The h Bar

Sir Cumference

12:55 AM

@JohnRennie Here you go

Secret

2:05 AM

Very blurry, the photon ring looks more symmetric though

Schilive

2:54 AM

Hello, is there anyone?

Akash. B

3:25 AM

yeah

Leaky Nun

4:38 AM

wavelength of light used = 1.3 mm
diameter of telescope = diameter of earth = 12742 km
therefore resolving power = 21 micro-arc-seconds

angular size of M87 event horizon = 42 micro-arc-seconds

so it should have been much worse?

@JohnRennie

John Rennie

@LeakyNun bear in mind the ring is at about 2.6 times the Schwarzschild radius

Leaky Nun

ikr

so could you refine my calculations :P

and why does other sources say that the angular size of M87 is 7 arc-minutes?

like what is the actual "resolving power" of that huge "telescope"

John Rennie

4:55 AM

@LeakyNun Googling suggests that your figure for the angular resolution is about right.

Leaky Nun

then why aren't we only seeing 2 dots?

John Rennie

I can't comment as I haven't read any of the EHT papers.

Leaky Nun

> The EHT observations use a technique called very-long-baseline interferometry (VLBI) which synchronises telescope facilities around the world and exploits the rotation of our planet to form one huge, Earth-size telescope observing at a wavelength of 1.3mm. VLBI allows the EHT to achieve an angular resolution of 20 micro-arcseconds — enough to read a newspaper in New York from a café in Paris. Each telescope of the EHT produced enormous amounts of data – roughly 350 terabytes per day — which was stored on high-performance helium-filled hard drives. These data were flown to highly specialis…

@JohnRennie the scientist in me wants to see a picture of the moon produced by the EHT as control...

John Rennie

The pictures I've seen look reasonable. the ring would be about 100 uas in diameter and the resolution looks about 20% of the ring diameter. Well, maybe slightly better than that but not loads better.

Leaky Nun

5:10 AM

but like control is important

John Rennie

@LeakyNun I would guess the EHT team have independently checked the resolution :-)

Leaky Nun

not just the resolution, but the method itself

since they're using Fourier transforms and whatnot

John Rennie

The method is tried and tested. It has been used for decades. Since at least the mid 80s when I was a student.

The advance with the EHT is how far they've pushed the method, not anything fundamental.

Luyw

6:02 AM

Good morning, I am confused about the electric potential energy, when we put a test charge near an fixed charge, it will be subject to a force which will make it move, how is it that applying a force in the opposite direction would make it move the other way? Wouldn't it just cancel the acceleration which just implies the velocity is now constant?

Akash. B

hi

user8718165

6:22 AM

hello @JohnRennie

John Rennie

@Akash.B @user8718165 hi

user8718165

are you free now?

John Rennie

Yes, for about half an hour

user8718165

@JohnRennie I'm still badly confused about yesterday's problem

John Rennie

Remind me what that was ...

user8718165

6:30 AM

@JohnRennie chat.stackexchange.com/transcript/message/49876078#49876078 this one

John Rennie

Shall we move to the problem solving room ...

user8718165

@JohnRennie sure

Bernardo Meurer

Yo @JohnRennie are you happy with the black hole pic?

I bet that got you half-chubbed at least!

John Rennie

6:57 AM

@BernardoMeurer I guess half chubbed is what we Brits would call a semi? :-)

Luyw

7:12 AM

Good morning, I am confused about the electric potential energy, when we put a test charge near an fixed charge, it will be subject to a force which will make it move, how is it that applying the same force in the opposite direction would make it move the other way? Wouldn't it just cancel the acceleration which just implies the velocity is now constant?

Bernardo Meurer

@JohnRennie You got it :P

Tell the truth, from 0 to 1 how chubbed were you when you saw that amazing pic?

John Rennie

@Luyw if you apply an equal and opposite force on the charge then the net force will be zero so the charge won't move, or more precisely the acceleration of the charge will be zero.

@BernardoMeurer the funny thing is that I was slightly disappointed. It's an amazing technical achievement, but it looks just like a fuzzier version of what we already knew it should look like.

The black hole looked a lot sexier in Interstellar :-)

Bernardo Meurer

@JohnRennie I now understand it when people complain that porn creates unrealistic expectations of sex for people :P

The same thing happened to you and black holes!

kyle campbell

I'd like to know about what axis we are viewing the accretion disc on, which I haven't been able to find anywhere.

