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00:00 - 05:0005:00 - 12:00

5:00 AM
@MAFIA36790 "Needy"...?
@SirCumference You don't need money?
@MAFIA36790 I don't know what I need.
I'm 18. No idea how the future's gonna play out.
My best advice is to join a gang in Mexico; it will be life changing ;)
@SirCumference Any eighteen-year-old who thinks differently is mistaken.
I don't think scholarships are easy to find; they are always rare.
5:02 AM
@MAFIA36790 It's not the scholarships that I've been worrying about, it's the jobs themselves.
But rob made a good point.
Dude!! Go back to study!!
You are a freshman ;(]
Why are you thinking of job now ?!
@MAFIA36790 Parents are worrying me.
@SirCumference They always worry.
But also just thinking about a future as an astronomer. Too much confusion...
@SirCumference At least complete your Bachelors first!!
Rest things later on ;/
5:04 AM
All right, if you say so...
Thanks tho
Morning @JohnRennie.
Morning all!
5:20 AM
Morning sir :-)
5:42 AM
@SirCumference See:
Q: Does any particle ever reach any singularity inside the black hole?

user1549I am not a professional physicist, so I may say something rubbish in here, but this question has always popped in my mind every time I read or hear anyone speak of particles hitting singularities and "weird things happen". Now to the question at hand, please follow my slow reasoning... As far a...

Although this has no really good answers. Even though the top answer is from a Nobel prize winner (!!) I don't think it really answers the question. My understanding is that you never cross the event horizon and never reach the singularity.
@JohnRennie Well, I've heard plenty of astronomers say that black holes grow as they consume matter. Doesn't that contradict the whole idea?
@SirCumference No, because the term black hole is used loosely by astronomers (and most of us). The objects described as black holes no not have an event horizon.
And indeed will never form one.
@JohnRennie How do you define "event horizon"?
The objects described as black holes (for example, Sagittarius A*) should indeed have strong enough gravity to prevent light from escaping, right?
@SirCumference No
@JohnRennie Then what gives? Why are they called "black holes"?
5:49 AM
@JohnRennie Do you mean "... as observed from the outside"?
Sag A* is an object that would inevitably form an event horizon if (a) we wait an infinite time and (b) Hawking radiation didn't exist. This is what the term black hole means in common usage.
@JohnRennie Yes, I think that's right.
I still don't follow. What would we see if we pointed a magic, incredibly powerful telescope at Sag A*'s location?
Most discussions of black hole dynamics are from the perspective of the material that falls into the hole.
You'd see the matter falling towards the surface where the event horizon would form if (a) we waited an infinite time and (b) Hawking radiation didn't exist.
5:52 AM
But those of us who live outside of the black hole see the infalling material redshifted, so that the infall process takes infinitely long
That light from that matter would be heavily red shifted but would still be visible in principle
And if you run that process backwards to include the formation of the black hole horizon, that's a process that would also take infinite time.
They would seem to be falling towards a seemingly arbitrary point?
And we could see the stars and material behind Sag A*?
@SirCumference I'm not sure what you're asking there. Do you mean we can see behind Sag A* because the light is lensed around it? We certainly can't see through Sag A* any more than we can see through any ball of matter like the Sun or Jupiter.
@JohnRennie Why's that?
5:55 AM
Why can't we see through matter?
If it doesn't have an event horizon, what's blocking the light?
The Sun doesn't have an event horizon, but we can't see through the Sun
Sag A* should be incredibly dense
@SirCumference No, Sag A* is a ball of matter slightly bigger than the event horizon radius.
@JohnRennie Huh?
5:57 AM
It's just a ball of matter
This is all new info to me, so bear my confusion
There's no horizon or singularity there (yet)
@JohnRennie Why not?
Because a horizon takes an infinite time to form
@JohnRennie All right, so let's assume I kept approaching Sag A*
Would I eventually reach its surface?
6:00 AM
Let's ignore the Hawking radiation for now. In that case an observer falling freely into the collapsing object reaches the singularity in a finite (usually quite short) time.
@JohnRennie Now let's take Hawking radiation into account. What would happen?
In fact that observer never crosses a horizon because in the observer's reference frame the horizon retreats before them. The observer only catches up with the horizon at the singularity.
