The h Bar - 2019-05-06 (page 1 of 2)

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12:13 AM

@heather I saw a YouTube video of someone building an SPDC setup at home

I thought you might find it interesting

Can you make quantum entangled photons using garbage and fertilizer? Probably.

I didn't have time to get all the way through it so I can't fully vouch for it

I can say, though: if you do want to build some of that stuff, be super careful. There's some stuff there with real potential to hurt you.

12:27 AM

DIY Cheap Optical Table Parts and Femtosecond Laser Lab

Relevant follow-up

1 hour later…

1:46 AM

ughhhhh my implementation of the transformer is not working....and I have no idea why

so frustrating -.-

2:22 AM

@EmilioPisanty you know, it's really funny - i just found and watched those the other day!

thank you very much for sharing them =)

@EmilioPisanty i appreciate the concern; my dad's supervising my experiments, and i'm making sure to obtain proper safety gear.

danielunderwood

@enumaris that's me pretty much every time I try to implement something from a paper

very annoying

the net outputs the same token at every timestep...

wtf...

danielunderwood

Like if you have word, the next logical word is word so you get word word word word...?

And if you wanted to group compare the similarity of documents, would you just learn an embedding then use cosine similarity or is there a better (and fairly quick) way?

2:39 AM

of a whole document?

You can try weighted average of word embeddings I suppose

right now I'm just telling it to output gibberish (i.e. random numbers)

but it outputs the exact same gibberish at each time step...

it's the self-attention heads...

self attention makes the outputs all the same...why...

wait...I turned off the self-attention heads and it still did this....

wuuuut

3:02 AM

gonna redo...most of the functions...into classes...

see if it does anything...

Oh oh. Matter is made, not from 3 Quarks as the Standard Model proposes, but from three separate Rings of (18), (12) and (6) Carrs. That's 20 points on the Baez crackpot index.

3:22 AM

welp, changed everything to classes

same problem

LOL

fml

3:44 AM

@enumaris Oh well. At least the classes will help to keep things more organised, so the effort wasn't totally wasted.

What on earth is the location of linear momentum? physics.stackexchange.com/a/477008/123208

>____>

2 hours later…

5:29 AM

@heather good stuff =). Both that you're working on stuff and that you're being careful.

@enumaris I'm just guessing here, but I suspect this might be because building a robot that transforms into a car is a bit too ambitious of a project.

3 hours later…

8:50 AM

@pss1 the easy way to approach this is to note that acceleration and a gravitational field are the same in many ways. So the tube accelerating sideways at an acceleration $a$ is equivalent to a tube standing on its end in a gravitational field with gravitational acceleration $a$.

Novalium Company

9:45 AM

In the case of a Young Slit Experiment, what exactly is a two-part wavefunction and what does it describe? Its ability to pass through both slits at the same, interfere with itself and then land as a single particle?

a two-part wavefunction ???

Novalium Company

Im reading a book... ill tell u more, one sec

The particle can go through both slits at the same time because it's a fuzzy cloud not a point mass.

Novalium Company

In the book they put a vertical and a horizontal polarizer at the slits, then send 45 degree photons.

What exactly is that orientation of the photons?

nvm I got it

Its really hard to imagine whats happening when a photons is 1000 things at the samw time.

@JohnRennie ||the easy way to approach this is to note that acceleration and a gravitational field are the same in many ways. So the tube accelerating sideways at an acceleration $a$ is equivalent to a tube standing on its end in a gravitational field with gravitational acceleration $a$.||.......fluid is accelerating right?.......should I have to now consider the tube is rest and fluids flowing in the situations you mentioned.....??

Novalium Company

9:58 AM

"A photon between a vertical and horizontal angle is in a superposition state" What the duck...?

Here is what I really dont understand: "The interference pattern that we see indicates that the photons are in a superposition state: the wavefunction describing each photon has two parts, one for the photon passing through the left slit, and the other for the photon passing through the right slit."

user image

quick one please guys

10:16 AM

@NovaliumCompany Take a look at this recent answer by Chiral Anomaly, which has a great intro level description of superposition.

