The h Bar

danielunderwood

12:02 AM

RL version of a local minimum?

I don't know enough about RL to even know if that's a sensible thing though

enumaris

12:40 AM

feels like it shouldn't be behaving this way lol

Shing

5:07 AM

I am trying to estimate the time computer takes to do matrix calculation.

but I am a bit confused

it seems to me the computer is dumb enough to do all the multiplication: so if i do: identical matrix times another identical matrix, the time it takes should be same as I multiply two very different matrix

right now, I am working on writing code to verify my thought

but I will need a very big matrix

and I am not so sure how I should right the 1000x1000 identical matrix...

I am not so sure if the build-in identical matrix is suitable for my experiment...

danielunderwood

5:26 AM

Maybe you could factor and manipulate the factorization to reduce the number of things that need to be multiplied? I don't know that the factorization will cost fewer operations than it saves though. Unless you're doing it for some specific reason, the easier path would be to just toss it into matlab/numpy

user157860

5:37 AM

@JohnRennie,Hi John, I have been studying your tips :"So suppose the have the shell observer velocity equal to $0.99c$ at $r = 10^8$m,
use that to work out what $\gamma$ is, then you can use the equation to find how $v$ varies with $r$.

let's call $r_A = 10^{10}$m and $r_B = 10^8$m.
Then the question is if an observer hovering at $r_A$ observes a comet to whizz past at 0.99c what speed will that comet have at $r_B$.

So we'll take $v=0.99c$ and $r = 10^{10}$m and substitute then into the equation to calculate $\gamma^2$ and the answer is $\gamma^2 = 50.23634618$ , (the mass is $2 \times …

John Rennie

@user157860 I'm busy in the JEE chat room right now (discussing RL circuits). Back in a bit.

user157860

@JohnRennie, OK, take your time!, can you specify up to what distance we can use the regular sqrt(2 PE) to calculate the velocity if a body starts at zero at infinity?

John Rennie

@user157860 we need to back up a bit. Every observer observes a different velocity. The observer at infinity measures one velocity, the shell observer at the psition of the falling object measures a different velocity and shell observers at different distances measure different velocities again.

So you need to be more precise about what you mean by velocity.

user157860

Iam asking at what distance we can use Newtonian parameter, where an observer would measure v = sqrt(2PE), but I discovered that the Schwarttzchild metric is just another way of using PE. AmI right?

John Rennie

Which observer would measure v = sqrt(2PE)? The observer at infinity? The shell observer?

user157860

5:48 AM

the observer at 10^10m from the star, for example , can he use the 2PE formula, the same he does on earth? if not, what about the one af 10*12 and so on?

John Rennie

A shell observer at a distance $r$ sees the infalling object pass them at a speed given by:

$$ v = \sqrt{\frac{r_s}{r}}c \tag{2} $$

And $r_s = 2GM/c^2$ so we get:

$$ v = \sqrt{\frac{2GM}{r}} $$

So the for the shell observer the Netwtonian expression applies everywhere. But, and it's an important but, the variable $r$ is not the radial distance.

user157860

@JohnRennie, can you clarify :"...but, the variable $r$ is not the radial distance."

John Rennie

The Schwarzschild coordinate $r$ is the circumference of a circle centred on the black hole divided by $2\pi$.

In flat spacetime this is obviously just the radial distance because the circumference of a circle in flat spacetime is just $2\pi r$.

But for a black hole the distance you would measure by letting down a tape measure towards the black hole is not the same as the $r$ coordinate you get by dividing the circumference if a circle by $2\pi$.

user157860

@JohnRennie, so , an observer at 10^10m from the center of the mass is at what distance in your coordinate, and does it affect the result of the observed velocity?

John Rennie

When you say an observer at 10^10m from the center of the mass do you mean if that observer let down a tape measure to the centre of the mass they would measure a distance of 10^10m?

user157860

6:03 AM

@JohnRennie, the actual measure is impossible, but we mean radial distance. Isnt a neutron star or a black hole spherical?

John Rennie

In this sort of calculation the variable $r$ is generally the Schwarzschild $r$ coordinate i.e. if your observer were to draw a circle centred on the mass and passing through their position the circumference of that circle would be $2\pi r$.

All observers agree on this definition of $r$ so it's a useful definition.

The actual distance measured by a tape measure turns out to be a rather complicated expression.

The point of all this is that $r$ is a useful way of describing the distance from the mass, but it isn't the same as the distance measured using a tape measure.

That's why I say that you need to be cautious about the physical interpretation of results like $$v = \sqrt{\frac{2GM}{r}} $$ for the shell observer.

