The h Bar

12:08 AM

Oh wow TIL fuzzy logic actually has applications

My friend was doing some work on fuzzy logic for health care applications with a professor once. TBH I hadn't even really heard of it until then, but it sounded like a really cool approach

skillpatrol

> Michael Hanss, 2005. Applied Fuzzy Arithmetic, An Introduction with Engineering Applications. Springer, ISBN 3-540-24201-5

Sir Cumference

+1 for non-boolean logic

there's an entire rabbit hole of weird kinds of logic

Vector logic

Vector logic is an algebraic model of elementary logic based on matrix algebra. Vector logic assumes that the truth values map on vectors, and that the monadic and dyadic operations are executed by matrix operators. "Vector logic" has also been used to refer to the representation of classical propositional logic as a vector space, in which the unit vectors are propositional variables. Predicate logic can be represented as a vector space of the same type in which the axes represent the predicate letters S {\displaystyle S} and P...

skillpatrol

+0.999... for weird kinds of logic :P

roobee

12:33 AM

so solar sails convert light to kinetic energy. but it they are very inefficient. is there some theoretical limitation or just we have not practically found a good material yet?

is there some mechanism that can reversibly convert light to kinetic energy? sorta like the piezoelectric effect except with light instead of electricity?

roobee

1:10 AM

if heat is atoms vibrating why can't we cool atoms to absolute zero by harvesting the thermal motions? i know thermodynamically this is impossible by not sure why kinetically it doesn't work.

JMac

@roobee Temperature basically tells you where the thermal energy is going to go when two objects are brought together. So to harness the "vibrations", or thermal energy, of those atoms you need to have something with weaker thermal motion for that energy to want to naturally transfer. To get something near absolute zero to give it's thermal energy to something warmer, you need to do work on the system, which costs energy.

roobee

@JMac That is the thermodynamic argument which I agree is experimentally correct. I am confused when thinking kinetically why I can't place tiny piezoelectric sensors (with rectifiers so the average current > 0) on a object to convert the thermal motions to electricity

JMac

1:29 AM

@roobee How would the piezoelectric sensors be able to get anything from that? From a "kinetic" perspective, unless the sensor material is colder, it has more average kinetic energy than the substance you're trying to take energy from

roobee

@JMac I was thinking the random motions of the material would bump into the piezoelectric sensor to convert into electricity?

JMac

@roobee But how would they take the energy from the atoms when the atoms trying to take the energy have more energy on average?

roobee

@JMac when thinking of energy conservation that is a good point. Although I'm not sure how the piezoelectric effect functions from a kinetic perspective. I was modeling it as a black box.

roobee

2:15 AM

So ya when thinking in terms of the microscopic movement of particles it's still confusing.

How does one deduce light power has 1 Watt originally without any pre-existing light standards? I can imagine one take a watt of electrical energy, feed it into a lightbulb, measure the heat dissipation, shine the light on an photodiode, measure the temperature increase and electrical power produced. Then set that light source as a standard for light power. But that seems a hassle. Is there an easier way?

Secret

10:12 AM

Secret

10:25 AM

arxiv.org/abs/1909.01343

bolbteppa

11:40 AM

There's always, always, a way to see some math trick as natural, even this whole $\chi_{k,l} = r \chi_{k,l+1}$ relation in the spherical bessel equation so often coming out of thin air in QM books!

ACuriousMind

2:05 PM

@AaronStevens I think that functionality exists, but it's not something moderators can turn on

Jake Rose

2:16 PM

@ACuriousMind you around?

ACuriousMind

sort of

Jake Rose

I’m trying to show $e^x e^y = e^{x+y} $ via the power series

but I’m getting a little stuck

Any help?

Slereah

Sounds like a job for the power expansion of sums

and Pascal's identity

Jake Rose

this is for a quantum question btw, I’m going to e land this to operators in place of x and y

*expand

Slereah

Errrr that's not correct then

You need the Lie-Trotter formula

Jake Rose

2:19 PM

Oh I know, but I’m trying to see where the ability to commute x and y is important

as in where I can’t do that for an operator exponential

Slereah

Put everything in the same form

For instance $x$ first and $y$ second

Then use the power expansion

Jake Rose

Already done it

Let me tackle this some more

havent put enough effort into it

ACuriousMind

@JakeRose I'd try it like this: Start with the r.h.s., expand with the binomial formula, then rearrange into the l.h.s.

But if you just want to know where it fails for operators, that's already in that first step - the binomial formula works only for commuting things.

Slereah

Yeah overall there is no expansion you can really do with operators

$$(x + y)^2 = x^2 + y^2 + xy + yx$$

To have the Lie-Trotter formula you have to sweep up excess operators in the commutator

Jake Rose

This is sort of a lemma for another question to do with differentiating an operator valued function, then deducing something else

thanks for the help guys

can see where it fails now which is nice

Slereah

2:28 PM

I had to eat up a lot of Lie-Trotter formula bc my master thesis was on path integrals

And boy do you use it a lot

Jake Rose

Mhmmm sounds juicy

you’ll have to send me a link and I’ll read it once we’ve finished this quantum course

Sir Cumference

2:51 PM

@JakeRose are $x$ and $y$ commuting operators?

if so use the Baker–Campbell–Hausdorff formula

i recall doing one of those problems in QM

Jake Rose

They commute with their commutator

Anybody got a hint for what I did wrong?

