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

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

lokar wrote:
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).

What lokar said.

Also, I'll repeat my previous offer: Give me a formula that takes the three detection times and spits out coordinates for the initial event. I'll implement it, and we'll see if it works or not.

Spoiler:
And because I can see it coming already, I've already implemented both the intersection of the circles and the intersection of the hyperbolas. It's the same math for both, and that's what the code is doing right now.
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steve waterman
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

lokar wrote:RED COMPUTATION numeric values and BLUE COMPUTATION numeric values, obtained through the SAME INTERSECTION METHOD (RED CIRCLES vs BLUE CIRCLES) [b]are different AS EXPECTED[/b]

In the readouts only...that use Relativity computations, the readouts will differ from one another..as I just explained to you...when either set of sensors employs v > 0.
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).[/b]

ALWAYS TRUE in the intersection point computational results...these identify the exact transmission site and exact transmission time, which are [ presently ] not shown/apparently missing on this critical applet page.

I so find the fact that the readouts may differ as completely irrelevant - as any one of the intersection methods do manifest the exact transmission site and exact transmission time and IS THE whole point here. So what, if the readouts disagree ? This has ZERO impact upon the fact that math of any intersection works to absolute perfection. So, to be clear, I 100 percent agree that the readouts do not necessarily equate their results to one another. This is no way negates, that the intersection methods work 100 percent of the time!!!
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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steve

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

Steve, you just used 3 exclamation marks in a row. That's a sign that you need to take a break for the sake of your health. See you tomorrow?
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Yeah, I think that is a good idea to take a break for a bit.
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steve waterman
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

SecondTalon wrote:Yeah, I think that is a good idea to take a break for a bit.

Because I used three exclamation points is certainly not a justified reason for a break.

As it turns out, I have now had my say on both major issues; my x' = x-vt challenge and my challenge to an event having a determinable transmission location and transmission time. There really is little left to say, either today or in the future, on this thread. I suspect that none of my recently posted suggested applet explanations/modifications will be deemed as being that important/pertinent/worthy and I see no need to keep flogging them at Schrollini as well. I possess little to no applet writing skills myself, nor am I even inclined to try and do such a complicated thing by undertaking that particular learning curve. So, under these conditions, I do indeed have plans for a very extended posting/writing break.
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steve

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

v > 0 wrt what?
my challenge to an event having a determinable transmission location and transmission time

I thought you were asserting that, not challenging that. And your assertion is false if the speed of light is observed to be constant independent of the reference frame. And the current experiments observe just that. See Scrollini's app for more info.
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Pfhorrest
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Steve, there's something I've been saying over and over again in this (and the other) thread and I don't think I've gotten a confirmation that you understood it.

Everyone understands what you're trying to prove, and says the Galilean transformation agrees with that.

You think the Galilean disagrees with that because you are misunderstanding what it is saying.

Can I please get an acknowledgement from you that you understand that:
• we all understand your point
• which is that the length between e.g. the (0,0) and (0,2) coordinates of two coordinate systems with the same scale don't differ just because the locations of the two systems differ
• and the Galilean transformation, as understood by those who created it and those who use it, agrees with this

Can I then get you to ask this question honestly and with an open mind:
• If the Galilean transformation agrees with that, what does it actually mean when it says things like x' = x - vt?

I may be jumping the gun a little bit here, but I'd also like you to try this, since you like building physical models. Get two clear pieces of plastic and draw a coordinate system on each one, to the same scale, one in red and one in blue. Hold them up away from your desk just in the air somewhere, and overlap them, slightly offset along their x axes. Pick one mark on the blue x axis. Now look through the clear plastic and see which mark on the red x axis lines up there. What is the relationship between the two numbers at those two marks? Write an equation describing it. Now without moving the sheets, pick another mark on the blue one, look through it and see which mark on the red one lines up with it. Write an equation describing the relationship between those two numbers as well. Do that for a couple other pairs of numbers. Now, can you write an equation describing the relationship in general between any blue mark and the red mark it lines up with? Does it look anything like "red mark number = blue mark number - offset between coordinate systems"? Does that look at all similar in form to another equation we've been talking about?
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steve waterman
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Pfhorrest wrote:Steve, there's something I've been saying over and over again in this (and the other) thread and I don't think I've gotten a confirmation that you understood it.

Everyone understands what you're trying to prove, and says the Galilean transformation agrees with that.

You think the Galilean disagrees with that because you are misunderstanding what it is saying.

Can I please get an acknowledgement from you that you understand that:
• we all understand your point
• which is that the length between e.g. the (0,0) and (0,2) coordinates of two coordinate systems with the same scale don't differ just because the locations of the two systems differ
• and the Galilean transformation, as understood by those who created it and those who use it, agrees with this

Can I then get you to ask this question honestly and with an open mind:
• If the Galilean transformation agrees with that, what does it actually mean when it says things like x' = x - vt?