@Luyw It's not clear what you're asking. I'm not even sure what the squiggles down the bottom mean. Are they supposed to be (1) and (2) ?

no no, (1) or (2)

I am basically asking how I should interpret the "bubble", as (1) or (2)

and that's an equal sign, so = this or that

Novalium Company

@PM2Ring Thanks, I read it but it's still confusing :D "In this way, the particle can "move" smoothly from being entirely at A to being entirely at B", how? By being at both places at the same time and when measured, it's wavefunction (with position probabilities) breaks down?

So when that photon that's in a superposition state between vertical and horizontal angle has a wavefunction with probabilities of it landing on one on the other when measured?

How do we know that things are really not defined and when we look at them, then their "wavefunction breaks down..."?

I mean, it can be a vertical angle all the time, why should I think it's in a superposition of horizontal and vertical?

Schrodinger's cat can always be dead, why do we think it was both at the same time?

10:34 AM

@NovaliumCompany Don't worry too much if it's confusing. This stuff has no classical counterpart, so it's not easy to make mental models of it. But eventually your mind will get used to it if you look at lots of examples of superposition. Feynman said this is the only hard thing in quantum physics. And it's often been said that if you haven't been confused by this then you haven't thought hard enough about it. ;)

But to try & answer your question. Let's say you have an ideal photon detector. If you send it a photon it will register it 100% of the time.

Now place a vertical polarizer in front of the detector. If you send in a vertically polarized photon, it will get detected with probability 1, if you send in a horizontally polarized photon, it will get detected with probability 0, that is, it won't get detected. Ok so far?

Novalium Company

Yep.

Great. Now we can represent the polarization of any photon as a kind of mixture of vertical & horizontal, like this V cos(theta) + H sin(theta), where V is the pure vertical state, H is the pure horizontal state, and theta is the photon's polarization angle, measured from the vertical axis, so theta = 0° for a vertical, theta = 90° for a horizontal. Ok?

Novalium Company

10:50 AM

Yep.

Now comes the tricky part. :) If we send a photon at our polarized detector, then it has a probability of $\cos^2(\theta)$ of being detected.

Novalium Company

The cos and sin parts are a bit confusing but that's bcs I'm retarded :P

I tried making a quick drawing on paint, but I still can't explain the sin and cos part.

user image

I try doing the soh cah toa thing but doesn't work :P

So if theta is 45° then the photon has probability .5 of being detected. If we send 1000 identical 45° photons at the detector, roughly 500 will be detected, 500 won't register, and there is no pattern to which ones get detected.

@NovaliumCompany Don't worry too much for now about where the sin & cos come from. The important part of that is that cos^2 + sin^2 = 1

Novalium Company

Alright then, what's next?

Macroscopically, if you send 45° light through a vertical detector, it dims the light. But at the quantum level, that dimming happens because each photon is either blocked or passed by the polarizer.

Novalium Company

11:02 AM

Got it, and each photon has a wavefunction with probability of passing?

And if theta is 30°, cos 30° is sqrt(3)/2, sin 30° is .5, so 3/4 of the photons get detected, 1/4 get blocked.

Novalium Company

nvm

Got it.

@NovaliumCompany Yes. And that equation I gave before tells us the probability if we know theta. So all 45° photons have the same state of sqrt(.5) V + sqrt(.5) H. There is nothing in their state that determines if they will actually get detected or not. There's just that probability.

Novalium Company

Got it.

Is that all?

Now here's the other tricky bit. ;) When we produce the photons with some polarization angle theta they don't "know" that they're going to be fired at a vertical polarizer. We might change our mind & rotate the polarizer on the detector to horizontal, or any angle in between. But we can still use that same equation to represent their state, and simple trig, to calculate the detection probability. Eg, if we make the detector horizontal, then sin^2(theta) tells us the detection probability.

Novalium Company

11:18 AM

Ok.

So the photon doesn't "know" in advance whether or not it will be detected by a given polarizer setting, unless it matches the polarizer exactly, i.e., a vertically polarized photon will get detected with probability 1 by a vertical polarizer.

Novalium Company

ok

So the core ideas are 1) that all the different polarization angles can be represented by a combination of vertical and horizontal, and 2) the outcome of passing a polarised photon through a detector is probabilistic. There's no hidden information anywhere that can be used to determine in advance what the actual result will be for any given photon.

J. S. Bell came up with an ingenious but simple way to demonstrate that, using 2 different polarization angles. At least, the Bell inequalities show mathematically that there's no information hiding locally in the photon generator, the photon itself, or the polarizer or detector, that determines the outcome.

