But it comes from turning Maxwell's equations into a wave equation and pulling out the wave speed factor, so it's the speed of an electromagnetic wave (or light). For more details, en.wikipedia.org/wiki/… is a decent place to start
@SillyGoose Texts are often imperfectly structured - something that is asserted on page 1 might become justified on page 35. I don't know to how many people doing this "initial deep reading" I've told "just keep reading, they'll explain that" over the years :P
that is a more complicated question than you might think because the modern SI definitions actually fix $c$ and $\epsilon_0$ as constants, not things to be measured
but the SI definition before that actually fixed $\mu_0$ as a constant
and the $\mu_0$ and $\epsilon_0$ are just some constants in there that may or may not be 1 in your chosen system of units
they are essentially conversion factors between the "strength" of the electric/magnetic fields and the "strength" of the action of those fields on a charge/dipole in vacuum
@Obliv there's not a lot of other electricity you can readily observe
it's not at all obvious that static electricity is the same kind of thing, and the only other thing I can think of is galvanic stuff and Galvani was approximately contemporaneous with Franklin
It really isn't, if I were born in the olden days I'd think lightning was separate from static charges in amber, electric fish, etc
It took thousands of years to get to the conclusion that what we all observed was the same thing
It just seemed absurd to me that when you search "How was electricity discovered" in google you get Benjamin Franklin flew a kite out in a thunderstorm and in all pop culture/cartoons I've seen he got struck lol.
A very absurd apple falling on newton's head moment
if gravity didn't exist until the apple fell, then what caused it to fall in the first place?!?!?!?!
/ s
on a more serious note, i think i remember a semi recent paper which showed ancient people collecting samples which had some optical effects when shaken together
maybe they associated electric effects with fire. which isn't that far off considering the plasma connections between fire and arc gaps etc
ugh i can't find the original paper, but from memory they suggested an interest in some hunter gatherers collecting Triboluminescent samples
What does this mean "..from heat created by the rapid movement of electrons, to brilliant flashes of visible light in the form of black-body radiation." via wiki for lightning
bb radiation wiki is more confusing to me calling it thermal electromagnetic radiation
I understand the concept of the voltage being greater than the breakdown voltage for the insulator of air so the electrons are stripped from the air molecules and shoot down at the speed of light
or something like that, but I don't understand the BB radiation part
Like the photoelectric effect experiments with mercury gas, you are exciting the gas to emit photons, but in this case it's not thermally excited but electrostatically?
I wish ACM were here lol I don't want to just say nonsense without his presence.
@antimony in an electric field any stray electrons that are hanging around in the air get accelerated by the field. Then they collide with the air molecules and ionise them. This creates more free electrons and they get accelerated in their turn, and so on. We get an avalanche effect where the number of free electrons increases rapidly.
There will also be some black body radiation because plasmas are efficient radiators of BB radiation, but I'm not sure how the emission from recombination and the BB emission compare.
Free electrons don't have a specific energy i.e. then can have any KE. So when it recombines with the ionised molecule there isn't a precisely defined energy change.
If you get transitions between electronic states of a molecule there is a sharp line, but even then there is line broadening due to collisions with other air molecules while the transition is happening.
oooh i see, does that collision essentially widen the possible interaction energy, therefore resulting in different gaps and therefore wider spectrum of emissions?
@TejasDahake I know that in India students are required to call any teacher "sir", but we are much more informal in Europe and calling ACM (or me, or anyone else here) "sir" feels wrong to us. It's probably best avoided.
We like to think of ourselves here as a big happy family! (i.e. we argue all the time :-)
Well a starting point would be that if we quantise the electromagnetic field we get states that are approximately infinite planes waves with an energy hf. You'll sometimes here these called modes.
These represent an infinitely delocalised photon i.e. Δp = 0 and Δx = ∞
These have a precise frequency, but they aren't much use since they are infinitely delocalised, plus they have other pathologies like they can't be normalised.
This would be what I think of as a photon.
