Discrete Space

>the universe is isomorphic to an infinite 3d grid spaced by Planck lengths
Mind showing me your peer-reviewed citations?

can you lay out your model in a more precise manner than just saying it is?

Yeah, daily reminder that any cartesian plane computers use to solve problems is discrete.

t. spent 10 minutes looking at random wikipedia pages and now is suddenly a "physicist"

>can you lay out your model in a more precise manner than just saying it is?
Precision in what sense? The universe is modeled (in a graph thereotic fashion) by a metric graph isomorphic to [math] \mathbb{Z}^3 [/math], where every edge has a Planck length

>the universe is isomorphic to an infinite 3d grid spaced by Planck lengths
In what category? Where can I read the proof?

can you for example show that your model is consistent?

you're such a goddamn fucking retard. stop posting about shit you know nothing about

>can you for example show that your model is consistent?
This is a meaningless notion.

Consistent with what?

>discrete space
>Cartesian coordinates
This is not well-defined though.

>In what category?
The category of metric graphs, see

>you're such a goddamn fucking retard. stop posting about shit you know nothing about
Can you try reformulating whatever your argument is supposed to be without being so emotional next time?

The category of metric graphs has not been shown to exist yet.

only a complete retard would follow your thought process
>Hmm... the universe is like R^3 definitely!
>But muh planck, so I guess it'z Z^3!
>Let's claim this is fact on Veeky Forums, great idea!

you don't have a fucking clue and don't deserve any hints or explanation beyond fuck off

what if space locally changes curvature :-DD

>>Hmm... the universe is like R^3 definitely!
This isn't a part of my though process, can you try again without strawmanning next time?

enlighten me then, idiot, why is the universe like Z^3 other than "hurr it's obviously 3d I guess"?

>Hmm... the universe is like R^3 definitely!
It can be shown that the universe isn't finitely generated as an [math]\mathbb{R}[/math]-module.
>But muh planck, so I guess it'z Z^3!
It can be shown that the universe isn't finitely generated as a [math]\mathbb{Z}[/math]-module.

This is actually true and it was proved by Einstein and Tesla.

>I don't know what green arrows are
either way, show some citation if you make a claim like that

>>I don't know what green arrows are
Who are you quoting?

It follows trivially from the definitions, no respectable scientist would even consider publishing it.

I'm not asking you for a publication, dumbass. I'm asking you for a sketch of proof or a note showing how, and what exactly you mean with your ambiguous shit.

>I'm not asking you for a publication, dumbass. I'm asking you for a sketch of proof or a note showing how, and what exactly you mean with your ambiguous shit.
Do you need to swear? It really invalidates your post.

>I don't actually know shit and I'm just pretending to be retarded
I figured. How about go fuck yourself now?
>>>/reddit/

In what way is it ambiguous? The proof uses the fact that if the universe is finitely generated as a module over either [math]\mathbb{Z}[/math] or [math]\mathbb{R}[/math], then light has non-trivial mass.

>>I don't actually know shit and I'm just pretending to be retarded
Who are you quoting?

what do you mean as "the universe"? the model that assigns to it a space-time manifold? are you then saying that one is not isomorphic to R^n as a manifold? with or without the metric? again, give some reference or you might as well have said nothing

>what do you mean as "the universe"?
I mean "the universe".
>are you then saying that one is not isomorphic to R^n as a manifold?
It's not isomorphic to any [math]\mathbb{R}[/math]-module which is free on a finite set, so it follows trivially that it's not isomorphic to it as a manifold.
>with or without the metric?
There actually exists no non-trivial metric on the universe, this follows (trivially) from it not being locally presentable as a category.
>again, give some reference or you might as well have said nothing
Any undergraduate course in physics should suffice.

>R-module which is free on a finite set
Those are the only kinds of R-modules.

>I don't actually know shit and I'm just pretending to be retarded
I figured. How about go fuck yourself now?
>>>/reddit/

>It's not isomorphic to any R-module which is free on a finite set
>There actually exists no non-trivial metric on the universe
>this follows (trivially) from it not being locally presentable as a category.
nigga what are you doing, stop

Consider the free [math]\mathbb{R}[/math]-module [math]\bigoplus_{n \in \mathbb{N}} \mathbb{R}[/math]. Is this "finitely generated"?
How high is your autism?
Sorry?

stop.
all R modules are free on a finite set.
there's no such thing as a "trivial metric", and all manifolds have a metric by whitney's embedding theorem. specifically, spacetime doesn't mean much if you don't specify a metric, asshole.
being "locally presentable as a category" doesn't mean shit, it's just words stuck together by some dumbass pretending to know math.