I may be jumping the gun a little bit here, but I'd also like you to try this, since you like building physical models. Get two clear pieces of plastic and draw a coordinate system on each one, to the same scale, one in red and one in blue. Hold them up away from your desk just in the air somewhere, and overlap them, slightly offset along their x axes. Pick one mark on the blue x axis. Now look through the clear plastic and see which mark on the red x axis lines up there. What is the relationship between the two numbers at those two marks? Write an equation describing it. Now without moving the sheets, pick another mark on the blue one, look through it and see which mark on the red one lines up with it. Write an equation describing the relationship between those two numbers as well. Do that for a couple other pairs of numbers. Now, can you write an equation describing the relationship in general between any blue mark and the red mark it lines up with? Does it look anything like "red mark number = blue mark number - offset between coordinate systems"? Does that look at all similar in form to another equation we've been talking about?

Very nicely expressed. Let me try my best to respond to all that you say.

There are two different types of comparisons that are relevant, as I understand this...

"red coordinate set = blue coordinate - offset between coordinate systems
i 100 percent totally agree with this assessment ..that is that the POINT at (2,0,0) in the red system lies opposite the point (-1,0,0) in the blue system when the separation distance between origins is 3.

My challenge is not NOT about what coordinate POINT lies opposite the number 2 on the red system when vt = 3. This is about EQUALITY of DISTANCE, in the blue system that the red 2 is from the red origin.

Pfhorrest wrote:[*]which is that the length between e.g. the (0,0) and (0,2) coordinates of two coordinate systems with the same scale don't differ just because the locations of the two systems differ

I am saying that x by itself and x' by itself refers to the DISTANCE from their own origin. I am AGREEING that the POINT Red(2,0,0) = POINT Blue(-1,0,0) - vt when vt = 3.

So, I repeat, you are comparing POINT locations between systems, whereas the Galilean as stated, actually mathematically compares the abscissa DISTANCE x from its origin to the abscissa distance x' from its origin. That is why, given x = x', remains x = x'.

Visually, you are looking across from your point to manifest equality...
Blue (x',0,0) = Red(x,0,0) - vt. whereas I am looking at the vector distance from origins to manifest equality ...so x = x' always, once given x = x'

The Galilean EQUATION x' = x-vt is unwittingly comparing abscissa distances although it professes to be comparing point locations....because if we let x = 2 that means abscissa distance. The Galilean coordinate transformation equation does not say, that S'(x',0,0) = S(x,0,0) -vt...does it? The Galilean says that x' = x-vt!

You think the Galilean disagrees with me because you are misunderstanding what their equation x' = x-vt is mathematically saying/comparing/equating.

Let me try this analogy, although i shall likely regret this and may wish to ask you to ignore this as I do not desire to open a new can of worms. If this helps...fine, if not...then I wish to delete this from any discussion.
You are comparing apples to apples. Which I agree with. [where apples are point locations]
I am comparing oranges to oranges. Which you agree with. [where oranges are abscissa lengths]
The Galilean EQUATION actually compares oranges to oranges, ( which you disagree with )
but uses this fact as a straw man ( which you also disagree with ) that it correctly compares apples to apples.
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steve

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

steve waterman wrote:The Galilean EQUATION x' = x-vt is unwittingly comparing abscissa distances

Go find some place, any place, where we said the Galilean was about comparing distances.

I dare ya.

I double dare ya.

I double dog dare ya.

You won't, because we didn't. We have always said that the Galilean is about comparing coordinates. We have specifically pointed out that you don't even need the concept of distance to get the Galilean transformation.

YOU are the the one who keeps bringing up distances. YOU are the the one who keeps insisting the Galilean is a statement about distances. No one else. So stop blaming us for your confusion.
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JudeMorrigan
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Really, steve, the "x' = x - vt" equation is just the relationship of the x-coordinate of a point in one system to the x-coordinate of a point in another system. No matter how much you may want it to be about the "abscissa distances", that's simply not how it's being used. Notation is, well, notional.

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

Ok Steve, let me try to play this in your language. (Mathy types - simply assume the overlay of suitable manifolds moving with respect to each other. Steve, just ignore the "mathy types" interjections to remain Stevian)

There is a point. I'll call it Eiffel. In the red system (R) it has coordinates of (4,0,0), which means that R(4,0,0) is the point called Eiffel.

We overlay a blue system (B) which is coincident with the red system. In the blue system our point Eiffel has coordinates of (4,0,0), so B(4,0,0) is also the point called Eiffel.

Now, I'm going to give names to coordinates. My name for the coordinate (4,0,0) as expressed in the red system will be Reiffel. With equal lack of creativity, I will call the coordinate (4,0,0) as expressed in the blue system Beiffel. (Mathy types, consider Beiffel and Reiffel to be points in appropriate new manifolds if you like)

At this, er... point... Eiffel, Beiffel, and Reiffel overlay each other.

Now, I'm going to name some distances. bx (this is one letter, pronounced bix) will be the distance from the origin of the blue system to Beiffel. Similarly, rx (one letter, pronounced rix) will be the distance from the origin of the red system to Reiffel. This essentially takes the place of Stevian x and x'

Right now, bx and rx are each equal to 4.

