The h Bar

Emilio Pisanty

3:00 PM

@Slereah see minor edits here, primarily as a response to this dupe.

rob

@EmilioPisanty Short answer: it depends.

Emilio Pisanty

3:27 PM

hey @JohnRennie I need a brit

can you explain what just happened in Parliament?

why so much fuss over a private sitting?

John Rennie

@EmilioPisanty Yesterday's vote on Brexit?

Emilio Pisanty

@JohnRennie I think it's from today

Scottish National Party walk out of Commons in protest

►

bbc.com/news/uk-politics-44467306

video's a bit too long for the actual content

John Rennie

There was a vote yesterday on whether the MPs have a right to veto the final Brexit agreement made by the Prime Minister and her cabinet.

Emilio Pisanty

basically, Blackford asks for a private sitting, Bercow says that a vote will take place at the end of the session, endless back-and-forth, Bercow expels Blackford for not sitting down, SNP leaves

@JohnRennie they already voted? how'd it go?

John Rennie

The vote was just on whether the MPs can throw out the final agreement if they don't like it, so it's just procedural.

The result was that MPs voted to leave the decision to the cabinet i.e. MPs will not have the right to veto the final agreement.

Emilio Pisanty

3:32 PM

@JohnRennie yeah, I'm on that page thus far, it's procedural but it's still important, innit

@JohnRennie srsly?

bejeesus

Sid

@JohnRennie ...why?

John Rennie

Yes, I was surprised

It feels a bit like turkeys voting for Christmas, but there you go

Emilio Pisanty

@Sid because Party Unity and Strong And Stable come before an actual exercise of the democratic powers of parliament

John Rennie

Anyhow, this has not gone down well with the anti-Brexiters. Since the SNP are vehemently anti-Brexit they are hopping mad.

And to be fair I agree completely with their point of view. But then I'm not a bigot living in my own personal cloud cuckoo land.

Emilio Pisanty

@JohnRennie How long until the next general election?

Sid

3:36 PM

The Conservatives had a Razor-thin majority, no?

Emilio Pisanty

I mean, I don't imagine all that Confidence And Supply business can survive the next twelve months

John Rennie

Three or four years. Actually I can't remember when the last election was. That nice Mr. Google will know ...

Emilio Pisanty

@Sid no, they came close to a majority but didn't get it

Sid

@EmilioPisanty with alliances, I meant

John Rennie

8th June 2017. Wow, we have four more years of this.

Emilio Pisanty

3:37 PM

@Sid then yes, razor thin, but only with a suitably weakened interpretation of 'alliance'

they agreed with a right-wing unionist Northern Irish party, the DUP, that the DUP would greenlight the formation of the government itself and would vote in favour of the budget but nothing more

John Rennie

The DUP are not a nice bunch. They are the main reason Northern Ireland's abortion laws are medieval.

Sid

3:54 PM

@JohnRennie What's their abortion laws? Punishable offence?

John Rennie

@Sid basically abortions are illegal except in the most extreme circumstances e.g. the mother will die if the pregnancy is allowed to come to term. Even then women have died because they were refused abortions.

Actually that was in Eire not Northern Ireland, but until recently they had the same abortion laws.

Now that Eire has voted to bring their abortion laws in line with the rest of Europe the pressure is on Northern Ireland to do the same. And the DUP are not happy about it. Nice people.

Actually UK abortion law isn't great either. It's very paternalistic.

enumaris

5:02 PM

gotta flex my SQL muscles today...

mmmmhm

John Rennie

Phew, got my first upvote of the day. I was beginning to think this would be my first day without any upvotes.

enumaris

lol

I think I have many days without upvotes

John Rennie

But then, to be fair, until just now I hadn't answered any questions :-)

enumaris

I go months without answering questions

Novalium Company

5:23 PM

Guys, how did humanity calculate that the universe was born 13.8 Billion Years ago? Is it complicated?

Semiclassical

wikipedia has a lot of discussion:

Age of the universe

In physical cosmology, the age of the universe is the time elapsed since the Big Bang. The current measurement of the age of the universe is 13.799±0.021 billion (109) years within the Lambda-CDM concordance model. The uncertainty has been narrowed down to 21 million years, based on a number of projects that all give extremely close figures for the age. These include studies of the microwave background radiation, and measurements by the Planck satellite, the Wilkinson Microwave Anisotropy Probe and other probes. Measurements of the cosmic background radiation give the cooling time of the universe...