Much more can be said on this topic, but I think that's probably enough to digest for now.

And I need to take a break. Typing stuff on the phone is not friendly on the fingers. ;)

Novalium Company

11:33 AM

Oh of course, thanks!

I'm famous for getting people into huge rabbit holes of explanations...

No worries.

11:48 AM

@NovaliumCompany I should mention that I've been using a simplified model of polarization. The more accurate model is a little more complex, because we need to work in 3D, not just a simple plane of directions, but the same basic principles apply.

1 hour later…

Novalium Company

12:55 PM

Whats the meaning of the uncertainty principle in the case of electron "orbiting" and interfering with itself to form the harmonic waves? I mean, whats position and momentum (velocity) in that case?

1:49 PM

Within the orthodox interpretation, at least, that picture of an electron "orbiting" is a classical one. It is meaningful to talk about what position/velocity we measure the electron as having; it is not meaningful to talk about what position/velocity it would have independent of that.

Novalium Company

2:04 PM

Ok but we know that the actual picture of an electron in the atom is in the form of harmonic oscillations right? How can that have velocity or position for them to be uncertain in the first place? And how does this picture of the electron collapse when we observe?

No, we do not have such a picture. At most, we have a picture which would be applicable in the classical limit.

That an electron experiences a harmonic potential does not imply that one can attribute to it some trajectory

What we do have is the Schrodinger equation as dictating how the wavefunction of that particle will evolve in time. The wavefunction dictates probability distributions for the observables of an electron (position/momentum), and the Schrodinger equation dictates how these distributions will evolve with time.

Novalium Company

So the electron is just a probability cloud of position and momentum?

@NovaliumCompany as a working mental model, yes

Novalium Company

What happens when we observe it and its probability collapses then?

as to whether the electron "is" the probability cloud, or whether the probability cloud has a more complicated relationship with the electron's ontology, that is essentially still a complicated open question.

Measurement problem

The measurement problem in quantum mechanics is the problem of how (or whether) wave function collapse occurs. The inability to observe such a collapse directly has given rise to different interpretations of quantum mechanics and poses a key set of questions that each interpretation must answer. The wave function in quantum mechanics evolves deterministically according to the Schrödinger equation as a linear superposition of different states. However, actual measurements always find the physical system in a definite state. Any future evolution of the wave function is based on the state the system...

@NovaliumCompany ditto. See above.

2:10 PM

Hi all!
I got a famliy of "vector fields" $f: \Bbb R^3 \to \Bbb R^3$ that satisy a certain physical property (that is not that relevant now) However they have all in common that theay are pointwise parallel to $\nabla\rho(\vec{r})$.
for $\rho: \Bbb R \to \Bbb R$.
parallel means now a multiple of any scalar including 0.
The question:
How to formulate that properly, shortly and concisely?
I thought of $f$ is the gradient of an arbitrary scalar function $g$ of $\rho$. Like $f=\nabla g(\rho)$. But I am getting unsure ...

@Rudi_Birnbaum No, that result is false.

It is interpretation-dependent, to be clear. For instance, in the Bohm interpretation you do have meaningful trajectories. But those trajectories come with some very strange properties, e.g. nonlocality, first-order dynamics, etc.

Novalium Company

@EmilioPisanty Simply said, we still dont know exactly what happens when we observe?

@EmilioPisanty please enlighten me, how can you write that?

If you know that $\vec f(\vec r) = h(\vec r) \nabla \rho(\vec r)$ it does not follow that there exists $\varphi(\vec r)$ such that $\vec f = \nabla \varphi$

As a simple counterexample, take $\rho(x,y) = x$ and $h(x,y) = y$

then $\vec f(x,y) = (y,0)$ has nonzero curl and is not equal to any gradient

If you have additional constraints on the $\vec f$'s which allow you to rule out that behaviour, then the story might change.

@NovaliumCompany It depends on what you mean by "what happens"

in a sense, we do. The wavefunction collapses.

but if you want a "deeper" explanation, then no, we don't.

2:14 PM

for any practical purpose, at least. whether wavefunction collapse is to be taken as "literal" is another interpretation-dependent point

@EmilioPisanty: No I haven't I have just that all are of the form $h(\vec{r})\nabla(\rho({r}))$

So it means, that is all I can say?