Like saying x=x', I will say bx=rx.

Now I'll move the red system by d units (where d=25). (Mathy types - along with its own little manifold, where Reiffel really lives) I'll move it without any rotation so that the origin of the red system is superimposed on the point B(25,0,0). This is equivalent to moving the red system twenty five units to the right, using standard graphical depiction.

Now, Eiffel and Beiffel still overlay each other, but Reiffel no longer does. It moved, and overlays B(29,0,0).

like saying x = x', in this case remember that we said bx = rx

This is still true. bx=4 and rx=4 the way they were originally defined.

bx is still the distance from Beiffel to B(0,0,0)
rx is still the distance from Reiffel to R(0,0,0)

So, with d=25, the equation bx = rx - d
is clearly false, just like Steve says!

Translated back into original Stevian notation,
x' = x - d
is also clearly false, given Steve's (similar) definition of x.

Before I go further into the implications of this, Steve, are you with me so far? Is this what you are saying?

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

Schrollini wrote:
steve waterman wrote:The Galilean EQUATION x' = x-vt is unwittingly comparing abscissa distances

Go find some place, any place, where we said the Galilean was about comparing distances.

I dare ya.

I double dare ya.

I double dog dare ya.

You won't, because we didn't. We have always said that the Galilean is about comparing coordinates. We have specifically pointed out that you don't even need the concept of distance to get the Galilean transformation.

YOU are the the one who keeps bringing up distances. YOU are the the one who keeps insisting the Galilean is a statement about distances. No one else. So stop blaming us for your confusion.

It's not that bad Schrollini.... it's worse.

You misunderstand - he's not saying that the people here are saying the Galilean is about comparing distances. He's saying that in spite of everything we're saying, in spite of all evidence to the contrary, and in spite of the fact that it actually successfully does what it purports to do, that the Galilean transformation is actually supposed to be doing what he claims, and it doesn't work, and that the rest of the world has only been deceived in to thinking it's supposed to calculate what it successfully calculates. It would be like ... actually... you know what? I don't have a metaphor for this one...

To Steve: is it really inconceivable to you that if the equation actually works when used to compare "apples to apples" as you put it, that maybe that's what it's supposed to be doing and you are misunderstanding the notation? Or are you actually fully aware, and just hoping that if you hold out long enough on imagined ambiguities you might convince someone else? Because now even you have made it clear that you understand that if used the way we say to use it, it works. Yet you're insisting that we're using it wrong. And your apparent "proof" of this is only your own interpretation of the equation, and that if you use it differently it doesn't work (which, really, supports the standard interpretation rather than yours).

If the options are
1) The equation is wrong, but happens to work for something it's not intended to do, and everyone in the world except you is confused about what the function does, and happens to be confused in to thinking it does the thing it coincidentally does successfully
or
2) You are misunderstanding the notation
well... one of those is rather more likely than the other, and I'm sorry to say it's not #1.

Why exactly are you so convinced your interpretation must be right, when even you can see the equation actually makes sense and works using the standard interpretation?

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

Schrollini wrote:
steve waterman wrote:The Galilean EQUATION x' = x-vt is unwittingly comparing abscissa distances

Go find some place, any place, where we said the Galilean was about comparing distances.

I dare ya.

I double dare ya.

I double dog dare ya.

You won't, because we didn't. We have always said that the Galilean is about comparing coordinates. We have specifically pointed out that you don't even need the concept of distance to get the Galilean transformation.

YOU are the the one who keeps bringing up distances. YOU are the the one who keeps insisting the Galilean is a statement about distances. No one else. So stop blaming us for your confusion.
I'm also going to say, Steve, that if you use anything I said in here as an argument supporting your "pro-distance Comparison" argument...

Well, we're all going to have a good hearty laugh, as I freely admit that I have no fucking clue what I'm talking about, and anything I say in this thread regarding it should probably be ignored completely.
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Schrollini
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Vetala wrote:You misunderstand - he's not saying that the people here are saying the Galilean is about comparing distances. He's saying that in spite of everything we're saying, in spite of all evidence to the contrary, and in spite of the fact that it actually successfully does what it purports to do, that the Galilean transformation is actually supposed to be doing what he claims, and it doesn't work, and that the rest of the world has only been deceived in to thinking it's supposed to calculate what it successfully calculates.

WAKE UP SHEEPLE!
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:The Galilean coordinate transformation equation does not say, that S'(x',0,0) = S(x,0,0) -vt...does it? The Galilean says that x' = x-vt!

It says that IF S'(x',0,0) = S(x,0,0) [meaning x',0,0 in S' lines up with x,0,0 in S] (and vt = the offset between S' and S) THEN x' = x - vt.

So IF S'(-1,0,0) = S(2,0,0) [meaning -1,0,0 in S' lines up with 2,0,0 in S] (and thus vt = 3) THEN -1 = 2 - 3. Which is true, isn't it?
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yurell
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Steve, you're taking symbols out of context and defining your own meanings again:
yurell wrote: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.