However for the outside world the freely falling oberver never reaches the horizon.
Observers outside see the freely falling observer slow down asymptotically as they approach the horizon and never quite reach it.
And redshift as he does so, right?
That means the part of the observers trajectory after crossing the observer doesn't exist for external observers.
@SirCumference Yes
So if I were to look at Sag A*, I'd see a run of the mill huge mass, right?
6:03 AM
This has caused endless confusion for GR students in the poast
@SirCumference Yes
I guess I'd find out whatever Sag A* is made of
Anyway, could Sag A* ever compress smaller than its Schwarzschild radius?
@SirCumference it would be just plasma like any star.
@JohnRennie Just like an everyday, gigantic neutron star?
@SirCumference are we still ignoring Hawking radiation?
@JohnRennie For now, yes
6:05 AM
OK. In that case Sag A* would take an infinite time to reach the Schwarzschild radius, and after that it would fall through that radius and form a singularity.
For students the sticking point here is that this process takes greater than infinite time, and what does that mean?
@JohnRennie Does Hawking radiation change this timeline?
The answer is that in general for a curved manifold any coordinate system only covers part of the manifold and so there will be parts of the manifold that our $(t,x,y,z)$ coordinates do not cover.
So there is not contradiction in having events happen at greater than infinite time according to our clocks.
Okay, so say we have a quasi-star (a hypothetical early star with a black hole in its center). In reality, we'd see a tiny star-like object form inside the even grander star, right?
Yes, it would look like a neutron star embedded in the gaseous envelope
@JohnRennie Sounds like a TŻO to me...
6:09 AM
Thorne-Żytkow object (a neutron star embedded in a red giant)
Which actually do exist
Ugh...this is all pretty confusing...
Yes it would basically be a TZO. Though if we measured the density of the central object it would be greater than the density of a neutron star.
As would the envelope be different, since...
So in a realistic (and nowadays) sense, black holes are quite bright?
That depends on how much matter is falling into it.
I mean in the sense that it's basically a neutron star?
6:14 AM
Remember that the light from the surface is heavily red shifted
So we'd see a redder neutron star, in a sense?
So in the absense of any fresh matter falling in a black hole is indeed pretty black. Not completely black, but close to it.
@JohnRennie Uh, why?
Wouldn't the matter just be redshifting?
Because red shifting lowers the energy of the light.
Unless we're talking about extremely hot and bright matter from an accretion disc...
6:16 AM
yes, an accretion disk is far enough from the surface for the light to be only slightly red shifted so it's basically full intensity.
And the accretion disk is very hot!
@JohnRennie Yes, in fact quasi-stars are postulated to have almost all of their brightness come from the accretion disc of the central black hole...
Well, essentially all the brightness
Yes indeed
All right, so why isn't it more known among non-general relativists that no event horizons have ever formed?
That seems like important news...
6:19 AM
Because there's an (apparent) contradiction here that's hard to understand unless you're really comfortable with GR.
The infalling observer reaches the singularity ina finite (and short!) time
But for someone watching outside the infalling observer never even reaches the horizon, let alone the singularity
Well, that's just due to how light reaches the outside observer, right?
@SirCumference No. What I've described is genuinely the case.
It has nothing to do with how long light takes to travel to the external observer.
Now is it getting freaky :-)
@JohnRennie All right, I'm clearly confused. Then how about regarding the cosmic event horizon...?
We should see galaxies redshift into oblivion once the light cone meets the event horizon, right?
@JohnRennie Well, consider this: I have a system that will eventually form a black hole, and which a normal person (including @SirCumference up until fifteen minutes ago) would have already called a black hole.
@rob OK
6:22 AM
It's a collection of highly-redshifted matter with some apparent temperature, which is warmer than but probably close to the Hawking temperature.
I drop some hot stuff on it.
As the new material falls in, its blackbody radiation redshifts and it looks cooler.
Is the time required for my new material to reach thermal equilibrium with the "black hole" finite?
Are we including the effect of Hawking radiation?
OK in that case as the collapsing object ages and evaporates away it gets hotter because Hawking radiation increases with decreasing mass of the object.
So your infalling hot matter is getting colder and the evaporating black hole is getting hotter and their temperatures will become equal in a finite time.
