 0 to 4.
4 input a value for the velocity ( which will represent a percentage of the speed of light ) = from 0 to 0.9.
5 hit APPLY button.

Notice that the three red hyperbola instantly intersect at the exact transmission site.

6 Move the slider to the right until the gold circle encompasses all the red and blue sensors
7 Move the slider to the left until the gold circle completely disappears.

Notice that both the red set of 3 circles intersects at the exact transmission site.
Notice that both the blue set of 3 circles intersects at the exact transmission site.
As well, in a similar fashion, a set of 3 blue hyperbola will also intersect at the the exact transmission site even if
the set of sensors has a speed of light value > 0. [ Note; blue hyperbola not shown in applet. ]

Notice that the values shown/calculated at the bottom do not necessarily agree with the mathematical results of the 3 different intersection methods. That is, Relativistic math was used instead to manifest those results...which is precisely what is being challenged.

Conclusions -
Any one of these 3 MATHEMATICAL methods is sufficient to independently manifest the exact transmission site and consequently also the exact transmission time.
These 3 methods would also work when extrapolated wrt a 3d setting.

In direct conceptual contradiction, Relativity believes that...
http://www.watermanpolyhedron.com/TALL.html
The reason is that the second postulate of special relativity -- that the speed of light is constant -- denies the existence of a universal time for all observers. Even the order of events in time can be different for observers moving relative to one another; there is no absolute meaning to such time orderings.

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

Posted: Mon Oct 07, 2013 1:54 pm UTC
What, Steve, no love for the latitude-schrollinitude coordinate system? I am disappointed.

Anyway, you're absolutely correct. If you start with the premise that time is not a coordinate but a universal constant for all observers, you can reach, through a series of complicated logical arguments (namely, restating the premise), the conclusion that time is a universal constant for all observers. With a bit more work, you can also reach a number of conclusions (the observer-dependence of the speed of light, the existence of the lumniferous aether, values for the Doppler shift) that contradict experimental results. Therefore, as scientists, we are forced to reject your premise and work instead with time as a coordinate. After all, as someone very intelligent said,
While premises and mathematics can be misleading, reality itself is not subjective.

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

Posted: Mon Oct 07, 2013 2:34 pm UTC
steve waterman wrote:Check this awesome Applet by Schrollini feeding it with x=4, y=4, t=4, v=0.9. Move the sliders full forth and then back till the gold circles disappear. Witness how blue circles and red circles intersect in the exact same point. Fuck math: using a compass and a ruler I got the NUMBERS for the gold circles origin, no matter what.***

*** DISCLAIMER
Quote snippets are artist's rendering only and may not accurately represent the original post.

Hi Steve, welcome back. I hope you are recovering well and wish you all the best.
You may remember me from the Pressure thread. I hope you will give me some credit for being with you for over 100 pages and bear with me, a little.

Up there in the quote snippet, UPPERCASE, bold and underlined you got your mistake on interpreting Schrollini's amazing applet results.

If you were "living" on the BLUE spaceship moving at 0.9 East, you would MATHEMATICALLY assign the values x=0.92, y=1.00, t=0.92 to the gold circles origin on earth and STILL "mean" the EXACT SAME origin of the gold circles that an earthling on RED would call x=4, y=4, t=4.
Those (0.92 et al.) are the numbers you would get from your calculations.

The only way for you (living on the BLUE spaceship) to know how an earthling would call that same event (point at a certain time, i.e. the origin of the gold circles) would be doing a transformation to "translate" BLUE coordinates (the only one YOU could MATHEMATICALLY calculate and thus know) to RED coordinates. Note how time itself is a coordinate. But choosing the red system over the blue system is kinda arbitrary, isn't it?

But let's go back to the "math" behind the formula used by the galilean transformation, shall we? Without talking about velocity or time or anything that displeases you.

LET
x be the distance (along the x-axis) of an arbitrary point P from the origin of the coordinate system S
x' be the distance (along the x-axis) of an arbitrary point P from the origin of the coordinate system S'
d be the distance (along the x-axis) between the origins of S and S'

The Galilean Transformation is a function that goes:

x = x' - d

Its sole purpose is to answer the question
"How do I get the VALUE of x if I know the VALUES of x' and d?