So let's use different symbols for a bit:
If S(x,t) = S'(f(x,t),g(x,t)), then in order for the co-ordinate systems S and S' to be related by a Galilean transform:
S(x,t) = S'(f(x,t),g(x,t)) = S'(x-vt,t)
This is by definition of the Galilean transform, and so inarguable.

So now that we have S'(f(x,t),g(x,t)) = S'(x-vt,t), is there any possible way we can break this down component wise, so that if I'm given a co-ordinate in S for a point, I can determine what the corresponding co-ordinate is in S'?
Yes! As I'm sure you can notice, if we use the equations:
(1) f(x,t)=x-vt and (2) g(x,t)=t

We have:
S'(f(x,t),g(x,t)) = S'(x-vt,t)

Sub in (1):
S'(x-vt,g(x,t)) = S'(x-vt,t)

Sub in 2:
S'(x-vt,t) = S'(x-vt,t)
As required.

In our dashed notation, x'=f(x,t), t'=g(x,t).
Last edited by yurell on Tue Oct 08, 2013 11:09 pm UTC, edited 1 time in total.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Pfhorrest wrote:
steve waterman wrote:The Galilean coordinate transformation equation does not say, that S'(x',0,0) = S(x,0,0) -vt...does it? The Galilean says that x' = x-vt!

It says that IF S'(x',0,0) = S(x,0,0) [meaning x',0,0 in S' lines up with x,0,0 in S] (and vt = the offset between S' and S) THEN x' = x - vt.

So IF S'(-1,0,0) = S(2,0,0) [meaning -1,0,0 in S' lines up with 2,0,0 in S] (and thus vt = 3) THEN -1 = 2 - 3. Which is true, isn't it?

Indeed, S'(-1,0,0) = S(2,0,0) [meaning -1,0,0 in S' lines up with 2,0,0 in S] (and thus vt = 3)...is TRUE.
That is about POINTS ALIGNING.
However,
the Galilean equation does not say that S'(x',0,0) = S(x,y,z) -vt...does it?

Look closely please, the Galilean equation in every textbook for over 100 years actually states that x' = x-vt.
There has never ever, ever been a single published textbook that states that the Galilean equation is S'(x',0,0) = S(x,y,z) -vt.

x' = x-vt is about abscissa DISTANCES having equal distances from their own origin...
2 ≠ 2-3 does it?

This is getting quite tedious...as we go round and round and you/all xkcders talk points and I talk about distances. I give up trying to convince people that x' = x-vt is about distances and not points. xkcders continue to believe that x' = x-vt is about a point in the manifold and not distances in the coordinate systems. So be it, as I no longer care if xkcders agree or not that x' = x-vt. I only wish my stance to be officially be placed on the record, and that seems to have been accomplished. I suspect that another 5000+ plus replies will do nothing to resolve the x' = x-vt debate, nor that there IS an determinable ordering of events using any one of my 3 intersection methods in the applet debate. So I am ( hopefully ) done posting here.

For the record, it will remain a mystery to me how Relativity can conclude that there is no event ordering that physically occurs just because their math cannot determine that ordered event sequence. Please, this previous statement is just a rhetorical statement and I am also not willing to enter into a counter-argument..as I highly suspect that all xkcders 100 percent disagree.

So, I AM going to take a posting break for several days if not for much longer.
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steve

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

It says that IF S'(x',0,0) = S(x,0,0) THEN x' = x - vt.

You ignored the part I've just bolded for emphasis.

The statement x' = x - vt is made in the context of some point with coordinates x',0,0 in S' and x,0,0 in S. It says if those coordinates line up, then the number x' in the coordinates S(x',0,0) has to equal the number x in the coordinates S(x,0,0) minus the offset between S' and S.

So IF S'(-1,0,0) lines up with S(2,0,0), THEN -1 = 2 - 3.

That's not saying that 2 equals 2 minus 3. It's saying -1 equals 2 minus 3.
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yurell
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:Look closely please, the Galilean equation in every textbook for over 100 years actually states that x' = x-vt.
There has never ever, ever been a single published textbook that states that the Galilean equation is S'(x',0,0) = S(x,y,z) -vt.

And never in those hundred years has there been a mathematical paper talking about magic distance changes! It has always been about co-ordinate transforms and you are just pulling symbols out of context! Most people are fine when they see x'=x-vt because they immediately understand that it means S'(x',t') = S'(x-vt,t) — so much so that such a level of formality is deemed to be unnecessary and even confusing for the high schoolers learning this stuff. The fact that we've gone through the difficulty of writing it out formally for you is because, as someone else in this thread put it, you have a Cargo Cult mentality involving symbols, treating them as some immutable thing irrespective of context.

steve waterman wrote:x' = x-vt is about abscissa DISTANCES having equal distances from their own origin...
2 ≠ 2-3 does it?