Novalium Company

Also, how do we know that the universe is expanding, since 'we see back in the past'? The unverse could be currently ending?

Sir Cumference

@NovaliumCompany You can get a good and simple estimate by assuming expansion is at a constant speed, and by using the Hubble law

Sid

@NovaliumCompany Hubble's Law

Novalium Company

Well, "the galaxy is moving away from Earth". How do we know that? We see the things that may have happend many years ago. Maybe the universe has already expanded and now is ending?

ACuriousMind

5:29 PM

@NovaliumCompany Look up cosmological redshift

@NovaliumCompany "now" is a trickier word there than you might think.

Cows

So yesterday was pretty interesting hehe

Watching the movie "sisters"

ACuriousMind

There is no absolute simultaneity, it is unclear what "now" refers to when talking about distant stars.

Novalium Company

@ACuriousMind Hmm, Einstein's stuff start to kick up. I've always had that idea that time is constant everywhere, but we think it's not because we see things in the past, but we only see it, that doesn't mean that everything is not simulated just because we see it late?

Anonymous

@ACuriousMind A bit of help needed: 1. Is disclosure necessary in case someone is citing their own paper in an answer without mentioning it? 2. Is there any discussion related to that on meta?

Sir Cumference

@ACuriousMind I think he's asking how we know expansion is accelerating

John Rennie

5:33 PM

@ACuriousMind You seem to think it's obvious why zero point energy doesn't contribute to the stress-energy tensor. Sadly it is not obvious to me.

Sir Cumference

Or at the very least, how we know expansion is not decelerating

ACuriousMind

@NovaliumCompany I'm not sure what the "simulated" part is about, but yeah - the notion of time is not absolute, and the 13.8 byrs are cosmological time, a special notion not completely equal to our "perceived" time on Earth.

Novalium Company

Ok, imagine some aliens far away are looking at earth, they may be seeing the old dinasours, but they are only seeing them, while now still exists and it's going. Get my point?

If the aliens start coming to us, they will essentially get to the point, where they see us.

Semiclassical

or they die in the meantime

Novalium Company

xD

Semiclassical

5:40 PM

which, given how long the meantime would be, is not an entirely silly comment

Novalium Company

Are my assumptions true? (lol, of course not but...)

Semiclassical

I think you're basically right, insofar as "if the signal can eventually reach you, then you can eventually see it happen"

Novalium Company

So everything happends simultaneously?

Semiclassical

What?

Novalium Company

Well, I just proved Eintein is wrong :D

xDDD

Semiclassical

5:42 PM

Or you said something nonsensical.

ACuriousMind

@JohnRennie A zero-point energy density caused by the quantum ground state does not show up as a term in the Lagrangian, and therefore has no way of contributing to the stress-energy tensor.

Novalium Company

Ok so, imagine there are aliens far far away that see the Earth, but they will see far far back in the Earth's existance (Example: dinasours). Well, they may be seeing the past, but that doesn't mean that we don't exist?

Well?

enumaris

We don't exist to them since we aren't in their past light cone.

Semiclassical

Too philosophical for me, tbh.

enumaris

None of our actions can affect them.

at that moment

Mohammad Areeb Siddiqui

5:47 PM

can someone explain to me the stationary phase argument? if its given that $\psi = \int_{-\infty}^{\infty} dk \phi(k) e^{ikx}$ and $\phi(k) = e^{-L^2(k-k_0)^2}$ where L is just a constant with length units. $\phi$ is sharply localized at $k = k_0$. Does this mean that the integral would exist only when $k = k_0$ ? Is this what is meant by stationary phase approximation?

Novalium Company

But we still exist.

enumaris

well we exist yes. Just because aliens far away can't be affected by us doesn't mean we don't exist.

Semiclassical

The point is more that the lack of simultaneity per Einstein has nothing to do with whether objects exist simultaneously or not

Novalium Company

Well, doesn't Eintein say that time is different everywhere?

Semiclassical

That's a gross oversimplification.

enumaris

5:49 PM

Einstein says there's no universal notion of a synchronized time.

Novalium Company

My point is that, one second is one second, everywhere?

