3:56 AM
Good Morning!

4:06 AM
@Azmuth GooD MorNinG...!

10 hours later…
1:56 PM
why QM in curved space doesn't work?

What you mean by curved space?

Or why does qm stuff does not respond to gravity effects
Can any body help me

@ACuriousMind ok I have read it, actually I have a confusion that does gravitational field interact with in the absence of gravitons

I'm sorry, I don't understand the question
there seems to be missing something between the "with" and the "in the absence" for that sentence for make sense
also I don't know what "absence of gravitons" is supposed to mean

2:06 PM
Sorry for typing error actually I was asking can gravitation field interact with matter in the absence of gravitons?if yes then does energy conservation hold?

Gravitons are already hypothetical assumptions of quantised matter for QM to roughly emulate Curved Spacetime
Quantised Gravity hasn't been yet practically proved.

@JackRod What do you mean by "in the absence of gravitons"?

@ACuriousMind Can gravitons be proved with QFT?
or rather derived ?

Isn't the graviton just a predicted particle in qft
to include a gravitational interaction

@ACuriousMind like real graviton
A quanta of gravitation field
If some quanta are absent

2:21 PM
@JackRod Gravitons simulate quantised gravity, you don't need matter, when you assume gravitational field without matter.

@Charlie indeed they are virtual but there existence is still there in our calculations

@JackRod I know what a graviton is, I don't know what you mean by "absence"

A quanta of gravitation field
If some quanta are absent

It doesn't help if you keep repeating the phrase I don't understand :P

Gravitons don't have to be virtual, right

2:24 PM
I don't know what it means for a particle to be "absent", it's not a pupil that can be absent from class or anything :P

a representative pupil?

@Charlie Like all particles, it could be both real and virtual, sure

yeah

@Charlie gravitons are hypothetical, they contradict GR.
@skullpatrol hey pal!

Sure, but "virtual" is not the same as "hypothetical"

2:26 PM
hi pal

I'm not even sure if "contradict gr" is the correct way of putting, don't they just have divergent feynman diagrams that can't be renormalised

@Azmuth Gravitons don't contradict GR. Did anyone actually read physics.stackexchange.com/q/387/50583 I already linked above?
Doing naive QFT for GR-type theories just gives you a non-renormalizable theory, which we don't like for various reasons

Okay, my fault, renormalisation problem.....

and he doesn't like to repeat himself :P

$\xor$
$\oplus$
> right-adjusted string of n – 1 1’s, where n is the size of the rightmost group of 1’s in x
@skullpatrol Do you understand that line?^
How are or and exclusive or operations differ?
Computer Algebra is so hard :'-(
WOW! Everyone Check this link on Writing an 16 Bit OS with C notice, how assembly used in C simplified everything so such a great extent!
Applause
Goosebumps!!

1 hour later…
3:56 PM
XD LMAO!

4:33 PM
Hi guys. Who's up for some physics chat?

5:00 PM
Uh: "As expected, most of the DNA recovered from the feces samples in the study belonged to frog species and plenty of lizards, but researchers also found evidence that the bats were eating other bats and even a hummingbird".

5:35 PM
Trump or Biden?
@Azmuth XXXXDDDDD

I've definitely never been a fan of Trump or his approach

5:54 PM
So one vote for Biden. Let's hear more and then we can generalize it and get a good approximation for the actual results.
Is Biden socialistic?

6:19 PM
I wouldn't say overly. The US Democrats lean more socialist than Republicans, but I don't think either are that socialist right now. Bernie Sanders for example leaned harder into that, but he didn't win the Democratic nomination.

Like most political terms, it's pretty malleable. Much of what US politics considers "socialist" (even as a self-description) would register as firmly center-left/social-democratic in many European countries.
And most of these Eurpoean social democrats would not describe themselves as socialists

Canada for example has generally more social programs than US (like universal health care) that even the right leaning side of Canada doesn't push back much on. The US seems to not really like that idea much, at least based on how reluctant they are to implement it. Even still, the most left leaning Canadian parties don't really describe themselves as "socialist" either.