Superficially it looks as if simple Newtonian gravity works, but this is only a superficial similarity because $r$ is not the Newtonian $r$ variable.

user157860

@JohnRennie, the question is : is there a distance at which a sitting, hovering observer can take the result of v=sqrt(2PE) as exact? when does relativity come in?

Semiclassical

I imagine that a specific example of this would be two observers along the same line emanating from the center, but with different Schwarzschild coordinates $r_1,r_2$. The tape-measure distance between them is presumably not $|r_2-r_1|$.

John Rennie

@user157860 For every shell observer that equation is exact for the velocity of the object as it passes them provided $r$ means the Schwarzschild $r$ coordinate.

@Semiclassical correct

@Semiclassical it's a straightforward integral, but as is so often the case the result is rather messy.

Semiclassical

Makes sense.

user157860

6:19 AM

@JohnRennie, is that true for a distance =2*r_s?, I thought that somewhere the similarity breaks down

Semiclassical

looking at wikipedia, I think it'd be $w(r_2)-w(r_1)$ where $w(r) = 2\sqrt{r_s(r-r_s)}$ and $r_s$ is the Schwarszchild radius

John Rennie

1

A: length contraction in a gravitational field

Your question isn't well defined, because there is no way to compare the two measuring sticks. I can take two clocks, put them at different heights, then bring them back together and I will find that they now show different times due to time dilation. However if I take two rulers, put them diffe...

Semiclassical

in particular one has $w(2r_s) = 2r_s$

John Rennie

@user157860 the expression is true everywhere outside the event horizon i.e. for all $r > r_s$.

It isn't true at or below the event horizon because you can't have a shell observer at or below the event horizon.

Semiclassical

presumably the more generic point is that, for $r\approx 2r_s$, one has $w(r)\approx w(2r_s)+w'(2r_s)(r-2r_s)=2r_s+(1)(r-2r_s) = r$

user157860

6:24 AM

@JohnRennie, please say it explicitly : at a distance of 2r_s (6*10^6) an observer can use the formula sqrt(2PE)?

Semiclassical

So for Schwarzschild coordinates near $2r_s$, one can approximately equate them to the actual tape measure distances

John Rennie

@user157860 Definitive statement: at $2r_s$ the shell observer can use the expression $v = \sqrt{2GM/r}$

But $GM/r$ is not the potential energy.

user157860

@JohnRennie, and what is it?but the value is the same, uh? Idon't get that

John Rennie

Potential energy is not a useful concept in GR. The problem is that gravitational time dilation is ubiquitous in GR and since different observers measure time in different ways this messes up any attempts to define energy ina simple way.

There's nothing to stop you defining a quantity equal to GM/r and calling it PE if you want to, but you'll find this is not useful for doing calculations. For example the observer at infinity will find that trying to use this "PE" to calculate velocity doesn't work.

user157860

@JohnRennie, OK the explanation is complicated, but an observer at 2 r_s will measure v = .2c? right

John Rennie

6:37 AM

Isn't that $v = \sqrt{1/2}c$ ?

$0.707c$

user157860

@JohnRennie, yes,but a r_s the velocity jumps to c, is that right?

John Rennie

It doesn't jump to $c$ - it rises smoothly to $c$ as $r \to r_s$

user157860

@JohnRennie, and that, as you explained, because it is near the singularity, and that has infinite force, right?

John Rennie

For a shell observer at a distance $r$ you can calculate the acceleration they need to generate to stay stationary at the distance $r$. That acceleration goes to infinity as $r \to r_s$.

So in that sense the force goes to infinity at the event horizon.

Though I must emphasise that the calculation becomes physically meaningless at the horizon since it is impossible to hover at the horizon.

user157860

@JohnRennie, now, what I do not understand is this: the blackhole has 1000 solar masses, isn't the potential gravitational force necessarily the same no matter in how big/smallvolume you confine it. In a singularity it is confined in a small space, it is conjectured, but why should its g-force change?

John Rennie

6:46 AM

Consider a classical point mass M: the force (per unit mass) is GM/r^2

So the closer you get to the mass the greater the force - all very straightforward.

Now for a black hole we have been considering the force at some fixed multiple of $r_s$.

But $r_s$ is bigger for a large mass than for a small mass. In fact $r_s \propto M$ so for a 1000 solar mass black hole the event horizon distance is a thousand times larger than it is for the Sun.

