Relativity or something [Split from "Pressures"]

[The Moomin]

*Edited because I realised I have no idea how to pronounce P', so couldn't be sure it was scanning or not.

I think it's usually "P-prime", so I think it retains the rhythm.

Thankyou kindly. In my head, I'd been calling it 'P dash', which is all kinds of wrong I guess.

[Pfhorrest]

Thankyou kindly. In my head, I'd been calling it 'P dash', which is all kinds of wrong I guess.

...dash? As in Le'a "the dash don't be silent" Ledasha?

[Max™]

Thankyou kindly. In my head, I'd been calling it 'P dash', which is all kinds of wrong I guess.

...dash? As in Le'a "the dash don't be silent" Ledasha?

Optimus', that is all.

[Kovacs88]

Red (x,y,z) = Blue (x',y',z')

FIXED POINT SPACE

http://www.watermanpolyhedron.com/FIXEDP.html

If nothing moves, what's your point?

[Radical_Initiator]

Red (x,y,z) = Blue (x',y',z')

FIXED POINT SPACE

http://www.watermanpolyhedron.com/FIXEDP.html

If nothing moves, what's your point?

P. And it is your point, too. In fact, it is The Point. Praise be to the One!

[Max™]

Red (x,y,z) = Blue (x',y',z')

FIXED POINT SPACE

http://www.watermanpolyhedron.com/FIXEDP.html

If nothing moves, what's your point?

P. And it is your point, too. In fact, it is The Point. Praise be to the One!

Hallelujah!

[Monika]

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA

Good summary

[steve waterman]

Red (x,y,z) = Blue (x',y',z')

FIXED POINT SPACE

http://www.watermanpolyhedron.com/FIXEDP.html

If nothing moves, what's your point?

The results are observer/system dependent, making this a variant transformation, since + 3 != -3.

[Max™]

Red (x,y,z) = Blue (x',y',z')

FIXED POINT SPACE

http://www.watermanpolyhedron.com/FIXEDP.html

If nothing moves, what's your point?

The results are observer/system dependent, making this a variant transformation, since + 3 != -3.

[Vetala]

The results are observer/system dependent, making this a variant transformation, since + 3 != -3.

The results are only observer/system dependent if the systems are doing something different relative to the point. If the point is fixed in space, then either observer will see the same as the other observer, assuming they are in the same place relative to the point. If the observers are in different places relative to the point, of course they won't see the same thing. They shouldn't and nobody but you even seems to think they should. And of course if they move opposite directions relative to the point, they definitely won't see the same thing.

The context in which you are trying to use "invariant" is meaningless. By the definition you seem to want to use, all transformations/functions are variant because if you put different inputs in to the function, you get different outputs. What a surprise that is! I can't think of too many functions for which that isn't the case with the exception of f(x) = x-x and f(x)=x*0.

Even if we give you the benefit of the doubt that you're trying to say something there that makes some modicum of sense (and after 50 pages, that's a pretty generous benefit), you're using a function designed on one set of definitions, while applying a completely different set of definitions.

If I say that distance = velocity * time, nobody would agrue. If I then claim that in d=v*t d means distance from the moon, that's only going to hold in a couple of specific cases - particularly those where the moving object started at the moon. Does this mean that d=v*t is wrong? Or does it mean my definitions are wrong for the formula? I suppose it could be either - but given the formula works until I redfine things, it's probably my definitions. This is what you are doing, in case I wasn't making that point clear enough for you. I could repeat that again, but hopefully it's getting through.

Beyond that, your entire argument appears to boil down to:
1) If x'=x-d is variant, then x'=x-d is wrong
2) If x'=x-d is wrong, then existing math is wrong
3) If x'=x-d is not equivalent to x'=x+d, then x'=x-d is variant
4) x'=x-d is not equivalent to x'=x+d
5) Therefore x'=x-d is variant
6) Therefore existing math is wrong
7) Therefore steve pseudo-math is right

Conclusion #7 doesn't follow from any premise. And in premises #3 and #4 you're falling for/relying on any of half a dozen informal logical fallacies (including, but not necessarily limited to argumentum ad nauseam, circular cause and consequence, false dilemma, straw man, and an association fallacy). Your entire side of this discussion wouldn't pass a first year philosophy course except in the fallacies unit - and as much as you want to claim that math isn't subjective, it certainly is when you start introducing errors like that in to it - because it's not math anymore, it's philosophy - and you did that, not us.

