The h Bar

Sir Cumference

5:00 AM

@MAFIA36790 "Needy"...?

user116211

@SirCumference You don't need money?

Sir Cumference

@MAFIA36790 I don't know what I need.

I'm 18. No idea how the future's gonna play out.

user116211

My best advice is to join a gang in Mexico; it will be life changing ;)

rob

@SirCumference Any eighteen-year-old who thinks differently is mistaken.

user116211

I don't think scholarships are easy to find; they are always rare.

Sir Cumference

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.

user116211

Dude!! Go back to study!!

user116211

You are a freshman ;(]

user116211

Why are you thinking of job now ?!

user116211

Ridiculous.

Sir Cumference

@MAFIA36790 Parents are worrying me.

user116211

5:03 AM

@SirCumference They always worry.

Sir Cumference

But also just thinking about a future as an astronomer. Too much confusion...

user116211

@SirCumference At least complete your Bachelors first!!

user116211

Rest things later on ;/

Sir Cumference

All right, if you say so...

Thanks tho

user116211

Morning @JohnRennie.

John Rennie

5:16 AM

Morning all!

user228700

Morning sir :-)

John Rennie

5:42 AM

@SirCumference See:

27

Q: Does any particle ever reach any singularity inside the black hole?

I 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.

2

Sir Cumference

@JohnRennie Well, I've heard plenty of astronomers say that black holes grow as they consume matter. Doesn't that contradict the whole idea?

John Rennie

@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.

Sir Cumference

@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?

John Rennie

I'm using the usual GR definition

@SirCumference No

Sir Cumference

@JohnRennie Then what gives? Why are they called "black holes"?

rob

5:49 AM

@JohnRennie Do you mean "... as observed from the outside"?

John Rennie

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.

rob

@JohnRennie Yes, I think that's right.

Sir Cumference

I still don't follow. What would we see if we pointed a magic, incredibly powerful telescope at Sag A*'s location?

rob

Most discussions of black hole dynamics are from the perspective of the material that falls into the hole.

John Rennie

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.

rob

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

John Rennie

That light from that matter would be heavily red shifted but would still be visible in principle

rob

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.

Yes?

Sir Cumference

They would seem to be falling towards a seemingly arbitrary point?

And we could see the stars and material behind Sag A*?

John Rennie

@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.

Sir Cumference

@JohnRennie Why's that?

John Rennie

5:55 AM

Why can't we see through matter?

Sir Cumference

If it doesn't have an event horizon, what's blocking the light?

John Rennie

The Sun doesn't have an event horizon, but we can't see through the Sun

Sir Cumference

Sag A* should be incredibly dense

John Rennie

@SirCumference No, Sag A* is a ball of matter slightly bigger than the event horizon radius.

Sir Cumference

@JohnRennie Huh?

John Rennie

5:57 AM

It's just a ball of matter

Sir Cumference

This is all new info to me, so bear my confusion

John Rennie

There's no horizon or singularity there (yet)

Sir Cumference

@JohnRennie Why not?

John Rennie

Because a horizon takes an infinite time to form

Sir Cumference

@JohnRennie All right, so let's assume I kept approaching Sag A*

Would I eventually reach its surface?

John Rennie

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.

Sir Cumference

@JohnRennie Now let's take Hawking radiation into account. What would happen?

John Rennie

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.

Sir Cumference

And redshift as he does so, right?

John Rennie

That means the part of the observers trajectory after crossing the observer doesn't exist for external observers.

@SirCumference Yes

Sir Cumference

So if I were to look at Sag A*, I'd see a run of the mill huge mass, right?

John Rennie

6:03 AM

This has caused endless confusion for GR students in the poast

@SirCumference Yes

Sir Cumference

I guess I'd find out whatever Sag A* is made of

Anyway, could Sag A* ever compress smaller than its Schwarzschild radius?

John Rennie

@SirCumference it would be just plasma like any star.

Sir Cumference

@JohnRennie Just like an everyday, gigantic neutron star?

John Rennie

@SirCumference are we still ignoring Hawking radiation?

Sir Cumference

@JohnRennie For now, yes

John Rennie

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?

rob

@JohnRennie Does Hawking radiation change this timeline?

John Rennie

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.

Sir Cumference

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?

John Rennie

Yes, it would look like a neutron star embedded in the gaseous envelope

Sir Cumference

@JohnRennie Sounds like a TŻO to me...