John Rennie

@BernardoMeurer So it's just a picture of a porn star then? :-)

2

Bernardo Meurer

7:16 AM

@JohnRennie The amazing thing, to me at least, is that it looks exactly like a potato-quality version of what we expected

@JohnRennie :O

black holes are porn stars!

John Rennie

@BernardoMeurer well yes, but then I don't think any physicists ever doubted it would look like the predictions. It's good that it has confirmed GR still works in the high curvature region near the event horizon because that had never been measured before. But at the end of the day it just proves what we all believed anyway.

As in it doesn't tell us anything new.

Bernardo Meurer

@JohnRennie I forget physicists are only excited when they are proven wrong :P

John Rennie

There will be lots of really detailed analysis to look for any deviations from GR, though it looks so far as if old Albert nailed it :-)

Bernardo Meurer

Damn Albert!

John Rennie

If you have five minutes spare there's a really good YouTube video explaining how the image is formed.

15 hours ago, by bolbteppa

How to Understand the Image of a Black Hole

►

Bernardo Meurer

7:20 AM

A bit of a shame Einstein, schwarzschild, and Hawking aren't around to see it :)

John Rennie

It's such a clear explanation

Bernardo Meurer

@JohnRennie Already watched it, super cool!

John Rennie

With gravitational waves LIGO also just proved something we all believed, but there was lots of new info e.g. no-one expected the masses of the merging holes to be so large.

Likewsie with the detection of the Higgs the Higgs mass turns out to be unexpectedly finely tuned. So again there was cool new stuff.

But with this measurement it's not clear to me what we've learned that we didn't already believe.

But I don't want to be a grinch - it's still an amazing technical achievement.

Leaky Nun

@kylecampbell the jet shoots almost directly towards us, and the jet is perpendicular to the accretion disc

> People are tired of seeing the image of a black hole within 24 hours of its debut, yet it is one of the peaks of modern space exploration.

John Rennie

@LeakyNun so what has EHT done for me lately? :-)

Leaky Nun

7:26 AM

what do you mean?

kyle campbell

Very nice

Leaky Nun

@JohnRennie where does the matter in the jet come from?

Bernardo Meurer

@LeakyNun Isn't it the gasses surrounding the hole?

Leaky Nun

what is the temperature of the matter in the accretion disc?

Bernardo Meurer

Hot is all I can say

Leaky Nun

7:28 AM

can we interpolate using redshift and the wavelength?

Bernardo Meurer

I do not know what tha means

Can you ask me some C99 questions so I feel smart instead?

Leaky Nun

well the wavelength of the light emitted depends on the temperature of the object

kyle campbell

Depends on the distance from the centre

Leaky Nun

but the light gets redshifted as it travels towards us

Bernardo Meurer

Oh, I see what you mean now!

@LeakyNun I'd guess so :)

Akash. B

7:42 AM

@BernardoMeurer what?

bolbteppa

8:35 AM

Scientists release first ever image of Brexit

GodotMisogi

Does anyone know how to obtain the maximum possible angular momenta for a given double pendulum configuration? I thought of taking Hamilton's equations and setting the $\dot{p_{\theta_i}}$ terms to $0$ and solving the equations, but they're coupled and nonlinear :/

bolbteppa

Thanks

Leaky Nun

@JohnRennie what happens if I shine a beam of light directly outward at the event horizon? where does the photon go?

John Rennie

@LeakyNun from inside the event horizon you mean?

Leaky Nun

no, at the event horizon

John Rennie

8:40 AM

Suppose you're hovering some small distance $d$ above the event horizon and you shine a laser outwards. You see the light speed head off outwards at $c$ as usual.

Leaky Nun

and if I'm $d$ below the event horizon?

John Rennie

Now, I'm far from the black hole watching you. In my frame you are experiencing gravitational time dilation, and that means in my frame the light moves slower than $c$. It's actually straightforward to calculate the speed of light in my frame and in fact I'm sure I've answered a question on exactly this subject.

Leaky Nun

wait, I thought light always moves at $c$

John Rennie

@LeakyNun in general relativity the local speed of light is always $c$, but away from your location the speed of light can be greater or less than $c$.

