The h Bar

@GlenTheUdderboat I don't know what you mean by that, light doesn't travel instantaneously

:O

who wrote that article

skill patrol

Nice evidence

old.stsl.ru/manuscripts/…

^if you want to see it used

5:31 PM

@xnor I mean that the light from A and the light from the end of the minus pole reach B simultaneously. Subtracting yields zero time from A to the end of the minus pole. Instantaneous.

@Slereah All the refs are written in Cyrillic, how do we know it's true?

Well you learn cyrillic

I can't do everything for you!

skill patrol

Good point.

ꙮ)

what did Einstein have to say on the matter

$ꙮ^{\mu\nu}$

I will use this in my dissertation, mark my words

2

5:33 PM

$ꙮ^{\mu\nu}_{\alpha \beta \gamma \kappa}$. Many eyes needs many indices.

Symbol for multiple observers

Or for an insect observer.

Quantum calculus

Quantum calculus, sometimes called calculus without limits, is equivalent to traditional infinitesimal calculus without the notion of limits. It defines "q-calculus" and "h-calculus", where h ostensibly stands for Planck's constant while q stands for quantum. The two parameters are related by the formula where is the reduced Planck constant. == Differentiation == In the q-calculus and h-calculus, differentials of functions are defined as and respectively. Derivatives of functions are then defined as fractions by the q-derivative and by In the limit, as h goes to 0, or equivalently as q goes...

@ACuriousMind why do I need this?

@0celo7 You don't.

7 is not many. Well, not that many.

5:35 PM

@GlenTheUdderboat I don't follow. A is moving with v. A generates a light flash in its origin when it overlaps with B. In A both ends of the 1 Ls long poles are reached after 1s. In B my calculations don't match what I'd expect, but it certainly is not the same for both ends

TanMath

@0celo7 cool, what are its applications?

@TanMath no idea!

I was asking ACM

and I have simplified my question to just one pole in positive x direction

@xnor I believe this is the first time that you mention that A and B are overlapping at some point. I assumed A was quite a bit away (more that the length of the pole).

7 is the number of spin 0 mesons for udc quarks

COINCIDENCE?

5:37 PM

@Slereah I don't think so!

nope

what do Einstein and the Evidence say about 7

Well

Go look Einstein's papers

Seven 0s stacked together. Ocelo7. COINCIDENCE?

Look for the number 7

7 is in my name :O

5:41 PM

You're one of THEM.

And ocelots have seven legs :o

@NeuroFuzzy ::counts::

er, no

@0celo7 Well, perhaps some of them.

@GlenTheUdderboat it could be away...
My example changed to this (including technical details): A has a photon source, detector and timer in its origin and from there extending a 1 Ls long pole in positive x direction with a mirror at the end.
B also has a detector and timer. B sees A moving with v=0.866c. When both timers line up they are reset and A generates a photon in positive x direction.

A detects the photon coming back after 2*1s. B detects it when and why?

@xnor What does "it" mean? What is B supposed to detect precisely?

5:46 PM

one of the reflected photons that A generated

B sees the light slowly catching up with the mirror (end of the pole of A) since A moves with 0.866c.

@xnor Which pole? There used to be two poles.

well, it doesn't actually see the light actually move, it's theoretical

A: x-----------------M where x=source, detector and timer and M=mirror
B: x where x=detector and timer

B sees A moving with v: A---> B, until they overlap, then they reset timers, A generates the flash and keeps moving

In A the light just takes 1s to reach the mirror, 1s back = 2s.

In B the light slowly catches up with the mirror in A, which is moving with 0.866c.

A explodes. The light of the explosion triggers a secondary explosion at the end of the "minus" pole. B is somewhat behind (so the end of the pole is between A and B). B sees both explosions simultaneously. That was my point.

no

I think that doesn't even involve relativity.

