The h Bar

so paradoxically, being curious for it's own sake, actually leads to better achievement of the objective compared to those AI's whose only goal was to achieve the objective

Balarka Sen

Well before that one has to stop talking in grandiose terms and actually write down a formalism for what curiosity means

Plebian philosophy won't do the math for you

Secret

it might actually bring us a step closer to a general purpose AI

Kenshin

@Secret, well I think curiosity can modify the subgoals, but the ultimate goals would surely ened to be defined from the beginning

Secret

but yeah, balarka has a point here, we will need some mathematical way to capture a notion of curiosity

Kenshin

@BalarkaSen, I could have a go at formalising curiosity

Balarka Sen

11:17 AM

@Kenshin Big if true

you're too deep in plebian philosophization

math is for Rick and Morty people

not for plebs

Kenshin

I think curiosity can be modelled using two processes:

(1) a feeling of pleasure when experiencing something that isn't consistent with existing models/predictions etc. such that one attempts to seek out such experiences that are inconsistent with current frameworks.

and

(2) a desire to modify one's models in light of (1) to better account for the new phenonemonon

after step (2) the phenenomon originally found to be pleasureable then becomes boring

since it can then be modelled

and the organism then seeks out new phenonemeno that cnanot be explained

and so forth, in turn buildnig a better and better model of reality

from which it can use to then achieve it's base objective function

So programming such an AI simply requires the AI to have a reward not just for achieveing objectives, but for also experiencing something new that cannot be explained using the AI's model so far

So the next step is simply to have a measure of novelity as described in (1)

this is fairly straightfoward, and the specifics of such will depend on the AI's particular modeling algorithms

Balarka Sen

>this is fairly straightforward
>can't translate my shit into math

Secret

I would imagine step (1) will require a lot of computational power to compare every set of environmental data (assume it is captured by n sensors which gives a vector v of real numbers). Then carry out step 1 is basically checking whether v can be fit by some model function m (which also need to compute something like least squares)

Balarka Sen

10/10

Kenshin

The main difficulty with step 1

is that an AI could get distracted by novel, but irrelevant phenenomon

Slereah

11:22 AM

@BalarkaSen mb

Kenshin

For instance, you may see a broken tv with lots of pixels flicerking about. A curious AI coudl fall into a trap where it jsut keeps staring at the tv enjoying the novelity of it

Secret

Suppose for illustration we have an AI with n sensors which picks up parameters like temperature, light intensity etc., and express that as some real scalar. Thus all the information experienced by the AI will be given by:

Kenshin

So the difficulty would be knowing how to switch off the AI's curiosity when it becomes apparant that the stimul is too unpredictible to bother with

Slereah

Topological details of thin shell wormholes are rather sparse

Secret

$$v = (x_1,x_2,x_3,...,x_n)$$

Kenshin

11:23 AM

indeed

Balarka Sen

@Slereah what are those

Kenshin

the AI for instance, may correlate temperature with a variety of external parameters (e.g. it may find that when it is indoors, it has a cooler temp or something)

this can be done through a neural network

then when the AI experiences that some indoor areas are in fact hotter

the error increases, and the AI enjoys the pleasure of the contradiction

Secret

Then the AI itself have some model, which is a function $\Bbb{R}^m \to \Bbb{R}^n$ that maps some internal parameters to some environment information vector $w$. We then do e.g. least squares of $v-w$ and find the $v$ which is smallest

Kenshin

it then updates the model

indeed

the least squares may be a good measure of the novelty

Slereah

@BalarkaSen wormholes where the matter propping up the throat is a shell of thickness 0

Kenshin

11:24 AM

and that is why I said the exact novelty measure will depend on the specific machine learnging algorithm used

but I think least squares would work well for most

Sir Cumference

@Secret I am concerned by how the components are $x$s instead of $v$s

Secret

@SirCumference ah, forgot that unspoken convention in physics that x,y,z are reserved for positions, ok then, rename those as v1,v2v3 etc.

Kenshin

arxiv.org/abs/0709.0674

^check this guy out

Secret

@Semiclassical: @EmilioPisanty needs you

Slereah

@BalarkaSen Basically you glue together two surfaces of a spacetime such that the metric is preserved outside of the gluing

Which usually means that the Riemann tensor is a distribution-valued tensor

Secret

11:27 AM

@Slereah How's your nonzero throat radius wormhole stuff going?

Balarka Sen

@Slereah Er, so it's not smooth at the surface you glued them along?