Leaky Nun

what

can I get a citation for this? :P

John Rennie

8:44 AM

@LeakyNun give me a moment and I'll find a link

Leaky Nun

thanks

John Rennie

14

Q: GR. Einstein's 1911 Paper: On the Influence of Gravitation on the Propagation of Light

Regarding the paper, what does Einstein means when he says: "If we call the velocity of light at the origin of co-ordinates $c_0$, then the velocity of light $c$ at a location with the gravitation potential $\Phi$ will be given by the relation: $c = c_0\cdot\left(1+\frac{\Phi}{c^2}\right).$ T...

See my answer to this question.

For a slightly more technical discussion see:

20

Q: Does light really travel more slowly near a massive body?

It is a routine problem for beginners in general relativity to calculate the coordinate velocity of light for the Schwarzschild metric. Starting from the metric: $$ ds^2 = -\left(1-\frac{r_s}{r}\right)c^2dt^2 + \frac{dr^2}{1-\frac{r_s}{r}} + d\Omega^2 $$ We use the fact that light travels on a ...

general-relativity black-holes speed-of-light coordinate-systems observers

Leaky Nun

so a photon at the event horizon should stay where it is?

John Rennie

@LeakyNun in whose coordinates?

Leaky Nun

I'm observing outside

John Rennie

8:53 AM

If you are outside the event horizon then in your coordinates a light beam emitted at the event horizon remains motionless.

Leaky Nun

that is very trippy

also I've been told that the event horizon is a fake singularity, i.e. a different coordinate should make it not a singularity?

and what does the photon see?

John Rennie

@LeakyNun correct. For example consider the rest frame of someone falling into the black hole. In their rest frame they are motionless in space and the black hole is accelerating towards them.

This observer sees an event horizon between them and the black hole, but as they approach the black hole the horizon retreats before them. The end result is that the falling observer never sees themselves cross the horizon.

So the two observers, the far away observer and the falling observer, see the horizon to be in different places.

Leaky Nun

I really learnt something new today

the speed of light is not constant :o

John Rennie

@LeakyNun careful!

What you are calling the speed of light is what relativists call a coordinate velocity, and by definition coordinate velocities depend on what coordinates you are using.

kyle campbell

that depends what you mean. the coordinate velocity of light is $>$ or $<$ but the local velocity is always $c.$

Leaky Nun

8:58 AM

@JohnRennie but the falling observer can always test if their out-going light beams reaches anywhere?

I think what I need is an animated penrose diagram of someone falling inside the black hole

all I can find online are penrose diagrams located outside the black hole

John Rennie

@LeakyNun yes, and if the outgoing beam hits some target then both observers will agree that the light beam has hit the target.

Leaky Nun

and eventually the singularity is in every direction of the observer?

that's when the observer crossed the event horizon?

John Rennie

In the frame of the free falling observer when they shine a light outwards at the event horizon the light moves away from them at $c$. But another way of interpreting this is that the light ray remains motionless and the observer is falling inwards at a speed $c$.

This is what we get if we use a choice of coordinates called the Gullstrand Painleve coordinates.

I answered a question on this ages ago. Let me have a look.

Here:

37

Q: Why is a black hole black?

In general relativity (ignoring Hawking radiation), why is a black hole black? Why nothing, not even light, can escape from inside a black hole? To make the question simpler, say, why is a Schwarzschild black hole black?

general-relativity gravity black-holes speed-of-light

Leaky Nun

@JohnRennie what is 2M?

and how is it consistent with the fact that the event horizon is always shrinking inwards for the free-falling observer?

John Rennie

@LeakyNun M is the mass of the black hole. Technically it's a geometric property called the ADM mass, but to keep things simple we just call it the mass of the black hole. This mass is an invariant i.e. all observers agree on its value.

Leaky Nun

9:10 AM

well you said event horizon is r=2M

are there sub-light-speed analogies that mortals can see?

like maybe with sound?

John Rennie

The event horizon radius is:

$$ r_s = \frac{2GM}{c^2} $$

But in GR it's common to use units in which $G = c = 1$. These are called geometrical units. When we use these units the equation for the event horizon radius simplifies to:

$$ r_s = 2M $$

Leaky Nun

so everyone agrees where the event horizon is?

John Rennie

Everyone agrees what $2M$ is, but only in the Schwarzschild coordinates doesthe event horizon exist at $2M$.

Leaky Nun

oh

does a photon think that the speed of light is c?

John Rennie

Photons do not have a rest frame, so that question has no meaning.