@Rudi_Birnbaum if that's all you've got, then yes

We know the laws of quantum mechanics and we know how to apply them. But (I would argue, and this is a subjective statement) we don't really "understand" those laws to an extent that'd be sufficient to merit the term "understand".

@Rudi_Birnbaum yep. With what you've given, that's the furthest you can go.

(I should think)

Note that, in a philosophical sense, we're talking about metaphysics (in the sense that we're talking about the properties of a system apart from they could conceivably be measured)

People tend to use that word pejoratively but I don't

You know its a gauge origin distribution, such that $\frac{1}{2}\vec{B}\times(h(\vec{r})\nabla(\rho(\vec{r}))$ is a current density (for some case)

2:19 PM

For me the point of QM and its interpretations is that it constrains the kind of metaphysics you can credibly adopt for the quantum world. But "constrains" is very different than "determines"

So there's a lot of freedom for different views on what's going on "beneath" QM, but they all have to be consistent with the QM formalism

(this is so long as we're not supposing that QM is merely an approximate description of something more fundamental. some people like that idea.)

Sorry, I mean: ($\frac{1}{2}\vec{B}\times(h(\vec{r})\nabla(\rho(\vec{r})))\rho(\vec{r})$

@Rudi_Birnbaum wait, what?

Abhas Kumar Sinha

$$
\dfrac{dL}{dt} = \sum \dfrac{\partial L}{\partial q_i}\dot q_i + \sum \dfrac{\partial L}{\partial \dot q_i} \ddot q_i
$$

proof:
$$
\begin{align}
\dfrac{dL}{dt} &= \sum \dfrac{d (K)}{dt} + \sum \dfrac{d(-U)}{dt} \\ &= \sum \dfrac{d(K)}{d \dot x} \dfrac{d \dot x}{dt} + \sum \dfrac{d(-U)}{dx} \dfrac{dx}{dt} \\ &= \sum\dfrac{\partial L}{\partial \dot x} \ddot x + \sum \dfrac{\partial L}{\partial x} \dot x \quad\dots \quad \left(\text{Because } \dfrac{d K(\dot x^2)}{d \dot x} = \dfrac{\partial L}{\partial \dot x} \text{ and } \dfrac{d(-U(x))}{d x} = \dfrac{\partial L}{\partial x} \right)

Is my proof correct?

Can anyone check it?

Also, $L$ is Lagrangian function...

$L = K(\dot x^2) - U(x)$

@EmilioPisanty Can you check it, please? I'm not sure about the mathematics here,

Novalium Company

@EmilioPisanty I mean, does the electron take a definite position and velocity after the collapsing of the wavefunction?

In the sense of it retaining those values after measurement, no.

Abhas Kumar Sinha

2:27 PM

@Semiclassical Are you talking about my proof?

I mean, if you measure position at one time and then measure it again soon after, then you should find the particle in a similar location

@NovaliumCompany It never simultaneously takes both a well-defined position and a well-defined velocity

But if (for a free particle) you measure position, then measure velocity, and then measure position again, you should not anticipate getting the same position back twice

if you choose to measure position, it will take a well-defined position immediately after you measure it (though it will then tend to spread as time goes on).

If you choose to measure velocity, then it will take a well-defined velocity, which it can retain or not depending on whether there is a nonzero force acting on it

@Semiclassical and that.

@AbhasKumarSinha This is not a "proof" because it does not start with hypotheses.

This is just a jumbled collection of formulae.

If it doesn't have text, it's not an argument, and if it's not an argument, it's not a proof.

Though, that said, the manipulations are roughly correct.

Abhas Kumar Sinha

@EmilioPisanty Sure?

Proof of the statement above the text "proof:"

2:30 PM

@AbhasKumarSinha To the (limited) extent that the jumbled collection of formulae can be made sense of.

If you want more, then provide an actual argument and explain what you're doing.

Abhas Kumar Sinha

To prove: $$\dfrac{dL}{dt} = \sum \dfrac{\partial L}{\partial q_i}\dot q_i + \sum \dfrac{\partial L}{\partial \dot q_i}\ddot q_i$$

@AbhasKumarSinha you shouldn't need help with something as basic as a simple chain rule derivative if you want to read a book this high level tbh, why not learn simpler material first

@EmilioPisanty strange?