The fact that you're still asserting this bullshit means that you haven't been reading what's been written for over the past one hundred and fifty pages! No one in this thread has ever made that assertion, and no one ever will! You're being mule-headed in your arrogance that you must be right, assigning to us assertions that no one has ever made in order for you to be correct! That's not how maths works — we have written out in painstaking detail exactly what the Galilean is, all because you weren't happy with how other sources established the notion. And what did you do? Immediately retreated to those sources to show a 'contradiction' (because we had the temerity to use SYMBOLS! differently to how they did) and then proceed to disregard everything that we wrote. Schrollini even went out of his way to teach you co-ordinate mathematics from the ground up, and you dismiss this with a sarcastic "2 ≠ 2-3 does it?"

We have told you repeatedly that the equation x'=x-vt;t'=t is simply a way to find the value of the co-ordinates in a transformed frame S'(x',t') such that they refer to the same point as S(x,t). If you think it means something else, you are wrong. If you've read all of this thread up to now and think it means something else, you are wilfully wrong.

steve waterman wrote:This is getting quite tedious...as we go round and round and you/all xkcders talk points and I talk about distances.

I agree — you've made this the Carousel of the Damned.

steve waterman wrote:I give up trying to convince people that x' = x-vt is about distances and not points.

And so you should, because such an assertion is wrong.

steve waterman wrote:So be it, as I no longer care if xkcders agree or not that x' = x-vt.

Goodbye then!
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

Pronouns: Feminine pronouns please!

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

I find Steve's latest flounce (don't worry, he'll be back) amusing, if for nothing more than the quote that steve has on his VERY FIRST POST IN THIS THREAD:

steve waterman wrote:Hopefully, GENERAL is the correct section to post my two questions re The Galilean equation x' = x -vt.
I will confine any possible future xkcd comments re this topic, to this one thread here in the General section.
wiki wrote:The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) of a single arbitrary event, as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and x’ directions, with their spatial origins coinciding at time t=t'=0:

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

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

eran_rathan wrote:I find Steve's latest flounce (don't worry, he'll be back) amusing, if for nothing more than the quote that steve has on his VERY FIRST POST IN THIS THREAD:

steve waterman wrote:Hopefully, GENERAL is the correct section to post my two questions re The Galilean equation x' = x -vt.
I will confine any possible future xkcd comments re this topic, to this one thread here in the General section.
wiki wrote:The notation below describes the relationship under the Galilean transformation between the coordinates (x,y,z,t) and (x′,y′,z′,t′) of a single arbitrary event, as measured in two coordinate systems S and S', in uniform relative motion (velocity v) in their common x and x’ directions, with their spatial origins coinciding at time t=t'=0:

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

I was tempted to point out when Steve referred SecondTalon to that page in defense of his "coordinate systems move" that if he did more than just look at the picture, and actually read the description he would have seen that it was actually treating time as a coordinate, and therefore in context was not actually moving anything. But I figured making sure SecondTalon realized that if reading the page was unnecessary, and pointing it out to Steve would have been useless.

I'd forgotten he'd actually quoted that part himself....

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

Wait, how is time anything but a axis/coordinate similar to x, y and z?

Isn't the entire real-world application of this to track an object moving in a three dimensional space? So time is.. I guess a spacial coordinate or something, but if you know that a object is at 0,0,0 at t=0, at 1,1,1 at t=1, at 3,3,3 at t=2 and at 10,10,10 at t=4, and 15,15,15 at t=5 we can safely assume the object is FUCKING METAL AS FUCK at t=3

....

because at t=3, it'd be at 6,6,6, see....

I have a confession. I am steve waterman. This entire elaborate ruse was for that joke.

I hope it was worth it for you. It was for me.

/no, I'm not Steve.
heuristically_alone wrote:I want to write a DnD campaign and play it by myself and DM it myself.
heuristically_alone wrote:I have been informed that this is called writing a book.

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

I recently gave my final stance upon various issues on this thread. It is only fair for xkcders to be given their own opportunity to register their last words too. Noting that, I plan no rebuttals to any of xkcders final statements/comments that deal with any of those unresolved items already covered in this thread or in the monster thread. http://forums.xkcd.com/viewtopic.php?f=2&t=96231

btw, I only noticed today that numerical inputs for this applet
http://rschroll.github.io/relativity/detectors-h.html
can indeed be n.xx for all the allowed inputs; x,y,t, and for v at 0.xx.

I must assume at this point, that since I have not heard back from Schrollini, that he has no plans to make any of my requested augmentations to this applet. [ re my post on Tue Oct 08, 2013 9:31 am UTC addressed to lokar ]
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

SecondTalon wrote:Wait, how is time anything but a axis/coordinate similar to x, y and z?

Isn't the entire real-world application of this to track an object moving in a three dimensional space? So time is.. I guess a spacial coordinate or something, but if you know that a object is at 0,0,0 at t=0, at 1,1,1 at t=1, at 3,3,3 at t=2 and at 10,10,10 at t=4, and 15,15,15 at t=5 we can safely assume the object is FUCKING METAL AS FUCK at t=3

....

because at t=3, it'd be at 6,6,6, see....