If you're thinking of a photon as a little packet if light then you need to construct a wavepacket as a superposition of the infinite plane waves i.e. take your packet and Fourier analyse it to get a spectrum of frequencies.
But now this has Δp > 0 because the packet is constructed as an integral over a range of frequencies.
@SillyGoose I risk wandering into areas that I don't fully understand, but the problem with the simple description is that it applies only to non-interacting particles, and photons are not non-interacting.
But as a starting point a single photon in QED is similar to the free particle you learned in quantum mechanics 101.
that photons with definite frequency would be non-normalizable has the same reason other particle with definite momentum are non-normalizable: The frequency/momentum operator has continuous spectrum and hence doesn't really have eigenstates
so are you saying there are three definitions: 1) the unnormalizable plane wave solutions, 2) the normalizable wave packets, 3) just as the energy transfered from EM fields to other things
@SillyGoose Yes, though I wouldn't swear there are other definitions I can't think of. The point is that our little photons turn out to be surprisingly complicated beasties.
the equation that governs photons is just the Maxwell equations (the classical equation of motion for a field is in general also the equation for 1-particle wavefunctions of the quanta of that field)
in this case it's literally just the same equation, you just interpret the solution as a wavefunction
the interpretational problem is that photons are inevitably relativistic and so this "wavefunction" doesn't really have the kind of nice positional probability density interpretation you have in non-rel QM
@SillyGoose It's less complicated than you think. If we take the electric and magnetic fields then Maxwell's equations give us expressions for their variation with position and time.
That's all we mean by the equations of motion for a field.
not sure what that means - the solutions to the vacuum Maxwell equations are waves
but I have a feeling we're discussing weird technical details here that are probably irrelevant for whatever concept you're actually trying to understand :P
@SillyGoose Any electric and/or magnetic field is a solution to Maxwell's equations. What makes the solutions different is the charge distribution. In a vacuum, where the charge density is zero everywhere, one of the solutions is an infinite plane wave.
if you just want to know why photons don't have a definite frequency then John's explanation above is good enough - it's exactly analogous to every other particle never having a definite momentum
well the main conceptual un-understanding for me i think is what it means to say "this photon has energy hf and thus wavelength hc/E". Are these quantities statements about the single photon itself or about other related properties (like a large amount of such photons as some answers on stack seem to suggest)
oh i see okay i guess that makes sense
so equations of motion of X are like the system of equations whose solutions give you X?
it's exactly as meaningful as for any other particle, i.e. you will never see a state with exact momentum p but often it's a good enough approximation that it doesn't matter - but when you look at an actual experiment you'll see line-widths that you can't explain with this approach
so it is an approximation that I make when i say "this photon has momentum p" because in reality only a wavepacket can have def momentum
err wait no
only wave packets can be physical which do nott have def momentum
@Relativisticcucumber you were right xD
is the hamilton-jacobi formulation of mechanics of any interesting use? I have never heard of it before randomly coming across it on some other wiki page. I have only heard of mech courses going into Lagrangian and Hamiltonian mechanics
also is each formulation of mechanics distinct from one another and they jsut produce the same results? or are they like trivially connected? and is there a formulation of QM for each formulation of mechanics?
Lagrangian and Hamiltonian mechanics are of course connected via Legendre transformation since the Hamiltonian is the Legendre transformation of the Lagrangian
Hamilton-Jacobi is essentially just Hamiltonian mechanics with a specific choice of variables - you choose the generalized coordinates to be constants of motion
The Hamiltonian/Lagrangian split in classical mechanics corresponds roughly to the operator/path integral split in quantum mechanics, but the equivalence of the quantum formulations is more subtle than in the classical context
In general, the classical equivalence is also not trivial - the Legendre transformation can be singular, this is the study of gauge theories
also from before since we were talking about the solution of a totally delocalized photon (which im not sure what the physical situation is--no other interactions with the photon?), does that necessarily preclude the possibility of having photon momentum eigenstates in general? Since in textbook quantum, sure there is the free particle, but also the harmonic oscillator and etc.
