The h Bar

John Rennie

04:46

@user76284 the CMB photons have a black body spectrum i.e. a Planck's law distribution of energies. The deep reason for this is that they were emitted from a plasma that was a good approximation to a black body.

The distribution of their momenta is isotropic and homogeneous because the medium they were emitted from was isotropic and homogenous.

Shuheng Zheng

05:46

Hi John Rennie. Do you mind taking a look at a question special relativity?

1

Q: Relative simultaneity and time going backwards as someone accelerates

I have an observer on Earth with an atomic clock, let's call this the unprimed frame with space coordinate x=0 and t. Universe is one dimensional. I have a rocket ship sitting stationary in empty space to the far "left" of the earth on the x-axis. Because it is stationary wrt the Earth, rocket...

special-relativity

John Rennie

The point Alfred Centauri makes is that "sees" and "observes" are different.

Shuheng Zheng

in what way?

John Rennie

To "see" something the light from that even has to cross the distance between you and the event then enter your eye. So any coordinate changes have to be combined with the travel time of the light.

Shuheng Zheng

Ah, I meant observe then.

John Rennie

But when we say "observe" we normally just mean "assign to some coordinate"

You're correct that as the moving observer accelerates spacetime points will move to an earlier coordinate time but this isn't things moving backwards in time.

It's still the same point in spacetime that it was before. You're just describing it in a different way i.e. the split between the spatial and time components changes.

When we talk about something "going back in time" we normally mean a process that results in a closed timelike curve, and that isn't happening in this case.

Shuheng Zheng

05:54

Hm.. So it is possible that before accelerating, some event's coordinate time is 0 but after accelerating the coordinate time becomes -5 or +5?

John Rennie

Yes. One of the things that's hard in relativity is that coordinates do not necessarily have any absolute physical meaning. They are just labels that we attach to points in spacetime to identify them.

Shuheng Zheng

In the case of 0 -> -5, would I observe events at that location rapidly speeding up? (Catching up as my plane of simultaneity swings)

John Rennie

@ShuhengZheng I think you're confusing "observe" and "see" again.

When you change coordinate systems (by accelerating) the time coordinate of events changes as you describe, and this will mean the time difference between pairs of events changes. But the proper time between those events is of course unchanged.

Shuheng Zheng

Yes, but as the moving observer accelerate and his frame undergoes Lorentz boosting, his proper time t' is "ticking forward", t' is also the coordinate time of a frame co-located with the moving observer. Now some events before acceleration would have t'=6 for example, but now would have t'=-5. So that means at that location this observer is rapidly seeing "history" going forward?

John Rennie

As the observer accelerates the time coordinates change, so in this sense yes events in the original frame move forward or backward in time.

But there is no physical significance to this. The moving observer is just labelling thise events in a different way. The order of events remains fixed i.e. there is no loss of causality and the proper time intervals between events is unchanged.

So you can say that history is indeed going forward, but so what?

Shuheng Zheng

06:11

Let me propose a half twin paradox. A rocket ship accelerates away and stops at Alpha Centauri. We record the whole process. We reconstruct a video depicting the rocket ship proper time with Earth's coordinate time.

John Rennie

You mean you are pointing a video camera on the rocket at the Earth and videoing what you see?

Shuheng Zheng

Yes, but doing some post processing after the trip is over so that the video camera has no transmission delay.

So the rocket people can observe Earth's clock with no delay.

Rocket crew would see Earth's clock slowing down. But we know that from Earth's point of view, when the rocket stops at Alpha Centauri, the rocket's proper time should be less than the Earth's elapsed time. Hence, I would think on the rocket ship, during the final stopping phase, the video camera would show Earth's clock rapidly advancing.

John Rennie

When you get to Alpha Centauri you still have to keep videoing because you have to wait another 4.2 years for the light from Earth to reach you so you can apply your correction to it ...

Shuheng Zheng

We could we post-processing.

i.e. run this whole thing in a physics simulator that gives us a God's eye view of what the rocket ship would see if the video has no delay.

So the magic camera would give the rocket ship crew a view of Earth's clock without transmission delay.

Relativity effects would still occur but just take out the light delay complications.

"We could do post-processing"

John Rennie

I need to drop out to work for an hour or so. I'll be around later.

Shuheng Zheng

06:24

thanks. i'm going to sleep for now and think about what you said.