(That's the stupidest thing I've said so far)

Semiclassical

I mean, there are some deep philosophical issues here

ACuriousMind

@JohnRennie Actually, of what I said the first part isn't wrong but the second part doesn't follow. It's more like this: When you want to talk about dark energy and the stress-energy tensor together with zero-point energies, you have to make a definitional choice of whether or not the value for the stress-energy tensor (which is now an operator!) that appears in Einstein's equations includes contributions from the vacuum state.

But since the contributions from the vacuum state are just a uniform energy density (Lorentz invariance of the vacuum state!), one canonically identifies them with the cosmological constant instead of with the stress-energy.

enumaris

A second is a unit of time that's well defined. As long as "everywhere" accepts this definition, then a second is a second "everywhere".

Novalium Company

I mean, the aliens and we on earth, we exist at the same time?

enumaris

5:52 PM

But I feel you're more running into issues with the andromeda paradox.

See here: en.wikipedia.org/wiki/…

There is no well defined universal notion of "same time"

ACuriousMind

When you say something like "it's unclear how the vacuum state contributes to dark energy but not the stress-energy", you essentially saying that you want to count this states contribution twice instead of once.

Novalium Company

Why?

enumaris

Special and General Relativity says there's no universal notion of a synchronized time.

Novalium Company

Maybe the universe started, BOOM, andnow, everywhere, the same amount of human seconds have passed?

enumaris

Different worldlines will have different "seconds" passing

Novalium Company

5:54 PM

well, according to human seconds.

ACuriousMind

@NovaliumCompany Relativity is experimentally proven correct: The existence of a universal speed limit forces non-universal notions of time upon us otherwise observers could not agree on light always having the same speed.

enumaris

what do you mean according to human seconds

Semiclassical

I think Einstein's train thought experiment is also pertinent here

Novalium Company

Well, for example maybe 1 alien second = 1.25 human seconds.\

ACuriousMind

@NovaliumCompany The notion of time is not a function of location! It is a function of how fast one observer is moving relative to another.

enumaris

5:55 PM

say you had two particles starting at the big bang. They each carry a clock. They run off somewhere, and then eventually meet back. They will find that their clocks don't match.

Novalium Company

^ I know that true, but it makes no sense. (to me)

enumaris

@NovaliumCompany The issue of no universal time is not a unit issue

Semiclassical

The order of the events (whether the two flashes of lightning occur simultaneously or not) depends on whether the observer is in motion.

ACuriousMind

You can't ask "are seconds at one place the same as seconds at another?". The correct question is "are seconds for one traveller the same as for another?".

enumaris

@ACuriousMind well it kind of is a function of location, since there's also gravitational time dilation

ACuriousMind

5:56 PM

@enumaris That's probably a needless complication in this case

enumaris

But I know you love technicalities

so u know...

Mohammad Areeb Siddiqui

gravity bends time???

what

ACuriousMind

It sure does

enumaris

gravity bends spacetime

Semiclassical

(heck, an observer moving left vs. right will see opposite orders for the flashes)

ACuriousMind

5:57 PM

@enumaris Well, since you started the technicalities war, I'll point out that there's no difference between "gravity" and "curvature of spacetime" in GR :P

So you're saying "gravity gravitates"

Novalium Company

Ok, in my head, it's like this: 2 particles boom of the big bang and they carry a human clock, which ticks 1 human second. They float away and after years, they will be in completely different parts of the universe, but if someone was to look at the clocks, they will be the same?

ACuriousMind

Which is at least not wrong, I guess

@NovaliumCompany That's an ill-defined question

Semiclassical

yes.

ACuriousMind

Where is this someone, and how are they looking at the clocks?

Semiclassical

in the following sense, at least

^

Mohammad Areeb Siddiqui

5:59 PM

is someone running?

enumaris

@ACuriousMind I am tautology man

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

hello people. is there anyone rich willing to put a bounty on physics.stackexchange.com/questions/284444/… because the current top voted answer does not deal with the question of whether there is an upper limit of wavelength size for a photon

enumaris

haven't you been paying attention to my conversations here lately? :P

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

30 upvotes for an answer that doesn't deal entirely with the question asked

leaving me in doubts

ACuriousMind

@enumaris Consider joining the tautology club, then

Novalium Company

5:59 PM

@ACuriousMind We don't need to look at the clocks, it's enough for us to know that they tick.