I have a confession. I am steve waterman. This entire elaborate ruse was for that joke.

I hope it was worth it for you. It was for me.

/no, I'm not Steve.

This was actually a saner explanation to whatever is happening here than the sad reality.

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

SecondTalon wrote:I have a confession. I am steve waterman

No -- I'm Steve Waterman! And so's my wife.

steve waterman wrote:I must assume at this point, that since I have not heard back from Schrollini, that he has no plans to make any of my requested augmentations to this applet. [ re my post on Tue Oct 08, 2013 9:31 am UTC addressed to lokar ]

I dunno why you'd trust me to implement anything else, when you obviously don't believe me when I say what I've already done. But here's a point-by-point rejection of your requests, if that makes you feel any better. (The code is released under the BSD license, so you or anyone is welcome to take it an modify it to their heart's content.)

steve waterman wrote:1 instructions for use

This is the one reasonable suggestion on the list. I'm resisting it, because I prefer to have something for people to play with and discover things on their own than a strict set of things to do.

steve waterman wrote: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.

Then the blue sensors would be working differently than the red sensors, which defeats the purpose of the applet.

Besides, the only way to do this would be to do one of those coordinate transformations that you have so rigorously proven to be wrong.

steve waterman wrote:3 mentioning that the red readouts and blue readouts are solely based upon the math for Relativity

This is untrue.

steve waterman wrote: a) that when v = 0, both the red and blue readouts manifest the exact transmission site and time

This is obvious.

steve waterman wrote: b) that when v > 0, only the red readouts manifest the exact transmission site and time

This is untrue.

steve waterman wrote: 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.

This is untrue.

steve waterman wrote: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

The math for the circle intersections and the hyperbola intersections is exactly the same, and it's what's implemented in the code. You can disbelieve it all you like, but it's true.

steve waterman wrote: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.

This assumes the existence of a universal coordinate system, which the applet clear demonstrates is a fallacy.

steve waterman wrote: b) any three sensors can be used ( 2 red and a blue ) or ( 2 blues and a red )

But only if you use relativistic coordinate transformations, which you take issue with.

steve waterman wrote:6 grid lines per unit distance of 1.00.

If we knew how to draw such grid lines, we wouldn't need to do this multilateralization.

steve waterman wrote:7 all values specified as n.xx or even n.xxx [ x,y,t ]

What purpose would this possibly serve? We're not arguing about effects on the 1% scale.

steve waterman wrote:8 some acknowledgment that the intersection methods/equation/logic were my contribution to this collaborative applet.

You are hardly the first person to come up with multilateraliztion. And despite my repeated requests for you to give me a formula to implement (still an open offer!), you never did.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Schrollini,

Now I am quite confused by your recent post.

Unfortunately, these are all still unclear to me regarding how your 5 results below are accomplished in the applet
by mutlilaterization math or
by using Relativity's math or
by some other math

1 for the red readouts
2 for the blue readouts
3 red circles intersection
4 blue circles intersection
5 red hyperbola intersection
6 do 3, 4 and 5 manifest the location of the transmission site?

i was under the impression that you were using my algorithms for the 3 intersection methods.
apparently not from your comments. So, I will post the three algorithms to this thread later today.
Please, do not let that get in the way of you responding to my above questions about what math is being used by your applet.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:Now I am quite confused by your recent post.

Unfortunately, these are all still unclear to me regarding how your 5 results below are accomplished in the applet
by mutlilaterization math or
by using Relativity's math or
by some other math

1 for the red readouts
2 for the blue readouts
3 red circles intersection
4 blue circles intersection
5 red hyperbola intersection
6 do 3, 4 and 5 manifest the location of the transmission site?

Well, he made it pretty clear that everything is the same for the blue and red. You made it pretty clear that you completely and totally missed that. So I'm going to go ahead and say things are only unclear to you, and it's because your reading comprehension skills are lacking and you get confused easily. It has nothing to do with Scrollini's app, which is open source so you can look at exactly what it does if you care too.

Anyway:
I only wish my stance to be officially be placed on the record, and that seems to have been accomplished. I

Well, I would like my stance to be officially placed on the record, too. My stance is that I am super awesome, and better at math than SexyTalon, although not as awesome. Good? Good. It's officially on the record now.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

My 2d algorithm to determine the exact transmission site in a brand new applet
If I knew how to write applet code, this is what I would try to manifest.

1 set up an (x,y) Cartesian system that goes from -5 units to +5 units having its origin at (0,0)

2 allow the user to input
a) sensor a at a random (x,y) location
b) sensor b at a different random (x,y) location
c) sensor c at a different random (x,y) location [ so that it is not co-lineal with sensor a and sensor b locations ]
d) random transmission site (x,y) and the corresponding random transmission time = t seconds.

3 have a slider that commences at t = 0 seconds...up to say 8 seconds.

4 generate a GOLD circle from the transmission site at the above user given transmission time.

5 expand gold circle's radius at one unit per second.