I think it's just not clear what you mean

Abhas Kumar Sinha

@bolbteppa Are you sure that my proof is correct?

2:34 PM

You are not proving the last statement that's literally just the chain rule, you're supposed to apply this to the given example of $L$ which in this special case is $L = T - V$

@Rudi_Birnbaum I'm not sure I understand the condition.

Abhas Kumar Sinha

@bolbteppa Are you prof. Balakrishnan?

@EmilioPisanty well the condition is that this thing should be divergenceless

Yes

...

Abhas Kumar Sinha

@bolbteppa Ask how did I know that?

2:36 PM

You're saying that there exists $\vec B$ such that $$\nabla \cdot \left[ \left(\vec{B}\times \bigg (h(\vec{r})\nabla\rho(\vec{r})\bigg) \right) \rho(\vec{r}) \right] = 0?$$

$j_d=(1/2 B x (r-d(r)))rho$

is the diamagnetic part of the current density

and I search for d(r) = gauge origin distributions s.t. div(j_d)=0.

then I obtained (r-(d(r))=grad rho(r)

@AbhasKumarSinha how

such that for d(r)=r-grad(rho) I obtain div(j_d)=0.

for example d(r)=r is called ipsocentric gauge origin districution j_d=0 and of course div(j_d)=0.

@Rudi_Birnbaum sorry, I'm having real trouble following your notation.

where trivially 0 || grad(rho)

with

user image

now you can use A=1/2 B x (r-d(r)) ..

with a so called "gauge origin distribution" d(r)

Abhas Kumar Sinha

2:41 PM

@bolbteppa 1. Because I've been in the video, that Prof. also writes $-V$ instead of $-U$. 2. Your nick rhymes with that 3. You write more about Math SE than Physics SE (which I've heard about that from your former students on Quora) 4. You teach very well, (explain even a bit complicacy easily) means that you have some teaching experience. 5. You linked me a video long ago, which was from some different professor. which everyone would do, instead of giving theirs (specially who are indian)

to be read as $\vec{d}(\vec{r})$

d then enters of course the perturbed density $\Psi$

Abhas Kumar Sinha

In addition to that, You wrote your name a long ago (those who have not written that long name before can't type that fast!), this part gave me hints about it.

@AbhasKumarSinha different people use $U$ or $V$ in the literature

@bolbteppa which is always a bit annoying to me, but so it goes

Abhas Kumar Sinha

@bolbteppa I know, that was one of the hints (although that was not to make 100% sure, but still that works)

2:44 PM

Kinetic energy is usually $K$ or $T$, potential energy is usually $U$ or $V$

Abhas Kumar Sinha

Yes

@bolbteppa So, you are still teaching or you are currently in research?

Mostly I don't like $V$ as potential energy when in electromagnetism it's potential

Abhas Kumar Sinha

@Semiclassical I too get confused over that matter :(

@AbhasKumarSinha maybe go through the introductory videos here then 'my' course here and then 'my' math course here and then 'my' quantum course here rather than the hardest books which casually use things that should be second nature to you

Novalium Company

So the only thing that counts as an observer is a polarization filter, no?

Has anyone tried to see if a cat counts as an observer?

Abhas Kumar Sinha

2:53 PM

@bolbteppa Okay sir...

@bolbteppa Sir, do I need quantum mechanics before Lagrangian mechanics?

@AbhasKumarSinha no you need Newtonian mechanics before Lagrangian mechanics

Novalium Company

Has anyone tried to shoot the pattern to the back of a human skin, would that break the wavefunction?

Abhas Kumar Sinha

@bolbteppa okay.

@AbhasKumarSinha what about A.C. Verma Concepts in Physics (discussed here and here)

Abhas Kumar Sinha

@bolbteppa Not sure, but I've completed other books like I Irodov's Problems in General Physics and SS Krotov

Novalium Company

3:01 PM

In which explanation of the collapse in QM do you guys believe?