Semiclassical

Actually, there's at least one sense in which it can definitely be no

Namely, the twin paradox

enumaris

@ACuriousMind no webcomics!

Semiclassical

Have one of the clocks remain at rest, while the other zooms away at relativistic speeds for a while, and then have them zoom back

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

if someone knows whether there is an upper limit of the wavelength of a photon, please let me know

Semiclassical

once the two clocks are in the same location again, compare them

Novalium Company

6:01 PM

Oh god, my head is going to boom like the big bang. Space is so weird...

Semiclassical

you'll find that more seconds will have ticked on the clock at rest than on the clock that was in motion.

ACuriousMind

@NovaliumCompany I'm not following you. My point is that your question is ill-defined, it is not clear what it means for the clocks to be "the same" without further information.

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP Why would there be?

enumaris

maybe the causal horizon?

but there's detection issues way before you get there.

ACuriousMind

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP The answer by Josephus is perfectly fine.

Novalium Company

Ahh, I'll be heading to watch Hawaii 5-0, see you later. We could continue later.

Mohammad Areeb Siddiqui

6:02 PM

did the time stop for the clock in motion because it was travelling at relativistic speed?

enumaris

clocks don't stop

Novalium Company

I think I just need to learn Eintein's stuff...

enumaris

they keep on ticking

otherwise they'd be broken

Semiclassical

according to the observer at rest: the clock in motion doesn't stop ticking, but it does tick slower

Novalium Company

Wow, I just realized teleportation is not possible.

actually,

enumaris

6:04 PM

sub light speed teleportation is kinda theoretically possible...

Novalium Company

it is

enumaris

though the no-clone theorem puts kind of a damper on it

Novalium Company

Electrons jump to energy states without covering the distance in between?

Mohammad Areeb Siddiqui

but if im sitting on the clock in motion, its ticking with the normal speed?

at*

ACuriousMind

@NovaliumCompany Energy states have no well-defined location, so "jump" is to be taken metaphorically.

enumaris

6:04 PM

if you're with the clock, it's ticking like normal

would be weird if somehow time-dilation affects clocks differently than you

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

@ACuriousMind but Josephus answers leaves the doubt about such limit. it is fine, but does not answer it

enumaris

heh

ACuriousMind

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP And how exactly do you expect anyone to prove the negative here?

Semiclassical

Plus, the causality of processes in quantum mechanics are sorta tricky

ACuriousMind

The best one can do is "I don't know of any limit of photon wavelength"

Novalium Company

6:05 PM

@ACuriousMind But they do teleport (without covering the distance)?

ACuriousMind

@NovaliumCompany That's an ill-defined question :P

Mohammad Areeb Siddiqui

so when i come back to the clock at rest, i'd see the same time in both the clocks?

Novalium Company

@ACuriousMind Ok, we know that electrons can change energy states, right?

Semiclassical

there's something I'm forgetting here.

ACuriousMind

@NovaliumCompany Yes.

Novalium Company

6:07 PM

@ACuriousMind If an electron receives energy, it will jump to a higher, more away from the nucleus energy level?

ACuriousMind

@NovaliumCompany It is not clear what "more away" means.

Quantum states usually do not have a precise location associated with them

Semiclassical

Emission/absorption of photons in quantum mechanics is not a spacetime process, at least in the orthodox interpretation

ACuriousMind

There is no notion of motion during a quantum process, at least not in the classical sense

Semiclassical

Just because we draw Feynman diagrams as though that's true, doesn't mean it's to be interpreted literally.

Novalium Company

@ACuriousMind Ok, but we know that if it receives a lot of energy, the electron will jump out of the atom?

Mohammad Areeb Siddiqui

6:08 PM

why isn't physics true for anything in whatever sense

Semiclassical

Because the universe doesn't care about making life easy for us :P

Novalium Company

The universe has created something for humanity to play around with (Physics) while we wait for our extinction.

ACuriousMind

@NovaliumCompany It can go to a state where it is no longer bound to the atom. In that case, any excess energy that it did not need to become unbound will increase the speed of the resulting free electron. (Speed makes sense for free particles because their wavefunction is a wavepacket and you can define the speed with which a packet travels just fine).