>say all R modules are free
>get a reply that not all R modules are finitely generated
you're such a fucking dumbass

>stop
>shit
>asshole
>dumbass
>fucking
>dumbass
Are you okay?

see

I'm just asking you out of genuine concern since you seem stressed out by something or woke up on the wrong side of the bed. How are you doing?

>someone just called me an idiot when I tried to pretend I knew math
>oh I know, I'll ask them about their stress levels and well-being :)
retard
you don't seem to understand that someone can cuss you out without being stressed, if you're acting like a dumbass

>all R modules are free on a finite set.
Are you genuinely retarded? By definition, anything in the image of the free [math]\mathbb{R}[/math]-module functor [math]F^\mathbb{R} : \mathbf{Set} \to \mathbb{R}\mathbf{Mod}[/math] is the free [math]\mathbb{R}[/math]-module on some set. In what way is [math]F^{\mathbb{R}}(\mathbb{N}) \cong \bigoplus_{n \in \mathbb{N}} \mathbb{R}[/math] free on a "finite" set?
>there's no such thing as a "trivial metric"
Any metric which is not non-trivial is a trivial.
>and all manifolds have a metric by whitney's embedding theorem
That metric need not be non-trivial in general.
>being "locally presentable as a category" doesn't mean shit
see ncatlab.org/nlab/show/locally presentable categories - introduction
Even someone of your intelligence should be able to understand.

>say all R modules are free
You said that every [math]\mathbb{R}[/math] is free on a finite set, which is precisely the same as saying that it is finitely generated. I suggest to get your brain checked.

>ncatlab.org
Have any non pop-sci source?

>it's the category guy who doesn't actually know any math
oh I thought it was someone else, go right ahead

No, since it's mainly a pop-sci level topic.
I'm not the one making trivial high-school level mistakes here.

>Any metric which is not non-trivial is a trivial.

>take a finite set, the R module generated there is free (like any R module). wow!

the category of you're a faggot is locally your mom hehehe
am I doing it right?

>⨁n∈N
This is a meaningless notion.

Yes?
>free (like any R module)
Not just free, but finitely generated.

hahaha oh it's you! such a comedian

Was it autism?

he did it again! haha you crack me up

You are genuienly retarded. Any R-module is free, but not necessarily finitely generated.

I take it you don't have high reading comprehension?
I can post a test to check your knowledge of scientific principles and you can share the link of your results, timestamped of course.
:D

You have offered zero counter-point, zero counter evidence.
Therefore I see no reason to continue with you if only I have something to intellectually contribute.

Your denialism is fallacious.

You are genuienly retarded. Any module is a group, but not necessarily free.

How is this even remotely relevant?

I suggest to get your brain checked. Even someone of your intelligence should be able to understand.

A retard being a retard is unsurprising.

>pwease take mi serius agen :(
hahaha you're worthless

Not being able to understand basic high-school level stuff is embarassing.

How is this even remotely relevant?

>all R modules are free on a finite set.
A free R-module is free (as an R-module) on a finite set if its basis is a finite set. Please at least try to understand that.

>pwease take mi serius agen :(

He's an engineer, no need to take him seriously.

He's an engineer, no need to take him seriously.

>this is what happens when CStards are allowed to post on Veeky Forums

Consistent with all physical laws, does relativity still work? That kind of thing.

Because a curved plank length is pretty hard to run the tensors on and arrive at the conclusions of relativity.

Einstein just used a few different coordinate systems to define his metrics. He then realized his metrics told him something about the space they describe. In the Planck length model would all Einstein's math still check out?

To reiterate a model is not the space it describes, it tells you some useful information about the space it describes. And my question is would the Planck length model be consistent with all current physical law?

>In the Planck length model would all Einstein's math still check out?
Not if you assume the axiom of choice.

Space is not discrete, but the different coordinate systems we have do a fine job at the level of analysis we usually look at.

Sometimes it's easier to switch to polar, oblique, cylindrical, or spherical but you could theoretically do everything in Cartesian.

>3d grid spaced by Planck lengths.
But is the grid crystaline in structure or amorphous? Does the grid flow?

those curves are still within the grid

>Consistent with all physical laws, does relativity still work?
Nothing is consistent with all physical "laws".

>Does the grid flow?
Define "flow"

O yeah? Your mother is