okay i guess that makes sense if momentum eigenstates look like plane waves then the free particle case is the only place you'd find them
maybe like "loosely reproduce". like we have this old theory. we have this new phenomena to explain. let's try and reproduce as much of the old theory as possible while still being able to explain this new phenomena
@SillyGoose What do you mean "find them"? Just because the energy eigenstates aren't plane waves anywhere but the free particle doesn't mean they're somehow less allowed in other contexts
systems don't need to be in energy eigenstates
@SillyGoose Since Hamiltonian and Lagrangian mechanics are both just classical mechanics, i.e. should predict the same thing for all physical quantities, I don't really get what you mean
i think the question i am asking (related to the correspondance between classical and quantum mech) is something like what % of quantum theory is just slightly modified classical theory
@TejasDahake It's nearly 50 years since I left school and I have no idea what pupils call teachers these days. In any case I'm not a teacher so you can just call me "John" :-)
@TejasDahake Europe is not a monolithic culture and standards for this vary quite a bit by country and personal preference. In Germany you usually just default to Mr/Ms <last name> + the polite form of address like with any other person you're not friendly with but some younger teachers - especially in university - prefer to be addressed by their first names + the familiar form of address. (German has two registers of politeness: polite/unfamiliar and familiar)
@SillyGoose another possible answer is "classical mechanics is just quantum mechanics with $\hbar = 0$" but the word "just" and the vagueness of what that actually technically means are doing a lot of hidden heavy lifting there
about that, when a text says "this $hbar$ turns out to be equal to the $hbar$ in Y equation", are they referring to $hbar$ not as the defined quantity, but just as an arbitrary factor in that moment?
@JohnRennie it actually feels wrong to me without calling them by 'Sir/Madam' to someone who is a mentor, more experienced, intelligent, veteran in some specific field etc.
@TejasDahake I chat to a lot of Indian students, and they all say the same as you. For them it's so normal to use "sir" that they feel uncomfortable not doing it.
and the ultimate question is whether u think there are great differences in reading a text in its native language vs. translation; particularly for faust
@SillyGoose but yes, there are great differences: I've read several books in both languages and there can be a lot of differences, particularly if the texts contain essentially untranslatable elements like puns or references to social norms or assume common knowledge that just aren't there in the target language
@ACuriousMind wow, you are from Germany, albert einstein was your countryman. I think it's really a nice place to live. I would love to be there once :-)
This can vary wildly depending on the function of the accent in the source material -accents usually have social connotations like a lower-class/upper-class or country/urban distinction and if you pick a target accent that doesn't match the intended implication in the source this can fundamentally change how a character is portrayed
@TejasDahake While Einstein was born in what is modern-day Germany, he understandably didn't have a particularly great view of Germany after he had to flee the Nazi regime. After WWII he explicitly preferred not to be associated with Germany (e.g. he tried to prevent publication of his works there) and never wanted to return, so it's not great to use him as reason to be impressed with Germany :P
@TejasDahake I think the problem is that from outside you tend to see only the good things about a country, because of course the country isn't eager to publicise its faults. For example there are a lot of us in the UK who are very unimpressed with the UK's decision to leave the EU.
And just at the moment our government has made itself so unpopular that it is the most unpopular government in the last hundred years!
@ACuriousMind No, I just recalled that albert einstein was your countryman but it's not actually the reason for which I'm impressed with Germany. There are a lot more other than that :-)
@JohnRennie Oh, I see, and I think it's not only for UK, this problem is with every other countries, that the respective governments tries to hide their faults in front of this world.
I know you're probably just trying to make small talk but German national identity - both for historic and more recent reasons - is not exactly a happy topic. It's fraught with history and being openly enthusiastic about Germany as a nation or being German is generally associated with opinions I really don't share.
By saying there's a symmetry, do we mean there's a unitary representation of a group acting on the Hilbert space? When writing the defining representation of the rotation group $SO(3)$, people inherently write the vector space as being real. Does it not make any difference? i.e we just have a complex value vector field that $SO(3)$ acts on. But then why don't people care about the defining representation of $SO(3)$ not being unitary?