5:54 PM

everything involves relativity

@0celo7 Well, this doesn't.

my thinking is this: v*t+1Ls - c*t = 0
So A moves with v for some time with the mirror an extra 1 Ls ahead.
=> t = -1Ls/(v-c) = 7.462s

I'm pretty sure Pythagoras or some of his friends would agree.

I missed to add: until the light catches up.

@xnor Please focus (for my pleasure) on my setup. From B's perspective the explosion of A and the end of the "minus" pole are simultaneous.

5:58 PM

how fast do explosions propagate?

I don't see how this matches anything I'm talking about

With the speed of light.

Once, a long time ago, you were having B measure the time between the release by A and it reaching the end of the "minus" pole. I say B measures zero.

or the forward facing pole, yes

B cannot measure zero

that would mean c=infinity

or the pole is not 1Ls long but 0 in length

@xnor Yes, s/he can. It is the time difference between the observations of the explosions. It is zero. Think! Think!

Try to visualise all the photons moving about. All of them will reach B at the same time.

no they don't

In B A moves. The initial light takes some time to travel to both ends, it takes a lot longer to travel to the positive x direction end, because that is where A moves with 0.866c

@xnor I'm only talking about the "minus" pole. The one of which I assumed the end to be between A and B. I'm currently not concerned with the other pole (which I think you might have gotten right).

6:07 PM

the minus pole initially is 1Ls in the negative x direction away from B (because A and B overlap when the flash is created)

it moves with 0.866c towards B while the flash travels in negative direction with c

@xnor Well, if you're going to insist on A and B being in the same spot at the time of the flash, then my example doesn't work. Too bad, because I liked it.

Well if they don't overlap at the instant of the flash then how would you synchronize them? You'd have no clue in B when A is flashing. You'd just receive some flashes after some random time.

@xnor Well, you would receive the flashes from A and the "minus" pole at the same time, and the flash from the "plus" pole some time later.

By putting B between A and the end of a pole, you seem to complicate things. Unnecessarily so.

@ACuriousMind what is quantum calculus used for

@0celo7 "The h-calculus is just the calculus of finite differences, which had been studied by George Boole and others, and has proven useful in a number of fields, among them combinatorics and fluid mechanics. The q-calculus, while dating in a sense back to Leonhard Euler and Carl Gustav Jacobi, is only recently beginning to see more usefulness in quantum mechanics, having an intimate connection with commutativity relations and Lie algebra."

6:25 PM

I can read the article. I was looking for specific uses.

@0celo7 My apologies.

@GlenTheUdderboat No need to apologize, I was looking for something like "this theorem/method is easier using QC"

@GlenTheUdderboat no, it simplifies things, because then you have an instant both agree on, can reset their timers on and agree on when the initial flash is created

So I still don't know why the lorentz factor doesn't match or what exactly is going on

TanMath

7:13 PM

@0celo7 why dont u ask a question on the main site on it?

@0celo7 : no. When spacetime is flat*, light goes straight. When space is curved, it doesn't.

* There's a little issue concerning flat as in not curved and flat as in horizontal, but I'll tell you about that another time.

7:28 PM

@GlenTheUdderboat : no, we aren't. See questions like this and this along with the answers for more.

@Slereah : yeah. A close run thing.

@xnor : the motion of light defines your space and time. The world is painted in light, and so is the canvas.

7:50 PM

@JohnDuffield that doesn't help me :s

Q: How Much Computing Power would be Required to Fully Simulate a Cubic Meter?

@xnor : maybe you need to ask a question and get some carefully explained answers that do help. Meanwhile try to imagine what kind of world it would be if light didn't move at all.

vzn

2

Imagine you want to simulate a cubic meter down to the particle. By following the Standard Model and other basic physical equations, how much computing power would be required to do this, in say, a day? Would a quantum computer help you in this task? Could you somehow directly simulate the parti...

@0celo7 Why do you suppose I know? Does "combinatorics" or "fluid dynamics" sound like stuff I care about to you? ;)

8:13 PM

@ACuriousMind no, "quantum mechanics" and "lie groups" does

Ah. Okay, that's legit, but I still have not heard of it.

vzn

8:42 PM

in Computer Science, 30 secs ago, by vzn

Ahmed Mohamed & his clock commotion, top links / TMachine blog

9:03 PM

Meh.