If you can demonstrate that x'=x-d is not a valid representation of relative position in the mathematical system it's intended to describe, without resorting to subtle changes to definitions of terms both in the equation and in the language, then perhaps some day someone might take you seriously (although, frankly, given how long you've been tilting at this windmill you're going to be fighting a lot of skepticism even if you do find a legitimate issue).

I do wish you the best of luck straightening out your logic and math quickly. The conclusion won't be what you want it to be, and I know that'll be painful. But you're clearly a .... passionate and motivated person, and it would be a shame to waste all that energy on windmills.

[edit, typos]

[J L]

I just returned from vacation and I must say I'm sad this thread has again reached a dead end ... despite the incredible efforts of the latest generation of volunteers trying to put into words what has been said so often before.

I think it was this slight hope that some kind of contact might be possible which kept people going ... like the Solarists on their station, riddling the ocean ... for it was never a problem of mathematics, or physics, but an inability to understand each other. Sometimes, with only few people speaking, a breakthrough seemed almost imminent; then Steve failed to reply in the expected way or hit the reset button once more, other people squeezed in again and the bubble burst. It is sad -- but I don't believe in the possibility of contact anymore. Even if some of you visited him and sat down at his table with him and went through the equations with him step by step. And certainly not on the internet.

[Kit.]

The results are observer/system dependent,

The coordinates are observer-dependent. By definition.

making this a variant transformation, since + 3 != -3.

You seem to miss the very basics.

Only scalar values (like vectors' scalar products) are transformation invarants. Coordinates aren't scalars, they are vector components.

[steve waterman]

[Monika]

Why do you write 0 = 0 - d (this means zero = zero - d, not origin = origin - d, right?), when clearly 0 = 3 - 3?

[Radical_Initiator]

And am I wrong, or does steve not recognize that he's comparing two different points? With either shift, red 0 now maps to blue -3, and blue 0 now maps to red 3. But once you move the systems from coincidence, you can't just demand that red 0 and blue 0 map to the same values anymore. Or did I miss something?

[jpers36]

And am I wrong, or does steve not recognize that he's comparing two different points? With either shift, red 0 now maps to blue -3, and blue 0 now maps to red 3. But once you move the systems from coincidence, you can't just demand that red 0 and blue 0 map to the same values anymore. Or did I miss something?

For Steve, the map is the territory. So Red 0 = 0 = Blue 0. Invariant.

[Radical_Initiator]

And am I wrong, or does steve not recognize that he's comparing two different points? With either shift, red 0 now maps to blue -3, and blue 0 now maps to red 3. But once you move the systems from coincidence, you can't just demand that red 0 and blue 0 map to the same values anymore. Or did I miss something?

For Steve, the map is the territory. So Red 0 = 0 = Blue 0. Invariant.

I suppose that's it. I keep seeing these examples of the "variant" transformation and thinking "You've just changed the order and moved things around."

[Kit.]

And am I wrong, or does steve not recognize that he's comparing two different points?

He compares points. That's enough to make him wrong.

He should have compared observables instead.

[victoertom]

Why only the swiss?

[Max™]

That's the first on-topic non-sequitur I think I've ever seen, bravo!


Einstein was a Swiss patent clerk, btw.

[eran_rathan]

That's the first on-topic non-sequitur I think I've ever seen, bravo!


Einstein was a SwissGerman patent clerk working in Switzerland, btw.

ftfy.

[steve waterman]

And am I wrong, or does steve not recognize that he's comparing two different points?

He compares points. That's enough to make him wrong.

