@ACuriousMind thanks

@acuriousmind en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics take a look at that table at the bottom. which interpretation do you think is the most popular and for what reasons?

it excludes a couple less popular ones like BST

Copenhagen, obviously

BST isn't really much of a QM interpretation

There's vague ideas but no math behind it

@NoahP it seems as if eqn (6) is just a matter of assembling bits from previous proofs.

Well, there is math, just none for QM

@JohnRennie Which would I start with?

We used the result $1+z=1/a$, and the values of $\omega$ for the three components.

Did you already know all those, or do you have to derive them?

why is the copenhagen interpretation more popular than the rest @slereah

@Obliv Of those named, Copenhagen is the most popular, but my impression is that a significant number of people actually doing quantum physics every day share my personal disdain for all the squabbles around "interpretations".

It's simple and generalizes well

The other popular interpretations tend to just be variations on copenhagen

@acuriousmind what is your disdain

@JohnRennie The redshift one is new - not sure if I should derive it. And $\omega$ is new

Interpretations is a topic that is most popular with laypeople and not so much physicists

@Obliv That it's pointless to argue about interpretations since they are, pretty much by definition, empirically indistinguishable.

There is room to test for interpretations but overall there aren't that many reasons to really do it much currently

the interpretations we have are pretty solid

@JohnRennie I understnad what $\omega$ corresponds to, so I think we can assume the values - theyre in my previous workings too

The motivation for it tends to be more on metaphysics than physics

@slereah @acuriousmind I disagree with the notion that choosing an interpretation is meaningless. I believe having an interpretation in mind when working on a theory will lead to different ideas to test for & overall a different evolution of a theory. If you are strictly a deterministic physicist working on QM, you will most likely choose a hidden variable interpretation in which if you make any progress with the interpretation it will be in favor of determinism, no?

So it seems to me that if you can prove the redshift result the rest follows. Does that seem so to you?

Sort of reverse engineering what we just did?

@Obliv It is not a scientific attitude to decide first that you believe in determinism and then go and try to make your theory conform to that.

And, anyway, believing in hidden variables changes nothing about quantum mechanics.

@ACuriousMind what do you know about jets, jet prolongations, and jet transversality?

The machinery of Hilbert spaces and operators will predict experimental results correctly no matter your interpretation

@0celo7 nothing

@JohnRennie I've got a derivation of the redshift result here, and it's way beyond the scope of anything else, so I think its ok to assume it regardless of a derivation

If you build a theory that deviates from quantum mechanics in its predictions, then you haven't got an interpretation, you've got a new theory

Interpretations are for people who believe that there is Truth(TM), that there is a useful notion of a thing being "real" as opposed to "purely theoretical construct", and that such truth is discernible by humans

@ACuriousMind Well that's not 100% true

But pretty close to accurate

@NoahP OK. Bearing in mind that I didn't set the paper so I'm not sure exactly what is bring asked, I think once you assemble all the bits equation (6) just drops out.

@JohnRennie Okay..? I'm not really sure how to derive such a thing then?

@ACuriousMind @ACuriousMind

OOps

@NoahP or maybe we're not supposed to assume specific types of matter/energy but just prove the result in general i.e. for any value of $\omega$. Maybe you could check with the chap running the course.

That would sound more likely @JohnRennie

@ACuriousMind I think it depends on context, though. It depends how justified your decision is. If your decision is for aesthetic reasons, then, no, it isn't very scientific, but, if your decision is based off of clues, say, an analogy with the theory with another theory, then there's nothing unscientific doing it

However, if you do construct your model, and it doesn't reveal any new information, or is inconsistent, and you continue to pursue it, that's when it becomes unscientific

@ACuriousMind I don't think attitude is relevant in the context of discovery & building theories that withstand experiment. If the theory is rigorous, there need not be any concern for the attitude that developed it. It may seem illogical & primitive to you, but that's how people function. When the human observes a phenomena, their conclusions are based on who they are (including past experiences and any predispositions to a certain way of thinking that follow)

but then again anything a human does seems illogical to you since you're an A.I.

@Obliv Well, duh. I'm not saying it leads to anything wrong per se to follow a particular interpretation, I'm saying it is pointless to argue which interpretation is "true". The content of the physics is in the formalism of quantum mechanics. What stories you tell yourself to be at ease with the formalism is irrelevant, and it is not very interesting to debate which of these stories are more popular or "true", at least not to me, and also not to others I've talked to.