Leaky Nun

9:20 AM

what is a rest frame?

John Rennie

@LeakyNun for any observer the rest frame is the frame in which the observer is motionless at the orign.

Leaky Nun

why doesn't photon have a rest frame?

kyle campbell

@LeakyNun can a photon ever be at rest?

John Rennie

For example in my rest frame I am motionless at my desk, despite the fact that the surface of the Earth is rotating about over a thousand km per hour.

Leaky Nun

@kylecampbell at their own frame, lol

Student404Mus

9:22 AM

Hello,

can you help me to understand this expression
$\inf _ { p _ { i } , V _ { i } \} } \sum _ { i } p _ { i } c \left( \Psi _ { i } \right)$

since we didn't take the courses of convex analysis !

what does it mean to take the infimum of a number ?

John Rennie

One of the fundamental principles in relativity is that the local speed of light is always $c$. But in the rest frame of light the speed of light would have to be zero, because that's what a rest frame means. Therefore light cannot have a rest frame.

kyle campbell

actually, it's not even defined

Leaky Nun

what does the origin even mean?

kyle campbell

@Student404Mus let A denote a set. then the infimum of A is the greatest lower bound of all the elements of A, or more precisely, the greatest element of A that is less than or equal to all elements of A.

Leaky Nun

why can't we define the position of light to be origin?

Student404Mus

9:30 AM

all elements of A means the sum, right?

Leaky Nun

there's such a frame, we just can't get to it using any Lorentz transformation, right? @JohnRennie

Student404Mus

by the way, the full expression is this
$c ( \underline { \varrho } ) = \inf _ { \left\{ p _ { i } , v _ { i } \right\} } \sum _ { i } p _ { i } c \left( \Psi _ { i } \right) ,$ with $p _ { i } > 0 ,$ s.t. $\varrho = \sum _ { i } p _ { i } | \Psi _ { i } \rangle \left\langle \Psi _ { i } |\right.$

kyle campbell

it would denote the infimum of the sequences $p_i$ and $v_i$

Student404Mus

$p_i$ and $\Psi_i$. I didnt write it well

and because $p_i$ and $\Psi_i$ are probabilities and states respectively, how we can say some state is greater than an or less than another ?

John Rennie

@LeakyNun no there is no such frame

Leaky Nun

9:38 AM

what is a frame?

Student404Mus

@kylecampbell if we apply the definition you said, we say this infimum equals the greatest lower bound $p_i$ of all elements of this sequence. Right?

kyle campbell

no, $inf \{p_i,v_i\}$ means you would have to find the greatest lower bound of both sequences together, taken as one set.

Leaky Nun

what if I did a change of coordinates with x' = x+ct? @JohnRennie

Student404Mus

10:36 AM

@kylecampbell I watched some videos about infimum and supremum, but my expression doesn't seem to fit what I have learned. Do you know what has the term $\sum _ { i } p _ { i } c \left( \Psi _ { i } \right)$ to do with infimum? because this term equals a number!

kyle campbell

it doesn't, the infimum is a separate operation on the sequences.

Student404Mus

what if we consider an example, $c ( \underline { \varrho } ) = \inf _ { \left\{ p _ { i } , v _ { i } \right\} } 0.1(0.3)+0.4(0.6)+0.5(0.1)$, how we could then consider the infimum of this?

kyle campbell

do you have an expression for the sequences defined by $p_i$ and $v_i$?

what is this problem in the context of?

Student404Mus

one thing to mention is $p_i$ are probabilities that sum to unity and $v_i$ is $\Psi_i$ instead which are states

Luyw

hello, do we understand the origin of electric forces or are we still considering them to be natural laws?

Student404Mus

10:47 AM

the context is from quantum information

$\Psi_i$ are states or vectors

kyle campbell

right, but is this from a paper or a textbook?