Slereah

08:17

morgen

Loong

huomenta

Slereah

08:43

So

what's the function space of a free scalar field

Is it $\mathscr{S}(\mathbb{R}^{n-1})$ or $\mathscr{S}(\mathbb{R}^{n})$

Is it only a Schwartz function at a given time, or is it just a Schwartz function

Novalium Company

09:33

@JohnRennie Genius! Is there a formula for dilutions?

John Rennie

@NovaliumCompany if you have a one molar solution and you dilute it by a factor of a hundred, e.g. by adding 10 cm^3 of it to a litre of water, then the concentration is 1/100.

Novalium Company

@JohnRennie Makes sense, it doesn't need a mathematical formula since it's not that complicated.

How do you know $10^{-4}M$ of NaOH is the right amount of NaOH for my goal?

John Rennie

We need a pH of around 10 because that gives the TiO2 particles a negative charge. I got this pH from reading that paper I linked back when we were first discussing this.

Slereah

pH of 10 doesn't sound very yummy

John Rennie

The pH is the -log10 of the H+ concentration, so pH 10 means the H+ concentration is $10^{-10}$ molar.

@NovaliumCompany OK so far?

Novalium Company

09:40

@JohnRennie Sure, I explain pH to myself as simply the concentration of hydrogen ions. Higher pH, more H+ atoms?

John Rennie

No, higher pH = fewer H+ ions

Novalium Company

@JohnRennie Oh, so it's reversed, alright then :D

What does the NaOH do to H2O? I mean, how does it "remove" H+ ions? (to make the pH 10)

John Rennie

In water the H+ concentration times the OH- concentration is always $10^{-14}$.

This is because water can disassociate into H+ and OH- ions, and it disassociates as much as is necessary to keep $[H+][OH-] = 10^{-14}$

Novalium Company

You mean $10^{-14}$M? (molarity)

John Rennie

Yes

Novalium Company

09:45

So the NaOH brings the water's molarity from $10^{-14}$ to $10^{-10}$?

John Rennie

So if pH 10 means $[H+] = 10^{-10}$M that means $[OH-] = 10^{-4}$M so that when you multiply the two together you get $10^{-14}$M.

Novalium Company

Got it

John Rennie

That's why you need $10^{-4}$M sodium hydroxide.

Novalium Company

I'm a bit confused. Why would add NaOH $10^{-4}$M to water with pH 10?

John Rennie

@NovaliumCompany Pure water is pH 7. You add $10^{-4}$M NaOH to increase the pH to 10.

Novalium Company

10:04

@JohnRennie pH 7 of pure water would mean $[H+] = 10^{-7}M$ and that means $[OH-] = 10^{-7}M$. (Meaning, there is equal concentration of H+ and OH-). To make the water pH 10 would mean to make $[H+] = 10^{-10}$, which means less H+. NaOH does that by adding more OH- ions? (Making the solution with less concentration of H+ and more OH+) Am I correct?

John Rennie

Yes.

Pure water dissociates in the reaction H20 -> H+ + OH- so the H+ and OH- concentrations are equal and both are equal to $10^{-7}$M.

When you add OH-, by adding NaOH, it makes the water dissociate less and that means less H+ is present.

Novalium Company

So how does adding TiO2 to the solution result in TiO2 having a charge?

John Rennie

In a vacuum a TiO2 particle has TiOH groups on it's surface. When you put it in water with alkali these groups dissociate to leave a TiO- at the surface. The result is that the surface of the TiO2 particle builds up a negative charge.

Novalium Company

Got it.

Time to eat now. Thanks @JohnRennie!

John Rennie

@NovaliumCompany bye

Slereah

11:53

Trying to find some paper that's in the sweet spot of in between "The inner product of the QFT Hilbert space is $\int \Psi[\phi] \Phi[\phi] \prod_x \phi(x)$" and "Consider the Borel set of a Frechet space etc etc"

A bit tricky

Schrodinger functionals are either a handwaving jamboree or a functional analysis nightmare

Slereah

13:18

Oh wait

We have that for a Gaussian measure $\mu$, we have the equality $$\int_X d\mu e^{if(\phi)} = e^{-\frac{1}{2} \langle f, Of \rangle }$$

I guess that might be the lynchpin to actually compute quantities

In the case of the vacuum state, that would just be $$\int_X d\mu e^{i0} = e^0 = 1$$

Which works out

also I'm not sure what $O$ is supposed to be

I assume some operator which depends on the specific theory

Hm

From what I know of the theory

I guess it could be $O = \Box$?