He should have compared observables instead.

btw, There is no "event" nor "point", just the tuplets, in my given example of two systems origin's separated by three abscissa units.

Even with a timed event, say an, exploding manhole, my COMPARISON is NOT squat to do with the manhole observations.
It compares the obsevarables in the two systems, to one another. Since we were coincident, and are observables were equal, then
one system one way observables must equal the other system the opposite way, observables, otherwise, the observables results
depend upon which arbitrary system the observer is in.




One observer, observers his observable as minus three, the other observer, observers their observable as plus three.

The observable value solely depends upon an arbitrary choice of which observer the system was in!

We again, come round to this same thing. Given x' = x, then a coordinate transformation that concludes that
x' does not equal x, is obviously mathematically wrong. The perpetual problem is that xkcd responders do not
agree that simply by mathematically allowing x' = x -d, actually manifests any change to the fixed/given systems.

The resolution to this thread, is then also, the resolution of this one issue. I am certain that it does manifest changing the coordinates values in the moved system, and xkcd is certain that is does not. Being math, this is only one valid answer here.

x' observes x -d is cool, but x' equals x -d is mathematically wrong, simply because x' = x was a given.

[philip1201]

That's the first on-topic non-sequitur I think I've ever seen, bravo!


Einstein was a German-born Swiss German patent clerk working in Switzerland, btw.

ftfy.

ftfy²

(Einstein resigned his German citizenship in 1896 in order to avoid military service, and was granted a Swiss passport in 1901).

[ivnja]

There once was this Waterman bloke
Who became the punchline to a joke.
We said to him, "Steve,
Why do you disbelieve
This transform obvious to normal folk?"

[jpers36]

btw, There is no "event" nor "point", just the tuplets, in my given example of two systems origin's separated by three abscissa units.

Either you're lying about your example, or you're lying about its applicability to the Galilean. The Galilean has points. We've shown this to you so many times that you have no further excuse.

[ivnja]

He can't see what the given means
And his logic is something obscene.
We wish we could help
But he'll shout and he'll yelp
About x' equals x less d.

[Monika]

There is no "event" nor "point", just the tuplets

See, and here we have the problem. A coordinate transformation transforms the coordinates of a point in one system to its coordinates in another system. You can't talk about coordinate transformation without talking about points.

One observer, observers his observable as minus three, the other observer, observers their observable as plus three.

This doesn't mean anything. Please rephrase.

Given x' = x, then a coordinate transformation that concludes that x' does not equal x, is obviously mathematically wrong.

I read a neat thing on Twitter today, I am going to quote it:
"That's it. I am ragequitting life."

Given x' = x at time 0 does not mean that x' = x at some other time. You are still mixing up vectors and points.

x' observes x -d is cool

This doesn't mean anything. x' does not observe anything. Please rephrase.

but x' equals x -d is mathematically wrong, simply because x' = x was a given.

It was given, and then one system was moved by 3 units, so now it does not hold true anymore. And we are comparing a point's coordinates, not two vectors.

Do you agree or disagree that the distance between (4,0,0)_S and (1,0,0)_S' is 0 (zero)?

[eran_rathan]

He can't see what the given means
And his logic is something obscene.
We wish we could help
But he'll shout and he'll yelp
About x' equals x less d.

+1 plot point for continued use of limericks.

[eran_rathan]

One observer, observers his observable as minus three, the other observer, observers their observable as plus three.

This doesn't mean anything. Please rephrase.

This marklar doesn't care about marklar marklar. He just wants to take your marklar and marklar his own marklar. The only marklar for this is to marklar.

[steve waterman]

One observer, observers his observable as minus three, the other observer, observers their observable as plus three.

This doesn't mean anything. Please rephrase.

This marklar doesn't care about marklar marklar. He just wants to take your marklar and marklar his own marklar. The only marklar for this is to marklar.

one observer sees -3, the other observer sees +3

[eran_rathan]

One observer, observers his observable as minus three, the other observer, observers their observable as plus three.

This doesn't mean anything. Please rephrase.