John Rennie

6:09 AM

TZO?

Sir Cumference

Thorne-Żytkow object (a neutron star embedded in a red giant)

Which actually do exist

Ugh...this is all pretty confusing...

John Rennie

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.

Sir Cumference

As would the envelope be different, since...

So in a realistic (and nowadays) sense, black holes are quite bright?

John Rennie

That depends on how much matter is falling into it.

Sir Cumference

I mean in the sense that it's basically a neutron star?

John Rennie

6:14 AM

Remember that the light from the surface is heavily red shifted

Sir Cumference

So we'd see a redder neutron star, in a sense?

John Rennie

So in the absense of any fresh matter falling in a black hole is indeed pretty black. Not completely black, but close to it.

Sir Cumference

@JohnRennie Uh, why?

Wouldn't the matter just be redshifting?

John Rennie

Because red shifting lowers the energy of the light.

Sir Cumference

Unless we're talking about extremely hot and bright matter from an accretion disc...

John Rennie

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!

Sir Cumference

@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...

John Rennie

Yes

Sir Cumference

Well, essentially all the brightness

John Rennie

Yes indeed

Sir Cumference

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...

John Rennie

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

Sir Cumference

Well, that's just due to how light reaches the outside observer, right?

John Rennie

@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 :-)

Sir Cumference

@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?

rob

@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.

John Rennie

@rob OK

rob

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.

John Rennie

OK

rob

Is the time required for my new material to reach thermal equilibrium with the "black hole" finite?

John Rennie

Are we including the effect of Hawking radiation?

rob

Sure

John Rennie

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.

rob

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.

John Rennie

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 :-(

rob

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

Sir Cumference

Oye...

John Rennie

@rob without Hawking radiation the infalling object will never equilibrate as observed by us

Anonymous

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? "

rob

6:39 AM

@S007 What do you think?

Anonymous

i think it should be distorted

rob

What's your reasoning?

Anonymous

repulsion

Anonymous

same charge

Anonymous

type of charge

rob

6:41 AM

What would be your reasoning for allowing the two orbits to touch?

Anonymous

I think they should not touch....they should just move far apart as soon as they are projected from the same place

rob

There's your answer, then.

Anonymous

@rob but do they follow any specific geometric pattern after that?

Anonymous

i guess so

Anonymous

btw thanks

rob

6:44 AM

@S007 I think so too ... what's your prediction?

Anonymous

prediction? maybe a tilted elongated circle type of thing

Anonymous

its complicated to imagine

Anonymous

maybe it can be mathematically calculated

Anonymous

the equation of trajectory

rob

@S007 Hmmm. "Tilted" suggests that the orbits would leave the plane perpendicular to the magnetic field, which they can't.

Anonymous

6:47 AM

tilted may not be perpendicular

rob

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.

Anonymous

i guess so @rob anyway thank you

Anonymous

gotta go

Anonymous

bbye

rob

@S007 Cheers

John Rennie

6:54 AM

I'm back. Are we bored with black holes yet? :-)

rob

@JohnRennie The subject has cooled but will heat up again as the Universe expands

John Rennie

@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.

Sir Cumference

@JohnRennie Nope

rob

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?

John Rennie

@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!

Sir Cumference

7:03 AM

What kind of objects emit Hawking radiation? Is it just everyday blackbody radiation?

John Rennie

@SirCumference That's a far more complicated question that you probably realise because it depends on what you mean by Hawking radiation.

Sir Cumference

@JohnRennie That's what's confusing me. You're applying "Hawking radiation" to objects like Sag A*

John Rennie

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.

rob

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.

John Rennie

Yes, but that's just regular black body radiation that would, given infinite time, go to zero.

rob

7:10 AM

@JohnRennie Do you mean "come into equilibrium with the CMB"?

John Rennie

@rob :-) Yes, though the CMB will also eventually go to zero

(it won't really but let's gloss over that :-)

rob

It's glossing over things that's causing all this damned trouble

John Rennie

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 :-)

rob

My approach in those cases is to start with thermodynamics.

Anytime you think you're going to hit zero temperature, an interesting conspiracy occurs.

Sir Cumference

7:26 AM

Sigh...it's 3:26am...

Secret

7:56 AM

theguardian.com/science/2016/oct/22/…

Plot twist: The conference itself is a scam source

Hopefully real journals will not be that stupid

rob

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.