I don't know who Joe Biden is but Trump has been entertaining surely

It's entertaining when you forget he's literally a world leader. Every time I remember that I generally just get disappointed and baffled.

He (the president) does have some power, but we cannot blame politicians for every problem the world has. Some people are evil, others are racists... it's easy to point a finger towards a single person but I think that even if we choose the best leaders in the world, everything would not suddenly get fixed. And the president is just a face chosen from the crowd, but a lot of the behind the scenes is run by many other people.

6:34 PM
@JingleBells I don't think anyone is blaming politicians for every problem the world has.

@ACuriousMind hmm, alright. Forget about my message then (cuz my premise is wrong I guess)
What are politicians blamed for?

We can't blame politicians for every problem, but it's hard to fix some problems without going through the political system. Even if choosing the best leaders won't fix everything, but choosing better leaders is still generally a better option. The president is just a face chosen; but the character of the choice represents something about the people. And yeah, a lot of behind the scenes is done by others; but the president chooses a lot of those people; which can also be problematic.

When the corona was starting in Bulgaria, we had like 3 sick a day, and the minister was like "yeah sure, stop kids from going to school" and after 4 or 5 months of this, schools reopened. Now we have around 1400 sick a day and the minister is like "Nah, we ain't closing shit"

The president has a lot of influence too, controlling the executive is a lot more than just being a public figure

@Charlie I agree

6:41 PM
People have short memories, as soon as cases started to slow most countries thought that was the all clear and are experiencing worse second waves of covid

I mean, if Trump is bad then why people elect/ed him? As JMac said, the president represents the people.

Trump did not even win the popular vote, there are a number of reasons he won though

My side of Canada has done really well. I'm still so surprised and happy by how it's going here. Look at how flat this curve has been since May. novascotia.ca/coronavirus/data

I would say Biden, since Trump denies global warming

Biden - 2 | Trump - 0

6:51 PM
@JingleBells People's voting behavior is usually a lot more varied and a lot less sophisticated than "this candidate/party/whatever's policies/character reflect what I want". Additionally, in many democracies without compulsory voting, voter turnout tends to be low enough that the non-voters make up the strongest "faction" - i.e. it is not ordinarily the case that the winners of a vote actually have the explicit support of a majority of the people they govern.

Got it.
In unbiased estimators, we're trying to estimate some statistical value (mean, variance...) by only having a part of X, right? But does that mean that we can also compute the probability distribution of the sample (part) of X? And we use this partial probability distribution built from a sample of the whole X (that we don't know) to estimate statistical values? Is my understanding of unbiased estimators correct?

that's the sort of question you'd need to ask an actual statistician :P

but... i thought you knew everything :(
get one of your friends pliz, I'll pay xDD
(jk i'm broke)

7:15 PM
when you realize the magenta color doesn't exist on the em spectrum

7:29 PM
Yeah that's a weird one to realize. Pretty cool demonstration of how weird our minds are though.

What's the problem with our brains perceiving the mix of red and violet as green? I'm a bit confused about why the brain has invented magenta.

A lack of green and green would do pretty different things to eye receptors as far as I understand, so it kinda makes sense that the brain sees them differently. Also I think colours and combinations would be really weird if that happened. There would be two completely different ways to make green.

Sorry, I don't understand why when red and purple enter our eyes, we don't see green. Is it because no green eye receptors are firing so our brain doesn't know what to do so it invents a color?

8:31 PM
Hi, all
Any one is fresh to help me out to understand physics mysteries.
I wonder in inelastic collision how momentum is conserved and Kinetic Energy is not conserved. What does it mean by that.

8:54 PM
@123 Probably that $mv_1 + mv_2 = m(v_1+v_2)$
but $0.5 mv_1^2 +0.5 mv_2^2 \neq 0.5 m(v_1+v_2)^2$
it's been a billion years since i've done that tho so i might be wrong

Hello, my question is that. how we know momentum conserved. and what happened K.E not conserved.