Abhas Kumar Sinha

@bolbteppa This is the latest book I've done (not completed, still a few questions on mechanics is left)

This one : archive.org/details/ProblemsInPhysicsSSKrotov

Bleh, I dunno. I'm personally sympathetic to the Bohm interpretation as providing a productive visualization, but there's real issues with viewing that as a "real" description of what's going on

It's a useful enough visualization for non-relativistic QM, but seems not to provide good lessons for how to understand relativistic QM or quantum fields

Novalium Company

@Semiclassical What exactly is it in simple terms?

How much have you worked with the Schrodinger equation?

Novalium Company

0%

That may be a problem :D

3:07 PM

yeeeah

the basic idea is: The Schrodinger equation dictates the evolution of the wavefunction, which by its nature is a field in configuration space

So you prepare the wavefunction in an initial form, and then let it evolve according to some potential. So far, that's just regular wave mechanics

In particular, this (time-dependent) wavefunction determines a (time-dependent) probability density $\rho(x,t)$ in configuration space.

@AbhasKumarSinha I'm not sure what your goal is so not sure what advice to give but the videos above would be a good guide for extra study out of interest

Abhas Kumar Sinha

Okay...

At the same time, there's also a time-dependent probability current density $\vec{j}(x,t)$ (as dictated by the form of the Schrodinger equation)

together, those imply a time-dependent velocity field $\vec{v}=\vec{j}/\rho$. you can interpret those as generating streamlines of the probability density

That's just math, and you can take it as a convenient visualization (or not)

Novalium Company

I want simple words, pleeease.

No.

Novalium Company

3:13 PM

Yes.

The Bohm interpretation goes a step further and interprets those streamlines as the set of allowed trajectories for the particle

Novalium Company

Im 16 years old, chill :D

The basic image is that the wavefunction evolves in time, subject to the Schrodinger equation, and this same wavefunction determines what paths the particle can take

Hence why it's also referred to as the "pilot-wave" interpretation. The wavefunction evolves according to the Schrodinger equation, and it pilots the particle.

Novalium Company

I thought its just a density cloud of probabilities?

In the standard interpretation, yes. But Bohm isn't the standard interpretation.

Novalium Company

3:17 PM

So many things to learn... in this QM..

and in the end, it turns out we were wrong and stupid.

in the Bohm interpretation, one does insist that the particles have meaningful trajectories even when you're not observing them

this doesn't come without cost, though. for instance the Bohm interpretation requires a measure of nonlocality, which seems to conflict badly with the spirit of special relativity

The more common view is that this wave function collapse stuff is vaguely like saying if we flip a fair coin, until we flip it and observe which outcome we got we can only describe the situation of flipping a fair coin with a 'wave function' $\psi = \psi_{up} + \psi_{down}$ but once we measure it can only be one of $\psi_{up}$ or $\psi_{down}$, the idea this 'collapse' means anything is pretty much just misundertanding what QM is

and IMO it's only a productive visualization when there's only a few degrees of freedom involved. more than that and your configuration space has high enough dimension as to elude visualization

@bolbteppa eh, i think it's hard to boil it down to just that. i mean, if all you have is one type of measurement, then attributing classical values isn't such a big deal

it's once you have more than one measurement that you run into non-commuting observables

As an example: Suppose you ran a Bell-type experiment on entangled (singlet state) electrons where one observer only had access to one Stern-Gerlach device and the other could change their device freely

In that case you'll never be able to observe a violation of the Bell inequality.

At a minimum, the first person needs to have access to measurement settings $a,b$ and the second to measurement settings $b,c$. Once you've got that, then you can test the Bell inequality

The classicalness of this situation doesn't matter here, it's just a way of saying: 'before we measure there are certain possibilities, the wave function simply captures this all along with the probabilities of them happening in a linear combination, but once we measure only one of those outcomes actually happens'

Sure, but my point is that you could equally well account for that setup with a local hidden variables model.

It's only once you've got multiple measurement types that you can run into situations which LHV can't account for

(This is not to say one can't think of the up/down situation like that, to be clear. It's just that the necessity of doing so isn't apparent in that situation)

3:28 PM

@NovaliumCompany you should be wary of your attitude with which you approach subjects.

Approaching everything with the attitude “this will fail eventually because everything else did in the past” isn’t a good attitude

hmm

anyone else having trouble getting to SMBC?

i mean, connecting to the site is fine, but they've got an advertisement for their upcoming book there which I can't close obscuring the screen

it ignores the incredible amount of experience that tells us how good our models currently are (in certain circumstances) and neglects more so, the spirit of science. Which is run with a model until you have something that contradicts it.