6:26 AM
@JohnRennie Is this true in a Universe with a finite-temperature CMB?
Currently our Universe is warm enough that, in the exchange between Hawking radiation and CMB, stellar-mass black holes are getting larger rather than smaller.
The black hole won't evaporate if the Hawking temperature is less than the temperature of the CMB, so you'll have to wait until the expansion of the universe has cooled the CMB. That would be a long wait, but still a finite one.
I have to work now for about half an hour - just when it was getting interesting as well :-(
Fair enough. Will my infalling blob ever equilibrate if we don't include Hawking radiation, or will it always be "catching up" to the material that's already near the horizon?
@JohnRennie Okay, cheers
@rob without Hawking radiation the infalling object will never equilibrate as observed by us
Can someone answer this? "Two ions have equal masses but one is singly-ionized and other is tripply-ionized. They are projected from
the same place in a uniform magnetic field with the same velocity perpendicular to the field. Do the two circles touch? Or are the circle distorted due to repulsion? "
6:39 AM
@S007 What do you think?
i think it should be distorted
What's your reasoning?
same charge
type of charge
What would be your reasoning for allowing the two orbits to touch?
I think they should not touch....they should just move far apart as soon as they are projected from the same place
6:43 AM
There's your answer, then.
@rob but do they follow any specific geometric pattern after that?
i guess so
btw thanks
@S007 I think so too ... what's your prediction?
prediction? maybe a tilted elongated circle type of thing
its complicated to imagine
maybe it can be mathematically calculated
the equation of trajectory
@S007 Hmmm. "Tilted" suggests that the orbits would leave the plane perpendicular to the magnetic field, which they can't.
tilted may not be perpendicular
6:48 AM
I'd expect the long-term solution would have the high-charge ion on its small-radius orbit, inside the larger-radius orbit of the less-charged ion.
But in real life they'd always have some out-of-plane repulsion, and move away from each other on helixes.
i guess so @rob anyway thank you
gotta go
@S007 Cheers
I'm back. Are we bored with black holes yet? :-)
@JohnRennie The subject has cooled but will heat up again as the Universe expands
@rob What I slyly didn't mention is that because of dark energy the universe will eventually develop a de Sitter horizon and that has a Hawking temperature of its own. So there is a minimum temp below which the CMB will never cool.
However that temperature is lower than the temperature of any black hole inside it, so those black holes will still eventually evaporate.
7:02 AM
@JohnRennie Nope
This de Sitter temperature is for an empty universe with "our" dark energy density, which is how things will look after the expansion has run away for a while and all the matter is beyond the horizon?
@rob Yes. Our universe will tend asymptotically to a de Sitter geometry but never quite get there. And the de Sitter horizon will also take an infinite time to form! Those pesky horizons!
What kind of objects emit Hawking radiation? Is it just everyday blackbody radiation?
@SirCumference That's a far more complicated question that you probably realise because it depends on what you mean by Hawking radiation.
@JohnRennie That's what's confusing me. You're applying "Hawking radiation" to objects like Sag A*
7:04 AM
When we say an object emits Hawking radiation we mean an observer at infinity i.e. an infinite distance away from the object sees a non-zero radiation flux.
When Hawking first did his calculation this required the presence of an event horizon - and I've just said these don't exist.
But Hawking was using a simplified calculation because that was the only way to handle the maths involved.
To be honest I'm not sure exactly what the criterion is, but we do know that objects like Sag A* do emit Hawking radiation and will eventually evaporate even though they contain not true event horizon.
The obvious question is whether an object like the Earth emits Hawking radiation, in principle at least. And I don't know the answer to that.
Well, if an external observer sees a black hole as a collection of hot plasma just larger than the Schwarzchild radius, you don't need a new mechanism to see radiation at the far field.
You just have ordinary blackbody radiation, redshifted.
Yes, but that's just regular black body radiation that would, given infinite time, go to zero.
@JohnRennie Do you mean "come into equilibrium with the CMB"?
@rob :-) Yes, though the CMB will also eventually go to zero
(it won't really but let's gloss over that :-)
It's glossing over things that's causing all this damned trouble
7:12 AM
My problem is that while I've managed to develop an intutive feel for standard GR I've failed to do this for Hawking radiation. Probably because I still have no feel for quantum field theory.