Semiclassical

in QM, it's not legitimate to talk about where a particle was in between measurements of its position.

all you can say is: I measured it at point X1 at time T1 and at point X2 at time T2.

enumaris

QM weird brah

Semiclassical

6:11 PM

In the Copenhagen interpretation, at least, one does not ask about the trajectory of the particle in the meantime.

Novalium Company

@ACuriousMind Ok, you said "It can go to a state". Does it cover that distance?

ACuriousMind

@NovaliumCompany There is no distance.

Don't take my "go" so literally

I could equally well have said that it will change its state.

Novalium Company

Ok I'll say it that way, I remember something about something not having to cover the distance betweens something. I am trying to prove teleportation exists, help me.

ACuriousMind

@NovaliumCompany It doesn't.

Novalium Company

OH

I remember now.

Semiclassical

6:12 PM

It doesn't exist in an operational sense, at least.

ACuriousMind

Especially with regards to "quantum" teleportation, I'll always sympathize with this scientist

Novalium Company

Entanglement, particles can send information to each other, without this information having to travel the distance in between?

ACuriousMind

(Sorry @enumaris :P)

@NovaliumCompany nononononononononono! :P

Novalium Company

xD

Semiclassical

In the sense of correlations, yes. In the sense of signalling, nooooo

ACuriousMind

6:14 PM

Entanglement is not transmission of signals

Novalium Company

I know xDDD

It's a spooky action at a distance. :D

ACuriousMind

::twitches::

Semiclassical

One should emphasize here, though, that the status of particle trajectories is very different in GR vs. QM

Novalium Company

So how are they entangled?

ACuriousMind

@NovaliumCompany Magic, basically

Novalium Company

6:15 PM

Hmm, so that's still a blur.

Semiclassical

"That's how the universe works."

Novalium Company

^ That's what a lazy person would say.

ACuriousMind

I'll be right back, gotta get something to eat.

Semiclassical

in GR, particle trajectories exist for all time. different observers will have different accounts of those trajectories, but they'll all agree on said trajectories existing.

Novalium Company

Me too, I'mma grab popcorn, watch Hawaii 5-0 and be back.

Semiclassical

6:16 PM

In QM, the same cannot be said.

The notion of a particle trajectory basically does not exist in QM.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

@ACuriousMind I do not know. Else I wouldn't have asked the question. I see Ana v and 2 other people on this website claiming that a photon cannot have a wavelength exceeding the size of the universe, but they give no justification of their claim

I see no boundary condition to be applied. I do not see clear edges of the universe either

Semiclassical

You can say that the particle was observed at a point X at a time T.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

I do not even know whether it's possible to have a photon with a single wavelength over such scales.

Semiclassical

You can then observe it at a different time. But it isn't taken as meaningful to ask what it was doing in the time between.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

intuitively, I see no upper limit of wavelength of photon

Semiclassical

6:19 PM

You can ask what the wavefunction was doing. But the position of the particle itself is not taken to be meaningful apart from what it is measured to be.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

ana v claims "The maximum possible wavelength cannot be larger than the size of the universe, and then we fall into cosmological models. "

but why???? where is the justification...?

it would mean we can't accelerate an electron a small amount, else it would emit a photon with a wavelength greater than the universe. so the electron in a way, is aware of the size of the current size of the universe for an observer at rest with him. this makes no sense

enumaris

hmmm

Semiclassical

6:34 PM

probably something along the lines of: the allowed wavelengths of a photon in a box must be some fraction of the box's width

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

but there is no box here

the universe is not a box with definite edges

Mohammad Areeb Siddiqui

isn't universe spherical?

Semiclassical

eh, a 3D box isn't so different in the details from a sphere

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP it's a crude analogy, but i think that is the analogy being made

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

but why would that matter? I did not mention a sphere

G. Bergeron

Bahaha Bott-Tu-lism!!!

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

6:41 PM

I said that there is no definite edges of the universe. so I see no way to apply a boundary condition to an EM wave or photon

G. Bergeron

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP Any finite volume will reject long enough modes

bolbteppa

Bott-Tu-ism := that which is supposed to be intuitive but isn't

G. Bergeron

It may not be as simple than a box or a sphere, but the same effect happens

It simply does not 'fit in'

bolbteppa

19

Q: Am I reading Bott - Tu right?