This marklar doesn't care about marklar marklar. He just wants to take your marklar and marklar his own marklar. The only marklar for this is to marklar.

one observer sees -3, the other observer sees +3

Because they are looking at two different situations/points/events/places/locations.

[ivnja]

but x' equals x -d is mathematically wrong, simply because x' = x was a given.

It was given, and then one system was moved by 3 units, so now it does not hold true anymore. And we are comparing a point's coordinates, not two vectors.

Do you agree or disagree that the distance between (4,0,0)_S and (1,0,0)_S' is 0 (zero)?

[Vetala]

simply because x' = x was a given.

"Given"??? There you go again. You aren't even using the term correctly for how you are trying to abuse it. How can I put this, and keep within the appropriate spirit of this thread now...

You're using the wrong definition
Of what's just the "initial condition"
An observer then moved
And all that you've proved
Is you don't grok observer position

x'=x is a joke.
And where I think you're trying to choke
is remembering the "d"
is important, you see
(understood by the real math folk)

Your so-called "given" is really
d=0... (and to put it civilly)
your meaning won't allow
any movement now
so your entire proof is just silly

[ivnja]

simply because x' = x was a given.

"Given"??? There you go again. You aren't even using the term correctly for how you are trying to abuse it. How can I put this, and keep within the appropriate spirit of this thread now...

You're using the wrong definition
Of what's just the "initial condition"
An observer then moved
And all that you've proved
Is you don't grok observer position

x'=x is a joke.
And where I think you're trying to choke
is remembering the "d"
is important, you see
(understood by the real math folk)

Your so-called "given" is really
d=0... (and to put it civilly)
your meaning won't allow
any movement now
so your entire proof is just silly

Bravo.

[AdmiralGreene]

This has pretty much proven my idea on how Steve handles changes in topics/proof that he is wrong.
He posts an image (sometimes old, sometimes all sparkly and new!) and says some nonsense about points and x this and x' that in an attempt to bring the topic back into a realm he is comfortable with.

So if managed to avoid de(or is it re-?)railment from steve, we might actually manage to begin moving on a linear instead of circular path.

[chenille]

one observer sees -3, the other observer sees +3

Absolutely: when you go one way relative to me, you see me go the opposite way relative to you. The funny thing is, we actually agreed to this difference a long time ago, but you ignored us because it wasn't helping your case. Because without being wrapped in a layer of enigmatic alternate definitions, it becomes obvious to the whole world that there's nothing unusual about this, let alone contradictory.

Steve Waterman moved 3 to my left
and proclaimed to all he knew best:
for when he looked at me
he saw me go right three.
But this "variance" left noone impressed.

[Max™]

That's the first on-topic non-sequitur I think I've ever seen, bravo!


Einstein was a Swiss patent clerk, btw.

ftfy.

ftfy²

(Einstein resigned his German citizenship in 1896 in order to avoid military service, and was granted a Swiss passport in 1901).

*knew this*

Ty for correcting the correction though.

[eran_rathan]

man, I suck at limericks.


"Aaaaannd writin' stuff would be easy, I'd be happy all the time;
because Nordic poetry alliterates, it does not have to rhyme."
- 'If I was a Viking,' Scott F. Vaughan

[Max™]

I thought I would write
A witty Pressures haiku
...ran out of syllab-.

[Kovacs88]

"Given"??? There you go again. You aren't even using the term correctly for how you are trying to abuse it. How can I put this, and keep within the appropriate spirit of this thread now...

You're using the wrong definition
Of what's just the "initial condition"
An observer then moved
And all that you've proved
Is you don't grok observer position

x'=x is a joke.
And where I think you're trying to choke
is remembering the "d"
is important, you see
(understood by the real math folk)

Your so-called "given" is really
d=0... (and to put it civilly)
your meaning won't allow
any movement now
so your entire proof is just silly

Love it.

Seriously Steve, your whole proof now seems to be "when I look at him he's west of me, and when he looks at me I'm east of him. Weird."