What was that equation you had to use in q4? The one that linked density and the Hubble parameter?

I've got:

$H \propto t^{-1}$

$\rho \propto t^{-2}$

$a \propto t^{1/3}$

Or,

$\dot{\rho}=-3H(\rho + \frac{p}{c^2})$

@Obliv : I agree. Without interpretation there's no understanding.

That last one.

$\rho \propto a^{-6}$

Ah ok

The continuity equation

Since $p = \omega\rho c^2$ that gives a relationship between $\rho$, $a$ and $\omega$. I wonder if that would help ...

@DavidZ Thank you very much! (For your edit)

@ACuriousMind But I don't think it's just about making yourself feel comfortable, though. Some people use it as a compass for pushing the boundaries of physics. If you are trying to add information to a theory, an interpretation might help you to know where to look.

How is the $\dot{\rho}$ going to work out though?

@JohnRennie

Mmm

I could integrate again

@ACuriousMind : Interpretations are for people who believe that there's some value in understanding the world. These people are called physicists.

@WilliamBulmer : well said that man.

$-3\frac{\dot{a}}{a}(1+\omega)=\frac{\dot{\rho}}{\rho}$

@JohnRennie

Hi all

Hi

hi

welcome

:-)

@acuriousmind Entertaining the idea of an interpretation of a theory gives one a different perspective and perhaps some motivation to explore an idea. But, that's besides the point. Your disdain stems from the idea that someone tries to conform a theory to their fixed views. I understand that, in general, this leads to crackpot theorists. Ideally, however, they are not fixed views and only act as a guiding force. In my original question, I ask because QM interpretations lead to

Is it true that given a conductor, if you introduce an equal amount of positive and negative charge simultaneously that they would then arrange at the surface of the conductor?

very different views of nature which in turn can motivate someone to work on different ideas (as I mentioned earlier)

So, @JohnRennie $-3a^{-1}(1+\omega)da=\rho^{-1}d\rho$

@acuriousmind In my viewpoint, it is mostly harmless to discuss such interpretations as choosing certain ones may lead to intuition on how nature works. I'm very out of my element though since I haven't taken any QM classes and maybe these interpretations don't actually help in this regard.

@NoahP so that would give you $\frac{\rho}{\rho_0} = \frac{1}{a^{3(1+\omega)}}$

\o @yuggib

o/

And there you go! Plug in $z+1 = 1/a$ and equation (6) falls out!

@JohnRennie How so? I'm playing catch up on the integration

@skillpatrol Is this true?

@NoahP I didn't do the integration. I guessed the equation for $\rho(a)$ then calculated $\frac{\dot{\rho}}{\rho}$ to check I was right, which I was :-)

So, I wonder if anyone could give me some advice. I am a physics student (currently independent, but I have 3 years of formal training under my belt) I am conversant in QM, EM, Statistical Mechanics, Lagrangians, Tensors, the Minkowski Formalism of SR, and so on, and so know just enough to be a dangerous crackpot. Lately, I have been monumentally distracted by my own ideas.

Should I do the integration then? @JohnRennie

The problem is that they keep working--I mean formally. But they are ideas about classical EM. On the other hand, I am gravely aware of my knowledge deficits

@NoahP No, why would you?

@JohnRennie A more solid argument than guessing? :P

And yet, when I pick up a book, say Rindler's Relativity, all I can think about is my own ideas, because they keep working. Is there any advice for goading my brain into being more productive in actually becoming conversant in what I need to?

Guessing the result then checking it is a well established mathematical technique that even has its own name. It is called Ansatz.

I'm not sure how you jumped to chat.stackexchange.com/transcript/message/31193368#31193368

@William I would guess that this is natural. My advice is to continue exploring this path until it possibly leads to different results, then see if you can verify these results. It also depends on what you mean by different ideas though

In solving question 4 you went through much pain to discover that the solution to $\frac{\dot{\rho}}{\rho} = -6 \frac{\dot{a}}{a}$ was $\rho = 1/a^6$. Yes?

Yes

@william Recognize that there is usually more than one way to model a theory mathematically. If your way also works, you aren't necessarily becoming a crackpot theorist. If they rely on radical untestable ideas, then you're possibly on your way.

So have a quick guess at the solution to $\frac{\dot{\rho}}{\rho} = -n \frac{\dot{a}}{a}$ i.e. where instead of 6 we have $n$.

