Recent content by Ravendark
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Integral with sine, cosine, and rational function
Homework Statement I would like to compute the following integral: I = \int\limits_0^\pi \mathrm{d}\theta \, \frac{\sin^2 \theta}{a^2 + b^2 - 2 \sqrt{ab} \cos \theta} where ##a,b \in \mathbb{R}_+##. 2. The attempt at a solution Substitution ##x = \cos \theta## yields I = \int\limits_{-1}^1...- Ravendark
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- Cosine Function Integral Rational Sine
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Is the Sign in the Covariant Derivative Important for Local Gauge Invariance?
Homework Statement Consider the fermionic part of the QCD Lagrangian: $$\mathcal{L} = \bar\psi (\mathrm{i} {\not{\!\partial}} - m) \psi \; ,$$ where I used a matrix notation to supress all the colour indices (i.e., ##\psi## is understood to be a three-component vector in colour space whilst...- Ravendark
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- Covariant Covariant derivative Derivative Gauge invariance Qcd Sign
- Replies: 2
- Forum: Advanced Physics Homework Help
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Second functional derivative of fermion action
Mhh, it seems you're right... Now I'm a bit confused...about which RHS/LHS are you talking about? You mean consistent on which argument ##\partial_\mu## actually acts?- Ravendark
- Post #7
- Forum: Advanced Physics Homework Help
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Second functional derivative of fermion action
Thank you very much, I got it...I totally forgot, that there exists an identity for the derivative of the delta function: ##\partial_\mu \delta^{(4)}(y-x)=-\partial_\mu \delta^{(4)}(x-y)##.- Ravendark
- Post #5
- Forum: Advanced Physics Homework Help
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Second functional derivative of fermion action
Right...instead of transfer the derivative to ##\bar\psi## I can perform the derivative directly since I know how the derivative w.r.t. ##\bar\psi## looks like. But why is it "forbidden" to use partial integration ((11) to (12) in my first post) at this point (apart from the fact that this...- Ravendark
- Post #3
- Forum: Advanced Physics Homework Help
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Second functional derivative of fermion action
Homework Statement [/B] Consider the following action: $$\begin{align}S = \int \mathrm{d}^4 z \; \bar\psi_i(z) \, (\mathrm{i} {\not{\!\partial}} - m)_{ij} \, \psi_j(z)\end{align}$$ where ##\psi_i## is a Dirac spinor with Dirac index ##i## (summation convention for repeated indices). Now I would...- Ravendark
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- Derivative Fermion Functional Functional derivative Grassmann
- Replies: 6
- Forum: Advanced Physics Homework Help