Since that would be the KG action

$\int d^nx \phi(x) \Box \phi(x)$

Ryan Unger

@Slereah the latter is nonsense

Slereah

13:33

@RyanUnger Probably

I can't read the math so good

Well $O= \Delta$, since it has to be positive definite

Euclidian path integral and all that

I'm not 100% sure if all my spaces are correct tho

I think $\phi$ belongs to $\mathcal{S}$ here, $f$ to $\mathcal{S}'$

So something like $$\int_{\mathcal{S}} d\gamma e^{if(\phi)} = e^{-\frac{1}{2} \int d^nx f \Box f}$$

Does it make sense to integrate tempered distributions like that tho

Hm

Ryan Unger

13:49

Uh what

Slereah

Trying to do the path integral properly

is what I am

Novalium Company

14:04

@JohnRennie Does it matter how much TiO2 I put in the final pH 10 water? (Is there some gram limit which I cannot exceed and which beyond the TiO2 particles do not get charged?)

Slereah

14:31

0

Q: Proper path integral of a field theory

I have been trying to find out the sweet middle ground of describing path integration of field theories, in between the physicist way and the mathematician way, but it seems hard to find something that is both rigorous and describes how to actually compute them. So far what I've been able to get...

quantum-field-theory mathematical-physics path-integral

bolbteppa

15:07

@Slereah Hatfield has 3 chapters on Schrodinger functionals which he then spends re-doing in the following four chapters with path integrals

Slereah

I did look at hatfield

He doesn't seem to do measures too much

Just the usual path integral as a limit thing

John Rennie

@NovaliumCompany If you put in more than 10% powder the suspension is going to get a bit thick and it won't separate well in an electric field. I would try 1% as a start and see how it goes.

1% by weight that is i.e. 10g TiO2 in a litre of water.

You could probably get away with less than that as TiO2 is strongly scattering and the suspension is very white.

vzn

16:13

← lol at new spacetime toothpaste TOE... physics research is defn making progress :P chat.stackexchange.com/transcript/71?m=51177116#51177116 + interesting stuff on sagnac effect physics.stackexchange.com/questions/493639/…

bolbteppa

@Slereah what does a measure on a function space of four-component spinors even mean

Slereah

@bolbteppa sticking to the scalar case for now my dude

Hard enough as it is

bolbteppa

boo

Captain Bohemian

I think the hotness hinders thinking; probably that's why school has summer vacation.

bolbteppa

Nobody knows how to deal with measures on function spaces, even measures in normal QM are a waste of time

Slereah

16:21

@bolbteppa I'm never deterred from wasting my time

Captain Bohemian

work outside school doesn't usually take much thinking; that's why workers outside school don't usually have summer vacation.

John Rennie

This got upvotes?

2

A: Heisenberg uncertainty principle in daily life

A very practical example-- My wife and I can only determine either the position or speed of the constituent parts the other's body, but not both at the same time. Therefore, it is quote impossible for us to dance together. The best we can do is to individually stimulate movement in rhythm to so...

Aaron Stevens

@JohnRennie I was thinking exactly the same thing.

@JohnRennie Well now it's down

enumaris

16:39

hmmm

Novalium Company

16:57

@JohnRennie Alright, thanks

Novalium Company

17:07

When I do the dilution process, when I first put 40g of NaOH in 1L of water, then get $10 cm^3$ of that solution, I should put it in a new 1L water? That would mean that the second solution is 1.001L?

John Rennie

17:56

Death by dark matter

What a lack of unexplained gory deaths tells us about dark matter

@NovaliumCompany you could add your 10cc to 990cc of water to get exactly one litre.

Novalium Company

@JohnRennie Got it, but wouldn't the molarity change a bit, since we are changing the volume?

John Rennie

@NovaliumCompany well yes, but the change would be negligible. If you were doing a really precise experiment it would matter. In this case I doubt it would make any difference.

Novalium Company

@JohnRennie Alright thanks.

Matthew Roh

Can electric fields pass through grounded surfaces?

Novalium Company

The black paint in the water shouldn't have any dangerous effects right? When should I add the black paint? At the end?

Matthew Roh

18:07

(well tbh I asked on here but I ain't getting any interest)

John Rennie

@NovaliumCompany I don't know. Try it and see.