Secret

Hmm, so the system used by APS kinda naturally select those guys out

rob

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.

user228700

8:45 AM

I've a quick question about permittivity.

user228700

If the absolute permittivity is related to the permittivity of free space as given by:

user228700

$\epsilon_r=\epsilon/\epsilon_o$, 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.

user228700

When stated that the "permittivity" of metals is infinite, they mean the absolute permittivity, right?

user228700

I'm unable to make much sense of that ^ relationship.

user228700

9:01 AM

Aah @JohnRennie: I've moved on to electrodynamics! Will u be able to help with this too?

user228700

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.

John Rennie

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 $\epsilon_r$.

user228700

Ohh :-( Damn.

John Rennie

I don't think the permittivity of metals is infinite. As I recall the permittivity of silver is a complex number.

user228700

@JohnRennie Yeah, I see that we can use this relation for different media, and write something like $\epsilon_{r1}\epsilon_1=\epsilon_{r2}\epsilon_2$

user228700

9:07 AM

@JohnRennie It's a complex number?! U mean, it's unreal?!

user228700

Well, my textbook says that the relative permittivity of metals is infinite.

John Rennie

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.

user228700

Wait, why are we talking about light waves, here?

John Rennie

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.

user228700

But why?! I'm talking about the electric permittivity...

John Rennie

9:10 AM

Evanescent field

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...

user228700

Oh, wow. OK, definitely not in my syllabus.

user228700

But look:

user228700

physics.stackexchange.com/questions/93542/…

user228700

(At the one answer to the question)

John Rennie

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:

refractiveindex.info/?shelf=main&book=Ag&page=Rakic

which gives the values of $\epsilon'$ and $\epsilon''$ for silver.

user228700

9:15 AM

I don't understand (and probably don't need to) the relationship b/w all this and light.

John Rennie

Because $\epsilon_r = n^2$ where $n$ is the refractive index.

Google, Google, aha see this article

user228700

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.

John Rennie

But this is all stuff you wouldn't cover until your degree. I'm surprised if you need this for the university entrance exam.

user228700

@JohnRennie Yeah, no, I don't.

John Rennie

Then I'd be cautious about stepping into what may be very deep waters :-)

user116211

9:19 AM

@Kaumudi They are very much related.

user116211

@JohnRennie, so you are talking about complex index of refraction.

John Rennie

@MAFIA36790 yes

user228700

Ah, yes, yes! I actually do have that relationship in my syllabus. Only, it's presented differently.

user228700

$c_o=1/(\sqrt{\mu_o\epsilon_o})$

user228700

^That's everything related to vacuum.

John Rennie

9:26 AM

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 $k=3.47$

user228700

For a given material, this is given as $c=1/(\sqrt{\mu\epsilon})$

user228700

With these two, we can arrive at this:

user228700

11 mins ago, by John Rennie

Because $\epsilon_r = n^2$ where $n$ is the refractive index.

user228700

OK, anyway, @JohnRennie: No, no complex refractive indices for me now, thanks! :-P

John Rennie

Yes, and if the refractive index is complex then so is $\epsilon_r$.

user228700

9:28 AM

Ahh, yes...

John Rennie

Note that the relative permittivity is often called the dielectric constant.

Just to be helpful :-)

user228700

Yes :-)

user228700

Oh, BTW, dyou know what my textbook is saying:

user228700

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.

John Rennie

That's misleading. It's the polarisibility of the material that is the key factor.

user228700

9:31 AM

Hm...

John Rennie

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.

user228700

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?

user228700

Wait.

John Rennie

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.

user228700

What do they mean by "charge induced"?

user228700

9:34 AM

They don't mean charge given by induction?

John Rennie

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:

Permittivity

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...

user228700

Omg omg, no! Our syllabus has nothing to do with light+electrostatics!

user228700

Yes, I did read that :/ But my textbook says something else so I was confused.

user228700

Or maybe I just didn't understand what my textbook was trying to say.

John Rennie

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.

user228700

9:37 AM

What about metals, then?

John Rennie

What about them?

user228700

They don't have these induced dipoles and all, correct?

John Rennie

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.

user228700

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?!

user228700

@JohnRennie No, Sir, let's forget about optics!

user228700

9:40 AM

OK, I'm having a breakdown. I'll do some Math for awhile and then I'll come back.

John Rennie

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.

DHMO

@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?

user228700

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.

user228700

@JohnRennie Huh, OK...

