The h Bar

8:45 AM

@schn The imaginary bit of the RI gives the lossy part doesn't it? i.e. it tells you how much of the light is absorbed.

A question for the panel:

The "dark sector" represents particles that interact with standard model particles only via gravity. I'm sure I've seen suggestions for an "even darker" sector that doesn't even interact via gravity. So effectively it is completely non-interacting with any particles we know of (though it still interacts with itself). However I've forgotten what the catchy name for this sector was.

Has anyone else come across this idea and if so can you remember what name it was given?

ACuriousMind

@JohnRennie If it doesn't even interact via gravity, what distinguishes it from not existing?

John Rennie

@ACuriousMind good question :-)

My recollection is that one of the many theorists with too little to do managed to construct the sector from (their interpretation of) string theory.

I mention this only because a (non-physicist) friend of mine has just posted this on Facebook:

ACuriousMind

9:01 AM

well, it's pretty easy to write down a theory like that, but I don't really see how its predictions could be meaningfully different from a theory that just doesn't have this, uh, irrelevant sector

I'd say doesn'tmatter is a pretty good name for this ;)

John Rennie

As I recall it was more recreational physics than a serious suggestion. I think it was posted on Arxiv, though I don't recall if it was ever published in a peer reviewed journal. I'm guessing not.

EVO

10:20 AM

When Bohr stated that the angular momentum of a electron is nh/2pi was he referring to the total angular momentum(sum of orbital and spin)

John Rennie

No, he was only talking about orbital angular momentum.

When he proposed the model the electron spin had not been discovered.

EVO

10:41 AM

@JohnRennie thanks

Mohammad Javanshiry

11:02 AM

@PhysicsMeta @JohnRennie Hello John. I read your answer but it could not help me much. The SR effects aside, I want to know if the gravitational equipotential lines are observer-independent. That is, as an interpretation to the fact that the Schwarzschild observer measures the rate of the satellite clocks to be the same is that everywhere on the circular orbit in which the satellites move the gravitational potential remains the same.

However, this circular orbit looks to be an ellipse from the standpoint of each satellite. If this elliptical orbit indicates an equipotential line of the gravitational field, each satellite claims that there is no change for the other orbiting clock due to gravitation (but a time dilation due to SR) since both satellites are located in the same gravitational potential line.

I want to say if your answer to my problem is correct, an equipotential line for one observer (the Schwarzschild observer) is no longer valid for the other observers (the satellites).

John Rennie

@MohammadJavanshiry Gravitational potential energy is a coordinate dependent quantity in GR, so different observers will disagree about the surfaces of equal GPE.

Mohammad Javanshiry

Aha, I got it, thanks. Is it the same for the electrical field around a point-like charge?

John Rennie

You can see this by considering Rindler metric. The Rindler metric describes the same flat spacetime as the Minkowski metric, but the Rindler observer sees what looks like a uniform gravitational field so they observe a GPE that changes with distance and also time dilation that changes with distance.

@MohammadJavanshiry I don't know. Electrodynamics in curved spacetime is too complicated for me I'm afraid.

Mohammad Javanshiry

Not in the curved spacetime. Assume that the charge is massless.

John Rennie

I'm not sure what you are asking then ...

Mohammad Javanshiry

11:15 AM

Assume that we have a tiny point charge in an interstellar space away from any gravitational field. The electric equipotential lines are thus circular around this charge. What about the observer who approaches the charge at a constant velocity? Does he see the potential line, which is now elliptical, to be equipotential?

John Rennie

I don't know ...

I would guess not because the electrostatic three potential is not Lorentz invariant.

The Lorentz invariant quantity is the vector four-potential.

Electromagnetic four-potential

An electromagnetic four-potential is a relativistic vector function from which the electromagnetic field can be derived. It combines both an electric scalar potential and a magnetic vector potential into a single four-vector.As measured in a given frame of reference, and for a given gauge, the first component of the electromagnetic four-potential is conventionally taken to be the electric scalar potential, and the other three components make up the magnetic vector potential. While both the scalar and vector potential depend upon the frame, the electromagnetic four-potential is Lorentz covariant...

@MohammadJavanshiry In fact, now I think about it I'm sure the answer is "no".

Mohammad Javanshiry

Then, which one is applicable to the point charge? A three- or four-potential?

John Rennie

If we start with a stationary charge, surrounded by a static electric field, then when we Lorentz transform this into the frame of a moving observer it looks like a combination of a (different) electric field and a magnetic field.

The Lorentz covariant quantity is the four-potential, but we don't observe that directly. What we observe are the regular electric and magnetic potentials in our frame.

Mohammad Javanshiry

Okay, thank you for your response.

John Rennie

:-)

Slereah

11:30 AM

Greetings

@JohnRennie Lorentz covariant*

John Rennie

Codeed.