9:22 PM

@0celo7 Did you become worse?

any one else experience an curious notion, yet struggle with formulating a coherent question around it?

@obe I have medicine now

I'm probably getting better

@ChrisWhite So the graduate algebra class is starting soon. I'm really curious about discrete subgroups of lie groups and that kind of thing, and I found undergraduate algebra to be pretty entertaining, but I'm not so sure it would be the class for me... any thoughts? I had actually already bought Hungerford's book (for being Rudin-like and fun) for cheap, but I'm not sure if I should focus on general relativity instead...

@0celo7 Cool, feel better soon.

thanks

9:26 PM

ex I could try to take grad quantum instead. Which might be more fun. Just looking for thoughts!

turns out the campus clinic is really cheap

Lie groups have been on a lot of people's mind lately, why is that?

@0celo7 It's free here. :(

@obe the clinic is free here as well

free = really cheap

I see lol.

9:28 PM

@punkerplunk Haha well I mean, groups are so general they come up everywhere, physics is mostly about calculus, so groups+calculus=lie groups=lotsa physics!

Plus they're cool in general.

How would Lie algebra/groups be applied to something . . . like relativity

don't answer that, it's a silly question

It's one of those topics I feel like I understand less the more I read about

@punkerplunk the Lorentz group is a lie group

also you have symmetry groups of spacetime

@obe problem set?

I need to re-learn qm which I am doing.

jesus

what problem are you stuck on

Surprisingly though I am able to take in all of the stuff in class without being serious.

I'm not stuck... I want to learn relatively rigorous qm for myself.

I'll do the problem set in a while.

(Which I will be stuck on later)

9:42 PM

rigorous QM?

when did you learn measure theory

when did you learn functional analysis

relatively rigorous.

I see

relatively compared to what?

sounds like your professor is guarding his tenure with defensive adjectives

@0celo7 shankar?

I find ballentine to be like 7% more difficult than shankar.

are those whiskeys?

9:53 PM

quantum mechanics textbooks

much harder to swallow

lol.

@punkerplunk nooo relativity should be one of those things you end up 1000% understanding!

You should save the "understanding less the more you read" for quantum stuff and higher :3

for example, I made this:

imgur.com/QSs9iW8

and WillO made this: i.stack.imgur.com/PIOiK.gif

what about it?

10:08 PM

Good thing there's no paradox :3

but wait

the difficulty I'm having, is when I attempt to frame that situation with Alice, aboard her ship, as the inertial frame of reference

people tell me I cannot do that

Yep, that's the difficulty everyone has.

Special relativity only works in inertial frames.

why would relativity dis-allow me from doing this?

@NeuroFuzzy that's the difficulty that everyone else has

okay, so maybe the word 'inertial' is not what I'm looking for

10:11 PM

It's like in classical mechanics. If you're in a rotating frame of reference Newton's laws look different and there's no galilean invariance and all that stuff, straight linesl ook like circles, it's ugly.

it's simply not true

Gourgoulhon's book does SR in accelerating, rotating frames

the math is hellish but it works perfectly fine

@0celo7 as it's introduced normalllyyyyyy dudeeee, I know you can do a transformation on the metric, and I know Landau defines the special relativistic metric as one which can be transformed to be the usual one everywhere.

You can but why would you

GR deals with it pretty naturally

I was attempting to make a statement that, because this (not) paradox holds true, it would be possible to travel to a location, even if at nearly the speed of light, and return OLDER than your twin

@0celo7 I just mean to say that the laws of physics need to be modified. Like going into a rotating frame in CM causes the addition of fictitious forces

10:13 PM

Though I would like to see a paper that does Born coordinates in details

Like I'm pretty sure that usually they only describe one coordinate patch

@NeuroFuzzy it's more complicated than transforming the metric

you have to do crazy stuff like Fermi derivatives

Well Fermi transport is just a transformation of the metric :p

@punkerplunk Right, sidetracked, sorry. The point I'm making is that regular, introductory special relativity assumes you're working in a frame that is inertial throughout all time. But in Alice's frame, objects without force acting on them would move in a line with a sharp corner.