I'm not asking you to work it out, just guess.

Ok, fair enough

But, if you're on a deadline to learn something maybe you should put your ideas on the backburner especially if you can determine if they're not meaningful in any way.

And your guessed solution is ... ?

@Obliv Well, they do sort of lead to different results. That's the problem. I am at the point where I actually have a set of mathematically candidate theories parametrized by one or two unknowns.

$\rho=1/a^n$

So, I have enough to occupy me for a long time, but not enough to justify it to a professional

Yes, that's what i guessed to. And you can take that guess, err ansatz, and use it to calculate $\frac{\dot{\rho}}{\rho}$.

And if you find that $\frac{\dot{\rho}}{\rho} = -n \frac{\dot{a}}{a}$ then you have proved your guess correct.

Okay

Rather annoying, I've got to go now - lift home from the library

I'll get back to you tomorrow probably, hopefully to say I've polished it off!

Thanks :)

@Obliv Yeah, I am trying not to rely on untestable ideas.

OK, see you tomorrow. Do the calculation to confirm th ansatz in your own time ...

@Obliv what I could really use is to be back in school to be forced to learn what is prescibed

What's stopping you?

@Obliv Err I meant mathematically consistent candidate theories

@skillpatrol Not having enough money, and being employed full time

@william I don't know what your goals are right now. If you have a job and physics is your hobby then work on this formulation in your spare time and see how it goes. Just try not to become obsessed with it. Odds are, the deficit you mentioned in your knowledge might come to prove your theory incorrect.

@Obliv Yes, I suspect that's true, hence my attempts at continuing to learn "the right way." I would not call physics my hobby. I am pretty serious about it. I have begun the process of preparing for a return to school

It's just that learning the right way is happening very slowly

Partly because I keep being distracted by these ideas

Well then you know what to do. Keep learning and building your arsenal of knowledge and eventually the problem will sort itself out. You may prove it to be true, find it is incorrect, or decide that it actually predicts the same results as modern theory, etc.

I mean, I AM learning new techniques. For example, I just found out about geometric algera from reading some of Hestene's work. I think that could be very useful to know

I would suggest staying away from the path less traveled, and look up the main stream course outlines for topics to self study :-)

@Obliv I do. I just am in the curious position of being simulltaneously frustrated and elated

@skillpatrol That's a good idea. I should try a regimented approach where I work on my ideas half the time and "real physics" the other half. Maybe I should jump directly to textbook problems to that I can give my mind those to chew on, instead of these ideas

The so called classic textbooks are full of problems to keep you busy.

Damn, I left Bott & Tu at home

No PDF?

@skillpatrol @Obliv thanks for listening

Thanks for asking :-)

@skillpatrol I do

But I like studying from a hard copy

Time to break into my advisor's office next door and grab his copy >:3

:O

@william no problem

@johnR how does one determine the mass of a particle?

There's several methods

You can look at the angle of curvature in a magnetic field, if it's charged

@Obliv loads of ways depending on the particle and circumstances. If it's charged you can measure e/m from the Lorentz force.

You can look at the threshold for decays and annihilations

etc

do they rely on the same fundamental notion of mass?

like $F = ma$ or whatever

Experimentally pretty much all mass notions are the same, assuming inertial = gravitational

in newtonian mechanics

@Obliv sort of. In a collider you look for a resonance at a certain energy, then you get the mass from $m = E/c^2$. That's why the Higgs mass is quoted in eV i.e. as an energy.

so then does this mean massless particles do not exhibit behaviors of inertia?

Depends what you mean by inertia

There's no direct analogue of $F = ma$ in QFT

resistance to change in motion

That's a term that doesn't have a precise meaning.

oh

Massless particles have a momentum if that's what you mean by inertia.

In QFT the mass of a particle is the first pole of its propagator

Basically

@JohnR that's because velocity gives mass to an object right? So even massive particles with velocity gain an additional mass in its momentum?

NOOOOOOOOOOOOOOOOOOOOOO! Aaaaaaaaaaaaaaaaaaaaaaaaaaaaargggggggggghh!!!

LOL

A photon has zero mass. Dividing its energy by $c^2$ gives a number with the dimensions of a mass, but it isn't the mass of the photon.

Momentum is not necessarily associated with mass. That's a misconception borne of everyday experience.

In the classical definition of momentum, it is necessarily associated with mass. @johnR can you give a more precise definition?