6 continue to expand gold circle radius of the gold circle until it intersects a sensor.

7 at the moment of intersection, generate a green circle from the (x,y) of that sensor, wherein its radius also expands at the rate of one unit per second.

8 generate a green circle when the gold circle intersects with a second sensor, and generate another green circle when the gold circle intersects the third sensor. Expand that radius at the same one unit per second, too.

9 expand the green circles until all three circles intersect at at one (x,y) point.

10 post that (x,y) intersection location, calling it the intersection point.

3d algorithm

1 same as 2d algorithm just replace -5 to + 5 from all three axis.

2 same as 2d algorithm just replace (x,y) 2d locations with (x,y,z) 3d locations.

3 same as 2d algorithm just replace all the 3 circles with 5 spheres. ( one extra sensor needed for 3d)

4 post that (x,y,z) intersection location, calling it the intersection point.

Note - you will notice no v value is user inputted nor are there any blue sensors at all.

That is by design...since, at the moment of sensor receipt, velocity of the sensor can be ignored as only the (x,y) is required to manifest the proper results. One might even imagine that the three sensors begin at (0,0) at t = 0, and then move from (0,0) to where they receive the transmission signal. Note as well, that the actual time in seconds is also not needed/to be known in order to achieve the final results; the intersection point which is also the user inputted transmission site.
Last edited by steve waterman on Wed Oct 09, 2013 6:55 pm UTC, edited 1 time in total.
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"Be careful of what you believe, you are likely to make it the truth."
steve

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

1. velocity is explicitly part of the Galilean, you can't just drop it for no reason.
2. this is multilateration, we get that. it also has nothing to do with the Galilean.
3. circular logic is circular.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

steve waterman wrote:Unfortunately, these are all still unclear to me regarding how your 5 results below are accomplished in the applet
by mutlilaterization math or
by using Relativity's math or
by some other math

1 for the red readouts
2 for the blue readouts
3 red circles intersection
4 blue circles intersection
5 red hyperbola intersection

I've explained this many times before, and if you really want to know, you can go find those other explanations. But briefly, I'm doing multilateralization. From each reception of the original signal, I project a light cone backwards and find the intersection of the three cones. (Sometimes, there are two intersections; I can't tell which is the correct one in this case.) The code that gives the readouts is just finding the intersection of these three cones given the locations of their vertices.

The circles are just constant-t slices through these cones. The place where all three circles intersect must be a place where all three cones intersect.

The hyperbolas are each the intersection of two of the cones. (There are three pairs, so there are three hyperbolas.) Since each hyperbola is on the surface of two cones, any place where at least two hyperbolas meet must lie on the surface of all three cones. (And for obvious reasons, the third hyperbola will also intersect this point.)

Now you could work out the intersections of the circles or the intersections of the hyperbolas separately. But you better get the same answer as from doing the cones directly, or else you've made a mistake.

steve waterman wrote:6 do 3, 4 and 5 manifest the location of the transmission site?

Yep. They give different coordinates, but the same event. Eventually, you may accept that there's a difference between the map and the territory. Then this statement will make sense, not before.

eran_rathan wrote:2. this is multilateration, we get that. it also has nothing to do with the Galilean

Not quite -- it's incompetent multilateraliztion. Steve seems to be looking for the intersection of the forward light cones, which will not be at the original event and may not even exist.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Schrollini,

Do you agree that some mathematical method* can be employed which can determine the exact transmission time;
given knowledge from 4 known locations and their own receipt time of a singular event?

* regardless of whatever that method may be called
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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

steve waterman wrote:Schrollini,

Do you agree that some mathematical method* can be employed which can determine the exact transmission time;
given knowledge from 4 known locations and their own receipt time of a singular event?

* regardless of whatever that method may be called

Steve,
do you agree that two observers* placed in different positions (i.e. RED vs BLUE) using some mathematical method (i.e. multilateralization) will get different spatial coordinates (because the calculated coordinates are wrt the observers' origins) for the same singular event?

*given knowledge from 4 known locations and their own receipt time of a singular event?
Last edited by lokar on Thu Oct 10, 2013 12:59 pm UTC, edited 1 time in total.

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

steve waterman wrote:Do you agree that some mathematical method* can be employed which can determine the exact transmission time;
given knowledge from 4 known locations and their own receipt time of a singular event?

Well, working only with arrival times, there may be an ambiguity between two transmission events. But ignoring this issue for the moment (as it can be easily solved), yes there is a way to determine the transmission time.

But the important point is, another such detector in another reference frame will also correctly determine the transmission time, and these two times may not be the same. This isn't because one of the detectors is broken. This is because this is how the universe works.

Because of this, we are forced to reject the notion of a universal time and a universal order of events. This doesn't mean that a detector can't measure the times of a series of events; it certainly can. Moreover, everyone will agree on the order of those events as measured in that reference frame. But another detector may come up with a different ordering, and this ordering is just as good as the first. (It's also worth remembering that not all pairs of events are subject to this ambiguity. All reference frames will agree on the ordering of causally-connected events; those do have a universal order.)