Summary: I'm finding Bott - Tu to be too brief and terse. I constantly have to look elsewhere to fill in details. This is not time-efficient. Am I missing something? If not - what other books do people recommend? First - some background; I study on my own. I've read Hatcher's Algebraic Topol...

G. Bergeron

Of course this assumes some particular boundary conditions...

enumaris

6:45 PM

I think the issue probably arises from the picture of a EM wave being a sinusoidal wave with some fixed definite wavelength. This is a good picture to use when solving a lot of problems, but it must be noted that instead, EM waves in practice are not these sinusoids of infinite extent, but, like a wave in a pond, they are periodic distortions of the background EM field. Think localized wave packets.

G. Bergeron

But at the same time, if the universe has an edge, it makes no sense to posit fields existing beyond it. This, combined with an assumptions on the continuity of fields implies a Dirichlet boundary with the field = 0.

Semiclassical

if the notion of having an edge bothers you, think instead about vibrations on a circular ring

G. Bergeron

@enumaris No but a simple change of basis leads to a picture of any possible waves in terms of a sum of pure sinusoids

Semiclassical

you can also just take it as a statement of dimensional analysis, though: the size of the universe is the scale of cosmological physics, so if a photon were to have wavelength at that scale then it's not obvious that the photon should behave in the same way

G. Bergeron

However, I don't see right now why electrons could simply not emit that too low energy photon.

enumaris

6:50 PM

well, you can certainly do fourier analysis yes

Semiclassical

these are only heuristic justifications, of course. but i don't thinkn you can do better

G. Bergeron

Two permuted indexes in a definition for quantum determinant caused me problems for months... -_-

Semiclassical

ew

Quantum determinant?

G. Bergeron

@Semiclassical $det_q= \sum\limits_{\sigma \in S_n} (-q)^{l(\sigma)} x_{1\sigma(1)} ... x_{n \sigma(n)} $

Semiclassical

ah, so quantum like 'quantum group'

G. Bergeron

6:54 PM

yeah

Mohammad Areeb Siddiqui

7:07 PM

if $\int_{-\infty}^{\infty} dk \phi (k)e^{ikx}$ has to be found exactly such that $\phi(k)$ is sharply localized at $k=k_0$. Does this mean the integral only exists when k is around $k_0$ ?

is this what is meant by stationary phase approximation?

Semiclassical

it's not the that the integral only exists when $k\approx k_0$. it's that the integrand is negligible away from $k\approx k_0$.

Mohammad Areeb Siddiqui

yes so the predicted value of the intergal would be

$\int_{-\infty}^{infty} dk k_0 e^{ikx}$

oh wait

Semiclassical

That's not a definite integral.

What?

Mohammad Areeb Siddiqui

im trying to

Semiclassical

$\phi(k)\approx \phi(k_0)$ at $k=k_0$

Mohammad Areeb Siddiqui

7:12 PM

predict the value of this integral

Semiclassical

not $\phi(k)\approx k_0$

Mohammad Areeb Siddiqui

oh yes

so that is it?

Semiclassical

for what you were saying, yes. but that doesn't sound like stationary phase to me.

Typically for stationary phase you'd moreover be approximating $\phi(k)$ as a Gaussian near $k=k_0$

You can make a box approximation of $\phi(k)$, though, e.g. $\phi(k)\approx \phi(k_0)$ if $|k-k_0|<\delta k/2$

in which case you'd get $$\int_{-\infty}^\infty dk\,\phi(k) e^{ik x}\approx
\int_{k_0-\delta k/2}^{k_0+\delta k/2} dk\, \phi(k_0) e^{i k x} = -\frac{i}{x}\phi(k_0)\left[e^{i k x}\right]_{k_0-\delta k/2}^{k_0+\delta k/2} = \frac{2}{x} \phi(k_0) e^{i k_0 x}\sin(\delta k \,x/2)$$

(I think.)

and that sin(x)/x form is what you'd expect for the Fourier transform of a box function

If you instead approximate the original function as a Gaussian, though, you get that the Fourier transform is itself approximately Gaussian

which is a really nice approximation in a lot of cases.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

7:29 PM

@G.Bergeron What do you say about the walls of a microwave oven? They have a near black body spectrum, right? are you saying that they cannot emit any wavelengths, but only those that "fits" into the oven?