So if you're calculating things at e.g. $r = 2r_s$ then this is a thousand times closer to the Sun than it is the 1000 solar mass black hole. Classically the 1/r^2 term would make the force 1000 times bigger for the Sun than for the 1000 mass object.

user157860

@JohnRennie, I don't think it works that way, we are considering PE GM/r, and then remember that even with a big body, it is calculated as if all the mass is at the center. so , as for the sun , it has GM=1.32*10*23 and that is the max value when distance is 1. You cant get a higher PE than that

John Rennie

I need to work now for about half an hour. Back later.

user157860

@JohnRennie, OK, I'll use that time to make a drawing!Thanks

Novalium Company

Anyone free to answer a quick Huygen's question?

As Huygen says, when we have a wavefront, infinite amount of wavelets spread out from it to make the second wavefront. When the second wavefront is made, will the wavelets keep growing? Is Huygens principle really what happens or just a neat way to visualize?

Nick

7:09 AM

@NovaliumCompany as a teenager, I could very much relate to a face growing a gajillion pimples that merge into another face that once again grows a gajillion pimples and thereby propogates (ie, moves in space)

also, here's a neat projectiles question i came here to share

14

Q: Is a network of spoon-launching trebuchets realistic?

On a comment to My carrier pigeons have been replaced by spoons: how can I send messages? I joke about creating a network of trebuchets to launch spoons. Is it realistic? My concerns: How far will it be launched? Would the spoon resist the impact? Is a trebuchet accurate enough? Is it affor...

Novalium Company

@Nick As a teenager, I can relate too. 😂 Great way of explaining!

You don't seem to have any pimples on your profile pic tho :D

John Rennie

7:30 AM

@Nick the problem is that spoons aren't very aerodynamic. Even if you launch the spoon at a high velocity the air resistance will rapidly bring it to a halt. I suppose you could melt down the spoons and make cannonballs ...

Secret

Actually, that's not magic

That's just correlations

It's just like, two bacteria came together to exchange their genetic material, and then swim away. Then of course the other bacteria will still be affected by the first bacteria, because it literally carries the genetic material of the first one

Likewise for people, when A and B talk to each other and than part ways, then B will still be affected by A in some way because B has memories of A.

Should you wipe out those synapse that stores that memory of A in B, I can dare you that B will no longer have traits related to A

As for entanglement and quantum discord, they are the only known thing that showed genuine nonlocal correlations, but the randomness of the measurement outcomes means you cannot send any messages with that.

bolbteppa

7:51 AM

One source I have has $i \{A,B \} \to [\hat{A},\hat{B}]$, another has $\frac{1}{i} \{A,B \} \to [\hat{A},\hat{B}]$ :\

Secret

well that's a difference of sign really

1/i = -i

I am used to seeing i{A,B} though

bolbteppa

If this really works in both cases, you heard it here first, 'east-coast quantization vs. west-coast quantization' ;)

user157860

8:30 AM

@JohnRennie, here is the drawing, ping me when you are ready and answer my previous query.Thanks

John Rennie

@user157860 yes?

user157860

@JohnRennie,I wrote: I don't think it works that way, we are considering PE GM/r, and then remember that even with a big body, it is calculated as if all the mass is at the center. so , as for the sun , it has GM=1.32*10*23 and that is the max value when distance is 1. You cant get a higher PE than that

John Rennie

Well consider Sun and a 1000 solar mass star.

And consider some distance $r$ from the Sun and $1000r$ from the big star.

The Newtonian PE has same value at $r$ from the Sun and $1000r$ from the big star, so an object falling from rest at infinity has the same speed $r$ from the Sun and $1000r$ from the big star.

If we define the event horizon distance from the Sun as $R$, then the event horizon distance from the big star will be $1000R$ because the event horizon radius is proportional to the mass.

So it seems entirely reasonable that an object falling into a solar mass black hole and a 1000 solar mass black hole will have the same velocity at the event horizon.

user157860

@JohnRennie, i follow..

but I made nother point

John Rennie

Yes?

user157860

8:42 AM

that when we give GM that is already the PE at the center of the mass, you use it up in one gulp if you are at 1 cm or so radius. If you conjecture a singularity ( and that conjecture cant be verified nor falsified) you are simply supposing that it is physically possible to stuff 10^63 electrons into 1 cm.

But whether it is really possible or not, you can never get more than GM from that mass.

John Rennie

Huh?

user157860

yeah

what's wrong?

John Rennie

The PE at the surface of the mass is GM/r. As you shrink the mass down $r \to 0$ so the PE goes to infinity.

So I don't understand what you mean by you can never get more than GM from that mass

user157860

That's your mistake, you can't to zero, if you did you'd be left with zero. Playing with zero and infinity is dangerous

GM of the sun is 1.327*10 ^20, that value is the PE energy of 2*10^60 electrons at the distance of 1, since the mass is already considered at the center of the sun

John Rennie

@user157860 it's true you can't divide by zero, but you can approach arbitrarily close to zero, and as you do so the potential energy gets arbitrarily large.