Let me be more clear on this: the galilean transformation is a "mathematical formula" (a function) that let you calculate an unknown value (x, namely the position of ANY point wrt the origin of the coordinate system S) in terms of other known values (x' and d, namely the distance of the SAME point from the origin of the coordinate system S' and the distance d between the origins of S and S'). Being a function/a formula OF COURSE when the x' value (or the d value) changes the x value CHANGES accordingly!

As you can see
WHEN d = 0 then x=x', else, OBVIOUSLY, x≠x' (i.e. x=x'-d)

You got it backwards, and it is THAT simple.

You may ask, why did they bother to specify when d=0 x=x', since it is implied in x=x'-d when=0?
There's the answer: because the two coordinate systems COULD have different metrics and by specifying x=x' when d=0 you can infer they share the same metric.

Lokar

By the way, x and x' ARE NOT (as you are trying to FORCE everyone to accept) the "fixed-length" segments going from S(0,0) to S(x,0) and S'(0,0) to S'(x',0) at t=0 being THUS immutable. That assumption is pointless, dull and wrong to any math extent. x and x' are VARIABLES. The VALUE of the first changes according to the VALUE of the other.

### Re: The wheels on the bus

Posted: Mon Oct 07, 2013 5:11 pm UTC
Hello, compulsive lurker here that has been around since the original pressure thread and still pretty confused on this whole situation(thingy), and I just want to clear some issues in my own head. I feel I should start by saying I'm still in the middle of my Math education so please keep it nice and more importantly in simple terms.
Spoiler:
At this point it seems to me that the Galilean Transformation is just a way to compare the coordinate systems of a something/object across two different points in time and finding out the velocity that way? I think?

Is Steve trying to move the manifold instead of comparing the difference in the two coordinate systems? Is he confusing the map with the territory. A map(in this case a coordinate system) as in an abstract simplification of a city(a manifold respectively) and is created as a easy way to communicate to others the layout of the city(manifold)?

A particular section of a blog-post I've read recently that I feel is relevant.
http://lesswrong.com/lw/j8/the_crackpot_offer/
Eliezer_Yudkowsky wrote:When I was very young—I think thirteen or maybe fourteen—I thought I had found a disproof of Cantor's Diagonal Argument, a famous theorem which demonstrates that the real numbers outnumber the rational numbers. Ah, the dreams of fame and glory that danced in my head!

My idea was that since each whole number can be decomposed into a bag of powers of 2, it was possible to map the whole numbers onto the set of subsets of whole numbers simply by writing out the binary expansion. 13, for example, 1101, would map onto {0, 2, 3}. It took a whole week before it occurred to me that perhaps I should apply Cantor's Diagonal Argument to my clever construction, and of course it found a counterexample—the binary number ...1111, which does not correspond to any finite whole number.

So I found this counterexample, and saw that my attempted disproof was false, along with my dreams of fame and glory.

I was initially a bit disappointed.

The thought went through my mind: "I'll get that theorem eventually! Someday I'll disprove Cantor's Diagonal Argument, even though my first try failed!" I resented the theorem for being obstinately true, for depriving me of my fame and fortune, and I began to look for other disproofs.

And then I realized something. I realized that I had made a mistake, and that, now that I'd spotted my mistake, there was absolutely no reason to suspect the strength of Cantor's Diagonal Argument any more than other major theorems of mathematics.

I saw then very clearly that I was being offered the opportunity to become a math crank, and to spend the rest of my life writing angry letters in green ink to math professors.

Glad to see you're feeling better Steve.

### Re: The wheels on the bus

Posted: Mon Oct 07, 2013 6:23 pm UTC
Yoshisummons wrote:At this point it seems to me that the Galilean Transformation is just a way to compare the coordinate systems of a something/object across two different points in time and finding out the velocity that way? I think?

A coordinate system is just a way to locate points on a manifold. (I'm not going to rigorously define a manifold here, but the plane, the sphere, and the torus are all examples of two-dimensional manifolds. If you keep that in mind, you won't go too wrong.) Since there can be multiple coordinate systems on a manifold, we want a way to switch between them. A coordinate transformation is a way to switch between coordinate systems. That is, given the label for a point in one coordinate system, the coordinate transformation will give you the label for that same point in another system.