Semiclassical

that's how Planck derives the blackbody distribution, I think.

with an approximation made to get a continuous radiation spectrum, though

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

im confused as hell

say you have a lightbulb in a microwave oven, i dont understand how it cannot emit say radio waves, but if we open the doors, suddenly it can

it's like if the electrons or atoms creating the photons get to know if the door is opened or closed

makes no sense to me

ACuriousMind

All frequencies can be emitted into a cavity, there's no boundary condition here, after all.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

if i remember well, there is a cutoff frequency

below which nothing fits . only TM and TE modes and their superpositions are allowed (no TEM)

ACuriousMind

Oh, I meant "cavity" in the literal sense of "shape with a hole in it", not that technical sense of cavity

I.e. I'm talking about the microwave :P

Mithrandir24601

7:45 PM

@ACuriousMind ... Now I know how mathematicians feel when I write down some maths :P

ACuriousMind

There's no requirement that a full wavelength of some "wave" must fit into the medium to be able to be emitted. It's debateable whether you want to call what is being emitted a "wave" though, since it will not even complete a full period before hitting the walls of the box

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

well if that counts as a photon.... and i dont care at all that it hasn't completed a full period if it's wavy

for me it's a wave

ACuriousMind

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP A photon and an electromagnetic wave are two very different things

A proper EM wave is a many-photon state with an indeterminate number of photons.

Mithrandir24601

@ACuriousMind I take it that you mean a classical EM wave?

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

indeterminate even in an ideal case?

by ideal I mean with a laser like beam where it has a dirac sharp wavelength

J Thomas

7:48 PM

I want to say classically, electromagnetic radiation happens when a charge gets accelerated.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

so I know exactly the energy of photons

J Thomas

It doesn't have to be accelerated in a wave motion.

Mithrandir24601

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP that's a coherent state, so there is an indeterminate number of photons

ACuriousMind

@Mithrandir24601 The very pedantic formulation of what I wanted to say would be "The classical correspondence of the many-photon state is the classical EM wave"

J Thomas

Just waves are plausible ways for charges to get accelerated.

probably_someone

7:49 PM

On an unrelated note, there seems to be a notational trend (i.e. I've seen at least 3 recent questions with this in them) of using $x$ for velocity. Does this notational choice come from a particular source? I've literally never seen it before, so I was just wondering if it was common someplace I haven't looked.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

oh yes, I know this @JThomas though it is unclear whether a uniformlly accelerated charge emits

Mithrandir24601

@ACuriousMind Hmm... It's all in the statistics

ACuriousMind

I'm getting the feeling this discussion is getting a bit off its original track :P

J Thomas

Given the math, how can a uniformly accelerated charge NOT emit? Do the LIenard Wiechert formula, you get the amount of force due to the radiation.

Mithrandir24601

@ACuriousMind Probably. I think you picked the one thing/topic/area in all of physics that I could actually be bothered to be pedantic about :P

J Thomas

7:51 PM

There's no wave while the charge is being uniformly accelerated. Just, you can't keep uniformly accelerating it for a real long time, you'll run away.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

@JThomas physics.stackexchange.com/questions/70915/…

one of the most famous PSE question

answer: undetermined yet

Semiclassical

@probably_someone :S that's bad

J Thomas

What is PSE?

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

physics stack exchange

probably_someone

@Semiclassical yeah, the main issue is that they don't explain what $x$ means until specifically asked, which usually reveals some gross misinterpretation of the physics. The fact that I've seen more than one of these recently makes me think that people are coming from the same source (e.g. lecture notes?)

ACuriousMind

7:55 PM

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP 7,000 views is not all that famous, actually :P

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

true, let's say well ranked then

Semiclassical

just ew

ACuriousMind

@probably_someone Got a link to an example?

Or two, for comparison?

J Thomas

If the equations actually fit the reality, then a uniformly accelerating charge creates a force on a second moving charge if the second charge is moving in the right direction, and that charge is perpendicular to the direction of the second charge's motion. You can calculate how big the force will be. That force gets called radiation.

If you want to call it something else when it isn't in waves then I don't mind.