I don't see your objection to that?

user157860

8:49 AM

that is sophistry,

John Rennie

No it isn't.

user157860

you cant play with hard matter, mass.

it is not math

epsilon doesnt exist in reality,

anyway, if you could, mass is considered at the center of the Sun, whatever that physically mens, if it's less than 1, OK

John Rennie

Matter is prevented from being compressed by exchange forces. Inside the event horizon the gravitational force inwards (speaking loosely) is greater than the exchange forces that prevent the matter from being compressed. So unless some new interaction crops up the matter will be squeezed to a point of infinite density.

user157860

@JohnRennie, OK john, that's GR and many people believe it's true, I thought there was an argument I ignored.

John Rennie

I, like all physicists I know, believe that there will be a new interaction due to quantum gravity effects so we we won't get an infinite density. However since we have no theory of quatum gravity no-one knows what actually happens.

user157860

8:54 AM

@JohnRennie, please consider now what is my real big concern, since involves not only GR but also Newton. Consider the picture above: two electrons/neutrinos are at .9c distance from the earth. A is still and B is travelling at.9c.

John Rennie

OK ...

user157860

A will gain delta PE (6.25*10^11-1.447*10^10) and reach v = 10^6 cm/s

it will take less than 3 days to reach the ground (can you tell me how to find the exact value)?

But for B even Newton does not apply, B can never gain same PE energy, so both in this case and in the case we examined lately it all goes wrong. right?

John Rennie

Calculating the time of fall is surprisingly messy because it requires an integral that doesn't have a nice simple form.

user157860

but a rough calc is good enough

the point is that if we add 6*10^11 Pe to B we are making a gross mistake

John Rennie

@user157860 if we ignore GR effects, which is a good approximation for the Earth's gravity, then we can do this easily. B does gain the same PE as A. You're presumably thinking it can't because it can't go faster than light, but the energy of a relativistic is not given by $\tfrac{1}{2}mv^2$.

The total energy of a relativistic object is:

$$ E^2 =p^2c^2 + m^2c^4 $$

and you subtract $mc^2$ off this to get the kinetic energy.

In this equation $p$ is the relativistic mometum $p = \gamma m v$

So B does gain the same PE as A but its velocity increases only by a tiny amount because of that factor of $\gamma$ in the equation for the energy.

user157860

9:04 AM

No, "...You're presumably thinking it can't because it can't go faster than light" I am thinking that it takes 3 days to absorb that much PE, B is exposed only for 3 seconds to gravity, so itcan absorb a lot less

John Rennie

Huh?

The PE change isn't time dependent ...

It doesn't matter how fast you go from the initial position to the surface. The PE change is always the same.

user157860

PE is GM/r because it's the integral of a =GM/r^2. What you really absorb is a and if you are exposed for 3 days it adds up to 10^11. if you are exposed for 3 second to gravity( considering an average of 50 cm/s^2, just to simplify calcs) all you get an acceleration of 50 cm/ s^2.

Just think about it

John Rennie

I am thinking about it :-)

PE is the work done by the gravitational force and it's equal to $\int F dr$ i.e. you integrate the force with respect to distance not time.

user157860

but even if we consider the max value of 980 as an average all acceleration you get is 30m/s^2

John Rennie

Time is irrelevant in this context.

user157860

9:12 AM

no, distance is covered in time

consider the case that g is constant,980 cm, how can you gain 10^11 PE in 3 seconds

John Rennie

Unless we've managed a fairly comprehensive miscommunication here I have to question your grasp of basic physics, in which case your chances of understanding GR are a little slim.

2

The integral of force wrt time gives you the impulse change not the energy change.

ACuriousMind

@user157860 It seems you somehow believe that the objects can only gain gravitational potential energy from falling, not because of prior motion. That's simply not true. If a thing moves lower/higher in a gravitational field it gains/loses the potential energy associated with that motion, regardless of whether the motion was due to the gravitational acceleration itself or not.

user157860

@ACuriousMind, I am asking if a mass exposed for 1 second, 3 seconds, three days to the same g force can acquire same KE, (i am leaving aside acceleration and PE)

ACuriousMind

Also, morning @JohnRennie

John Rennie

@ACuriousMind morning :-)

How's the ankle?

ACuriousMind

9:20 AM

@user157860 I'm afraid I don't understand what that has to do with anything else. You were talking about potential energy all the time, not kinetic energy.

@JohnRennie Still broken :P

Thankfully it doesn't hurt much

John Rennie

I've never broken an ankle though I did once sprain my ankle falling down a cliff in a cave !!!