In physics, we often deal with the manifold called spacetime. This is the manifold that contains events, things that have a location both in space and in time. A ball hitting a bat, your high school graduation, and a dentist appointment are all events, since they have distinct places in space and time. (The ball, your high school, and the dentists office are not events by themselves, since they don't have a distinct location in time.) A coordinate system for spacetime, also called a reference frame, is a way to label events. And of course there are coordinate transformations between reference frames.

One such transformation is the Galilean transformation, which converts the coordinates in one reference frame to the coordinates in another frame whose spatial part is moving at a constant speed relative to the first. It tells you that if you see an event with coordinates (x, y, z, t) in one frame, that same event has coordinates (x-vt, y, z, t) in a frame moving in the x direction at velocity v.

Now what you say is true. If you have two coordinate systems related by a Galilean transformation, you can take the coordinates of several events to work out what the relative velocity of the two systems is. But that's not really the purpose of it; it's really more of a side effect.

Yoshisummons wrote:Is Steve trying to move the manifold instead of comparing the difference in the two coordinate systems? Is he confusing the map with the territory. A map(in this case a coordinate system) as in an abstract simplification of a city(a manifold respectively) and is created as a easy way to communicate to others the layout of the city(manifold)?

At this point, I'm not sure what Steve is trying to do, and frankly I don't care. But he has often confused the map for the territory, while you have the analogy exactly correct.

Eliezer_Yudkowsky wrote:And then I realized something. I realized that I had made a mistake, and that, now that I'd spotted my mistake, there was absolutely no reason to suspect the strength of Cantor's Diagonal Argument any more than other major theorems of mathematics.

Celebrate your mistakes! They are your chance to learn something. Pity those who never make mistakes; they can never learn anything.

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

Posted: Mon Oct 07, 2013 6:28 pm UTC
lokar,

I believe that you wrote the following paragraph?

"Check this awesome Applet by Schrollini feeding it with x = 4, y = 4, t = 4, v = 0.9. Move the sliders full forth and then back till the gold circles disappear. Witness how blue circles and red circles intersect in the exact same point."

You apparently agree that that three stationary red sensors return the exact transmission site and exact transmission as is shown in the red thingie at the screen's bottom.
You also agree the three blue circles intersect the red circles at the exact same time and location as the red circles do. Hence the blue circles also can be used to determine the exact transmission site and time.

You are overly focused upon what the values that Relativity comes up with in the blue detectors in the bottom of the screen. I am disagreeing with those results. That is, I am challenging those results as valid...since the blue circles and the red circles and the red hyperbola and the blue hyperbola all show the correct transmission time and exact transmission site. Of course, using Relativity's equations/math DOES indeed yield those results as itemized in the blue thingie at the screen's bottom.

What would be nice is for this applet to have a numbered grid with lines, so that locations were more visually obvious!

btw, we already have the slider that keeps time ( but only accurate to n.0, whereas the thingies at that bottom are accurate to n.00).

Obviously, there blue thingies at the bottom do NOT properly manifest the actual transmission site nor transmission time since the blue circles intersect the red circles at the instant that the gold circle disappears, and the red circles DO replicate the inputted data.

Recap - of course the blue thingie at the bottom do not agree when x = 4, y = 4,t = 4 and v = 0.9 to the transmission point and transmission time determined by either of the 4 methods that DO manifest the inputted values -
as it uses Relativity's math to manifest its numeric results.

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

Posted: Mon Oct 07, 2013 6:39 pm UTC
steve waterman wrote:lokar,

I believe that you wrote the following paragraph?

"Check this awesome Applet by Schrollini feeding it with x = 4, y = 4, t = 4, v = 0.9. Move the sliders full forth and then back till the gold circles disappear. Witness how blue circles and red circles intersect in the exact same point."

Steve- if you look up about three posts, i.e. the one you quoted, you'll see that lokar quoted you.

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

Posted: Tue Oct 08, 2013 1:11 am UTC
steve waterman wrote:You are overly focused upon what the values that Relativity comes up with in the blue detectors in the bottom of the screen. I am disagreeing with those results. That is, I am challenging those results as valid...since the blue circles and the red circles and the red hyperbola and the blue hyperbola all show the correct transmission time and exact transmission site. Of course, using Relativity's equations/math DOES indeed yield those results as itemized in the blue thingie at the screen's bottom.