@ACuriousMind I read about the history of Germany back in my childhood years, that was so depressing, I just thought from what kind of scenarios the Germans have been through. And I can understand your feelings.
@DIRAC1930 You have to distinguish between the representation on the target space of the classical field and the representation on the quantum space of states. See physics.stackexchange.com/a/215548/50583 (replace SO(1,3) by SO(3), the specific group doesn't matter)
@JohnRennie Feeling really happy to hear this from you John, but as you said, the faults are always kept hidden, same thing goes with india :-)
You might have been through the most popular places to visit in Pune but to explore more reality about this country, you can visit the other places which are not so popular in Pune itself.
@ACuriousMind So in terms of the wavefunction in QM, are you saying $\hat{U}(\Lambda) \psi_i(X) = \sum_j \rho_{ij}\psi_j(\Lambda^{-1}X)$ where $\rho$ here is the defining representation of the group?
@DIRAC1930 oh, no, if you're looking at wavefunctions the rule is just $\psi(\Lambda^{-1}x) = U(\Lambda)\psi(x)$
since the wavefunction takes values in the spin representations corresponding to its spin
e.g. a spin-1/2 wavefunction is $\mathbb{C}^2$ valued, a spin-1 wavefunction is $\mathbb{C}^3$ valued, etc.
if you're just bothered by the "defining" representation of SO(3) being on $\mathbb{R}^3$, the corresponding unitary representation is just the complexification on $\mathbb{C}^3$ - i.e. you just allow the vectors to be complex instead of real with the default inner product
@DIRAC1930 the complex representations of a real Lie algebra are equivalent to the complex representations of the complexification of that algebra, so complexification doesn't hurt anything and it's usually easier to figure out the irreps of the complexification. E.g. in the case of $\mathfrak{su}(2)$, this allows you to define the ladder operators $J_\pm = J_1 \pm\mathrm{i} J_2$ which are not part of the real algebra.
Okay, so just so I'm clear, is it correct to say that $su(2)$ is a real Lie algebra (even though it has complex generators) because when you exponentiate it, the parameters in the exponent are real?
What is the proof KE and PE conserved. Is there any systematic simple way in newtonian mechanics at which we can say energy conserved. Not other quantity . Is there any testing procedure to check conservation?
@DIRAC1930 No, what we mean by a "real" Lie algebra is that you consider the algebra to be a vector space over the reals. I.e. when we say it has generators $J_i$, we mean that all its elements can be written as $\sum_i c_i J_i$ where $c_i\in\mathbb{R}$. A complex Lie algebra has $c_i\in\mathbb{C}$ (and hence twice the number of generators if considered as a real algebra)
...maybe that's what you mean by "the parameters in the exponent"
but you can state this without ever thinking about the associated group, it's purely a property of the algbera
So when you say 'the complex representations of a real Lie algebra are equivalent to the complex representations of the complexification of that algebra', are you saying that for many cases e.g. $su(2)$, we are looking for complex unitary representations of the real Lie algebra?
Oh sorry, I'm mixing up the algebra and the group
A complex representation of a Lie algebra when exponentiated (with complex coefficients) should give the same Lie group as the real representation of the Lie algebra (with real coefficients) right?
I'm basically a fresh ug student in physics, i was wondering what path should I take so that I can understand, to the level of the prerequisites needed to study string theory
In my university, not many people doing string or even theory research
And many are inaccessible
I'm basically on my own here
Basically i just wanna know what the fuss about string theory is
If you take time translation symmetry, you have $\hat{U}(\Lambda)\psi(t) = \psi(\Lambda^{-1}t)$. We have a real Lie algebra that exponentiates to form $\hat{U}$ since $\hat{U} = e^{\imath t' \hat{H}}$ (since $t'$ is real) but it acts on a complex vector space $\mathcal{H}$. The representation $(U,\mathcal{H})$ is a complex unitary representation (since $\mathcal{H}$ is complex) but the Lie algebra is real because of the $t'$ being real
There must be something I'm missing because the Lie algebra also acts on the complex vector space $\mathcal{H}$
basically worrying about string theory before you have a firm understanding of at least basic quantum field theory is pointless, and if you're just starting undergrad that alone will take you quite a while
@DIRAC1930 why do you think you're missing something?