So my only recourse is the maths, and that's over my head. Under the circumstances I'm stumbling around in the dark to a large extent.
So it's the blind leading the blind :-)
My approach in those cases is to start with thermodynamics.
Anytime you think you're going to hit zero temperature, an interesting conspiracy occurs.
7:26 AM
Sigh...it's 3:26am...
7:56 AM
Plot twist: The conference itself is a scam source
Hopefully real journals will not be that stupid
8:23 AM
@Secret Interesting. I have met some of the people on the conference organizing committee, but don't know any of them well.
@Secret Note that the APS has a policy of accepting all conference presentation abstracts; the incomprehensible ones tend to end up in their own session, and any of those presenters who attend mostly just talk to each other.
Hmm, so the system used by APS kinda naturally select those guys out
Most of them don't show --- they just want their "publication" in the Bulletin of the APS, which is the fancy name for the "journal" that prints all the conference programs.
8:45 AM
I've a quick question about permittivity.
If the absolute permittivity is related to the permittivity of free space as given by:
$\epsilon_r=\epsilon/\epsilon_o$, where $\epsilon_o$ is the permittivity of free space, $\epsilon_r$ is the relative permittivity of the material and $\epsilon$ is the absolute permittivity of the material.
When stated that the "permittivity" of metals is infinite, they mean the absolute permittivity, right?
I'm unable to make much sense of that ^ relationship.
Aah @JohnRennie: I've moved on to electrodynamics! Will u be able to help with this too?
Oh, also, given that the value of $\epsilon_r$ is constant for a given medium, seriously, what is the point of that relationship? All it does, is relate a bunch of constants.
9:04 AM
Electrodynamics is one of the areas of physics that never interested me so I quickly forgot everything I learned about it. So I doubt I'll be much help. However I recall that optics can get complicated with materials that absorb or like metal behave as a free electron gas.
@Kaumudi $\epsilon_0$ is a fundamental property of free space. We relate the permittivity of any material to the permittivity of free space using the relative permittivity $\epsilon_r$.
Ohh :-( Damn.
I don't think the permittivity of metals is infinite. As I recall the permittivity of silver is a complex number.
@JohnRennie Yeah, I see that we can use this relation for different media, and write something like $\epsilon_{r1}\epsilon_1=\epsilon_{r2}\epsilon_2$
@JohnRennie It's a complex number?! U mean, it's unreal?!
Well, my textbook says that the relative permittivity of metals is infinite.
I'm grasping at vague memories here, so this may be complete rubbish. But a light wave incident on silver does not propagate into the silver as a wave. I think it forms an evanescent wave so it's not oscillating and falling exponentially with distance.
Wait, why are we talking about light waves, here?
9:09 AM
When we write a refractive index we're comparing an incident wave to a refarcted wave. However when the refracted wave is an evanescent wave that only works if the refractive index is complex ... or something like that.
Yes, light waves.
But why?! I'm talking about the electric permittivity...
In electromagnetics, an evanescent field, or evanescent wave, is an oscillating electric and/or magnetic field which does not propagate as an electromagnetic wave but whose energy is spatially concentrated in the vicinity of the source (oscillating charges and currents). Even when there in fact is an electromagnetic wave produced (e.g., by a transmitting antenna) one can still identify as an evanescent field the component of the electric or magnetic field that cannot be attributed to the propagating wave observed at a distance of many wavelengths (such as the far field of a transmitting antenna...
Oh, wow. OK, definitely not in my syllabus.
But look:
(At the one answer to the question)
I don't think that's true. I had to do a bit of work with silver as part of my PhD and part of that involved it's optical properties. My recollection is that we write the permittivity as $\epsilon = \epsilon' +i\epsilon''$ i.e. as a complex number.
In fact, some Googling just found this:
which gives the values of $\epsilon'$ and $\epsilon''$ for silver.
I don't understand (and probably don't need to) the relationship b/w all this and light.
9:16 AM
Because $\epsilon_r = n^2$ where $n$ is the refractive index.
Google, Google, aha see this article
But I do think that I'm supposed to take it as infinity :/ My textbook says this: "For the material in which more charge can be induced, $\epsilon_r$ will be higher.