For those of you following along, please know that this is entirely untrue. The same calculation is being done for the arrival times at the blue detector (which Steve doesn't believe) and at the red detector (which Steve is apparently perfectly fine with). These calculations are not at all relativistic, beyond assuming the constancy of the speed of light. You can check for yourself; the calculation of the emission coordinates is done in lines 135-146. (This code runs once per detector.) This has been pointed out to Steve on several occasions.

The irony here is that I could make the blue detector return the same numbers as the red detector only by using those relativistic equations that Steve doesn't believe.

eran_rathan wrote:Steve- if you look up about three posts, i.e. the one you quoted, you'll see that lokar quoted you.

To be fair, lokar paraphrased Steve. But lokar pointed this out, so I'm not sure what Steve's on about. Of course, that is true generally.

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

Posted: Tue Oct 08, 2013 9:50 am UTC
steve waterman wrote:lokar,
I believe that you wrote the following paragraph?
lokar wrote:
steve waterman wrote:Check this awesome Applet by Schrollini feeding it with x = 4, y = 4, t = 4, v = 0.9. Move the sliders full forth and then back till the gold circles disappear. Witness how blue circles and red circles intersect in the exact same point.

You apparently agree that that three stationary red sensors return the exact transmission site and exact transmission as is shown in the red thingie at the screen's bottom.
You also agree the three blue circles intersect the red circles at the exact same time and location as the red circles do. Hence the blue circles also can be used to determine the exact transmission site and time.

Actually I was paraphrasing your words. Anyhow EVERYBODY will agree that the blue circles and the red circles will identify the same exact event (a certain point at a specific time). As I pointed out, though, your mistake lies in the fact that you believe that the RED values to identify that point are the "right" ones (as opposite to wrong). You insist on "labeling" that event using the RED coordinate system. But what if you could make your calculation only through the blue circles?
As Schrollini pointed out, RED AND BLUE USE EXACTLY THE SAME FORMULA TO GET THE ORIGIN OF THE GOLDEN CIRCLES, but they get different values, and that will be ALWAYS true (relativity or not) since RED and BLUE positions are different!

If I was standing 5 steps ahead of a tree and you were standing 3 steps back of the same tree and we call x our distance from the tree

You would say x = 3
I would say x = -5

Since we measured the distance from ourselves, our reference frame (coordinate system) is obviously different (has a different origin) thus the difference in our x's.

To be more accurate we could have called our x's xSTEVE and xLOKAR thus stating:
xSTEVE=3
xLOKAR=-5

If you knew how distant I was from you (i.e. distance=8) you could infer how distant is the tree from me, just using your own observation/calculation (using your own circles):

xLOKAR = xSTEVE - distance
xLOKAR = 3 - 8 = -5

BTW there's the Galilean transformation applied.

Talking about an "absolute position" of the tree is totally arbitrary: if its position is calculated starting from you it will be (3,0). If it is calculated starting from me it will be (-5,0). To avoid confusion we will usually label the position of the tree using (3,0)STEVE or S(3,0) when referring to your calculation and (-5,0)LOKAR or L(-5,0) and we could say, avoiding any confusion, that x = (3,0)STEVE = (-5,0)LOKAR.
Saying that the position of the tree is (3,0) is NONSENSE since I would ask: ACCORDING TO WHO? (or REFERRED TO WHO?).

Now.
The equations used by RED to draw (and intersect) its circles at the origin of the golden circles are the exact same equations used by BLUE but they will obviously get a different value.

You are overly focused upon what is the "graphical" result (all the circles intersect on the same event - i.e. point in space at a certain time) and you are not focused enough on the "values" of the calculation which WILL BE OBVIOUSLY DIFFERENT.
If you are not convinced think about the position of the tree from you and from me: even without any motion involved, the position will be -5 for me and 3 for you!

The only way for us to get the position of the tree ACCORDING TO THE OTHER would be using a simple calculation (aka coordinate transformation).
The galilean transformation will accomplish the task like a charm if BLUE speed (relative to RED) is very small (v<<c).
But in REALITY if BLUE moves very fast relatively to RED, the "moving circles equation" will give results that will NOT transform correctly using the galilean transformation because they don't account for time as being a variable of the transformation. But this is another story...