I was under the impression that energy is lost in the form of heat to one of the bodies in the collision

I term of physics

@Charlie ah yes that's what it was

An experiment that shows this would be placing a bouncy ball in a vacuum and letting it bounce up and down under gravity. Eventually it will stop since each time it hits the ground some kinetic energy is lost in the form of heat to the ground.

8:58 PM
@Charlie yes i can understand energy lost may be by heat. What is meant by momentum conserved. when heat lost in term of energy lost
@Charlie Ookay, but if heat is lost means energy not conserved. But there is no effect of heat lost in momentum this is my question.

In an isolated system it's not possible to lose overall momentum.

are you asking how momentum can possibly stay the same when we've lost KE, and therefore speed?

Energy isn't being lost either in an isolated inelastic collision, it's being transferred from bulk kinetic energy to microscopic kinetic energy if you like

@SirCumference Yes
@Charlie Pls explain this phenomenon in physics way.

That is "a physics way", when the two objects are travelling towards each other their kinetic energy is due to their bulk motion, $\frac{1}{2}mv^2$. Once they collide, if the collision is inelastic, some of that bulk kinetic energy is "lost" as it is transferred in the form of heat (which is microscopic movement) to the atoms of the two bodies.
So after the collision they two objects have less bulk kinetic energy than when they started, but the total energy hasn't changed, assuming the system is isolated.

9:04 PM
@Charlie Ookayie... But we don't take potential energy here. Total Energy = Kinetic Energy. How it is possible?
I want to connect loss of kinetic energy with momentum conservation. If heat lost how can possibly momentum conserved.

Yes, what you're getting at there is that the total energy isn't actually changing, that would violate the conservation of energy.

mechanical energy isn't the same as total energy
mechanical = KE + PE
total = mechanical + other types of energy, e.g. heat or electrical energy

@SirCumference Ahaa.. Hm.. What is total energy here?

Mechanical energy + heat in this case

@Charlie I see...

9:07 PM
Momentum isn't a form of energy so you can't "lose" momentum in the same way you can "lose" bulk kinetic energy in the form of heat

mechanical energy is most useful when the other types of energy are absent, in which case mechanical = total

the microscopic movement corresponding to heat is unordered movement where the particles move randomly in all directions, so the net momentum associated with thermal motion is zero.

But if heat lost there is no effect on change in speed?

There is a change in speed, $\text{KE}=\frac{1}{2}mv^2$
The speed is reduced since the kinetic energy is reduced and the mass isn't changing

@Charlie If speed change. How can possibly momentum didn't change. $\vec{P} = m\vec{v}$
But momentum is also depend on speed. this was the problem i struggled.

9:11 PM
you're asking for an intuition

Momentum and K.E both has speed dependent.

it's a good question, i'll think about it

Also in term of math. Because this is the problem with book author did not explain the topic.
@SirCumference Thank you. Pls take your time.

@123 Note that there are two bodies involved here, $p = m_1 v_1 + m_2 v_2$. That overall kinetic energy is reduced does not mean you can't find $v_1,v_2$ so that the momentum is still the same.

Ok I thought about it for a second because what you're saying did actually make sense. The momentum from the bulk motion of the objects will decrease
wait am I wrong?

9:15 PM
@123 One thing to notice is that KE depends on $v^2$, not just $v$, so the values of $v$ where initial KE = final KE so you shouldn't expect them to align fully with the values where initital momentum = final momentum

This is a bit troubling

@ACuriousMind Hi, Yes i know but for kinetic energy there is also same two bodies involved.

There's still momentum associated with the internal thermal motion

@Charlie No

oh but it's net has to be zero

9:16 PM
8 mins ago, by ACuriousMind
the microscopic movement corresponding to heat is unordered movement where the particles move randomly in all directions, so the net momentum associated with thermal motion is zero.

ah yeah

@ACuriousMind he's saying that intuitively we should have KE decrease => speed decrease => momentum decrease, yet momentum is somehow conserved

@123 so?

it's more of an intuition question

@SirCumference Yeah, but that's just not true when more than one body is involved
A loss in total KE doesn't mean a decrease in speed at all. If one body is like a hundred times more massive than the other and I take 50% of the energy and give it to the second body and throw the other half away, the second body will be a lot faster than the first was with the full energy.
(note that these specific numbers are likely not a valid solution for conserved momentum :P)

9:19 PM
@ACuriousMind K.E and P both has mass and velocity dependent only. If K.E change means either mass or speed would change. mass is constant. There must be speed reduce in term of heat so KE change.
How it is possible momentum conserved if speed changed.