But this is all stuff you wouldn't cover until your degree. I'm surprised if you need this for the university entrance exam.
@JohnRennie Yeah, no, I don't.
Then I'd be cautious about stepping into what may be very deep waters :-)
@Kaumudi They are very much related.
@JohnRennie, so you are talking about complex index of refraction.
9:20 AM
@MAFIA36790 yes
Ah, yes, yes! I actually do have that relationship in my syllabus. Only, it's presented differently.
^That's everything related to vacuum.
Some Googling and racking of brain cells later, the refractive index of a metal like silver is complex and given by $n = n_1 + ik$ where $k$ is the extinction coefficient and $n_1$ is confusuingly referred to as the refractive index.
If you look at the link I posted then for silver $n_1=0.15$ and $k=3.47$
For a given material, this is given as $c=1/(\sqrt{\mu\epsilon})$
With these two, we can arrive at this:
11 mins ago, by John Rennie
Because $\epsilon_r = n^2$ where $n$ is the refractive index.
OK, anyway, @JohnRennie: No, no complex refractive indices for me now, thanks! :-P
Yes, and if the refractive index is complex then so is $\epsilon_r$.
Ahh, yes...
9:28 AM
Note that the relative permittivity is often called the dielectric constant.
Just to be helpful :-)
Yes :-)
Oh, BTW, dyou know what my textbook is saying:
13 mins ago, by Kaumudi
But I do think that I'm supposed to take it as infinity :/ My textbook says this: "For the material in which more charge can be induced, $\epsilon_r$ will be higher.
That's misleading. It's the polarisibility of the material that is the key factor.
When light is travelling through a dielectric the electric field of the light can polarise the dielectric i.e. induce a charge separation in it. And it's this polarisation that causes the refractive index to be greater than one.
But I still don't get it. If the molecules constituting the material become polarized, it's not like the two poles move far away from each other, right?
9:33 AM
The theory behind it is quite hard and you won't do it until your degree. Basically the oscillating field of the light causes an induced oscillating dipole in the dielectric.
What do they mean by "charge induced"?
They don't mean charge given by induction?
But oscillating dipoles radiate, so the induced dipoles emit radiation that interferes with the original light. The end result of this is that the speed at which the wave propagates is reduced.
They must mean more charge separation can be induced
Actually the Wikipedia article covers this quite nicely:
In electromagnetism, permittivity or absolute permittivity is the measure of resistance that is encountered when forming an electric field in a medium. In other words, permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium. The permittivity of a medium describes how much electric field (more correctly, flux) is 'generated' per unit charge in that medium. More electric flux exists in a medium with a low permittivity (per unit charge) because of polarization effects. Permittivity is directly related to electric susceptibility, which is a measure of how...
Omg omg, no! Our syllabus has nothing to do with light+electrostatics!
Yes, I did read that :/ But my textbook says something else so I was confused.
Or maybe I just didn't understand what my textbook was trying to say.
The textbook isn't exactly wrong, but it's misleading.
I suspect they're trying to keep it simple and that means being a bit misleading.
What about metals, then?
9:37 AM
What about them?
They don't have these induced dipoles and all, correct?
Metals behave as if they contain a free electron gas i.e. the electrons can respond to the electric field of the light with no obstructions.
So metals are very very polarisible
But if you're asking me how exactly this corresponds to their optical properties then I'm afraiud i've long ago forgotten.
When an electric field is applied, the electrons align themselves in the direction of the electric field...and hang on, I read somewhere that the electric field produced by the alignment of these electrons completely cancel the outside electric field...how does that work?!
@JohnRennie No, Sir, let's forget about optics!
OK, I'm having a breakdown. I'll do some Math for awhile and then I'll come back.
In the static, i.e. constant field case, the net field inside a conductor (e.g. metal) is zero because the electrons flow so they cancel out any external field.
But electrons have a mass so they can't move infinitely fast.
So for an oscillating field there isn't complete cancellation.
@JohnRennie I have searched everywhere including this site; I found an answer on this site but I still do not understand; how is the formula for Planck length derived?
9:41 AM
BTW sir, in my course, we don't have to learn how to relate optics and electrostatics. So every time I ask u a question about one, it is safe to not at all mention the other.