A specific one, sure

well, let's say she's at an orbital velocity (high altitude, granted) and thus appears to move perfectly straight

locally anyway

10:17 PM

the single best thing in SRGF is that he does the twin paradox with an actual smooth trajectory for the "moving" twin

@punkerplunk even better, could we forget the earth and work with two spaceships? ^^

no bullshit "corner"

Twin paradox with GR is easy peasy lemon squeezy

explain

I've never seen it done

hold on Neuro, I want to go somewhere else with this

10:18 PM

It was actually done in a paper done by none other than

EINSTEIN

(and the evidence)

Einstein and the Evidence? I must see this.

Because blanch returns to Alice in the illistration, and is OLDER than alice when she returns - because the entire question was framed as the earth as the inertal frame

Basically in one frame, you're in Minkowski space with a non-geodesic trajectory

why blanch

And in the other frame, you're in some accelerated coordinates

10:19 PM

why not bob

That's why the situation isn't symmetric

or steve

or albert

cause I'm steve

cause girls

@punkerplunk Okay. And?

AND

I attempted to make the case that the choice of the earth is arbitrary. That I could have chosen nearly any frame of reference as the 'inertial' frame . . .

10:21 PM

no, you can choose any inertial frame of reference as the inertial frame :p

en.wikisource.org/wiki/…

Might be that one?

most frames would be non-inertial

But really once you do it in GR there is basically no ambiguity

But how can there even be an inertial frame? Isn't that just a device to help describe the conjectures of the theory.

The difference between the two frames is that one is Minkowski + a 4-force

The other is some kind of Rindler metric

10:23 PM

@punkerplunk No. It isn't, not even in Newtonian mechanics

there is a real, physical, definite, intrinsic difference between inertial and noninertial frames.

It's sort of discomforting, I agree.

hence the two twins do not move in their respective frames for the same reason

Isn't the earth just a space ship, though?

Well the point isn't the spaceship part

It's the speed part

Okay, I get that

that's not my issue

@punkerplunk Yeah, but if you consider a short enough period of time in its path around the sun, objects in motion will continue moving in a straight line, etc.

so it will be inertial.

It's like the issue of angular velocity being absolute in Newtonian mechanics.

10:26 PM

I'm not even really sure if the motion and gravity of the earth matters all that much

Maybe the best thing to do would be to compute it with an arbitrary metric

@Slereah but then my lines won't be straight :(

I find that particular aspect of relativity theory disgusting, philosophically.

@punkerplunk What particular aspect? That reference frames are absolute?

yes

Okay... well the same holds in classical mechanics so

nuthin' new.

10:28 PM

I mean, isn't that simply saying the earth is the center of the universe?

By absolute I mean "inertial" and "noninertial" frames are different*

No

oh crud

too late for my correction*

Also if you don't like it switch to GR :p

GR has no preffered frame at all!

oh

10:30 PM

@punkerplunk The Earth's frame, and any frame that moves with respect to the Earth's frame with a constant velocity, is inertial. Inertial frames are the simplest frames in special relativity, so yes, in that sense, special relativity prefers inertial frames.

But we're not talking about one frame, we're talking about "Earth's [momentary rest] frame and any frame that moves with respect to that frame with a constant velocity"

-sigh- I'm making a bunch of tiny mistakes in my wording, sorry.

no no

I don't want to talk about that

because that's just the twin paradox

and that I understand

I want to talk about blanch, leaving alice, and returning older

okay.... well you know you can't go into Alice's frame to do your time dilation calculations then.

at least not until she's safe and sound in an inertial frame for good.

I'm fine with that

I don't care

Great!