@ACuriousMind ahh, that's true
@123 He's saying that when we have multiple bodies, total kinetic energy doesn't necessarily match up with total speed

@123 If both bodies have the same mass $m$, and one is initially at rest while the other moves with $v$, then in a perfectly inelastic collision they'll stick together after the collision and both move with $v/2$. The initial momentum is $m \cdot v + m\cdot 0$, the final momentum is $m \cdot v/2 + m \cdot v/2$, so momentum is conserved even though the speeds of both bodies changed!

@ACuriousMind If other body gain the speed it means total momentum is conserved. But it also conserved KE.

@123 You have $\frac 1 2 m v_{1i}^2 + \frac 1 2 m v_{2i}^2 = \frac 1 2 m v_{1f}^2 + \frac 1 2 m v_{2f}^2$ compared to $mv_{1i} + mv_{2i} = mv_{1f} + mv_{2f}$ The combinations of $v_f$ and $v_i$ that make each side equal can be different for momentum than for kinetic energy, so from the math it should kinda make sense that momentum can be the same on each side while kinetic energy no longer is.

For simplicity we take two bodies collision

9:22 PM
Intuitively let's say we have a very heavy body and a very light body. If we reduce the heavy body's speed, we've greatly reduced the system's KE (since it's so massive). But if we reduce the light body's speed by the same amount, we haven't reduced the system's KE by as much
I could keep the KE the same by decreasing the heavy body's speed by a little, and consequently increasing the light body's speed by a lot

but if you compute the KE for this scenario, it's initially $\frac{1}{2} m v^2$ and after the collision it's $\frac{1}{2}m v^2/4 + \frac{1}{2}m v^2/4 = \frac{1}{4}mv^2$, so KE was not conserved

The total KE is kept the same but the total speed has changed

@ACuriousMind L.H.S (before collision) = R.H.S (after collision) in your KE equation and this should be equal, but you take before collision $\frac{1}{2} mv^2$ separately.

I don't know what you mean
It should not be equal, this is an inelastic collision after all!
Both bodies have mass $m$, one is moving with $v$, the other is at rest. Afterwards, both have velocity $v/2$. My computations are correct.

Yeah the point he was showing is that when momentum is equal before and after, kinetic energy isn't necessarily equal as well
Or one point at least

9:31 PM
It is possible for you guys. Give me one example with calculation where momentum and KE along with in same example. and also explain the physics. Thank you
I wonder the equality sign in KE equation before and after collision.

What's insufficient about the example I just gave?

@ACuriousMind In your KE comment. I assumed you take $\frac{1}{2}mv^2 + \frac{1}{2}m(0^2) = \frac{1}{2}mv^2$ because first body is moving with speed v and second body is at rest before collision.
This is what i am thinking.

you're not wrong

After collision both moving with same speed.
According to my calculation in your example energy is also conserved.
In case if i accept by experiment, if bot bodies stick together total mass becomes $m = m_1 + m_2$ but $m_1 = m_2$ , Total mass $= 2m$ So, $\frac{1}{2}mv^2 + \frac{1}{2}m(0^2) \neq mv^2$ , The equality does not hold.

the bodies don't stop moving, they move with $\frac{1}{2}v$ after the collision.

9:46 PM
Also i accept by experiment we observed, We take $\frac{1}{2}v$. Then, 
Also i accept by experiment we observed, We take $\frac{1}{2}v$. Then, $\frac{1}{2}mv^2 + \frac{1}{2}m(0^2) \neq \frac{1}{2}m (\frac{1}{2}v)^2 + \frac{1}{2}m(\frac{1}{2}v)^2$