Slereah

@BalarkaSen The metric is only $C^0$, yes

@Secret The geodesics are problematic

but then again, some dude warned me about this a few years ago

Balarka Sen

Why are you dropping regularity? Is there any physical reason for that?

Slereah

I found a random paper (which was just a Tex file) saying that it was easy to build thin-shell wormholes with no closed null geodesics

@BalarkaSen It simplifies the analysis of curves

Emilio Pisanty

> The fact that some geniuses were laughed at does not imply that all who are laughed at are geniuses. They laughed at Columbus, they laughed at Fulton, they laughed at the Wright brothers. But they also laughed at Bozo the Clown. ─Carl Sagan

5

Slereah

11:29 AM

Since then you can just pick Minkowski space outside of the mouth

Hence all geodesics are just lines, so that you can solve a fair amount with just Euclidian geometry

Emilio Pisanty

(just something to keep in mind)

Slereah

So a lot of the problems just become line-sphere intersection problems

Kenshin

arxiv.org/pdf/0812.4360.pdf

Slereah

of course, the tough part is showing what geodesics that cross the wormhole look like

Since the connection isn't continuous there

Balarka Sen

Aha I see

Slereah

11:31 AM

Though I do plan a later paper where the metric is actually smooth

But then I have to deal with something awful

Kenshin

What does a wavefunction look like that passes through a wormhole?

Slereah

Bispherical coordinates!

@Kenshin Oh man

I tried to work out the wave equation through a Morris-Thorne wormhole once

It's pretty bad

It's a modified version of the prolate ellipsoidal wavefunction

Kenshin

I wonder what it means, like there is a probability the particle is in the past and a probability it is found in the future

Slereah

if you want the time-travelling case, you have to read Friedman on the topic

Balarka Sen

I think Slereah is just talking about a spatial wormhole

Slereah

11:33 AM

arxiv.org/abs/gr-qc/0401004

Kenshin

@BalarkaSen space-time is one big fabric tho

Balarka Sen

@Kenshin So?

Kenshin

@BalarkaSen, well space and time aren't really separate so space wormholes are time wormholes

you can't separate them

it all depends on the observer

Balarka Sen

It's a 4-manifold with a well-defined foliation by the spatial hypersurfaces (spacelike 3-submanifolds, I think you call them?)

Kenshin

see time and space they merge into each other depending on the speed of the observer

Slereah

11:34 AM

a wormhole having closed timelike curves is independant of the observer, though

Kenshin

but can you jsut have a space wormhole?

or would any space wormhole but a space-time wormhole to other observers?

Balarka Sen

@Kenshin That's just what happens to the metric, not the topology

Slereah

@BalarkaSen We call them achronal spacelike surfaces :p

Balarka Sen

Or at least so I think

Slereah

Or Cauchy surfaces, if they're very nice

Sir Cumference

11:35 AM

@Slereah Oh what the hell. Why is that a thing

Balarka Sen

@Slereah Nice, true GR people

Slereah

@SirCumference Well in my case I am very glad they exist

that way the boundary conditions are easy to do

On the other hand, the solution of the wave equation in bispherical coordinates is very bad

You have to use the toroidal wavefunction modes

I thought it was a great innovative idea to use bispherical coordinates, but so it turns out Wheeler did that 60 years ago

Balarka Sen

Wheeler seems like a loony

But so are most physicists

Slereah

wheeler's fine

he just had a lot of handwavy arguments

Kenshin

Penrose is my favourite physicist

Balarka Sen

11:39 AM

One-electron universe

The one-electron universe postulate, proposed by John Wheeler in a telephone call to Richard Feynman in the spring of 1940, hypothesises that all electrons and positrons are actually manifestations of a single entity moving backwards and forwards in time. According to Feynman: == Overview == The idea is based on the world lines traced out across spacetime by every electron. Rather than have myriad such lines, Wheeler suggested that they could all be parts of one single line like a huge tangled knot, traced out by the one electron. Any given moment in time is represented by a slice across spacetime...

^ loony material

Kenshin

I thought of the one electron universe myself in high school

Slereah

To be fair, that was a phone conversation, not a paper :p

Kenshin

i thought of the idea doing washing at home

Slereah

[1] Private phone conversation

Balarka Sen

@Kenshin You should be in 4chan/sci/, not here

Kenshin

11:40 AM

y

Balarka Sen

an absolute madlad

Kenshin

really

i'll check it out bud

i love mad ideas

if you can think of something mad that fits the facts, you can't prove it wrong

Secret

You know what, I am gonna just integrate this and see what happens...