Student404Mus

the framework is used in both of resources

it was mentioned in an article about the definition of convex roofs, "The proper, unambiguous generalization of a pure state monotone, that also we will use in
the following, therefore uses the infimum over all decompositions into pure
states - the so-called convex roof
$\mathcal { M } ( \varrho ) = \inf _ { \left\{ p _ { i } , \mathbf { V } _ { i } \right\} } \sum _ { i } p _ { i } \mathcal { M } \left( \Psi _ { i } \right) ,$ with $p _ { i } > 0 ,$ s.t. $\varrho = \sum _ { i } p _ { i } | \Psi _ { i } \rangle \left\langle \Psi _ { i } |\right.$"

it is mentioned in an article about the definition of convex roofs, "The proper, unambiguous generalization of a pure state monotone, that also we will use in
the following, therefore uses the infimum over all decompositions into pure
states - the so-called convex roof
$\mathcal { M } ( \varrho ) = \inf _ { \left\{ p _ { i } , \mathbf { \Psi } _ { i } \right\} } \sum _ { i } p _ { i } \mathcal { M } \left( \Psi _ { i } \right) ,$ with $p _ { i } > 0 ,$ s.t. $\varrho = \sum _ { i } p _ { i } | \Psi _ { i } \rangle \left\langle \Psi _ { i } |\right.$"

Slereah

Not the cauchy demons D:

kyle campbell

interesting. well, if my idea of the notation is correct, you would then need to apply the infimum to the sequences prior to taking the sum. so find the greatest lower bound of both $p_i$ and $\Psi_i$ and then take the sum.

Student404Mus

11:02 AM

@kylecampbell So we have two operations here, taking the infimum then we multiply it by the sum. Right?

kyle campbell

that's what I'm reading

Slereah

I like how he uses ð for demons

Eldritch symbols for spacetime demons

bolbteppa

Lets say you have a representation of a group and you know it's irreps, and then you find a different rep of that group - does this different rep always possess irreps that are isomorphic to the irreps of the first rep?

Student404Mus

@kylecampbell Thanks for your time and help

kyle campbell

no problem

John Rennie

12:32 PM

@TheEastWind I have not received an alert from signing in to my Google account from an incognito window. You might consider changing your password.

Leaky Nun

1:03 PM

@bolbteppa no: any rep can be broken down into (repeated) sums of irreps; and any sums of irreps is a rep

so if your group has irrep A and B, then A and B are counter-examples to your question

Lozansky

In general, if I have a polarized EM wave where the electrical field is written as $\mathbb{E} = A(\cos(u(t,\mathbb{r})) \hat{x} + \sin(u(t,\mathbb{r})) \hat{y})$, where $A$ is even in $z$, then the field is RCP for $z>0$ and LCP for $z<0$?

Kyle Kanos

1:36 PM

For a guy who spent ~8 years using Fortran for his MS & PhD degree, it feels weird having to give a talk on object oriented programming after working professionally with it for ~3 years.

Sir Cumference

@LeakyNun IIRC the worldline of light is parameterized by the eigenvectors of the Lorentz transform. Therefore it's the only part you can't move.

Though it does get scaled during a Lorentz transform iirc

Welp to be more specific light moves along the span of the eigenvectors

Leaky Nun

but there are also frames you can't access by doing Lorentz transforms?

Sir Cumference

Hmm? Mathematically anything beyond the worldline of light can be transformed to, but physically we are only concerned with ones within the light cone

The only problem is light, because its coordinates are eigenvectors, and therefore it can't be transformed to move along the vertical axis

bolbteppa

2:52 PM

@LeakyNun so if $G$ has irrep $A$ and a different non-isomorphic irrep $B$ then $A \oplus A$ is a rep of $G$ which does not contain $B$ as an irrep i.e. the rep $A \oplus A$ of $G$does not possess irreps isomorphic to $B$, right

Leaky Nun

right

bolbteppa

Great, thanks a lot

DanielSank

3:22 PM

I would like to know if anyone here is personally familiar with the project that produced the now-popular black hole image.

It seems to me that the attention being focused on one particular individual may be a case of hero worship, but before I form an opinion I'd like to know more.

@DavidZ you're in astro physics, right?

vzn

Einstein was RIGHT: How first ever black hole picture CONFIRMS 100-year-old cosmic theory / express UK

Black Holes Were Such an Extreme Concept, Even Einstein Had His Doubts / history channel

↑ lol!

EverydayFoolish

I have a question for everybody, a rather philosophical discussion.
Has theoretical physics overtaken experimental physics?

Does theory guide experiments or are we completely depend on experimental work, even today?

vzn

3:38 PM

@EverydayFoolish hi EF a recent book by Hossenfelder focuses on this question. its in line with 2 other books by Woit and Smolin critical of string theory over 1 decade old now.

