Recent content by Pouramat

I started from eq(3.113) and (3.114) of Peskin and merge them with upper relation for $S_F$, as following: \begin{align} S_F(x-y) &= \theta(x^0-y^0)(i \partial_x +m) D(x-y) -\theta(y^0-x^0)(i \partial_x -m) D(y-x) \\ &= \theta(x^0-y^0)(i \partial_x +m) < 0| \phi(x) \phi(y)|0 >...

suppose I should evaluate $$Qa_{p1}^{r \dagger}a_{p2}^{s \dagger} b_{p3}^{t \dagger}$$ I get lost in the commutator relation. Any help?

Can you give me another reference except Carroll?

Summary:: Can anyone introduce an informative resource with solved examples for learning variation principle? For example I cannot do the variation for the electromagnetic lagrangian when ##A_\mu J^\mu## added to the free lagrangian and also some other terms which are possible: $$ L =...

Einstein's vacuum solution metric: $$ ds^2 = -(1-\frac{2GM}{r})dt^2 +(1-\frac{2GM}{r})^{-1}dr^2+r^2 d\Omega^2 $$ which ##g_{\mu \nu}## can be read off easily. metric Killing vectors are: $$ K = \partial_t $$$$ R = \partial_\phi $$ How can I relate these to Maxwell equation?

It seems ok till here, why don't you try to integrate this? $$ dE = \frac{k \lambda \, d\theta}{R} $$

$$\bar u(p') \gamma^i u(p) = u^\dagger(p') \gamma^0 \gamma^i u(p)$$ if ##p = p'## we can use $$u^\dagger(p) u(p) = 2m \xi^\dagger \xi$$ but how can we conclude the statement?

dear @vanhees71 And 1 more question, do you have any idea to explicitly show that ##\sigma^2 \psi^*_L## transforms like a right-handed spinor? using the identity.

thank you @vanhees71. Yes you are right.

Ohhh, Yes, it was a typo :) the LHS: $$ \large \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & i \\ -i & 0 \\ \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 1+i \\ 1-i & -1 \\ \end{pmatrix} $$ which should be equal to RHS: $$...

\begin{align} \psi_L \rightarrow (1-i \vec{\theta} . \frac{{\vec\sigma}}{2} - \vec\beta . \frac{\vec\sigma}{2}) \psi_L \\ \psi_R \rightarrow (1-i \vec{\theta} . \frac{{\vec\sigma}}{2} + \vec\beta . \frac{\vec\sigma}{2}) \psi_R \end{align} I really cannot evaluate these from boost and rotation...

:star:Merry X-mas:star: Ok, let's start: $$ \begin{align} {\omega^1}_2 &= -{\omega^2}_1 = -\cot \psi \, e^2 = - \cos \psi \, d\theta\\ {\omega^3}_1 &= -{\omega^1}_3 = \frac{cos \psi}{\sin^2 \psi} \, e^3 = \cot \psi \, \sin \theta \, d\phi\\ {\omega^3}_2 &= -{\omega^2}_3 = \frac{\cot...

Dear TSny; I revised my notes and got the mistake in it, as you mentioned: $$ \begin{align} {\omega^2}_1 &= -{\omega^1}_2 = -\cot \psi \, e^2\\ {\omega^3}_1 &= -{\omega^1}_3 = \frac{cos \psi}{\sin^2 \psi} \, e^3 \\ {\omega^3}_2 &= -{\omega^2}_3 = \frac{\cot \theta}{\sin \psi} \, e^3 \end{align} $$

My attempt at solution: in tetrad formalism: $$ds^2=e^1e^1+e^2e^2+e^3e^3≡e^ae^a$$ so we can read vielbeins as following: $$ \begin{align} e^1 &=d \psi;\\ e^2 &= \sin \psi \, d\theta;\\ e^3 &= \sin⁡ \psi \,\sin⁡ \theta \, d\phi \end{align} $$ componets of spin connection could be written by using...

I think you are right. Thanks for being such a star.