But if it is not there, that says something important.

probably_someone

https://physics.stackexchange.com/questions/411362/what-does-this-imaginary-number-mean-for-time-and-velocity

https://physics.stackexchange.com/questions/410245/closed-gravitational-orbits-and-gradient-systems?noredirect=1#comment918861_410245

those were the two most recent ones I could find. I distinctly remember a third, but I can't locate it at the moment

(It appears the first one was edited since I last saw it. check the comment-chat for evidence)

ACuriousMind

8:01 PM

@probably_someone thanks

@probably_someone I can just check the edit history...scrolling past 20 revisions... ::sigh::

probably_someone

@ACuriousMind yeah, that first one's a mess in general

@ACuriousMind and apparently the second one says $x$ is momentum, not velocity, so I may just be mis-remembering things haha

Semiclassical

eh, x as momentum isn't much better

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

8:19 PM

on explaning Gibbs paradox, Gibbs used a complicated sentence, not sure if this qualifies as English : Again,when such gases have been mixed, there is no more impossibility of the separation of the two kinds of molecules in virtue of their ordinary motions in the gaseous mass without any especial external infuence than there is of the separation of a homogeneous gas into the same two parts into which it has once been divided, after these have once been mixed

No wonder the paradox was intringuing... if that's supposed to explain the resolution....

enumaris

lol

isn't that just saying mixing two different kinds of molecules together is just as hard to separate as mixing a homogenous gas together

and by hard to separate I mean the tendency of the gasses to separate spontaneously

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

not sure what he really meant

ACuriousMind

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP He's saying that the following two things are equally unlikely: 1. Two different gases spontaneously ("by virtue of their ordinary motions") separate into two separate gases from a mixture. 2. The particles of a single gas, as in the box in the paradox, spontaneously return to the half they were in before the separator was lifted.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

oh... thank you very much

you're right, now that I read your sentence and his, they match

ACuriousMind

As a German, I felt right at home reading that run-on sentence ;P

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

8:34 PM

:D

G. Bergeron

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP Indeed, this applies to a microwave oven! However, it's not really a question of emitting something, but rather of which modes are allowed in the cavity (ideal one here).

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

TM and TE modes are allowed, above a frequency cutoff

and any sum of these of course

I cannot form radio waves for instance, with those, as far as I know, because the wavelength is too big

the freq. is lower than the cutoff freq. so I cannot reconciliate the blackbody or grey body emission inside a cavity

it looks like the property of a body to be a black body depends on whether it is put inside a cavity or not. take a black hole, it would stop to emit Hawking's radiation as it does outside the microwave, it doesn't make any sense

ACuriousMind

8:51 PM

If you actually put the radiator inside the cavity, of course it will radiate the frequency! The presence of the cavity cannot alter the radiation the radiator emits, now that would be spooky action at a distance.

You will see very erratic behaviour, though: Since the frequency is not resonant, there'll be loads of constructive and destructive interference

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

I agree. hence there should be no limit of wavelength of photon despite a finite size of the universe

ACuriousMind

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP That's a spurious argument, anyway, even if there was some sort of limit in cavities: A real photon is not an infinitely extended wave, so the argument wouldn't hold anyway.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

ok. and with the expansion of the universe, im not sure it would make sense to speak about a wave of a well defined wavelength

Mithrandir24601

@ACuriousMind Have you ever heard of the Purcell effect?

May or may not be exactly what you're referring to, but it's related anyway

ACuriousMind

@Mithrandir24601 Well, yes, but in that case we're treating both the radiator and the cavity as a quantum system. In which case all the talk about "blackbodies" and so on become moot since that's a semi-classical concept, anyway. I feel this whole discussion is rather confused because @ofhe_iAgDWolbuuTZO_5X1L6uuwfVP hasn't really nailed down whether we're thinking classically or quantumly here.

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

8:58 PM

let's say classically first, we forget photons and we speak about EM waves

Mithrandir24601

@ACuriousMind Ah. Fair enough. I should probably read the whole conversation before randomly interrupting...

ofhe_iAgDWolbuuTZO_5X1L6uuwfVP

TM and TE modes are the only modes allowed and they have a cutoff. they form a basis from which we can form any wave with linear combinantion of those. problem is, the cutoff. I cannot form any wave in a cavity. but apparently I should be able to do that

ACuriousMind

@ofhe_iAgDWolbuuTZO_5X1L6uuwfVP In that case, it absolutely makes no difference whether you put an emitter inside a box or not

Transcript for