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1d
comment Graphene has a honeycomb lattice - true or false?
Ok. Well than wh
2d
comment Graphene has a honeycomb lattice - true or false?
Funny isn't it. I just feel i am going to be unsatisfied with a response "you cant call that a lattice" until somebody explains "because it doesnt satisfy property X or Y according to the definition of a lattice".
2d
comment Graphene has a honeycomb lattice - true or false?
Could I ask why it is not technically correct? Or maybe I should say, going by the mathematical definition of a lattice (not a Bravais lattice) why can't I say honeycomb lattice?
2d
comment Graphene has a honeycomb lattice - true or false?
Agreed. This is the cause of much (what I'd say are petty) arguments within my group whenever someone delivers a talk on graphene and uses the phrase "honeycomb lattice" in passing. Everyone knows exactly what that person means but people get upset because they assume that by lattice the speaker means 'Bravais lattice'. I understand it may not follow convention, but is it still technically correct to say that graphene has a honeycomb lattice?
Oct
22
asked Graphene has a honeycomb lattice - true or false?
Mar
22
awarded  Notable Question
Mar
12
accepted Bogoliubov transformation with a slight twist
Mar
12
answered Bogoliubov transformation with a slight twist
Mar
11
asked Bogoliubov transformation with a slight twist
Jan
20
awarded  Quorum
Jan
13
accepted Missing terms in Hamiltonian after Legendre transformation of Lagrangian
Jan
9
comment Missing terms in Hamiltonian after Legendre transformation of Lagrangian
Another question if you will. I was advised to try the problem in the coulomb gauge (which is where I'm ultimately heading), and in this case where $\nabla \cdot \vec{A}=0$ and $\phi = 0$ I have a similar situation where my derived Hamiltonian is the same as the Lagrangian apart from sign flips and the fact I am 'missing' the term $\frac{1}{c}\vec{A}\cdot\frac{\partial \vec{P}}{\partial t}$. However in this gauge there is no $\phi$ field and thus no conjugate momenta which are equal to zero. In this situation can we still use similar arguments as you presented above? Thanks.
Jan
9
comment Missing terms in Hamiltonian after Legendre transformation of Lagrangian
Thank you once again @Qmechanic, Dirac_Bergmann analysis is certainly new to me, I don't suppose there are any texts you'd particularly recommend on the subject? My internet hunt hasn't returned anything that good so far.
Jan
8
awarded  Commentator
Jan
8
comment Missing terms in Hamiltonian after Legendre transformation of Lagrangian
Hmm yes, as is the case in my example here, the 'missing' terms are linear in time derivatives of quantities that are functions of spatial coordinates, so kind of like a velocity. I could believe my Hamiltonian is correct, but I am doubting myself due to these 'missing' terms, something I have not before encountered.
Jan
8
revised Missing terms in Hamiltonian after Legendre transformation of Lagrangian
edited body
Jan
8
asked Missing terms in Hamiltonian after Legendre transformation of Lagrangian
Jan
8
comment Is it physically realistic to have an electric field and polarisation density but no displacement field?
Ah you make a good point, yes as $\vec{D} = \epsilon \vec{E}$ if $\vec{D}$ is zero then the permittivity must be zero, which it isn't as I derived it earlier. Ok back to the start I suppose!
Jan
7
asked Is it physically realistic to have an electric field and polarisation density but no displacement field?
Dec
31
accepted Derivation of Lagrangian density for an infinite classical dielectric in interaction with the EM field