I haven't been caving since - that kind of eliminated my interest in the sport.

ACuriousMind

Heh, I can understand that ;)

user157860

@ACuriousMind, that makes thing more complicated , let's talk of plain KE, isn't time of exposure relevant, vital?

ACuriousMind

@user157860 Yes, of course the kinetic energy gained through acceleration is dependent on the time of exposure.

John Rennie

@user157860 the problem with your question is that the three objects will fall different distances in the different times.

And since work is $\int F.dr$ the PE change of the three will be different.

user157860

9:22 AM

in the picture there are 2 bodies, A and B. Can B get same KE as A?

John Rennie

But in this case the distance is the distance to the surface of the Earth so the two particles fall the same distance.

ACuriousMind

@JohnRennie Apparently it's usually really painful, but I managed a very neat break that - with the cast on - is just annoying :P

John Rennie

@user157860 the KE change of B will be the same as the KE change of A.

user157860

@JohnRennie, yes the increase is the same

John Rennie

Since they start with different KEs they end with different KEs.

@ACuriousMind just tedious then i.e. stopping you from doing stuff.

user157860

9:25 AM

@JohnRennie, yes, the increase of KE is the same even if one has been exposed some 300000 times less than the other? Is this the conclusion? @ACuriousMind

John Rennie

Correct. As long as they fall the same distance the KE change is the same.

ACuriousMind

@user157860 Observe that kinetic energy is not linear in velocity, but quadratic, but that the increase in velocity is linear in time. So if velocity is already large, you need much less time to effect the same total increase in kinetic energy

@JohnRennie Yeah, the most annoying thing is that I don't have a free hand while using crutches

John Rennie

@ACuriousMind I was on crutches for a bit (after an operation on the ankle that got damaged when I sprained it) and the thing I remember most is the difficulty of pushing supermarket trollies.

You had to swing along on the crutches and keep butting the trolley with your belly. Actually quite entertaining but hard to steer.

ACuriousMind

Hm, I don't think I'll try shopping, my flatmate is nice and buys stuff for me too :P

John Rennie

Aha, the Parker Solar Observatory has been successfully launched. Hot stuff! :-)

Secret

9:33 AM

in Mathematics, 4 mins ago, by Secret

What happens in the joke is basically you have a map that maps an element generated by induction, and is invertible to an absorber. Thus in order for such rule be compatible, some symmetry breaking is needed to accommodate this discrepancy. In our case, associativity breaks

not sure if I have used "symmetry breaking" correctly

Anonymous

@JohnRennie Ah, good news :D I remember tons of people submitting their names

Anonymous

(Though I still don't understand their agenda behind that :P)

ACuriousMind

Submitting their names for what?

Anonymous

@ACuriousMind nasa.gov/feature/goddard/2018/…

John Rennie

I suppose it's a way of encouraging public interest

ACuriousMind

9:39 AM

Ah.

John Rennie

You can imagine it going down well in schools.

@Blue it still has a looooooooong way to go though!

Anonymous

@JohnRennie Seems legit :D (I thought they might sell the data)

Anonymous

@JohnRennie Ah, true

Anonymous

Parker Solar Probe

Parker Solar Probe (previously Solar Probe, Solar Probe Plus, or Solar Probe+, abbreviated PSP) is a NASA robotic spacecraft en route to probe the outer corona of the Sun. It will approach to within 8.86 solar radii (6.2 million kilometers or 3.85 million miles) from the "surface" (photosphere) of the Sun and will travel, at closest approach, as fast as 700,000 km/h (430,000 mph).The project was announced in the fiscal 2009 budget year. Johns Hopkins University Applied Physics Laboratory designed and built the spacecraft, which was originally scheduled to launch in 2015. A successful launch occurred...

John Rennie

Anonymous

9:43 AM

"The spacecraft trajectory will include seven Venus flybys over nearly seven years to gradually shrink its elliptical orbit around the Sun, for a total of 24 orbits."

Anonymous

whew

John Rennie

It's amazing really. It's got so that we take this stuff for granted, but it's a fantastic technical feat.

Anonymous

@JohnRennie Yes, no doubt. It really is an engineering marvel :) We've come a long way since the 1960's

Anonymous

@JohnRennie Where did you find that one?

John Rennie

@Blue damn, I've lost the link ...

Aha, here

Anonymous

9:48 AM

@JohnRennie Thanks!

Anonymous

Oh, the "Planned Timeline" section on Wikipedia is pretty comprehensive

Emilio Pisanty

10:31 AM

@JohnRennie more importantly: when does the science start coming out?

John Rennie

@EmilioPisanty from November I guess, though to actually get into the solar atmosphere will take six years.

