Recent content by ManDay

I don't know your particular choice for a, b, and c, but while ∫01 lies on the end of the left, horizontal arm, ∫0exp(iπ/12) is likely to lie on the same edge, which means they have the same real part. In general, however, ∫1exp(iπr) with r ∈ ℝ, does not point strictly vertically and thus has...

You seem to conveniently forget that there is a derivative involved. How would that be written with your "easy" notation? $$\frac{\mathrm{d}f}{\mathrm{d}\left[\frac1x\right]}\left(\frac1x\right)$$ \begin{aligned} \frac1{x^2}\frac{\mathrm{d}\zeta}{\mathrm{d}z}\left(z=\frac1x\right) &=...

This is meant in functional notation, not system notation. Two functions f(x) and f(y) have the same form (actually, they are the same function). There are no globally bound variables or any such hocus-pocus that is popular among physicists :-P $$ \frac { \mathrm{d}\zeta } { \mathrm{d} y } (y)...

I suspect this is a misunderstanding: Yes, the last line you quoted is indeed obtained by just inserting z* = 1/z into the definition, so to speak, of ζ. The actual contradiction is shown in the following (after where you stopped quoting me) and is concluded where I say Did this clarify it?

The Schwarz-Christoffel mapping (a Riemann-mapping) from the unit disk (z-plane) to a twice-symmtric area (a cross, ζ-plane) $$ \zeta : \mathbf C \to \mathbf C $$ is given by: $$\frac{ \mathrm{d}\zeta }{ \mathrm{d} z} = \left( \frac{ ( z^2-b^2 ) ( z^2-\frac 1 {b^2} ) }{ ( z^2-a^2 ) (...

Thanks for the reply, it answers my question. I'm aware of the implications of integrating a non-holomorphic function and, admitted, the way I phrased the contour-independence was obviously wrong, now that you mention it. I do see your point and how it relates to the two examples you gave. But...

No, I'm not really "new" to the subject. I don't understand why you neglect to quote the questions I asked and then claim you wouldn't understand what I was asking, to be honest.

I'm struggling to reconcile that a complex integrable function may be multi-branched with the statement that its integral is contour-independent. Consider f(z) = z^(1/n), n natural, n-branched and its integral from z_0 to z_1. On the way from z_0 to z_1 I can take a few detours near the...

You should ask yourself what exactly is "q" in that equation going to be for your capacitor with a given charge density (*cough*) ρ

Hello DimReg, I looked at my derivation and found a few mistakes, indeed. Thank you for the heads up. Here is my derivation | tail: With multiindex α := (α₁,...,αₙ) and K := 1 - iλ/24∫d⁴z(-3Δ²(z,z) + i6Δ(z,z)(∫d⁴xΔ(z,x)J(x))² + (∫d⁴xΔ(z,x)J(x))⁴), which is the interacting part due to λϕ⁴...

Fourier Transform on the "connected part" of QFT transition prob. Homework Statement Calculate ⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ up to order λ from the generating functional Z[J] of λϕ⁴-theory. Using the connected part, derive the T-matrixelement for the reaction a(p₁) + a(p₂) → a(p₃) +...

E is defined as the force per unit of charge on a charge (or density thereof). That's my very point. By that definition, it should account for the Lorentz force due to a magnetic field as well. But as illustrated, it does not. Maxwell equations which determine E uniquely do not account for the...

Is "E" in Maxwell-Equations really "E"? Consider a perfectly static and spatially bound magnetic field B ∈ ℝ³ such that B ≠ 0 ; ∂/∂t B = 0 further, a continuous, time-stationary, and spatially bound current j ∈ ℝ³ through that magnetic field j ≠ 0 ; ∂/∂t j = 0 ; ∇ j = 0 which...

What does "curl E = const." on Ω say about E on ∂Ω? Assume I have a simply connected domain Ω and a twice differentiable vector field E for which I know that "∇×E = const." (1) and "∇E = 0" (2) on Ω - I am interested in solving a BC Problem on ∏ = (Ʃ ⊃ Ω)\Ω, the remainder of Ʃ less Ω. (1) and...

I wouldn't know how to solve the PDE analytically by any other means than using an ansatz. If I find a solution to the problem - even if I can prove the solution is unique within the space allowed for by the ansatz - I still don't know whether it's globally unique - which is what I need to know...