I'm attempting to describe a location, or a dimension, or an aspect of space time

such that:

{X | A -> B : E -> X}

does that make sense or no?

10:35 PM

Not to me no. I mean that's not really a set, is it...?

The set of all X such that E to X?

cana frame of reference be a set?

Ehmmmm

Sure. a point in spacetime to be your origin, plus four spacetime basis vectors.

and a journey in a space ship described as an operation upon that set?

So if I call A and say that is the set of states describing Alice's motion, B for blanch, and E for emma.

No, a journey in a spaceship is a line in spacetime. If $e_i$ are your basis vectors and $x_i(s)$ are your positions as functions of time, a journey in a spaceship is a line that can be parameterized as $\sum e_i x_i(s)$. It's independent of the basis you work in.

For the basis transformation, you may be interested in this physics.stackexchange.com/questions/91262/…

A change of reference transformation is a change of basis plus a translation.

but a path is not an action on the basis. It's just a line/set of points.

There's another thing I think you might be getting at: Are you concerned whether proper time (the time elapsed for an observer) is an axis of spacetime or a scalar?

well I KNOW it is an axis

because I exist

and can see and think

10:42 PM

@punkerplunk aha! There it is. physics.stackexchange.com/questions/116870/…

proper time of a spacetime path is like the length of a regular euclidean (squiggly/not-nice) line.

okay. But do you understand what I was attempting to describe with {X | A -> B : E -> X}

you actually have to integrate $c^{-1} \sqrt{(c dt)^2-dx^2+dy^2+dz^2}$ to get the proper time.

I'm sorry, I don't.

@NeuroFuzzy The top voted answer is...actually false. One should describe proper time as the length of a worldline, not as some "dot product".

X, being a fourth sister, say Xandra who leaves the earth on a separate mission, and returns OLDER than all three sisters

X is to E as B is to A

@punkerplunk What is this meant to be? A set? An observer? What are A,B,E,X?

10:45 PM

A,B,E,X are explained here: imgur.com/QSs9iW8

@punkerplunk The set of Xandras such that the ratio of proper times of alice to bob equals the ratio of proper times of the Earth to Xandra?

An more accurate description is provided by: i.stack.imgur.com/PIOiK.gif

yes, Neuro

@punkerplunk isn't that just the set of all paths with a given proper time?

Sure, I'll allow that

alice/bob=earth/xandra, xandra=earth*bob/alice (referring to their elapsed proper times)

Okay... now we have a bunch of zig zag paths through space with the same Lorentzian length. So?

(It's like the set of paths length L in Euclidean space)

(and we can also assume that they're timelike [slower than light] and start and end at Earth?)

10:49 PM

So, would that suggest it would be possible for an object to leave the earth and nearly the speed of light, go to 'some location X' and return OLDER - the opposite effect of the normal twin paradox

@ACuriousMind indeed.

once again, SRGF does this very well

really great book

@punkerplunk no. That's equivalent to this: en.wikipedia.org/wiki/…

for instance, we have the following general expression for the Lorentz factor $$\gamma=\sqrt{(1+\vec a\cdot\vec{OM})^2-(\vec V+\vec \omega\times_u \vec{OM})^2}^{-1}$$

this works for a rotating, accelerated frame

@punkerplunk $v$ is a vector that is one leg of the trip, $w$ returns to Earth. $v+w$ is just a line going in a straight path. The given inequality implies that the time elapsed for Earth, $\|v+w\|$, is always greater than the sum of the time elapsed on the two legs, $\|v\|+\|w\|$.

@0celo7 : But why

That is madness

Why not use quaternions while you're at it

10:53 PM

you can use quaternions if that's your fancy

why did "ridin' dirty" just come up on my pandora electronic playlist?

Good song.

I can't keep up with Krayzie Bone tho

@0celo7 Krazy bones? en.wikipedia.org/wiki/Gogo%27s_Crazy_Bones

@NeuroFuzzy only if the earth is the 'inertial' frame