It may turn out that our models are wrong, and that we just keep going and going with better approximations. But possibly we arrive at a theory that is as good as it can be (not likely imo, but I doubt the future theories in 100 years won’t hold at there core principles from today). Even so, the theories we have now work ‘good enough’ in a lot of cases. For example. Classical dynamics explains an unbelievable amount of phenomena, and with additions of quantum theory explains even more.

My point is this, if you want to learn physics it’s also about understanding the stuff that works ‘good enough’ and knowing that as best as you can. Ultimately it’s going to be a part of the bigger puzzle.

Thanks for coming to my ted talk

Novalium Company

4:15 PM

@JakeRose Im not talking that physics will turn out to be wrong, Im saying that it may all be interpretation of our brain ans sensors. I mean, everything could break down to our brain, consciousness and interpretations...

The brain of a rat < the brain of a cat < the brain of a human < enchanced human brain

That’s more philosophy. Which physicists don’t take that well to. And fundamentally yes, everything is a measurement +interpretation of some sort. If you can’t measure it or at least measure something related to it then there’s little point even entertaining the idea

string theorist enters

But the point is, if you want to understand physics stop worrying about all that. Just learn!

Novalium Company

I want to learn little of everything :)

But physics is interesting.

This is avnish kabaj's sock

What's up with your elections

Novalium Company

Why not enchance the brain to better make sense of the world... actually no one knows what happens

@NovaliumCompany if every time you try and learn physics you stop learning because you talk about how it’s all gonna be trivial in 100 years then I’m afraid you won’t learn it very well

the way you enhance the brain is by studying and thinking.

Novalium Company

4:28 PM

... ok :)

0

Q: Asking about the origin of an equation?

The other some grad colleagues in a local forum were asking about a certain system of equations of unknown origin, and I was wondering if I would be allowed to ask about it in the Physics Exchange. My concern is that I'm not even sure if this system belongs in physics, and I was just hoping someo...

discussion questions

Novalium Company

@JakeRose It makes no sense for me not to be able learn physics properly despite my views on it and the world.

I’m not saying you opinions are wrong

I’m saying your attitude is

Novalium Company

ok :)

@JakeRose my own view is that the various interpretations of QM (I'm excluding modifications of QM, e.g. GRW) show that it's perfectly legit to have intuitions/concepts about what's going on with QM "behind" measurement

but any such metaphysics of quantum is at the very least constrained by the formalism.

So one shouldn't expect to be able to say much about what's going on in QM without having a rock-solid understanding of the bare formalism.

(one can of course critique such interpretations on grounds other than "is it compatible with the QM formalism". but if they can't do that much, they're not interpretations)

4:57 PM

@JakeRose Hey! ::sobs::

5:47 PM

@NovaliumCompany the wavefunction collapses when the microscopic particle interacts with a macroscopic object

thats all

the cat business is all humbug

Novalium Company

The microscopic particle is interacting with another microscopic particle, which is part of a macroscopic particle... soo wtf?

measurement is = the interaction of the microscopic particle (=the quantum object) with a macroscopic object (measurement device).

Novalium Company

The macroscopic devices are made up of microscopic particles, soo??

@NovaliumCompany irrelevant

nothing like that is required or appears in the theory

Novalium Company

I don't care about the theory, I care about the logic, and to me, there is no logic...

It's microscopic particles measured by other microscopic particles, which collectivley are named "macroscopic".

5:55 PM

you possibly don't mean logic.

You cant read the meter if its not something that obeys the laws of classical physics.

Novalium Company

I have no idea what you are trying to tell?

@NovaliumCompany a correct interpretation of quantum mechanics

that does not require any observer or consciousness

Novalium Company

As far as I know, there isn't yet a universally accepted interpretation of QM.

thats not the point

Novalium Company

There are many interpretations, and you choose in which one to believe.

Mithrandir24601

5:58 PM

@NovaliumCompany Not really, no

well you don't have to choose

its enough to know them

if they are equivalent it doesn't matter

its "just" interpreatations

physcis is more about predictions of numbers

than about "interpretations"

Novalium Company

What are you talking about buddy?

about your question on the interpretation of QM.

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