@Obliv Not at all. With what mass would the momentum of an electromagnetic field be associated?

There's two definitions of momentum

fields.. have momentum? @acuriousmind

$p = \int_\Sigma \frac{\partial \mathcal L}{\partial \dot \varphi} d^3x$

what is $\varphi$

@Obliv Yes, otherwise momentum conservation fails in classical electromagnetics

$p_i = \int_\Sigma T_{ii}d^3x$

Where $\varphi$ is the field.

The first is the canonical momentum

The second is the regular old momentum

They are often the same

@Slereah [citation needed]

It all comes from Gut Feeling

You can quote my gut

They're the same if you have simple classical mechanics systems, but that already fails in basic EM (the canonical momentum is gauge variant)

@Slereah is $T_{ii}$ the stress-energy tensor?

@acuriousmind How can you provide experimental evidence of the momentum of an E-field, based on the definitions of momentum used by @slereah? (which, I can't comprehend with my level of math :D)

Some components of it, yes

@Obliv Light has pressure

@JohnRennie The better way to recall that formula is to say that momentum is the conserved quantity associated to spatial translations by Noether's theorem.

@Obliv The second of Slereah's definition imposes that momentum is conserved. You can shoot light at a reflective metal plate and it will move slightly. By conservation of momentum, the light must have carried the momentum imparted to the plate.

@Obliv the Poynting vector carries momentum in classical EM

For instance

Light sails rely on the momentum of light

You know what's harder though?

Momentum of light in a medium

@slereah That is true for light, but isn't an EM-field an abstract concept in general? It spans to an infinite distance and relays information instantly, no? How can you associate a momentum to this definition of an em-field?

No?

You're thinking of a monochromatic wave

@Obliv relays information instantly - no.

Momentum and energy are really just two sides of the same coin, so if you understand how a field can have kinetic energy without mass, then you understand how it can have momentum without mass

Those do not indeed exist and carry infinite momentum

But that's not really a problem

Momentum and energy are unified under the concept of 4-momentum.

@Obliv You "associate" momentum to the EM field by writing down the function of $E$ and $B$ that makes conservation of momentum hold.

Claiming that photons have a mass is sort of like kicking a beehive. The bees physicists all come flying out and start stinging you :-)

Which you may deduce by Noether's theorem, or guess, or deduce in some special case and then generalize it.

@JohnRennie haha

there's an old Feynman argument about why photons mass must be under $10^{-N} eV$

but this definition of momentum doesn't rely on mass? .. what is it then?

@JohnRennie Unlike bees, physicists can sting you more that once, though :)

@Obliv in a variety of ways

@Obliv Momentum is the conserved quantity associated to spatial translations through Noether's theorem, like energy is the conserved quantity associated to time translations.

His best estimate was that it must be under $10^{-17} eV$ because otherwise, the earth's magnetic field would decay too fast to push off some solar radiation

@ACuriousMind Though certain members of the SE (naming no names :-) appear to have developed an immunity

@JohnRennie Yes, we clearly need to evolve better poison

@acuriousmind you gave me that definition of energy many months ago and I still don't understand that definition intuitively

@Obliv Why do you think you should understand that definition "intuitively"?

@Slereah that is the most physicist thing I've ever heard

Physics has no requirement to conform to our intuition.

I don't understand it period, is what I meant to say

What definition @Obliv ?

@Obliv For one, you can say that momentum is velocity*KE/c^2

Just remember that $E^2 = p^2c^2 + m^2c^4$, where $E$ is the energy, $p$ the momentum and $m$ the rest mass. For a photon $m=0$ but the equation is otherwise unchanged.

@0celo7 definition of momentum via noether's theorem

or energy for that matter

translation invariance

seems pretty simple

God, I need to learn how to use Noether's theorem

@Obliv Well, of course you have to learn about Noether's theorem to understand it. Which you should, because Noether's theorem is one of the most profound results classical mechanics has to offer.

@JohnRennie my high school chemistry teacher disagrees...

@JohnR Stop stuffing equations down my noggin D: I don't understand the motivation for associating the old definition of momentum with energy multiplied by the speed of light squared, etc.

we were told a photon has mass $h\nu/c^2$

@0celo7 and your point is ... ?

@Obliv Special Relativity gives some pretty good explanations for how you can have momentum without mass

You don't need Noether's theorem

Though the latter is more elegant

@JohnRennie clearly there is disagreement in the scientific community on this point