Matthew Roh

@NovaliumCompany Maybe you can see if the black paint has any dangerous effects itself? There should be no dangerous effects if the black paint doesnt have dangerous effects

unless, maybe, the black paint forms a toxic compound with water

Novalium Company

@JohnRennie Alright. I'll put it at the end, before the TiO2

Matthew Roh

but thats a different story

Novalium Company

@MatthewRoh It doesn't have dangerous effects with water

I'll be using paint for kids drawings kit....

Matthew Roh

18:11

Ahh, then it shouldn't have any problems

oh

so... do electric fields pass through grounded surfaces?

(I know I'm too attention-hungry, but I just can't help it)

Novalium Company

@MatthewRoh believe me, you are not

Matthew Roh

oh

daysofsnow

18:28

Any chance there's any crystallographers or group theorists lurking here: can you confirm that the notion of star{k} is modulo reciprocal lattice vectors {Gi}? Then, if I have a k-vector (G1 + G2 + G3)/2 that vector is equivalent to another vector, say, (-G1 - G2 + G3)/2?

I'm confused because the Bilbao crystallographic server seems to think otherwise in its KVEC program.

user76284

18:46

Is the metric of a static and axially-symmetric spacetime of the following form?

$$c^2 d\tau^2 = c^2 dt^2 - dz^2 - r^2 d \theta^2 - f(r) dr^2$$

bolbteppa

Assuming it's just the 'cylindrical coordinates' version of Schwarzschild you should probably have a function $g(r)$ in front of $c^2 dt^2$

Static means that $f$ and $g$ are not also functions of $t$

user76284

The Schwarzschild metric isn't axially-symmetric though?

You mean you can add a $g(r)$ in front of $c^2 dt^2$ and it's still static and axially-symmetric?

bolbteppa

Schwarzschild uses the spherical coordinate 3-line element $dr^2 + r^2 d \theta^2 + r^2 \sin^2 \theta d \phi^2$ in $g(r) c^2 dt^2 - f(r) dr^2 - r^2 d \theta^2 - r^2 \sin^2 \theta d \phi^2$ , your 4-line element has the cylindrical coordinate 3-line element $dr^2 + r^2 d \theta^2 + d z^2$ in $g(r) c^2 dt^2 - f(r) dr^2 - r^2 d \theta^2 - dz^2$.

user76284

"Schwarzschild metric isn't axially-symmetric" Derp, disregard this.

Should've said cylindrically-symmetric, not axially-symmetric (the former is a special case of the latter, as is spherical symmetry).

So should it be
$$c^2 d\tau^2 = g(r) c^2 dt^2 - f(r)^2 dr^2 - r^2 d\theta^2 - dz^2$$

For a spacetime that's static and cylindrically-symmetric?

I see now

bolbteppa

19:02

I would think so, but there is a way to formally derive that it should take this form to be 100% sure, just mimic the way one goes from $ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$ to $ds^2 = g(r) c^2 dt^2 - f(r)dr^2 - ...$

Deriving the Schwarzschild solution

The Schwarzschild solution describes spacetime in the vicinity of a non-rotating massive spherically-symmetric object. Of the solutions to the Einstein field equations, it is considered by some to be one of the simplest and most useful. == Assumptions and notation == Working in a coordinate chart with coordinates ( r , θ , ϕ , t ) {\displaystyle \left(r,\theta ,\phi ,t\right)} labelled 1 to 4 respectively, we begin with...

user76284

Yeah. I'm trying to find the metric for an infinite massive cylinder. Maybe it looks like the Schwarzschild metric but with the cylindrical line element rather than the spherical line element.

Are there any additional constraints/symmetries in that situation that could simplify the metric even further (i.e. $g(r)$ and $f(r)$)?

bolbteppa

This is a similar way to get from $ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$ to the simpler form, probably more general (using Killing vectors). As you see for the Schwarzschild case, any further constraints on $f(r)$ and $g(r)$ are found by plugging them into the Einstein Field equations

Ryan Unger

@bolbteppa we're back to this?

bolbteppa

19:17

Game on

Sir Cumference

@RyanUnger did you not play it as a kid?

mario 64 was awesome

Ryan Unger

I didn't have video games until I was like 10

my first mario was super mario galaxy

well I take that back

first TV console when I was 10

I had a gameboy and a DS before that

JMac

man I didn't even own a N64 but my friend did, and Mario64 was like the coolest thing ever