Opartunity

Can i know..

vzn

This was humanity’s 236-year path to seeing a black hole / alternet

Opartunity

Why does red light travels far distances and not blue light . I know that the usual explanatuon is that red light has longer wavelength. Agreed...... but what if i say that blue light has got more enegy than red one. Shoudnt the light with more energy trael farther distance.

Leaky Nun

@Opartunity source for "red light travels far distances and not blue light"?

Opartunity

@LeakyNun I dont understand your point

Leaky Nun

3:52 PM

who told you "red light travels far distances and not blue light"?

Opartunity

So lets say in a foggy day.. Red light can be seen from farther distances.

Leaky Nun

because blue light is scattered by the atmosphere

vzn

in theory salon, 30 secs ago, by vzn

The creation of the algorithm that made the first black hole image possible was led by MIT grad student Katie Bouman / techcrunch

Opartunity

@LeakyNun I know but.. thats the point, why does it had to be scattered more. It has got more energy. So it can easily penetrate though atmospheric dust

Leaky Nun

4:09 PM

@Opartunity all I can find on the internet is people saying that others are wrong and only their theory is right :')

Opartunity

When did i say that othres are wrong..

Leaky Nun

that's not what I meant

Opartunity

@LeakyNun but isn't my thinking somewhat different. And moreover why is Energy of the photon not considered to explain it

Leaky Nun

@Semiclassical what do you think?

John Rennie

@Opartunity there are three different scattering regimes, Rayleigh scattering, Mie scattering and macroscopic scattering. Only Rayleigh scattering is strongly wavelength dependent.

Opartunity

4:19 PM

@JohnRennie is Rayleigh scattering dominant for visible part of the Em spectrum

John Rennie

Rayleigh scattering increases with decreasing wavelength (i.e. bluer light scattered more) because the primary mechanism is a dipole interaction and that is stronger with higher frequencies.

Opartunity

@JohnRennie thanks !! i got it

John Rennie

@Opartunity the scattering mechanism depends on the size of the particles doing the scattering. For particles smaller than the wavelength of the light Rayleigh scattering dominates. For particles of around the wavelength of the light it's Mie scattering and for large particles macroscopic scattering.

For example the blue of the sky is due to Rayleigh scattering by air molecules. The white of clouds is due to Mie and macroscopic scattering.

Clouds look white because Mie and macroscopic scattering aren't heavily wavelength dependent.

The sky looks blue because Rayleigh scattering is heavily wavelength dependent.

Leaky Nun

@JohnRennie could you elaborate on this?

also can we find a material that can produce a "reverse rainbow", i.e. in which red refracts more than blue?

John Rennie

@LeakyNun no, the rainbow is not due to scattering. It's due to refraction of light rays by raindrops. The refractive index of the water changes with wavelength, so the angle of refraction depends on the wavelength, and that's what causes the rainbow.

Leaky Nun

4:25 PM

oh I mean, rainbows produced by glass prism

in glass blue refracts more than red

experimental data for silica glass over a wide range of wavelengths (Kitamura 2007)

John Rennie

@LeakyNun yes, as a general rule for visible light the refractive index increases with decreasing wavelength.

Leaky Nun

it seems that "blue refracts more than red" is material-dependent

John Rennie

@LeakyNun the refractive index increases as you approach an absorption line i.e. a wavelength at which the energy of the photons matches a transition energy of the material.

Leaky Nun

Metamaterial prism creates a reverse rainbow

JMac

@LeakyNun A prisim produces a spectrum, not really a rainbow. Rainbows are called rainbows because of the bowing of the light

Leaky Nun

4:29 PM

@JMac let's not be pedantic

John Rennie

If you plotted the optical absorptance of the glass alongside your RI plot you'd see absorption lines at all the spikes in the RI plot.

Leaky Nun

why?

you mentioned dipoles; are they relevant?

John Rennie

There are loads of questions about that on the main site.

JMac

@LeakyNun They are different things. Rainbows have pretty specific properties because of how they are formed; they are a specific instance of a spectrum; if you just want to make a reverse spectrum that's probably far less complex

Leaky Nun

@JohnRennie I've been trying to find a question to no avail...

John Rennie

4:31 PM

10

Q: Why does the refractive index depend on wavelength?

Why do different wavelength get impeded more or less when in different materials? Moving with the same speed, but a longer physical distance would imply that the fields oscillate less times in the material, but I don't know why a difference in the number of oscillations would impede the wave- I d...

quantum-mechanics electromagnetism optics quantum-optics

39

Q: Why do prisms work (why is refraction frequency dependent)?