Emilio Pisanty

11:48 AM

earthobservatory.nasa.gov/tournament-earth

(just dropping this off, it looks cool)

Nihar Karve

2:07 PM

Permutation Orbifolds of Vertex Operator Algebras

►

this is absolutely fantastic (and in this video he literally starts from the axioms of a VOA)

Charlie

3:07 PM

Have I got this right, the reason we get "ghost" states in string theory (similar to QED) is because of the gauge symmetry, and the point of lightcone gauge is to completely gauge fix the worldsheet before quantising so you don't have to deal with this straight out of the gate?

the worldsheet "gauge" freedom being diffeomorphism and Weyl symmetry I believe

ACuriousMind

3:22 PM

@Charlie Yes, the appearance of ghosts is a general feature of the quantization of gauge theories, but sometimes you can get away with an ad hoc method (like Gupta-Bleuler in QED) so you don't have to do full BRST quantization.

Charlie

I see, defining the equivalence classes of states related by the addition of a null state (which I think is the gupta-bleuler stuff) looked suspiciously similar to constructing the de Rham cohomology, is this just a coincidence?

it looks like something interesting from a mathematical perspective, just dressed up in physics language

ACuriousMind

@Charlie No, you're computing the (0th) cohomology (="image modulo kernel") of the BRST operator, just like deRham computes the cohomology of the exterior derivative operator

Charlie

ah interesting

ACuriousMind

as usual, the best technical reference is QoGS that includes things like the "main theorem of homological perturbation theory"

Charlie

I will definitely have a look at the h/t book at some point

Nihar Karve

3:56 PM

So there was this recent question about whether instantons are "valid" in Minkowski spacetime

Why is the answer not simply that the way the principal bundle classification works, instantons are topological and so it doesn't make a difference

And the only reason Euclidean space is used in the standard derivation is that it's easy to take the one-point compactification to impose the pure gauge condition at infinity

ACuriousMind

@NiharKarve Link to the question? I'd say that's exactly the answer, but it might depend on what exactly you mean by "instanton"

Nihar Karve

@ACuriousMind Here it is: physics.stackexchange.com/q/619214

ACuriousMind

4:20 PM

@NiharKarve I left a comment

you're right of course that the instanton configuration as such doesn't care about the sign of the metric, but the reason why the instanton is relevant in the Euclidean version is that it's a classical solution to the e.o.m. I think OP is confused because it is not a solution in the Minkowskian version, and thinks it "should" be.

it's also possible that OP is confused by the regrettable terminology - an instanton does not correspond to a (quasi-)particle state, which one might be tempted to think from the -on suffix (and also the way some people talk about it)

Nihar Karve

4:37 PM

nice, we'll see what OP has to say

ACuriousMind

5:19 PM

Hm, I left more comments in the last hour than in the weeks before that and now I'm remembering that the major downside of comments is that people reply to them :P

Faded Giant

Oh, yes, that's why I often write comments but don't send them.

And if I really send a comment, I never read the replies.

You can use AdBlock to hide the recent inbox messages icon. That makes SE a much nicer place.

ACuriousMind

eh, I'm not sure what your inbox looks like but I actually want to read some of the responses!

it's just that often I would expect people to either ignore my comment or edit their posts, and instead I get some comment-reply that's neither here nor there

John Rennie

5:35 PM

@FadedGiant you're in an unusually cynical mood today.

Physics Meta

5:54 PM

0

Q: How to Answer homework like questions?

Is it a bad way to hint at the solution and, ask them to do what they understood and edit the original question to include their steps, and then they follow it up pointing out the mistakes and correcting them? I am asking this question because two of my solutions got closed even though I was exac...

schn

6:13 PM

@JohnRennie Yes, the imaginary bit of RI gives the lossy part, but is this the same as the imaginary bit of the complex wave number? According to k=(n\omega)/c that shouldn’t be, but yet the lossy part is also the imaginary part of the complex wave number in Griffith’s if I remember correctly.

schn

6:25 PM

Since if one has exp(i(kx-t\omega)) and replaces k with the complex version, one gets exp(-\kappa)exp(i(kx-t\omega)), where \kappa is the lossy part, but is this also the extinction coefficient referred to in the imaginary part of RI?

Charlie

mathjax works in the chat btw @schn if you have the chrome plugin

schn

8:24 PM

@Charlie not on phone :)

Or maybe...but for those accessing via a PC, it is best to write with $$.

It would have been no biggie adding $$ to the above, sorry.

Physics Meta

11:29 PM

0

Q: Should we have a one-way speed of light tag?

We have numerous questions on the site about the speed of light, and we have the speed-of-light tag for them. However, since the end of October 2020, when the video Why The Speed Of Light* Can't Be Measured was posted by Veratasium on YouTube, we have had a constant stream of questions about the...

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