John Rennie

@DHMO the Planck length is one of the Planck units.

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...

DHMO

9:43 AM

@JohnRennie isn't this an impact rather than a cause?

John Rennie

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

DHMO

I mean, after we know Planck length, we derived the Planck units?

John Rennie

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.

DHMO

like, how do we know that GR fails beyond Planck length?

John Rennie

@DHMO We don't, we have no theory of quantum gravity.

user228700

9:46 AM

Oh, @DHMO: Dyou think u could maybe help me with some geometry?

DHMO

@JohnRennie I am not convinced by dimension analysis

@JohnRennie how do we know that we need quantum gravity after that link?

@Kaumudi yes

user228700

MSE chat please?

DHMO

ok

user228700

I'll link u to the question I posted over there yesterday.

John Rennie

@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.

DHMO

9:50 AM

@JohnRennie why does a blackhole implies the failure of our theories?

Swapnil Das

10:46 AM

Impulse is in the direction of force, true or false?

rob

@SwapnilDas Definition of impulse?

user228700

@SwapnilDas True.

Swapnil Das

@rob Change in momentum.

rob

@SwapnilDas Newton's second law makes a statement about change in momentum.

Swapnil Das

@Kaumudi It is so astonishing in physics that even small concepts matter, thanks!

user228700

10:48 AM

U.welc. :-)

Swapnil Das

Does it mention direction, I don't think so? @rob

rob

@SwapnilDas Newton's second law is a vector equation.

user228700

^

Swapnil Das

God.

I took carelessness to the next level :P

rob

Happens to all of us.

Swapnil Das

10:51 AM

@rob Are you a physicist?

rob

@SwapnilDas I am.

Swapnil Das

Wow! what's your field of research?

@rob

DHMO

@SwapnilDas $\vec{F} = t \Delta\vec{p}$

rob

Symmetries, weak interactions, precision measurements.

Swapnil Das

Yup :)

DHMO

10:54 AM

since $t$ is a scalar, it must follow that $\vec{F} \parallel \Delta\vec{p}$

heather

11:28 AM

hello

DHMO

@heather greetings

heather

@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.)

DHMO

yes, so lnx needs to be >0 also

heather

wait could you say $\ln^2x>0$?

DHMO

ambiguous notation

heather

11:41 AM

right

DHMO

but no, ln(ln(x)) can have any value

heather

?

DHMO

ln(ln(x)) can be -1

heather

the answer to it? oh, yeah that makes sense

DHMO

domain and range

heather

11:44 AM

hmm, would the domain just be x>0?

DHMO

not here

heather

I don't know, then. How would you figure out the domain here?

DHMO

you know that the input to ln must be >0

here, the input is ln(x)

heather

right

DHMO

so ln(x)>0

x>1

heather

11:47 AM

wait, okay, new question: what are you doing to the left and right side? to get rid of the $\ln$?

DHMO

we raise e to the power of it

e^(lnx) > e^0

x > 1

heather

wait, don't you multiply both sides by $e^x$? I thought that was the inverse of the natural logarithm.

DHMO

no, inverse means you get back the input if you apply the function

heather

oh...

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.

DHMO

e^(ln(x)) is just x

heather

11:54 AM

because e to a natural logarithm...right. Okay, so the domain is $\{x\in\mathbb{R}|x>1\}$.

DHMO

yes

heather

Thanks so much for putting up with my not-getting-it-ness. =)

DHMO

you are welcome

0celo7

12:52 PM

Doing PhD analysis in here?

@JohnRennie certainly not...multiply by 2 or something

2 is just as fundamental as those other things lol.

user228700

1:08 PM

@0celo7 Someone should keep telling u this when u're in the middle of one of those omg-what-are-they-talking-about discussions with bloo.

Qmechanic

1:28 PM

Last Monday October 17th, 2016 (a week after the election), Phys.SE has for the first time reached more than 100000 visits per day! (100885 to be precise). Also as a first, the total number of page views per day were more than 140000. (147651 to be precise). Pat yourself on the shoulder! New records are expected later this fall.

9

Secret

1:53 PM

This is why I want to develop a type 3 model (Mutable histories)
Because this one cannot be explained by any type 4 models (Branching timelines)

http://zeroescape.wikia.com/
https://en.wikipedia.org/wiki/Your_Name
The true complexity of time travel is only revealed in a body swap type time travel

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