The h Bar

Emilio Pisanty

12:42 AM

@DanielSank what?

what do you mean by "mean"?

basically

remember this one?

16

Q: Photons with half integer angular momentum - what's happening?

I have just read this article - what is happening? Analysing these beams within the theory of quantum mechanics they predicted that the angular momentum of the photon would be half-integer, and devised an experiment to test their prediction. Using a specially constructed device they were able...

and how everybody was like, "yeah, there's this interesting new type of angular momentum $J_\gamma$, though it only makes sense if $\gamma\in\frac12 \mathbb Z$"

well

about that last bit

I beg to differ.

danielunderwood

I know it's not the case, but it sounds like you wrote the paper to prove people on the internet wrong :D

Emilio Pisanty

well

that's not entirely inaccurate

danielunderwood

Or maybe we've reached the point where scientific discoveries come from someone on the internet saying something wrong lol

"You're wrong, just give me a couple years to prove it"

Emilio Pisanty

Though really it's more symptomatic to the broader fact that the singular optics literature's treatment of polarization orientation singularities tends to be somewhat hamstrung by an undue over-emphasis on monochromatic fields.

I want to break them free from that cage.

@danielunderwood well, this could've been published in 2017, but we decided to wait for the experiment. It makes the paper a whole heck of a lot nicer.

danielunderwood

Is that something that would normally be done? I was under the impression that the thing to do was to publish when it's ready then publish again after the experiment. I suppose that's why arxiv gets so many submissions though

Although I suppose a lot of those aren't going to be supported/denied by experiment anytime soon

Emilio Pisanty

12:57 AM

@danielunderwood it depends

in this case we're after a glossy-cover outlet so that wouldn't have been a great idea

DanielSank

1:32 AM

@EmilioPisanty No I do not :-P

Emilio Pisanty

@DanielSank fair enough

vzn

2:14 AM

@DanielSank mean returning random walk eh? sounds like this en.wikipedia.org/wiki/Hurst_exponent

Akash. B

5:43 AM

@Secret what is that?

Secret

7:07 AM

A slide that showed how to take over The World

This is because the Eternals will Return and the Illuminatibwilk rise again

user157860

7:31 AM

@JohnRennie, Hi John, can I ask you something re your [answer:](https://physics.stackexchange.com/questions/419651/if-gravity-reaches-infinite-intensity-on-the-all-event-horizon-surface-doesnt/422828#422828) you say:"..**this is due to the time dilation experienced by an observer hovering close to the event horizon. If we label the time measured by the hovering observer by $\tau$, while the time we measure far from the black hole is $t$, then the time dilation is given by**:

$$ \frac{\tau}{t} = \sqrt{1 - \frac{r_s}{r}} \tag{4} $$

John Rennie

The only acceleration that has a coordinate independent meaning is the proper acceleration. For a shell observer this is equal to the acceleration experienced by that observer, and for the shell observer this does to infinity at the horizon.

For a freely falling observer the proper acceleration is zero, and in fact that's why freely falling observers are weightless.

So the two observers are completely different.

user157860

8:08 AM

@JohnRennie, what do you call the acceleration g (980 cm/s^2) a frefalling body gets near the surface of the earth? if time where dilated by1/2, wouldn't g (acceleration of the body) decrease by 1/4? if so, why wouldn't that compensate the other factors?

John Rennie

That's not a well defined question.

user157860

sorry about that

John Rennie

You can calculate the coordinate acceleration in the coordinates of the distant observer or a shell observer (the two will be different). You do that using the geodesic equation. Things like time dilation and length contraction are automatically included in the calculation.

The calculation is easy in shell coordinates, but I'm not sure there is a closed form solution for the distant observer.

Captain Bohemian

if a university type as shown in Wikipedia is independent/free catholic university, doesn't that mean that university is a private university which doesn't get funds from the government?

user157860

@JohnRennie, why must we always refer to an observer? if an asteroid falls on the earth its acceleration is independent of the fact whether any observer is there?

John Rennie

8:15 AM

In GR when we talk about observers what we really mean is a coordinate system.

user157860

Can we camculate what is g at 100 m from r_s?

we might consider a black hole with 1000 solar masses

John Rennie

In what coordinate system? The coordinate system corresponding to a shell observer, the coordinate system of the distant observer or the coordinate system of the freely falling body?

user157860

and r_s = 2967*10^6

the falling body

Captain Bohemian

@JohnRennie it's like even outside GR, an observer in physics usually refers to a coordinate system, too.

user157860

@JohnRennie the falling body

John Rennie

8:22 AM

In the coordinates of the falling body the acceleration is zero because in those coordinates the body is at rest at the origin.

user157860

@JohnRennie, doesn't it feel any a? So why does it accelerate?

what's the difference with an asteroid approaching the earth or the sun?

John Rennie

So why does it accelerate - this is the point. Acceleration is $d^2x/dt^2$ for some $x$ and $t$, but every observer has different definitions of $x$ and $t$ so they all disagree about the acceleration.

user157860

But an observer sitting on the asteroid does not feel the acceleration, but he can see the earth approaching faster and can calc the real acceleration, right?