Shedding that much gravitational potential energy isn't easy :-)

Emilio Pisanty

that whole "it's easier to escape the solar system than it is to get to the Sun"

hmmmm. On second thought, I'm not sure grayscale English gentlemen of a certain age would recognize the hard-faced, scarred, indeed-ing young man of that picture.

John Rennie

Google, Google - Omar from The Wire?

Emilio Pisanty

@JohnRennie indeed ;-)

did you just google "oh indeed"?

John Rennie

oh indeed meme

After much tutoring by Balarka I know understand what memes are :-)

Emilio Pisanty

10:41 AM

@JohnRennie excellent

have you watched The Wire?

Anonymous

@EmilioPisanty Google's got some crappy image recognition which works fairly well on that image

John Rennie

@EmilioPisanty I think I've heard of The Wire but I know nothing about it.

Emilio Pisanty

@JohnRennie tsssk tsssk

watch it

seriously

you'll thank yourself by the time you're at episode two

John Rennie

That isn't an obscure form of age based snobbery it's just that I've never been keen on visual arts. I generally only watch films when I'm dragged along to one. My primary form of entertainment is reading. Though sadly reading SF rather than anything educationally enhancing.

I don't even have a TV licence.

Anonymous

Is TV license a British thing?

Anonymous

10:44 AM

Never heard of that term before

ACuriousMind

@JohnRennie Psssst. Nobody watches TV these days :P

Emilio Pisanty

@JohnRennie ¯\ _(ツ)_/¯

Anonymous

@ACuriousMind Nah, most watch TV shows on the internet instead :P

Anonymous

(At least the "under 35" age group)

ACuriousMind

@Blue Yes, that's what I meant.

John Rennie

10:45 AM

@ACuriousMind presumably people do watch televisions. They just don't watch broadcast TV but instead watch Netflix, etc.

Emilio Pisanty

@Blue yep. It allows the UK to fund the BBC *gasp* without much advertising.

speaking of Google, though

it seems you lot have been denied the joy of today's Doodle

I feel sincerely sorry for y'all

John Rennie

@EmilioPisanty to be fair the TV licence has allowed the BBC to build up a well deserved reputation for quality broadcasting. I don't mean to criticise the requirement for a licence.

ACuriousMind

@JohnRennie Sadly, state funded TV is not a guarantee for quality ::glances at German state TV::

Anonymous

@ACuriousMind You know, I sometimes think the whole TV broadcasting thing can be replaced with "updating of TV shows on a weekly basis", so that people can watch their TV shows at their convenience, without having to spend time trying to stream online on their small PC screens

Anonymous

Netflix n all is good, but I'd rather watch my TV in the dining room on the larger screen

John Rennie

10:50 AM

@ACuriousMind That seems a shame. I wonder why that is. Maybe Germans generally attach less importance to TV, in which case good for them! :-)

ACuriousMind

@Blue Smart TVs - i.e. those with access to the internet - are fairly common here, so what you want is already possible.

Anonymous

@ACuriousMind Smart TVs stream over the internet...

Anonymous

And also consumes a hell lotta data

ACuriousMind

@Blue So? What does the mode of data transportation matter to you?

Anonymous

If you have a large screen and want good quality images

John Rennie

10:51 AM

@Blue in Europe that's rarely an issue.

My data allowance is something silly - 100s of GB.

ACuriousMind

@Blue Uh...no one pays for data anymore here, unless you're consuming multiple TB a month

John Rennie

I forget the exact figure but you'd have to be a real couch potato to use it all up.

ACuriousMind

@JohnRennie I think we have lower standards for TV because we all grow up with terribly dubbed TV :P

John Rennie

(That's with a fibre connection not mobile)

Emilio Pisanty

@ACuriousMind shhhhhh, there are countries where 'reasonable' is still not a thing in internet connectivity.

Anonymous

10:54 AM

@ACuriousMind "unless you're streaming multiple TBs a month"....how do they even afford to give so much data allowance for free? O_o

Emilio Pisanty

@Blue it's not 'for free'. you're paying for it.

Anonymous

"no one pays for data anymore here"

ACuriousMind

@Blue Well, you do pay a base price for the connection (we're talking a physical, wired connection here, not a mobile one).

@Blue I meant you do not pay for the amount of data. Of course you do pay for the basic fact that you have an internet connection at all

Emilio Pisanty

@ACuriousMind and presumably the price increases with the speed of the connection

John Rennie

I pay £22.99 a month, and that gives me effectively unlimited data. There is a limit but it's so high I never reach it

That's for a fibre connection with 80MHz download speed.

