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seen Sep 27 at 17:40

Sep
22
comment Packing fraction of atoms in a HCP structure
Hi Jon. Could you elaborate on the geometry? It is probably simple, but I have stared at it for some time and can't get it right. Thanks.
Sep
22
comment Packing fraction of atoms in a HCP structure
yes, that is what I assumed. For some reason, the geometry I can't get quite right.
Sep
22
comment Packing fraction of atoms in a HCP structure
ah cool thanks ;) Yes, it is their $c$ that I am struggling to derive. Do you have any pointers?
Sep
22
asked Packing fraction of atoms in a HCP structure
Sep
18
comment Question about derivation of tensor in Di Francesco's CFT
For clarity, it was uncovered that there is indeed no trace anomaly and that the vanishing of the trace still holds everywhere even when we include the addition of $A_5$. This may be the reason why the author neglected its addition, viewing it as perhaps a superfluous term or maybe there is more to it.... So the question is still open for answers to anybody else who is interested. Thanks.
Sep
16
comment Question about derivation of tensor in Di Francesco's CFT
Hi Dilaton, thanks for commenting here, I do not remember posting my question on overflow. I could not understand the argument very well, what does it mean to say a 'trace anomaly'?
Sep
13
comment Demostrating possible equivalence of two tensors
I made another thread, but it doesn't seem to be attracting a lot of attention. Do you think it would be okay to contact the author about this?
Sep
13
accepted Demostrating possible equivalence of two tensors
Sep
12
asked Question about derivation of tensor in Di Francesco's CFT
Sep
10
accepted From Paris to … London
Sep
10
comment From Paris to … London
Just a quick last question, in the equation $$\frac{mgR}{2} = \frac{1}{2}mv(r)^2 + \frac{mgr^2}{2R}$$ if I rearrange and maximize, I get that $v_{max}$ occurs when $r=0$. How from this do I infer that the maximum speed is in fact at $r=h$?
Sep
10
comment From Paris to … London
Many thanks! Did you miss a minus in your expression for $F_{grav}$? Then I would solve $$\int_0^v \tilde v d \tilde v = -\frac{g}{R} \int_{-d/2}^{d/2+x} \tilde x d \tilde x$$
Sep
10
comment From Paris to … London
Yes, I believe I am. It would have been nice to have got the same answer using the D.E. I could do the integral using a hyperbolic sub but then when I put in the limits, it looks really messy. But thanks for the alternative approach.
Sep
10
comment From Paris to … London
Sorry, $L$ is $d$ here. Yes, I see they are equivalent thanks :). And yes, I was back and forth on the D.E deciding what variable to express things in.
Sep
10
comment From Paris to … London
The answer given is $$v^2 = \frac{g}{R} x(L-x)$$ where $x$ is defined in the question. I tried, but I can't quite see if the results are equivalent. It requires $R^2 - h^2 = x ( L-x )$. Thanks. Also, could you tell me whether the D.E I set up was in fact correct, just a more difficult method?
Sep
9
comment Demostrating possible equivalence of two tensors
Ok, thanks anyway. Yeah, it is not the first time I have found mistakes in his book, I'll check with one of my professors, thanks ;)
Sep
9
comment Demostrating possible equivalence of two tensors
@ACuriousMind: I was hoping they would be equivalent :) Are you familiar with Di Francesco's book on CFT? On P.108, he writes a general form for a tensor but I don't see why he neglects the equation above with four terms? I thought he did it because of the equivalence with one of the others, but this is now not the case.
Sep
9
comment Demostrating possible equivalence of two tensors
Ok, thank you. Are you familiar with Di Francesco's book on Conformal field theory at all?
Sep
8
comment Demostrating possible equivalence of two tensors
@ACuriousMind The permutation $1111$ yields the constraint $2a=b$ and so does the permutation $1112$ but $1221$ gives $a=0$, which means $b$ would have to be zero from the first constraint. What am I to make of this?
Sep
8
comment From Paris to … London
Oops, I made an error in my expression, it should be $U = mg r^2/2R$ on dimensional grounds, so that $$\frac{mgR}{2} = \frac{1}{2}mv(r)^2 + mgr^2/2R $$