The changes in the EM field propagate outwards at the speed of light, there's no instantaneous change in the field at any nonzero distance @schn

@Charlie there's no instantaneous change in the field at any nonzero distance...could you elaborate?

What change could there be?

A bit like disturbing some stationary water, the disturbance propagates out into the water at the speed of sound, same happens here, no?

@Charlie do check it

Hello Guys

Hi

whoa, I just got serially upvoted

I just hope it wasn't malicious (i.e. in an attempt to trick the autobot into reporting me for voting irregularities)

That happens to me occasionally. The upvotes will be removed automatically in the next 24 hours.

@JohnRennie but that's because you're famous! :-) I wonder why I got hit - perhaps someone who knows me IRL chanced upon my account

Interestingly there are only 11 people that could have done it, according to the weekly reputation tables

@NiharKarve never underestimate caching :P

ah yeah

(and I meant voting tables)

@NiharKarve Since there are no consequences merely for being the "victim" of serial votes, that would be malicious but stupid and ineffective :P

@ACuriousMind I wasn't concerned about the repercussions so much as having offended someone on the website

maybe I should drop the cynicism and assume someone was particularly overjoyed at one of my characteristically mundane comments

Studying the irreducible representations of a finite group. Must there always be the trivial irrep of dimension one? I am considering the irreps of a high symmetry point and there are only two two-dimensional irreps and it confuses me

the 1D trivial irrep always exists

or have I misunderstood your question

@B.Brekke Some people might take "irrep" to mean "non-trivial irrep"

@ACuriousMind Yes, I thought about that, but I have only eight group elements and two, two-dimensional irreps. So I feel like there is no "room" for the trivial irrep as well.

@B.Brekke The trivial irrep always exists since there is no obstruction to the representation map that just maps every group element to the identity.

I'm not sure what you mean by "no room". The number of distinct complex irreps of a finite group is the number of conjugacy classes of the group. The number of elements is irrelevant (but of course an upper bound for the number of conjugacy classes)

And the trivial rep is the one corresponding to the character that's zero on all conjugacy classes except that of the identity, since every group has an identity, every group has a trivial rep

@ACuriousMind I agree, so there must be something else that I don't understand here. It is in regard to nonsymmorphic space groups. I was thinking that the order of the group = the sum of squares of dimensions of irreps, and 2^2 + 2^2 = 8

@B.Brekke Ah, right. What is the group of order 8 here?

@B.Brekke Are you sure you haven't made a mistake somewhere? I'm pretty sure you have to include the 1^2 for the trivial irrep in the SoS formula

The only thing you can have overlooked here is really that the two 2d representations are either isomorphic or one of them is reducible

@ACuriousMind it must be either D4 or the quaternion group, and both only have one 2d irrep

@ACuriousMind I am using an online database to consider the irreps of the symmetries of a crystal. The manual is quite long and maybe not accessible, but I don't know any other way to show my case. Figure 5 shows an example which I am perfectly comfortable with, while figure 2 shows the case where the irreps look weird. I provide the link to the manual onlinelibrary.wiley.com/doi/epdf/10.1107/…

So common for the examples which I don't understand is that the symmetry elements contain translations, and I think my lack of understanding of this is what causes my confusion

unfortunately that link isn't accessible for people without a subscription

@ACuriousMind Ah okay sorry. It should be easy for me to check if my "irreps" are isomorphic right? I just check if they have identical characters. And If the irreps are both diagonal and off-diagonal over the conjugancy classes, they can't be reducible or? I mean in this case of 2-d matrices it should be that simple?

@B.Brekke I'm not sure what you mean by the irreps being (off-)diagonal "over the conjugacy classes"

But yes, the two reps are isomorphic iff their characters are the same

And a rep is irreducible iff the norm of its character in the space of characters is 1.

(assuming we're talking complex representations, for real representations it's more complicated)

Idk if I'm getting at anything here, but is there any relationship between the exterior/interior derivative on the Grassmann algebra that takes you between the various exterior tensor spaces, and the creation/annihilation operators in QFT that take you between the various tensored Hilbert spaces?

Are these two things related in any way? Or do they just do similar things but aren't two specific examples of a more general operation/structure

@Charlie does $(a^\dagger)^2 = 0$

No but isn't that just a consequence of the anti-symmetry of the exterior algebra? Maybe it's possible to define something similar on regular tensor algebras

that being said I've never seen such a definition

Maybe the covariant derivative, though that is reaching the boundary of what I actually know about this

@Charlie I don't see how they're supposed to be related. The c/a operators do not act as derivations, they have no Leibniz rule

the only similarity is that both act on a graded vector space by increasing/decreasing the degree by 1

That's true, maybe there are better examples I could have chosen but I don't know any that aren't derivations

yeah that similarity is essentially the only connection I could actually see bewteen them

I just wondered because in QFT an explicit construction of the c/a operators isn't really given, they're just defined as operators that give you a particular state in the next level of the Fock space

I'm not sure what you mean by "explicit construction"

if you pick a concrete realization of the Fock space (e.g. the Segal-Bargmann space), then of course you also have explicit formulae for the c/a operators

Ah yeah I assumed something more precise must exist, but at least in introductory texts their definition is essentially just a physical one, "it creates/removes a particle of momentum $p$"

it's not less precise, it's just more abstract!

The abstract definition is a perfectly fine definition for an operator on Fock space

I guess so, maybe my complaining isn't justified

hello everyone

I have been using Griffith's electrodynamics (4th edition) for a while now, and I have observed that quite a few problems have the statement:

"this problem was intentionally deleted from this edition"

What could be some reasons for the publishers/Griffiths to remove problems from a specific edition?

I dont think it's the case that the problems were "wrong", since then there would be far too many "wrong questions" in a book of such huge reputation

perhaps people in the past complained that the question was not well asked

Hi @JohnRennie Sir..

Hey. Anyone here good with Mathematica plotting?

I'm looking to draw 2 plots. Each plot has 3 curves. I want the 2 plots to be one below the other, with the same x-axis. And a horizontal line separating the two plots. A vertical line to mark the origin. How can I do this?