Emilio Pisanty

10:58 AM

@JohnRennie MHz?

John Rennie

So I get download speeds of around ten megabytes per second.

@EmilioPisanty megabits per second.

ACuriousMind

@EmilioPisanty Used to be the case, but these days it's my impression your location (whether new high-speed cables are available or not) is the actual limiting factor.

@EmilioPisanty Since bits are dimensionless, he's technically correct ;)

Anonymous

@ACuriousMind Ah, even that's pretty cool. We have to pay for every 100GB here

Emilio Pisanty

@ACuriousMind ¯\ _(ツ)_/¯

Anonymous

And a 2 hour HD movies for the large screen is about 10-15 GB in size

ACuriousMind

11:00 AM

I think the last time I had an internet connection with a data limit one could reasonably reach was some time in the mid 00s

John Rennie

@Blue hmm, that's a low compression. I generally find 720p looks fine at a compression of around 1GB per hour.

For 1080p 2GB per hour.

Anonymous

@JohnRennie 720 wouldn't work on a 65 inch screen :/ The image "cracks"

John Rennie

Well even 1080p is only 2GB per hour.

4GB per film.

Anonymous

I think the TV app only allows 4k downloads

Anonymous

I don't use it much though

Anonymous

11:03 AM

Maybe I can run 1080 there

John Rennie

OK, OK, if you insist on 4k the data rates are going to be much higher :-)

ACuriousMind

You also have to account for streaming not allowing for optimal compression - you can't wait till the end of the download in that case, so you can't take full advantage of compression.

Anonymous

@JohnRennie Heh, yeah. And it's much easier to stream them over a TV cable rather the broadband, at least for now. But these days those high speed net connections are coming up (fibre optic n all) and those might be a lot faster albeit much more expensive

Anonymous

So until the prices for net connection goes down, it wouldn't be very practical to stream 4k movies on the NetFlix (or equivalent) TV app

Anonymous

@ACuriousMind Yup, exactly!

Anonymous

11:12 AM

"streaming not allowing for optimal compression"

Anonymous

Also cable TV has a lot more bandwidth allocated for each channel

Emilio Pisanty

11:46 AM

great, SE has started messing with the CSS

that looks ugly as hell

but maybe it means that the switch to the new site-design scheme is imminent?

John Rennie

Speaking as a physics nerd my view is ... meh

Emilio Pisanty

@JohnRennie it did change, though, right?

mostly I care because of this one

31

Q: Ch-ch-ch-changing the site logo to cut out the weird arrow thing

Apparently, Ch-ch-ch-changes are coming to the site navigation and theming. Given that that's the case (and that therefore there are designer work-hours already scheduled to attend to our site design), can I suggest that maybe now is a good time to drop the super-awkward-once-you-notice-it (exce...

discussion feature-request design

hmmmm, or maybe it didn't change

and all that happened is that knzhou edit-bumped a bunch of high-viewcount threads?

David Z

Seems that way.

Secret

12:10 PM

A glimpse of the new world

knzhou

Dang, I was trying to find a quiet time to rapid-fire edits.

Over ~10k views has always been in orange!

user54826

12:27 PM

any exclusive Linux users in here?

ACuriousMind

@user54826 I use it, but not exclusively. Why?

David Z

@user54826 me, almost

user54826

just curious. I do feel a bit alone then! I'm using Linux exclusively since a few years now

David Z

@knzhou You probably picked the best time, but even so there's always some activity

@user54826 99% of my computer time is using Linux, I'm not sure if that comes close enough for you

user54826

it's enough

David Z

12:32 PM

The only time I don't use it is when work needs call for a laptop, and my work laptop is a Mac

user54826

I see

David Z

oh and I guess Windows for some games, but it's probably been months since I booted up Windows

@EmilioPisanty There will be a meta post (on our meta) when the switch is imminent for us

Anyway I have to be off now, I'll be back later

user54826

here's a screenshot of my cpu screen imgur.com/a/ikF0kOh

user54826

12:46 PM

rainbow cpu colored info displayed in top screen starting with entropy of /dev/random i mean what else any physicist could dream of? :D

Nick

@NovaliumCompany I'm 21 and that was a good day :)

Emilio Pisanty

@DavidZ good to know.

though I imagine we'll be the last to get re-designs, as usual =P

Nick

@user54826 firstly, no one calls it a "cpu screen" and secondly, what is that terminal?

user54826

xrxvt is the name of the terminal (it's written on the top of the window)

Nick

@Secret I am in the mod elections for MSE. Crazy, right.

