/sci/ - Science & Math

A relevant number is the numer of atoms you're dealing with. If you treat big molecules, or even bigger systems, then it's faily conventional to just consider classical fields.

>dat feel when complex fermion mass

Anyway,
>>4946472
that sceme will work out fine.

>>4946486
not much

>>4946481
It wouldn't be uninteresting to see someone rant about the Yang Mills mass problem (e.g. the need for the higgs mechanism) so go on. I'm actually not sure by heart if the problem applies for the commutative gauge group (as OP is impling U(1)).


But I mean you know that the propagator will just turn out to be the green function, so you can go streight ahead and apply your formal inverse differential operator magic for DG=1
See e.g.
http://en.wikipedia.org/wiki/Pseudo_differential_operator

That is (D+m^2)G=1 <=> G=int exp(-ipx)*/(p^2+m^2)

where D is the d'alambert.

There must be a \mu\nu somewhere, so I figure you plug in another \eta

The fact that both are dark/black comes from the fact that both don't emit much light. In the shadow case, it's because light doesn't even go there. See the difference?

I really have no idea about the subject, but what do you mean by "apply"? The expression <x^2-1> will probably also just denote the polynomials, generated by "(x+1)(x-1)" right?

>>4216460
>>4216482
I don't get what you are talking about.
>tan(x)=f'(x)
is x on the left hand side the angle of the slope of f?
How does this equal x?
I don't see if you're impying you can compute the slop of any function at any point without using calculus.

Galois Theory --> Representation Theory

Real Analysis --> Meassure Theory

what is this.

I also like how the person who drew this really hates geometry.

Let <span class="math">F[/spoiler] be an "anti-derivative" of <span class="math">f[/spoiler].
Then

<span class="math">I[f](x):=int_c^{g(x)} f(t) dt = F(g(x))-F(c).[/spoiler]

Hence

<span class="math">\frac{d}{dx}I[f](x)=\frac{d}{dx}F(g(x))=\frac{d}{dg}F(g)g'(x)=f(g(x))g'(x)[/spoiler]

(I[f](x):=int_c^{g(x)} f(t) dt = F(g(x))-F(c).
Hence
\frac{d}{dx}I[f](x)=\frac{d}{dx}F(g(x))=\frac{d}{dg}F(g)g'(x)=f(g(x))g'(x))

I'll ask Mathematica what it thinks about it...

>>4052441
You're missunderstanding something OP, it does violate causality. With such a walkie talkie you could receive a message and send a message back which says "never send me that first message!!"

(1), or (4) if I can trigger it if I'm 80, just to live again.

your options

a) Guess the answer
b) know a lot about modular spaces and Galois theory and feel the answer
c) use the general formula
http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function
which exists for polynomials up to 4th oder
d) use da computa
http://www.wolframalpha.com/input/?i=Solve%5Bx%3D%3Dx%5E3%2B1%2Cx%5D

Lets call your object A-A^T by the name B, i.e.

B:=A-A^T

Since for any A

A=(1/2)(A-A^T)+(1/2)(A+A^T)

your B is just two times the anti-symmetric part of the Matrix A. Then the sum over i of (A-A^T)_ij=0 is the sum of a row of that matrix (or column if you like - it's the same quantity time -1).

Starting from here
http://en.wikipedia.org/wiki/Trace_%28linear_algebra%29
going to here
http://en.wikipedia.org/wiki/Chern_class
you could learn a lot about invariants, but you sum over i is non of them. This condition you want is not basis invariant, see so there is probably not much classification to get from there

take for example this

http://www.wolframalpha.com/input/?i=JordanDecomposition[{{0%2C1}%2C{-1%2C0}}]&asynchronous=fals
e&equal=Submit

in two bases, there are two different values of the sum over i.

Furthermore, since every skew matrix can be written like this

http://upload.wikimedia.org/math/4/b/6/4b6cc9790ab544e8b070295c48f28ced.png

there is always a basis in which your property is not zero for all rows (well unless if A is symmetric such that A-A^T is the zero-matrix.)

there are certainly such matrices in certain bases in existence, but atm I don't know what this might imply.