Lokar

Also, aside from skipping the part of this post where I pointed out that your calculation, if you were living on the BLUE spaceship, would result in different values, you also skipped the part of the same post where I talk about the galilean transformation equation as a simple function to answer a simple question ("How do I get the VALUE of x if I know the VALUES of x' and d?"). I'd like you to hear your objections (if any) to that part. Do you find anything incoherent or simply hard to grasp?

EDIT for clarity and typos.

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

Posted: Tue Oct 08, 2013 1:31 pm UTC
lokar wrote: Anyhow EVERYBODY will agree that the blue circles and the red circles will identify the same exact event (a certain point at a specific time).

Yes. That has always been my focus. I assumed that was sufficient proof that since all the intersection methods yielded the same results ( the exact transmission site and exact transmission time ) that the numeric results in the "blue readouts" was well, irrelevant.
lokar wrote:As Schrollini pointed out, RED AND BLUE USE EXACTLY THE SAME FORMULA TO GET THE ORIGIN OF THE GOLDEN CIRCLES, but they get different values, and that will be ALWAYS true (relativity or not) since RED and BLUE positions are different!

Only now do i comprehend how the "red readouts were achieved". This is mostly due to the almost complete lack of explanation that is present on the applet page itself. http://rschroll.github.io/relativity/detectors-h.html
i had incorrectly assumed the red readouts were actually manifested from computation of one of the intersection methods. Indeed, this was never the case. BOTH readouts are exclusive to the Relativity equation and concepts!

Here is what would be ever so nice/pertinent to have included on the applet page.

1 instructions for use
2 the blue sensors ( not with their own origin and coordinates ) with coordinates that also use the origin as do the red sensors. Thus, they would be at (3,0) (4,0) and (3,1) wrt to red origin.
3 mentioning that the red readouts and blue readouts are solely based upon the math for Relativity
a) that when v = 0, both the red and blue readouts manifest the exact transmission site and time
b) that when v > 0, only the red readouts manifest the exact transmission site and time
c) if both the red and the blue sensors are assigned v > 0,
then neither readout will manifest the exact transmission time and exact transmission time.
4 I would LOVE to see the intersection method calculated (x,y,t) results shown/posted on the applet page for the following TOO!...
a) the red circles
b) the blue circles
c) the red hyperbola
d) the blue hyperbola
5 a note stating the following parameters
a) while the above depiction only allows for either set of sensors to remain in relative position to one another, this
is not a requirement. All that IS required is that at the INSTANT of transmission receipt, we know the present
universal location and the present universal time for each of any three sensors at THEIR OWN moment of receipt.
b) any three sensors can be used ( 2 red and a blue ) or ( 2 blues and a red )
6 grid lines per unit distance of 1.00.
7 all values specified as n.xx or even n.xxx [ x,y,t ]
8 some acknowledgment that the intersection methods/equation/logic were my contribution to this collaborative applet.

Unfortunately, i suspect this is far too much to request as explanation for this/your/our applet page. It is however, as of now, quite difficult for Joe Q User to grasp either how to use it or how the values in the readouts were achieved. Indeed, even I thought the red readouts were reflecting one of the intersection methods and not simply Relativity's calculation when v is assigned a value equal to 0.

Again, my own focus was in that the intersection methods all met at the transmission site and at the exact transmission time and that I believed this to be the only proof I needed that my proposed intersection methods all independently worked to perfection.

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

Posted: Tue Oct 08, 2013 2:01 pm UTC
steve waterman wrote:i had incorrectly assumed the red readouts were actually manifested from computation of one of the intersection methods

YOU HAD CORRECTLY ASSUMED IT!!! THE RED READOUTS ARE ACTUALLY MANIFESTED FROM COMPUTATION OF ONE OF THE INTERSECTION METHODS.
THE BLUE ONE ARE ACTUALLY MANIFESTED FROM COMPUTATION OF ONE OF THE INTERSECTION METHODS TOO.
RED COMPUTATION numeric values and BLUE COMPUTATION numeric values, obtained through the SAME INTERSECTION METHOD (RED CIRCLES vs BLUE CIRCLES) are different AS EXPECTED (go back to the tree example).