@user54826 this?

user54826

12:50 PM

urxvt*

Nick

URXVT is the Best Terminal. (And some configuration notes)

►

user54826

not really, your link is just a dot file associated to that terminal. it's extremely customizable @Nick

I do have my own dot file

Nick

I still didn't find the installer for Win10x64

user54826

I don't think it exist for Windows but I'm not 100% sure

For laptops I don't really understand why don't they use the battery(ies) of the strong core i7 laptops to power the celeron or pentium. that would make enormous battery life laptops like 15hours or more

coz currently from what I read, most batteries rarely exceed 8 hours which is just a working day. most would die before that, if not plugged.

Nick

i have 2.5hrs unplugged

i3 lenovo ideapad 100

user54826

1:04 PM

nice laptop. but sad battery life

do you have an idea on which laptop to choose if you upgrade?

i think I'll wait for 10 nm processors

Anonymous

@user54826 Better than a battery life of 0 hrs (my situation ~3 months back) :P

user54826

should eat less power for superior speeds

wow @Blue really? and now what is your battery life?

Anonymous

@user54826 Yeah, the laptop battery had stopped charging (probably due to age). I bought a new one and it lasts for ~1.5 hrs when fully charged

Anonymous

My laptop is pretty old (~5 years), so....it's not very efficient any more

Anonymous

in the battery domain

user54826

1:07 PM

I see

mine lasts for maybe 2 hours, but gets discharged by 10 to 15% overnight, even when powered off

but is has always been like that, even when I bought it

Anonymous

But after installing a SSD the performance has become a lot smoother

Anonymous

So that's one good thing

user54826

I thought about repllacing the slow 5400 rpm HD by an SSD, but it's an old celeron so it might not change its speed too much anyway. Plus, it's a nightmare to maintain, because it was made for Windows, not Linux

I will just wait for 10nm laptops. I want a single SSD with 120 GB. I dont care about RAM and anythong of the sort. Only want a 12 to 15 hours battery. So even after 2 or 3 years I can expect 9 hours, i.e. a working day

Anonymous

What was "made for Windows"?

Anonymous

Your PC?

Anonymous

1:11 PM

It shouldn't matter which OS you use

user54826

my laptop

it does. it's an HP. they messed up the BIOS and made it compatible with windows only. when it boots up, it will look for a specific file on the HD. if you erase that file, it cannot boot up at all

and it cannot even boot up Grub or anything Linux related

it can only boot windows.... unless you use the hack I used. I replaced that specific file for a Linux Grub file. so when it boots, it believes it's booting Windows while it's booting a Grub-like program

ACuriousMind

@Blue While the situation is much better than it was years ago, finding working drivers for specific pieces of hardware can be a nightmare under Linux

user54826

took me over a full week of googling and seeking help on IRC to install Linux on that machine. It's been like 4 years now, and some people still remember me from that

they suggested me to not bother with an SSD, since the laptop is made for Windows, it might be a trouble to deal with the installation if I plan to use only Linux (as I do)

I've read there are a few other laptops, aside HP (like Sony's) that are made to boot up Windows only

Anonymous

@ACuriousMind That's true. Luckily I didn't have to worry about it since the install version was completely compatible with my system. I remember some people having troubles with activating their WiFi card after installing Ubuntu

Anonymous

and some had problem with their speakers

user54826

1:19 PM

in the past, I think printers were a nightmare. Now, I think they are ok

I bought one recently (HP), and it seems to fully work

Physics Meta

1:40 PM

0

Q: Why are answers by high rep people not downvoted?

Recently I noticed something. When those with high reputation post answers, they are often upvoted. On the other hand, if a low-reputation guy posts an answer, it is immediately downvoted and voted for closure. Check this one out: Why must the converging lens have a shorter focal length than t...

feature-request down-votes reputation voting up-votes

Emilio Pisanty

1:54 PM

@PhysicsMeta $\displaystyle\huge 🙄$

Abcd

Can someone please help me spot the 4th nearest neighbour of Body centred cubic unit cell using the atom at the corner?

Anonymous

@PhysicsMeta While it might be simply a case of confirmation bias in this instance, rep points do often correlate with quality or at least quantity of answers. It might have been an interesting discussion had the OP provided some solid examples.

Anonymous

90

Q: The problem with reputation: does high reputation attract too many upvotes?

It's not that I feel bitter when somebody else's answer is upvoted above my own, but it's when a user with a high rep answers a question and their answer seems to attract an immediate barrage of upvotes. Now I'm not saying that these votes are always unwarranted, but by simply observing this beh...

discussion reputation voting high-rep-users

Abcd

The answer given is" go to a distance of $\frac 32 a + \frac a{\sqrt 2}$ and spot the 4th nearest neigbour" ...But I am unable to identify where to go using this.

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