@JohnRennie Huh, OK...
@DHMO the Planck length is one of the Planck units.
In particle physics and physical cosmology, Planck units are a set of units of measurement defined exclusively in terms of five universal physical constants, in such a manner that these five physical constants take on the numerical value of 1 when expressed in terms of these units. Originally proposed in 1899 by German physicist Max Planck, these units are also known as natural units because the origin of their definition comes only from properties of nature and not from any human construct. Planck units are only one system of several systems of natural units, but Planck units are not based on...
@JohnRennie isn't this an impact rather than a cause?
If we take the three constants $\hbar$, $G$ and $c$ we can use them to define three constants with the dimensions of length, time and mass, and these are the Planck units.
I'm not sure what you mean by an impact
I mean, after we know Planck length, we derived the Planck units?
No, starting with $\hbar$, $G$ and $c$ there is only one way to end up with units of length, time and mass, and that gives the Planck length, the Planck time and the Planck mass.
9:46 AM
like, how do we know that GR fails beyond Planck length?
@DHMO We don't, we have no theory of quantum gravity.
Oh, @DHMO: Dyou think u could maybe help me with some geometry?
@JohnRennie I am not convinced by dimension analysis
@JohnRennie how do we know that we need quantum gravity after that link?
@Kaumudi yes
MSE chat please?
9:47 AM
I'll link u to the question I posted over there yesterday.
@DHMO for any object smaller than a Planck length the uncertainty priciple suggests the energy of that object will so so high it must form a black hole. So that means our tehories become suspect for distances shorter than a Planck length.
Though we don't know exactly what happens and how.
Anyway, I have to go and make the bed (it's all action here this morning! :-). Back in a few minutes.
@JohnRennie why does a blackhole implies the failure of our theories?
10:46 AM
Impulse is in the direction of force, true or false?
@SwapnilDas Definition of impulse?
@SwapnilDas True.
@rob Change in momentum.
@SwapnilDas Newton's second law makes a statement about change in momentum.
@Kaumudi It is so astonishing in physics that even small concepts matter, thanks!
10:48 AM
U.welc. :-)
Does it mention direction, I don't think so? @rob
@SwapnilDas Newton's second law is a vector equation.
I took carelessness to the next level :P
Happens to all of us.
10:51 AM
@rob Are you a physicist?
@SwapnilDas I am.
Wow! what's your field of research?
@SwapnilDas $\vec{F} = t \Delta\vec{p}$
Symmetries, weak interactions, precision measurements.
Yup :)
10:54 AM
since $t$ is a scalar, it must follow that $\vec{F} \parallel \Delta\vec{p}$
11:28 AM
@heather greetings
@DHMO, I have a bit of a question. I'm trying to find the domain of the function $V(x) = \ln (\ln x)$. No complex numbers are allowed. I know that for $\ln x$ the domain would be all numbers $>0$ but I know that this has to be true again as those numbers are put into another logarithm. So I don't quite know how to confirm that. (It probably doesn't help that I'm not super familiar with logarithms or what they output.)
yes, so lnx needs to be >0 also
wait could you say $\ln^2x>0$?
ambiguous notation
11:41 AM
but no, ln(ln(x)) can have any value
ln(ln(x)) can be -1
the answer to it? oh, yeah that makes sense
domain and range
11:44 AM
hmm, would the domain just be x>0?
not here
I don't know, then. How would you figure out the domain here?
you know that the input to ln must be >0
here, the input is ln(x)
so ln(x)>0
11:47 AM
wait, okay, new question: what are you doing to the left and right side? to get rid of the $\ln$?
we raise e to the power of it
e^(lnx) > e^0
x > 1
wait, don't you multiply both sides by $e^x$? I thought that was the inverse of the natural logarithm.
no, inverse means you get back the input if you apply the function
so the inverse here is actually raising e to the power of the expression on both sides, leaving it with $e^{\ln x}>1$. Okay, then how do you simplify that?
oh, wait, duh.
e^(ln(x)) is just x
11:54 AM
because e to a natural logarithm...right. Okay, so the domain is $\{x\in\mathbb{R}|x>1\}$.
Thanks so much for putting up with my not-getting-it-ness. =)
you are welcome
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