It is well known that a prism can "split light" by separating different frequencies of light: Many sources state that the reason this happens is that the index of refraction is different for different frequencies. This is known as dispersion. My question is about why dispersion exists. Is f...

electromagnetic-radiation waves frequency refraction dispersion

Leaky Nun

@JohnRennie do you have a demonstration of n<1?

John Rennie

@LeakyNun I can't think of one offhand.

Leaky Nun

ok this is very complicated :P

@JohnRennie what does the origin mean in coordinates of a frame?

aren't I allowed to use any coordinate system I like?

Abhas Kumar Sinha

@ACuriousMind @bolbteppa I've written an answer must be having a few errors rectify please: physics.stackexchange.com/a/472044/208739

John Rennie

@LeakyNun a frame basically just means a choice of coordinates. The origin can be anywhere.

Leaky Nun

4:42 PM

@JohnRennie then why isn't there a frame in which the photon is always at the origin?

John Rennie

@LeakyNun it's too late in the day for me to want to get into debates about frames

Leaky Nun

ok thanks

ACuriousMind

@LeakyNun A more careful statement would be: There is no inertial frame in which something that is moving at the speed of light in any other inertial frame is stationary.

The reason for this is deceptively simple: All inertial frames are connected to one another by Lorentz transformations, but the formula for Lorentz boosts into an inertial frame with velocity $v$ diverges as $v\to c$.

Leaky Nun

hey @ACuriousMind

ACuriousMind

Also, hey :)

Slereah

4:49 PM

Lorentz transforms don't form a group anymore if you try to include speeds of $c$!

Abhas Kumar Sinha

I've downloaded the Irodov's Relativistic Mechanics Part, seems quite well written, but I never read up to Lorentz Transformations...

Leaky Nun

@ACuriousMind so if I change the coordinates in some peculiar manner then I no longer get an inertial frame?

ACuriousMind

@LeakyNun Yes. What you get are null (or light-cone) coordinates in which two coordinates are null and two are space-like.

Leaky Nun

but in any frame the local velocity of light is c, as I've been told? @ACuriousMind

ACuriousMind

@LeakyNun The local velocity is still $c$, but in these coordinates the velocity is no longer a simple derivative of the space-like coordinate w.r.t. time-like coordinate, because there is no time-like coordinate!

Leaky Nun

4:57 PM

so I need to figure out the Lorentz metric?

and set it equal to zero?

ACuriousMind

@LeakyNun See e.g. physics.stackexchange.com/a/413077/50583 for an explicit formula for light-cone coordinates and the metric

Leaky Nun

$x^0$ is time-like?

ACuriousMind

Yes

Leaky Nun

> All the particles move forward with the time and all of them fall inside the light cone

what is "the time"

ACuriousMind

@LeakyNun There is no "the time". I suspect the author didn't mean anything with the definite article and simply meant any time-like coordinate.

Leaky Nun

5:09 PM

@ACuriousMind and what is "inside the light cone"?

ACuriousMind

@LeakyNun Do you know what a light cone is in general?

Leaky Nun

I think so

ACuriousMind

So what's unclear about what being "inside the cone" is?

Leaky Nun

because I don't know where the cone is in the new coordinate system

ACuriousMind

@LeakyNun The null coordinates are basically coordinates in which the cone is aligned with the axes. If you draw a typical 2d space-time diagram with $x$ and $t$ on the axes, then the light cone is the diagonals. The transformation to null coordinates is rotating the axes by 45°.

Leaky Nun

5:12 PM

ok thanks

but can we have a frame in which a particular photon is always at the origin?

ACuriousMind

Anyway, null coordinates are neat tools for some computations, but they don't really carry any specific meaning - they're not an inertial frame and don't represent the "viewpoint" of any observer.

@LeakyNun In a system such as null coordinates, it is unclear what "always" means since there is no time coordinate

Leaky Nun

let's say the worldline of this photon is the origin

ACuriousMind

@LeakyNun A point is not a line.

Leaky Nun

oh right...

@ACuriousMind but an observer can be accelerating

ACuriousMind

@LeakyNun Sure, but that doesn't make their coordinates null. There are interesting effects in accelerating frames but their natural coordinates are Rindler coordinates.