This is why everyone scoffed at your suggestion to build a simulation of only a single detector. The problem isn't whether a single detector works, it's what different detectors in different reference frames measure.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Schrollini,

So, if we have just ONE reference frame ( only one origin ),
then any set of four sensors has the ability to manifest identical transmission times for the same event?
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
"Be careful of what you believe, you are likely to make it the truth."
steve

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

Surely if there are four detectors they're all in different reference frames by definition?
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Schrollini
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

ahammel wrote:Surely if there are four detectors they're all in different reference frames by definition?

Yes, but I'm willing to be sloppy and say that all comoving observers are in the same reference frame. After all, they are related by a translation, which is simple enough that ....

Oh, crap.

steve waterman wrote:So, if we have just ONE reference frame ( only one origin ),
then any set of four sensors has the ability to manifest identical transmission times for the same event?

All comoving detectors (those having no relative velocity) will measure the same times for events, up to a constant offset, just as they measure the same position in space for those events, up to an offset.

However, your question as posed is counterfactual. We don't have just one reference frame. We can never have just one reference frame. As soon as we consider spacetime, we must consider that there are an infinitude of reference frames. And the principle of relativity tells us that none of them are special.
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Schrollini wrote:However, your question as posed is counterfactual. We don't have just one reference frame. We can never have just one reference frame. As soon as we consider spacetime, we must consider that there are an infinitude of reference frames. And the principle of relativity tells us that none of them are special.

And I'll add that experiemental evidence supports the principle of relativity on this. The universe is what it is, regardless of whether or not we like it.

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

I am off to another doctor's/hospital appointment now.
I will be posting again perhaps only around 4:00.
"While statistics and measurements can be misleading, mathematics itself, is not subjective."
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

And since you'll be on Doctor Time, we'll expect you around 7-8 instead of 4.

(Not that I know what time zone you're in, so... keep that in mind. Time is subjective. +/- GMT times or stating a timezone is always nifty if your location isn't correctly filled out)
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### Re: Galilean:x' with respect to S'? AND SPECIAL BONUS x' = x

Because we're getting into similar territory here, I want to copy one of my last posts from the old thread which never got a response before it was locked:

Steve, the scenario we need to be discussing is about:
• what two theories (two mathematical-models-of-reality), relativity and classical/Newtonian/"Cartesian" physics,
• that make different predictions about reality,
• will say about what two observers (two sets-of-enough-sensors-to-do-multilateration),
• that are in different frames of reference (moving at different velocities),
• will say about two events (two spots-in-space-and-time),
• that are space-like separated (further apart in space than they are in time)

Same predictions? No problem to discuss.
If the two theories made the same predictions, there would be no disagreement and no problem to discuss. Thankfully we all agree (it seems, I hope) that they make different predictions, because they do.

Only one theory? No problem to discuss.
If there was only one theory, of course it would agree with itself and there would be no problem to discuss. So there's no point in just looking at "what if" one theory was true; yes, it would agree with itself, that's trivial, nobody questions it.

Observers moving at the same velocity? No problem to discuss.
If the two observers are moving at the same velocity, then the two theories will say that both observers will agree on when and where each event is, and there is no problem to discuss.

Only one observer? No problem to discuss.
If there is only one observer then of course it agrees with itself, and there is no problem to discuss.

Events closer in space than time? Not much problem to discuss.
If the two events are closer in space than they are in time, then the two theories will say that both observers will agree on the order of the events, and there is not much problem to discuss. The two theories will disagree about how far away in space and time the two observers will say the two events are, but the theories will still agree that the observers will agree on the order of the events.

Only one event? Not much problem to discuss.
If there is only one event then of course it comes at the same time as itself, and there is not much problem to discuss. The two theories will disagree about how far away in space and time the two observers will say the event is, but with only one event to speak of there can be no question of order.

You keep asserting, and nobody disagrees, that a set-of-enough-sensors-to-do-multilateration can compute the spot-in-space-and-time that something happened according to the classical/Newtonian/"Cartesian" mathematical-model-of-reality. That is not at issue at all; everybody agrees with that. Everybody agrees that relativity would agree with the classical/Newtonian/"Cartesian" theory there, even. And everybody agrees that both theories will agree that two observers moving at the same velocity would agree on the event, too.

What is at issue is what the two different mathematical-models-of-reality predict will happens when there are two different sets-of-enough-sensors-to-do-multilateration, and whether they will agree on the spot-in-space-and-time that something happened; and most prominently, whether they will agree on the order in time of two different things that happened.

Can you agree that that is the scenario we need to be discussing? What two theories with different predictions say two observers in different reference frames would say about the order of two events separated by a space-like interval. Because about any other scenario, nobody disagrees about anything, so what are we even discussing at all?

And it is very important that you understand what all the words in those bold bits mean, if we're going to discuss that:
• theory,
• prediction,
• observer,
• reference frame,
• event, and
• space-like interval.
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