John Rennie

If you're asking about the acceleration of an asteroid falling to Earth you are implicitly asking what is the acceleration of the asteroid in the coordinates of the shell observer at rest on the surface of the Earth.

the body does accelerate only in one way - the only coordinate invariant acceleration is the proper acceleration.

But this isn't what you intuitively think of as acceleration because intuitively we tend to think of coordinate acceleration i.e. $d^2x/dt^2$ in our coordinate system.

user157860

@JohnRennie, an observer sitting on the falling asteroid cannot feel the acceleration, but sees the earth getting bigger and surely can calculate the real acceleration, so, why is the observer on the earth priviledged?

John Rennie

8:30 AM

We're going round in circles. The observer on Earth is measuring the coordinate acceleration not the proper acceleration.

The proper acceleration is the same for all observers and is zero, but the observer on Earth isn't measuring the proper acceleration. They are measuring the coordinate acceleration. They would have to calculate the proper acceleration from their measurements and they too would calculate it to be zero.

user157860

@JohnRennie, sorry what do you call the acceleration of 9.8 m/s^2 near the earth?

John Rennie

That's the coordinate acceleration.

user157860

can any other observer find the same value?

can we call it the real acceleration or not?

what am I missing?

John Rennie

No, other observers will measure different values for the acceleration at the surface of the Earth i.e. their measurement wouldn't be 9.8 m/s^2.

user157860

why an observer onthe asteroid with good instruments and computers can't find it?

John Rennie

8:38 AM

Because coordinate acceleration is different for different observers, and that's because they are using different coordinate systems i.e. the times and distances they measure differ from measurements made by the other observers.

user157860

@JohnRennie ,what I do not understand is this: whatever observers observe, the asteroid is getting 9.8 a, isn't that so inGR?

John Rennie

No

The figure of 9.8 applies only to a single coordinate system.

user157860

@JohnRennie you mean even outside GR?

can't an observer on the moon find out that value?

I though cosmologist calculate speed and a of distant bodies?

John Rennie

@user157860 yes they do. But they first choose a coordinate system then do the calculations using that coordinate system. We generally use the average rest frame of the universe i.e. the frame in which the cosmic microwave background is isotropic.

The Earth isn't actually at rest in this frame, but its speed is slow enough to be approximately at rest.

So in effect we use the rest frame of the Solar System.

user157860

9:00 AM

@JohnRennie, so any observer who is enough smart can find the 9,8 value? isn't it objective (relative to the earh in anappropriate rest frame)? if the asteroid is a approaching a black hole, can't any smart astronomer find its acceleration?

John Rennie

@user157860 Suppose I stand far from a black hole and drop in a stone then measure the acceleration. What I will measure is that the acceleration rises at first as Newton would expect, but then the acceleration peaks and starts to fall again. In fact the acceleration becomes negative i.e. the stone actually slows down as it approaches the event horizon.

I have a army of friends hovering at various distances from the horizon, and I ask them to measure the acceleration of the stone as it passes them. My friends report that acceleration rises smoothly as the stone goes in and goes to infinity as the stone approaches the horizon.

So my friends and I measure two completely different things. Who is right?

We are all right. What we've measured is the coordinate acceleration, and that is different for every observer. So the acceleration near the horizon is both small and negative and large and positive depending on what observer you ask.

If you ask "yes but what is the real acceleration" the answer is that there is no real acceleration. That is a meaningless concept in GR.

There is a proper acceleration, but that isn't directly observable - it is a quantity that has to be calculated from our observations. And the proper acceleration of the stone is zero, and when all observers do the calcualtion they will all get the same result zero.

user157860

9:27 AM

@JohnRennie, thanks, but what will a distant astronomer measure? can we say that the rest frame of the universe is absolute?

Physics Meta

0

Q: Is it OK to change commas to narrow no-break spaces in numbers when editing a post?

As a non-native English speaker, I find it utterly confusing to read text containing figures like "1,234 m", which in English means 1234 m while for a person used to read e.g. French or German (person's native language) it'll mean 1234 mm at first, only to be re-read to reinterpret correctly. Th...

discussion editing formatting

John Rennie

@user157860 there are no absolute frames, but the frame in which the CMB is isotropic is certainly convenient since all observers in the universe can measure the CMB and determine their velocity relative to it.

However this frame is more complicated that you think.

For example we know distant galaxies are moving away from us in accordance with Hubble's law. However those galaxies are also (approximately) at rest relative to the CMB. In the coordinates you would naively use for measuring the universe the CMB is not homogeneous. The CMB actually defines a coordinate system called comoving coordinates.

user157860

@JohnRennie, thanks, John, you are extremely kind and clear. One last query, if I may: in our original example (r_s= 3*10^6, A 10^10, B 10^8) we calculated V at B knowing v at A, using the Schartzchild metric, if I'd calculate it adding delta PE (a- B) to the KE at A, should I get the same rsult or not?

John Rennie

No, because you can't do that calculation. PE isn't a well defined quantity.

user157860

@JohnRennie, (I meant GM/r_B - GM/r_A),..... so the result is different. Thanks, again

danielunderwood

3:23 PM

happy Friday dudes

Slereah

3:33 PM

Transcript for