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.
@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 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.
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 :-(
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 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.
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.
@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.
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.
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 $\epsilon_r$.
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.
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.
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...
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.
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 $k=3.47$
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.
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...
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 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 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, 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.)