you are correct that this is a complex representation of the real Lie algebra $\mathbb{R}$
Where does the complex representation of the real Lie algebra come from? I only mentioned the group being a complex unitary representation of the translation group
it can be worthwhile to go look at the actual historical papers after you understand how modern physics understands something, but I would never recommend that as a starting point
So from Wikipedia (in reference to Lie algebra representations)
'In the language of physics, one looks for a vector space $V$ together with a collection of operators on $V$ satisfying some fixed set of commutation relations, such as the relations satisfied by the angular momentum operators'
@DIRAC1930 No, what we mean by a "real" Lie algebra is that you consider the algebra to be a vector space over the reals. I.e. when we say it has generators $J_i$, we mean that all its elements can be written as $\sum_i c_i J_i$ where $c_i\in\mathbb{R}$. A complex Lie algebra has $c_i\in\mathbb{C}$ (and hence twice the number of generators if considered as a real algebra)
A Lie algebra is a vector space. If it's a real vector space it's a real algebra, if it's a complex vector space it's a complex algebra, nothing more. This has a priori nothing to do with whether representations of the algebra are on real or complex vector spaces.
But the reason we often complexify real algebras is that the representations of real algebras on complex vector spaces are always also representations of the complexification of that real algebra on the same complex vector space, so we can just always look at the complexification
And in turn the reason we're only interested in representations on complex vector spaces is that the quantum space of states is a complex vector space
When you say different representations of algebras, is this sating that the different representations satisfy the same commutation relations acting on their respective vector spaces?
representations of $\mathbb{R}$/$\mathbb{C}$ aren't interesting, all irreps are one-dimensional (spanned by an eigenvector of the sole generator)
we don't complexify e.g. $\mathfrak{su}(2)$ because we somehow need to, but because it is simpler to deduce the irreps of $\mathfrak{su}(2)_\mathbb{C}$ than doing it directly for the real version - the ladder operators we use to get the standard $2s+1$ irrep for spin-$s$ live in the complexification
@DIRAC1930 Yes. If you're wondering about the $\mathrm{i}$, you have to remove the $\mathrm{i}\hbar$ on the r.h.s. to get an actual real algebra but this is just the mathematician/physicist difference between wanting generators of the algebra to be anti-hermitian/hermitian
@ACuriousMind Hii there, how are you....I'm back here with another question
Okay, I was reading about radioactivity today, at some point of time it focuses on neutrinos and anti-neutrinos it says that they get out of the nucleus with variable energy and If i heard right then they are like photon, so my question here is why do they exist if only beta negative and beta positive particles can alone take whole energy and just decay out like how alpha particles decays? What is the significance of their existence?
Aa okay. I think I will need some time to digest all these things. They are so unrelatable :,)
If we talk about the existence of something then i think the question gets more complicated
Would you like to give me some tips to how to understand the set of quantum mechanics, which you personally used to make it more effectively understandable?
I don't think there's any sort of secret to understanding quantum mechanics. In fact, if anything it's the opposite: I believe the boring reality is that the better you understand classical mechanics beforehand (especially Hamiltonian mechanics) and the slower you take it - i.e. not insisting on unraveling all of the things you ever wanted to know about QM at once - the easier it is to learn.
really, the point of learning is just that you need the time to do so - I learnt most of what I know when I was a student and had few other obligations
@ACuriousMind the time when we are student is precious what I think and as a student in present I don't want to waste it, rather I would invest this time to learn these kind of stuff which I actually love to do.
I am at my very initial stage to learn QM and I think it'll take a very massive amount of time to absorb these things at your level. I'm just a student of general physics which I thought would help me to understand QM as well but that was not actually the case. QM needs different kind of mechanics :-)