Leaky Nun

5:16 PM

I'm just saying that the viewpoint of an observer doesn't need to be inertial

ACuriousMind

@LeakyNun That's right. The 'and' in my sentence wasn't supposed to imply that.

Leaky Nun

so how do you find velocity?

ACuriousMind

What "velocity"?

Apart from the speed of light, velocity is a frame-dependent notion

Leaky Nun

yeah I mean how does velocity work in a frame

ACuriousMind

There is no unique notion of velocity. Our naive notion of velocity is the "coordinate velocity", which is simply the derivative of the space-like coordinates of the position of an object w.r.t. the time-like coordinate in that frame

Sir Cumference

5:29 PM

@LeakyNun Not sure if you're using a textbook for spec rel, but if not I highly recommend using one. The subject is really more clear when it's taught with structure, as opposed to asking specific questions. Once ya get a strong intuition on the basics, a lot more becomes clear.

Plus it only takes less than a month to get the same grasp on it as a typical physics undergrad.

ACuriousMind

But there's also e.g. proper velocity

Sir Cumference

My spec rel class went through the subject in 3 weeks (hell but doable)

ACuriousMind

And it is the local proper speed of light that is always $c$ even in non-inertial frames, not its coordinate speed.

Pernk Dernets

5:51 PM

Would you classify evaporative cooling as convection?

i meen, it's causing two liquids to mix..

JMac

@PernkDernets I wouldn't count it as the same thing. The evaporative cooling doesn't require you to mix fluids. You could form a vacuum and cause evaporation without mixture. Convection is usuallly extremely beneficial to evaporation cooling, as it usually helps reduce the partial pressure of the evaporating fluid, leading to easier evaporation, instead of just reaching equilibrium with surroundings

Pernk Dernets

It's a good point @JMac thank you

JMac

That's actually a kinda interesting question I had never really thought of, so thanks for asking

Pernk Dernets

Haha np. I was looking at the highest voted question here, about how to effectively cool down a cup of coffee and a friend and me vent through the different cooling mechanisms.

I'm a bit sad the question also adds blowing on the cup of coffee. I would love to see an analysis of it

JMac

@PernkDernets Blowing adds some pretty hard to account for variables. You could probably make blowing win hands down, especially if you were blowing a very high volume of extremely dry air, then evaporation would dominate with convection keeping the evaporation rate as high as possible. Removing blowing eliminates a lot of the harder to control variables

Pernk Dernets

6:11 PM

That's a good point :D

Curio

Why can't I use p=f/s to find the force that a particle of a gas applies to the surface?

I know the medium pression and the surface (face of a square), but the force isn't right

Anonymous

Interesting read: How weird was Gödel?

Anonymous

(Not sure whether everything in that answer is true, but this is one of those rare excellent Quora answers.)

Secret

hmm... I wonder if one can talk about the density of Cauchy demons...

say... a spacetime outside of the cauchy horizon that actually has fundemental constraints on the possible demon trajectories that are possible

Curio

A cube of 15cm length contains 300K Helium. What is the medium force applied by a molecule on the surface?

Secret

6:26 PM

Thus it does seems that, if one managed to produce an event that lies on the demon worldline, then it must be true that the whole demon worldline is produced

That seemed to be a generalisation of the concept of time loop creation in scifi stories, one either does not make the loop, or make it all at once, but can never made one single event of it because causality will not allow it

So it seems, in a spacetime beyond the cauchy horizon, potentially any event can be caused by a family of worldlines, most are cauchy demons, and which of these worldlines describes the history of events is in a sense unpredictable

But, can we take this to the extreme:

Picture a spacetime where global hyperbolicity breaks so badly such that any demon worldline that is produced, forms a dense set, and said demons together form an evolution of infinite multiplicity.

Then we have a scanerio where the trajectory of some observer within it is completely unpredictable in both the past and future light cones

I think the experience of this observer will be quite interesting, as they literally do not have a future nor past no matter how much the proper time is elapsed

almost like a perfect model of presentism

8 hours ago, by Slereah

hmm... perhaps, a more well behaved subset of these spacetimes where cauchy demons exists can be used to model back to the future time travel models, as one can describe every change of the past as a production or annihilation of cauchy demon and demon worldlines

PM 2Ring

10:36 PM

Is this comment too harsh? Or not harsh enough? ;)

This answer sounds highly speculative. Do you have any citations for the many non-mainstream claims you make here? — PM 2Ring 1 min ago

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