Sir Cumference

gba was the best

good times

Ryan Unger

first mario I played would have been Super Mario 3 on the GBA

Sir Cumference

19:24

@RyanUnger yeah that was my first

Ryan Unger

jMac I don't even remember N64 until I was like 7

JMac

oh man, I still remember getting the yellow GBC with pokemon yellow that like everyone including myself got for christmas one year

Ryan Unger

first console I ever saw was a gamecube and xbox

neighbors in texas had those

JMac

first one I remember seeing was a NES; my friends uncle owned one and lent it to him when we were like 5, it was great

playstation was my real jam anyways. When Crash Bandicoot came out last year on PC it was a huge nostalgia trip for me

Sir Cumference

@RyanUnger every gamecube game was great tho

how do you say pre-2003 games were bad

Ryan Unger

19:31

when was the gamecube released

2001?

Semiclassical

@JMac yeah, I think I had that too

user76284

20:31

I'm probably doing something dumb, but isn't the Ricci tensor supposed to vanish for a vacuum solution like Minkowski space?

Ryan Unger

that doesn't look like the Minkowski metric

what is -Indexed[x,2]^2

user76284

$-r^2$

Ryan Unger

yeah that's not the Minkowski metric

if you want polar coordinates you need some angular stuff

user76284

Since the metric is $dt^2 - r^2 d\theta - dr^2 - dz^2$

Ryan Unger

oh cylindrical coordinates

it should be $r^2\sin^2\theta\,d\theta^2$

user76284

20:34

Isn't that spherical coordinates?

Cylindrical coordinate system

A cylindrical coordinate system is a three-dimensional coordinate system that specifies point positions by the distance from a chosen reference axis, the direction from the axis relative to a chosen reference direction, and the distance from a chosen reference plane perpendicular to the axis. The latter distance is given as a positive or negative number depending on which side of the reference plane faces the point. The origin of the system is the point where all three coordinates can be given as zero. This is the intersection between the reference plane and the axis. The axis is variously called...

I'm taking the inner product of the line element with itself.

Ryan Unger

oh yah for the plane it's just $dr^2+r^2d\theta^2$ right

then there must be some issue with the mathematica implementation

are you sure that's the right formatting for inputting the metric

user76284

What do you mean?

Wait, hold on

Ryan Unger

like $r$ is a single coordinate in this system

why do you need a vector

user76284

I'm interpreting $x$ as $(t,r,z,\theta)$

I think I might've found the error though

I'm supposed to use the metric with lower indices in Γ1 and the one with upper indices in Γ2.

But I'm using the same matrix unaltered for both.

List of formulas in Riemannian geometry

This is a list of formulas encountered in Riemannian geometry. == Christoffel symbols, covariant derivative == In a smooth coordinate chart, the Christoffel symbols of the first kind are given by Γ k i j = 1 2 ( ∂ ∂ x j...

Ryan Unger

ohhh does Indexed take out the second coordinate

user76284

20:40

Yep

Ryan Unger

ok

well yeah if you didn't distinguish between the metric and its inverse you're gonna be in trouble

user76284

I think the $g$ I have is correct if it's interpreted as $g^{ij}$, right?

So I was supposed to take its inverse when calculating the Christoffel symbols of the first kind.

Ryan Unger

no, you have $-g_{\mu\nu}dx^\mu dx^\nu=dt^2-dr^2-dz^2-r^2d\theta^2$

so it's $g_{\mu\nu}$

or $-g_{\mu\nu}$

depending on conventions

user76284

Oops you're right.

Well it seems to work now.

enumaris

bowchicawowow

user76284

20:43

Gonna prettify the code a little bit

enumaris

Good news everyone! Simply performing a coordinate transform between Cartesian and Polar coordinates did not warp the spacetime fabric!

user76284

@enumaris That's good :P

enumaris

:)

user76284

Well it seems cylindrical coordinates works fine but there's a problem with the spherical coordinates.

Wait, nevermind. It's still 0.

enumaris

Good News!

Ryan Unger

20:58

thank god the universe isn't collapsing

user76284

It also vanishes for the Schwarzschild metric, yay

enumaris

now try it for Kerr

user76284

Good idea, testing with nondiagonal elements.

Hmm

Ryan Unger

lol

user76284

21:18

Oh wait I forgot to define Sigma and Delta in terms of r, rs, and a.

enumaris

fun times

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