Sir Cumference

i never proved it with a Taylor series iirc, but if you're allowed to use that formula than it'd make the problem very easy

Jake Rose

Nope, got to show it

if I use a firmula I generally need to have shown that too

Semiclassical

3:08 PM

@JakeRose Another way that often works with those problems is to make it a function, e.g., let $F(t)=e^{-t X}e^{t(X+Y)}e^{-t Y}$. (I'm not sure why I chose that order, but w/e)

and then consider derivatives.

That doesn't seem so effective here, though. Hrm. I'm more used to the case $e^{-t X}Z e^{t X}$

oh. that's what you already did. ffs why can't I read

Sir Cumference

3:34 PM

looking back at my QM book, 90% of the problems were crappy proofs using bad notation

i don't get the point of that but whatever

Slereah

@SirCumference Welcome to physics

Sir Cumference

@Slereah physics is just bad math? :O

Slereah

@SirCumference As you well know

I prefer to use the term "good enough math"

Sir Cumference

i mean i guess proofs like these are good if you want to be an algebraist

but no one's gonna use them as an e.g. experimental physicist

Aladdin

@ACuriousMind can you open my frozen room.

ACuriousMind

3:44 PM

@Aladdin yup, done :)

Aladdin

Thanks a lot

Novalium Company

Hello sentient things.

I, Thor Almighty, come with peace and love.

@JakeRose equations.

Novalium Company

3:59 PM

No one chats with me here anymore :/

Avnish Kabaj

4:17 PM

Sir this is a physics chatroom

7

Basj

6:06 PM

Hi! My question physics.stackexchange.com/questions/509140/… has been closed as a homework/exercise question, however it was not. It was a real question that I asked myself since weeks, do you think it could be reopened?

JMac

@Basj Looking for a formula is generally off-topic, the homework policy is the generic reason for that. I think AccidentalFourierTransform explained it really well here physics.meta.stackexchange.com/a/10215/127931

Slereah

6:29 PM

Manchak still hasn't answered my email response to him

Very rude

I only took 7 months to answer his mail!

Hopefully he is still alive and I didn't answer TOO LATE

PM 2Ring

7:23 PM

@Basj I assume you've done the calculation now. That's a lot of mass to lift, even if you had an ideal 100% generator. I suppose you could lift it with a block & tackle. And perhaps consider getting a less power-hungry laptop. Are you sure it actually consumes 60 watts? It's probably around half that.

Basj

8:17 PM

@PM2Ring Let's say it consumes 30 W only.

I see 100kg and 2 meters height gives 0.5 Wh assuming 100% efficiency

this is really few energy @PM2Ring

so it could power the computer during 1 minute if it consumes 30 W only

PM 2Ring

8:31 PM

@Basj Correct. Using Google Calculator, I get about 65 seconds, assuming 100% efficiency.

Basj

8:47 PM

@PM2Ring hummm this is a quite bad idea then :/

@PM2Ring Isn't there a trick to *10 the energy when lifting weights?

with a pulley or something?

PM 2Ring

9:03 PM

@Basj Several pulleys. Google "block and tackle".

Basj

@PM2Ring this would allow to lift it more easily

but could this offer n times more energy when the weight is going down, and making turn an alternator?

in order to power a laptop for more than 1 minute with 100 kg?

PM 2Ring

9:20 PM

@Basj A block & tackle gives you mechanical advantage, that is, it reduces the force you need to apply to lift the load. But it doesn't give you free energy, the law of conservation of energy must be obeyed! So if your pulleys give you a mechanical advantage of 5, you can lift 500 newtons with the same tension that would be required for 100 newtons without the pulleys. But you must exert that effort for 5 times longer, and to lift the load by 2 metres you must pull the rope 10 metres.

Basj

Yes indeed @PM2Ring. So do you think there's no way to turn this grandmother clock technique with weights to produce, says 30 W at home?

PM 2Ring

@Basj You could do it, but it will be strenuous work. If you want to do that much exercise, I guess that's ok. :) But humans are not as efficient as internal combustion engines. You'll probably end up spending more money on food than you save on electricity. :)

Basj

Haha probably :)

PM 2Ring

FWIW, I used to know a guy on a (now closed) science forum who rigged up an exercise bike with an alternator to power his computer.

Basj

Haha fun! So in terms of efficiency riding a bike gives more energy then having a 100kg weight going down in 15 minutes from ceiling to ground?

PM 2Ring

9:26 PM

That way, he didn't feel so guilty spending time on the forum.

Basj

lol

@PM2Ring How much watt do you think we can do with a medium-speed riding on a apartment bike + alternator?

PM 2Ring

Well, legs are more powerful than arms. And he was basically pedalling continuously. If you were really fit, you could easily produce 30 W, but I think it'd be pretty tiring to do more than that over a long time span. In short bursts, a fit adult male can probably do over 100 watts.

There's a science museum in my city (Sydney) that has a bike generator setup that tells you how much power you're producing, and powers an incandescent light bulb. But the last time I tried it was a few decades ago, and my memory's a little fuzzy on the details. ;)

This site says amateur cyclists can do about 3 W/kg (watts per kg of body mass) and if they really push it 5.5 W / kg, but that's pretty gruelling. Professionals can achieve rates like that without so much metabolic stress.

@Basj ^ Please see the linked article for the full details.

Scientist Smith YT

9:53 PM

I've got a physics energy and transference question. This is open to everyone who is willing to give input. I want to make an electromagnetic dirt compactor that I can use to compact the ground before putting a building in. My questions is, with the in and out motion would that compactor lose only 2 times the amount of energy it gets? Considering it uses electricity.

Basj

10:11 PM

@PM2Ring interesting! cyclinganalytics.com/blog/2018/06/…

Transcript for