Hopefully the universe does not implode

$$ i \hbar \int_{t_0}^{t_1} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t} dt = \int_{t_0}^{t_1} \hat{H} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t}dt$$

Kenshin

don't do it you'll cause a black hole

I protest

Secret

Didn't grandma taught you that only outstanding questions will cause black holes?:P

$$ i \hbar (\lvert \psi (x,t_1)\rangle - \lvert \psi (x,t_0)\rangle) = \int_{t_0}^{t_1} \hat{H} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t}dt$$

Slereah

11:43 AM

Abcd

@Blue Yeah, I did it...It was easy ...It's from HCV though.

Secret

damn, cannot integrate the RHS, fail

Semiclassical

11:59 AM

@EmilioPisanty what’s up?

Emilio Pisanty

@Semiclassical nothing, same as last week =P

Semiclassical

lol

Emilio Pisanty

whatcha think? link

Secret

Emilo and Semi, do you happen to know whether there is physical meaning in integrating the SE wrt time?

Semiclassical

In local news, my laptop stopped booting last night

@Secret not sure what you mean by that. You can use the Schrodinger equation to get the time evolution, but that’s done by integrating the propagatir against the wavefunction

Secret

12:02 PM

Well, I am sometimes wondering about this:

$$i \hbar \int_{t_0}^{t_1} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t} dt = \int_{t_0}^{t_1} \hat{H} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t}dt$$

it seems sensible but it looks nothing like the time evolution operator

I suspect I might be instead computing some sort of time average when doing this

but I am not sure

Semiclassical

Well, note that that’s an integral equation

Akash. B

Why heat is coming with the light?

Secret

hmm...

Semiclassical

@Secret tho actually that looks wrong: there should be no time-derivative on the right

Slereah

cds.cern.ch/record/557642/files/0206037.pdf

Secret

12:06 PM

@Semiclassical o I screw up the the copy paste, but you get what I mean, anyway

Slereah

neat

Secret

$$i \hbar \int_{t_0}^{t_1} \frac{\partial \lvert \psi (x,t)\rangle}{\partial t} dt = \int_{t_0}^{t_1} \hat{H} \lvert \psi (x,t)\rangle dt$$

ok... this looks even nastier...

Emilio Pisanty

@Secret the term you want is probably "Dyson series"

also "Schwinger series"

(same thing, different communities)

cf. p. 48 of my thesis

Nehal Samee

@JohnRennie ... You there....?

John Rennie

@NehalSamee I'm around, though I'm a bit busy.

Semiclassical

12:13 PM

@Secret well, you can integrate the left side explicitly. Rearranged slightly, it’s $$|\psi(x,t_1)\rangle = |\psi(x,t_0)\rangle -\frac{i}{\hbar}\int_{t_0}^{t_1} \hat{H}|\psi(x,t_1)\rangle \, dt$$

Nehal Samee

Hmmm...Can you look at the problem I gave in the chat earlier ...

Semiclassical

So that second term can be interpreted as the change in the wavefunction over that time interval. Problem is that it seemingly requires you to already know what the wavefunction is to start with

John Rennie

@NehalSamee Link?

Semiclassical

However, you can get an iterative solution if you start with an initial guess for the wavefunction

Plug that into the RHS and use that to compute the LHS

Nehal Samee

OK then ... I'm trying to post the question here...
A spring is hanging at 20m height from the ground . A ball of mass 0.2kg is shot towards the spring with velocity 49m/s such that extension of spring is 3m . What will be the rebound velocity of ball on the ground ?
I found v at height 20m and then its kinetic energy ... This kinetic energy is equal to stored potential energy of spring . Then I found k equating the energies and consequently F...

@JohnRennie ...

Semiclassical

12:17 PM

Then use that as your new guess, and repeat

Emilio Pisanty

@Semiclassical should read $$|\psi(x,t_1)\rangle = |\psi(x,t_0)\rangle -\frac{i}{\hbar}\int_{t_0}^{t_1} \hat{H}|\psi(x,t)\rangle \, dt$$

($t$ inside the integrand)

Nehal Samee

Once I approached with force and again with conservation of energy ...But ended into different answers. @JohnRennie ...

Semiclassical

Good catch

Emilio Pisanty

@Secret if you're going down this route, the trick is to then use that approximation for $|\psi(x,t_1)\rangle$ inside the integrand, i.e. for the $|\psi(x,t)\rangle$ after the $\hat H$

Nehal Samee

@Secret ... Have you found any solution ...? To your satisfaction , the spring hangs towards ground ...

Emilio Pisanty

12:19 PM

which gives you a term in $\hat H$ and a term in $\hat H^2$

and you hope that (i) the series converges, or (ii) that it is useful at some truncated stage

Semiclassical

Yeah. Formally this gives the Dyson series which is durn important

Emilio Pisanty

quite often the latter and not the former

Secret

It does looks like the Dyson series, but I am not sure whether it is actually the Dyson series expressed in the schroedinger picture. I might check later

Emilio Pisanty

i.e. the Dyson series diverges and if you want to use it in full you need to go down a very scary rabbitt hole of renormalization and interacting vs free-theory vacua and so on

Semiclassical

It is.

Secret

12:21 PM

@NehalSamee ok, so the ball is hitting the spring form below I am guessing

Semiclassical

Any discussion of the time-dependent Schrodinger equation should have this

Note that you really only go down this route if H ie time-dependent

Secret

well I have seen derivations of the Dyson series, but it is introduced only via the interaction picture (at least in my old uni), which is why I don't immediately recognise it in the schroedinger picture. I think I understood now, thanks guys

@NehalSamee So that means, the KE of the ball is converted into the gravitational potential energy and the elastic energy of the spring (which compresses it), and thus the KE the ball gains as it left the spring will be form these two sources

@Akash.B in thermodynamics, heat is a transfer of energy, thus energy from electromagnetic waves(light) can dissipate as heat when it interacts with matter (e.g. air molecules)

Nehal Samee

@Secret ...Yes...

Abcd

What's the mathjax code for mean <> ?

Nehal Samee

Now , I've found the final velocity at height 20m and used it to find kinetic energy ... That kinetic energy is the elastic potential energy ...I find elastic constant from it ...Then I find restoring force F and divide by m to find acceleration...

Then I use $v=\sqrt{2(a+g)} $ ...

Secret

12:33 PM

the initial KE should be converted to both the GPE and elastic PE (since your ball is moving higher AND compressing the spring)

Nehal Samee

@Secret ...Now I want to find v using energy conservation ... But I'm confused ...

Secret

so the initial KE should funnel into two reservoirs and then converted back to KE again, this time sending the ball pointing down. That is final KE = GPE + EPE assuming the initial KE is just right

If the initial KE is larger than the sum of GPE and EPE, then I think the problem will become a bit more complicated because the ball will be pushed by part of the spring as it fell

@NehalSamee actually, are you on desktop now, I think a diagram will help to see if we are on the right track?

Nehal Samee

I'm on Android ...

Secret

hmm.. give me a sec when I drew the diagram...

Nehal Samee

@Secret ... Why should I use KE =GPE +elastic PE ... ? Cause I'm given the compression of spring 3m ...

Secret

12:40 PM

O wait, so your spring is already compressed before the ball hits it?

Slereah

-1

Q: Observer problem in Relativity

(Please bear in mind that each observer/clock independently concludes that one second has lasped according to their lightclock)

special-relativity

Nehal Samee

No...It is compressed and again rebounds. That rebounding velocity is what's required...

Secret

@NehalSamee I have a diagram drawn, is my diagram reflecting the question?

Nehal Samee

@Secret ... You got it absolutely right...But 20m is distance up to the free end ...

And 3m compression is after 20m ...

Secret

so ball is projected from some unknown initial height, to 20 m and then 20+3 m which the spring become compressed (thus KE is converted into GPE mg(20+3-h_i) and then the compressed spring storing EPE = 1/2k9. This is then all released thus final KE 1/2mv_f^2 = 1/2k9 + mg23) ?

and thus v_f should be the rebound velocity

Nehal Samee

12:50 PM

@Secret ...Yes...

Secret

I have pretty much wrote the whole formula so you should be able to calculate v_f

1/2mv_f^2 = 1/2k9 + mg23 gives the velocity when the ball just hit the ground

Nehal Samee

But the answer does not match with the one I tried with force ...

@Secret ...

Secret

When the question asked the rebound velocity, is it when the spring rebounds but the ball is still at height 20 m or the velocity when the ball landed on the floor?

If it's the former, then the equation will be:
1/2mv_f^2 = 1/2k9 + mg3

Nehal Samee

The velocity on the ground ... @Secret...

Semiclassical

@Abcd I use \langle and \rangle

Abcd

12:58 PM

$\langle M\rangle$

Okay.

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